Set Theory (Pure and Applied Mathematics: A Series of Monographs and Textbooks Vol 79)

SET THEORY

This is a volume in PURE AND APPLIED MATHEMATICS A Series of Monographs and Textbooks Editors: SAMUEL EILENBERG AND HYMAN BASS A list of recent titles in this series appears at the end of this volume.

SET THEORY THOMAS JECH Department of Mathematics The Pennsylvania State University University Park, Pennsylvania

ACADEMIC PRESS

New York San Francisco London

A Subsidiary of Harcourt Brace Jovanovich, Publishers

1978

COPYRIGHT @ 1978, n Y ACADEMIC PRESS,INC. ALL RIGHTS RESERVED. NO PART OF THIS PUBLICATION MAY BE REPRODUCED OR TRANSMITTED IN ANY FORM OR BY ANY MEANS, ELECTRONIC OR MECHANICAL, INCLUDING PHOTOCOPY, RECORDING, OR ANY INFORMATION STORAGE AND RETRIEVAL SYSTEM, WITHOUT PERMISSION IN WRITING FROM T H E PUBLISHER.

ACADEMIC PRESS, INC. 111 Fifth Avenue, New

York. New York 10003

United Kingdom Edition published by ACADEMIC PRESS, INC. (LONDON) LTD. 24/28 Oval Road, London N W l

Library of Congress Cataloging in Publication Data Jech. Tliomas J Set theory.

(Pure and applied mathematics, a series of nionographs and textbooks : ) Bibliography: p. Includes index. 1. Set theory. I. Title. 11. Series. QA3.P8 [QA248] 5 1 0 ’ . 8 ~ (511’.3] 77-11214 ISBN 0-12-381950-4

PRINTED IN T H E UNITED STATES OF AMERICA

For. Paula, Pavel, and Susanna

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Aus dem Paradies, das Cantor uns geschaffen,

sol1 uns niemand vertreiben konnen.

David Hilbert

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CONTENTS Preface

xi

PART I SETS Chapter 1 AXIOMATIC SET THEORY 1 . Axioms of Set Theory

2. 3. 4. 5. 6. 7. 8. 9.

Ordinal Numbers Cardinal Numbers Real Numbers The Axiom of Choice Cardinal Arithmetic Filters and Ideals. Closed Unbounded Sets Singular Cardinals The Axiom of Regularity Appendix: Bernays-Godel Axiomatic Set Theory

1 12 22 29 38 42 52 61 70 76

Chapter 2 TRANSITIVE MODELS OF SET THEORY Models of Set Theory Transitive Models of ZF Constructible Sets Consistency of the Axiom of Choice and the Generalized Continuum Hypothesis 14. The H, Hierarchy of Classes, Relations, and Functions IS. Relative Constructibility and Ordinal Definability 10. I I. 12. 13.

78 87 99

108 1 I4

126

PART 11 MORE SETS Chapter 3 FORCING AND GENERIC MODELS 16. Generic Models

17. Complete Boolean Algebras 18. Forcing and Boolean-Valued Models 19. Independence of the Continuum Hypothesis and the Axiom of Choice 20. More Generic Models 21. Symmetric Submodels of Generic Models

ix

137 144 159

176 187 1 97

CONTENTS

X

Chapter 4 SOME APPLICATIONS OF FORCING 22. 23. 24. 25. 26.

Suslin’s Problem Martin’s Axiom and Iterated Forcing Some Combinatorial Problems Forcing and Complete Boolean Algebras More Applications of Forcing

216 229 244 261 283

PART I11 LARGE SETS

Chapter 5 MEASURABLE CARDINALS 27. 28. 29. 30. 31. 32.

The Measure Problem Ultrapowers and Elementary Embeddings Infinitary Combinatorics Silver lndiscernibles The Model L[Ul Large Cardinals below a Measurable Cardinal

295 305 32 I 337 359 38 1

Chapter 6 OTHER LARGE CARDINALS Compact Cardinals Real-Valued Measurable Cardinals Saturation of Ideals and Generic Ultrapowers Measurable Cardinals and the Generalized Continuum Hypothesis 37. Some Applications of Forcing in the Theory of Large Cardinals 38. More on Ultrafilters

33. 34. 35. 36.

398 416 427 450 465 478

PART IV SETS OF REALS

Chapter 7 DESCRIPTIVE SET THEORY 39. Bore1 and Analytic Sets 40. X: and II: Sets and Relations in the Baire Space 41. Projective Sets in the Constructible Universe 42. A Model Where All Sets Are Lebesgue Measurable 43. The Axiom of Determinacy 44. Some Applications of Forcing in Descriptive Set Theory

493 509 527 537 550 563

HISTORICAL NOTES AND GUIDE TO THE BIBLIOGRAPHY

579

BIBLIOGRAPHY

5%

NOTATION

61 I

Index

615

PREFACE

The main body of this book consists of 106 numbered theorems and a dozen of examples of models of set theory. A large number of additional results is given in the exercises, which are scattered throughout the text. Most exercises are provided with an outline of proof in square brackets [ 1, and the more difficult ones are indicated by an asterisk. I am greatly indebted to all those mathematicians, too numerous to mention by name, who in their letters, preprints, handwritten notes, lectures, seminars, and many conversations over the past decade shared with me their insight into this exciting subject.

xi

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PART I

Sets

CHAPTER 1

AXIOMATIC SET THEORY

1. AXIOMS OF SET THEORY

Axioms of ZermelwFraenkel:

I. Axiom of Extensionality. I f X and Y have the same elements, then X = Y .

II. Axiom of Pairing. For any a and b there exists a set {a, b} that contains exactly a and b. III. Axiom Schema of Separation. If cp is a property (with parameter p ) , thenfor any X and p there exists a set Y = {u E X : q(u, p ) } that contains all those u E X that have the property cp.

fV. Axiom of Union. For any X there exists a set Y elements of X .

=

uX,

the union of all

V. Axiom of Power Set. For any X there exists a set Y = P(X),the set of all subsets of X . VI. Axiom of Infinity. There exists an injinite set. VII. Axiom Schema of Replacement. If F is a function, then for any X there exists a set Y = F [ X ] = { F ( x ): x E X } . VIM Axiom of Regularity. Every nonempty set has an €-minimal element. IX. Axiom of Choice. Every family of nonempty sets has a choice function.

The theory with axioms I-VIII is ZF, Zermelo-Fraenkel axiomatic set theory; ZFC denotes the theory ZF with the axiom of choice.

Why Axiomatic Set Theory? Intuitively, a set is a collection of all elements that satisfy a certain given property. In other words, we might be tempted to postulate the following rule of formation for sets. 1

2

1. AXIOMATIC SET THEORY

Axiom Schema of Comprehension (false). If rp is a property, then there exists set Y = {x : rp(x)}.

Q

This principle, however, is false:

Rwell’s Paradox. Consider the set S whose elements are all those (and only those) sets that are not members of themselves: S = {X : X # X}. Question: Does S belong to S? If S belongs to S, then S is not a member of itself, and so S 4 S. On the other hand, if S 4 S, then S belongs to S. In either case, we have a contradiction. Thus we must conclude that

{X : x 4 X } is not a set, and we must somewhat revise the intuitive notion of a set. The safe way to eliminate paradoxes of this type is to abandon the schema of comprehension and keep its weak version, the schema ofseparation: Ifrp is a property, then for any X there exists a set Y = { x E X : rp(x)}. Once we give up the full comprehension schema, Russell’s paradox is no longer a threat; moreover, it provides useful information: The set of all sets does not exist. (Otherwise, apply the separation schema to the property x # x.) In other words, it is the concept of the set of all sets that is paradoxical, not the idea of comprehension itself. Replacing full comprehension by separation presents us with a new problem. The separation axioms are too weak to develop set theory with its usual operations and constructions. Notably, these axioms are not sufficient to prove that, e.g., the union X u Y of two sets exists, or to define the notion of a real number. Thus we have to add further construction principles that postulate the existence of sets obtained from other sets by means of certain operations. The axioms of ZFC are generally accepted as a correct formalization of those principles that mathematicians apply when dealing with sets.

Language of Set Theory, Formulas The axiom schema of separation as formulated above uses the vague notion of a property. To give the axioms a precise form, we develop axiomatic set theory in the framework of the first order predicate calculus. Apart from the equality predicate =, the language of set theory consists of the binary predicate E , the membership relation. The formulas of set theory are built up from the atomic formulas xEy,

x=y

1.

3

AXIOMS OF SET THEORY

by means of connectives cpA4k

cpvbk

cp+*,

l c p 9

cpw$

(conjunction, disjunction, negation, implication, equivalence), and quantiJers 3x cp

vx cp,

In practice, we shall use in formulas other symbols, namely defined predicates, operations, and constants, and even use formulas informally; but it will be tacitly understood that each such formula can be written in a form that only involves E and = as nonlogical symbols. Concerning formulas with free variables, we adopt the notational convention that all free variables of a formula c~(u1,. . ., un)

are among ulr . . . , u,, (possibly some uiare not free, or even d o not occur, in cp). A formula without free variables is called a sentence. Classes Although we work in ZFC which, unlike alternative axiomatic set theories, has only one type of object, namely sets, we introduce the informal notion of a class. We do this for practical reasons: It is easier to manipulate classes than formulas. If cp(x, pl, . . . , pn) is a formula, we call

C = :1. ..., pn)) a class. Members of the class C are all those sets x that satisfy cp(x, pl, . . . ,p,,): iff x EC c ~ ( xP,I , . . . , pn) ~p(x9

We say that C is dejnable from pl, . . . , p,,; if cp has just one free variable, then the class C is definable. We shall use boldface capital letters to denote classes. We shall, however, make departures from this general rule in cases when the standard notation is different (e.g., V , L , Ord, N,etc.) Two classes are considered equal if they have the same elements: If

C = {X :

.

~ 1 *% *

~p(x9

7

then C = D iff for all x ~ ( xP I,,

D = {X IL(x, 41,... qm))

pn))?

...( pn)

3

-

V(X* 41,

...?

The universal class, or uniuerse, is the class of all sets:

v = {x : x = x}

qm)

1.

4

AXIOMATIC SET THEORY

We define inclusion of classes CED

VX(XEC-XED)

++

and the following operations on classes: CnD={x:xECAxED} CU D={x:xECVXED} C - D = {X : x

E

CAx

# D}

IJ C = { x : x E S for some S E C} =

u

{ S : S E C}

Every set can be considered a class. If S is a set, consider the formula x the class

E

S and

{x : x E S}

That the set S is uniquely determined by its elements follows from the axiom of extensionality. A class that is not a set is a proper class. Extensionality

ff X and Y have the same elements, then X = Y:

vu(uEX++UEY ) + X = Y The converse, namely, if X = Y,then u E X u E Y,is an axiom of predicate calculus. Thus we have

X=Y

-

++

Vu(uEX-uEX)

The axiom expresses the basic idea of a set: A set is determined by its elements. Pairing For any a and b there exists a set {a, b} that contains exactly a and b: Va Vb 3c Vx(x E c-x

=a

v x = b)

By Extensionality, the set c is unique, and we can define the pair {a, b} = the unique c such that Vx(x E c t)x = a v x = b )

The singleton {a} is the set {a} = {a, a}

Since {a, b} = {b, a}, we further define an ordered pair (a, b )

.

5

1. AXIOMS O F SET THEORY so as to satisfy the following condition: (a, b ) = (c,d)

(1.1)

iff

a = c and

b=d

For the formal definition of an ordered pair, we take (a, b ) = { { a } ,{a, b}}

Exercise 1.1. Verify that the definition of an ordered pair satisfies (1.1).

We further define ordered triples, quadruples, etc., as follows: (a, b, c ) = ((a, b), c ) (a, b, c, d ) = ((a, b, c), d)

(al, . . . r a n t , ) = ( ( a i , . . . ,

an)?an+l)

It follows that two ordered n-tuples (a,, . . . , a,) and (bl, . . . , b,) are equal iff a, = b , , ..., a, = b,. Separation Schema Let q ( u , p ) be a formula. For any X and p , there exists a set Y = {u E : q ( u , p ) } :

x

V X vp 3Y v U ( U E Y H U E x A C p ( U , p ) ) (1.2) For each formula q ( u , p), the formula (1.2) is an axiom (of separation). The set Y in (1.2) is unique by Extensionality. Note that a more general version of separation axioms can be proved using ordered n-tuples: Let $(u, p , , . . . , p , ) be a formula. Then

vx V p , " ' v p , 3Y vu(U E (1.3) Simply let q ( u , p) be the formula 3 ~ * 1* .

Yt+U E XA$(U,

3Pn(P = (pi, . . ., P n ) and $(u, P I ,

p1,

..

. 3

..., p , ) ) pn))

and then, given X and p , , ..., p , , let

Y = {u E X : P(U,( P I , * . . )P n ) ) } We can give the separation axioms the following form: Consider the class

C = {u : q ( u , p l , ..., p,)}; then by (1.3) VX 3Y(C n X = Y) Thus the intersection of a class C with any set is a set; or, we can say even more informally a subclass of a set is a set

1.

6

AXIOMATIC SET THEORY

One consequence of the separation axioms is that the intersection and the difference of two sets is a set, and so we can define the operations

X

n

Y = ( U E X : U Y}, E

X - Y = { u E X : U Y) $

Similarly, it follows that the empty class

0 = {u:u #

u)

is a set-the empty set; this, of course, only under the assumption that at least one set X exists (because 0E X):

3X(X = X )

(1.4)

We have not included (1.4) among the axioms, but it follows from the axiom of infinity. Two sets X, Y are called disjoint if X n Y = 0. If C is a nonempty class of sets, we let

n C n {x: x

C}= {U : u E x for every x E C} C is a set (it is a subset of any X E C).Also, X n Y = =

E

n

Note that {X,Y). Another consequence of the separation axioms is that the universal class I/ is a proper class; otherwise, S={x€I/:x$x)

would be a set.

Union For any X there exists a set Y =

u X:

vx 3Y v U ( U E Y++ 3 Z ( Z E x A U E Z))

(1.5)

Let us introduce the abbreviations (32 E

X )cp

for

3z(z

E

X

~ c p )

(Vz

X )cp

for

Vz(z

E

X

+ cp)

and E

By (lS), for every X there is a unique set

Y = { u : (32 E X ) [u E z]}

=

u

{z: z

E

X }=

ux

the union of X . Now we can define

x

u Y=

{x,Y}, x

u Yu

z = (X u Y)u Z,

and also {a, b, c) = {a, b} u {c)

etc.

7

1. AXIOMS OF SET THEORY

and in general {Ul, ..., u,} = {Ul} u * * . u {a,}

Power Set For any X there exists a set Y = P(X):

VX 3Y Vu(u E

Y W U 5

X)

A set U is a subset of X, U G X, if Vz(z

E

u

-b

z E X)

If U E X and U # X, then U is a proper subset, U c X. The set of all subsets of X P ( X ) = { u : u c X} is called the power set of X. Exercise 1.2. There is no set X such that P ( X ) C _ X [Use Russell's paradox.]

Using the power set axiom we can define other basic notions of set theory. The Cartesian product of X and Y is the set of all pairs (x, y) such that xEXandyEY:

X

(1.6)

The notation {(x,y) :

. a * }

X

Y = { ( X , y ) : X E X A y E Y)

in (1.6) is justified because

{(x, y ) : cp(.% y ) } = {u : 3x 3y(u = (x. y ) A cp(x, Y ) ) } The product X x Y is a set because

xx

Y c PP(X u Y)

Further, we define

xx

Y xZ=(X x Y)xZ

and in general

x, x ... x

X , + , = (X, x ... x X,) x X,+l

Thus X , x ... x X, = {(x,, ..., x,): x1 E X,

A--*Ax, E

We also let X" = X x ... x X (n times)

X,,}

8

1. AXIOMATIC SET THEORY

An n-ary relation R is a set of n-tuples. R is a relation over X if R customary to write R(x,, . . . , x,) instead of

G

X". It is

(xi, - .. , x,) E R and in case that R is binary, then we also use XRY for (x, y) E R. If R is a binary relation, then the domain of R is the set dom(R) = { u : 3u[(u, 11) E R]} and the range of R is the set ran(R) = { u : 3u[(u, u ) E R]} Note that dom(R) and ran(R) are sets because dom(R) C _

uu

ran(R) c

R,

uu

R.

Thefield of a relation R is the set field (R) = dom(R) u ran(R). In general, we call a class R an n-ary relation if all its elements are n-tuples; in other words, if R G v" = the class of all n-tuples where C" (and C x D) is defined in the obvious way. A binary relation f is a function if (x, y) ~f

(x, z ) ~f

and

implies

y =z

The unique y such that (x, y) Efis the value offat x ; we use the standard notation Y =f (4 or its variations

f:

XHY,

y =f,,

etc.

for (x, y) € 5 f is a function on X if X = dom(J). If dom(f) = X", then f is an n-ary function on X. f i s a functionfrom X to Y, f X - Y ifdom(f) = X and ran(f) E Y. The set of all functions from X to Y is denoted by ' Y . Note that *Y is a set:

'Y

E P(X x

Y)

1.

9

AXIOMS OF SET THEORY

If Y = ran(f), thenfis a function onto Y. A functionfis one-to-one if f ( x )= f ( y )

implies

x =y

An n-ary operation over X is a functionf X" + X. The restriction of a functionfto a set X (usually a subset of dom(f)) is the function

f

I x = { ( x , y) E f :

x E X)

ZL i.e., dom(f) E dom(g) and

A function g is an extension of a functionfif g g(x) =/(x) for all x E dom(f). We denote the image of X by f by

f [ X I = {Y (3x E X) [Y =f(x)I) and the inverse image by

f- AX) = 1.

: f W E XI Iff is one-to-one, thenf-' denotes the inverse off iff

f-'(x)= y

x =f(y)

These definitions also apply t o functions that are classes, i.e., a relation F such that (x. y ) E F and

( x , z) E F

implies

y =z

For instance, F[C] denotes the image of the class C under the function F. It should be noted that a function is often called a mapping or a correspondence (and similarly, a set is called ajhmily or a collection). An equivalence relation over a set X is a binary relation = which is rejexive, symmetric, and transitive:

= x for all x E X x = y implies y = x x = y and y = z implies x = z x

A family of sets is disjoint if any two of its members are disjoint. A partition of a set X is a disjoint family P of nonempty sets such that

X=U{Y:YEP} Let

= be an equivalence relation over X. For every x E X, let [XI

= {y E

x :y = x )

(the equivalence class of x ) . The set

x/=

= {[XI :x

E

X}

1.

10

AXIOMATIC SET THEORY

is a partition of X (the quotient of X by = ). Conversely, each partition P of X defines an equivalence relation over X : ( 3 Y E P ) [ X € Y A Y € Y]

XEJJ

If an equivalence relation is a class, then its equivalence classes may be proper classes. In Section 9 we shall introduce a trick that enables us to handle equivalence classes as if they were sets. Infinity There exists an infinite set.

To give a precise formulation of the axiom of infinity,we have to define first the notion of finiteness. The most obvious definition of finiteness uses the notion of a natural number, which is as yet undefined. We shall define natural numbers (as finite ordinals) in Section 2 and give only a quick treatment of natural numbers and finiteness in the exercises below. In principle, it is possible to give a definition of finiteness that does not mention numbers, but such definitions necessarily look artificial. We give the most successful version below. We therefore formulate the axiom of infinity differently: 3s[@E

s A (VX E s)[X

U {X} E

s]]

We call a set S with the above property inductioe. Thus we have: Axiom of Injinity. There exists an inductioe set.

The idea is that an inductive set is infinite. We note that using the replacement schema, one can show that an inductive set exists if there exists an infinite set (see Section 2). A set S is T-fnite if every nonempty X E P ( S ) has a c-maximal element, i.e., u E X such that there is no u E X with u c o. Let N= {X : X is inductive}

n

N is the smallest inductive set. Let us use the following notation:

O=@, If n E N , let n

2={0,1},

1={0},

...

+ 1 = n u {n}. Let us define < (over N) by n<m

c--)

nEm

A set T is transitioe if x E T implies x G T Exercise 1.3. If X is inductive, then the set {x E X : x c X} is inductive. Hence N is transitive, and for each n, n = {m E N : m < n}

1.

11

AXIOMS OF SET THEORY

Exercise 1.4. If X is inductive, then the set {x E X : x is transitive} is inductive. Hence every n E N is transitive. Exercise 1.5. If X is inductive, then the set {x E X : x is transitive and x $ x} is inductive. Hence n $ n and n # n + 1 for each n E N. Exercise 1.6. N is T-infinite; the set N E P ( N ) has no c-maximal element. Exercise 1.7. If X is inductive, then {x E X : x is transitive and every nonempty z c x has an E-minimal element} is inductive (t is €-minimal in z if there is no s E z such that s E t). Exercise 1.8. Every nonempty X E N has an E-minimal element. [Pick n E X and look at X n n.] Exercise 1.9. If X is inductive then so is {x E X : x = 0 or x = y u {y} for some y}. Hence each n # 0 is m + 1 for some m. Exercise 1.10. Induction. Let A be a subset of N such that 0 E A, and n Then A = N. Exercise 1.11. Each n

E

E A +n

+ 1 E A.

N is T-finite.

A set X is said to have n elements (n E N ) if there is a one-to-one mapping of n onto X. A set is finite if it has n elements for some n E N. Exercise 1.12. Every finite set is T-finite. Exercise 1.13. Every infinite set is T-infinite. [If S is infinite, consider X = { u c S : u is finite}.]

Replacement Schema

If’ F is a function, then for every X, F[X] is a set.

For each formula q ( x , y, p ) , the formula (1.7) is an axiom (of replacement):

As in case of separation axioms, we can prove the version of replacement axioms with more than one parameter: Replace p by pl, . . ., pn. If F = {(x, y ) : q ( x , y, p)}, then the premise of (1.7) says that F is a function, and we get the formulation above. We can also formulate the axioms in the following ways:

IfF is afunction and dom(F) is a set, then ran(F) is a set. If F is afunction, then VX 3f[F I X =f]. Exercise 1.1 4. The separation axioms follow from the replacement schema. [Given cp. let F = {(.x, x): cp(x)}. Then {x E X: cp(x)} = F[X], for every X.]

12

1.

AXIOMATIC SET THEORY

Exercise 1.15. Instead of union, power set, and replacement axioms consider the following weaker versions: (1.8)

VX 3Y[U X E Y]

(i.e., VX 3Y(Vx E X ) ( V u E x ) [u E Y])

(1.9)

vx 3Y[P(X) E Y]

(VX 3Y V U [ U E x + u E Y])

(1.10) If F is a function, then

VX 3Y(F[X] E Y)

Then axioms IV, V, and VII can be proved from (1.8). (1.9). and (1.10), using the separation schema (1.3).

The remaining two axioms, choice and regularity, will be dealt with later. 2. ORDINAL NUMBERS

In this section we introduce ordinal numbers and prove the theorems on transfinite induction and transfinite recursion.

Linear and Partial Ordering A binary relation < over a set P is a partial ordering o f P if:

(i) p 4: p for any p E P ; (ii) if p < q and q < r, then p < r. ( P , < )-and, by abuse of notation, P - is called a partially ordered set. A partial ordering < of P is a linear ordering if moreover

(iii) p < q or p = q or q < p for all p, q E P. If c is a partial (linear) ordering, then the relation s is also called a partial (linear) ordering (and < is sometimes called a strict ordering). If P is a partially ordered set and X is a nonempty subset of P , and a E P, then:

a is a maximal element of X if a E X A (Vx E X )[ a 4: X I ; a is a minimal element of X if a E X A (Vx E X) [x 4: a ] ; a is the greatest element of X if a E X A (Vx E X) [x I a ] ; a is the least element of X if a E X A (Vx E X ) [a IXI; a is an upper bound of X if (Vx E X ) [x I a ] ; a is a lower bound of X if (Vx E X) [a x ] ; a is the supremum of X if a is the least upper bound of X ; a is the infimum of X if a is the greatest lower bound of X. The supremum (infimum) of X (if it exists) is denoted by sup X (inf X).Note that if X is linearly ordered by c ,then a maximal element of X is its greatest element (similarly for a minimal element).

2.

13

ORDINAL NUMBERS

If ( P , < ) and (Q, < ) are partially ordered sets and f: P + Q, then f is order-preserving if x
-

f ( 4< f ( Y )

If P and Q are linearly ordered, then an order-preserving function is also said to be increasing. A one-to-one order-preserving function of P onto Q is an isomorphism of P and Q; ( P , < ) is then isomorphic to (Q, < ). Exercise 2.1. The relation “ ( P , < ) is isomorphic to ( Q , < ) ” is an equivalence relation

(over the class of all partially ordered sets).

Well-Ordering A linear ordering < of a set P is a well-ordering if every nonempty subset of P has a least element. The concept of well-ordering is of fundamental importance in view of the induction theorems. It is shown below that well-ordered sets can be compared by their lengths; ordinal numbers will be introduced as order-types of wellordered sets.

Lemma 2.1. If ( W , < ) is a well-ordered set and f: W + W is an increasing function, then f ( x ) 2 x for each x E W. Proof. Assume that the set X = {x E W :f ( x ) < x} is nonempty and let z be the least element of X . If w =f(z), thenf(w) < w, a contradiction.

Corollary 1. The only isomorphism of a well-ordered set onto itselfis the identity. Proof. By Lemma 2 . l , f ( x ) 2 x for all x, andf-’(x) 2 x for all x.

Corollary 2. If two well-ordered sets W,, W, are isomorphic, then the isomorphism of W, onto W, is unique. W If W is a well-ordered set and u E W , then {x E W : x < u} is an initial segment of W (given by u).

Lemma 2.2. N o well-ordered set is isomorphic to an initial segment of itselJ Proof: I f ran(f ) = {x : x < u}, thenf(u) < u, contrary to Lemma 2.1.

Theorem 1. If W, and W, are well-ordered sets, then exactly one of thefollowing three cases holds:

(a) W, is isomorphic to W, ; (b) W, is isomorphic to an initial segment of W, ; (c) W, is isomorphic to an initial segment of W,.

14

1.

AXIOMATIC SET THEORY

Proqf: For u E M.: (i = 1, 2), let K(u)denote the initial segment of by u. Let

given

f= { ( x , y ) E W , x W, : W , ( x )is isomorphic to W,(y)} Using Lemma 2.2, it is easy to see that f is a one-to-one function. If h is an isomorphism between W , ( x ) and W2(y)and x' < x, then W,(x')and W,(h(x')) are isomorphic. It follows that f is order-preserving. If dom(f) = W, and ran(f) = W, , then case (a) holds. If y , < y, and y , E ran(f), then y , E ran(f). Thus if ran(f) # W, and y o is the least element of W, - ran(f), we have ran({) = W,(y,). Necessarily, dom(f) = W,, for otherwise we would have ( x o , yo) ES, where x o = the least element of W, - dom(f). Thus case (b) holds. Similarly, if dom(f) # W,, then case (c) holds. In view of Lemma 2.2, the three cases are mutually exclusive. If W, and W, are isomorphic, we say that they have the same order-type: 0.t.

w,= 0.t. w,

0.t.

w, < 0.t. w,

Similarly, we define if W, is isomorphic to an initial segment of W, . Exercise 2.2. Assume that each well-ordered set W is assigned an ordinal a = 0.t. W . Show that the relation a < fl is a linear ordering of the class of all ordinals and that

every nonempty set of ordinals has a least element.

We shall now give a formal definition of ordinal numbers.

Ordinals The idea is to define ordinal numbers so that a
and

iff

a={/3:/3
A set T is transitive if

u

Vx(x E T - t x E T )

In other words, T c T , or T c P( T). A set is an ordinal number (an ordinal) if it is transitive and well-ordered by E. We shall denote ordinals by lowercase Greek letters a, /3, y , . . . . The class of all ordinals is denoted by Ord. We define a
iff

a ~ / 3

2.

15

ORDINAL NUMBERS

Lemma 2.3. (a) 0 = 0 is an ordinal. (b) lfa is an ordinal and /3 E u, then p is an ordinal. (c) lju,p are ordinals and a c /?, then u E p. (d) lja,p are ordinals, then either a c p or p E u. Proqf (a), (b) by definition. (c) If u c p, let 7 be the least element of the set p - u. Since c1 is transitive, it follows that u is the initial segment of given by y. Thus u = {lE p : 4 < y} = y, and so u E p. (d) Clearly, 2 n is an ordinal, u n p = y. We have y = u or y = /I, for otherwise 11 E a,and 7 E /I,by (c). Then y E 11, which contradicts the definition of an ordinal (namely that E is a strict ordering of k). Using Lemma 2.3, one gets the following facts about ordinal numbers (the proofs are routine):

(2.1) u < p is a linear ordering of the class Ord. (2.2) For each a, u = { p : < u}. (2.3) If C is a nonempty class of ordinals, then C is an ordinal, C E C and C = inf C. (2.4) If X is a nonempty set of ordinals, then X is an ordinal, and IJ x = sup x. (2.5) For every u, a u ( u } is an ordinal and ct u {u} = inf{P : p > a}.

n

0

u

n

We thus define z -t 1 = u u { u ) (the successor of u). In view of (2.4), the class Ord is a proper class; otherwise, consider sup(0rd) 1. We can now prove that the above definition of ordinals provides us with order-types of well-ordered sets.

+

Theorem 2. E i w y well-ordered set is isomorphic to a unique ordinal number. Proof: The uniqueness follows from Lemma 2.2. Given a well-ordered set W , we find an isomorphic ordinal as follows: Let F(x) = u if u is isomorphic to the initial segment of W given by x . If such an u exists, then it is unique. By the replacement axioms, F[ W ]is a set. For each x E W , such an a exists (otherwise consider the least .Y for which such an u does not exist). If y is the least y 4 F[ W ] .then F[ W ] = 1’ and we have an isomorphism of W onto y. H

If a = p

+ 1, then u is a successor ordinal. If a is not a successor ordinal, then

u = sup{p : fi

u

< u ) = a;a is called a limit ordinal. We also consider 0 a limit ordinal and define sup 0= 0. Exercise 2.3. a is a limit ordinal

if and only if fi < Q implies fi

+ 1 < a, for every 8.

The existence of limit ordinals other than 0 follows from the axiom of infinity:

16

1. AXIOMATIC SET THEORY

Exercise 2.4. If a set X is inductive, then X n Ord is inductive. The set N = inductive} is the least limit ordinal # 0.

{ X : X is

We denote the least nonzero limit ordinal o (or N). The ordinals less than o (elements of N) are called finite ordinals, or natural numbers. Specifically, 0=0,

1=0+1,

2=1+1,

etc.

We use n, m, I, k, j , i (usually) to denote natural numbers. Exercise 2.5. (without the axiom of infinity). Let w = least limit c( # 0 if it exists, w = Ord otherwise. Call a set X finite if X is in a one-to-one correspondence with some n E w. Show:

3(inductive set) --t ](infinite set) + w is a set -, ](inductive set) [For the second implication, apply replacement to the set {all finite subsets of X}.]

Induction Theorems

Theorem 3 (Transfinite Induction). Let C be a class of ordinals and assume that: (i) 0 E C ; (ii) i f a E C,then a + 1 E C; (iii) f a is a nonzero limit ordinal and

B E C.for all fl < a, then a E C.

Then C is the class of all ordinals.

Proof: Otherwise, let a = inf(0rd-C) and apply (i), (ii), or (iii). A function whose domain is the set N is called an (infinite) sequence. (A sequence in X is a function$ N + X.)The standard notation for a sequence is

( a , : n < o) or variants thereof. A finite sequence is a function s such that dom(s) = { i :i < n} for some n E N; then s is a sequence of length n. A transfinite sequence is a function whose domain is an ordinal:

r < a> It is also called an a-sequence, or a sequence of length a. We also say that a sequence ( a t : 5 < a ) is an enumeration of its range {a: : t < a ) . I f s is a sequence of length a, then snx or simplysx denotes the sequence of length a + 1 that extends s and whose ath term is x : s n~ = sx = s

u (a, x)

Sometimes we shall call a "sequence" (a, :a a function (a proper class) on Ord.

E

Ord)

2.

17

ORDINAL NUMBERS

Definition by transfinite induction ” usually takes the following form: Given a function G (on the class of transfinite sequences), then for every 0 there exists a unique 0-sequence “

(a, :a < 6 ) such that a, = G ( ( a , : t < a ) ) for every a < 8. We shall give a general version of this theorem, so that we can also construct sequences ( a , : a E Ord).

Theorem 4 (Transfinite Recursion ; Definition by Induction). For every G there is F such that: I f G is a,/unction on V . then F is the uniquefunction on Ord such that F ( a )= G ( F ( a ) jor each a. In other words, if we let a, = F(a), then for each a, a, = G ( ( a t : 5 < a ) ) (Note that we tacitly use replacement: F I a is a set for each a.) Proof: We let

F ( a ) = x ++there is a sequence ( a , : t < a ) such that: (a) (V< < a)[., = G ( ( a , rl < <>)I; (b) x = G ( ( a , : 5 < a)).

For every a, if there is an a-sequence that satisfies (a), then such a sequence is unique: If ( a , : 5 < a ) and ( b ; : 5 < a ) are two a-sequences satisfying (a), one shows a t = b , by induction on (. Thus F ( a ) is determined uniquely by (b), and therefore F is a function. It follows, again by induction, that for each a there is an a-sequence that satisfies (a) (at limit steps, we use replacement to get the a-sequence as the union of all the <-sequences, < a). Thus F is defined for all a E Ord. It obviously satisfies

<

F ( a ) = G(F I a )

If F’ is any function on Ord that satisfies

F’(a) = G(F’I a ) then it follows by induction that F’(a) = F ( a ) for all a. W Exercise 2.6. If W is a well-ordered set, then there exists no sequence ( a , : n E N) such that a. > a , > a2 > ....

in W

18

1. AXIOMATIC SET THEORY

(Using the axiom of choice, this condition implies that W is well-ordered.)

< <

Let a > 0 be a limit ordinal and let ( y c : < a ) be a nondecreasing sequence of ordinals (i.e., c q implies ys I y,,). We define the limit of the sequence by lim y e = sup{y, : < a}

<

{--'a

Exercise 2.7. There are arbitrarily large limit ordinals; i.e., Va 38 > a (/I is a limit). [Consider limn+wa,, where a,, = a, 1.1

+

A sequence of ordinals (yo, : a E Ord) is normal if it is increasing and continuous, i.e., if for every limit a, yo, = lims,, y e . Exercise 2.8. Every normal sequence has arbitrarily largefixed points, i.e., a such that y. = a. a, .]

[Let a,+ = y G , and a =

Ordinal Sums and Products We shall briefly touch the subject of addition and multiplication of ordertypes, and define ordinal sum and product, a + p and a * /I. Let us assume that we define in some way the order-type of linearly ordered sets. That is, two linearly ordered sets have the same order-type just in case they are isomorphic. The order-type of a well-ordered set W is the unique ordinal isomorphic to W. If T, and 7 2 are order-types, then T + r 2 is defined as the order-type of a set X u Y where X and Y are disjoint, 0.t. X = 71, 0.t. Y = 7 2 , and X u Y is ordered so that all x E X precede all y E Y. In case of ordinal numbers, the ordinal sum a+B

is the unique ordinal y 2 a such that /3 is isomorphic to the set y - a = {< : a I < y } .

<

Exercise 2.9. Prove that a

+ O = a, a

Thus a

a

+ (B + 1 ) =

+ B = lim (a + t), c -1

(a

+ B) + 1

if fl is a limit

+ fl is a normal function of B.

The ordinal sum is associative, (a + fl) + y = a + (p + y), but not commutative : l+o=o#fw+l

To define ordinal product, we consider lexicographic ordering of Cartesian products. If A and Bare linearly ordered sets, we define the (vertical) lexicogra-

2.

19

ORDINAL NUMBERS

-

phic ordering of A x B as follows:

(2.6) ( a , , b , ) < ( a , , h,)

either a , < a ,

or ( a , = a , and b , < b 2 ) It is easy to see that the relation < defined by (2.6) is a linear ordering of A x B. See Fig. 2.1.

B

B

A

A

FIGURE 2.1 Vertical and horizontal lexicographic ordering.

-

Similarly, the horizontal lexicographic ordering is defined as follows : (2.7) ( a , . h , ) < ( a , , b,)

either b , < b,

or ( b , = b, and a , < a , ) The producr of order-types t, . T~ is defined as the order-type of the Cartesian product A x B of sets of order-types t , and T ~ respectively, , under the horizontal lexicographic ordering. If both A and B are well-ordered, it is easy to verify that A x B is wellordered by either lexicographic ordering. Thus the ordinal product a.B

is the unique ordinal isomorphic to a x ordering. Note that a . 2 = o! + a.

fl under the horizontal lexicographic

Exercise 2.10. Prove that

u.O=O, tl

a.(B+ l)=a.P+a

. B = lim u . (,

if

is a limit

;-P

Thus for u # 0, u . B is a normal function of

b.

Exercise 2.11. An ordinal a is a limit ordinal if and only if u = (I) . fl for some

The ordinal product is associative, but not commutative: 2 . w =0 # 0 . 2 =Q +w

B.

1. AXIOMATIC SET THEORY

20 As for distributivity, we have a

while, e.g.,

*

(/I+ y ) =

a.

+ 1). 2 = (w + 1 ) +

(0

Exercise 2.12.

/I+ a . y

(w

+ 1)=

0 '2

+1

Define by induction: a0

a6 =

a@+1= q B . a

= 1,

Iim

if fl is a limit

at

i'-B

The function Exercise 2.13.

UP

is a normal function of = aB

p (for a > 1).

. ay, (d)Y= a@''.

The Canonical Well-Ordering of a x a We define a well-ordering of the class Ord x Ord of ordinal pairs. Under this well-ordering, each a x a is an initial segment of Ord'; the induced wellordering of a2 is called the canonical well-ordering of a'. Moreover, the wellordered class Ord2 is isomorphic to the class Ord, and we have a one-to-one function r of Ord2 onto Ord. As a bonus, for many a's the order-type of a x x is a. If a, p are ordinals, let max(a, p) be the greater ordinal of the two. We define : (2.8) (a, p ) < (y, 6)

-

either max(a, p) < max(y, 6)

or (max(a, p) = max(y, 6) and a < y )

or (max(a, p ) = max(y, 6) and a = y and f l < 6) The relation < defined in (2.8) is obviously a linear ordering of the class Ord x Ord. Moreover, if X c Ord x Ord is nonempty, then X has a least element (verify!). Also, for each a, a x a is the initial segment given by (0, x ) . See Fig. 2.2. If we let T(a, p ) = the order

a

a

FIGURE 2.2 The canonical well-ordering of a x a.

2.

21

ORDINAL NUMBERS

-

r is a one-to-one mapping of Ord2 onto Ord, and (2.9) (a, P ) < hJ7d) r(a,B ) < T(Y9 6) Note that r [ w x w ] = w and since ?(a) = r[a x a] is an increasing function

then

of a, we have y(a) 2 a for every a. However, y(a) is easily shown to be normal, and so r[a x a] = a for arbitrarily large a. Exercise 2.f4. Show that y(u u = Q", we have x a] = u.

r[a

+ 1) = y(u) + u + a + 1,

and hence y(u)

w'. Thus if

Well-Founded Relations Now we shall define an important generalization of well-ordered sets. A binary relation E over a set P is well-founded if every nonempty X G P has an E-minimal element, that is a E X such that there is no x E X with x E a. Clearly, a well-ordering of P is a well-founded relation. Exercise (a, : n E

2.15. If E is a well-founded relation over P, then there is no sequence N ) in P such that a, E

a2

UO.

E a,,

a3

E a2,

...

(Using the axiom of choice, this condition implies that E is well-founded.) Given a well-founded relation E over a set P , we can define the length of E, and assign each x E P an ordinal number, the rank of x in E. Theorem 5. If E is a well-jounded relation over P, then there exists a function f from P into the ordinals such that-for any x, y E P, x E I'

+

f ( x )
Moreoiler, there is u unique ordinulfunction p E such that for each x E P, (2.10)

p E ( x )= fhc least a such that a > p E ( y ) f b rall y E .x

+ 1 :y E

= SUP{~)E(~)

X)

The range of p E is an initial segment of the ordinals (thus an ordinal). This ordinal is called the length of E : (2.11)

length@) = sup{p,(x)

+ 1:x E P)

Proof: We shall define a function p E satisfying (2.10) and then prove its uniqueness. By induction, let P,

=

0,

P a + , = {x E P : Vy(y E x -+ y

P, =

u P,

E

Pa)}

if a is a limit

<
Let 0 be the least ordinal such that P o + = P o (such 0 exists by Replacement). First, it should be easy to see that P a E P a + for each a (by induction). Thus

22

1. AXIOMATIC SET THEORY

Po c P , c c Po. We claim that Po = P: Otherwise, let a be an E-minimal element of P - P o . It follows that each x E a is in Po, and so a E Po+ I, a contradiction. Now we define (2.12)

,

p E ( x ) = the least a such that x E Pa+

It is obvious that if x E y, then p E ( x ) < p E ( y ) ;also, (2.10) is easily verified. The ordinal 8 is the length of E. The uniqueness of p E is established as follows: Let p' be another function satisfying (2.10) and consider an E-minimal element of the set {x E P P E ( X ) # P ' ( X ) } . 3. CARDINAL NUMBERS

Two sets X, Y have the same cardinality (cardinal number, cardinal), if there exists a one-to-one mapping of X onto Y. The relation (3.1) is an equivalence relation over the universe. We assume that we can assign each set X its cardinal number I X I so that two sets are assigned the same cardinal just in case they satisfy the condition of (3.1). The cardinal numbers can be defined either using the axiom of regularity (via equivalence classes of (3.1)), or using the axiom of choice. In this section we shall define cardinal numbers of well-orderable sets; and it follows from the axiom of choice that every set can be well-ordered. We recall that a set X isfinite if I X 1 = In I for some n E N ;then X is said to have n elements. Clearly, In I = I m I if and only if n = m, and so we define finite cardinals as natural numbers, i.e., I n I = n for all n E N. Exercise 3.1.

(a) A subset of a finite set is finite. (b) The union of a finite set of finite sets is finite. (c) The power set of a finite set is finite. (d) The image of a finite set (under a mapping) is finite.

The relation I X I = I Y I for infinite sets is the most important notion of set theory. Most of set theory is practically the study of infinite cardinals. The following theorem tells us that the concept of a cardinal is not trivial.

Theorem 6 (Cantor). For every set X ,

I W ) l f 1x1 Proof: Let f be a function from X into P ( X ) . The set Y = {x

E

x :x # f ( x ) }

3.

23

CARDINAL NUMBERS

is not in the range off: If z E X were such thatf(z) = Y, then z E Y if and only if z @ Y, a contradiction. Thus f is not a function of X onto P(X). Hence IP(X)I f 1x1. The ordering of cardinal numbers is defined as follows:

1x1 IPI if there exists a one-to-one mapping of X into Y.

1x1 < I Y I * 1x1 I I Y I ~ I X fI I Y I The relation I is clearly transitive. In view of the following theorem, c is a partial ordering of cardinal numbers. Theorem 7 (Cantor-Bernstein). If

PI = PI*

IA I II BI

and

IBI I IA 1,

then

-,B andf,: B + A are one-to-one, then if we let B’ = f , [ B ] and A , =fi[f,[A]],we have A , E B’ E A and [ A , I = [ A I. Thus we may assume that A c B E A and that f is a one-to-one function of A onto A we wish to show that I A I = I B 1. See Fig. 3.1. Proof: Iff,: A

,;

FIGURE 3.1

We define (by induction)

Let g be the function on A defined as follows: g(x) =

(y

if x E An - Bn for some n otherwise

Then g is a one-to-one mapping of A onto B, as the reader will gladly verify. Thus [ A ( = IBI.

24

1.

AXIOMATIC SET THEORY

The arithmetic operations on cardinals are defined as follows:

K + A = IAuBI (3.2)

K.L= K’=

I A x BI IBA[

where

IAl = K ,

where

I A J= K ,

where

[ A [= K ,

IBI = A , and A , B are disjoint IBI = 1

IB1 = A

Naturally, definition (3.2)is meaningful only if it is independent of the choice of A and B. Thus one has to check that, e.g., if I A I = I A’ I and I B I = I B 1, then [ A x BI = I A ’ X BI. Lemma 3.1. If I A I = K, then I P ( A )I = 2“. Proot For every X E A, let x x be the function

The mapping f: X ”{O, 1).

t-t

x x is a one-to-one correspondence between

P ( A ) and

Thus Cantor’s theorem can be formulated as follows: K

< 2“

for every cardinal

K

(Clearly, I A I II P ( A )1, for x H {x} is one-to-one.) A few simple facts about cardinal arithmetic:

(3.3) + and . are associative, commutative and distributive. (3.4) ( K . 1y = K” . 1’. (3.5) KA+’ = K’ . K’. (3.6) (K’)” = K’ p. (3.7) If K I 1,then K” IA’. (3.8) If 0 < 1Ip, then K’ IK’. (3.9) K O = 1; 1“ = 1; OK = O if K > 0. ’

To prove (3.3t(3.9),one has only to find the appropriate one-to-one functions. Alephs

An ordinal a is called a cardinal number (a cardinal) if I a I # I fi I for all a. We shall use K, 1,p, . . . to denote cardinal numbers and Card the class of all cardinals. If W is a well-ordered set, then there exists an ordinal a such that I W I = I a I. Thus we !et

fl<

I W I = the least ordinal such that I W I = I a 1 Obviously, 1 W I is a cardinal.

3.

25

CARDINAL NUMBERS

Every natural number is a cardinal (ajn ite cardinal); and if S is a finite set, then 1 S I = n for some n. The ordinal w is the least infinite cardinal. Note that all infinite cardinals are limit ordinals.

Lemma 3.2. (a) For every a there is a cardinal greater than a. (b) IfX is a set ofcardinals, then sup(X) is a cardinal.

For every a, we let a+ denote the least cardinal greater than a (the cardinal successor of I a I ). Proof: (a) For any set X, let the Hartogs number of X be (3.10)

h ( X ) = the least a such that there is no one-to-one function of a into X

There is only a set of possible well-orderings of subsets of X. Hence there is only a set of ordinals for which a one-to-one function of a into X exists. Thus h(X) exists. If a is an ordinal, then (a1 < Ih(a)l by (3.10). That proves (a). (In fact, h(a) = a + . )

(b) Let a = sup X. I f f is one-to-one mapping of cx onto some p < a, let X be such that p < K Ia. Then I K I = I f [ ~1 ]= I order type ( f [ ~ ]1 )I /3, a contradiction. Thus a is a cardinal. K E

Using Lemma 3.2, we define the normal sequence of all infinite cardinals (alephs).We usually use K, to denote the cardinal number, and w , to denote the order-type: KO = WO = (0,

K,,,

= w,+1=

K,+

if a is a limit K, = w, = sup{wa: /3 < a} Sets whose cardinality is KO are called countable; a set is at most countable if it is either finite or countable. Set of other cardinalities are uncountable. A cardinal K, + is a successor cardinal. A cardinal K, whose index is a limit ordinal is a limit cardinal. Addition and multiplication of infinite cardinal numbers is a trivial matter, due to the following fact:

,

Theorem 8. K, . N, = K,. ProoJ Consider the canonical one-to-one mapping r of Ord x Ord onto Ord defined in Section 2. We shall show that T[w, x w,] = 0,. This is true for a = 0. Thus let CI be the least ordinal such that T[w, x w,] 3 w , . Let /3, y < w , be such that r(p,y ) = a,.Pick 6 < w,such that 6 > p and 6 > y. Since 6 x 6 is an initial segment of Ord x Ord in the canonical well-ordering and contains

26

1.

AXIOMATIC SET THEORY

(/I, y), we have r[6 x 61 3 w , , and so 16 x 6 I

2 8,.However, 16 x 6 I = 16 I . 16 1, and by the induction hypothesis, ( 6 I . 16 I = ( 6 I < K,. A contradiction.

As a corollary, we have

K, + K, = K, . K,

(3.11)

= max(K,, K,)

Exponentiation of cardinals will be dealt with in Section 6. Without the axiom of choice, one cannot prove that 2Nmis an aleph (or that P(w,) can be well-ordered), and there is very little one can prove about 2Nsor K,Kp. Exercise 3.2. There is a canonical well-ordering of Ord3 such that for each a, (u; is an

initial segment and its order-type is 0,. Hence K:

=

K..

Exercise 3.3. There is a well-ordering of the class of all finite sequences of ordinals such that for each a, the set of all finite sequences in w , is an initial segment and its order type

is

0..

We say that a set B is a projection of a set A if there is a mapping of A onto

B. Note that B is a projection of A if and only if there is a partition P of A such that I PI = I BI. If I A 1 2 I BI > 0, then B is a projection of A. Conversely, using the axiom of choice, one shows that if B is a projection of A , then I A I 2 1 B 1. This, however, cannot be proved without the axiom of choice. Exercise 3.4. If

B is a projection of cum,then I B 1 2 K,.

of w, has cardinality K.. [The set is a projection of the set of finite sequences.]

Exercise 3.5. The set of all finite subsets

I P (B) I < I P ( A )I. [Consider g(X) =f- l[X], wherefmaps A onto B.]

Exercise 3.6. If B is a projection of A, then

Exercise 3.7. w.+ is a projection of P(w,). [Use w, x w , = w , and project P ( o , x a,): If R E w , x w , is a well-ordering. let

I

I

f(R)be its order-type.]

< 2'"". [Use Exercises 3.7 and 3.6.1

Exercise 3.8. K, +

Cofinality Let u > 0 be a limit ordinal. We say that an increasing /?-sequence < /I), fi a limit ordinal, is cojinal in a if limg+B.ue= u. Similarly,A E a is cofinal in a if sup A = u. If u is an infinite limit ordinal, the cofinality of u is ( u C:

<

cf u = the least limit ordinal fi such that there is an increasing /?-sequence ( a e : < f i ) with lim,,B u5 = u

<

Obviously, cf a is a cf(w w ) = cf K, = w.

+

limit

ordinal,

and

cf a I a.

Examples:

3.

27

CARDINAL NUMBERS

Lemma 3.3. cf(cf u ) = cf a. Proof: If ( u ; : ( < p ) is cofinal in u and ( ( ( I ) :I < y) is cofinal in (u,(,) : 1 < y ) is cofinal in u. I

fl, then

Two useful facts about cofinality :

Lemma 3.4. Lat u > 0 be a limit ordinal. (a) I f A c a and sup A = u, then o.t.(A)2 cf u. (b) l f Po I fll I '. . I P, I ..., 5 < y, is a nondecreasing y-sequence of ordinals in a and 1imthyfl, = u, then cf y = cf u. Proof: (a) Let ( u , : ( < 0.t. A ) be the increasing enumeration of A. (b) If ;' = limv+cf ((v), then u = 1imv+ i' /I,(,,,, and the nondecreasing sequence (p<,,, : v < cf y) has an increasing subsequence of length I cf y,with the same limit. Thus cf u Icf y. To show that cf y I cf u, let a = limv--rcfau v . For each v < cf u, let ((v) be the least ( greater than all ( ( I ) I. < v, such that /I; > u,,. Since limv-.cfa/?,(,, = u, it follows that lim,,,,,, ((v) = y. and so cf y I cf a. I ~

An infinite cardinal K, is regular if cf(W,) = Q,. It is singular if cf(o,) < 0,.

Lemma 3.5. cf(a) is always a regular cardinal. Proof: It is easy to see that if u is not a cardinal, then using a mapping of I u I onto a, one can construct a cofinal sequence in u of length I I u 1, and therefore cf a < u. Since cf(cf u ) = cf u, it follows that cf u is a cardinal, and is regular.

There are arbitrarily large singular cardinals. For each a, N,,, is a singular cardinal of cofinality o. Using the axiom of choice, we shall show that every K,, is regular. It is an open problem whether one can prove without the axiom of choice that there exists a regular uncountable K,. (The informed guess is that one cannot.)

K. is an uncountable limit cardinal, then cf w. = cf ct; w, is the limit of a cofinal sequence ( w , : 5 < cf m) of cardinals.

Exercise 3.9. If

Lemma 3.6. An infinite cardinal

K

is singular if and only ifthere exists a cardinal

1 < K and afamily { S , : ( < 1)ofsubsets of K such that IS, I < K for each ( < A, and K = S , . (And the least cardinal equal to cf K . )

A that satis-es the condition is in fact

Prooj: (a) If K is singular, then there is an increasing sequence ( u , : ( < cf K) with lim, a, = K. Let I = cf K , and S , = u5 for all ( < A. (b) If the condition holds, let 1< K be the least cardinal for which there is a family { S , : ( < A} such that K = S , and IS, I K for each t < I . For every 5 < I , let fl, be the order-type of U,.< S , . The sequence ( f l s : < A> is

uc
-=

<

28

1. AXIOMATIC SET THEORY

<

nondecreasing, and by the minimality of I , p, < K for all < 1.We shall show that lim, /3< = K, thus proving that cf K I A. . is a one-to-one mappingfof K = S, into Let = limc+A/I,There 1 x p: If a E K, let f ( a ) = (<, y), where { is the least such that a E S, and y is the order-type of S , n a. Since A < ti and I I x p I = A . 1 j?1, it follows that

uc
<

p = K .

As mentioned earlier, one cannot prove without the axiom of choice that w 1

is not a countable union of countable sets.

u.
Exercise 3.10. Show that w 2 is not a countable union of countable sets. [If w 2 = S , , then sup(0.t. S. : n < w } = w 2 .]

The only cardinal inequality we have proved so far is the Cantor theorem K < I" for every A > 1, and in particular ti < tiK (for K # 1). The following theorem gives a better inequality. This and other cardinal inequalities will also follow from Konig's theorem, to be proved in Section 6. K

< 2". It follows that

Theorem 9. If

ti

is an infinite cardinal, then K

< ticfK.

froofi Let F be a collection of K functions from cf ti to K : F = {fu : 01 < ti}. I t is enough to findJ cf ti + ti that is different from all theJ&.Let ti = lim+cfz 0 1 ~ . For < cf ti, let

<

I(<) = least y such that y #fu({)for all 01 < ag Such a y exists since Z
rn

I {h({): a < at} I I I ucI < K. Obviously, f # f ,

for all

Consequently, K' > ti whenever 12 cf K. An uncountable cardinal ti is weukly inaccessihle if it is a limit cardinal and is regular. There will be more about inaccessible cardinals later, but let me mention at this point that existence of (weakly) inaccessible cardinals is not provable in ZFC. To get an idea of the size of an inaccessible cardinal, note that if N u > No is a limit and regular, then K, = cf h', = cf a I 01, and so K, = a. Since the sequence of alephs is a normal sequence, it has arbitrarily large fixed points; the problem is whether some of them are regular cardinals. For instance, the least fixed point Nu= a has cofinality (0: ti

= lim (w, o w , ow", .. .) =

lim

ti,,

where

tio =

w,

K,+

= cox"

n-w

A set S is Dedekind-finite (D-finite) if there is no one-to-one mapping of S onto a proper subset of S. Every finite set is D-finite. Using the axiom of choice, one proves that every infinite set is D-infinite,and so D-finiteness is the same as

4. REAL

29

NUMBERS

finiteness. Without the axiom of choice, however, one cannot prove that every D-finite set is finite. The set N of all natural numbers is obviously D-infinite and hence every S such that 1 S 1 2 No, is D-infinite. Exercise 3.11. S is D-infinite if and only if S has a countable subset. [If S is D-infinite, let$ S + X c S be one-to-one. Let x,, E S - X and x,+ Then S 2 {x, : n < w).]

=f(x.).

Exercise 3.12.

(a) If A and B are D-finite, then A u B and A x B are D-finite. (b) The set of all finite one-to-one sequences in a D-finite set is D-finite. (c) The union of a disjoint D-finite family of D-finite sets is D-finite.

On the other hand, one cannot prove without the axiom of choice that a projection, power set, or the set of all finite subsets of a D-finite set is D-finite, or that the union of a D-finite family of D-finite sets is D-finite. Exercise 3.13. If A is an infinite set. then P P ( A ) is D-infinite. [Consider the set { { X c A : 1 X 1 = n) : n < 04.1

4. REAL NUMBERS

Theorem 10 (Cantor). The set ofall real nunibers is uncountable. Proof: Let us assume that the set R of all reals is countable, and let co, cl, . . . , c,, . . . , n E N , be an enumeration of R. We shall find a real number different from each c,. We let a, = c, , h, = ck where k iS the least k such that a, < ck , a,, = ck where k is the least k such that a , < c k < b,, b,+ = ck where k is the least k such that a,+ < ck < h, (see Fig. 4.1). I f we let a = sup{a, : n E N } , then a # ck for all k . H 1

a0

I

a1

I

a2

...

I I

I

bi

bi

bo

R

FIGURE 4.1

I t is the complrtenrss of the real line that accounts for uncountability of R. The real numbers are usually defined as Dedekind cuts in the set of rational numbers. We will not bother with a formal definition of rational numbers, and note in passing only that the set Q of all rationals is countable (since h’,, . No = No)and include another theorem of Cantor, which shows that the order-type of (Q, < ) is the unique order-type of a countable dense unbounded linearly ordered set.

1. AXIOMATIC SET THEORY

30

A linearly ordered set P is dense if for any a < b there exists c such that a < c < b. P is unbounded if it has no least and no greatest element.

Theorem 11 (Cantor). Any two countable linearly ordered sets that are dense and unbounded are isomorphic. Proof: Let P, = {a, : n E N) and P2 = {b, : n E N) be dense unbounded linearly ordered sets. We construct an isomorphism& P, -,P2in the following way (see Fig. 4.2): We first definef(ao), thenf- '(b0), thenf(a,), thenf- '(b,), etc., so as to keepforder-preserving. For example, to definef(a,), if it is not yet defined, we letf(a,) = b, where k is the least index such thatfremains order-preserving (such a k exists becausefhas been defined for only finitely many a E P,, and because P, is dense and unbounded). W

FIGURE 4.2

Exercise 4.1. Every countable linearly ordered set P can be embedded in the rationals; i.e., P is isomorphic to a set of rationals.

The real line is the unique dense, unbounded, linearly ordered set (up to isomorphism) that is complete and contains the rationals as a dense subset. A dense linearly ordered set is called complete if every nonempty bounded subset has a supremum. Let P be a dense linearly ordered set; a set D G P is dense in P if for any a < b there is d E D such that a < d < b.

'lleorem 12 (Completion ofdense linear ordering). Let (P, < ) be a dense unbounded linearly ordered set. Then there is a complete unbounded linearly ordered set (C, <) such that: (i) P E C, and < and 4 agree on P; (ii) P is dense in C.

The set C is unique up to isomorphism; moreover, if C, and C2are two such sets, then there is an isomorphism h between C and C2such that h(p) = p for all p E P.

,

Proof: A Dedekind cut in P is a pair (A, B) of disjoint nonempty subsets of P such that: (a) A u B = P; (b) a < b for any a E A and b E B; (c) If inf B exists, then inf B E B (note that sup A = inf B).

31

4. REAL N U M B E R S

,,

Let C be the set of all Dedekind cuts in P and let ( A B,) 4 ( A , , B2)if 3 8,).The set C is complete: If { ( A i ,Bi) : i E I } is a nonempty I Ai, I Bi) is its supremum. For p E P, let bounded subset of C,then

(ui ni

A , c A2 (and B ,

A, = {x

E

P :x < p},

B,

= {x E

P : x 2 p}

Then P' = {(A,, B,) : p E P} is isomorphic to P and is dense in C. To prove uniqueness of C,let C and C'be two complete dense unbounded linearly ordered sets, let P and P' be dense in C and C', respectively, and letfbe an isomorphism of P onto P'. Thenfcan be extended (uniquely) to an isomorphismf* of C and C': For x E C,letf*(x) = sup{f(p) : p E P and p I x}. H The real line R, or the continuum, is defined as the completion of the set Q of all rational numbers. It follows from the construction by Dedekind cuts that I RI I2'O. To show that I RI = 2'O, it is enough to exhibit a set of reals of cardinality 2'O. The Cantor Set

,

Let C be the set of all reals of the form a,/3", where each a, = 0 or 2. C is obtained by removing from the closed interval [0, 11, the open intervals

(3, t),(4, $), (3, $), etc. (the middle-third intervals). See Fig. 4.3. Cis in a one-toone correspondence with the set of all w-sequences of 0's and 2's and so (CI = 2'0. O

i

4

4

4

4

1

'

FIGURE 4.3 The Cantor set.

The Continuum Hypothesis The cardinal 2'O is also called the continuum. The continuum hypothesis (CH) is the conjecture that 2'0 = K, Neither CH nor its negation is provable in ZFC. Typical sets of cardinality 2'O are the set of all sequences of natural numbers, the set of all sequences of real numbers, the set of all complex numbers. This is because K,NO = (2'0)N0= 2'O, 2% * 2'O = 2"". R + R has cardinality 2& (while the set of all functions has cardinality 22"0). [A continuous function on R is determined by its values at rational points.]

Exercise 4.2. The set of all continuous functionst

Exercise 4.3. There are at least 2'" countable order-types of linearly ordered sets. [For every sequence a = (an : n E N ) of natural numbers consider the order type f,

= a0

+ t + a1 + t + + ... a2

where t is the order-type of integers. Show that if a # b, then

f,

# fb.]

32

1. AXIOMATIC SET THEORY

A real number is called algebraic if it is a root of a polynomial whose coefficients are integers. Otherwise, it is transcendental. Exercise 4.4. The set of all algebraic reals is countable.

Lemma 4.1. r f S is a countable set of reals, then I R

I

- S = 2'O.

If Proof: A trick: Use R x R rather than R (because IR x RI = 2'O). S E R x R is countable, then for some x E R, S n ( R x {x}) = 0 and IR x =2'O.

{}.I

Corollary. (a) There are 2'O irrational numbers. (b) There are 2"O transcendental numbers (Cantor).

Suslin's Problem The real line is, up to isomorphism, the unique linearly ordered set that is dense, unbounded, complete and contains a countable dense subset. Since Q is dense in R, every nonempty open interval of R contains a rational number. Hence if S is a disjoint collection of open intervals, S is at most countable. (Let (rn : n E A') be an enumeration of the rationals. To each J E S assign r, E J with the least possible index n.) Let P be a dense linearly ordered set. P is called separable if P has a countable dense subset. (Thus the real line is separable.) If every disjoint collection of open intervals in P is at most countable, then we say that P satisfies the countable chain condition (c.c.c.). Clearly, every separable linear ordering satisfies the C.C.C.

Suflin's Problem. Let P be a complete dense unbounded linearly ordered set that satisfies the countable chain condition. Is P isomorphic to the real line? This question cannot be decided in ZFC; we shall return to the problem in Section 22.

The Topology of the Real Line Open sets of reals are unions of open intervals; closed sets are complements of open sets. Since every nonempty interval has cardinality 2'9 it follows that every nonempty open set has cardinality 2'O. Every finite set is closed, and there are countable closed sets; it will follow from the Cantor-Bendixson theorem that every uncountable closed set has cardinality 2'O. The open intervals with rational endpoints (the rational intervals) form a basis of the topology of the real line. Thus every open set G is the union of all rational intervals (4,r ) such that (4,r) E G (and every closed set is F = R ((4,r ) : F n (4,r) = 0)). Hence the number of all open sets (= the number of all closed sets) is 2N0.

u

4. REAL

33

NUMBERS

A set of reals is compact if it is closed and bounded. A function8 R -+ R is continuous if and only iff- ,[GI is open for every open set G. Exercise 4.5. (a) If Co 2 C , 2 . . . 2 C, 2 ..., n E N is a descending sequence of nonempty compact sets then C, # 0. (b) If 1: R 4 R is continuous, and C is compact, then f [ C ] is compact.

n.'=o

A dense set ofreals means a set that is dense in R.

Theorem 13 (Baire Category Theorem). IfDo, D,,. . . , D,, . . . ,n E N , aredense open sets ofreals. then the intcmrction D = D, is dense in R.

n:=o

Proof: We show that D intersects every tmnempty open interval I . First note that for each n, Do n ... n D, is dense and open. Thus let (Jk : k E N) be an enumeration of rational intervals. Let lo = I , and let, for each n, I,, = J k = (qk, rk), where k is the least k such that the closed interval [ q k ,rk]is included in I , n D,.Then a E D n I, where a = limk-x q k . H

,

Corollary. The set Q of all rationals is not the intersection of a countable collection of open sets.

n;=o

G , , where G, are open sets, then each G, is dense in R. Prooj: If Q E Thus each G, - {q,,},where qn is the nth rational, is dense and open, and so G, - Q is nonempty.

Perfect Sets Let A be a set of reals. A real number a is a limit point of A if every neighborhood of a contains some x E A other than a ; a E A is an isolated point of A if there is a neighborhood U of a such that a is the only element of A in U . A set of reals is perfect if it is nonempty, closed, and has no isolated points. Exercise 4.6. The Cantor set is a perfect set.

Lemma 4.2. Every perfect set has cardinality 2'O. Proof: Given a perfect set P , we want to find a one-to-one mapping F of "'(0, I} into P . The construction of F resembles the construction of the Cantor set. We construct, by induction on length, for every finite (0, 1)-sequence s: n + (0, l}, elements a, of P such that: (4.1) if length(s) = length(t) and s # t , then a, # a,; (4.2) i f f € "'(0, l), then a/ = limsc, a, = limn+mall,, exists; (4.3) i f f # g, then a/ # a,. Since P is closed, each a/ is in P and F : f ~ + ais~a one-to-one mapping of N{O, 1) into f.

34

1. AXIOMATIC SET THEORY

Let a0 E P be arbitrary. If s = (s(O), . . . , s(n - l)), let SO and sl denote So = (s(O), . . . , s(n - l), 0) and sl = (s(O), . . . , s(n - l), 1). To satisfy (4.2) and (4.3) it is enough to make sure that for each s, a,, and a,, are close to a,. For instance, it suffices to let (4.4)

as, = as

and (4.51

[as, -as1 5 E n ,

where co = 1 and E,

=

min(la, - a , [ : s z t ~ " ( 0I}) ,

We can find a,, as follows: if it exists sup{x E P : a, < x Ia, + E,} (4.6) = infjx E P : as - E, I x < a,} otherwise Since a, is a limit point of P and P is closed, there is some a,, that satisfies (4.6). H

I

We shall show that every closed set is either at most countable or else contains a perfect subset and thus has cardinality 2'O. Lemma 4.3. Let F , 3 F , 3 ... 3 F , 2 ..., tl < 8, be a strictly descending sequence of closed sets. Then I9 is countable. Proof: For each o! < 19,there is a rational interval (q, r ) that is disjoint from F a + 1 but not from F a . However, there are only countably many rational intervals. H

Lemma 4.4. Every set of reds has at most countably many isolated points. Proof. Let A c R. Let J , , n E N, be an enumeration of rational intervals. For every isolated point a of A, let &(a)be the least k such that a is the only element of A in J , . Clearly &(a)# k ( b ) if a, b are distinct isolated points of A. H

Theorem 14 (Cantor-Bendixson). If F is an uncountable closed set, then F = P v S , where P is perfect and S is countable. Corollary. If F is a closed set, then either I F I I No or I F I = 2'O. Proof: For every A c R, let A' = the set of all limit points of A

(the derivative of A). It is easy to see that A' is closed, and if A is closed then s A. Thus we let Fo = F, F3+1 = Fl

A'

Fa = (I F , Y
if a > 0 is a limit

4.

35

REAL NUMBERS

Since Fo 2 F , 2 ... 2 F, 2 ..., there exists an ordinal 8 such that F, = F Bfor all a 2 8. (In fact, the least 8 with this property is countable, by Lemma 4.3.) We let P = F o . If P # 0,then P' = P and so it is perfect. Thus the proof is completed by showing that F - P is at most countable. We have F - P = UaGB (Fa - Fh); hence if a E F - P, then there is a unique a such that a is an isolated point of F a . As in Lemma 4.4, we let k(a) denote the least k such that a is the only point of Fain the interval J k . Note that if a I p and b E F , - F b , then b $ Jk(o),and hence k(b) # k(a).Thus the correspondence at+k(a) is one-to-one, and it follows that F - P is at most countable. H Exercise 4.7. If P is a perfect set and (a9b ) is an open interval such that P n (a, b ) # @, then ( P n (a, b)l = 2'0. Exercise 4.8. If P z $ P I are perfect sets, then 1 P - P [Use Exercise 4.7.1

,)= 2H?

If A is a set of reals, a real number a is called a condensation point of A if every neighborhood of a contains uncountably many elements of A. Let A* denote the set of all condensation points of A. Exercise 4.9. If P is perfect then P* = P . [Use Exercise 4.7.1 Exercise 4.10. If F is closed and P [ P = P* s F*.]

EF

is perfect, then P

C_

F*.

Exercise 4.11. If F is an uncountable closed set and P is the perfect set constructed in Theorem 14, then F* E P ; thus F* = P. [Every a E F* is a limit point of P since 1 F - P 1 2 No.] Exercise 4.12. If F is an uncountable closed set, then F = F* u (F - F*) is the unique partition of F into a perfect and an at most countable set. [Use Exercise 4.8.1

'Ihe Baire Space The Baire space is the space ,.v^ = "N of all infinite sequences of natural numbers, (a, : n E N), with the following topology: For every finite sequence s = (ak : k < n), let

us= { f E

N> : (Vk < n ) [ C ) = ak]) The sets (4.7) form a basis for the topology of JV. Note that each U s is also (4.7)

.N" : S

C f }=

{(ck : k

E

closed. A sequence ( a , : n E N ) is euentually constant if there is no such that a, = a,,, for all n 2 n o . The set of all eventually constant sequences is countable and is dense in ,A'", i.e., every nonempty open set in JV contains an eventuallv constant sequence.

1. AXIOMATIC SET THEORY

36

If ( a , : n E N) is a sequence of positive integers, let 1

bo=--,

1

b , =-

a0

+ a11 ’

a.

-

etc. Notice that (4.8)

b1 < b, < b, < ... < b4 < b2 < bo

It can be shown that {b,}F=oconverges; let ii = fraction (4.9)

b,; ii is the continued

1

a= a0

1

+ a,

+-a2 +1 ...

Obviously, 0 < ii < 1. It can be verified that if a and a’ are two different sequences, then ii # a’. Exercise 4.13. Every continued fraction is irrational. [Assume that ii = p / q . Then q = p . a.

1

+p .

1

‘

Remember the Euclidean algorithm.] Exercise 4.14. If c is an irrational number, 0 < c < 1, then there is a continued fraction such that c = ii. [Construct the sequence (ao, a l , .. .), using (4.8).] -

Thus if we assign to each a E A’”the continued fraction F ( a ) = a + 1, where a + 1 = ( a , + 1 : n E N), we have a one-to-one correspondence F between the Bake space and the set of all irrational numbers in the unit interval [0, 11. Exercise 4.15. The function F: JV -+ [0, 11 is continuous, and its inverse is also continuous. Thus the Baire space is homeomorphic to the space of all irrational numbers in [0, 11 (and hence to the set of all irrational numbers) with the topology of the real line.

Perfect Sets in the Baire Space Let Seq denote the set of all finite sequences of natural numbers:

u “N m

Seq=

n=O

4. REAL

37

NUMBERS

A (sequential) tree is a set T E Seq that satisfies (4.10) if f

E

T and s = t I n for some n, then s

E

T.

If T C Seq is a tree, let [TI be the set of all injnite paths through T : (4.11)

[TI = { f . ~ N : f / nE Tfor alln

E

N)

The set [TI is a closed set in the Baire space: LetfE .N be such thatfg [TI. Then there is n E N such thatf 1 n = s is not in T. In other words, the open set U s = {g E . : g 3 s}, a neighborhood of 1 1is disjoint from [TI. Hence [TI is closed. Conversely, if F is a closed set in - f ,then the set (4.12)

TF = {s E Seq : s cffor somef’E F}

is a tree, and it is easy to verify that [TF] = F: I f f € JV is such thatf In E Tfor. all n E N , then for each n there is some g E F such that g 1 n =f In; and since F is closed, it follows thatfe F. Exercise 4.16. The tree TFin (4.12)has no maximal node, i.e.,s E Tsuch that there is no t E T with s c t . The map F H T, is a one-to-one correspondence between closed sets in

. I ‘ and sequential trees without maximal nodes.

Iffis an isolated point of a closed set F in A”,then there is n E Nsuch that there is no g E F, g # J such that g In =f In. Thus the following definition: A nonempty sequential tree T is perfect if for every t E T there exist s 1 3 t and s2 3 t, both in T, that are incompatible, i.e., neither s1 2 s2 nor s2 2 sl. Lemma 4.5. A closed set F E .Af is perfect i f and only ifthe tree TF is a per$ect tree. H

The Cantor-Bendixson analysis for closed sets in the Baire space is carried out as follows: For each tree T E Seq, we let

T’= {t E T : there exist incompatible s1 3 t and s2 3 t in T} (Thus T is perfect if and only if 0# T = T.) (4.13)

The set [TI - [TI is at most countable: For eachfE [TI such thatfg [TI, let s = ,f In where n is the least number such that f I n 4 T.IfJ g E [TI - [TI, then sJ # sg, by (4.13). Hence the mappingfHsf is one-to-one and [TI - [TI is at most countable. Now we let

To= T,

(4.14)

T. =

f7

T,

T,+, = T,‘ if a > 0 is a limit

,
Since To 2 TI 2 . . . 2 T, 2 . . . , and To is at most countable, there is an ordinal 0 < o1such that &+,= &. If & # 0, then it is perfect.

38

1. AXIOMATIC SET THEORY

(4.15)

hence (4.15) is at most countable. Thus if [qis an uncountable closed set in X , the sets [To]and - [TO] constitute the decomposition of [qinto a perfect and an at most countable set.

[a

5. THE AXIOM OF CHOICE

Axiom of Choice (AC). Eoery family of nonempty sets has a choicefunction.

If S is a family of sets and 0 4 S, then a choicefunction for S is a functionf on S such that (54

f (XIE x

for every X E S. The axiom of choice postulates that for every S such that 0 @ S there exists a function f on S that satisfies (5.1). The axiom of choice differs from other axioms of ZF by stating existence of a set (i.e., a choice function) without defining it (unlike, for instance, the axiom of pairing or the axiom of power set). Thus it is often interesting to know whether a mathematical statement can be proved without using the axiom of choice. It turns out that the axiom of choice is independent of the other axioms of set theory and that many mathematical theorems are unprovable in ZF without AC. In some trivial cases, the existence of a choice function can be proved outright in ZF: (a) when every X E S is a singleton X = {x}; (b) when S is finite: the existence of a choice function for S is proved by induction on the size of S; (c) when every X E S is a finite set of real numbers: letf(X) = the least element of X. On the other hand, one cannot prove existence of a choice function (in ZF)just from the assumption that the sets in S are finite; even when every X E S has just two elements (e.g.,sets of reals), we cannot necessarily prove that S has a choice function. Using the axiom of choice, one proves that every set can be well-ordered, and therefore every infinite set has cardinality equal to some Nu.In particular, any two sets have comparable cardinals, and the ordering is a well-ordering of the class of all cardinals.

39

5. THE AXIOM OF CHOICE

Theorem 15 (Zermelo’s Well-Ordering Theorem). Every set can be well ordered. Proof: Let A be a set. To well-order A, it suffices to construct a transfinite one-to-one sequence ( a , : a < 0) that enumerates A. That we can do by induction, using a choice functionjfor the family S of all nonempty subsets of A. We let a, =f(A - {at : t < a } ) a. =f(A), if A - { a s : < a } is nonempty. Let 0 be the least ordinal such that A = (as : t < 0). Clearly, ( a , : tl < 0) enumerates A. H

<

In fact, Zermelo’s theorem is equivalent to the axiom of choice: If every set can be well-ordered,then every family S of nonempty sets has a choice function. To see this, well-order S and let f(X)be the least element of X for every

u

x E s.

Of particular importance is the fact that the set of all real numbers can be well-ordered. It follows that 2N0is an aleph and so 2’O 2 K,. The possibility of well-ordering the reals leads to some “unpleasant ” constructions of sets of reals. Well known is Vitali’s construction of a nonmeasurable set; another example is an uncountable set of reals without a perfect subset. Exercise 5.1. There exists a set of reals of cardinality 2N0without a perfect subset. [Let ( P , : a < 2No)be an enumeration of all perfect sets of reals. Construct disjoint A = {a, : a < 2’O} and E = {b, : a < 2No}as follows: Pick a,such that a, $ {a<: { < a} u {b, : < a}, and b, such that b, E P, - {ac: { 2 a}. Then A is the set.]

If every set can be well-ordered, then every infinite set has a countable subset: Well-order the set and take the first o elements. Thus every infinite set is Dedekind-infinite, and so finiteness and Dedekind finiteness coincide. Dealing with cardinalities of sets is much easier when we have the axiom of choice. In the first place, any two sets have comparable cardinals. Another consequence is : (5.2)

if f [ A ] = B,

then

IBI I IAl

To show (5.2), we have to find a one-to-one function from B to A. This is done by choosing one element fromJ’- , ( { b } )for each b E B. Another thing one can show using the axiom of choice is:

(5.3) The union of a countable family of countable sets is countable. (By the way, this often used fact cannot be proved in ZF alone.) To prove (5.3) let A, be a countable set for each n E N. For each n, let us choose an enumeraA,: tion (ank: k E N) of A,. That gives us a projection of N x N onto

u:=o

Thus

u;=,, is countable. A,

(n* k ) H a n k

1.

40

AXlOMATlC SET THEORY

In a similar fashion, one can prove a more general statement. ~emma5.1. [US1I IS\ . s u p { I x I : x E S } . Proof: Let K = IS I and 3, = sup{ I X I : X E S}. We have S = { X u : a < K} and for each a < K, we choose an enumeration X, = { a a a :p < I,},where I, I 1. Again, we have a projection (a9 aaa of K x 3, onto S , and so SI IK I. W

u

Iu

In particular, the union of K, sets, each of cardinality K,, has cardinality K,. Corollary. Every K,, is a regular cardinal. This is because otherwise cardinality at most K,.

would be the union of at most K, sets of

Exercise 5.2. If X is an infinite set and S is the set of all finite subsets of X , then

1st = 1x1. [Use

1x1 =K..] Axiom of Choice in Mathematics

In algebra and point set topology, one often uses the following version of the axiom of choice. We recall that if (P, < ) is a partially ordered set, then a E P is called maximal in P if there is no x E P such that a < x. If X is a nonempty subset of P, then c E P is an upper bound of X if x I c for every x E X. We say that a nonempty C c P is a chain in P if C is linearly ordered by < . Theorem 16 (Kuratowski-Zorn Lemma). If ( P , < ) is a nonempty partially ordered set such that every chain in P has an upper bound, then P has a maximal element. Proof: We construct (using a choice function for nonempty subsets of P), a chain in P that leads to a maximal element of P. We let, by induction, a, = an element of P such that a, > at for every 5 < a if there is one

<

Clearly, if a > 0 is a limit ordinal, then C, = {ac: < a} is a chain in P and a , exists by the assumption. Eventually, there is 8 such that there is no ao+ E P, a,+ > a,. Thus a, is a maximal element of P. Like Zermelo’s theorem, the Kuratowski-Zorn lemma is equivalent to the axiom of choice: Exercise 5.3. Using Theorem 16, prove the axiom of choice.

[Let S be a family of nonempty sets. To find a choice function on S,let P = { j : f i s a choice function on some Z E S } , and apply the Kuratowski-Zorn lemma to the partially ordered set (f,c).]

5.

41

THE AXIOM OF CHOICE

There are numerous examples of proofs using the Kuratowski-Zorn lemma. To mention only a few: (a) Every vector space has a basis. (b) Existence and uniqueness of algebraic closure of a field. (c) Hahn -Banach extension theorem. (d) Tychonoff product theorem for compact spaces.

Countable Choice and Dependent Choices Many important consequences of the axiom of choice, particularly many concerning the real numbers, can be proved from a weaker version of the axiom of choice. Countable Axiom of Choice. Every countable family of nonempty sets has a choice function.

For instance, the countable A C implies that the union of countably many countable sets is countable. In particular, the real line is not a countable union of countable sets. Similarly, it follows that K is a regular cardinal. O n the other hand, the countable A C does not imply that the set of all reals can be well-ordered. The countable AC implies that every infinite set has a countable subset. [If A is infinite, let A, = {s : s is a one-to-one sequence in A of length n } for each n. Use a choice function for S = { A , : n E N) to obtain a countable subset of A.]

Exercise 5.4.

In the next two exercises use the countable AC and note that the converse implication holds in ZF. Incidentally, neither of the two statements is provable in ZF alone. A counterexample is provided by an infinite, Dedekind-finite set of reals (in some model of ZF in which A C fails). Exercise 5.5. If A is a set of reals and a ( a , : n E N) in A such that a = lim a,.

is in the closure of A, then there is a sequence

Hence a set A E R is closed whenever an arbitrary convergent sequence (a,, : n E N ) of numbers in A converges to a E A. In order thatf: R -,R be continuous at a, it suffices thatf(a) for every sequence that converges to a. Exercise 5.6.

= limn f(a,)

We briefly mention a related principle which follows from the axiom of choice. Principle of Dependent Choices (DC). If p is a binary relation over a nonempty set A , and if for every a E A there exists b E A such that b p a, then there is a sequence a,, a , , . . . , a,,, .. . in A such that (5.4)

a , , , p a,

for all n E N

42

1.

AXIOMATIC SET THEORY

The principle of dependent choices is stronger than the countable axiom of choice. Exercise 5.7. Use DC to prove the countable AC. [Given S = { A n :n E NJ, consider the set A of all S, = { A i : i I n}, with the binary relation

2.1

choice functions on some

The import of the principle of dependent choices lies in the following characterization of well-founded relations and well-orderings :

Lemma 5.2. (a) A linear ordering < ofa set P is a well-ordering ofP ifand only ifthere is no injnite descending sequence a0

> a, >

. a *

> a, > ..'

in A . (b) A relation E over P is well-founded ifand only ifthere is no infinite sequence (a, : n E N) in P such that (5-5)

a,,

, E a,

for all n E N

Proof: Note that (a) is a special case of (b) since a well-ordering is a wellfounded linear ordering. If a,, a,, . . . , a,, . . . is a sequence that satisfies ( 5 . 9 , then the set {a, : n E N) has no E-minimal element and hence E is not well-founded. Conversely, if E is not well-founded, then there is a nonempty set A E P with no E-minimal element. Using the principle of dependent choices we construct a sequence a o ,a,, . .., a,, . . . that satisfies (5.5). 6. CARDINAL ARITHMETIC

In the presence of the axiom of choice, every set can be well-ordered and so every infinite set has the cardinality of some K,. Thus addition and multiplication of cardinal numbers is simple:

K, + K, = K, . K, = max(K,, K,) (6.1) The exponentiation of cardinals is more interesting. This section is concerned with the operations 2'n and Krn. Lemma 6.1. rfa I b, then

K,NS

= 2'8.

Proof:

2P' I h',Nn (2'n)'p 5s'2 (6.2) If K is a cardinal less than K,, then by (6.2)

'

'I,

= 2%

Ec," I K,K, = 2N3= 1 P ( 0 , ) I

6.

43

CARDINAL ARITHMETIC

In fact, we can express K,"as the cardinality of a certain subfamily of P(ma). I f A i s a s e t a n d ~ sIA1,wedefine [A]" = the set of all X

c A such that I X I = K

Similarly, [A]<"={XcA:

Lemma 6.2. If

K I

K,, then K,"=

1x1 < K }

I [ma]" I.

Proof. On the one hand, every8 K

ma is a subset of

K

1 I = K.

x ma, and f

Thus

K,"= IKmaI II[.

= I[ma]"I.

x

On the other hand, we construct a one-to-one function F: [m.]" + "ma to show that I [coalKI I K,".If X c ma and I X I = K, let F(X) =f be some function f on K whose range is X. Clearly, F is one-to-one. Infinite Sums and Products

Let

{ K ~: i E

I } be an indexed set of cardinal numbers. We define

where {Xi : i E I} is a disjoint family of sets such that 1 X i I = xi for each i E I. Exercise 6.1. If (Xi : i E I} and IXiI = X I for each i E I, then

I

{ :i

E

I} are two disjoint families such that

lUirl X i [ = lUirl X I .

[Use AC.]

Note that if

K

and I are cardinals and

K~ = K

for each i < I, then

In general, we have the following

Lemma 63. If I is an injinite cardinal and

c

i
Ki

=I

K~

> 0 for each i < I , then

. sup K i i
xi<*

Proof: Let K = supi<*K~ and cr = K ~ .On the one hand, since K~ I K for all i , we have (T I < K = I . K. On the other hand, since K~ 2 1 for all i, we have I= 1 Icr, and since (T 2 K~ for all i, we have cr 2 supi< K~ = K. Therefore

ci

0 2 I . K .

In particular, if I I

K ~ we ,

have

44

1. AXIOMATIC SET THEORY

Thus we can characterize singular cardinals as follows: An infinite cardinal K is singular just in case

where I < K and each K~ .< K. An infinite product of cardinals is defined using infinite Cartesian products. If { X i : i E I} is a family of sets, then the Cartesian product is defined as follows: (6.5)

n Xi

= {f:fis

a function on I andf(i) E X i for each i E I}

icl

In other words, elements of

H i e l X i are functions (xi : i E I)

such that x i E X i for all i E I. Note that if some X iis empty, then the product is empty. If all the X i are nonempty, then AC implies that the product is nonempty. If { K :~ i E I} is a family of cardinal numbers, we define

where {Xi : i E I} is a family of sets such that I X i ( = tii for each i E I. (We abuse the notation by using both for the Cartesian product and the product of cardinals.)

fl

Exercise 6.2. If { X i : i E I } and { y i : i E I} are such that I X i I = I yi I for each i E 1, then lniel X i ( = (ni61yi(. [Use AC.]

If K~ = K for each i E I, and 1 I I = 1,then f l i pI K~ = .'it Also, infinite sums and products satisfy some of the rules satisfied by finite sums and products. For K; = (Hi xi)A, or tiii = ) i T i A i . Or if I is a disjoint union instance, I= A j , then

ujEni

ni

(the associative law). See Fig. 6.1. If k i 2 2 for each i E I, then

(The assumption K~ 2 2 is necessary: 1 + 1 > 1 1.) If I is finite, then (6.8) is certainly true; thus assume that I is infinite. Since fli 1 t i i 2 I 2 = 2'" > [ I [ , it suffices to show that X i K~ I( I I . K ~ .If { X i :i E I} is a disjoint

ni

ni

6.

45

CARDINAL ARITHMETIC

I

Aj

FIGURE 6.1

family,weassigntoeachxE uiXiapair(i,f)suchthatx E X i , f ~ f l i X i a n d f(i) = x. Thus we have (6.8). The infinite product of cardinals can be evaluated using the following lemma : Lemma 6.4. 1/'1is an infinite cardinal and ( K :~ i < 1) is a nondecreasing sequence of' nonzero cardinals, then

fl

(6.9) Prooj: Let

Ki

= (SUP K i )1

i
K = sup, K ~ Since . tii

i

5 K for each i < A, we have

fl

To prove that 'K I xi, we consider a partition of 1into A disjoint sets A j , each of cardinality 1:

A = IJ A j

(6.10)

j < l

(To get a partition (6.10),we can, e.g., use the canonical one-to-one mapping r: I x I -+ I, and let A j = T[1 x { j } ] . )Since a product of nonzero cardinals is greater than or equal to each factor, we have K~ 2 supiEAj K~ = K. for each j < I. Thus, by (6.7),

niEAj

Exercise 6.3.

flo
Exercise 6.4.

n.<wK.

Exercise 6.5.

n.

+

=

K?.

K, = h'"O 'I,

+

,i,

'

The few strict inequalities in cardinal arithmetic proved in Section 3 can be obtained as special cases of the following general theorem.

'Iheorem 17 (Konig). Iftii < Aifor eoery i

E

I, then

1.

46

z.

AXIOMATIC SET THEORY

ni

Proof: We shall show that xi 8 l i .Let T , i E I, be such that I Ti I = l i for each i E I. See Fig. 6.2. It suffices to show that if Zi, i E I, are subsets of T = HiGI '&,and lZil I K~ for each i E I, then Vie,Zi # T.

I

i

FIGURE 6 2

For every i E I, let Si be the projection of Zi into the ith coordinate: Si = { f ( i ) :

fZi} ~

Since I Z i I c I '& I, we have Si c '&. Now let f E T be a function such that # Si for every i E I. Obviously,fdoes not belong to any Zi, i E I, and so U i e , zi # T* Corollaries

f(i)

1.

K

c 2 " f i r eoery 1

K:

+ 1 + ...

(K

times) c 2 . 2

(K

times)

2. cf (2'9 > K, .

It suffices to show that if K~ c 2'a for i c o,, then

c

3. cf (K,nt) > K, .

Ki

n

<

i
l j = (2Na)N. = 2'"

, then

4. uCf > K for eoery infinite cardinal K~

c K, i c cf K, be such that K =

< 2's Let l i= 2's

i
We show that if tci c K,Np for i <

Let

tci

c

=

K

<08

K~

Xi

K


fl

K ~ Then . K = Kcfn

i
"he Continuum Function

By Cantor's theorem, 2'" 2 K,,

for every a.

Generalized Continuum Hypothesis (GCH)

2& = K,,

c K:p. Let li = KZp.

K.

Ki<

i
xi

for all a

6.

47

CARDINAL ARITHMETIC

The generalized continuum hypothesis is independent of the axioms of ZFC. At this point, we shall give some properties of the function

K, H 2U" called the continuum function. To start, it is obvious that if a I B then 2 u a2'@ ~

(6.11)

By Cantor's theorem and by Corollary 2 of Konig's theorem, we have 2H" > K , ,

(6.12)

cf(2Na) > K,

It can be shown (Easton's theorem) that for regular cardinals, conditions (6.11) and (6.12)are the only restrictions on the continuum function provable in ZFC. The case of singular cardinals is more interesting. For example, in the absence of large cardinals (namely if 0' does not exist), the continuum function is determined by its values at regular cardinals (Jensen). Some of the recent results are presented in Section 8. In the following case, the value of 2" is determined by the values of the continuum function below K . Theorem 18 (Bukovsky-Hechler). Let

K be a singular cardinal such that the continuum function below K is eventually constant, i.e., there exists y o c K such that 2y = 2y0 for all 7, yo Iy < K Then 2" = 2y0.

The theorem is a consequence of the following lemma which gives us a formula for 2" at a singular cardinal K . If K is a limit cardinal, let us define 2'" = sup{2A : I, < K} Note that if the GCH holds, then 2'" = K for every limit cardinal Lemma 15. If

K.

is a limit cardinal, then 2" = (2'K)CfK Proof: Let K = K ~ where , K~ < K for each i. We have 2" = 2LKi = 2"' I 2'" = (2'K)Cr" 5 (2K)ErK= 2" K

n

fl

i

i

Proof of Theorem 18. If K is a singular cardinal that satisfies the assumption of the theorem, then there is yo such that cf K I yo < K and that 2'" = 2y0.Thus

2" = (2CK)EfK= (2YO)EfK= 2 Y O If K is a limit cardinal and if the continuum function below K is not eventually constant, then the cardinal I = 2'" is a limit of a nondecreasing sequence

1.

48

of length

K.

AXIOMATIC SET THEORY

By Lemma 3.4b, we have Cf

1 = Cf K

Using Lemma 6.5, we get 2" = ( 2 C K ) C f K = J C f l

(6.13)

If

ti

is a regular cardinal, then

K

= cf ti; and since 2" = ti", we have

2" = KcfK

(6.14)

Thus (6.13) and (6.14) show that the continuum function can be defined in terms of the function 2(ti)

= ticfK(the gimel function).

ti is a successor cardinal, then 2" = X (ti). (b) If ti is a limit cardinal and if the continuum function below ti is eventually constant, then 2" = 2'" . X(K). (c) If ti is a limit cardinal and if the continuum function below ti is not eventually constant, then 2" = X(2'").

(6.15) (a) If

Exercise 6.6. If fl is such that [Let fl 2 w. Let a be least K = N.+.; since cf ti = cf a 5 2Hrn'i = K a + S + B = K.+B. By K,+p < K.+.+B.I

2Ha= K a + Bfor every a, then fl < w . such a + fl > fl. We have 0 < a 5 fl, and a is a limit. Let a < K, K is singular. For each 5 < a, ( + fl = fl, and so Theorem 18, 2" = K a + B , a contradiction, since

Exponentiation of Cardinals We are interested in the operation NrB. If u 5 j,then Nro = 2'8 and so we shall concentrate on the case u > j. We start with the following observation: If ti is a regular cardinal and I < K, then every function f: A -P ti is bounded (i.e., supf[l] < K). Thus (6.16)

and so (6.17) a<"

In particular, if

K

is a successor cardinal, we obtain the Hausdorffforrnula

6.

49

CARDINAL ARITHMETIC

Exercise 6.8. If K is a limit cardinal and I, < cf

K,

then

K'

=

1. I a 1'
[Note that (6.16) holds in this case.]

In general, we can compute Lemma 6.6. I f

K

K'

using the following lemma:

is a limit cardinal, and I 2 cf 'K =

Proof: Let K l I

K =

(n

xi<

cf

then

(lim a a ) c f R U+K

I(

K ~ where , K~ < K

n (lim

A

xi)

K,

=

K? I

i

i
o1-K

for each i. We have

a') =

(lim a a y f K 5 ( K a y f K = Ka 01-K

Theorem 19 (Inductivr computation qf'hr:~). Let K, he j x e d . Then K.';'p can be computed as follows: (i) I f a fl, then K,Kp = 2'0. (ii) If there is y < a such that K,Np 2 K,, then K,K"= K,Np. (iii) l f a > p and lfK,Np < K, for all y < a, then: (a) if K, is regular or cf K, > K,, then K,Np = K,; (b) [f cf K, IK,,< K, , then K,KP = Kzf ". ProoL (i) has been already proved. (ii) K,Np Ih',Kp I (h'?)'~ = h'y. (iii) If K, is a successor cardinal, use the Hausdorff formula. If K, is a limit cardinal, then note that (6.19)

If cf K, > K,, then every 1: (0, + to, is bounded, and (6.19) gives K:P = HE.If cf K, 5 K, < K,. then use Lemma 6.6 and (6.19) to conclude that K,HP = KCf 'a 1 .

.

Corollary 1. For all a,fl, the talue of K,"' is always either: (a) 2'11 0 I' (b) Kz nr /or. sottie y I s(, whrrr X , is such t h t cf X , s S, < X, . (c) X;" (If neither (a) nor (b) holds, let K, be the least K such that K'P = K:") Corollary 2. I f the GCH holds, then (6.20)

h',Na=

1

1%.

1

if K, < cf K, if c f K , < K , < K , if K, I K,

(6.20) holds because if we assume the GCH, then

K,"'" for all a.

=p 7 = Nu+

50

1. AXIOMATIC

SET THEORY

In the course of investigation of the functions 2". and S\;J~ we have proved the following theorem.

Theorem 20 (Bukovsky). The continuum function can be dejined from the gimel function. The exponentiation function K,Ka can be defined in terms of the gimel function and the cofinality function cf K. This theorem shows the importance of the gimel function. We shall return to it below. Exercise 6.9.

Ktfl

=

K? . 2'l.

Exercise 6.10. If u < (ol, then Kr' = K.N" 2". Exercise 6.11. If a < ( 0 2 , then

h't2= K,"

. 2N2.

A cardinal K is a strong limit cardinal if

2A< ti

A
for every

Obviously, every strong limit cardinal is a limit cardinal. If the GCH holds, then every limit cardinal is a strong limit. It should be easy to see that if K is a strong limit, then ,Iv< K

for all A, v < K

An example of a strong limit cardinal is K O .Actually, the strong limit cardinals form a proper class: If o! is an arbitrary cardinal, then the cardinal K = sup(o!, 2", 22'.

. . .)

(of cofinality w ) is a strong limit. Another fact worth noticing is

(6.21) If

K

is a strong limit, then 2" = ticr

This need not be true if K is a singular cardinal that is not a strong limit. If limit cardinal, let us define

(6.22)

K
-

I[K]'"I

K

is a

= lim 'K "+K

Obviously, K
Exercise 6.12. If

limit then

ti<''

2 2'" 2 K

is regular and a limit, then

ti

ticK=

2'". If

ti

is regular and a strong

= ti. = 2"' >

Exercise 6.13. If

ti

is singular and is not a strong limit, then

Exercise 6.14. If

ti

is singular and a strong limit, then 2'" = ti and

ti<''

ti.

'.

ti
6.

51

CARDINAL ARITHMETIC

For the sake of completeness, we define infinite cardinal K : 1'" = s u p ( 2 :

A<" for any cardinal A and any

In particular, we have A < H + = 1".

Properties of the Gimel Function Theorem 20 shows the significance of the function 3 ( ~ = ) K" " in cardinal arithmetic. We shall discuss here some properties of the gimel function. As shown earlier, we have ](K) > K

(6.23)

cf I(K) > cf

and

K

for all K. Also, if K is regular then I ( K ) = 2". If K is singular, then in the following case, J(K) is determined by the behavior of 3 below K: A singular cardinal K is bound if (6.24)

1C'"

2K

for some 1< K. Using the computation from Theorem 19, it follows that ti is bound just in case K I l(1) for some 1< K of cofinality I cf K (see Exercise 6.15 below). A strong limit singular cardinal is not bound, but there may exist singular cardinals that are neither strong limits nor bound (e.g., K, if 2N0= K 1 and 2'' > K,). If ti is a bound singular cardinal and (6.24) holds for some 1< ti, then (6.25) - 1'' K and the value of >(ti) is determined by exponentiation of smaller cardinals. ((6.25) is proved as follows: A"'" I K" I (1"K)C' " = Acr ".) Exercise 6.15. If K is a bound singular cardinal, then I(K)is either 2"', or I(1)where 1is the least cardinal 1 2 cf K that satisfies (6.24). [Use Corollary 1 of Theorem 19.1 Exercise 6.16. If

2'O

Exercise 6.17. If 2'l

> K,, then Kto= 2% =

K2 and K? > K,,, then K t ; = K?.

Exercise 6.18. If 2'O 2 K,,, then X(K,) = 2'O and X(K,,) = 2".

If a singular cardinal K is not a strong limit, then 3 ( ~ is) subject to the following condition: If K I X(A) for some 1< K such that cf K I cf 1, then (6.26)

I@) I

>(A)

(This is clear since K" " I(1'' ')cf " = IZ"'.) Apart from this condition (6.26) and a further condition on cofinality (Exercise 6.19), cardinal arithmetic alone does not seem to impose further restrictions on the behavior of 3 (or of the continuum function). In Section 8 we present a result of Silver that gives additional substantial

52

1. AXIOMATIC SET THEORY

information on the continuum function of singular cardinals of uncountable cofinality. And a recent result of Jensen shows that unless 0" exists (see Section 30 for a formulation of this large cardinal axiom), then whenever K is a singular cardinal such that 2" < K , we have K" = K', and so all cardinal exponentiation is determined by the continuum function on regular cardinals and the cofinality function). If K is singular and not bound, then c f ( l ( K ) ) 2 least a such that A" 2 K for some 1 < K. In particular, if K is a strong limit, then cf(I(K)) > K. [Since K is not bound, we have a > cf K. Let cf li I y < a. Since 1' < K for all I < li, we have K~ = K" '. However, c f ( K y ) > y.]

Exercise 6.19.

Exercise 6.20.

If 2" = K 2 ,then K$

+ Nu,.

Inaccessible Cardinals We recall that K is weakly inaccessible if it is uncountable, regular, and a limit. We say that a cardinal K is inaccessible (strongly) if K > KO,K is regular, and K is a strong limit. Every inaccessible cardinal is weakly inaccessible. If the GCH holds, then every weakly inaccessible K is inaccessible. The inaccessible cardinals owe their name to the fact that they cannot be obtained by the usual set theoretical operations (a precise statement will be proved in Chapter 2): If K is inaccessible and IX I < K , then I P ( X ) l < K. If IS1 < K and if I X I < K for every X E S, then I S I < K . In fact, KO has this property too. Thus we can say that in a sense an inaccessible cardinal is to smaller cardinals as No is to finite cardinals. This is one of the main themes of the theory of large cardinals in Part 111.

u

7. FILTERS AND IDEALS. CLOSED UNBOUNDED SETS

Filters and ideals play an important role in abstract mathematical disciplines (algebra, logic, measure theory). In this section we introduce the notion of filter (and ideal) over a given set. The notion of ideal extrapolates the notion of small sets: Given an ideal I over S, a set X c S is deemed to be small if it belongs to I. Afilter over a set S is a collection F of subsets of S such that: (7.1)

(i) S E F ; (ii) if X E F and Y E F, then X n Y E F; (iii) if X, Y E S , X E F, and X c Y, then Y E F.

53

7. FILTERS A N D IDEALS. CLOSED UNBOUNDED SETS

A filter F is proper if 125 $ F . We shall consider only proper filters; thus we always assume S # 125 and

( i d 0 !# F. An ideal over a set S is a collection 1 of subsets of S such that:

(7.2)

(i) 0E 1; (ii) i f X E I a n d Y E 1 , t h e n X u Y E I ; (iii) if X, Y c S, X E 1, and Y c X, then Y

E

1.

We shall consider only proper ideals:

(iv) S !# 1 If F is a filter over S, then the set 1 = {S - X : X E FJ (7.3) is an ideal over S; and conversely, if 1 is an ideal, then

F = {S -

(7.4)

x :x E I}

is a filter. If (7.3), or (7.4), holds, we say that F and I are dual. Examples 1. A triiial filter: F = { S } . 2. A principal filter. Let X,,be a nonempty subset of S. The filter

F = { X G S : X 2 X,} (7.5) is a principal filter. The dual notions are a trivial ideal and a principal ideal. 3. The FrPcketJiltrr. Let S be an infinite set, and let 1 be the ideal of all finite subsets of S. The dual filter F

=

{X E S : S - X is finite)

is called the Frechet filter over S. Note that the Frechet filter is not principal. Exercise 7.1. Exercise

Every filter over a finite set is principal.

7.2. If F is a filter and X E F, then P ( X ) n F is a filter over X

7.3. Let K be an infinite cardinal. I S I 2 nonprincipal ideal over S . Exercise

K.

The set {X c S : I X

I < K } is

a

Exercise 7.4. Let A be an infinite set and let S = [A]<"' be the set of all finite subsets of A . F o r each P E S, let P = ( Q E S : P G Q ) . Let F be the set of all X c S such that X 2 for some P E S. Then F is a nonprincipal filter over S.

1.

54

AXIOMATIC SET THEORY

Let {Ai : i E I } be an indexed family of nonempty sets and let us consider the Cartesian product

Let further F be a filter over the set I of indices. We define the following relation over P:

f =Fg

(7.6)

iff

{i

E

I :f ( i ) = g(i)}E F

(that is, if there is only a small set of i's such that f ( i ) # g ( i ) ) . It follows easily from the properties (7.1) of F that = is an equivalence relation over P. The set of all = equivalence classes in P is called the reduced product of A i , i E I , mod F

n

(7.7)

Ai =

F

In case that A i = A for each i E I, the reduced product (7.7) is called the reduced power of A mod F. A family G of sets has the Jinite intersection property if every finite H = {Xl,. . . , X,) c_ G has a nonempty intersection:

X 1n . . . n X , # f Z I Every filter has the finite intersection property. Lemma 7.1.

n u P ( S ) has thefinite intersection property, there is aJilter F over S such

(a) I f 9 is a nonempty family offlters over S , then 9 is afilter over S . (b) I f % is a c-chain ofJilters over S , then %? is aJilter over S .

(c) If G c that G E F .

Proof: (a), (b) left to the reader; verify (7.1). (c) Let F be the set of all X E S such that there is a finite H = {XI, ..., X,} E G such that

X 1 n ... n X, E X (7.8) It is easy to verify that F is a filter. U Since every filter F 2 G must contain all finite intersections of sets in G, it follows that the filter F constructed in the proof of Lemma 7.lc is the least filter over S that extends G: F=

n {D: D is a filter over S and G E D}

The filter F is generated by G. Exercise 7.5.

The filter from Exercise 7.4 is generated by the sets {a}.,

II E

A.

7.

FILTERS A N D IDEALS. CLOSED UNBOUNDED SETS

55

A filter U over a set S is an ultrafilter if

(7.9) for every X c S, either X

E

U or S - X

E

U

(but not both: X n (S - X )= 0, and 0$ U). The dual notion is a prime ideal: For every X zS , either X E I or S - X E I. Note that I = P ( S ) - U . Exercise 7.6. If U is an ultrafilter and

X u Y E U ,then either X

E U

or Y E U .

A filter F over S is maximal if there is no filter F' over S such that F c F'. Lemma 1.2. A filter F over S is an ultrafilter i f and only i f it is maximal. Proof: (a) An ultrafilter U is clearly a maximal filter: Assume that U c F and X E F - U . Then S - X E U , and so both S - X E F and X E F , a contradiction. (b) Let F be a filter that is not an ultrafilter. We find F' 3 F: Let Y c S be such that neither Y nor S - Y is in F . Consider the family G = F u { Y } ; we claim that G has the finite intersection property. If X E F, then X n Y # 0, for otherwise we would have S - Y 2 X and S - Y E F. Thus, if X I , . . . , X , E F , we have X , n ... n X, E F and so

YnX,n.-.nX,,+@ Hence G has the finite intersection property, and by Lemma 7.lc there is a filter F' 2 G . Since Y E F' - F, we have F c F'. Theorem 21 (Tarski's Ultrafilter Theorem). Every filter over S can be extended to an ultrafilter. Proofi Let Fo be a filter over S. Let P be the set of all filters F over S such that F 2 F o and consider the partially ordered set (P, c).If V is a chain in P, then V is a filter and hence an upper bound of V in P. By the by Lemma 7.lb, Kuratowski-Zorn lemma there exists a maximal element U in P. This U is an ultrafilter by Lemma 7.2.

u

For every a E S, the principal filter (7.10)

{X Es:a E X }

is an ultrafilter. If S is finite, then every ultrafilter over S is of the form (7.10). If S is infinite, then there is a nonprincipal ultrafilter over S: If U extends the Frechet filter, then U is nonprincipal. A filter F over S is countably complete (a-complete) if whenever { X , : n E N) is a countable family of subsets of S and X, E F for every n, then (7.11)

fi X , E F

n =O

1. AXIOMATIC SET THEORY

56

A countably complete ideal (a a-ideal) is such that if X,, E 1 for every n, then

fi X , E I

n=O

More generally, if K is a regular uncountable cardinal, and F a filter over S, then F is called K-complete if F is closed under intersections of less than K sets, i.e., if whenever {Xg: a < y } is a family of subsets of S, y < K, and X, E F for every a < y, then (7.12)

The dual notion is a K-complete ideal. A o-complete filter is the same as K,-complete. Every filter is closed under finite intersections. Every principal filter is closed under arbitrary intersections. Exercise 7.7. If w, is singular, then there is no nonprincipal w.-complete ideal over w,. [Let X = {a : {a} E I}. X is the union of < K.sets of cardinality < K,, and so X E I.]

There is no nonprincipal o-complete filter over a countable set S. If S is uncountable, then { X E S : 1x1 I N o } is a a-ideal over S. Similarly, if K > w is regular and IS I 2 K , then { X E S : 1x1


is the least ti-complete ideal over S containing all singletons { a } . The question whether a nonprincipal ultrajlter over a set can be o-complete gives rise to deep investigations of the metamathematics of set theory. In particular, if such ultrafilters exist, then there exist many large cardinals (inaccessible, etc.). We shall return to this problem in Part 111.

Closed Unbounded Sets Let ti be a regular uncountable cardinal. We call a set C bounded in K if (7.13)

GK

closed un-

(i) for every sequence a. < a1 < *.. < a; < ...(( < 7 ) of elements of C , of length y < K , we have liml+y a; E C (closed); (ii) for every a < ti, there is j3 > a such that j3 E C (unbounded).

Lemma 7.3. lf C and D are closed unbounded, then C n D is closed unbounded. Proof: It is immediate that C n D is closed. To show that C n D is unbounded, let a < K. Since C is unbounded, there exists a > a, a E C. Similarly, there is a2 > a,, u2 E D. In this fashion, we construct an increasing sequence

,

(7.14)

a

< a1 < u2 < ... < a,, < ...

,

7.

57

FILTERS A N D IDEALS. CLOSED UNBOUNDED SETS

such that a ] , a 3 ,a s , ... E C , and a 2 , a4, a 6 , .. . E D. If we let the sequence (7.14), then f l < ti, and fl E C and /3 E D. H

be the limit of

The collection of all closed unbounded subsets of K has the finite intersection property. The filter generated by the closed unbounded sets consists of all X E K that contain a closed unbounded subset. We call this filter the closed unbounded filter over ti. The set of all limit ordinals a < ti is closed unbounded in K. If A is a subset of ti, we say that a limit ordinal a < ti is a limit point of A if 0 < a = sup@ n a). If A is an unbounded set of K, then the set of all limit points of A is closed unbounded. A function j ti ti is normal if it is increasing and continuous (f( a ) = lim,+, f ( l )for every nonzero limit a < K ) . The range of a normal function is a closed unbounded set. Conversely, if C is closed unbounded, there is a unique normal function that enumerates C .

-.

Exercise 7.8. The set of all fixed points (i.e..f(a) = a ) of a normal function is closed unbounded. Exercise 7.9. Iff

K

-, K , then the set of all a < K such that/[a]

E a is closed

unbounded.

The closed unbounded filter over ti is K-complete: Lemma 7.4. The intersection of less than K closed unbounded subsets of closed unbounded.

ti

is

Proof: We prove, by induction on y < K, that the intersection of a sequence {C, : a < y} of closed unbounded subsets of ti is closed unbounded. The induction step works at successor ordinals because of Lemma 7.3. If y is a limit ordinal, we assume that the assertion is true for every a < y; then we can replace each C, by (7,suC , and obtain a decreasing sequence with the same intersection. Thus assume that Co 2 C1 2 ... 2 C , 2 ... (a < y) are closed unbounded, and let C = C,. It is easy to see that C is closed. To show that C is unbounded, let a < ti. We construct a y-sequence P o < P I < ’ . . < P, < ... (7.15) (t < Y)

0,.

as follows: We let Po E C o be such Po > a, and for each 5 < y , let P, E C , be such that P, > sup{P, : I < t). Since ti is regular and y < ti, such a sequence (7.15) exists and its limit P is less than K. For each q < y, /? is the limit of a sequence ( P , : r] I 5 < y ) in C , , and so P E C,. Hence j? E C. H Let ( X u: a < K ) be a sequence of subsets of K. The diagonal intersection of X u ,a < ti, is defined as follows: (7.16) A{X,:a
1.

58

AXIOMATIC SET THEORY

Note that AX, = AY, where Y, = {t E X, : 5 > a } ; see Fig. 7.1. Note also that

AX, =

n. (x,u (5 : 5 I

a}).

Lemma 7.5. The diagonal intersection of a K-sequence of closed unbounded sets is closed unbounded. Pro05 Let ( C , : a < K ) be a sequence of closed unbounded sets. It is clear from the definition that if we replace each C , by (7esaC,, the diagonal intersection is the same. In view of Lemma 7.4 we may thus assume that CIJ2 c1 3 " ' 3 c, 2

*"

('Z

C K)

FIGURE 7.1

Let C = A{C, : a < K}. To show that C is closed, let a be a limit point of C. We want to show that a E C, or that a E C , for all 5 < a. If 5 < a, let X = {v E C : 5 < v < a}. Every v E X is in C , , by (7.16). Hence X c C , and a = sup X E C , . Therefore a E C and C is closed. To show that C is unbounded, let a < K. We construct a sequence (B,, : n < o) as follows: Let B0 > a be such that Bo E C o , and for each n, let Bn+ > B,, be such that B,+ E Cpn.Let us show that /3 = limn B, is in C: If 5 < /?,let us show that B E C , . Since 5 < B, there is an n such that 5 < fin.Each BL,k > n, belongs to C , and so B E C,". Therefore B E C , . Thus /? E C , and C is unbounded. W Corollary. The closed unbounded filter over intersections.

K

is closed under diagonal

We say that a set S E K is stationary in K if S n C # 0 for every closed unbounded subset C of K . A set T c K that is not stationary will be called thin. Note that the collection of all thin sets is the ideal dual to the closed unbounded filter; we shall call this ideal the thin ideal over K. Exumple, The set S = {a c K : cf a = o}is stationary in K. Similarly, if K > ol, the set S = {a -= K : cf a = ol} is stationary in K. Exercise 7.10. If S is stationary and C is closed unbounded, then S n C is stationary.

An ordinal function f on a set S is called regressive iff(.) nonzero a E S.

< a for every

7.

FILTERS AND IDEALS. CLOSED UNBOUNDED SETS

59

Theorem 22 (Fodor). I f f is a regressioe function on a stationary set S o c K , then there exists a stationary set S c S o and some y < K such that f ( a ) = y for all a E S. Proof. Let us assume that for each y < K, the set { a E So :f ( a ) = y} is thin, and choose a closed unbounded set C , such that f ( a ) # y for each a E S o n C,. Let C = A{C, : y < K } . The set So n C is stationary and if a E S o n C,we have f ( u ) # y for every y < a ; in other words,f ( a ) 2 a. This is a contradiction. Exercise 7.11. If T c K is thin, then there exists a regressive functionfon { a : f ( a ) 5 y} is bounded, for every y < K. [Let C n T = 0,and let f ( a ) = sup(C n a).]

7'such that

The closed unbounded filter over K is not an ultrafilter. This is because there is a stationary subset of K whose complement is stationary. If K > wl, this is clear: The sets { a : cf a = o}and { a : cf a = w l } are disjoint. If K = wl. the decomposition of w l into disjoint stationary sets uses the axiom ofchoice. It is an interesting problem (related to large cardinals) to show that the use of AC is necessary. In fact, every stationary S c K can be decomposed into K disjoint stationary sets. We shall prove this in Section 35. Here is a weaker statement:

Lemma 7.6. Every regular uncountable sets.

K

is the union of

K

disjoint stationary

Proof: Let W = { a < K : cf a = (03. For every u E W , we choose an increasing sequence (a: : I I E N } such that limna,' = a. See Fig. 7.2. First we show that there is an n such that for all q < K, the set

(7.17)

{ a E W : a : 2 q}

is stationary. Otherwise there is q,, and a closed unbounded set C , such that a: < q, for all a E C, n W, for every n. If we let q be the supremum of the q, and C the intersection of the C,, we have a: < q for all n, all a E C n W. This is a contradiction. Now let n be such that (7.17) is stationary for every q < K. Let ,f be the following function on W :

f (a) = a::

FIGURE 7.2

The functionf is regressive; and so for every q < K, we find by Fodor's theorem a stationary subset S, of (7.17) and y, 2 q such that f ( a ) = yq on S,. If y, # yq., then S, n S,. = 0, and since K is regular, we have l{S,: q < K } [ = I{Y,:q
60

1.

AXIOMATIC SET THEORY

Let a be an ordinal. A set A is a closed set of ordinals of length a if it is the range of a normal sequence of length a (note that sup A is not in A if a is a limit ordinal). Exercise 7.12. For every stationary S E w I and every a < w , there is a closed set of ordinals A of length a such that A c S . [By induction on a: Vy Llosed A c S of length a such that y < min(A). The nontrivial step: If true for a limit a, find a closed A c S of length a such that sup A E S. Let A ; , 5 c wl, be closed subsets of S, of length a such that 1, = sup A i < min A,. There is such that A, E S . Let t = limn (,, Pick initial segments Bcnc A , of length a. 1 where limn a, = a. Let A = EL.]

.

Ui<,

+

u."=o

Exercise 7.12 does not generalize to closed sets of uncountable length. It is not provable in ZFC that given X c w 2 , either X or w 2 - X contains a closed set of length wl. It is an open problem whether this statement is consistent. Let F be a filter over a cardinal K ; F is normal if it is closed under diagonal intersections: (7.18) if XuE F, a < K , then A { X u : a < K} E F An ideal 1 over K is normal if the dual filter is normal. Exercise 7.13. A K-complete ideal / over K is normal if and only if for every So 4 / and any regressivefon So there is S c S o , S 4 1, such that/is constant on S. [One direction is like Fodor's theorem. For the other direction, let X. E F for each a < K. If AX, $ F, let So = K - AX, and let f ( a ) = some ( < a such that a $ X,. If f(a) = y for all a E S, then X, n S = 0,a contradiction.] Exercise 7.14. There is no normal nonprincipal filter over w. [Use the regressive functionf(n 1) = n.]

+

Exercise 7.15. If

K

bounded subsets of

is singular, then there is n o normal ideal over

K

that contains all

K.

The closed unbounded filter is K-complete and normal, and contains all complements of bounded sets. It is the smallest such filter over K : Lemma 7.7. lf K is regular and uncountable and is F is a normalfilter over K that contains allfinal segments { a : a. < a < K}, then F contains all closed unbounded sets.

Proof. First we note that the set Co of all Limit ordinals is in F: Co is the diagonal intersection of the sets

X,={<:a+ 1< < < K } Now let C be a closed unbounded set, and let C = {a, : a < K} be its increasing enumeration. We let xu= {< : a, < < K } Then C z Co n A{X, : a < K } .

8.

61

SINGULAR CARDINALS

Mahlo Cardinals Let K be an inaccessible cardinal. The set of all cardinals below K is a closed unbounded subset of K, and so is the set of its limit points, the set of all limit cardinals. In fact, the set of all strong limit cardinals below K is closed unbounded (to find the next one, take the limit of A, 2', 22A,. ..). If K is the least inaccessible cardinal, then all strong limit cardinals below K are singular, and so the set of all singular strong limit cardinals below K is closed unbounded. If K is the ath inaccessible, where a < K, then still the set of all regular cardinals below K is thin. An inaccessible cardinal K is called a Mahlo cardinal if the set of all regular cardinals below K is stationary. (Then the set of all inaccessibles below K is stationary, and K is the Kth inaccessible cardinal.) Exercise 7.16. Let K be the least inaccessible cardinal such that cardinal. Then K is not Mahlo. [Use/.@) = a where 1is the ath inaccessible.]

K

is the Kth inaccessible

Similarly, we define a weakly Mahlo cardinal as a cardinal K that is weakly inaccessible and the set of all regular cardinals below K is stationary (then the set of all weakly inaccessibles is stationary in K). Exercise 7.17. If K is a limit (weakly inaccessible, weakly Mahlo) cardinal and the set of

all strong limit cardinals below sible, Mahlo).

K

is unbounded in

K,

then

K

is a strong limit (inacces-

8. SINGULAR CARDINALS

We have shown in Section 6 that if K is singular, then the size of P(K) depends on the continuum function below K (Theorem 18). Namely, if 2'" is eventually constant for K, < K, then 2" = 2'". This is just one of many examples when exponentiation involving singular cardinals depends substantially on the values of the continuum function at regular cardinals. In fact, unless strong axioms of infinity (so-called large cardinal axioms) are assumed, then all exponentiation can be computed in terms of 2'" for regular K,. Let us consider the following hypothesis. The Singular Cardinals Hypothesis. For every singular cardinal then xcr = K + .

K,

if2" " < K ,

Obviously, the singular cardinals hypothesis follows from the GCH. If 2Crh2 x then x'" = 2crh. If Zcrh < K , then x + is the least possible value of x E f h . If we know the continuum function on regular cardinals and if we know the values of K"" on singular cardinals, then we can compute 'K for all K and I (Theorem 20). In case the singular cardinals hypothesis holds, cardinal arithmetic is particularly simple:

1. AXIOMATIC SET THEORY

62

Lemma 8.1. Assume the singular cardinals hypothesis. (a) If

K

is a singular cardinal, then

(8.1) 2'" 2"= ((2<")'

if the continuum function is eventually constant below otherwise

K

(b) l f ~1,are infinite cardinals, then 2'

(8-2)

K K+

if 2°K if 2 A < K a n d A < C f K if 2'CfK

[Note that the size of K' in (8.2) is least possible: It is necessary that K' 2 K and K' 2 2'; moreover, since cf(K') > 1, it is necessary that K' > K whenever 1 2 Cf K.]

Corollary. lf the singular cardinals hypothesis holds, then cardinal exponentiation is completely determined by the continuumfunction on regular cardinals (and the cofinality function). [Note that by (b),thesingular cardinals hypothesis is equivalent to the assertion that K' I 2i * K + for all infinite cardinals K and 1.1

'

Proof: (a) If K is a singular cardinal, then by Lemma 6.5,2"is either 1 or 1" where 1 = 2'". The latter occurs if 2" is not eventually constant below K. Then ' by the singular carcf 1 = cf K, and since 2"' < 2'" = 1, we have ACf = 1 dinals hypothesis.

'

(b) We prove (8.2) by induction on K, for a fixed 1.Let K > 2'. If K is a successor cardinal, K = v ' , then v' I K (by the induction hypothesis), and K' = (v')' = v + . v' = K, by the Hausdorff formula. If K is a limit cardinal, then v' < K for all v < K. By Theorem 19, K' = K if 1< cf K, and K' = K" " if L 2 cf K. In the latter case, 2"" 5 2' < K, and by the singular cardinals hypothesis, K" = K'. It follows from a recent result of Jensen that the singular cardinals hypothesis holds unless a large cardinal axiom is true (namely, if the singular cardinals hypothesis fails, then 0" exists). On the other hand, assuming a large cardinal axiom, one can show that the singular cardinals hypothesis is independent. In this section we present a theorem of Silver and related results concerning the values 2" and K ' ~ " for singular cardinals of uncountable cofinality.

8.

63

SINGULAR CARDINALS

'lleorem 23 (Silver).

(a) Let K be a singular cardinal of uncountable cofinality, cf K 2 K , . If the generalized continuum hypothesis holds below K , then it holds at K : if

,

= K,+ for all

2'"

K, < K ,

then 2" = K +

(b) If the singular cardinals hypothesis holds for all singular cardinals of cofinality w, then it holds for all singular cardinals. Theorem 24 (Galvin-Hajnal). Let K be a strong limit singular cardinal of uncountable cofinality such that K < K, . Then 2" < K, . [A better estimate: If K = N,,,then 2 " ~< N, where y = (21ql)+.] These two theorems are samples of a number of theorems that can be proved about singular cardinals of uncountable cofinality. We shall give a few more examples in the exercises. The assumption that cf K is uncountable is necessary. It has been shown that the following is consistent (relative to a large cardinal assumption): 2"n = N,+ for all n, and 2 ' ~= Both theorems can be proved in the same way; in fact, they are special cases of a more general theorem. We first consider a more general version of Theorem 23a: Lemma 8.2. Let K be a singular cardinal, let cf K 2 0,. I f ( K , : a < cf K ) is a normal sequence of cardinals such that lim K , = K, and i f the set { a < cf

is stationary in cf

K,

K

: 2"- = K:}

then 2" = K + .

Part (b) of Theorem 23 follows from the next lemma, which is a generalization of Lemma 8.2: Lemma 8.3. Let K be a singular cardinal, let cf K 2 wl, and assume that Acr " < K for all 1 < K. r f (K,: a < cf K ) is a normal sequence of cardinals such that lim K , = K, and ifthe set {a < Cf K : K E r r E = K : } is stationary, then

K""

= K+.

Proof of Theorem 23b. We prove by induction on the cofinality of K that 2" " < K implies K ~ =~ K"+ . The assumption of the theorem is that this holds for each K of cofinality w. Thus let K be of uncountable cofhality and let 2"" < K. Using the induction hypothesis and formulas (8.2),one easily verifies that 2'" < K for all 1 < K. Let (K, : a < cf K ) be any normal sequence of cardinals such that lim K , = K . The set

S = { a < cf K : cf K,

=w

and 2'O < K,}

1. AXIOMATIC SET THEORY

64

is clearly stationary in cf K, and for every a E S, K Hence xcf'= K + . W

: ~ = K:

by the assumption.

We shall first prove Theorem 23 via Lemmas 8.2 and 8.3 and then generalize the proof to obtain Theorem 24. To simplify the notation, we shall consider the special case when K = K,,

The general case is proved in exactly the same way. Letfand g be two functions on wl. We say thatfand g are almost disjoint if there is a. < w1 such thatf(a) # g(a) for all a 2 ao. A family 9of functions on w1 is an almost disjoint family if any two distinct S, g E 9 are almost disjoint. Lemma 8.3 follows from Lemma 8.4. Assume that Krl < H,, for all a < wl. Let 9 be an almost disjoint family of functions

9=

n

A,

a<,,

such that the set {a < 0 1 : (8.3) 9I IK,, + is stationary. Then 1

,.

p.1 5 K , + l 3

[In the general case, we consider almost disjoint functions on cf

K.]

Proof of Lemmas 8.2 and 8.3from Lemma 8.4. We assume that 2" = K,+ for a + 1 . For every X G K ,, ,we stationary set ofa's; we want to show that 2Nml = KO,, a < w l ) where Xu= X n K,, and let 9= { f x : X c K,,). If let f x = ( X u : X # Y, then f x and fy are almost disjoint. Moreover,

9s

fl

P(%)

U < O ,

Since for a stationary set of a's, I P(w,)I = Nu+ we have 1 9I I K,, so

p(%,)l =

+ 1,

and

%,+1.

To prove Lemma 8.3, we assume that K," < K,, and that He' = K,+ for a stationary set of a's; we want to show that NE;= Nu,+,.For every h: w 1 3 K,,, we letfh = (ha : a < a,), where dom ha = w 1 and

and let 9= {fh: h Moreover,

E

"'K,,}. If h # g, then 9E

fl a<,,

"1K,

fh

and

fg

are almost disjoint.

8.

65

SINGULAR CARDINALS

Since for a stationary set of a's, KtI = K,+ (namely for all a such that K, > 2 H l and K> = N,+ 1), we have 1 st I N,,,+ and so [culN,,I = N,,,+ 1 . [In the general case of Lemma 8.3 we have to show that (a
= {a : a

is closed unbounded in cf K; if a E C, then cf K = K,Ef a

Kzf

1" c K,}

is a limit ordinal and (VA < K,) K,


K

and we have

-1

The first step in the proof of Lemma 8.4 is Lemma 8.5. Assume that K,K1< K,, for all a < wl. Let 9be an almost disjoint family of functions 9 c A, a<,, such that the set

fl

(8.4)

{ a < w l : 1A.l S K U }

I

is stationary. Then 1 9 I Kwl.

(The assumption (8.3) is replaced by (8.4) and the bound for 1 9 1is N,, rather than K , , + l . ) Proof. We may as well assume that each A,is a set of ordinals and that A, for all a in some stationary subset of K,. Let

c w,

So = { a < w1 : a is a limit ordinal and A, E 0,)

Thus i f f € 9, then f ( a ) < w, for all a E So. GivenfE 9, we can find for each a > 0 in So some p < a such thatf(a) < ma; call this p = g(a). The function g is regressive on S, and by Fodor's theorem there is a stationary S E Sosuch that g is constant on S. In other words, the function f is bounded on S, by some wy < %II. We assign to eachfa pair (S,f IS) where S E So is a stationary set andf IS is a bounded function. For any S. i f j IS = g IS, thenf= g since any two distinct functions in S are almost disjoint. Thus the correspondence

fH(Sl f I S) is one-to-one. For a given S, the number of bounded functions on S is at most

c Kp

Y <,I

= sup

K;'

= K,,

Y

Since I P ( w l )1 = 2Kl < KW1, the number of pairs (S,f I S) is at most Kwl. Hence 1 9 1 IN,,.

66

1.

AXIOMATIC SET THEORY

flu<,,

Proofof Lemma 8.4. Let 9 c A, be an almost disjoint family of functions satisfying (8.3). We may assume that each A, is a set of ordinals, and that the set

S o = { a < o1: A, E ma+

is stationary. the set First we shall prove that for everyfe 9, ,Ff= { g E F : 3 stationary S G So(Va E S ) g(a) s f ( a ) }

has cardinality I Nu1.See Fig. 8.1. If S E So is stationary, let 9f.s

= {g E

9 : (Va E S ) g(a) I f ( a ) }

fla<,l

The family 9,,sis almost disjoint, and since E B, where B, =f(a) + 1 for all a E S, we use the fact that I f ( a ) + 1 I 5 K, and apply Lemma 8.5 to conclude that I9,,sI I Nu,.Then we have

9,= U {9f,s : S.G So stationary} and so )9,1 I Nu,.

f 01

“1

FIGURE 8 1

To complete the proof, we shall construct a sequence (8.5)

(&:5<8>

of functions in 9, such that 8 IK,,+

and

4 = (J {qC : 5 < e} (8.6) We construct (fc : t < 8) by induction. Given f,, v < 5, we look for fcE 9 such that ft # FfV, for all v < 5. If there is no such&, then we let 8 = 5 and have (8.6). If there is suchf, then the set {a E s o

:f&) sf&)}

is thin, and so fy E 9,,, for each v < 5. Since ISfc I IK,, and 9,, 2 {f,: v < t},it follows that ft exists only if 5 < K,, + That shows that the sequence (8.5) has length 8 IK,, + and using ( 8 4 we get 1 9 1IK,,+l.

8.

67

SINGULAR CARDINALS

This sequence of lemmas completes the proof of Theorem 23. Now we shall generalize Lemma 8.4 in order to derive Theorem 24. As before, we shall consider the special case K = K,,

Lemma 8.6. Assume that K,H1c K o , i o r all a < wl. Let 9be an almost disjoint familv of functions 9 c

fl

A,

a
such that 1 A, I < Kolfor all a < wl. Then

1% I < K, where y = (2")'.

Theorem 24 (for K = Kol) follows from Lemma 8.6 in exactly the same way as Lemma 8.2 follows from Lemma 8.4 (see the proof). And the general case of Theorem 24 follows from the general version of Lemma 8.6 that is obtained by making only notational changes (namely, considering almost disjoint functions on cf K). To prove Lemma 8.6, we first introduce the following relation among functions (P: w1 -+ w1 (8.7)

(P

<#

iff

{a < w1 : q ( a ) 2 #(a)} is thin

Since the closed unbounded filter is o-complete, it follows that there is no infinite descending sequence (Po > c p 1 > (P2 ". Otherwise, the set { a < w 1 : cp,(a) Iq n +l(a) for some n} is thin and so there is a such that > cpl(4 > %(a) > . * * which is a contradiction. Hence the relation (P < $ is well-founded and by Theorem 5, we can define the rank ll(~llof (P in this relation (called the norm of cp) such that

'

%(co

llcpll = SUP{ll$Il + 1 : II/ < cpl (In fact, the relation < induces a well-founded partial ordering of the reduced power modulo the closed unbounded filter.) Note that IIq I( = 0 if and only if cp(a)= 0 for a stationary set of a's (iff there is no such that I// < cp). Lemma 8.6 follows from

%I,

Lemma 8.7. Assume that K r l < K,, for all a < wl. Let cp: w1 -,w1 and let 9 be an almost disjoint family offunctions such that for every a < wl. Then 191 I ~

~ , + l l ~ l l .

68

1. AXIOMATIC SET THEORY

To prove Lemma 8.6 from Lemma 8.7, we let cp be such that If 8 is the length of the well-founded relation cp < $, then

[ A I I [I

+

certainly 101 Ilolwl I = 2'' and so 8 < (2")'. Hencew, llcpll < (2")' for every cp and Lemma 8.6 follows. Note also that Lemmas 8.4 and 8.5 are special cases of Lemma 8.7, namely when JIcpJIis 1 or 0 respectively.

Proof of Lemma 8.7. By induction on l cpll. If IIcpII = 0, then cp(a) = 0 on a stationary set and the statement is precisely Lemma 8.5. To handle the case llcpll > 0, we first generalize the definition of cp < $. Let S E w1 be a stationary set. We define (8.8)

cp < s $

{a E S : cp(a) 2 $(a)} is thin

iff

The same argument as before shows that cp < $ is a well-founded relation and so we define thenorm I(cp(l,accordingly.Note that ifS E T, then llcpll IIlcplls. In particular, (IcpJI IIIcplls, for any stationary S. Moreover,

(8.9) llcplls u T = min(llcplls, IIcpIIT) as can easily be verified. For every cp: w 1 + w , , we let I, be the collection of all thin sets along with those stationary S such that )Jcp1) < )Icp)ls. If S is stationary and X is thin, then IIcp 1, = llcplls. This and (8.9) imply that I, is a (proper) ideal over 0,. If llcpll is a limit ordinal, then S = {a < w1 : cp(a) is a successor} E I,

because if S 4 I,, then a E S . Hence

l cppll = IIqlls = ll$lls + 1, where $(a) = q(a) - 1 for all {a < w1 : cp(a) is a limit} $ I,

Similarly, if I(cp(( is a successor ordinal, then {a < w1 : cp(a) is a successor}6 I , Now we are ready to proceed with the induction. (a) Let IIcpII be a limit ordinal, llcpll > 0. Let S = {a < w1 : cp(a) > 0 and is a limit ordinal}

It follows that S $ I,. We may assume that A, E K,+@,)for every a, and so we havef(a) < K,+@,) for every f c 9. Givenfc 9, we can find for each a E S some < q(a) such thatf(a) < w e + ) ;call this /3 = $(a). For a 6 S, let $(a) = cp(a).Since S $ I,, we where have llJlll I Il$lls < Ilqlls = llJ1l. We also havefE .F+. 5, = { f 5~:f(a) < o , + + ,for ~ )all a)

and so -9=

u

{

S

F

+

IIIc/II < IIVII}

69

8. SINGULAR CARDINALS

By the induction hypothesis, 1 9 , I IK,, + llJIII < K,, + 1,1 for every 1(/ such that 111(/11 c llqll. Since the number of functions I): wl+ a l is 2", and 2'l c K,,, we have 1 9I I K,, + ll?ll . (b) Let llqll be a successor ordinal, llqll = y + 1. Let So = {a < w1 : q(a) is a successor)

It follows that So 4 I,. Again, we may assume that A, E o , + ~ for , ) each a c ol. The proof will follow closely the proof of Lemma 8.4. the set First we prove that for every f~ 9,

9,= {g E 9: 3s E S o , S 4 I,, (Va E S) g(a)
Let 1(/: w1 -+ w1 be such that $(a) = q(a) - 1 for a E S, and $(a) = q(a)otherwise. SinceS 4 I,, we have 111(/11 I Il$lls < Ilqlls = Ilqll = y 1 and so 111(/11 = y. Since .Fr,s c B,, where 1B,1 I K,+#(,,for all a, we use the induction hypothesis to conclude that [ 9 , , s I I K,,,,,. Then it follows that

+

nz.wl

1 9 1 ,

I%,+Y,

To complete the proof, we construct a sequence (8.10)

such that 0 I K w l + y + land

u

9 = {Fj< : 6 < e} Givenf",, v < 5, we letft E 9 (if it exists) be such thatft 4 F r v ,for all v < 6. Then the set (8.11)

{ a E s o :&(a) Ifv(a)l

belongs to I,, and so f v E 9,,; for each v < 5. Since I9,, I IK,, +,, and 9,, 2 { f v : v < t},it follows that 6 < K,,+,+ if fsexists. Thus the sequence (8.10) has length 0 5 Kwl+y+l. Then we have

9=

u

{ P j ;: 5 < 6)

andso 1 9 1I K , , + y + l . In the following exercises, apply Lemma 8.7 (generalized, if necessary).

fl < w I and if 2" < N,,+@for a stationary set of a's, then [By induction on fl: If q ( a ) Ifl on a stationary set, then llqll I8.3

Exercise 8.1. If

Exercise 8.2. If

; : N

IN,,

+j3.

2Ne,11

INu,, +8.

fl < w,, if 2'1 < N,, and if N,NO IN.+# for a stationary set of a's. then

70

1. AXIOMATIC SET THEORY

Exercise 8.3. If 2NaI Na holds for all cardinals of cofinality w, then the same holds for all singular cardinals. Exercise 8.4. If 2Hq

K1 I cf K, < K,, if f l < cf N,, and if 2'- 5 KatBfor all

a c q, then

I Kv+fl.

Exercise 8.5. If 2" I Ratatfor a stationary set of a [If cp(a) = a for all a < wl, then llq~ll= wl.] Exercise 8.6. If 2 N w ~ + -c KOI+u+ufor all a [Use the sets A,, = w,,

+,.I

< wl, then

< wl, then

IN,, + ,,

2'wl

+

< N,,+,,+,,.

2nul+w1

Exercise 8.7. If

2" c K,, and if K."o 5 Katatl for all a < w l r then K;:

Exercise 8.8. If

K

IK w l + w l + l .

is a strong limit cardinal, K = N,, and d K 2 HI, then 2" < K,,where

Y =(I~lCfK)+. Exercise 8.9. If

K 1 Icf

K

< K and if 1"' c K for all A < K, then

K"'

c H,, where

Y = (Idcf"r It is an open problem to find a bound on 2" as in Theorem 24 if K, = K (but see (35.14e)). THE AXIOM OF REGULARITY Axiom of regularity. Every nonempty set has an eminimal element. 9.

vS[S

z 0 + (&

ES

)S nx = 0 3

In other words, the relation E over any family of sets is well-founded. As a consequence, there is no infinite sequence x o 3 x 1 3x2

3""

(Consider the set S = {xo,xl, x 2 , . . .} and apply the axiom.) In particular, there is no set x such that XEX and there are no "cycles" xo E X I E

* * *

E

x,

E

xo

Thus the axiom of regularity postulates that sets of certain type d o not exist. This restriction on the universe of sets is not contradictory (i.e., the axiom is consistent with the other axioms) and is irrelevant for the development of ordinal and cardinal numbers, natural and real numbers, and in fact of all ordinary mathematics. However, it is extremely useful in the metamathematics of set theory, in construction of models. In particular, all sets can be assigned ranks and can be arranged in a cumulative hierarchy. We recall that a set T is transitive if x

E

T

implies x G T

Lemma 9.1. For every set S there exists a transitive set T 2 S.

9.

71

T H E AXIOM OF REGULARITY

Proof: We define by induction

and

u m

T = TC(S) =

S,

n=O

Clearly, T is transitive and T 2 S. Since every transitive set must satisfy TC(S) is the smallest transitive T 2 S, i.e.,

uT

G

T, it follows that the set

TC(S) = (I{ T : T 2 S and T is transitive} We call TC(S) the transitioe closure of S . Lemma 9.2. Every nonempty class C has an +minimal element. then S is a minimal element of C; Proof: Let S E C be arbitrary. If S n C = 0, if S n C # $3, we let X = T n C where T = TC(S). X is a nonempty set and by the axiom of regularity, there is x E X such that x n X = 0.It follows otherwise, if y E x and y E C, then y E T since T is transitive, that x n C = 0: and so y E x n T n C = x n X. Hence x is a minimal element of C. 'Ihe Cumulative Hierarchy of Sets We define, by transfinite induction,

vo = 0 V, =

u Vs

if a is a limit ordinal

K) The sets V, have the following properties (by induction): K+l

=P

(a) Each V, is transitive. (b) If a < /?,then V, c V,. (c) a s V,. The axiom of regularity implies that every set is in some V,: Lemma 9.3. For every x there is a such that x E V,:

u

V,=v

a E Old

Proof: Let C be the class of all x that are not in any V,. If C is nonempty, then C has an eminimal element x. That is, x E C, and z E V, for every z E x. Hence x, E UaE V ,.

ua

1.

72

AXIOMATIC SET THEORY

By Replacement, there exists an ordinal y such that x G x E V, and so x E V,,,. Thus C is empty and we have (9.1).

u,<, Hence V,.

Since every x is in some V,, we may define the rank of x: . rank(x) = least a such that x E V,, (9.2) Thus each V, is the collection of all sets of rank less than a, and we have (a) If x E y, then rank(x) < rank(y). (b) rank(a) = a. Exercise 9.1. rank x = sup{rank(z) + 1 : z E x}.

a,, = I V , + m \is a normal sequence: a. = KO. 1,= 2H0, a2= 2 Z W o , ... and so there are K such that K = 3, = I V, I.

Exercise 9.2. The transfinite sequence

Exercise 9.3. If 3, = K, then Exercise 9.4. If

K

K

is a strong limit and K, = K.

is inaccessible, then I V, I = K.

One of the uses of the rank function is a definition of equivalence classes in case of an equivalence relation over a proper class. The basic trick is the following: Given a class C,let

2:= {x E C : (Vz E C)rank x I rank z ) (9.3) is always a set, and if C is nonempty, then C is nonempty. Moreover, (9.3)can be applied uniformly. Thus, for example, if = is an equivalence over a proper class C,we apply (9.3) to each (ordinary) equivalence class of =, or define outright

[XI = { y E C : y 3 x and Vz E C ( z = x -, rank y I rank z)) and

c/== {[XI : x E C) In particular, this trick enables us to define isomorphism types for a given isomorphism. For instance, one can define order-types of linearly ordered sets, or cardinal numbers (even without AC). We use the same argument to prove the following. Collection Principle

vx 3Y(VU E X )[3u q ( u , u, p ) --* (30 E Y) cp(U, 0, P)] (9.4) ( p is a parameter). The collection principle is a schema of formulas. We can formulate it as follows:

9.

73

T H E AXIOM OF REGULARITY

Given a “collection of classes” C,, u E X (X is a set), then there is a set Y such that for every u E X,

if C, #

0,

To prove (9.4), we let Y=

-

then C, n Y #

0

u t”

,EX

(where C, = { u : q ( u , u, p ) } ) ; i.e., v

E

Y

( 3E X ) [ q ( u , u, p) and V z ( q ( u , z, p) -+ rank u Irank z)]

That Y is a set follows from the replacement schema. Note that the collection principle implies the replacement schema: Given a function F, then for every set X we let Y be such that (VU E X ) (30 E Y ) F(u) = u

Then

FIX=Fn(Xx Y)

is a set by the separation schema. E-induction

The method of transfinite induction can be extended to an arbitrary transitive class (instead of Ord), for both proof and definition by induction: Theorem 25 (A) (Proof by E-induction). Lut T be a transitive class, let 0 he a property. Assume that

(i) W0); (ii) ifx E T and @ ( z )holdsfor every z Then every x E T has the property 0.

E

x, then O(x).

(B) (Definition hy E-induction). Let T be a transitive class and let G be afunction (defined.for all x). Then there is a .function F on T such that F(x) = G ( F J x )

(9.5)

for eoery x E T. Moreover, F is the unique function that satisfies (9.5). Proof: (A) Let C be the class of all x E T that d o not have the property a. If C is nonempty, then it has an €-minimal element x; apply (i) or (ii). H

-

(B) We let, for every x

F(x) = y

E

T,

there exists a function f such that dom(f) is a transitive subset of T and: (a) (b)

(VZ E

d o m ( f ))”

f ( 4= y

= G(f

I41

1.

74

AXIOMATIC SET THEORY

That F is a (unique) function on T satisfying (9.5) is proved by E-induction. Example. Let A be a class. There is a unique class B such that

B=

(9.6) Proof: Let

F(x) =

1 0

{X E

A:x

5

B}

if x E A and F(z) = 1 for all z E x otherwise

Let B = {x : F(x) = 1). The uniqueness of B is proved by E-induction. We denote the class in (9.6) by

B=XA and say that each x

E

B is hereditarily in A.

Exercise 9.5. X A = {x E A : TC(x) E A}. Exercise 9.6. I’A is the largest transitive

class B such that B G A

One consequence of the axiom of regularity is that the universe does not admit nontrivial E-automorphisms. More generally:

Theorem 26 (Isomorpltism Throrem). No two distillet rrarisitiiv classes are isomorphic. More specifically: Let T , , T2 be traitsitii-e classes and let n be a11 ~-isornorphismof T, onto T2 : ix., n is one-to-one, arid (9.7) Tlim T,

LIEU

= T2 and

nu

= u for

t*

every u

E

IIUEnv

T,.

ProoJ: We show, by E-induction, that nx = s for every x E T I . Assume that nr = z for each z E .x and let y = n . ~ . We have x c y because if z E x, then z = nz E n . = ~ y. We also have y zx: Let t E y. Since y c T 2 , there is z E T I such that n r = t . Since nz E y , we have 2 E x, and so t = nr = z. Thus t E x. Therefore nx = x for all .x E TI, and T, = T,. H

Well-Founded Relations The notion of well-founded relations that was introduced in Section 2 can be generalized to relations over proper classes, and one can extend the method of induction to well-founded relations. Let E be a binary relation over a class P. For each .x E P, we let ext,(x) = ( z E P : z E x) the extension of x.

9.

75

THE AXIOM OF REGULARITY

A relation E over P is well-founded, i f

(9.8)

(i) every nonempty set X E P has an E-minimal element; (ii) ext,(x) is a set, for every x E P.

(Condition (ii) is vacuous if P is a set.) Note that the relation E is well-founded over any class, by the axiom of regularity.

Lemma 9.4. If E is a well-founded relation over P, then every nonempty C G P has an E-minimal element. Proof: We follow the proof of Lemma 9.2; we are looking for x E C such that ex+) n C = 0. Let S E C be arbitrary and assume that ext(S) n C # 0. We let X = T n C where

u W

T=

S,

n=O

and So = ext S,

Sn+l =

u {ext(z):z

E

S,}

(See Fig. 9.1.) As in Lemma 9.2, it follows that an E-minimal element x of X is E-minimal in C.

FIGURE 9.1

Theorem 25' (Well$ounded Induction). Let E be a well-founded relation over P. (A) Let 0 be a property. Assume that: (i) every E-minimal element x has the property 0: (ii) i f x E P and 0 ( z ) holdsfor every z such that z E x, then

a(.).

Then every x E P has the property 0. ( B ) Let G be afunction (on V x V). Then there is a (unique)function F on P such that

(9.9) for every x

F(x) = G(x, F 1 ext(x)) E

P.

76

1. AXIOMATIC SET THEORY

Proof: A modification of the proof of Theorem 25.

(Note that if F satisfies (9.9),then F(x) = F(y) whenever ext(x) = ext(y); in particular, F(x) is the same for all minimal elements.) Examples. 1 (The rank function). We define, by induction, for all x E P:

p(x) = SUP{~(Z) + 1 :z E

X}

(compare with (2.10)).The range of p is either an ordinal or the class Ord. For all x, y E P, x E Y P ( X ) < P(Y) 2 (The transitioe collapse). By induction, let +

n(x) = n[ext(x)] = {n(z): z E x}

for every x E P.The range of n is a transitive class, and for all x, y xEY

+

4.1

E

E

P,

n(y)

APPENDIX: BERNAYS-GODEL AXIOMATIC SET THEORY

There is an alternative axiomatization of set theory. We consider two types of objects: sets and classes.

Axioms: A. 1. Extensionality (for classes): Vu(u E X c)u E Y)+ X = Y 2. Every set is a class. 3. If X E Y,then X is a set. 4. Pairing: For any sets x and y there exists a set {x, y}. B. Comprehension

where cp is a formula in which only set variables are quantified (no class variables). C. 1. Infinity: There is an infinite set.

u

2. Union I * For every set X there exists a set II P(Xx. 3. Power set * ). 4. Replacement: If a class F is a function and X is a set, then F[X] is a set. D. Regularity. E. Choice: There is a function F such that F(X) E X for every nonempty set X. Let BG denote the axiomatic theory A-D, and let BGC denote BG

+ choice.

APPENDIX: BERNAYS-GdDEL AXIOMATIC SET THEORY

77

Exercise. If a set-theoretical statement is provable in ZF (ZFC), then it is provable in

BG (BGC).

On the other hand, a theorem of Shoenfield (using proof-theoretic methods) states that if a sentence involving only set variables is provable in BG, then it is provable in ZF. This result can be extended to BGC/ZFC using the method of forcing; cf. the remark at the end of Section 20.

CHAPTER 2

TRANSITIVE MODELS OF SET THEORY

10. MODELS OF SET THEORY

Let E be a binary relation over a class M. We call 91 = (M, E)* a model (of the language of set theory), and define, for every formula cp(xl, . . . ,x,), cp(a1, . . ., a,)

‘9.R

(10.1)

(a,,

. . . , a,

E

M)

(9.R satisfies cp(a,, . . . , a,), cp holds in M),as follows:

(a) If cp is an atomic formula: ‘9.Rka~b

if a E b

‘9.Rka=b

if a = b

(b) If cp is a negation, conjunction, etc.: ‘9.R k -I+(u,, ..., a,)

if not ‘9.R t +(a,, ..., a,)

m t AX(^,,

if ‘9.R k +(al, ..., a,) and 91 t ~ ( a ,..., , a,)

‘9.R t:

. a * ,

an)

+ v ~ ( a , .,. ., a,)

if ‘9.R t +(a,, . . . , a,) or 91 k ~ ( a ,.,. . , a,)

‘9.R I. ++~(a,,...,an)

if ‘9.R k +(al, ..., a,) implies 91 k ~ ( a ,..., , a,)

‘9.R I. + + + X ( Q ~ ? . . * ,an)

if ‘9.R t: +(al, ..., a,) iff 91 t: ~ ( a ,..., , a,)

(c) If cp is 3x+ or Vx+ . . * ,an)

if (3a E M) ‘BI k +(a, a,, . . . , a,)

m t Vx+(x, a,, . . . , a,)

if (Va E M) 9R t $(a, a,, ..., a,)

‘9.R

*$(x,

The above definition (“by induction on complexity of cp”) gives us instructions on how to “decade” (10.1) for any particular formula cp. For instance, ‘BI t= Vu(u E

xcru E Y)- x = Y

* The sophisticated reader will not object to our use of a pair of classes. 78

10. MODELS OF SET THEORY

79

is the formula

VUEM(UEXHUEY) + X = Y If o is a sentence and !Dl k 0, then we say that 9.l is a model ofa. A model ofZF (ZFC) is an 91 that satisfies every axiom of ZF (ZFC).

Relative Consistency An axiomatic theory is consistent if the axioms are not contradictory-if there is no proof of a contradiction in the theory. Gijdel’s second incompleteness theorem states that no axiomatic theory at least as powerful as the axiomatic arithmetic can be proven consistent by methods that are formalizable in that theory itself. In particular, it is impossible to show consistency of ZF (or related theories) by means limited to ZF alone. O n the other hand, once we assume that ZF (or ZFC) is consistent, we may ask whether the theory remains consistent if we add an additional axiom A. Let T be a mathematical theory (in our case, T is either ZF or ZFC), and let A be an additional axiom. We say that T + A is consistent relatiue t o T (or that A is consistent with T) if the following implication holds: if T is consistent, then so is T

+A

If both A and 1 A are consistent with T, we say that A is independent of T. The question whether A is consistent with T is equivalent to the question whether the negation of A is provable in T (provided T is consistent); this is because T + A is consistent if and only if 1 A is not provable in T. The way to show that an axiom A is consistent with ZF (ZFC) is to use models. For assume that we have a model !Dl of ZF such that 9.l k A. Then A is consistent with ZF: If it were not, then 1 A would be provable in ZF, and since !Dl is a model of ZF, 91 would satisfy 1 A . (Here we tacitly use the fact that if !Dl is a model of a theory T and 0 is a sentence provable in T, then 9.l is a model of o.)However, 91 k 1 A contradicts !Dl k A since !Dl I=1 A iff 9.l F A. The models introduced here are closely related to models as defined in model theory. They are in a sense more general since they are classes rather than just sets, but also more special since proper classes cannot be manipulated as freely as sets. There is another remark that is in order. By Giidel’s second incompleteness theorem, one cannot show consistency of ZFC within ZFC. Thus if Con(ZFC) is the statement that ZFC is consistent, we have Con(ZFC) is not provable in ZFC By Giidel’s completeness theorem, an axiomatic theory is consistent if and only if it has a model (in the sense of model theory-thus a set). Therefore the sentence 32I(2I is a model of ZFC) is not provable in ZFC.

2.

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TRANSITIVE MODELS OF SET THEORY

When dealing with models, we use standard terminology of model theory. Also, some results t o be presented are related to standard theorems of model

theory. For this reason, we insert a brief review of basic notions and facts from model theory. Review of Model lheory

A language is a set of symbols: relation symbols, function symbols, and constant symbols : Y = { P , ..., F , ..., C, ...)

Each P is assumed to be an n-placed relation for some integer n 2 1 ; each F is an m-place function symbol for some m 2 1. Terms and formulas of a language Y are certain finite sequences of symbols and of logical symbols (identity symbol, parentheses, variables, connecof 9, tives, and quantifiers). The set of all terms and the set of all formulas are defined by induction. If the language is countable (i.e., if 1 9 1 I No),then we may as well as the logical symbols, with some hereditarily identify the symbols of 9, finite sets (elements of Vu); then formulas of Y are also hereditarily finite (arithmetization of language). A model for a given language 9 is a pair PI = ( A , Y), where A is the universe of PI and 9 is the interpretation function which maps the symbols of Y to appropriate relations, functions, and constants in A. A model for 9 is usually written in displayed form as PI = ( A , Pa, . . ,, F", ..., c', . . .)

or just PI

=

( A , P , ..., F , ..., C, ...)

By induction on length of terms and formulas one defines the value of a term t"[a,, . . . , a,]

and satisfaction PI k cp[a,, . . . , a,]

(t is a term, cp is a formula, n E N, and (al, . . . , a,) is an n-termed sequence in A). The relation of these models to the models defined in (10.1)is the following: Let Y be a language consisting of one binary predicate symbol E:

9= {€} and let Form c V, be the set of all formulas of 9. As with any actual (metamathematical) natural number we can associate the corresponding element of N, we can similarly associate with any given formula of set theory the corre-

10.

81

MODELS OF SET THEORY

sponding element of the set Form. To make the distinction, if cp is a formula, let rcp1 denote the corresponding element of Form. If M is a set and E is a binary relation over M and if a . . . ,a, are elements of M , then

,,

(10.2)

91 k cp(a,, . . . , a,)

iff

(M,E ) 1

~p

[ a , , * . . *a,]

as can easily be verified. Thus in the case when M is a set and cp a particular (metamathematical) formula, we shall not make a distinction between the two meanings of the symbol k. We note however that the left-hand side of (10.2) is not defined for cp E Form, and the right-hand side is not defined if M is a proper class. To proceed with our review of model theory, let us recall Gijdel's completeness theorem: A set C of sentences of 6" is consistent if and only if it has a model. Construction of a model of Z requires in general the axiom of choice; in fact, the ultrafilter theorem (Theorem 21) suffices. The compacrness theorem states that a set of sentences X has a model if and only if every finite S G Z has a model. The Lbwenheim-Skolem theorem says max(lZ:(,No). In that if C has a model, then it has a model $I such that 1 A I I particular, if Y is countable, then every consistent set of sentences has an at most countable model. These theorems again use the axiom of choice. Two models $1 = ( A , P , . . . , F , . . . , c, . . .) and 91' = ( A ' , I", . . . , F', . . . , c', ...) are isomorphic if there is an isomorphism between PI and PI', that is a one-to-one function fof A onto A' such that:

( i ) P(.u,, . . .. s,) iff P'(f(.u,), ...,.f(x,)), (ii) f(F(.ul, . . . , x,)) = F ' ( f ( s , ) , . . . ,,f(.u,)), (iii) f ( c ) = c', for all relations, functions, and constants of PI. I f f is an isomorphism, then . . . ,,f(a,)] for each term, and

f(r"[a,, . . . , a,]) = t"[f'(a,),

PI k

cp[a,, ..., a,]

iff

PI'

cp[f(a,),

...,f(a,)]

A submodel of PI is a subset B G A endowed with the relations P" n B", . . . , functions F" I B", . . . , and constants c,' . . . ; B has to be such that all c' belong to B, and that B is closed under all F" (if (.Y,. ...,.urn) E B", then F'(.u,, .. ., .urn)E B). An embedding of 8 into PI is an isomorphism between 8 and a submodel 8' c PI. A submodel '8 c 91 is an elementary submodel

8 < PI if for every formula cp, and every a , , . . ., a,

E

B,

82

2.

TKANSlTlVE MODELS OF SET THEORY

The key lemma in construction of elementary submodels is: A subset B E A forms an elementary submodel of PI if and only if for every formula cp, and every a,, . . . , a, E B,

if 3a E A such that PI k cp[a, a , , ..., a,]. then 3a E B such that PI k cp[a, a,, .. ., a,]

(10.4)

A function h : " A -+ A is a Skolem.function for cp if* 3a E A PI k ~ [ u a, , , ..., a,]

PI k cp[h(al,..., a,), a,, ..., a,]

implies

for every a,, ..., a,. Using the axiom of choice, one can construct a Skolem function for every cp. If a subset B E A is closed under (some)Skolem functions for all formulas, then B satisfies (10.4) and hence forms an elementary submodel of PI. An elementary embedding is an embedding whose range is an elementary su bmodel. A set X E A is definable over PI if there is a formula cp and a . . . , a, E A such that

,,

X

=

{X E

A : PI k cp[x, a,,

. . ., a,,]!

( X is definable in PI .from a,, ..., an). If cp is a formula of x only, without parameters a,, . . . , a,, then X is definable in PI. An element a E A is definable (from a,, . . ., a,) if the set {a}c A is definable (from a,, . . . , a,).

E-Models and Transitive Models A model 91 = (M, E) is an €-model if the relation E is just the erelation, i.e., if E = E n M2. If 'N is an €-model, then we call the class M a model and write

(M, E) k cp

or just

M k cp

Note that the formula M 1 cp is obtained from cp by replacing all 3x and Vx by 3x E M and Vx E M. In particular, if cp is quantifier-free, M k cp is the same as cp. If M is a transitive class, then the €-model M is called a transitive model. An important property of transitive models is the following: Call 3 4 E X and Vu E X restricted quantifiers, and say that a formula cp is a restricted formula if it has no other quantifiers than restricted quantifiers.

Lemma 10.1. If M is a transitive class and cp a restricted formula, then for all x,, . . . , x, E M,

(10.5)

M

~p(x,,...,x,)

* I f n = 0 then a Skolem function for implies PI b cp[h].

cp

iff

~p(x,,...,xn)

is an element h E A such that 31E A PIC q[a]

83

10. MODELS OF SET THEORY

Prooj: If cp is an atomic formula, then (10.5) holds. If (10.5)holds for cp and $, then it holds for l c p , cp A $, cp v $, cp -, $, and cp-$. Let cp be the formula (3u E x)$(u, x, . . .)and assume that (10.5)is true for $. We show that (10.5) is true for cp (the proof for Vu E x is similar). If M k cp, then M 1 3u(u E x A $) i.e., ( ~ U E M ) ( U EA XM I = $ )

Since M 1 $ if and only if $, we have (3u

E

x) $

Conversely, if (3u E x ) $, then since M is transitive, u belongs to M, and since $(u, x, ...)- M k $(u, x, ...), we have 3 u ( u ~ MA

and so

U E X A

M b (3u

Mk$)

EX)$

Example. Every transitive model satisfies the axiom of extensionality:

Proof: [(Vu E X ) ( u E Y ) A (Vu E Y ) (u E X ) ] --* X = Y is a restricted formula.

Consistency of the Axiom of Regularity As the first application of models we show that the axiom of regularity is consistent with ZF minus Regularity (axioms I-VII). We recall the construction of the sets V ,. The construction is performed without using Regularity; the axiom of regularity is used to show that V = V,. We shall show, using only axioms I-VII, that the class P = ord V, is a model of ZF. This proves consistency of the axiom of regularity relative to the theory ZF minus Regularity. We shall make use of the fact that if X is a subset of P, then X E P. I. P k Extensionality This is so because P is transitive. 11. P k Pairing

uord

(10.6)

P b Va V h 3c[a

E

ua

c ~ E bC A ( V XE c) (x = a v x = b)]

Given a, b E P, let c = {a, h}. Clearly, c E P and P k [. . .] since the formula [. . .] is restricted and holds (in the universe; it says: c = {a, b}). 111. P k Separation

Pk Given X, p

E

Vx Vp

3Y

VU(U E

Y-U

E xACp(U,

P, we let Y = {u E

x : P k q(u, p ) }

p))

2.

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TRANSITIVE MODELS OF SET THEORY

Since Y s X and X E P, we have Y E P, and

P k Vu(u E

Y-U

E

x Arp(U, p ) )

IV. P t Union (10.7) P I=VX 3Y[(Vz E X ) (Vu E z ) u E Y A (Vu E Y) E

u

X )u E 23

E P, we let Y = X. It is easy to see that Y E P (if X E V,, then V,). Moreover, P t [ . . I since [...I is a restricted formula, saying

Given X

X

(32 E

Y=UX. V. P t Power set

P k VX 3Y Vu[u E

(10.8)

Y W U

c XI

Given X E P, we let Y = P ( X ) . Again, Y E P. However, the formula V u [ ..] is not restricted; but we argue as follows: [u E Y++(VzE u)z E XI is restricted, and for every u E P, P k [. * since [ . .] is true for all u. VI. P t Infinity a]

P I=3S[0 E S

(10.9)

A

(Vx E S)x u {x} E S]

The formula in (10.9) contains defined notions, { }, u, and 0; and strictly speaking, we should first eliminate these symbols and use a formula in which they are replaced by their definitions, using only E and =. However, we have already proved that both pairing and union are the same in the universe as in P, and similarly one shows that X E P is empty iff P t X is empty. In other words, {a, b)' = {a, b},

up,

upX

=

X,

fzlp= 0

where {a,b}', and 0 ' denote pairing, union, and empty set in the model P. Since w E P, we let S = w and easily verify that (10.9) holds for this S . VII. P t Replacement (10.10)

p

VX -P

Vy

VZ[Cp(X.

Y,

P) A V(X.

VX gy VY[Y E Y - P x

Z,

P) Y = 23

E

X ) d x , y, P)]

-+

Given p E P, assume that P I= Vx Vy Vz[...]. Thus

F = {(x, y ) E P : P t V(X, Y, P)) is a function, and we let Y = F[X]. Since Y c P, we have Y verifies that for every y E P,

PkyEy - p

E

X ) cp(x, y , P)

VIII. P I: Regularity

(10.11)

P t VS[S # 0 - (hE S ) S n x = 0 1

E

P, and one

10. MODELS OF SET THEORY

85

If S E P is nonempty, then let x E S be of least rank; then S n x = 0.Hence [...I is true for any S ; moreover, (S n x)'= S n x , and [ . . I has only a restricted quantifier. Thus P k [...]. Exercise 10.1. If the axiom of choice holds, then P k axiom of choice. Exercise 10.2. If a is a limit ordinal, then V, is a model of: Extensionality, Pairing, Separation, Union, Power set. and Regularity. If AC holds, then V, is a model of AC. Exercise 10.3. If a > w , then V, is a model of Infinity. Exercise 10.4. V,, the set of all hereditarily finite sets, is a model of ZFC minus Infinity.

Inaccessibility of Inaccessible Cardinals Theorem 27. Existence of inaccessible cardinals is not provable in ZFC.

Moreover, it cannot be shown that the existence of inaccessible cardinals is consistent with ZFC.

We shall prove the first assertion and invoke Giidel's second incompleteness theorem to obtain the second part. First we prove (in ZFC):

Lemma 10.2. I ~ isKan inaccessible cardinal, then V, is a model of ZFC. Proof: The proof of all axioms of ZFC except Replacement is as in the proof of consistency of the axiom of regularity (see Exercises 10.1-10.3). To show that V, k Replacement, it is enough to show:

(10.12) If F is a function from some X

E

V, into V,, then F

E

V,.

Since K is inaccessible, we have I V, I = K and I X I < K for every X E V,. If F is a function from X E V, into V,, then I F[X] I I I X I K and (since K is regular) F[X] c V, for some a < K. It follows that F E V,. W

-=

Proofof Theorem 27. If K is an inaccessible cardinal, then not only is V, a model of ZFC, but in addition V, V, V, V,

k k k k

a is an ordinal a is a cardinal a is a regular cardinal a

is an inaccessible cardinal

iff iff iff ifT

a is an ordinal. Q

is a cardinal.

a is a regular cardinal. a is an inaccessible cardinal.

(We leave the details to the reader.) In particular, if K is the least inaccessible cardinal, then V, k there is no inaccessible cardinal

Thus we have a model of ZFC + " there is no inaccessible cardinal" (if there is no inaccessible cardinal, we take the universe as the model). Hence it cannot be proved in ZFC that inaccessible cardinals exist.

2.

86

TRANSITIVE MODELS OF SET THEORY

To prove the second part, assume that it can be shown that the existence of inaccessible cardinals is consistent with ZFC; in other words, we assume if ZFC is consistent, then so is ZFC + I

where I is the statement “ there is an inaccessible cardinal.” We naturally assume that ZFC is consistent. Since I is consistent with ZFC, we conclude that ZFC + I is consistent. It is provable in ZFC + I that there is a model of ZFC (Lemma 10.2). Thus the sentence “ Z F C is consistent” is provable in ZFC + I. However, we have assumed that “ I is consistent with ZFC” is provable, and so “ZFC + I is consistent” is provable in ZFC + I. This contradicts Gdel’s second incompleteness theorem. W The wording of the second part of Theorem 27 (and its proof) is somewhat vague; it cannot be shown” means: It cannot be shown by methods formalizable in ZFC. “

K is an inaccessible cardinal then V. I= there is a countable model of ZFC. [Since ( V . , E ) is a model of ZFC, there is a countable model (by the LowenheimSkolem theorem). Thus there is E c w x w such that YI = (u,E) is a model of ZFC. Verify that V, k (91 is a countable model of ZFC).]

Exercise 10.5. If

Exercise 10.6. If K is an inaccessible cardinal, then there is a < K such that ( V , , E) < ( V , , E). Moreover, the set { a < K : ( V , , E) < ( V , , E)} is closed unbounded. [Construct Skolem functions h for V,, and let a = limn a,, where a,+ < K is such

that h I Ka E Van+I for each h.]

We recall the definition (9.6): &‘A = {X E A : x c XA}

(the class of all x hereditarily in A). If A is the class of all finite sets, then X A is the set of all hereditarily finite sets, denoted by HF. In fact, H F = V,. Let HC denote the class of all hereditarily countable sets: HC=&‘({X: 1x1 IN,}) = {X : every y E TC({X}) is at most countable} Exercise 10.7. For each a, let H a = HC n V,. Show that H o = $3,H , = a limit ordinal, and H,+ = (X G H a : X 5 KO}.

I 1

up<.H , if

a is

HC is a set: Exercise 10.8. (In ZFC) HC = Hu1. Exercise 10.9.* (In ZF) HC = H,,

.

[Show that I rank(A)I K 1 for every A E HC. Since rank(A) = {rank(x): x E A}, it suffices t o find, uniformly for all A E HC, functions F; (n E (0)such that each F; maps w’i’ onto the set {rank(x): x E A}. Let F $ ( a ) = the

u:=o

u”

u“

11.

87

TRANSITIVE MODELS OF ZF

ath element of the set {rank(x): x E A } , and F;, l(ao, . . . , a,+ the set {Ff(ao, . . . , a,) : X E A } . ]

= the a,+ Ith element

I)

of

Exercise 10.10. (In ZFC) HC is a transitive model of ZFC minus Power set. [AC is needed to show that X E HC if X E HC (the countable axiom of choice

u

suffices).]

11. TRANSITIVE MODELS OF ZF

In this section we shall investigate the transitive classes that satisfy the axioms of ZF. In view of Giidel's second incompleteness theorem, we shall concentrate on models of ZF that are proper classes. First we prove the Mostowski collapsing theorem, the import of which is that every well-founded model of Z F is isomorphic to a transitive model. (We leave the subject of non-wellfounded models for a later time.) The main theorem of this section (Theorem 31) gives a complete description of transitive models of Z F that are proper classes. Throughout the section (and in fact throughout most of Chapter 11) we work in ZF; that is, we d o not use the axiom of choice.

Isomorphism of Models

e)

Let 111 = (M, E) and !Ill=(M, be models. 111 and !Illare isomorphic if there is an isomorphism between YJl and !Ill,that is, a one-to-one function n of M onto M such that (11.1)

x

E y iff n ( x ) E n(y)

for all x, y

E

M

It follows easily from the definition (11.1) that if cp is a formula, then (11.2)

9.JI k 44x1, ... x,) I

iff

k cp(nx,, . . . , nx,)

We say that two classes M and M are isomorphic (or €-isomorphic) if the €-models M and M are isomorphic.

Extensional Classes and Relations We say that a class M is extensional if for any distinct X, Y E M,

XnM#YnM

(11.3)

That is, if X, Y

E

M and U€X

*--)

U € Y

holds for every u E M, then X = Y. A relation E over a class M is called extensional if extE(X) # e x t m whenever X, Y E M and X # Y. (11.4)

2.

88

TRANSITIVE MODELS OF SET THEORY

Thus a class is extensional iff the relation Clearly, (11.4) is equivalent to

E

over M is extensional.

tlu~M(uEXc-,uEY) +

X=Y

which is just the statement that (M, E) satisfies the axiom of extensionality. Thus we have:

Lemma 11.1. A relation E over M is extensional i f f ( M , E) is a model ofExtensionality. A class M is extensional f l i t is a model of Extensionality. Note that every transitive class is extensional: If X E M, then X n M = X and (1 1.3) is trivially satisfied. The following theorem shows that every extensional class is isomorphic to a transitive class and the isomorphism is unique (the transitive collapse).

Theorem 28 (Mostowski’s Collapsing Theorem).

If E is a wellfounded and extensional relation over a class P, then there is a transitive class M and an isomorphism A between (P, E) and ( M , E). The transitive class M and the isomorphism n are unique. ( B ) In particular, every extensional class P is isomorphic to a transitive class M . The transitive class M and the isomorphism n are unique. ( C ) In case ( B ) , ifT C P is transitive, then AX = x f o r every x E T.

(A)

Proof: Since (B) is a special case of (A) (E = E in case (B)), we shall prove the existence of an isomorphism in the general case. Since E is a well-founded relation, we can define n by well-founded induction (Theorem 25’), i.e., n ( x ) can be defined in terms of the n(z)’s,where z E x . We let, for each x E P,

(11.5)

n(x)= n[exts(x)]

= { A ( z ): z

E x}

In particular, in the case E = E , (11.5) becomes (1 1.6)

n(x)= n[x n PI

The function n maps P onto a class M = n[P], and it is immediate from the definition (11.5) that M is transitive. We use extensionality of E to show that n is one-to-one. Let z E M be of least rank such that z = n ( x )= n(y) for some x # y. Then ext(x) # ext(y) and there is, e.g., u E ext(x) such that u $ ext(y). Let t = n(u). Since t E z = n(y), there is v E ext(y) such that t = A(v). Thus we have t = n(u) = n(v),u # v, and t is of lesser rank than z (since t E z). A contradiction. Now it follows easily that (11.7)

xEy

c-,

n(x)~ny

11. TRANSITIVE MODELS OF ZF

89

If x E y, then n ( x ) E n(y) by definition (1 1.5). O n the other hand, if n ( x ) E ~ ( y ) , then by (1lS), n ( x ) = R ( Z ) for some z E y. Since R is one-to-one, we have x = z and so x E y. Uniqueness of the isomorphism R (and the transitive class M = n[P]) follows from the isomorphism theorem (Theorem 26). If R , and K, are two isomorphisms of P and M,, M,, respectively, then n2n;’ is an isomorphism between M , and M 2 , and therefore the identity mapping. Hence K , = n2. It remains to prove (C). If T E P is transitive, then we first observe that x c P for every x E T and so x n P = x, and we have .(x) = .[XI

for all x E T. It follows easily by €-induction that

RX = x

for all x

E

T. H

It is worth noting that Theorem 28 is a generalization of Theorem 2 (every well-ordered set is isomorphic to a unique ordinal number). While dealing with transitive models (most of Chapter II), we shall use the letter 9l to denote a transitive class (possibly a proper class).

Reflection Principle The theorem to be proved below is the counterpart of the LowenheimSkolem theorem. While that theorem states that every model has a small elementary submodel, the reflection principle provides, for any Jinite number of formulas, a set M that is like an “elementary submodel” of the universe, with respect to the given formulas. The theorem is proved without the use of the axiom of choice, but using AC, one can obtain a countable M.

‘Ibeorem 29 (Rejection Principle). (A) Let cp(x,, . . . , x,) be a formula. For each M , there exists a set M 2 M,such that M

(1 1.8)

..., x,)

~p(x1,

iff

cp(x,, * . . , x,)

for every x,, . . . , x , E M . ( W e say that M reflects cp.) (B) Moreover, there is a transitive M 2 M o that reflects cp; moreover, there is a limit ordinal u such that M , E V, and V, reflects cp. ( C ) Assuming AC, there is M 2 M , such that M reflects cp and 1 M 1 I max( I M o 1, No). In particular, there is a countable M that rejects cp.

Remarks. 1. We may require either that M be transitive or that ( M I I max( I M , 1, KO)but not both. 2. The proof works equally for any finite number of formulas, not just one. Thus if cp,, . . . , cp, are formulas, then there exists a set M that reflects each of Cp,,

*.*,

cpn.

2.

90

TRANSITIVE MODELS OF SET THEORY

3. If 0 is a true sentence, then the reflection principle yields a set M that is a model of 0 ;using AC, one can get a countable transitive model of 6. 4. As a consequence of the reflection principle, and Godel's second incompleteness theorem, it follows that the theory ZF is not finitely axiomatizable. For, any finite number of theorems of ZF have a model (a set) by the reflection principle, while existence of a model of ZF is not provable. (By the same argument, no consistent extension of ZF is finitely axiomatizable.) The key step in the proof of Theorem 29 is the following lemma, which we prove first.

Lemma 11.2. (i) Let cp(ul, ..., u,, x) be a jormula. For each set M , there exists a set M 3 M, such that (11.9)

if

3x cp(u,, . .., u,, x)

for every ul, . . , , u,

E

then (3x

E

M)(p(ul, ..., u,, x )

M.

Assuming AC, there is M' IM'I I m a x ( I M o I , K o ) .

2

M , such that (11.9) holds for M' and

are formulas, then for each M , there is M 3 M o such that (ii) If q l , . . ., (11.9) holds for each cp 1, . . ., qk . Proof: We shall give a detailed proof of (i). An obvious modification of the proof gives (ii); we leave that to the reader. Note that the operation A?(u1, . . . , u,) defined below has the same role as Skolem functions in the Lowenheim-Skolem theorem. Let us recall the definition (9.3): = {x E C : (Vz E

(1 1.10)

For every ul,

. . ., u,,

C) rank x I rank z )

let X ( u l , ..., u n ) =

(1 1.11)

c

where

c = {x : cp(u,, ... , u,, x)}

(11.12)

Thus A?(ul, . . . , u,) is a set with the property (11.13) if

h q ( u , , ..., u,, x), then

3x

E

X(u,

..., u,)q(u, ..., u,,

We construct the set M by induction. We let (11.14)

Mi+l

= Mi u

u

. . . , u,) : ~ 1 .,. . , U, E Mi}

{%(~l,

and

M=

u m

i=O

Mi

x)

11.

91

TRANSITIVE MODELS OF ZF

Now, if u l , ..., u, E M, then there is i E N such that u l , ..., u, E M i and if cp(u, ..., u,,x)holdsforsomex,thenit holdsforsomex E M i + , by(ll.13)and (1 1.14). Assuming AC, let F be a choice function on P(M). For every u . . . ,u, E M , let

,,

h(u,, . . ., u,) = F(.W(u,,. . . , U"))

and let us define Mb = Mo MI,, = MI u {h(u,, ..., u , ) : ul, ..., u, E MI} m

M'=

U MI i=O

(1 1.9)can be verified for M' in the same way as for M. Moreover, each MI has cardinality at most max( I Mo 1, Xo), and so does M'.

,,

Proof of Theorem 29. Let cp(x . . . , x,) be a formula. We may assume that the universal quantifier does not occur in cp (Vx . . . can be replaced by 1 3 x 1 . . *). Let cp . . . , cpk be all the subformulas of the formula cp. Given a set M,, there exists, by Lemma 11.2(ii), a set M 2 M o , such that

,,

(11.15)

3xcpj(u, . . . , x ) + ( 3 x € M ) c p j ( u,..., x),

j = 1 , ..., k

for all u, . . . E M. We claim that M reflects each c p j , j = 1, .. . , k,and in particular M reflects cp. This is proved by induction on the complexity of c p j . It is easy to see that (every) M reflects atomic formulas, and that if M reflects formulas $ and x, then M reflects -I$, t+h A x, $ v x, $ + x, and $ H x. Thus assume that M reflects cpj(u,, . . . , urn,x) and let us prove that M reflects 3xcpj . If u l , . . . , urnE M, then M b 3x cpj(ul, ..., urn,x)

H

(3x

E

M ) M 1 cpj(u,, ..., urn,x)

( 3 E~M ) cpj(U1, . . ., urn x) 9

3x

...) urn, X )

~ p j ( ~ 1 ,

The last equivalence holds by (1 1.15). This proves part (A) of the theorem. Part (C) is proved by taking M of size 5 max( I M, 1, No). To prove (B), one has to modify the proof of Lemma 11.2 so that the set M used in (11.15) is transitive (or M = V.). This is done as follows: In (11.14), we replace M i + , by its transitive closure (or by the least V, 2 M i +,). Then M is transitive (or M = V,). H Exercise 11.1. Find a formula p(x) such that every nonempty transitive set that reflects ~ ( x is) uncountable.

2.

92

TRANSITIVE MODELS OF SET THEORY

Exercise 11.2. For every formula cp, there is a closed unbounded class C, c Ord such that for each a E C,, V, reflects cp. [C, = C, n C,, C, = C, n K,, where K, is the closed unbounded class K, = {a E O r d : Vx, ... x, E K[3x cp(x, xl, ..., x,)- (3x E V.) cp(x, x,. ...,x.)]}.]

,

Exercise 11.3. Let 1 ' 11 be a transitive class and let cp be a formula. For each

there exists M

c 1' 11 such that M o E M and that 1 XI, ..., x,) M 1 xi. ++

for all xI,..., x,

E

...?

Mo

E 'N

x,)

M.

A transfinite sequence ( H a : a E O r d ) is called a cumulatioe hierarchy if

(i) H a c_ H , , , c_ P ( H a ) (ii) if a is a Limit, then H,= (Note that each H a is transitive and H,

(11.16)

uB
UDEOId

Exercise 11.4. Let ( H a : a E Ord) be a cumulative hierarchy, and let H = H,. Let cp be a formula. Show that there are arbitrary large limit ordinals a such that

H 1 V ( X I , . . . , x,) for all xl,

.

Ha 1 XI, . . . x,)

++

.... x, E H a . Ciidel Operations

The axiom schema of separation states that, given a formula cp(x),then for every X there exists a set Y = {u E x : cp(u)}

(1 1.17)

It turns out that for restricted formulas, the construction of Y from X in (11.17) can be described uniformly for all restricted formulas, by means of a finite number of elementary operations.

Theorem 30 (Normal Forp Theorem). There exist operations 8 1, . . . , 8 such that if cp(u,, . . ., u,) is a restricted formula, then there is an operation 8, a composition of B1,..., ijl0such that for all X I , .. ., X,, (11.18)

8(Xl, ..., X,) = {(ul, .. ., u , ) : u1 E X1,. . . , u,

E

X,

and cp(uI, . .., u,)}

In case n = 1, we define ( u ) = u, and we have in particular

S ( X ) = (11 E

x : cp(u)}

The operations Fjl, . .. , Bl0 will be defined below. Compositions of B1,. . . ,Bl0 are called G a e l operations. We call the following sentence an instance of restricted separation: (11.19)

VP1

**.

VP,,

t/x 3Y V U ( U E y * U

E x A V ( U , pi,

..., p,))

11.

93

TRANSITIVE MODELS OF ZF

where cp is a restricted formula. We say that a transitive class 93 satisfies Restricted Separation if for every restricted cp, 9JI satisfies (11.19). A class C is closed under an operation 5 if g(x 1, . . . , x,) E C whenever xl, . . . , x, E C. Let us call a class 9JI (Giidel) closed if it is closed under the operations g,, . . . , gl0;equivalently, if 93 is closed under all Gadel operations.

Corollary. If 9JI is a closed transitive class, then 93 satisjies Restricted Separation.

Pro05 Let cp(u, p l , .. ., p,,) be a restricted formula, and let X , pl, .. ., p,, E 91. Let Y={~EX:(P(K

P ~ . . . . T P ~ ) }

By Lemma 10.1, it suffices to show that Y E 91, in order that 9 1 1 satisfy (1 1.19). By the normal form theorem, there is a Giidel operation 5 such that S(X, {PI}, . . . ) { p n } ) = {(u,P I , . . . ) P n ) : u

E

X A C P ( U~ , 1

pn)l

.. 7 .?

It follows that Y = { u :3

... ~ u , ( u ~,

~ 1

1 . . 9* )

u,) E iJ(X, {PI}, ...) { p n } ) }

-... dom S(X, {pl}, .. . , {P,})

= dom

n

Since both {x, y) and dom(x) are G d e l operations (see below) and since 91 is closed. we have Y E 91. H Godrl Oprrutions

81(X, Y ) = { X , Y} &(X, Y )= x x Y

g3(x,Y ) = & ( X ,Y ) = {(U, U ) : U E XAU E g4(x,Y ) = x - Y

YAU

E U}

g 5 ( X ,Y )= X n Y =

ux

iw)= dom(X) & ( X ) = {(U, 0) : (0, ).

&(X) = {(u, v.

E

X}

w ) : (4w, u ) E X}

SlO(X) = {(u, 0, w ) : (0,w, U ) E X ) Proof of Theorem 30. The theorem is proved by induction on the complexity of restricted formulas. To simplify matters, we consider only normalformulas: A formula ‘p(ul, . . . , u,) is normal if:

2.

94

TRANSITIVE MODELS OF SET THEORY

(i) the only logical symbols in cp are 1,A , and restricted 3; (ii) = does not occur; (iii) the only occurrence of E is ui E u j where i f j ; (iv) the only occurrence of 3 is

(%+

1

E

Ui)lCl(ul,

. * . -urn+ 1 )

where i Im. Every restricted formula can be rewritten as a normal formula: The use of logical symbols can be restricted to 1,A , and 3; x = y can be replaced by a restricted formula (VU E X ) U E y A ( v U E

x

y)

U EX

x can be replaced by

E

(3u

E

x) u = x

and the bound variables in cp(ul, .. . , u,) can be renamed so that the variable with the highest index is quantified. Note that we allow “dummy” variables, so that for instance cp(ul, ..., u s ) = u3 E u2 and cp(u,, . . . , u 6 ) = u3 E u2 are considered separately. Thus let cp(u,, ..., u,) be a normal formula and let us assume that the theorem holds for all subformulas of cp. Case I. cp(ul, . . . , u,) is an atomic formula, ui E u j ( i # j ) . We prove this case by induction on n. Case la. n = 2. Here we have {(Ul, U 2 ) : U 1 E

xl A U 2 E x 2A U 1 E U 2 } = &(XI,x 2 )

and u 2 ) : U l E X 1 A u2 E x2 A U 2 E

@19

4)= 5 * ( & ( X 2 X l ) ) t

Case Ib. n > 2 and i, j # n. By the induction hypothesis, there is ((U1, . .

. , U,-

1)

: U1

E

X I , . . ., U,-

1 E

x,-

1A

Ui E Uj} =

5 such that

G(x1,. . . , x,-1 )

Obviously, ((U1, ..

. , U,) : U1 E X I ,. . . , U, E x,A U i E u j } = s(x1,. . . , x,-1 ) X

x,

Case Ic. n > 2 and i, j # n - 1. By the induction hypothesis (Case Ib) there is 5 such that ((u,,

..., u , - ~ u,, , u , , - ~ )u1 : E X,,..., u,

E

X, and ui E u j } = S ( X , , .. ., X , )

Noting that ( ~ l r. . . r ~ n - 2 r ~ , , ~ n - 1 ) = ( ( ~ l r . . . t ~ n - 2 ) r u n t u n - l )

11.

95

TRANSITIVE MODELS OF ZF

we get {(ul,

..., u,): u l E X,,..., u,

E

X, and ui E u j } = S9(S(X1, X,)) ...?

Case Id. i = n - 1, j = n. By Ia, we have {(~,-1,U,,):U,,-1

E X , - l ~ ~ , E X , ~ E4,}=E(X,-i,Xn) ~U,-l

and so { ( ( u , - ~u,), , (ul, ..., u , - ~ ) ) u: 1 E Xl,..., un E =

E(X,_l. X,) x (X,x ".

=

S(X1,.... X,)

x

X, and

u,-,

E

u,}

X,-2)

Now we note that and and thus {(ul,

. . ., u,): U , E X,,..., u,

Case Ie. i

= n, j = n -

E

X, and

u,-

, E u,} = SlO(S(X1, ..., X,))

1. Similar to Id.

Case 11. cp(u,, . . . , u,) is a negation, l$(ul, . . . , u,). By the induction hypothesis, there is 8 such that {(u1, . . . , u,): u1 E X,,..., u,

Clearly,

E

X, and $(ul, ..., u,)}

{(u,,.. ., u,): u1 E X,,..., u,

=

S(Xl,..., X,)

X, and cp(u,, ..., u,)} = (X,x ... x X,) - S(X,,.. ., X,)

Case I l l . cp is a conjunction,

{(ul, . . . , u,) : u ,

E

X,,. . ., u,

$1A $ 2 .

X, and

E

E

By the induction hypothesis, $i(~1,

. . ., u,)}

= Qji(X1,.. .

X,) ( i = 1, 2)

Hence {(u1, . . ., u,) : u , E

X1,. . . , u,

E

X, and

cp(u,,

.. . , u,)}

@,(X,,..., X,) n O2(X1, ..., X,) Case IV. cp(u,, . . ., u,) is the formula (3u,+ E ui)l(l(ul,. . . , u,+ ,). Let ~ ( u , .,.., u , + , ) be the formula $(u,, ..., u , + , ) A u , , + , E u i . By the induction hypothesis (we consider x less complex than q), there is 8 such that =

,

(1 1.20)

{ ( u , , ..., u.+1): =

U, E

S(X1,

. . a ,

Xl,..., U.+l

L + l )

E

Xn+l and

X(U1,

... h+J} 9

2.

96

for all X,, .. ., X,,

(11.21)

TRANSITIVE MODELS OF SET THEORY

,. We claim that {(u,,

..., u,):

u1 E X,,

E

X, and

. . . , u,) and X = X , x ... x

c ~ ( u ) c~

cp(u,,

..., u,)}

... x X,) n dom(S(X,, ..., X,,

= (X, x

Let us denote u = (u,,

.. ., u,

30 E ui$(u, 0)

u Xi))

X,. For all u E X, we have

u xi) u

*

3U(U E Ui A $(U, U ) A U E

++

u E dom{(u, u ) E X x

X i : ~ ( uu)} ,

and (11.21)follows. This completes the proof of Theorem 30. H We recall that a set W is called Gbdel closed if it is closed under all G d e l operations. For every set M , there is a smallest closed W 3 M : Let W= W , , where W, = M and?

u."=,

w,,,= w,u {&(X, Y ) : x E W,,

YE

w,,i =

1, ..., lo}

Let us denote W = cl(M)

and call W the (Gadel) closure of M. Exercise 11.5. If M is a transitive set, then cl(M) is transitive.

[Show (by induction on W.) that if X

E

W , then TC(X) E W.]

Exercise 11.6. If M is closed and extensional and if X E particular, if (x, y) E M ,then x E M and y E M.

M is finite, then X c M. In

Exercise 11.7.* If M is closed and extensional, and if n is the transitive collapse of

M,

then n(&(X, Y)) = &(nX, nY),

i = 1, ..., 10

for all X, Y E M . [Use the normal form theorem.]

85 and g8 are compositions of the remaining [ W X ) = dom(510(810(89(S10(X x X))))).]

Exercise lf.8. The operations

Si.

Exercise 11.9.* The axioms of comprehension in Bernays-Gdel set theory can be

proved from a finite number of axioms of the form

vx V Y 3 2 z = q x , Y) where the fii are like the operations Si. Thus Bernays-Gijdel set theory is finitely axiomatizable. [Formulate and prove a n analog of the normal form theorem.]

t We shall treat the operations i = 6 . .... 10.

. . . , g,oas binary operations; we let &(X, Y )= &(X)for

97

11. TRANSITIVE MODELS OF ZF

Transitive Models of ZF The main theorem of Section 11 gives sufficient conditions for a transitive class to be a model of ZF. In fact, the conditions are necessary for proper classes, and we have a characterization of transitive models of ZF that contain all ordinals. We say that a class 91 is almost universal if every subset X E W is included in some Y E W. (Note that 91 has to be a proper class.)

Theorem 31. Let W be a transitive, closed, almost universal class. Then W is a model of ZF. Remark. The conditions are necessary for a transitive model 911 of ZF that is a proper class: We shall show in Section 12 that the notion of ordinals and rank, as well as G d e l operations are absolute for transitive models of ZF, that is Ordm = Ord n W, rankm(x)= rank(x), and s " X , Y ) = &(X,Y ) .Hence 91 is closed under the &; and if it is a proper class, then 91 2 Ord; and if X c 91 then X c Y E 91, where Y = V y for some a. The only difficulty in the proof of Theorem 31 is the proof of the separation axioms. Half of the proof has been provided by Theorem 30; the other part is contained in the following lemma.

Lemma 11.3 (Reduction Lemma). For any formula cp(u,, ..., u,) with k quant$ers, let @ ( u l .. . . ,u,, Y,, . . ., Yk) denote the restrictedformula that results i f w e r e p l a c e e a c h 3 x , V x i n c p b y 3 x E q , V x E Y , , j = 1, ..., k . Let 9 3 1 be a transitiire almost uniuersal class. Thenfor every X E W there exist Y,, . . . , Y, E 911 such that (11.22)

iff

91 k (p(u,,..., u,)

for all u , , . . . , u,

E

q(u1, . . . v un, Y1, . . . ) Y,)

X.

Proof: By induction on the complexity of cp. We may assume that cp has only existential quantifiers. If cp is quantifier-free,then (p = cp and (11.22) holds. If cp is -I$, Ic/ A x, + x, etc., and if (11.22) holds for $ and x, then it also holds for cp. Thus let cp(u) be 3u $(u, u ) and assume that (1 1.22) holds for $. Assuming that II/ has k quantifiers, q is ( 3 u E & +,) $(u, u, Y,, .. . , Y,). Let X E 91. We look for Y,, . . . , Y, , Yk+ E 91 such that for every u E X ,

,

(11.23)

9l k 3uII/(u,u )

iff

( 3 u E Y,,,) $(u, v , Y,, ..., Y,)

By the collection principle (9.4) (applied to the formula u E 91 A 91 k $(u, u ) ) , there exists a set M such that X E M c 91 and that for every u E x, (11.24)

(30 E 91)91 k $(u, u )

iff

(3u E M ) W k $(u, v )

2.

98

TRANSITIVE MODELS OF SET THEORY

Since %I is almost universal, there exists Y E 9JI such that M from (11.24) that for every u E X, (11.25)

(3E ‘m)

iff

k $(u, u )

By the induction hypothesis, given Y for all u, u E Y , (1 1.26)

%I k $(u, u )

E

k $(u,

91,there exist Y,, . . . , yk

iff

Thus we let U,, = Y, and since X %I k 3u$(u, u )

E

(3u E Y )

E

$(u,

u, Yl, ...,

0)

E !M such that

5)

Y, we have for all u

E

iff

(3u E ‘m) Iu1 k $(u, u )

iff

(30 E

iff

(30 E Y ) $(u, u, Y,, ...,

Y) %I k

Y. It follows

X,

Q(u, u )

5) W

Proof of Theorem 31. Let %I be transitive, closed and almost universal. We shall show that every axiom of ZF holds in %I.

I. Extensionality Every transitive class satisfies Extensionality. 11. Pairing “ z = {x, y } ” can be expressed by a restricted formula, see (10.6). %I is closed under { , } and so it satisfies Pairing. 111. Separation If q ( u , p) is a formula, we wish to show that for every X,p E !M, the set Y = {u E X : !llk q ( u , p ) } is in %I. By the reduction lemma, there are Y,, .. . , YkE %I such that Y = {u E

x : q(u, p , Y1, ..., y,))

(use a “parametric” version of Lemma 11.3 or replace X by X u { p ) ) . Since @ is a restricted formula, Y belongs to ’331 by the corollary to Theorem 30. IV. Union “ Y = X ” is a restricted formula, see (10.7). Since !llis closed under it satisfies Union. V. Power set Since 9R is a model of Separation, it suffices to prove

u

(11.27)

u,

mkvx3Yvu(uGx-+uEY)

Since “ u E X *’ is a restricted formula, (11.27) is equivalent to

(VXE ~ R (3Y ) ~ ! l l m ) P ( Xn) 9Rc Y This follows from almost universality of 9.l. VI. Infinity It suffices to show that o E 9.l (see the proof of consistency of Regularity in Section 10, (10.9)). Since 9R is closed, it can be shown by induction that %I

12.

99

CONSTRUCTIBLE SETS

contains all natural numbers; and since '31is almost universal, we have w E X for some X E 911. Thus it is enough to show that "U

is a natural number"

is a restricted formula, for then w = {u E

X : 91 1 u is a natural number}

By the axiom of regularity, E is well-founded and so we have (11.28) u is an ordinal iff u is transitive and u is linearly ordered by E

It is fairly easy to verify that the right-hand side of (11.28) is a restricted formula. Moreover, we have a E w iff a E Ord and a # 0 + (38E a ) a = p + 1, and (VY E a ) [Y # 0

and a = p

+

(38 E Y) Y = B + 11

+ 1 is a restricted formula: BEa

~ fEi ar\(V< E a )

[r E B v t = 83

VII. Replacement Let cp be a formula and assume that 'u1 1 vx, Y , z[cp(x, Y ) A cp(x, z)

+

Y = .I

Let F be the function F = {(x, Y ) E 9J : 93 1 ~ ( xY,) } It is enough to show that F[X] E 'u1 for every X E Iu1. In fact, since W satisfies Separation, it is enough to show that F[X] c Y for some Y E W.Such Y exists by almost universality of W. VIII. Regularity As in (10.1l), every nonempty S E 'u1 satisfies (in ID1) (3x

E

S )S nx = 0

since (S n x)" = S n x. If Illl is transitive, almost universal, and closed under gl, then 101 satisfiesthe axioms of Extensionality, Regularity, Pairing and weak Union, Power set, and Replacement (( 1.8)-(1.10)). Exercise 11.10.

12. CONSTRUCTIBLE SETS

We shall now give an example of a transitive model of ZF. This model, the constructible universe of Giidel, is of particular importance: (i) It is the least transitive model containing all ordinals; (ii) it satisfies the axiom of choice and the generalized continuum hypothesis, thus providing a proof of consistency of AC and GCH with ZF.

2.

100

TRANSITIVE MODELS OF SET THEORY

We recall that cl(X) denotes the G a e l closure of X. Let us denote def(U) = cl(U u {U}) n P ( U )

(12.1)

i.e., the family of all X E U obtained by a finite number of successive applications of G d e l operations to U and elements of U . Exercise 12.1. U k cp(x,, ..., x,) is a restricted formula (with variables U, xl, ..., x,). Exercise 12.2. Let cp be a formula. There is a Godel operation transitive U and all xl. . . ., x, E U,

U :U k

{u E

cp(u, X I . ..., x,)} =

8 such that for every

5 ( U , X I , ..., x,).

[Use the normal form theorem.] Exercise 12.3.* If U is a transitive set and X c U is definable over 11 = ( U , then X E def (U). [Use induction on cp E Form.]

E n

U2)

We shall show in Section 14 that if 5 is a G d e l operation, then u E g(x,, . . . , x,) can be written as a restricted formula. Use this fact in the next exercise. Exercise 12.4. If 5 is a Godel operation, then there is a formula cp such that for every transitive set U and all x,, ..., x, E U , if X = 5 ( U , x l , . . . , x,) c U, then

x = {u E U : U 1

cp(x1,

..., x,)}.

Exercise 12.5.* If U is a transitive set and X definable over U = ( U , E n U 2 ) .

E

cl(U u {U))is a subset of U ,then X is

In view of Exercises 12.3 and 12.5 we call the set def(U) (for any transitive set U ) the set of all subsets definable over U . Exercise 12.6. If U is a finite transitive set, then def(U) = P ( U ) .

We now define the constructible hierarchy: (12.2)

r,=0

u L+ def(L) L= u L, = 1

L,

if a is a limit ordinal

#
=

L,

a E Ord

Note that U c def( U ) c P( U ) for every U and so the L.form a commulative hierarchy (see (11.16)). Thus each La is transitive and La E V,. Using the fact that “ a is an ordinal” is a restricted formula (see (11.28)), and Exercise 12.2, we can easily prove by induction that a E I,, and so

L,

n Ord = a

Elements of the class L are called constructible sets.

12.

101

CONSTRUCTIBLE SETS

Axiom of Consrrucribiliry (I/ = L). Every set is constructible.

Theorem 32 (Gael). (i) L is a model of ZF. (ii) L satisfies the axiom of constructibility. (iii) If91is a transitive model of ZF containing all ordinals, then L

E 911.

Consequently, the axiom I / = L is consistent with ZF. It will be shown in Section 13 that the axiom of constructibility implies the axiom of choice and the generalized continuum hypothesis. Proof of (i). We show that L is transitive, closed, and almost universal. L is transitive since (L, : a E Ord) is a cumulative hierarchy. To show that L is closed, it suffices to verify that for each limit a, L, is closed. We leave the verification to the reader. And finally, L is obviously almost universal since any subset of L is included in some L, and L, E L W Exercise 12.7. L, = V, Exercise 12.8. Let a,

H a+

, 2 def(H.).

.

( H a: a E O r d ) be a cummulative hierarchy and assume that for each Then H = U b E O,d Ha is a model of ZF.

To prove the rest of Theorem 32, we shall introduce the notion of absoluteness of set theoretical concepts. Absoluteness Let 9 1 1 be a transitive class and let cp(x is absolute for 9.R if 9 1 1

c ~ ( x ,..., , xn)

,,. . . ,x,) be a formula. We say that cp

iff

&,, . . . , x,)

for all x l r . . . , xn E 9.R. We have proved in Lemma 10.1 that if cp is a restricted formula and 9.R is transitive, then cp is absolute for 91. In fact, the proof of Lemma 10.1 gives a little more general result: If cp and I// are absolute for a transitive W,then l c p , cp A I//, cp v I//, cp -+ I//, and (Vx E y ) q, ( 3 x E y ) cp are absolute for 9.R. In the following lemma we present a number of restricted formulas, which are therefore absolute for any transitive class W.

Lemma 12.1. Thefollowing expressions can be written as restricted formulas and thus are absolute for all transitive classes. (a) x = {u, v}, x = (u, v ) , x is empty, x s y , x is transitive, x is an ordinal, x is a limit ordinal, x is a natural number, x = w. (b) 2 = s i ( X , Y), i = 1, ..., 10. (c) X is a relation, y = dom(x), y = ran(x),f is a function, y = f ( x ) , Y =f[X], s = f IX.

102

2.

TRANSITIVE MODELS OF SET THEORY

Proof.

(a) x

= {u, u } - u

X = ( U , U)*

X A ( V W E X)(W = u v w = U ) (3W E X)(3Z E X)[W = {U} A Z = {U, U}] A EXAUE

(Vw E X)(W = { u } v w = fu, u } ) x is empty-(Vu E x)u # u x E y o ( V u E x)u E y x is transitive (Vu E x)u E x x is an ordinal c)x is transitive A (Vu E x)(Vu E X)[U E u v u E u v u = u ) (Vu E x)(Vu E x)(Vw E X)[U E u E w + u E w ] x is a limit ordinal c)x is an ordinal A (Vu E x)(3u E x)u E u x is a natural number x is an ordinal A (x is not a limit v x = 0) A (Vu E x)[u = 0 v u is not a limit] x = o - x is a limit ordinal A x # 0 A (Vu E x ) x is a natural number

-

-

{X, Y } ... see above Y*(VZEZ)(k€X)(3yE Y ) Z=(X,y) A (Vx E X)(Vy E Y ) ( 3 zE Z ) z = (x, y ) Z = E ( X ,Y ) . . . similarly z=x - Y c - , ( V Z E Z ) [ Z E X A Z $ !Y ] A ( V Z E X ) [ Z $ !Y + Z E Z ] Z = X n Y . . . similarly z = X-(VZ E z)(3X E x)Z E X A (VX E x ) ( V Z E X)Z E z Z = dom(X)t, (Vz E Z ) z E dom X A (Vz E dom X ) z E X

(b) Z

=

z=xX

u

and we show that:

(12.3)

(i) z E dom X is a restricted formula; (ii) if cp is restricted, then (Vz E dom X ) q is restricted.

(i) z E dom X o ( 3 x E X ) ( L E x)(3u E u)x = ( z , 0). (ii) (Vz E dom X ) c p o ( V x E X ) (Vu E x) (Vz, u E u ) [x = (2, u ) + cp]. An assertion similar to (12.3) holds for ran(X), and for 3. Z = & ( X ) c)(Vz E Z)(3u E ran X)(3u E dom X)z = (u, u ) (Vu E ran X)(Vu E dom X ) ( 3 E Z ) z = (u, u ) Z = 59(X). . . similarly Z = B l 0 ( X )... similarly

A

-

(c) X is a relation (Vx E X)(3u E dom X)(3u E ran X ) x = (u, u ) .f is a function -,f is a relation A (V-x E dom/)(Vy, z E ranf)[(.x, y ) ~ .Af (x, 2) ~ f y + = z] where (x, y ) Ef" (3u E f ) U = (x, y ) g =,f I X ++ g is a function A g G .f A (Vx E dom g)x E X (Vx E X)([.YE domJ'+ x E dom g ]

A

A

12.

103

CONSTRUCTIBLE SETS

Exercise 12.9. The following can be written as restricted formulas: x is an ordered pair, x is a partial (linear) ordering of y, x and y are disjoint, z = x u y , y = x u {x}, x is an inductive set, z E & ( X , Y) (i = 1,. . . , lo), j' is a one-to-one function of X into

(onto) Y,f is an increasing ordinal function, f i s a normal function.

It should be emphasized that cardinal concepts are generally not absolute. In particular, the following expressions are known not to be absolute:

Y = P(X),

IYI

=

I X 1,

/3 = cf(a), a is regular

a is a cardinal,

Exercise 12.10. Let 'W be a transitive class.

(a) I f W k 1x1 5 IYI,then 1x1 2 IYI. (b) If a E 9 u 1 and if a is a cardinal, then 'M t a is a cardinal.

[ I X 1 5 I Y I c* 3fcp(f, X, Y); a is a cardinal and i+h are absolute formulas.]

-

13f(3jE a A $(a, b , / ) ) , where cp

Absolute Operations Let !IN be a model (of the language of set theory). Strictly speaking, we defined 91 k cp only for formulas whose only nonlogical symbols are E and =. However, we shall frequently use familiar set theoretical concepts such as classes and operations in the context of a model 9Jl,and use !IN k cp even if cp is an expression that contains other set theoretical symbols. We have introduced classes in order to use the notation

c = {x : cp(x)} We shall thus use the symbol C" to denote corresponding class in 91:

c" = {x E 91 : 91 k cp(x)} For instance, V'* = 911 and 0"= 0. Using the absoluteness of ordinals, one can easily see that if 91 is a transitive class, then (12.4)

Ord" = Ord n 91 =

i

either Ord or the least a # 91

-

Let 91 be a transitive class, and let F be afunction: Y = F(x)

cp(.X?

Y)

Then {(x, y) E 911 : 911 1 cp(x, y)} is a relation in 91 (the notion of an ordered pair is absolute). If, moreover, (12.5)

911 k

v.X

'Jy v Z ( C p ( X , y )

A Cp(X. Z ) -+

y = Z)

then the relation is a function in 911 and we denote it by F": (12.6)

y = F"(x)

t*

91 k P(X,y )

2.

104

TRANSITIVE MODELS OF SET THEORY

An operation is usually defined by a defining formula S(x) = {u : $(& 4) (Our restriction t o one variable x is, of course, inessential.) Thus we let, for xE

W,

(12.7)

p ( x ) = {u E 91 : 9 1 k $(x, u ) ]

Since an operation is a function and since y = 8(x) * y = {. : $(x,

.)I

(12.7) and (12.6) (with cp(x, y) being the formula y = {u : $(x, u ) } ) give the same definition of Gm(x), provided S"(X) E 91i. However, we define (12.7) even if g'*"(x).$ 91i. A function F is said to be absolute for W if (i) (12.5) holds, i.e., if F" is a function in W,and (ii) y = F(x) is an absolute formula, i.e., (12.8) I=c p k Y ) iff Y) for all x, y E W.An operation 5 is absolute if Sm(x) = S(x) for all x for an operation 8 defined by cp(X3

S(x) = .{ : $(X?

E

W.Thus

4)

to be absolute for W it suffices that (i) S(x) E W for all x E W,and (ii) $ is absolute for W. In general, if the defining formula $ is absolute, then we have S"R(x) =

S(x) n W

An example is the operation P ( X ) :

P"(x)

= P(X) n9 1

We say that an operation 8 is defined in W if g"(x) E W for all x

E

W.

Lemma 12.2. (a) If 8 is an absolute operation (for W)and defined in W,and ifq is absolute, then q ( Z ( x ) ) is absolute. (b) If 8 is absolute and defined in m, and if0 is absolute, then 0 0 5 is absolute. Proof: (a) Given x E W,we have y = S(x) = g"(x)

E

W,and

dY) (b) Apply (a) to the formula z = 0 ( y ) . H kdy)

iff

Lemma 12.3. (a) The operations S i ( X , Y ) , i = 1, . . . , 10, are absolutefor all transitiue classes. (b) If W is transitiue and closed and 8 is a C d e l operation, then 8is absolutefor

m.

12.

105

CONSTRUCTIBLE SETS

Proqf (a)follows from Lemma 12.1b, and (b) follows from (a) and Lemma 12.2. (In Section 14 we shall prove a stronger result.)

Absoluteness for Models of ZF

Lemma 12.4. Let 111 be a transitive model of ZF, let G be afunction on V , such that C is absolute for W and dom(Gm)= W.Let F be defined by induction: F(a) = G(F 1 a) Then F is absolute for 111 and dom(Fm)= 0rdm.

(12.9)

-

Proot The function F is defined as follows: (12.10)

y = F(a)

a is an ordinal A 3 f q ( f , a)

where q(f,a) is the formula (12.11)

f i s a functionAdomf= aA(V<E a ) f ( r ) = G ( f l < )

The operation g ( J () =f I( is absolute for 311; and since 311 is a model of ZF, is defined in 311, and so q(,f,a) is absolute for 311. Therefore, if (y, a) E Fm, then YE 111 q(f,a), and so y = F(a). This also implies that P is a function in 111. It remains to show that if a E 111 and y = F(a), then y E 111 and y = Fm(a).We have proved (in ZF) that (given G), for every a there exists a functionfon a such thatf(r) = C ( f It)for all < a. Since 111 is a model of ZF, we have

<

111 I=

Yq(f, a)

and if y = G ( f ) ,then y = Fm(a)= F(a). W In particular, if 111 is a transitive model of ZF and G is absolute and

G[U] G W,then the functionfon w defined by

f (0) = a0

f ( n + 1) = G(f(n))

9

is in 111. Exercise 12.11. The function r: Ord' tive models of ZF. [Define r- ' by induction.] Exercise 12.12. The operations a

--*

Ord defined in Section 2 is absolute for transi-

+ /3 and a . /3 are absolute for transitive models of ZF.

Exercise 12.13. TC(S) is an absolute operation for transitive models of ZF.

I have already mentioned that the operation P(X) is not absolute: Pm(X)= P(X)n YJI. If YJI is a transitive model of ZF, then the function V, is defined in I1T1, and one can prove by induction (12.12) C=KnI1T1 (because U is an absolute operation).

106

2. TRANSITIVE MODELS OF SET THEORY

We shall now prove absoluteness of the notion of constructibility and complete the proof of Theorem 32.

Lemma 12.5. (i) The function a++& is absolute for transitive models of ZF. (ii) If 'ill is a transitive model of ZF containing all ordinals, then '' x is constructible" is absolute for 'ill and in fact L?" = L As a consequence of (ii), L?" = L and thus L E 'ill,for any transitive model of ZF containing all ordinals. Also by (ii), if x E I, then

L k x is constructible

and so L satisfies the axiom of constructibility. Thus the statements (ii) and (iii) of Theorem 32 follow from Lemma 12.5. f'roo& (ii) follows from (i); since !Ul is a model of ZF, the function a- L y is defined in 91;and by absoluteness, we have for all x, x

E

~ - 3 4E L~) - 3 a ( x

E

e)-x

E

Lm

To prove (i), let 911 be a transitive model of ZF. Throughout the proof, we tacitly use the fact that each operation in question is defined in 911 (since 91 is a model of ZF). The function a HL, is defined by induction: (12.2)

=

0,

L, =

u

L a + , = def(&)

(if a is a limit),

L,

LJ
where (12.1)

u E def(U)-u

E

cl(U u {U)) A u

and cl(A4) is defined by induction: (12.13)

cI(M)

=

u ran(

C_

U

W)

where (12.14)

W ( 0 )= M ,

W(n + 1) = W(n)u { & ( X , Y ) : X

E

W(n),Y

To start, we show that in (12.14) W(n absolute operation. For all U , Z, we have

E

W(n),i = 1, . . . , 10)

+ 1) is obtained from W(n)by an

12.

107

CONSTRUCTIBLE SETS

Since z' = &(x, y) is a restricted formula (i = 1, . .. , lo), the right-hand side of (12.15) is also restricted and thus absolute. By Lemma 12.4, the function Win (12.14) is absolute and so, by (12.13), cl"(M) = cl(M)

(12.16)

for all M E W.Using (12.1), we see that the formula u E def(U)

is absolute for W,and moreover, def(U) c W for all U tion def(U) is absolute for W:

E

W.Thus the opera-

defSR( V) = def (V)

(12.17)

for all U E W. Now we apply Lemma 12.4 to the inductive definition of L, (12.2)and since both u and def are absolute, we conclude that

e=L. for all ordinals a E W. H Exercise 12.14. If 91 is a transitive model of ZF and 91 is a set, then L* = L, where u is the least ordinal not in Zn. Exercise 12.15. Let us define a sequence (J, : a E Ord) as follows: J O = 0, = cl(J. u {J,}). Prove:

J, = Up.. J , if a is limit, and

(i) Each J , is transitive. (ii) J, is a model of ZF (and so L E old J,). (iii) The function U H J , is absolute for L and thus L = U m G O r d J.. (iv) The class C = { a : J, = I}is closed and unbounded.

u.

UmE

K

Using the fact that L k AC (Section 13) one can show that for every regular > No,C n K is closed unbounded in K .

Exercise 12.16. If Zn is a transitive model of ZF, then V, c Zn and so " x is finite" is absolute for 91.

We conclude this section by showing that the notion of well-foundedness is absolute for transitive models of ZF.

Lemma 12.6. If W is a transitiue model of ZF, then the formula E is a well-founded relation over P

(12.18) is absolute for

W.

ProoJ: On the one hand, we can write (12.18) as

(12.19) E is a relation over P and ( V X ) [12/ # X c P -+ (3a E X ) a is E-minimal in XI.

2. TRANSITIVE MODELS OF SET THEORY

108

If (12.19) holds (in the universe), then it holds in 9.N since both " E is a relation over P" and [..I are restricted formulas, and so [ . - I holds in 91 for every

x E m.

On the other hand, E is well-founded if and only if there exists a function on P into Ord such thatf(x) < f ( y ) whenever x E y . Thus (12.18) is equivalent to (12.20) E is a relation over P and 3f[f is a function A (Vu E ran(f)) u is an ordinal A (Vx, y B P)(x E y + f ( x )
Since .] is a restricted formula, if [ holds in 91 for somef holds; and so if (12.20) holds in m, it holds in V. [a.

a]

E

'91, then 3f[. . .]

13. CONSISTENCY OF THE AXIOM OF CHOICE AND THE GENERALIZED CONTINUUM HYPOTHESIS

In the present section we show that the constructible universe L satisfies the axiom of choice and the generalized continuum hypothesis. Consequently,both AC and GCH are consistent with ZF. "leorem 33 (Godel). There exists a well-ordering ofthe class L Thus if V = L we have a well-ordering of the universe, and so the axiom of constructibility

implies the axiom of choice. Proof: We shall in fact show that the class L has a dejinable well ordering: we shall define a well-ordering c of the constructible universe. In turn, we can define a function F on L - (0)by F(X) = the <-least element of X and have a definable choice function on L By induction, we construct for each a a well-ordering < a of L,. We do it in such a way that if a < fl, then < is an end-extension of < a , i.e., (13.1) (a) if x c ay, then x y; (b) if x E L, and y E L, - L,then x <@

<@

y.

Notice that (13.1) implies

if X E ~ E L , ,then x < , y

(13.2)

First let us assume that a is a limit ordinal and that we have constructed c for all fl < a and that if fl, < f12 < a, then < 82 is an end extension of < In this case we simply let @,.

(13.3)

i.e., if x , y E L,we let (13.4)

x<,y

iff

(3P
109

13. CONSISTENCY OF THE AXIOM OF CHOICE AND THE GCH

Thus assume that we have defined <. and let us construct ordering of La+1. We recall the definition of L+ :

,

L+1 = wu)n c l ( L u

(13.5)

<e+l,

a well-

uw m

wa.))

=P(L)

n=O

where

wo = L u {L}. %+, = {ai@,Y): X, Y E L, u {L), i = 1, ..., 10)

(13.6)

The idea of the construction of < a + l is now as follows: First we take the elements of L,,then L,, then the remaining elements of W,, then the remaining elements of W, , etc. To order the elements of W,+ we use the already defined well-ordering of W, since every x E W,+ is equal to gi(u, u ) for some i = 1, . . . , 10 and some u, u E W,. We let

,

(13.7)

<:+ is the well-ordering of L, u {L} that extends < a and such that La is the last element (ii) <:I is the following well-ordering of W,, : x y iff either: x <:+, y or: x E W, and y 4 W, or: x 4 W, and y 4 W, and (a) the least i such that 3u, u E W,(x= &(u, 0)) c the least j such that 3, t E W, (y = 3j(s,t ) ) (i)

or

(b) the least i = the least j and [the <"-least u E W, such that 3u E W,(x = &(u, u ) ] en [the <"-least s E W, such that 3t E W, (y = &(s, t ) ]

or

(c) the least i = the least j and the least u = the least s and [the <"-least u E W, such that x = &(u, u ) ] <" [the <"-least t E W, such that y = &(u, t ) ]

Now we let

u m

(13.8)

=

<.+I

n ( P ( L )x

<:+I

P(L))

n=O

,

and it is clear that <.+ is an end-extension of e a and is a well-ordering of la+ 1. Having defined -=a for all a, we let x<,y

iff

3axe,y

The relation e Lis a well-ordering of L We call e Lthe canonical well-ordering of L

2.

110

TRANSITIVE MODELS OF SET THEORY

Exercise 13.1. The sequence ( < : a E O r d ) is absolute for models of ZF; consequently, the well-ordering < of L is absolute. Exercise 13.2. For all a 5 o,IL.I = I a I .

The proof of the generalized continuum hypothesis in L is a combination of an absoluteness argument and the reflection principle.

Theorem 34 (Giidel). If V = I., then 2Nm = K,+ for every a. The key element of the proof is the following lemma:

Lemma 13.1. Assume that V = L If X E a,,then there exists y < o,+ such that X E L,. Proofof Theorem 34. Assume that V = L By Lemma 13.1, every subset of o,is constructible before the stage i.e., P(w,) c L,a+I. It is easy to show by induction that for every y 2 w, 1 L , I = Iy 1, and so I L,=+I = h’, + Therefore, [ P(o,) I = X z+ w

I

To prove Lemma 13.1 we need a more subtle absoluteness argument than Lemma 12.5. We say that a transitive set M is adequate if (13.9) (a) M is closed; (b) if U E M ,then cl(U) E M ; (c) if a E M, then ( L B: /3 < a ) E M.

Lemma 13.2. (i) The function a H L,, is absolute for eoery adequate transitive set M . (ii) 1f M is an adequate transitive set, then M satisfies the axiom of constructibility ifand only if M = L, for some a. Proof. (i) We follow closely the proof of Lemma 12.5. Since we do not assume that M is a model of ZF, we have to prove not only that certain operations are absolute for M but also that these operations are defined in M. To start, for each i = 1, . . . , 10, the formula

z = Si(x, Y ) is restricted; since M is closed, every Giidel operation is absolute for M and defined in M (see Lemma 12.3b). Furthermore, the formula (13.10)

(13.11)

Z = { S i ( x , y ) : x ~ U , y ~ U , ,..., i = l lo}

is restricted and thus absolute for M. Also, if U E M ,then Z C = cl(U), then

z = { u E c : 3x, y E U[U = S,(x, y) v

E

M :If we denote

... v u = S,,(x)]}

13. CONSISTENCYOF THE AXIOM OF CHOICE AND THE GCH

111

and so, by the normal form theorem, we have 2 = g(V, C) where 5 is a Giidel operation. Next we show that (13.12)

x

is absolute for M.Given V

E

Cl(U)

M ,let W be the function on w defined as follows: W(O)= u,

(13.13)

W(n

E

+ 1) = W ( n )u {&(x, y ) : x E W(n),y E W(n),i = 1, ... , 10)

Since M is closed, it contains all natural numbers, and one can see (using induction) that W I n E M for all n. Thus the function W is absolute for M and defined in M (although W itself does not have to be in M), and since x EcI(U)

iff

3n x

E

W,

the formula (13.12) is absolute for M. Consequently, the operation def(V) is absolute for M ,and also it is defined in M since def(V) = {X E C : X E V} = g(C, V ) where C = cl(U u { V}) and 8 is a Godel operation. It remains to show that the function a++ La is absolute for M. Since the function is defined by induction, we follow Lemma 12.14. We have (13.14) y = L, iff a is an ordinal and 3fsuch that: f i s a function on a and (Vt < a ) [t 1 < a -f(< 1) = def(f(<))]

+

+

u

and (“5 < a ) [< is a limit -f(t) = ran(f It)] The operations def(X), X, ran(f), andf 1 are absolute for M and defined in M (because ran(f) and f 1 t are Godel operations, as one can easily verify). Thus, using (13.9c), it follows that the formula y = t,is absolute for M. We conclude the proof of (i) with two remarks. First, notice that if a E M then La E M ;this is because a + 1 E M and L. is in the transitive closure of ( L , : /3 < a + l ) , and M is transitive. Second, the formula x E La is also absolute for M : x E La iff 3y(x E Y A Y = L.)

u

<

(ii) Let M be an adequate transitive set. If M = L, for some a, then a is a limit ordinal since M is closed and Ord n La = a. Thus we have (13.15)

and since x

(Vx E

E

M ) (38 E M ) x

E

L, is absolute for M ,

(13.16)

i.e., M satisfies V = L

M C

Vx 38 x

E

L,

L,

2.

112

TRANSITIVE MODELS OF SET THEORY

On the other hand, if M satisfies V = I, then by absoluteness of x E L, we : /I E OrdM}= L, where a = Ord'. W get (13.15) and so M = (4

u

An almost immediate consequence of the proof of part (i) of the preceding lemma is the following: Lemma 13.3. There is a sentence 0 (prouable in ZF) such thatfor eoery transitioe set M, M is adequate if and only if M k 0.

Proof: Let 0 be the conjunction of the sentences

(13.17) (a) V X , Y 3 2 2 = &(X, Y ) , (b) VLI 3C C = cl(U); f l < a). (c) Va 3f f= ( 4 :

i = 1, ..., 10;

If M is an adequate transitive set, then the above operations are absolute for M and defined in M and so M k 0. On the other hand, if M is transitive and satisfies 0, then we proceed exactly as in the proof of Lemma 13.2(i)and show that the operations in (a), (b), (c) above are absolute for M and defined in M. It follows that M is adequate. H Exercise 13.3. The canonical well-ordering

of L is absolute for all adequate transitive

sets M. Exercise 13.4. If { M i :i E I} is a chain of adequate transitive sets (i.e., M, G M i ) , then M i is transitive and adequate.

Uial

M ic M jor

The following two exercises show that many L,, in particular all L, where is an uncountable cardinal, are adequate. Later we shall show (Lemma 30.1) that by carefully writing the definition of La, one can prove that the function at+& is absolute for every LAwhere 1 is a Limit ordinal. Similarly, Exercise 13.13 holds for every Limit ordinal a. K

For every f l 2 w, there is a 2 fl such that l a I = I fl 1 and La is adequate. [By the reflection principle, let S 2 fl be such that S satisfies Extensionality, V = L and 0 , and that IS 1 = I fl I. Let M be the transitive collapse of S.Then M is adequate and M = 1 . 1 Exercise 13.5.

Exercise 13.6. If K is a regular uncountable cardinal, then {a < K : La is adequate} is closed unbounded in K. Exercise 13.7.* Let M be a transitive model of ZF-,set theory without the power-set axiom. Then M is adequate. [The theorem on transfinite induction, Lemmas 12.4 and 12.5 work as well for the theory ZF-.]

Now we are ready to prove the crucial lemma for Theorem 34: Proof of Lemma 13.1. We assume that V = L and let X E w,. Since X is constructible, there exists /I such that X E L,.

13. CONSISTENCY OF THE AXIOM OF CHOICE AND THE GCH

113

By the reflection principle, there exists a set S such that: (13.18)

() (i) (ii) (iii) (iv)

S k Extensionality; S k 0; o,G S, X E S, fl E S;

IS I = K,; S k fl is an ordinal and X

E

4.

(We can find S satisfying (iii) since the axiom of choice holds.) Since S is extensional, there exists an isomorphism a of S onto a transitive set M = n[S] and we have

(i) M is adequate; (ii) a ( t ) = t for all 5 < o,,n ( X ) = X ; (iii) I M I = K,; (iv) M k a @ ) is an ordinal and n ( X ) E LNB,. Ordinals are absolute and so y = a @ ) is an ordinal and we have M ~ X E L , Since M is adequate, the formula X E L, is absolute for M and it follows that X E L,. However, y belongs to M and M is transitive and I M 1 = K,. Thus necessarily y < om+and we are done. (13.19)

Exercise 13.8. If a 2 w and X is a constructible subset of a, then X least cardinal in L greater than a. [Do the construction (13.18) in L, and choose S 2 a.] Exercise 13.9. If X is a constructible set, then X dinal in L greater than 1 TC(X) [In (13.18) choose S 2 TC(X).]

IL.

E

E

Lg,where /?is the

La where a is the least infinite car-

Exercise 13.10. The canonical well-ordering of L, restricted to the set RL = R n L of all constructible reals, has order type mk. [R n L c lo,^.] Exercise 13.11. If K is a regular uncountable cardinal in L, then L. is a model of ZF-. [Prove it in L. Replacement: (i) If X E L,, then 1x1 < K ; (ii) if Y E L, and I Y I < K, then Y E L..] Exercise 13.12. If ZFC + ( V = L).

K

is inaccessible in L, then L, = Vfr = V, n L and L, is a model of

Exercise 13.13. If L. is adequate and X fl I a.

< ( L a ,E),

then X is isomorphic to some Lg,

Exercise 13.14.* If L. is adequate, then the model ( L a ,E) has definable Skolem functions. Therefore, for every X c La, there exists a smallest M < ( L a , E) such that X EM. [The well-ordering <,, is definable in (L, E). Let h,(x) = the <,-least y such that

(L, E) c d X 9 Y1.1

2.

114

TRANSITIVE MODELS OF SET THEORY

Exercise 13.15.* Assume V = L If M i (L,, E), then M = L,, for some a. [Show that M is transitive. Let X E M. Let f be the <-least mapping of w onto X. Sincefis definable in (L,, E) from XJis in M.Hencef(n) E M for each n and we get

X c M.]

Exercise l3.16.* Assume V = L If M < (L2, E). then u1n M = a for some a 5 wI. [Same argument as in Exercise 13.15: If y < wIand y E M ,then 7 = M.]

14. THE ZmHIERARCHY OF CLASSES, RELATIONS, AND FUNCTIONS

We shall digress slightly from our investigation of transitive models and look more closely into the complexity of definition of the fundamental notions of set theory. Our starting point is the fact that restricted formulas are absolute for all transitive models, and the understanding that the more unrestricted quantifiers are needed to define a concept, the more complex is the concept. A formula cp is a Z,-formula if it has only restricted quantifiers 3x E y and Vx E y (i.e., if cp is restricted). A Z,-formula is a formula of the form 3xcp(x, . . .)

where cp(x, . . .) is a X,-formula. A n,-formula is a formula of the form V.ucp(.u, . . .)

where cp(x, ...) is a E,-formula. Similarly, &, Z,,Z4, .. ., formulas are of the form

n2,H , , n4,...,

3.u v y cp,

3x vy 32 cp,

3.u v y

v u cp,

. ..

v.u 3y cp.

v.u 3y v z cp,

vx 3y v z 3u cp.

. ..

32

where cp is a X,-formula. A class C (a property P,a relation R) is a &,-class (&,-property,&,-relation) if x E C (P(x),R(x, y)) can be written as a &,-formula. Similarly for n,. A function F is a &-function if the relation Y = F(x)

is a &,-relation. [Note that while Z,-ness of a formula is a purely syntactical notion, the C:,-ness of a class or a function depends on the axioms of ZF: Using the axioms, we may find an equivalent way of expressing x E C.] A property P is a A,-property if it is both X, and n, (similarly for relations and functions). That is, there is a &-formula cp and a n,-formula IIlsuch that

P(x1

+ +

cp(x 16 - b

It should be noted that if a concept is Z,, (or n,),then it is both C,, and n,, (just add a “dummy ” quantifier). Also, every definable concept is C, for some n

14. THE Z,, HIERARCHYOF CLASSES,

115

RELATIONS, A N D FUNCTIONS

(write the definition in the prenex normal form). Thus we talk about the Z,-hierarchy." We have already shown '(Lemma 12.1, Exercise 12.9) that the following notions are Zo : &(X,Y ) , i = 1, . . . , 10, x G y, x is empty, x is an ordered pair, x is transitive, x is an ordinal, x is a limit ordinal, x is a natural number, x = w, one-to-one, x is a relation, j' is a function, y =f(x), g =.f 1 X , Y =f[X], f IS ' X is an ordering.

"

Lemma 14.1. Every Godel operation is a Z,-function. f r o 0 5 We already know that Z = & ( X , Y) is Zo for i = 1, . . ., 10. We want t o show that Z = s ( X , , . .., X,) is X o whenever 5 is a composition of 5,... .,

510.

We show it by induction on the complexity of 5. In fact, we simultaneously show the following: (14.1) u E

5 ( X , ...) is Zo.

(14.2) If cp is Z o , then so are (Vu

E

B ( X , ...))q and (3u E S ( X , ...))q.

(14.3) Z = B ( X , . . .) is Zo . (14.4) If cp is Z o , then so is

cp(5(X, ...)).

We shall prove (14.1) and (14.2) only for a typical example and entrust the execution of the full proof to the reader. In (14.1) consider the formula 11

E

B ( X , ...) x @ ( X 9 ...)

This can be written as 3X E

5 ( x .. . .) 3y E @(x,. . .) 14 = (X, y)

In (14.2), consider the formula vu E

This can be written as

{ S ( X , ...Iq x ,..)I

cp(u)

Cp(s(x....))ACp(@( x,...))

(14.3) then follows from (14.1) and (14.2):

Z

=

B ( X . ...)

iff

(Vu

E

Z ) 11

E

s ( X , ...) A V U E s ( X , ...) u

E

Z

To prove (14.4). let cp be a &-formula. Then in cp(?j(X, . . .)), B ( X , . . .)occurs in the form 11

E

vu

S ( X , ...), E

B ( X , ...),

S ( X . ...) E u, 3u E S ( X , ...)

z= S ( X , ...),

Since s ( X , . . .) E u can be replaced by (3r E u ) c = B ( X , . . .), we use (14.1)-(14.3) to show that c p ( s ( X , ...)) is a Zo-property. H

2.

116

TRANSITIVE MODELS OF SET THEORY

The following lemma is of practical value-using relation or function is C,, ll,, or A,.

it, one can tell when a

Lemma 14.2. Let n 2 1.

I f P, Q are &-properties, then so are 3sP, P

A Q, P v Q, (3u E x ) P, (Vu E x ) P. (ii) If P, Q are ll,-properties, then so are VxP, P A Q, P v Q, (Vu E .Y) P, (3u E x ) P. (iii) IfP is C , , then 1 P is ll,; ifP is then 1 P is C,. (iv 1 I f P i s l l , a n d Q i s ~ , , t h e n P - , Q i s C , ; i f P i s ~ , a n d Q i s l l , , t h e n P + Q i s ll,. I f P and Q are A,, then so are l P , P A Q, P v Q, P -+ Q, P-Q, (Vu E x) P,(3u E x) P.

n,,

IfF is a Xn-function, then dom(F) is a &-class. lf F is a X,,+nction and dom(F) is A,, then F is A,, . IfF and G are Cn-functions,then so is F G. If F is a &,-function and if P i s a &property, then P(F(x)) i s Z,, . 0

Proog Let us prove the lemma for n = 1. The general case follows easily by

induction. (i) Let

P(x, ...) Q(x, ...)

where cp and

I )

-

-

c)

cp(z, x,

...)

3u $(u, x,

...)

32

are Z,-formulas. We have

(14.5) 3x P(x, ...)

c)

3x 32 cp(z, x, ...) 3U 3W E U 3 E W , 3 Z E

W[U

= (X,

2 ) A p ( Z , X,

...)]

The right-hand side of (14.5) is a Z,-formula. Furthermore, P(x, ...) AQ(x, ...) P(x, ...) vQ(x, ...) (3U E X) P(X,

. . .)

--

c)

32

3u[cp(z, x, ...) A$(u, x, ...)]

3z 3u[cp(z, x, ...) v $ ( u , x, 32 3 U [ U E X A q ( Z , X,

...)I

. . .)]

To show that (Vu E x) P is a C ,-property, we use the collection principle (9.4):

-

(Vu E x)P(x, .. .) * (vu E x) 32 q ( z , x,

3y(Vu E x) (3u E y ) q ( 2 , x,

(ii) follows from (i) and (iii). (iii) 1 3 2 q(z, x, ...) 1VZ

cp(z,x, ...)

*

v z lcp(z, x, ...) 32

.. .)

lcp(z,x, ...)

. . .)

14. THE C, HIERARCHY OF

CLASSES, RELATIONS, A N D FUNCTIONS

117

(iv)

(P-Q)

c*

(1PvQ)

(v) follows from (i)-(iv), (vi 1 x E dom(F)++ 3y y = F(x) (vii) Since F is a function, we have (14.6)

y = F(x)

If z = F(x) is X, and x (viii)

E

+-+

x

E

dom(F) A Vz[z

=

F(x) + y = z]

dom(F) is H, , then the right-hand side of (14.6) is ll,.

y = F(C(x))

(ix ) P(F(x))

c*

-

3z[z = G(x)

A

y = F(z)]

3 Y b = F(XbP(Y)l

Since C,-properties are absolute for all transitive models, it is clear that X,-properties are upward absolute: If P(x) is XI,if 911 is a transitive model of ZF and if 911 k P(.x), then P(x). Similarly, the HI-properties are downward absolute, and consequently, A,-properties are absolute for transitive models of ZF. In Lemma 12.6, we showed that well-roundedness is absolute for transitive models ZF. In fact, the proof shows that the property (12.18)

E is a well-founded relation over P

is both X I (12.20) and H I (12.19). Thus we have

Lemma 14.3. “E is a well-founded relation over P” is A l , is finite” is A1. [To get a nl-formulation, use T-finiteness from Section 2.1

Exercise 14.1. Show that ‘‘ X

The power-set operation P ( X ) is obviously I l l ; since it is not absolute as we shall have the opportunity to see in Chapter 3, it is not C, . Similarly, the cardinal concepts are Ill but not X l :

Lemma 14.4. “ a is a cardinal,” ‘ ‘ a is a regular cardinal,” and “ a is a limit cardinal ” are ll Pro05 (a) 13f[fis a function and dom(f) E a and ran(f) = a]. (b) u > 0 is a limit ordinal and 13f[fis a function and dom(f) E tl and U ran(f) = a]. (c) (Vg < a) (3y < a ) [B < y and y is a cardinal]. W

2.

118

TRANSITIVE MODELS OF SET THEORY

Exercise 14.2. “ X is countable” is Z1. Exercise 14.3.

1 X I 5 I Y 1, I X I = I Y I

are X I .

Lemma 14.5. Let n 2 1, let G be a A,,-function on V, and let F be defined by induction:

I

F(a) = G(F a)

Then F is a A,,-jiunctionon Ord. Pro$ Since Ord is a Co (and hence following expression is En: (14.7) y = F(a)

n,,)class, it is enough to verify that the

iff

3f[fisafunctionAdom(f)=aA(V< < a ) f ( < ) = G ( f ( < ) ~ y = G ( f ) ]

All the properties and operations in (14.7) are &, and G is A,,, and hence y = F(a) is C,,. Exercise 14.4. The functions a

+ fl and a . b are A l .

Exercise 14.5. The canonical and well-ordering of Ord x Ord is a &-relation. The func-

tion

r is A l .

Exercise 14.6. The function S H TC(S) is A l . Exercise 14.7. If G is a &-function (n 2 1) and F is defined by €-induction using G, then F is a &function. Exercise 14.8. The function x H rank(x) is A,.

Theorem 35. (a) Thefunction a- L, is A l . (b) “ x is constructible” is a C1-property. (c) The canonical well-ordering of L is a Cl-relation. ProoJ: (a) We intend to show that the relation

Y=L

(14.8)

is El. Since (14.8) is defined by induction, it suffices to show that the induction step is C,. Obviously, it suffices to verify that the function (14.9)

Y = cl(M)

is C,. We recall that by (12.15), the relation

Z = {&(x, y ) : x

E

U , y E U ,i = 1, ..., lo}

14. THE Z,,HIERARCHY OF CLASSES, RELATIONS,A N D FUNCTIONS

119

is Zo. By (12.13) we have (14.10) Y = cl(A4)

++

3W[W is a function

A

A

dom(W) = w

(Vn E dom( W ) ) (W ( n

A

W ( 0 )= M

+ 1) = W ( n )u { &(x, y ) : x E W(n),y E W(n), i = 1, . . . , 10)) A

Y = (J ran(w)]

and so cl(M) is a A,-operation. (b) x E L iff 3a 3y(x E y A Y = La). (c) Since x
iff

3a(x


y)

it suffices to prove that the function a-

<,,

which assigns to each a the canonical well-ordering of L, is A,. The function a- < a is defined by induction and thus it suffices to show that the induction step is Z In fact, < a + is defined by induction from < a (see (13.8), (13.7)). It suffices to verify that <.+ is obtained from a by means of a A,-operation (similar to the way in which L a + , is obtained from L, by L,+,= def(La)). The operation that yields <.+ when applied to <. is described in detail in (13.7). It can be written in a C ,-fashion in very much the same way as cl(A4) is in (14.10). The only potential difficulty might be the use of the words " the <-least," and that can be overcome as follows: For example, in (13.7)(ii)(c)

,.

,

,

-=

,

the <"-least u E W, such that x = &(u, u ) <" the <"-least r E W, such that y = &(u, t ) can be written as (3U E K ) [ X = a i ( U , U ) A (vf E

K) ( y = s i ( U , f ) + U <" l ) ]

Exercise 14.9. Let u be a theorem of ZF and let P(x)be a property that is upward absolute for transitive models of u. Then P(x)is X I . [P(x)- 3 M ( M is transitive, x E M , and M 1 u AP(x)).]

This gives another proof of the Z,-ness of ''x is constructible." A function F(X,, . . .,X,) is primitiue recursive if it is obtained by the following rules: (i) Every G d e l operation is a primitive recursive function. (ii) F(X,, . . ., X,) = w is a primitive recursive function.

2.

120

TRANSITIVE MODELS OF SET THEORY

(iii) The composition of primitive recursive functions is primitive recursive. (iv) Induction (Recursion Schema): If G is primitive recursive, then the function F defined by F(X, X I , ... , X,) = G(X, XI, .. ., X,, FIX)

is primitive recursive. Exercise 14.10. Every primitive recursive function is A 1 . Exercise 14.11. The following functions are primitive recursive: a + fl, a rank(x), a H La, a w <=.

. fl, TC(S),

Exercise 24.12.* The following function is primitive recursive and hence A l : (M, E ) ~ ( ( c p a, l , . . a,,) : cp . ?

E

Form, (al, . . . , a,,) E “ M and ( M ,E )

cp[a1, . . ., an]}

The LevyShoenfield Absoluteness Lemma It follows easily from the definition that Z,-properties are upward absolute: If P ( x ) is a X,-property and 91 a transitive model of ZF, and if P ( x )holds in Iu1, then P ( x ) holds in the universe. We shall now present a very powerful theorem which states that if a C,-sentence a is true in the universe, then it is true in L, and so in every transitive model containing all ordinals.

Theorem 36 (Levy-Shoenfield Absoluteness Lemma). Let a be a C,-sentence, u = 3xx(p(x),where q ( x ) is a C,-formula of one free variable. If there is an x such that q ( x ) , then there is a constructible x such that ~ ( x ) : 3x 4 x 1

+

( 3E

L)x(p(X)

and so a

t-,

Lka

In fact, if0 = mk, we have 3x d x )

+

( b a dcp(x)

Consequently, a is absolute for all transitive M such that M

2

l+ .

Remark. It is important that u is a C ,-sentence; an analogous theorem would be false for a Z,-formula $(x): For example, “ x is countab1e”can be written as a C,-formula, and it is by no means provable that a countable constructible set is countable in L We shall use methods of model theory to prove Theorem 36. The theorem follows from Lemma 14.6. We say that a model 2l= (A, E, R,, ..., Rk) is well-founded if E is a well-founded relation over A.

14.

THE

&, HIERARCHY

OF CLASSES, RELATIONS, A N D FUNCTIONS

121

Lemma 14.6. h t 0 he u sentence true in a welLfounded model VI = ( A , E , R , , ..., R , ) . Then 0 has a well-founded constructible model 6,and moreover L k 6 is countable. Proof of’ Theorem 36. Let q ( x ) be a C,-formula and assume that 3x cp(x)

By the reflection principle, there exists a transitive set A such that A k 3x q ( x )

Let 0 be the conjunction of the axiom of extensionality and 3x cp(x). Since A k 0,there exists, by Lemma 14.6, a well-founded constructible model (B, E) of 0.and B is countable in L. Applying the collapsing mapping, we obtain, in L, a countable transitive model M of b cp(x). If we denote 0 = 04, then clearly M E &, and since cp is &, we have

L, k

3x cp(x)

Now the model theoretical methods. A sentence (of any language) is a V3sentence if it has the form (14.11)

VX,

*

*

a

3 ~ ... 1 yn

X,

...*yn)

$(XI* * * * , xm, ~ 1 ,

where $ is quantifier-free.

,

Lemma 14.7. Let ‘&,E PI E . . . G 91, G . . . be a chain of models, and let 0 be a V3-sentence (14.11). Assume that f o r each k, if x,, ..., x, E A, then there are y,, . . . , yn

E

,

A, + such that

(14.12)

Then O is true in PI

v(k+l

=

xrn,

...?

yl,

. . * ?

yn]

%k.

Proof: If x,, .. . , x, E up=,-, A,, then there exists k such that x,, . . ., x, E A,. Then we find y,, . . . ,y n E A,+ such that (14.12) holds and since $ is quantifierfree, we get PL k 0. H

,

Lemma 14.8 (Skolem’s Normal Form Theorem). Let cp be a sentence of a language 9. Then there is a Vhentence @ with additional predicates R , , . . . ,R k such that:

(a) If9I k cp, then there are relations R , , . . . , Rk such that (21, R 1, . . . ,R , ) k @. (b) If(v1, R , , ..., R , ) k @,then 91 k cp. Proof. We give the proof of a special case. (In the general case, first write cp in the prenex normal form, i.e., quantifiers first, and then proceed as in the following example.) Let cp be the sentence vu 3u vx 3y

I&,

v, x, y )

I22

2.

We replace 3y $ by R(u, u. x ) and V x 3y

TRANSITIVE MODELS OF SET THEORY

+ by S(u, u). i.e., we replace cp by

(14.13) Vu 3u S(u, u ) A V U U [ S ( Uu)+-+V?r , R(u, u, x ) ] A

Vuux[R(u, u, x ) - 3y $4~4,u, x, y ) ]

To show that the sentence (14.13) can be written in a V3-way, it suffices to express it as a conjunction of V3-sentences. The second part of (14.13) is equivalent to Vuux[S(u, u ) --* R(u, u, x ) ] A Vuu 3x[R(u, u, x ) + S(u, u)]

The third part of (14.13) is equivalent to Vuux 3y[R(u, u, x ) + $(u, u, x, y ) ]A Vuuxy[$(u,

I),

x, y )

+

R(u, 0, x ) ]

Proofofkmma 14.6. Let PI = ( A , E , R 1, .. . ,R k ) and let 0 be a sentence true in PI. In view of Lemma 14.8, we may assume that 0 is a V3-sentence, vx,

... x , 3y, ... y ,

..., x , ,

+(XI.

y1,

..., y.)

($ is quantifier-free), We also assume that 0 does not have a finite model (otherwise we are done). We shall construct % as the union of a constructible o-chain of finite models satisfying (14.12). Clearly, such '13 is constructibly countable, and by Lemma 14.7, '13 is a model of @. To be sure that % is well-founded, we also construct a witness to it, a mapping of B into ordinals. We say that a function$ B + Ord demonstrates that B = (€3, E", ...) is well-founded if u ESnu

implies

f(u) c f ( u )

for all u, u E B. Let p be the function on A defined by

d.1

= SUP{P(Y)

+ 1 : YEXI

and let a be the length of E (cf. (2.10), (2.11)). Let P be the set of all pairs (23,f) such that: (14.14)

(i) $'3 is a finite model of the language of 91 and B c w ; (ii) S: B -,a demonstrates that 23 is well-founded.

Let us define a partial ordering of P as follows: (14.15)

(8,f) < (23',f')

if:

(i) 23' is a submodel of 23; (ii) f ' ~ f ; (iii) for all xt, . , ., x, E B', there exist yl, . . . , y,, E B such that

23 I= +[XI, -.*)

x m , Y,,

YJ

14.

THE

1,HIERARCHY OF CLASSES, RELATIONS,A N D

123

FUNCTIONS

It is obvious from the definition (14.14), (14.15) that the partially ordered set ( P , < ) is constructible. We shall show that ( P , < ) is not well-founded. Let us first see how that gives a proof of the lemma. Since being well-founded is absolute, L satisfies that ( P , < ) is not well-founded, and so there exists a constructible descending chain

u:=,

> ... > ( q d > )..*

( S 0 , f O )> ( S I J - 1 )

U:=,fn,

If we let 23 = 23, a n d f = then it is easy to see that 23 1 0 and that .f demonstrates that 23 is well-founded. Thus it suffices t o show that ( P , < )is not well founded. This can be done by exhibiting a nonempty X G P without a <-minimal element. Let X be the set of all (23,,f) E P such that: (a) 23 is isomorphic to a finite submodel 91' of PI; (b) there is an isomorphism h of 23 and YI' such that for all 6 (14.16)

E

B,

.f'(h)= P(h@))

It is clear that given an isomorphism h of 23 and some PI' c PIJis determined by (14.16). Therefore in order to show that X has no <-minimal element, it suffices to show that given a finite YI, c PI, there exists a finite P I , such that PI, c QI, c PI and for allx,, ..., x, E A , , thereexist y,, ..., y, E A , such that PI, 1

$[XI,

. . .. x,,

y1,

.. * , y,]

is quantifier-free and PI satisfies 0.

However, this is certainly true since

The &,-Hierarchy of Definable Sets in Models In analogy with the &,-hierarchy of formulas and classes, we can classify the elements of the set Form of all formulas of the language 9 = {E},and the sets definable in models of the language Y . Thus for each n E o we denote by C, (n,) the set of all cp E Form that have the appropriate form. Given a model 91 = ( A , E ) of 9, we say that a set X c A is En(n,)over PI if there is cp E C, (cp E n,) such that

x = {x

E

A : PI =! cp[x]}

In general, a relation R over A is &, (n,)over PI if R

= {(x,,

.. ., x,)

E

" A : PI 1 c ~ [ x ,..., , x,]}

for some cp E &, (cp E l-I,,). A relation is A, over PI if it is both C, and ll, over PI. A function F in PI is C, (n,, A,) over PI if the relation y = F ( x ) is. In general, the collection of C, relations over a model PI does not have the closure properties proved in Lemma 14.2, unless the model 21 satisfies some

2. TRANSITIVE MODELS OF SET THEORY

124

axioms of set theory. It can be easily verified that if M is a transitive set closed under the operation { }, then C, and Il, relations over (M, E) have all the closure properties from Lemma 14.2, except closure under (Vu E X ) and (3u E x). Note that in the proof of Lemma 14.2 we used the collection principle in this case. For instance, to show that (Vu E x) P(x,. . .) is C if P is C we used the following equivalence :

,,

,

(VU E x) 32 q ( 2 , x,

. ..)-

3y(Vu E x) (30 E y ) q ( 2 , x,

. . .)

Similarly, a A,-relation restricted to a transitive model (M, E) is A1 over M, provided M satisfies the theorem of ZF that is used to show that the relation in question is Al. Thus we have: Exercise 14.13. If M is a transitive model of Theorem 5, then “ E is a well-founded relation over P” is A, over M. Exercise 14.14. If M is a transitive set closed under the Godel operations and the function TC(S), then TC(S) is a A,-function over M .

If M is a transitive set closed under Godel operations and the function rank(.x). then rank(x) is a A,-function over M.

Exercise 14.15.

And most notably: Exercise 14.16. If M is a transitive model of ZF-, then the functions a- L, and a++ <, are A I over M ; and ‘*x is constructible” and “.x < y” are Z,over M .

The result of Exercise 14.16 can be considerably improved. It has been shown that there is a sentence 0 whose transitive models are exactly all the sets L,, a an ordinal number. Consequently, the function a- L, is Co. We recall that assuming the countable axiom of choice, the set HC is a model of ZF-. We include two more exercises: the first one is a variation of Exercise 14.9, and the second one is a consequence of the Levy-Shoenfield absoluteness lemma: Exercise 14.17. Assume AC. Let D be a theorem of ZFC such that D implies Extensionality, and let P(.x)be a property upward absolute for transitive models of 0.Then P(x) is Z, over HC. [P(x)iff 3 countable M such that x E M and M 1 Q A P(x)iff 3 countable transitive M 3 .x such that M 1 Q A P(x).] Exercise

14.18. If

Q

E

X I is a sentence true in (HC,

E),

then

Q

is true in (HC’,,

E).

Truth Definition for C,-Sentences

Closely related to Giidel’s second incompleteness theorem is a theorem of Tarski, stating that the notion of truth in a theory containing arithmetic is not expressible in the theory. In particular, there is n o set-theoretical property T(x)

14. THE En HIERARCHY OF CLASSES, RELATIONS, A N D FUNCTIONS

125

such that if o is a sentence then T( r d ) holds if and only if o holds. However, for every (metamathematical) natural number n 2 1, truth for X,,-sentences is definable. We shall sketch a proof of Tarski's theorem and present a truth definition for &,-sentences. As is customary, we arithmetize the syntax and consider some fixed effective enumeration of all expressions by natural numbers (Godel numbering). In particular, if u is a sentence, then #o is the GMel number of o,a natural number. We say that T(x) is a truth definition if (14.17)

(i) Vx(T(x)-,x E a); (ii) if o is a sentence, then

u *T( # o) We shall show that a truth definition does not exist. Let us assume that there is a formula T(x) satisfying (14.17). Let

(Po,

(P11

(P27

be an enumeration of all formulas with one free variable. Let $(x) be the formula There is a natural number k such that $ is (Pk. Let u be the sentence $(k). Then we have which contradicts (14.17). For every (metamathematical) natural number n, we define a binary predicate I,as ,follows: (14.18) ko(P[?L.1,...r?(k]iff

xo,

...,.yk) is a k-sequence and 3 M [ M is transitive, (.Y1, ..., S k ) E M and ( M , E) k Cp[x1, ... , Xk]] k 1 3ucp[u, xl. . . . , xk] iff cp E c 0 ,<xl, . . . , xk) is a k-sequence and 3u such that I=,-,q [ u , xl, . . . ,xk] k 2 3u Vucp[u, u, xl, ..., xk] iff cp E Eo, (xl, . . . , xk) is a k-sequence and 3u V u ko q[u, u, xl, . . . , xk] cp E

etc.

(x1,

It is obvious that if q(xl, . . . , x,) is a C,,-formula, then

c~(x,,...,xm)obn(~[x1* ....xm]

Thus if we define T,, as follows:

x E w and 3u E &[a is a sentence, x = #o and k,, o] T,,(x) iff then T,, is a truth definition for &,-sentences.

I26

2.

TRANSITIVE MODELS OF SET THEORY

The predicates C,, are defined for all n, but in view of Tarski's theorem, it is impossible to define b,, uniformly for all n E o,i.e., as a property of n. Exercise 14.19. The relation Co is

XI;for each n 2 I ,

1" is

Xn

If 91 is a transitive class and n a natural number, then we can similarly define the predicate 91 k,, cp[x,, ..., xk] for cp E Z,,. We leave the details to the reader. We can further use C, to define the notion of a Z,,-elementary submodel ofthe universe. For every n, we define

M <,, V

iff

. .., xk) E 'M q[xl, * . . *X k l W k n q[xi,

V q E EnV(x,,

((M, E,

*.-)

xk])

Exercise 1420. M i V holds for every transitive set M . Exercise 14.21. Let n be a natural number. For every M o there exists M 2 M o such that M 4. V .

[Reflection principle.]

One can similarly define M <,, 91 for a transitive class 91. K is an uncountable cardinal then L. i , L [Proceed as in the proof of GCH in L]

Exercise 1422. If

There is of course no problem in defining X any n E o. Exercise 1 4 2 3 . I f L, is a model of ZF- and if X

<,,(M, E) for any set M

<, ( L , , E ) ,

and

then X is isomorphic to

some L,, j3 I a.

Exercise 1424.* Let 0 = mi. If X < I ( b , E ) , then X = L, for some a. [A stronger version of Exercise 13.15.1

15. RELATIVE CONSTRUCTIBILITY AND ORDINAL DEFINABILITY

A Few Remarks on Transitive Models

We have dealt quite extensively with the notion of absoluteness and shown that many fundamental concepts are absolute for transitive models of ZF. The cardinal concepts are in general not absolute. If 9 l is a transitive model of ZF, then we have: (15.1)

P"(X) = P ( X ) n 9Jl,

(15.2)

IfW C

= V, n 9Jl.

1x1 = IY1,then 1x1 = I Y I .

(15.3) If a is a cardinal, then a is a cardinal in '3l; if a is a limit cardinal, then a is a limit cardinal in W.

15.

RELATIVE CONSTRUCTIBILITYA N D ORDINAL DEFINABILITY

127

(15.4)

Ia I

(15.5)

If a is a regular cardinal, then a is a regular cardinal in 91;if a is weakly inaccessible, then a is weakly inaccessible in 91.

5 I a lm, cf(a) Icfm(a).

(15.6) If 91 is a model of ZFC and K is inaccessible, then K is inaccessible in 91.

Concerning the last statement, if a < K, then since 91 k AC, we must have either (2")yR< I\' or (2")lm 2 K and the latter is impossible since 2" < K. If 91 is a transitive model of ZFC, then the axiom of choice in 91 enables us to code all sets in 91 by sets of ordinals and the model is determined by its sets of ordinals. The precise statement of this fact is: If 91 and % are two transitive models of ZFC with the same sets of ordinals, then W = %. In fact, a slightly stronger assertion is true. (On the other hand, one cannot prove that W = % if neither model satisfies AC.)

Lemma 15.1. Lat 91 and 91 be transitive models of ZF and assume that the axiom of choice hoMs in 91.rf91and 91 have the same sets of ordinals, then 91 = a. Proo/: We start with a rather trivial remark: 91 and % have the same sets of pairs of ordinals. To see this, use the absolute canonical one-to-one function r: Ord x Ord + Ord. If X c Ord2 and X E 91, then T[X] is both in 91 and in 91, and we have X = r- I[r[X]] E 91. First we prove that W E 9.Let X E W.Since W satisfies AC, there is a one-to-one mappingfE 91 of some ordinal 9 onto TC({X}). Let E E 91 be the following relation over 9: EB iff f ( 4E f ( B ) E is a set of pairs of ordinals and thus we have E E %. In W,E is well-founded and extensional. However, these properties are absolute and so E is wellfounded and extensional in %. Applying the collapsing theorem (in %), we get a transitive set T E 91 such that (T, E) is isomorphic to (0, E). Hence T is isomorphic to TC({X})and since both are transitive, we have T = TC({X}). It follows that TC({X})E % and so X E 9. Now we prove 91 = % by €-induction. Let X E % and assume that X c 91; we prove that X E W.Let Y E W be such that X E Y (for instance let Y = where a = rank X; the rank function is absolute). LetfE W be a one-to-one and so f[X] E %. function of Y into the ordinals. Since W c %, f is in However, f[X] is a set of ordinals and so f[X] E W, and we have

x =f- l[f[x]]E 91. Relative Constructibility The notion of constructible sets will be generalized to define sets constructible from a given set A . If A is an arbitrary set and U is a transitive set, let def,(U) be the set (15.7)

def,(U)= cl(U u { U } u { A n U})n P ( U )

I28

2.

TRANSITIVE MODELS OF SET THEORY

The set def,(U) is transitive, and we have

U E def,(U) E P ( U ) Note also that

(15.8)

def,(U) = def,

,(U)

Similarly to the case ofdef(U) (Exercises 12.3 and 12.5), the set def,(U) consists of subsets of U definable over U with A considered a unary predicate on U : Exercise 15.1.* IT U is a transitive set and A E U, then def,(U) is the set of all subsets of U definable over the model U = ( U , E n U 2 ,A ) .

The class of all sets constructiblefrom A is defined as follows: (15.9)

u 4 1 = fa La[A]=

u

LB[A]

if a is a limit ordinal

D
,[A] = def,(La[Al)

44= UOld L[AI aE

The following theorem is the appropriate generalization of Theorems 32-34 on constructible sets:

Theorem 37. kt A be an arbitrary set. (i) L[A] is a model of ZFC. (ii) q A ] satisfies the axiom 3X(V = UX]). (iii) If 91 is a transitive model of ZF containing all ordinals and such that A n 91 E 911, then L[A] E 91. (iv) There exists a. such that for all a 2 a. , L[A] 1 2K1= Hat 1

The proof that L[A] is a model of ZF is exactly the same as for L: the class L[A] is transitive, closed, and almost universal. The proof of the rest of Theorem 37 follows closely the corresponding proofs for I!, but some additional arguments are needed.

Lemma 15.2. k t A = A

n L[A].Then

L [ a = Y A ] and moreover A E L [ q .

ProoJ: We show by induction on a that Lda = La[A].The induction step is obvious if ct is a limit ordinal; thus assume that &[A7 = L,[A] and let us prove

L+,[A7 = Lt , [ A ] .

If we denote U = La[A].then we have A n t r = A n U n L[A]=

A

nU

15.

129

RELATIVE CONSTRUCTIBILITYA N D ORDINAL DEFINABILITY

and by (15.8)

L+, [ A ] = def,(U) = def,

n

"(U) = def,W) = L+

,[A7

Thus L [ a = L[A]. Moreover, there is a such that A n q A ] = A n L J A ] and thus A E I,,+ , [ A ] . As was the case with constructibility, the key factor in the proof is the absoluteness of relative constructibility:

Lemma 15.3. If 91 is a transitive model of Z F and A E 91, then the function L [ A ] is absolute for 91.

a-

ProoJ: As in Lemma 12.5.

In particular, L[A] satisfies V = L [ q and we have (ii). Following Theorem 33, we observe that V = Y X ] implies the axiom of choice; in fact there is a well-ordering of the universe definable from X . Hence L [ A ] is a model of ZFC. Part (iii) of Theorem 37 follows from the following lemma: Lemma 15.4. If 91 is a transitive model of Z F containing all ordinals and if A n 91 E 911, then L [ A ] = L[A n 911.

Proqf By induction on a,

L,[A] = L,[A n 9111 The proof of the induction step is exactly as in Lemma 15.2,once we know that (15.10)

A n L,[A] = A n 91 n & [ A ]

(15.10) is true because by absoluteness, L,[A] = e [ A n 911 is a subset of 911.

For part (iv) we need a stronger version of Lemma 15.3. Call a transitive set M A-adequate if: (15.11)

(a) M is closed, and if X E M, then A n X (b) if LI E M ,then cl(U) E M; (c) if a E M ,then (Lp[A]: p < a ) E M.

E

M;

Lemma 15.5. The.function a H L,[A] is ahsolute.for every A-adequate transitive set M. Proqf As in Lemma 13.2.

We can similarly improve Lemma 13.3:

Lemma 15.6. There is a sentence O* (provable in ZF) such that for every transitive set M , M t= O* if and only if M is A d e q u a t e for all A E M . H Part (iv) of Theorem 37 now follows from

2.

130

TRANSITIVE MODELS OF SET THEORY

Lemma 15.7. l f I/ = L[A] and A c L,,[A], then every X i’ < Q,+1. Proof: Let (15.12)

p be such that X

() (i) (ii) (iii) (iv)

E

E (0, is

in some L,[A],

L p [ A ] .Let S be such that:

S k Extensionality; S k O*; L J A ] G S, X E S , p E S; IS I = 8,; S 1 /I is an ordinal and X E &[A].

(Compare with (13.18).) As in Lemma 13.1, we collapse S onto a transitive set M = n[S],and use absoluteness to show that X E L,[A] where y = .(/I). This completes the proof of Theorem 37. A consequence of Theorem 37(iv) is that if I/ = L [ A ] and A E o,then the generalized continuum hypothesis holds. For a slightly better result, see the following exercises. Exercise 15.2. Prove (in ZFC) the following variant of the reflection principle: For a given formula cp, there exists a countable M that reflects cp and, moreover, a E M implies a c M for every a < wl. Exercise 15.3. If V = 4 A ] where A E w I Tthen 2N0= K1. (Consequently, the GCH holds.) [Modify the proof of Lemma 15.7 using Exercise 15.2; show that if X c m, then X E I [ A n { ] for some a < wl, 5 < wI.It follows that IP(w)l = K,.]

Although in practical applications of relative constructibility one often uses Y X ] where X is not necessarily a set of ordinals, for theoretical purposes it is sufficient to consider only L [ X ] where X c Ord: Exercise 15.4. For every X there is a set of ordinals A such that L [ X ] = L [ A ] . [Let X = X n L.[X], and let (0, E) be isomorphic to TC{X}(in 4x1).Let A = T [ E ] where r is the canonical mapping of Ord’ onto Ord. Then A E L [ X ] and X E Y A ] , and hence 4 A ] = L [ X ] . ] Exercise 15.5. Let a 2 w be a countable ordinal. There exists A E w such that a is countable in L [ A ] . [Let W c w x w be a well-ordering of w of order type a; let A E w be such that Y A I = 4W.1 Exercise 15.6. If w I(in V ) is not a limit cardinal in L, then there exists A E o such that w1 = m y . [There exists a < w I such that in L, w l is the successor of a. Let A be such that a is countable in LIA].] Exercise 15.7. (In ZFC) There exists A E w 1 such that w1 = w?”’. [For each a < w l , choose A . E w such that a is countable in q A , ] . Let A E w I x w I be such that A, = : (a, E A } for all a ; then wfl”] = wl.]

{r r )

15.

131

RELATIVE CONSTRUCTIBILITYA N D ORDINAL DEFINARILITY

Exercise 15.8. (In ZFC) I f w 2 is not inaccessible in L. then there exists A (i,;lA1 = (oI and (o?I”l = ( u 2 .

c w 1 such that

Some generalizations of L[A]:I f A is a class, let us define L[A] as in (15.9), where def,( U ) is defined as in (15.7). Exercise 15.9. L[A] = L[A], where A = A n L[A], and L[A] is a model of ZFC. Moreover, L[A] is the least model 911 2 Ord such that VT n A E ‘111 for all a. Exercise 15.10.* Assume that there exists a choice function A (even a class A E Ord) such that V = L[A].

F on V . Then there is a class

[Combine the various coding tricks used in the preceding exercises.]

If T is a transitive set, let us define L ( T )as follows

h ( T )= 0

(15.13)

Lu(T)= Lu+

U L~(T)

if

o!

is a limit

P<@

u m =u =

defA(La(T))

AET

u E Ord

Exercise 15.11. L( T) is a model

L(T)

of ZF. Moreover, L( T) is the least model 211 2 Ord such

that T c 211.

L ( T )is not necessarily a model of ZFC: If T = P(w),then P (w )c L(T)and is not necessarily well-ordered. Exercise 15.12. If T is a transitive class, then (15.13)defines a model L(T) of ZF, the least model ’111 2 Ord such that T c 211. Exercise 15.13. Let 211 be a transitive model of ZF, M 2 Ord, and let X be a subset of 211. Then there is a least model M[X] of ZF such that 211 E 211[X] and X E M[X]. If 211 k AC, then M[X] k AC. [Let 211[X] = L(211 u {X}).]

Relativization of the UvyShoenfield Absoluteness Lemma Theorem 36’Let q ( x , Z ) be a &,-formula. If2 E w, then (15.14)

% d x l2)

-,

(hE Wl) d x , 2 )

ProoJ: (Sketch) Rather than giving the proof in full, we shall describe how to generalize each step of the proof of Theorem 36 so as to obtain the relativized version of the theorem. To prove the theorem from Lemma 14.6, we considered a transitive model A of h q ( x ) and obtained a constructible transitive model M of h cp(x). To generalize, we add to the language constant symbols for each n E w, and for w, and a unary predicate symbol 2. Instead of one sentence 0 we consider a set of

132

2.

TRANSITIVE MODELS OF SET THEORY

sentences S; in addition to the sentence 3x cp(x, Z), S contains sentences describing the natural numbers, the set w and the set Z (i.e., sentences like Z ( n ) (or lZ(n)), 0 is the empty set, 1 = {0},2 = {0,l}, . . . , w is the set of all natural numbers). Notice that US] = U Z ] . Thus in the generalized Lemma 14.60 is replaced by a set of sentences S, and the language of the model VI is assumed to contain infinitely many constant symbols; the lemma asserts that S has a well-founded model 8 E L[S]. By the Skolem normal form theorem we may assume that all sentences in S are V3-sentences. (In fact, the sentences Z ( n ) (or l Z ( n ) ) and the sentences describing the natural numbers are already V3, so that we add only finitely many predicate symbols R . . . , R , to the language.) It remains to construct a well-founded model 8 of S such that B E L[S]. This is done as in (14.14)and (14.15). We consider finite models whose universe is a subset of a given countable set and functions that demonstrate that the models are well-founded. We let P be the set of triples (8,Jk), where k is a natural number, and define ( S , J k ) < (b',f',k') as in (14.15)with the additional condition (0)

k'

-= k

and with the requirement that (iii) holds for the first k sentences in S. With these modifications, the proof proceeds as in Lemma 14.6 and we obtain a well-founded model B of S such that S E L[S]. Upon collapsing 8 onto a transitive model M, the predicate Z becomes the set Z and M satisfies 3x cp(x, 2).Since cp is C o , we have U Z ] k 3x cp(x, 2). Exercise 15.14. If u ( Z ) is X I and Z E w , then u ( Z ) is absolute for all transitive M 2 &[Z] where 0 = w : [ ~ ~ . Exercise 15.15. ( HCL[". E).

If o ( Z )E X I , Z

G w,

and u ( Z ) holds in (HC. E). then u ( Z ) holds in

[Relativized Exercise 14.19.1

A set of reals A is E l over HC if A = {x E R : HC k cr(x)). for some El. If 91 is a transitive model of ZF, we denote A" the set {x E R": HC" k ~ ( x ) } .

G(X)E

Exercise 15.16. Let 91 be a transitive model of ZF containing all countable ordinals. If A is a set of reals X I over HC. then A" = A n 91t. [If x E RmA,then 91 a where 0 = wf[xl, and HC k u ( x ) iff HC" k u(x).]

&[XI,

Ordinal-Definable Sets A set X is ordinal-definable if there is a formula cp such that

(15.15)

X

= { u : cp(u, a l ,

for some ordinal numbers a l , . . . , a,, .

. . . , a,,))

15.

RELATIVE CONSTRUCTIBILITYA N D ORDINAL DEFINABILITY

133

It is not immediately clear that the property “ordinal definable” is expressible in the language of set theory. Thus we give a different definition of ordinal definable sets and show that it is equivalent to (15.15). We recall that cl(M) denotes the closure of a set M under Giidel operations. The class OD of all ordinal-definable sets is defined as follows: (15.16)

OD =

u

4E

cl{V’:

/3 < a }

Ord

In other words, OD is the Giidel closure of { V, : a E Ord}, that is, ordinal definable sets are obtained from the V, by application of Giidel operations. We shall show that the elements of the class OD are exactly the sets satisfying ( 15.15).

Lemma 15.8. There exists a definable well ordering of the class OD (and a definable mapping F of Ord onto OD). Proof: In (13.7) we described how to construct from a given well-ordering of set M , a well-ordering of the set cl(M). For every a, the set { V, : /3< a } has an obvious well-ordering, and we call the canonical well-ordering of cl{V, : /3 < a } the well-ordering obtained from it. Thus we get a well-ordering of the class OD, and denote F the corresponding (definable) one-to-one mapping of Ord onto OD. Now it follows that every X E OD has the form (15.15). There exists a such that X is the ath set in the canonical well-ordering of OD, or, in other words, X = { u : q(u, a ) } where q ( u , a ) is the formula u E F(a). We shall show that on the other hand, if cp is a formula and X is the set in (15.15), then X E OD. By the reflection principle, let /3 be such that X E Vs, xl, . . . , a,, < /3 and that V, reflects q. Then we have

X

= {u E Vs : V,

1 q ( u , a ] , . .., a,,)}

Since U k cp is a Zo-formula,we may apply the normal form theorem and find a G d e l operation 3 such that

X

=

3(V,, a l , ...,a,,)

(see also Exercise 12.2). Using the normal form theorem once again, we find a Giidel operation (5 such that for every a.

a = { x E V , : x is an ordinal} = 6(V,)

6 such that x = $(V4,, . . ., K”

and so there is a Godel operation

9

and therefore X belongs to OD.

Vfl)

134

2.

TRANSITIVE MODELS OF SET THEORY

Exercise 15.17. Assume that not every ordinal number is definable and show that the

property X is definable" is not expressible in the language of set theory. [Consider the least a that is not definable.] "

Exercise 15.18. If X E OD, then there exists y such that X is a definable subset of

(v,,E) (without parameters). Hence OD is the class of all X definable in some K, . [If X

=

(u E V,: V, k cp(w.

a)}, consider y =

r(a, p).]

Exercise 15.19. If F is a definable function on Ord, then ran F E OD. Thus: OD is the largest class for which there exists a definable one-to-one correspondence with the class of all ordinals.

For any class A, we have defined in (9.6)the class of all ?I hereditarily in A: %A = {X

E

A : x E %A} = {X E A : TC(x) E A }

Thus let HOD denote the class of hereditarily ordinal-deJnable sets HOD = {X : TC({x})c OD} The class HOD is transitive and contains all ordinals. Theorem 38. The class HOD is a transitive model of ZFC.

This theorem provides an alternative proof of consistency of the axiom of choice. Proof. The class HOD is transitive, and it is easy to see that it is closed under Gijdel operations. Thus to show that HOD is a model of ZF, it suffices to show that HOD is almost universal. For that, it is enough to verify that V, n HOD E HOD, for all a. For any a, the set V, n HOD is a subset of HOD, and so it is sufficient to prove that V, n HOD is ordinal-definable. This is indeed true because V, n HOD is the set of all u satisfying the formula u E V, A (Vz E TC({u}))3 s [ z E cl{V, : y

< b}]

and thus V , n HOD E OD. It remains to prove that HOD satisfies the axiom of choice. We shall show that for each a there exists a one-to-one function g E HOD of V, n HOD into the ordinals. Since every such function is a subset of HOD, it suffices to find g E OD. By Lemma 15.8, there is a definable one-to-one mapping G of the class OD into the ordinals. If we let g be the restriction of G to the ordinal-definableset V, n HOD, then g is clearly ordinal-definable. H The model HOD contains all constructible sets; HOD 1 L It is consistent that HOD # L Also, unlike constructibility,ordinal definability is not an absolute property, and it is consistent that HODHoD# HOD. I shall return to this in Section 25.

15. RELATIVE CONSTRUCTIBILITY A N D ORDINAL DEFINABILITY

135

Exercise 15.20. HOD is the largest transitive model of ZF for which there exists a definable one-to-one correspondence with the class of all ordinals. [Use Exercise 1519.1

We conclude this section with several generalizations of ordinal definability. A set X is ordinal-defnablefrorn A, X E OD[A], if there is a formula cp such that

X

(15.17)

= {u : cp(u, a l , . .., a,,, A)}

for some ordinal numbers a l , . . . , a,, . As above, this notion is expressible in the language of set theory: OD[A] = cl({V, : a E Ord} u { A } )

(15.18)

The class OD[A] has a well-ordering definable from A and thus every set in OD[A] is of the form (15.17). Conversely (using the reflection principle), every set X in (15.17) belongs to OD[A]. The proof of Theorem 38 generalizes easily to the case of HOD[A]. Thus HOD[A], the class of all set hereditarily-ordinal-definable from A, is a transitive model of ZFC. As a further generalization, we call X ordinal-dejnable over A, X E OD(A), if it belongs to the G d e l closure of { V m :a E Ord} u { A } u A. If X E OD(A), then X E cl({V, : a E Ord} u { A } u E), where E = {xo, . . , ,xk}, a finite subset of A. Hence there is a finite sequences = (x,, , . . . ,X J in A such that X is ordinaldefinable from A and s. On the other hand, if s is a finite sequence in A, then obviously s E OD(A) and thus we have (15.19) OD(A) = {X : X E OD[A, s] for some finite sequence s in A }

In other words, X E OD(A) iff there is a formula cp such that (15.20)

X

.

= {u : cp(u, a l , . . . , a,, , A, ( X O , . . , xk))}

for some ordinal numbers a l , . . ., a,, and a finite sequence (x,,, . .., xk) in A. The class HOD(A) of all sets hereditarily ordinal-definable over A is a transitive model of ZF. To show that HOD(A) is almost universal, it suffices to verify that V, n HOD@) E OD(A). In fact, V, n HOD(A) is ordinal-definable from A : It is the set {u E V, : (Vz E TC({u}))[z E cl({Vm : a E Ord} u { A } u A ] } Exercise 15.21. Let C be a definable class. Let OD(C), the class of all sets ordinal definable over C, be the G d e l closure of {V. : a E Ord} u C. Show that HOD(C) is a model of ZF.

136

2.

TRANSITIVE MODELS OF SET THEORY

Let 91 be a transitive model of ZF, 9.1 2 Ord. A set X is 9.l-deJinahle, X E WD,if there is a formula q such that

x = {u : q ( u , x)} for some x E W. Exercise 15.22. Show that “ X is 9Ldefinable” is expressible in the language of set theory, and prove that HUD is a model of ZF. Exercise 15.23. LD = OD.

PART I1

More Sets

CHAPTER 3

FORCING AND GENERIC MODELS

16. GENERIC MODELS

One of the consequences of Giidel’s proof of consistency of the axiom of constructibility is that one cannot prove in ZFC that there is more than one transitive model of set theory, namely the constructible universe L And L satisfies the generalized continuum hypothesis. Thus if we wish t o show, e.g., that the continuum hypothesis is unprovable in ZFC we cannot just look for a transitive model 9JI that would satisfy 2N0> K,.What is needed is a different way of constructing models of set theory. There are several possible approaches to this problem: (a) to construct a model 911 = (M, E)where E is not well-founded (a nonstandard model); (b) to make an assumption that there exists a countable transitive model 9JI of ZFC and extend 91 by adjoining a set G 4 9JI to obtain a model W[G] of ZFC such that Ordm = Ordmmrcl; (c) to introduce “imaginary” sets and adjoin an imaginary set G to the universe to obtain an extension V [ G ] that is a model of ZFC; (d) to generalize the notions of model and satisfaction and construct a “generalized ” model of ZFC. There is an inherent disadvantage in approach (a): Unlike in the case of transitive models, one cannot use absoluteness of fundamental notions of set theory to establish truth or falsity of various statements in the model. In case (b) the main problem is to choose the set G properly, so that W[G] is a model of ZFC with the same ordinals as 91. For instance, one has to avoid the case when G c w x w is a well-ordering of order-type Ordgn. It turns out that a combination of the ideas in (c) and (d) is most fruitful. The key to the problem is the method of forcing, invented by Cohen. This method not only establishes independence of the continuum hypothesis and of the axiom of choice, but proves to be a powerful method of construction of models of set theory, providing a number of consistency results. 137

138

3.

FORCING A N D GENERIC MODELS

The idea of extending a mathematical structure by adjoining new “ideal” objects is very frequent. Typical is the construction of the ring of polynominals over a given field. Of course, to extend the universe of set theory to a model of set theory is more difficult, first because there are no sets outside the universe, and secondly because the extension has to satisfy the axioms of ZFC. Cohen’s method of extending the universe of set theory is general enough to apply to any model of set theory. We are however interested primarily in transitive models. Thus let 91 be an arbitrary but fixed transitive model of ZFC (possibly V , possibly a countable model). We shall call 91 the ground model. We shall describe how to adjoin to a generic set G to obtain a generic extension 91[G]. We shall do it in such a way that: (i) 91[G] is a model of ZFC with the same ordinals as 91; and (ii) 1UI[G]can be described inside 911.

Before defining the notion of generic set, let us consider a typical example. Suppose that we want to enlarge the ground model 91 in such a way that the ordinal w? becomes countable (or, as we say, is collapsed). We want to describe a model 3J1[G] that contains a mappingfof w onto 07. Letfbe such a-hypothetical-mapping. For every n, the initial segment f In = (f(O), . . . , f ( n - 1)) lies in 911. Thus let P be the set of all finite sequences ( a o , . . . , a,of countable ordinals in 91. We shall call the elements of P,forcing conditions. We may look at forcing conditions as giving us information aboutf, namely telling us that ( a o , ..., a n - , ) is an initial segment of .f We say that ,forces that f(0) = a o , . . . ,f ( n - 1) = a,- Apparently, if p = ( a o , . . . , a,p c q. then q gives us more information than p; we say that q is stronger than p. The relation “ q is stronger than p ” is a partial ordering of P. The convention is to use 4

u

to have the following properties: (16.1)

(i) if p , q E G , then p and q are compatihle, i.e., p ( k ) = q ( k ) for all k E dom(p) n dom(q); (ii) for every n there exists p E G such that n E dom(p); (iii) for every a < w? there exists p E G such that a E ran(p).

We shall return to this example after giving the general definition of generic sets to verify that an 91-generic set G has the properties (ik(iii) in (16.1).

16.

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GENERIC MODELS

Forcing Conditions and Generic Sets In the ground model 91, let us consider a nonempty partially ordered set (P, <). We call ( P , < ) a notion offorcing and the elements of Pforcing conditions. We say that p is stronger than 4 if p < q. If p and q are conditions and there exists r such that both r I p and r I q, then p and q are compatible; otherwisethey are incompatible. A set W E P is incompatible if its elements are pairwise incompatible. A set D C P is dense in P if for every p E P there is q E D such that q I p. A set F G P is aJilter on P if: (16.2) (i) F is nonempty; (ii) if p I q and p E F, then q E F; (iii) if p , q E F, then there exists r E F such that r I p and Y I q. A set of conditions G c P is generic (over 91)if: (16.3)

(i) G is a filter on P ; (ii) if D is dense in P and D E 91, then G n D #

0.

Note how (16.3)(ii) depends on the ground model 911: What matters is which dense subsets of P are in 91.Thus if 9 is any collection of sets, let us say that a set G E P is a 9-generic filter on P if it is a filter and if G n D # 521 for every dense subset of P that is in P. Then G is generic over 91 just in case it is P-generic where 9 is the collection of all D E 911 dense in P. Exercise 16.1. Show that in the definition ofgenericset one can replace (16.2)(iii) by the following weaker property: If p, q E G. then p and q are compatible. [To prove (16.2)(iii). show that D = {r E P : either r is incompatible with p o r r is incompatible with q. or r 5 p and r 5 4)is dense.]

A set D E P is dense below p if for every p' I p there is q q
E

D such that

Exercise 16.2. A filter G on P is generic over 91 if and only if for every p E G, if D E 91 is dense below p then G n D # 0. Exercise 16.3. A filter G on P is generic over 91 if and only if G n D # D E 91 is open and dense in P.

0 whenever

A set C E P is a cover of P if every p E P is compatible with some q E C. A set W E P is a partition of P if it is a cover of P and if any two distinct p , 4 E W are incompatible. Exercise 16.4. A filter G on P is generic over 91 if and only if G n C C E 9t is a cover of P.

+ 0 whenever

Exercise 16.5. A filter G on P is generic over 91 if and only if G n W # 0 whenever W E 9t is a partition of P.

We shall now return to our first example: we consider the partially ordered set ( P , 3 ) of finite sequences of ordinals less than m?. Let G E P be generic

3.

140

over 91. Since G is a filter, it is clear that f = to see that for every n E (0,the set

FORCING A N D GENERIC MODELS

u

G is a function. It is rather easy

D, = { p E P : n E dom(p)) is dense in P. Similarly, for every a < coy, the set

E,

= {p E

P : a E ran(p)}

is dense in P. Therefore, G n D, # 0 for every n and G n E , # 0for every a, and so dom(f)= (0 and ran(f )= 07. Thus we have proved that any transitive model 91 =I 91 that contains G satisfies that coy is a countable ordinal. If the ground model 91 is countable, then for every notion of forcing ( P , < ) in 911 there exists a generic set. This follows from the next lemma:

Lemma 161. If ( P , < ) is a partially ordered set and 9 is a countable collection of dense subsets of P. then there exists a %genericfilter on P. In fact, for every p E P there exists a 9-generic filter G on P such that p E G. Proof. Let D,,D, , . .. be the sets in 9.Let po = p, and for each n, let p, be such that p, 2 pn - and p , E D, . The set G = {q E P : q 2 p, for some n is a %generic filter on P and p

E

E

N)

G.

We shall now state the first of two main theorems on generic models.

'Iheorem 39 (Generic Model Theorem). Let 91 be a transitive model of ZFC and let (P, < ) be a notion offorcing in 91. If G c P is generic over P, then there exists a transitive model N [ G ] such that: (a) (b) (c) (d)

YJl[G] is a model of ZFC; YJl E 91[G] and G E 9l[G]; OrdW''' = Ordm; if '31is a transitive model of ZF such chat 91 c 91 and G E %, then %31[G] c '31.

The model 931[G] is called a generic extension of 91. The proof of Theorem 39 (as well as of the other theorem of this section) will be given in Section 18. We shall now give another example of a generic set; in fact, this is the simplest possible nontrivial example of a generic set. Let P be the set of all finite 0-1 sequences ( p ( O ) , .. . ,p(n - l)), ordered by inverse inclusion: ( P , 3).If G is a generic set of conditions (over the ground thenf= , G is a 0-1 function on o;as in the first example, f is a model !Ill) function because G is a filter on P, and dom(f) = o because for every n, the set

u

D = { p E P : n E dom(p)} is dense in P and therefore D n G # $3.

16.

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GENERlC MODELS

The function J': w + (0, 1) is the characteristic function of a unique set A G W . We shall show that A is not in the ground model '9R. Equivalently, let us show that f $ 91. Let g : w + (0, 1) be a function in '9R, and let us consider the set

+

D, = {P E p : P 9 ) It is easy to verify that D,is dense in P and since D, E 91,we have G n D, # 0 and hence f # g . The above example describes the simplest way of adjoining a new set of natural numbers to the ground model. A set A c w obtained in this way is usually called a Cohen generic set (or a Cohen generic real). Exercise 16.6. Let (P,< ) be a notion of forcing in 91 with the following property: For every p E P there exist q and r such that q Ip . r < p and such that q and r are incompatible. Show that if G C _ P is generic over 91, then G 4 911. [If F is a filter on P,then { p E P : p $ F } is dense in P.]

The last example (adding a generic real) suggests how to attack the problem of independence of the continuum hypothesis: One may attempt to adjoin to the ground model 91 a large number of generic reals. Let P be (in 91)the set of all functions p such that (16.4)

(i) dom(p) is a finite subset of w2 x o, (ii) ran(p) E (0, l),

and let p be stronger than q just in case p 3 q. If G is a generic set of conditions, then if we let f= same argument as before to show that (16.5)

u

G, we can use the

(i) f i s a function; (ii) dom(f) = w? x w.

Now, for each a < w?, let f,:w + {O, 1) be the function defined as follows: (That is, we may look a t l a s coding a collection of 0-1 functions on w :

(f,:a < w'4")) Again, it is easy to see that eachf, is outside 91;for any g E 91, the set {p E P : p(a, n) # g ( n ) for some n} is dense in P and thusf, # g . Moreover, if a # fl, thenf, #fp;this is because the set

D = {p E P : p(a, n) # p(B, n) for some n} is dense in P and hence G n D # 0. Thus in W[G] we have a one-to-one mapping a-f,

of o? into "(0, 1). However, in order to conclude that the

142

3. FORCING AND GENERIC MODELS

continuum hypothesis is violated in W[G], we have to know that co? is not collapsed in W[G]. To prove that, one has to use more subtle arguments. This leads us to the other important feature of Cohen’s method, namely that the generic model W[G] can be described within the ground model. Associated with the notion of forcing (P, < ) is a forcing language. This forcing language as well as the forcing relation IF are defined in the ground model W.The forcing language contains a name for every element of W,and a constant G (the canonical name for a generic set). In fact, the way the forcing language will be set up will provide for a name for every potential element of the generic extension W[G]. Once we select a generic set G, then every constant of the forcing language is interpreted as an element of the model W[G]. The forcing relation is a relation between the forcing conditions and sentences of the forcing language: p

Il- 0

( pforces a). The forcing relation, which is defined in W,is in a sense a generalization of the notion of satisfaction (in the generic model). For instance, if p IF 0 and if 0‘ is a logical consequence of 0, then p IF 6’. In due course, we shall prove various rules for the forcing relation. For instance,

and q Ip ,

then q

(16.6)

if p

IF

cp

(16.7)

P

IF

PA*

iff

p

P

IF

lcp

iff

no q 5 p forces q

1 1 cp IF

cp

and p

IF

i,b

or (16.8)

In view of the forthcoming theorem, these rules will enable us to decide (within 91) whether a given set theoretical statement holds in W[G]. Let (P, < ) be a fixed notion of forcing in the ground model 91.

Theorem 40 (Forcing Theorem). If for every G E P generic over W, ( 16.9)

91[G] k a

0

is a sentence oftheforcing language, then

ifand only

if

(3p E G ) p IF

0

[In the left-hand-side 0 one interprets the constants of the forcing language according to G.] Theorem 40 will also be proved in Section 18. Prior to proving Theorems 39 and 40 we shall develop the theory of forcing and prove the basic rules for the forcing relation. In the following exercises, we give a definition of the forcing relation under an assumption that is true if 91 is countable, thus giving a definition of IF for countable models. However, this definition is in terms of generic sets, and there

16.

143

GENERICMODELS

is no straightforward way of showing that IF is definable inside 111. O n the other hand, using the forcing theorem one can readily derive the basic rules for lk. In the following exercises, we assume that for each p E P there is G E P generic over 111 containing p. Let us define

(16.10)

p

Ik 0

iff

111[G] k

for all G such that p E G

(J

Exercise 16.7.

(a) if p It u and q 5 p , then q Ik u; (b) no p forces both u and l u ; (c) for every p there is q I p such that either q

IF CT

or q

10.

[Use (16.9) to prove (c).] Exercise 16.8.

(a) p (b) p

lb c p ~ $iff p 1 1 rp and p 1 1 $; 11 Vxcp iff p 11- q ( a )for every constant

a of the forcing language.

[For (b), use: 'iUUI[G] t Vxcp iff %I[G] C rp(a) for every constant of the forcing language.] Exercise 16.9. p Ib c iff no q 5 p forces lu. [Use Exercise 16.7c.I Exercise 16.10.

(a) p (b) p (c) p

11cp + $

iff (Vq I p ) q IF rp implies q ~k $; v $ iff (Vq I p ) (3r I q ) r Ik cp or r 11 $; IF 3xcp iff (Vq Ip ) ( 3 r I q ) 3a r 1 1 rp(a). IF cp

[Express

-+, v ,

and 3 in terms of

A,

1, and V.]

To conclude this section, I shall briefly touch the question of existence of generic sets. Of course, if '111 is countable, then every P E 331 has an 331-generic filter. On the other hand, in general one can at least prove something like: Given P E '111, then it is consistent that P has an 331-generic filter. (A theorem of this sort will be proved in Section 18.) Even when formulated correctly, one has to be careful to exclude the obvious false cases (like when 331= the universal class). In fact, it seems necessary to formulate the theorem only for 331 having an absolute definition (like '111 = the constructible universe). However, the problem of consistency of existence of generic sets in general is not really a problem since in practice, when dealing with the consistency of a particular set theoretical statement, the correct existence theorem can be easily formulated. Moreover, the use of a generic extension can be converted directly into a consistency proof, avoiding the question of existence of a generic set altogether. I shall elaborate on this in Section 18.

3.

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FORCING A N D GENERIC MODELS

17. COMPLETE BOOLEAN ALGEBRAS

Boolean Algebras

A field of sets (an algebra of sets) is a collection 9 of subsets of a given nonempty set S with the following properties:

(17.1)

(i) S E 9; (ii) if X, Y E 9, then X u Y E 9, X nY

Obviously, P ( S ) is a field of sets, and also, trivial examples:

9, and X - Y E 9. S} is a field of sets. Some less E

{a,

(a) Let S be an infinite set and let B be the collection of all X such that either X is finite or S - X is finite. (b) Let S be a topological space and let 9 be the collection of all clopen (closed and open) sets in S. (c) The collection of all Bore1 sets of reals. (d) The collection of all Lebesgue measurable sets of reals. The algebraic properties of fields of sets are abstracted in the notion of Boolean algebra. A Boolean algebra is a set B with at least two elements, 0 and 1 (zero and unit), endowed with binary operations + and and a unary operation - . The Boolean operations satisfy the following axioms: (17.2)

u .u =u .u

u+u=u+u u

+ (u + w )= ( u + u ) + w

(commutativity)

u . ( u . w ) = (u . u ) . w

(associativity) u . (u

+w)=

u

.u +u . w

u

+ (u . w ) =

(u

+ 0 ) . (u + w ) (distributivity)

u+u=u u

u*u=u

. (u + u ) = u

+ (u . u ) = u

x+o=x

x*o=o

x + l = l

x.l=x

x+(-x)= -(-x)= -(x

u

1

x.(-x)=O

x

+ Y ) = (4 ( - Y )

-(x

*

*

Y ) = (-4 + ( - Y ) (deMorgan laws)

Two elements u, u of B are disjoint if u u = 0. Let us write u-u

17.

145

COMPLETE BOOLEAN ALGEBRAS

instead of u . ( - u), and let us define

(1 7.3)

U ~ L '

iff

u - ~ = 0

It is easy to see that I is a partial ordering of B and that

(17.4)

iff

usc

u+r=u

iff

u'u=u

Moreover. 1 is the greatest element of B and 0 is the least element. Also, for any E B, u + L) is the least upper bound of u, u, and u . u is the greatest lower bound of {u, 0).Since - u is the unique u such that u + u = 1 and u . u = 0, it follows that all Boolean-algebraic operations can be defined in terms of the partial ordering of B. For any field of sets 9, the set 9 with Boolean operations X u Y , X n Y , and S - X is a Boolean algebra (the unit is S and the zero is 0).The Boolean algebraic I is the same as inclusion. u, u

Exercise 17.1. Let B be a Boolean algebra and define 11

0 D = ( u - 1 ' ) + (1. - u )

Then B with operations 0 and . is a ring (with zero 0 and unit 1).

We shall now give an example showing the relation of Boolean algebras and logic. This relation will later be utilized in Boolean-valued models. Let 9be a first order language and let S be the set of all sentences of 9'.We consider the equivalence relation kcp $ over S . The set B of all equivalence classes [cp] is a Boolean algebra under the following operations:

-

[cpl + [+I = ['p v $1 [cpl . [$I = [cp . $1 - [cpl = [ 7cpl

1 = [cp v 1471

0 = [cp A

7cpl

This algebra is called the Lndenbaum algebra. Another important example of a Boolean algebra is this: Let S be a topological space. An open set U in S is called regular open if U = interior (closure(U ) ) .Let B be the collection of all regular open sets in the space S , and let (17.5)

U . V = U n V

U

+ I / = int cl(U u -U = s - cl(U)

V)

Although tedious to verify, it is nevertheless true that B with the operations defined above is a Boolean algebra. Note that the complement of U in B is not the set-theoretical complement. However, I is again the same as inclusion.

146

3.

FORCING A N D GENERIC MODELS

Exercise 17.2. I f U is an open set, let U' = S - cl(U). Show that:

(i) U1l1 = ul; (ii) U is regular open iff U" = U ; (iii) U l l is the unique regular open set V such that cl(U) = cl(V); it is maximal among all open V such that cl( V) = cl(U).

A subset A of a Boolean algebra B is a subalgebra if it contains 0 and 1 and is closed under the Boolean operations: (17.6)

(i) O E A , 1 E A ; (ii) if u, 11 E A , then u

+ u E A , u . u E A, - u

E

A.

If X E B, then there is a least subalgebra A of B that contains X ; A can be described either as (7 { A : X c A E Band A is a subalgebra}, or as the set of all Boolean combinations in B of elements of X. The subalgebra A is said to be generated by X. As for other abstract algebras, if X is infinite, then 1 A 1 = I X 1. However, if X is finite, we can do better in case of Boolean algebras: Exercise 17.3. A finitely generated Boolean algebra is finite. In fact, if A has k generators, then A I 22A.

l l

[Using distributivity and the deMorgan laws, show that the following finite sums constitute a subalgebra of B: us,

+ + US" "'

where u s , s E '{O, I are pairwise disjoint elements of the form us =

*XI

.

*.xz

. ".

'

fxl,

where xl, . . . , x k are the generators and +.xi is either xi or - x i according to whether s(i) = 0 or s(i) = 1.1

Let B and C be two Boolean algebras. A mapping h: B phism if it preserves the operations: (17.7)

C is a homomor-

(i) h(O)= 0, h ( l ) = 1; (ii) h(u + u ) = h(u) + h(u), h(u . u ) = h(u) . h(u), h ( - u ) = -h(u).

Note that the range of a homomorphism is a subalgebra of C and that h(u) I h ( u ) whenever u I u. A one-to-one homomorphism of B onto C is called an isomorphism. An embedding of B in C is an isomorphism of B onto a subalgebra of C. Exercise 17.4. I f h: B + C is an order-isomorphism of B and C (i.e., a one-to-one mapping such that u I iff hu 5 htl), then h is an isomorphism. 11

An isomorphism of a Boolean algebra onto itself is called an automorphism. A nonzero element a E B is called an atom if it is a minimal element of B - (0);

17.

147

COMPLETE BOOLEAN ALGEBRAS

equivalently, if there is no x such that 0 < .x < a. A Boolean algebra is atomic if for every nonzero u E B there is an atom a Iu ; B is atomless if it has no atoms. If A = {al, . . . , an} are the atoms of B. then B is isomorphic to the field of sets P ( A ) . Hence B has 2" elements.

Exercise li.5. Every finite Boolean algebra B is atomic.

Exercise 17.6. Any two countable atomless Boolean algebras are isomorphic.

[Use the method of Cantor's Theorem 11.1

Ideals and Filters The definition of filter (and ideal) given in Section 7 generalizes to arbitrary Boolean algebras. Let B be a Boolean algebra. An ideal on B is a subset 1 of B such that: (17.8)

(i) 0 E I, 1 # I ; (ii) if u E I and u E I, then u u E I; (iii) if u, u E B, u E I and u Iu, then u

+

E

1.

Afilter on B is a subset F of B such that (i) 1 E F , 0 # F ; (ii) if u E F and u E F, then u . u E F ; (iii) if u, u E B,u E F, and u 5 u, then u E F. In (16.2),we defined the notion of filter on a partially ordered set P. Note that if B is a Boolean algebra, then if we consider the partially ordered set B - (0)for all nonzero elements of B then the present definition of filter on B coincides with the definition (16.2). The trivial ideal is the ideal (0); an ideal is principal if I = {u E B : u I uo} for some uo # 1. Similarly for filters. Note that two nonzero elements u, u of B are disjoint just in case they are incompatible in the partially ordered set B - {O}. Thus we may call u, u compatible if u * # 0. A subset G of B - {0}has the finite intersection property if for every finite { u l , . . . , u,) E G , u , . ... . u, # 0 Every G c B that has the finite intersection property generates a filter on B; this and the other two clauses of Lemma 7.1 hold also for Boolean algebras. There are two important examples of ideals on the Boolean algebra of all Borel sets of reals:

(17.9)

1. The Borel sets of Lebesgue measure zero form an ideal. 2. A set X of reals is nowhere dense if the closure of X does not contain a nonempty open subset: int(c1 X )= 0.A union of countably many nowhere dense sets is called meager. Exercise 17.7. The meager Borel sets form an ideal on the algebra

[The Baire category theorem states that R is not meager.]

of all Borel sets.

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FORCING A N D GENERIC MODELS

There is a relation between ideals and homomorphisms. If h : B + C is a homomorphism, then

I

(1 7.10)

= {U E

B : h(u) = 0)

is an ideal on B (the kernel of the homomorphism). On the other hand, let I be an ideal on B. Let us consider the following equivalence relation over B : (17.11)

u-v

iff

uAvrz1

where

u A u = (U - U ) + (U

- U)

Let C be the set of all equivalence classes, C = B / - , and endow C with the following operations: (17.12)

[u]

+ [u] = [ U + u],

[u] . [u] = [U . U ]

-[u] = [-u] 0 = [O],

1 = [l]

(Of course, the conscientious reader will verify that (17.11) is an equivalence relation and that (17.12) does not depend on the choice of representatives.) Then C is a Boolean algebra, the quotient of B mod I, and is a homomorphic image of B under the homomorphism h(u) = [.I The quotient algebra is denoted B/I. (17.13)

Exercise 17.8. Regarding a Boolean algebra as a ring (with operations @ and ., see

Exercise 17.1). show that the definition of ideal coincides with the ring theoretical definition of ideal.

An ideal I on B is a prime ideal if (17.14)

for every u E B,

either u E I or - u E I

The dual of a prime ideal is an ultrafilter. Lemma 7.2 holds in general: An ideal is a prime ideal (and a filter is an ultrafilter) iff it is maximal. Also, an ideal I on B is prime iff the quotient of B modulo I is the trivial algebra (0, 1). The ultrafilter theorem easily generalizes to Boolean algebras: Theorem 21’ (Prime Ideal Theorem). Every ideal on B can be extended t o a prime ideal.

The proof of the prime ideal theorem (PIT) uses the axiom of choice. It is known that PIT cannot be proved without using the axiom of choice (see Example V in Section 21). However, it is also known that PIT is weaker than

17.

149

COMPLETE BOOLEAN ALGEBRAS

full AC: It does not imply the axiom of choice. Still, in many proofs in algebra and topology, one can do with PIT in place of full AC (examples: compactification theorems, Hahn-Banach theorem). Exercise 17.9. Show that PIT follows from its weaker version: Every Boolean algebra has a prime ideal. [Given an ideal I on B, consider a prime ideal K on B / / and use K to construct a prime ideal J 2 / on B.]

The prime ideal theorem is of particular importance in logic. We recall the basic two theorems of the first order predicate calculus:

Completeness Theorem. Every consistent set of sentences has a model. Compactness Theorem. A set Z of sentences has a model ifeachjnite subset of E has a model. The usual proofs of the completeness theorem use the axiom of choice; namely to show that a consistent set ofsentences can be extended to a complete consistent set of sentences. Since this can be proved using an ultrafilter on the Lindenbaum algebra, it follows that the completeness theorem can be proved using PIT. The compactness theorem is an easy corollary of the completeness theorem. On the other hand, one can give a proof of PIT using the compactness theorem, thus showing that they are in fact equivalent: Exercise 17.10. Using the compactness theorem, prove that every Boolean algebra has a prime ideal. [Given B, consider the language containing +. ., -, names for all u E B, and a unary predicate I . Let Z be the set of all quantifier-free sentences true in B, plus the sentences u,~~ stating that /isaprimeideal:/(O). l / ( l ) , / ( u ) v / ( - u ) , / ( u ) ~ / ( u ) ~ / ( u + i ~ ) .B.]

The prime ideal theorem also gives the following important representation theorem for Boolean algebras.

Theorem 41 (Stone’s Representation Theorem). Eaery Boolean algebra is isomorphic to a j e l d of’ sets. Proof: Let B be a Boolean algebra. We let S = {p : p is an ultrafilter on B} (1 7.15)

For every II E B. let Xu be the set of all p E S such that u E p. Let .JF = { X u :u E B ) (17.16) Let us consider the mapping A : U H X u . Clearly, n( 1) = S and n(0)= fa. It follows from the definition of ultrafilter that

n(u . u ) = A ( U ) n

~ ( 2 9 ) .

A(U

+ t i ) = n(u) u n(v),

A(-.)

=S

- A(.)

Thus n is a homomorphism of B onto 9 (and 9is a field of sets). I t remains to show that A is one-to-one.

3.

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FORCING AND GENERIC MODELS

If u # v, then using PIT, one can find an ultrafilter p on B containing one of these two elements but not the other. Thus R is an isomorphism. Stone's representation theorem is usually considered in a topological form : Call the Stone space of B the space S of (17.15) with the topology given by the base P of (17.16). The Stone space is a compact Hausdorff space with a base of clopen (closed and open) sets. [To prove compactness: Every subset of B with the finite intersection property can be extended to an ultrafilter.]

Exercise 17.11.*

B is isomorphic to the algebra of all clopen sets of its Stone space. [Use compactness of S to show that every clopen set is in .9.]

Exercise 17.12.*

The following two exercises show that if S is the Stone space of an infinite Boolean algebra B then IS 2 I B 1. Let A be a subalgebra of a Boolean algebra B and let u E B - A . Then there exist ultrafilters F, G on B such that u E F, u 6 G, and F n A = G n A.

Exercise 17.13.

Let B be an infinite Boolean algebra, I B1 = K. There are at least K ultrafilters on B. [Assume otherwise. For each pair (F, G) E S x S pick u E F - G, and let these u's generate a subalgebra A . Since 1 A 1 5 1 S 1 < K, let u E B - A. Use Exercise 17.13 to get a contradiction.]

Exercise 17.14.

Complete Boolean Algebras The partial ordering I of a Boolean algebra can be used to define infinitary operations on B, generalizing + and .. Let us recall that u

+ 1' = sup{u, 14.

r

=

inf{u,

13;

IT)

in the partial ordering of B. Thus for any nonempty X E B, we define

c

(17.17)

{ u : U E X) = sup

x

n{u:ueX\=infX

n

provided that the least upper bound (the greatest lower bound) exists. We also = 0 and 0 = 1. define If the infinitary sum and product is defined for all X c B, the Boolean algebra is called compfere.Similarly, we call B K-complete (where K is a regular uncountable cardinal) if sums and products exist for all X of cardinality < K. As usual, an HI-complete Boolean algebra is called o-complete or countably complete.

2

17.

151

COMPLETE BOOLEAN ALGEBRAS

An ideal I on a k--complete Boolean algebra is h--complete if

1 { u : u E X} E I whenever X c I and I X I < K. A ri-complete filter is the dual notion. A field of sets 9 is h--completeif it is closed under unions and intersections A h--completefield of sets is a h--complete of collections of less that K sets in 9. Boolean algebra, and X = X for every X E 9, I X I < K. A trivial example of a complete Boolean algebra is the complete field of sets P ( S ) . For a nontrivial example, let S be a topological space and let B be the collection of all regular open sets of the space S. B is a Boolean algebra, with the operations defined in (17.5). With a little additional effort the reader can verify that B is complete and that

u

1

C(U:U~~}=intcl(Jx n { U : ~ ~ ~ } = i n t c l n x Exercise 17.15. Let B be a complete Boolean algebra.

(a) Verify the distributive laws:

. C {u : u E X } = 1{a . u : u E X } a + { u : u E X } = fl { a + u : u E X } a

n

(b) Verify the deMorgan laws:

-2 { u : u E X } = fl { - u :

-n

{ u :u E X } =

1{ - u :

uE X} uE

X}

Exercise 17.16. If I is a K-complete ideal on a K-complete Boolean algebra B, then B/I is x-complete, and

1 { [ u ]: u E X } = [{I u :u E X } ] I I

for every X G B, X

< K. Similarly for products.

Exercise 17.17. For B to be complete it is sufficient that all the sums [fl X = { u : u Ix for all x E X}.]

1

X exist.

Let B be a complete Boolean algebra. A subalgebra A of B is a complete subalgebra if X E A and X E A for all X E A. (Caution: A subalgebra A of B that is itself complete is not necessarily a complete subalgebra of B.) Similarly,a complete homomorphism is a homomorphism h of B into C such that for all X G B,

n

c

n

h (C X) = h[X17 4rI X) = h[XI A complete embedding is an embedding that satisfies (17.18). Note that every isomorphism is complete. (17.18)

152

3.

FORCING AND GENERIC MODELS

Since the intersection of any collection of complete subalgebras of B is a complete subalgebra, every X c B is included in a least complete subalgebra of B. This algebra is called the complete subalgebra of B (completely) generated by X.

Completion of Boolean Algebras Let A be a subalgebra of a Boolean algebra B. A is said to be dense in B if for every nonzero u E B there is a nonzero u E A such that u I u. (Note that this is just in case that A - (0) is dense in the partially ordered set B - {O}. We therefore call a set D c B - (0) dense in B if D is dense in B - (03.) A completion of a Boolean algebra B is a complete Boolean algebra C such that B is a dense subalgebra of C.

Lemma 17.1. The completion of a Boolean algebra B is unique up to isomorphism. ProoJ: Let C and D be completions of B. We define an isomorphism A : C + D as (1 7.19)

A(C) =

1”{U E B : u IC}

To verify that A is an isomorphism, one uses the fact that B is a dense subalgebra of both C and D. For example, to show that AC # 0 whenever c # 0: There is u E B such that 0 < u I c, and we have 0 < u 5 AC. Theorem 42 (Completion of Boolean algebras). Every Boolean algebra has a completion. We shall prove a somewhat more general theorem (Lemma 17.2), using the method of Dedekind cuts. For a topological proof of Theorem 42, see the following exercise: Exercise 17.18.* Let S be the Stone space of a Boolean algebra B. The algebra of regular

open sets of S is (isomorphic to) the completion of B.

If P is a dense subset of a Boolean algebra B (i.e., dense in B - {0}),then it has the following property : (17.20) If p $ q, then there is r I p that is incompatible with 4.

We shall call a partially ordered set P satisfying (17.20) separative. Lemma 17.2. Let (P, < ) be a separative partially ordered set. Then there is a complete Boolean algebra B such that: (i) P c B - {0}and < agrees with the partial ordering of B ; (ii) P is dense in B. The algebra B is unique up to isomorphism.

153

17. COMPLETE BOOLEAN ALGEBRAS

ProoJ: We use a construction similar to the method of Dedekind cuts

(Theorem 12). Let us call a set U (17.21)

p

~

EP

and q qEU

a cut if implies

p~ U

For every p E P, let U pdenote the cut {x : x I p}. A cut Lr is regular if (17.22) whenever p

4 U , then

3q 5 p such that Lr, n U = 0.

Note that (17.20) defines ( P , < ) to be separative just in case every U pis regular. We let B be the set of all regular cuts in P. We claim that B, under the partial ordering by inclusion, is a complete Boolean algebra. Note that the intersection of any collection of regular cuts is a regular cut, and hence each cut U is included in a least regular cut U . In fact, O={p:(Vqsp)Un U , # 0 )

Thus for u, u E B we have U * V = U A V ,

u+v=uvv

The complement of u E B is the regular cut - u = { p : U pn u =

0)

And, of course, 0 and P are the zero and the unit of B. It is not difficult to verify that B is a complete Boolean algebra, and we leave the verification to the reader. Since (P, < ) is separative, every U pis a regular cut and so the function e : PH U p maps P into B - (0).The function e is one-to-one and orderpreserving. Moreover, the set of all U p is dense in B and thus we have an embedding of P onto a dense subset of B. Uniqueness of the complete Boolean algebra B is proved essentially by the same argument as in Lemma 17.1: If B and C are two complete Boolean algebras, both having P as a dense subset, then a: B + C defined is an isomorphism: a(b)=

( p E P : p I b}

The above proof of Lemma 17.2 can be regarded as a topological proof in disguise. When one defines a topology on P by taking the U pas basic open sets then regular cuts are in fact regular open sets, and B is the algebra of regular open sets. Thus we shall use the following notation:

B = r.o.(P) When (P, < ) is an arbitrary partially ordered set, we can still define the above topology on P and associate with P the algebra of regular open sets. However,

3.

154

FORCING AND GENERIC MODELS

if P is not separative, we do not get an embedding of P into B; the exact relation is described in the corollary below.

Lemma 17.3. Let ( P , < ) he a partially ordered set. There exists a separative partially ordered set (Q, <)and a mapping h of P onto Q such that (i) x I y implies h(x)4 h(y); (ii) x and 1' are compatible in P i f h ( x )and h(y) are compatible in Q . Proqf Let us define the following equivalence relation over P:

-

iff Vz(z is compatible with x * z is compatible with y ) y Let Q = PI- and let us define x

[XI < [ y ]

iff

(Vz

I x ) [z and y are compatible]

<

The relation over Q is a partial ordering, and it does not take long to verify that (Q, <) is separative. The mapping h ( x )= [ X I

satisfies (i) and (ii). Exercise 17.19. The partially ordered set Q in Lemma 17.3 is unique up to isomorphism. [Use this: If ( Q , i) is separative, then can be defined in terms of compatibility. x y iff Vz(z and x are compatible + z and y are compatible.]

<

<

Corollary. For every partially ordered set (P, < ) there is a complete Boolean algebra B = r.o.(P) and a mapping e: P -, B - (0) (the canonical embedding?) o f P in B) such that: (17.23) (i) i f p I q, then e(p) I e(q); (ii) p and q are compatible i f f e ( p ) * e(q) # 0; (iii) {e(p): p E P } is dense in B.

Generic Ultrafilters on Complete Boolean Algebras An important consequence of Lemma 17.2 is that in theoretical considerations dealing with forcing and generic sets we can restrict ourselves to complete Boolean algebras (see Lemma 17.4). In the ground model 'Ill, let B be a complete Boolean algebra. In the universe, B is still a Boolean algebra, though not necessarily complete. An ultrafilter G on B is called generic (over 'Ill)if (17.24)

X

EG

and X

E

'Ill implies

fl X E G

(G is also called 'Ill-complete since it is closed under products of collections in 'Ill.)

t e is not necessarily one-to-one.

17.

155

COMPLETE BOOLEAN ALGEBRAS

Lemma 17.4. Lrt P be a partially ordered set, let B = r.o.(P) and let e : P the canonical embedding of P in B (all this in the ground model W). If G is generic on B, then

-,B be

G‘ = { p E P : e ( p ) E G } is generic on P . Conuersely, if G’ is generic on P , then G = {u E B : ( 3 p E G’) e ( p ) 5 u } is generic on B. In either case, %Jl[G]= 91[G’]. Proof: (a) Let G be a generic ultrafilter on B. First we notice that G meets then every dense set D i n B (ifD E W).Let X , = { - u : u E D } . If G n D = 0, X G G and hence 0 = X, E G, a contradiction. Next we observe that if D is a dense set in P , then e[P] is a dense set in B. Thus G‘ = { p E P : e ( p ) E G } meets every dense set D in P which is in 91,and it suffices to show that G’ is a filter. If p and q are arbitrary elements of P, then the set

n

Ep,q= { r E P : either r 5 p and r Iq, or r is incompatible with p , or r is incompatible with q} is dense in P: Given s E P, if s is compatible with p, then there is t Is such that t Ip ; if t is compatible with q, then there is r It such that r I q. Let p and q be in G’. Then because E , , is dense, there exists r E Ep,qsuch that e ( r ) E G. It is obvious that r cannot be incompatible with either p or q and hence r Ip and r Iq. Hence G’ is a filter. (b) Let G’ be a generic filter on P and let G = { u E B : ( 3 p E G ‘ ) e ( p ) I u}. Since G’ is a filter on P, G is a filter on B. If D is a dense set in B, then the set E D = { p E P : e ( p ) Id for some d E D} is a dense set in P. Given q E P, there is d E D such that e ( q ) * d # 0; then there is r E P such that e ( r ) 5 d and that e ( q ) . e ( r )# 0. It follows that q and r are compatible and hence there is p E P such that p Iq and e ( p ) I d. Now i f u E B, then the set {u # 0 : u 5 u or u I- u } isdensein Band hence D, = { p E P : e ( p ) I u or e(p)I- u } is dense in P; thus D, n G’ # 0, and therefore either u or - u is in G. Hence G is an ultrafilter. For every subset X of B (in W),let D x = { p E P : either e(p) I X or e(p) x = 0 for some x E X}

fl

The set D x is dense in P and hence meets G’. It follows that if X G G and X E W,then X E G. Thus G is generic over tm.

n

3.

156

FORCING AND GENERIC MODELS

(c) It is clear that in either case p

E

G’

c-*

e(p)E G

and that the filters G and G’ can be obtained from one another in every model 9 2 9Jl that contains either G or G’. Hence m[G] = m[G’].

Corollary. G c B is a generic ultrafilter on B ifand only ifG is a genericfilter on the partially ordered set B - (0). A partition of B is a pairwise disjoint subset W of B such that W = 1. In general, a partition of u E B is a pairwise disjoint W such that W = u.

1

Exercise 17.20. An ultrafilter G is generic iff for every partition W of B such that W E Zn, there exists a unique u E G n W .

The following version of Lemma 16.1 will be useful later. If B is a Boolean algebra and X a collection of subsets of B, we call an ultrafilter F on B X-complete if

nX

E

F

whenever

X

X

c F,

E

X , and

n X exists

Lemma 17.5. If B is a Boolean algebra and X is a countable family of subsets of B, then there exists an X-complete ultrafilter on B.

n

Proof. Let X = {X,,XI,. . ., X,,...} and assume that each X, exists. We construct a sequence a , 2 a , 2 ... of nonzero elements as follows: Let a , = 1. Having a,, let a , , = a , fl X , if this is nonzero, or else let a , , = a , - x for some x E X , . Then let F be any ultrafilter that contains all the a,.

,

A Few Facts on Complete Boolean Algebras

Given a Boolean algebra B and a nonzero a

E

B, consider the set

B,={uE B : u l a } endowed with the following operations: u ,* u = u u +, 0 = u + u,

-u B, is a Boolean algebra. (Its Stone space is a subspace of the Stone space of B, namely the clopen set X,.) If W is a partition of a complete Boolean algebra B, then B can be viewed at as a product of the algebras B,, u E W; see Fig. 17.1. Each b E B corresponds to the function ( b . u : u E W ) , and b = { b . u : u E W ) . *

FIGURE 17.1

0,

-,u = a

17.

157

COMPLETE BOOLEAN ALGEBRAS

I

Exercise 17.21. If B is an infinite complete Boolean algebra, then IEIKo= B I . [First consider thc case when IB,,)= IBI for all L/ # 0: There is a partition W such that J W J= x,).atid J B I= n ; I B , , j : t / E W )= 111 general, call # o ,srtrh/e if JB,I = JB,,J for all Y 5 I ) . .Y # 0. The set of all stable L I E B is dcnse. and (B,,l = 2 or IB,,(N"= (B,,I i f u is stnblc. Let W bea pnrtitionofBsuch thatenchrrE Wisstablc:wehave(B( =nilB,,(:tr~ W) mid the theorem rollows.]

1~1'".

The Chain Condition Let K be an infinite cardinal. A Boolean algebra B is K-saturated if there is no partition W of B such that I W I = K, and

(17.25)

sat@) = least K such that B is K-saturated

B is also said to satisfy the K-chain condition; this is because if B is complete, B is K-saturated if and only if there exists no descending K-sequence uo > u 1 > . * .> u, > -.., a < K, of elements of B. The K,-chain condition is called the countable chain Condition (c.c.c.). Lemma 17.6. I f B is an infinite complete Boolean algebra, then sat(B) is a regular uncountable cardinal. Proof: Let K = sat(B). It is clear that K is uncountable. Let us assume that K is singular; we shall obtain a contradiction by constructing a partition of size K. For u E B, u # 0.let sat(u) denote sat(B,). Let us call u E B stable if sat(u) = sat@) for every nonzero u I u. The set S of stable elements is dense in B; otherwise, there would be a descending sequence uo > u 1 > u 2 > ... with decreasing cardinals sat(uo)> sat(ul) > .... Let T be a maximal incompatible subset of S. Thus T is a partition of B, and 1 T 1 < ti. First we show that sup{sat(u):u E T ) = K. For every regular 1 < K such that 1> I T 1, consider a partition W of B of size 1.Then at least one u E T is partitioned by W into 1 pieces. Thus we consider two cases: Case I . There is u E T such that sat(u) = K. Since cf K < K, there is a partition W of u of size cf K : W = {u, : a < cf K } . Let K,, a < cf K , be an increasing sequence with limit K. For each a, sat(u,) = sat(u) = K and so let W, be a partition of u, of size K , . Then UaccfW, is a partition of u of size ti. Case If. For all u E T , sat(u) < K , but sup{sat(u): u E T} = K. Again, let K, + K , a < cf K . For each a < cf ti (by induction), we find u, E T , distinct from all u p , p < a, which admits a partition W, of size K,. Then W, is a pairwise disjoint subset of B of size K. A partially ordered set ( P , < ) is k--saturated (satisfies the ti-chain condition) if there is no set W G P of pairwise incompatible elements such that 1 W 1 = K . It can be easily observed that if P and Q are partially ordered sets and P is dense in Q, then sat(P) = sat(Q). In particular, if B is a Boolean algebra and C is its completion, then sat(B) = sat(C). Moreover, if a partially ordered set P is

Ua
3.

158

FORCING AND GENERIC MODELS

-

mapped into the complete Boolean algebra B = r.o.(P),then the embedding e: P B preserves compatibility (see (17.23)) and thus we have (17.26)

sat(P) = sat(r.0. P)

Therefore, by Lemma 17.6, sat(P) is always a regular uncountable cardinal (provided P admits infinite incompatible subsets). Exercise 17.22. If B = r.o.(P) and K = sat(P), then 1 B I 5 1 P [For each nonzero u E B, there is an incompatible W G P such that u =

2 W.]

Exercise 17.23. If B is a K-complete, K-saturated Boolean algebra, then B is complete. [It suffices to show that X exists for every open X (i.e., u u E X implies u E X). If X E B is open, show that X = W where W is a maximal incompatible subset of X.]

2 1 1

Distributivity of Complete Boolean Algebras The following distributive law holds for every complete Boolean algebra:

(and dually, a similar law for a sum of two infinite products). To formulate a general distributive law, let K be a cardinal, and let us call B K-distributive if (1 7.27)

fl 1

a < i iEl.

uai

=

1

f E & < ~ f z

n

~ a , ~ ( z )

z
(Every complete field of sets satisfies (17.27).) We shall see later that distributivity plays an important role in generic models. For now, let us give two equivalent formulations of K-distributivity. If W and Z are partitions of B, then W is a refinement of Z if for every w E W there is z E Z such that w 5 z. A set D E B is open dense if it is dense in B and 0 # u I L' E D implies u E D.

Lemma 17.7. The,following are equivalent,for any complete Boolean algebra B: (i) B is k--distributive. (ii) The intersection of K open dense subsets of B is open dense. (iii) Every collection of x partitions of B has a common refinement.

nu
xin, If

Proof: (i)- (ii). Let D,, a < K , be open dense, D = D,. D is certainly open; thus let u # 0. If we let {uai: i E I,} = { u . u : u E D,}, then uui= u for every a and the left-hand side of (17.27) is u. For each f~ I,, let uf = u , , ~ ( , , ;clearly, each nonzero uf is in D. However, uf = u, by (17.27), and so some uf is nonzero. (ii) + (iii). Let W,, a < K be partitions of B. For each a, let D, = { u : u I v for some u E W,}; each D , is open dense. Let D = D,,and let W be a

fl,

nil<.

18. FORCING

159

A N D BOOLEAN-VALUED MODELS

maximal incompatible subset of D.Since D is dense, W is a partition of B, and clearly, W is a refinement of each W,. (iii) + (i). Let {uUi: a < K, i E I,} be a collection of elements of B. First we show that the right-hand side of (17.27)is always Ithe left-hand side. For each . f ~ I,, let uf = u,,f(,);we have u f I~ , , f ( ~and ) so u f I 1. uai for each a. Thus, for each a,

n,<,

and so

nu<,

xn f

ua,f(a)=

a

xi

n

Cf u/ I 1 a

uai

nu

i

To prove (17.27), assume that (iii) holds, and let u = Ci uUi;we want to show that u , , ~ ( ,= ) u. Without loss of generality, we can assume that u = 1 (otherwise we argue in the algebra BJ. For each a, let us replace {uai : i E I,} by pairwise disjoint {vai : i E I,} = W, such that vUiIuUi and uUi = C uUi (some of the uUimay be 0).Clearly u , , ~ ( , )I u,,~(~). Each W, is a partition of B and so there is a partition W that is a refinement of each Now for each w E W there existsf such that w I v,,f(,),and so 1 ,n u = 1.

Ifflu

x nu

xi

w,.

x nu

nu

%f(U)

18. FORCING AND BOOLEAN-VALUED MODELS

Boolean-Valued Models Let B be a complete Boolean algebra. A Boolean-valued model (of the language of set theory) U = (U, I, E) consists of a Boolean universe U and binary functions I and E on U with values in B. Instead of I(x, y ) and E(x, y ) , we write IIX

= YlL

IIX E

Yll

(the Boolean values of = and E), and assume that I and E satisfy the following: (i) IIx = XI[ = 1 (ii) IIX = Yll = IlY = XI1 (iii) I1x = yll . Ily = zll IIIx = zll (iv) IIX E Yll . IIU = XI1 . IIW = Yll 5 110 E WII For every formula cp(x,, . . . , x,), we define the Boolean value of cp

(18.1)

(18.2)

Ilq(a1, . .

* ?

%)ll

(01,

* ' * 9

a, E U)

as follows: (a) If cp is an atomic formula,

[la = blJ = I(a, b),

IJaE b(J= E(a, b )

3.

160

FORCING A N D GENERIC MODELS

Note how the notion of a Boolean-valued model generalizes the notion of a model (Section 10); the Boolean value of cp is a generalization of the satisfaction predicatc b. If B is the trivial algebra (0,I}, then a Boolean-valued model is just a (two-valued) model; i.e., consider U/= where x = y iff IIx = yl( = 1. We say that cp(a,, ..., a,) is valid in U, if llcp(ul, .. ., a,)ll = 1. Note that an implication cp + $ is valid just in case II~III I II$II. Hence it is postulated in (18.1) that the axioms for the equality predicate = are valid in a Booleanvalued model. It can be easily verified that all the other axioms of predicate calculus are valid, and that the rules of inference applied to valid sentences result in valid sentences. Thus every sentence provable in predicate calculus has Boolean value 1, and if two formulas cp, $ are provably equivalent, we have llcpll = ) 1 $1) . For example, we have

IIcp(4ll

5 IIcp(Y)II The Boolean-valued models can therefore be used in consistency proofs in much the same way as two-valued models. Let U be a Boolean-valued model such that all the axioms of ZFC are valid in U. (We say that U is Booleanvalued model of ZFC.) Let A be set theoretical statement and assume that 1 All # 0. Then we can conclude that A is consistent relative to ZFC; otherwise, 1 A would be provable in ZFC and therefore valid in U : 11 l A l ( = - IlAll = 1. There is an important special case of Boolean-valued models, and in this special case, the Boolean valued model can be transformed into a two valued model. We say that a Boolean valued model U isfull if for any formula q ( x , x . .. , x,) the following holds: IIX

= Yll

*

,,

(18.3) for all a,, . . . , an E U, there exists a E U such that I(cp(a, a,,

* * *

9

an)ll =

I(

V(X, a,,

* * *

9

an)ll

161

18. FORCING A N D BOOLEAN-VALUED MODELS

Let F be an ultrafilter on B. We define an equivalence relation over U by ( 18.4)

iff

x -y

IIx=yll

E

F

and a binary relation E over U/= by (18.5)

That = is an equivalence, and that (18.5) does not depend on the choice of representatives are easy consequences of (18.1) and the fact that F is a filter (cf. the definition of reduced power in Section 7). Thus U/F = (U/=, E)is a model. Moreover, we have the following relationship between the Boolean valued model U and the model U/F:

Lemma 18.1. Let U befull. For anyformula q ( x l , ..., xn), (18.6)

U/F

q([a11, ...) [an])

iff

Ilq(a1,

.**,

an111 E

F,

for all a l , ..., a n € U.

Proof: (a) If q is atomic, then (18.6) is true by definition. (b) If cp is a negation, conjunction, etc., we use the basic properties of an ultrafilter, and the definition of 1) 1 ; e.g., we use IIlI(IlI

E

iff

F

lII(Il1 $ F

III(I%dE F iff 11I(11E F and llxll E F (c) If q is &I(I(x, . . .), we use fullness of U to prove (18.6), assuming it holds for +. By (18.3), we pick a E U such that Ilq(a, ...)[I = I(3x q ( x , ...)I/ and then we have ) ( 3 xq ( x ,

...)I[

E

F

(31 E U) ((q(a,...)/I

iff

E

F

which enables us to do the induction step in this case. 'Ihe Boolean-Valued Model V B

We shall now define the Boolean-valued model V Bthat we shall be using in subsequent independence proofs. Let B be a complete Boolean algebra. Our intention is to define a Boolean-valued model in which all the axioms of ZFC are valid. In particular, we want V Bto be extensional, i.e., the Axiom of Extensionality to be valid in V B : (18.7)

IIVu(u E

x-u

E

Y)" IIJX= YI(

We shall define V B as a generalization of V : Instead of (two-valued) sets, we consider " Boolean-valued " sets, i.e., functions that assign Boolean values to its

I 62

3.

FORCING AND GENERIC MODELS

“elements.” Thus we define V B as follows: (18.8)

v:=

QI

Vl=

u

if a is a limit ordinal

V;

P
V,”+I = the set of all functions x such that dom(x) E V,” and ran(x) G B V B = (J V,” a E Ord

The definition of JJxE y J J and JIx= yJ1 is motivated by (18.7), and the requirement (18.9)

x(t) 5 Ill E

XI1

We define the Boolean values by induction. Each x E V Bis assigned the rank in VB, p ( x ) = least a such that x

E

V,”+

The forthcoming definition is by induction on pairs (p(x), p ( y ) ) , under the canonical well ordering. To make the notation more suggestive, we introduce the following Boolean operation (corresponding to the implication): (18.10)

u*v=

-u+v

(Compare with cp -+ $ iff l c p v $.) (18.11)

IIX E

c

Yll =

fcdomy

(18.12) 1I.x (18.13)

c Yll

n n

=

(1.

= tlI

(x(t)*

Ill E YII)

1 {p

B : either p . x ( t ) = 0 or

fEdomx

=

.Y ( 4

E

p I (It E yII)

redomx

(18.14)

Ilx = Yll =

IIX

E Yll . IlY 5 XI1

We are going to show that V Bis a Boolean-valued model. To do that, we have to verify (18.1). To start, the clause (ii) in (18.1) is trivially satisfied since the definition of (Ix = y((is symmetric in x and y: IIX = Yll

Lemma 18.2. I(x = XI(

=

1,for all x

E

= IlY = XI1

VB.

Proof: By induction on p ( x ) . Clearly, it suffices to show that IIx G xII = 1, i.e., by (18.12) we wish to show that (18.15)

x(t)=. Jlt E XI1 = I

18.

163

FORCING A N D BOOLEAN-VALUED MODELS

for all t

E dom(x). If t E dom(x), then by induction hypothesis we have Ilt = tJI= 1 and hence, by definition of [It E

xJI,

(18.16)

x ( t ) = I(t = ti(

. x(r) I [It E XI1

Now (1 8.15) follows immediately from (18.16). Incidentally, the proof of Lemma 18.2 also proves (18.9).To prove (18.l)(iii) and (iv), we prove them simultaneously by induction:

Lemma 18.3. For all x, y ,

iE

VB,

IIx = zll, (a) 11.x =yll . Ily = 211 I (b) 1l.y E Yll 1l.X = zll I llz E Yll? (4 IlY E XI1 . 1I.X = 4 IIlY E ill. ’

Proof: By induction on triples ( p ( x ) , p(y), p(z)), under the canonical well-ordering. (a) It suffices to prove IIX YII . IIY = 4 I IIX c ZII r E dom(.u) be arbitrary; we wish to show that

(18.17) Let

IlY = ;I1

(18.18)

. (4)* (It E YII) 5 .+)*

(It E

;I1

(Then we get (18.17) from (18.18) using the definition (18.12) of JIxG zll.)Note that (18.19) J ( y= :(I . (.u(t)*

Ilf E yII) = 1 {p 5 I(y= zll : either p . x ( t ) = 0

By the induction hypothesis, we have Yll . IIY = =I1 5 Ilt E

Ilt E

=I1

Thus if p I Ily = =I/ satisfies the condition on the right-hand side of (18.19), it also satisfies the condition in (18.13) and we obtain (18.18). (b) Let t E dom(y) be arbitrary. By the induction hypothesis we have 1I.X = 211

.

1I.X

= tll I(Iz = t(l

and so (18.20)

Ijx = 211 .

= tll

IIX

Taking the sum of (18.20) over all t Ily

=

=I1

. IE

c

(1l-y

dom y

. y(t)I llz = tll . y ( t ) E

dom(y), we get

= tll . Y ( 4 ) 5

c

(Ilz =

redomy

that is. IIX =

=I)

. 1I-X E Yll

5 llz

E

Yll

4. Y(t))

3. FORCING A N D GENERIC MODELS

164

(c) Let r

E dom(x). By

the definition of Ilt

E 211

we have

x ( t ) . JIX = 211 I (It E 211

and so IlY = tll . x ( t ) . llx = ZII 5 IIY = tll * (It E By the induction hypothesis, (18.21)

IlY = tll . Ill E

(18.22)

211

5 IlY E

211

2113

and therefore (18.23)

IlY = tll * x ( t ) . Ilx = 211 5 IlY

E 211

Taking the sum of the left-hand side of (18.23) over all t E dom(y), we get

c

redomy

(IIY = tll

*

x ( N . Ilx = 4 5 IIY E

4

that is, IIY E

XI1

. Ilx = zll 5 IIY E zll

Thus V” is a Boolean-valued model. We shall later show that all axioms of ZFC are valid in V”. For the moment we shall just show that V Bis extensional, and also that V Bis full. To show that V B is extensional, let X, Y E V”. By the definition of a b (18.10), we observe that if a I a’, then ( a ’ - b) I (a b). Thus for any u E V B we have, using (18.9),

-

-

IIU E

and therefore (18.24)

n

,€Ye

(Ilu

XI1 => IIU E YII I X(U)=> Ilu E YII

XI1 * IbJ Yll) 5

n (X(+

uove

Ib E Yll)

While the left-hand side of (18.24) is equal to IIVu(u E X u E Y)II, the righthand side is easily seen to equal I(Y c XI1 (cf. (18.12)). Consequently, IIVU(U E

x-u

E

Y)II I IIX = YII

and we have:

Lemma 184. V Bis extensionai. We shall now show that V” is full.

Lemma 18.5. If W is a set of pairwise disjoint elements of B and ifa,, u E W ,are elements of VB,then there exists a E V Bsuch that (18.25)

for all u E W.

18.

165

FORCING A N D BOOLEAN-VALUED MODELS

Proqf: Let D =

uuedom(a,). and for every t E D, let a @ )=

1{ u . a , ( t ) : u

E

W).

Since the u’s are pairwise disjoint, we have for each u

E

W and each t E D,

u . a ( t ) = u . a&) In other words, I (a@)* a&)),

J-4

I (a&) * a @ ) )

and by (18.12) u

I ( ( a= a,I(

Lemma 18.6. V B is full. Given aformula cp(x, . . .), there exists a E V Bsuch that (18.3)

IIda1 . .

.)I1 = Il3x cpk ...)I1

ProoJ In (18.3), I holds for every a. We wish to find a Let uo = 1 3x&, . ..)/I. Let

E

V’such that 2 holds.

D = {u E B : there is a, such that u S Ilcp(au, . ..)I]} It is clear that D is open and dense below u,,. Let W be a maximal set of pairwise disjoint elements of D; clearly, { u : u E W }2 u o . By Lemma 18.5 there exists a E V B such that u I / a = a, ( 1 for all u E W. Thus for each u E W we have u 5 ...)It, and hence uo 5 ((cp(a,...)I\. Exercise 18.1.

Exercise 18.2. Lemma 18.6 was the only place in the present section where we used the axiom of choice.

Forcing Let ( P , < ) be a notion of forcing, i.e., a partially ordered set. As I promised in Section 16, I shall now define the forcing language associated with (P, <), and the forcing relation

P

‘k cp

(pforces cp), where p E P and cp is a sentence of the forcing language. As proved in Section 17, there exists a complete Boolean algebra B = r.o.(P),and a canonical embedding e: P + B - (0) such that

(17.23)

(1) if p I q, then e ( p ) I e(q); (ii) p and q are compatible iff e ( p ) . e(q) # 0; (iii) { e ( p ): p E P} is dense in B.

3.

166

FORCING AND GENERIC MODELS

Exercise 18.3. For each S E P , let S' = { q E P : q is incompatible with every p E S } . Let B = { u E P : u l l = u), and partially ordered by inclusion. Then B = r.o.(P) and:

(a) 4 P ) = b)"; (b) - u = u l , u . u = u n u, u

+ u = (u u u)".

[Compare with Exercise 17.2 and Lemma 17.2.1 The forcing language associated with the notion of forcing (P, < ) is the language of set theory with additional constants (names),namely the elements of the Boolean-valued model V B (where B = r.o.(P)).The forcing relation is defined as follows

(18.26)

P

IF

cp(a,, . . . ,an)

iff

e ( ~5) II~o(a1,..., an)ll

where cp is a formula and a , , . . . , a, E V B(i.e., cp(al, . . . ,a,) is a sentence of the forcing language). The following two lemmas are easy consequences of the definition of 1 11 and (18.26):

Lemma 18.7. (a) l f p forces cp and q I p , then q 11- cp. (b) No p,forces both cp and l c p . (c) For every p there is q 5 p such that q decides cp, i.e., either q q IF l c p . rn

1 1 cp

or

Lemma 18.8. (a) p 11l c p iff no q Ip forces cp. (b) p IF c p ~ l ( /i f f p Ik cp and p 11 t+h. p IF ~xcpI f p 11cp(a)for every a E I/! (c) p I F c p v l ( / i f f V q 1 p 3 r < q ( r IFcporr p IF 3xcp if V q I p 3r 5 q 3a r 11 cp(a).

IF$).

Exercise 18.4. When does p force cp + J/? When cp Exercise 18.5. If { q : q

11 cp}

-

J/?

is dense below p . then p forces cp.

Proof of the Forcing Theorem We are now ready to prove Theorem 40 stated in Section 16. Let us recall the background: We are considering a fixed transitive model 911 of ZFC (the ground model), and in Vli, a notion of forcing ( P , <). To define the forcing relation associated with P , we let (in 91)B = r.o.(P) and construct the Boolean-valued model VB.Since all this is done inside 331, we denote this Boolean-valued model by %IB. !IllE is a class in 331, and

18.

167

FORCING A N D BOOLEAN-VALUED MODELS

W E=

ua

or,j n ‘ ~ fl:, 1 where

W:,

= {x E

W : x is a function, dom(x) 5 W:, ran(x) C B}

etc. Still in !Vl, we define the Boolean values 11cp11 and the forcing relation p If cp. Just for completeness, let us state explicitly that for any formula cp, the relation p ~t- cp(al, . . ., a,,),

p

E

P, a l , . . . , a,

E

9IB

is in the model 91. Now let G be a filter on P generic over 91.We shall define a transitive model 911[G],give the elements of 9J[G] names in W E ,and prove:

Forcing Theorem. For any formula cp, where x,, . . . , x, E 91’ are namesjor x,, . . . , x,.

T o define 91[G], we construct, by induction on p ( x ) , a function iG on 91’ called the G-interpretation of 9JB: (18.28)

(i) i@)

(4

=

0;

ic(x) = { i G ( Y ) : (3P E G )

4)5 X(Y)I.

Here e is the canonical embedding of P in B;in (ii), iG(y)is already defined for all y E dom(x). Now we define 9lJ1[G] = { i c ( x ) : x

E

911’}

It is clear from (18.28) that 911[G] is a transitive class. For each x E 911[G], there exists x E 911’ such that .Y = i c ( x ) . We call (any such) x a name for x. We showed in Lemma 17.4 that if we let G’ = { U

E

B : ( 3 p E G ) e(p)IU )

then G’ is an 911-generic ultrafilter on B. Note that for every x E 911’, (18.29)

iG,(.K)

= { i c . ( y ): X(Y) E

G’}

Moreover. it follows easily by induction that i G , ( x )= i c ( x ) for all x E 911’, and consequently, 911[G’]= 911[G]. Similarly, it is easy to verify that the right-hand side of (18.27) is equivalent to ((cp((E G’. Thus we can restate the Forcing Theorem as follows: If G is an 91Lgeneric ultrafilter on B, then where xl, . . . , x, E 911’ are names for x l , . . . , x,. We prove the forcing theorem (18.30) in two steps; first for atomic formulas, and then for arbitrary cp.

168

3.

FORCING A N D GENERIC MODELS

Lemma 18.9. Let G be a generic ultrajlter on B and let i interpretation of !VIE. Then

= i(; be the G-

(a) i(x) E i(Y) ifs llx E Yll E G, (b) i(x) = i(y) iff IIx = yll E G, for all x, y E %IB. Proof: We prove (a) and (b) simultaneously, by induction on pairs ( p ( x ) ,p(y)).

0holds by the genericity of G, 0by the induction hypothesis. (a)

IJx

E

yll E G iff z e d o m ) . (Y([)

@ @

IIx

(3E dom(y))[y(t) E G and (Ix = fll

iff

3rb(t)E G and i(x) = i(r)]

iff

i(x) E {i(t): y(t) E G ) i(x) E i(y)

Yll

@

a

IIx = tll) E G.

iff

iff (b)

*

iff

fliedom(x)

(x(t)*

1'

E

GI

G.

YII)

iff

(Vr E dom(x))[x(t) E G implies Ilt E yll E GI

iff

Vt[x(t) E G implies i(t) E i(y)]

iff iff

{i(t): x ( t ) E G} G i(y) i(x) G i(y)

Exercise 18.6. Let C be a generic ultrafilter on B and let 91B/G be the model defined as

in (18.4) and (18.5). Then 9XIIB/G is isomorphic to 91[G]. [Let h ( [ x ] )= iG(x).]

Once we have proved (18.30) for atomic formulas, we finish the proof for arbitrary cp by induction on the complexity of cp. (a) cp is -I$, $ A 1,$ v x, etc. Assuming (18.30) for Ic/ and step works because G is an ultrafilter. For instance,

U[G] b

II/ A X

Ic/ and 9X[G]

iff

9X[G] l=

iff

iff

11$11 E G and llxll E G l l ~ l l llxll E G

iff

IlIc/AXll

E

l=

x, the induction x

G

Similarly for 1,v , etc. (b) cp is 3x $(x, ...) or Vx $(x, ...). We assume (18.30) for $ and use the

18.

169

FORCING A N D BOOLEAN-VALUED MODELS

genericity of G : 9lI[G] k 3.x $(x, ...)

iff

(3E 91[G])91[G]k $(x, ...)

iff

(3x E 9UIfl)91[G] k $ ( i ( x ) ,

iff

(3x

iff

11 3x$(x, ...)I[

E

. . .)

91fl)IIII/(x,...)[I E G

E

G

The equivalence @ holds because if we let A = { Il$(x, . . .)I[ : x A E B and A E 91, and since G is generic we have

E

91"},then

( ~ ~ E A ) ~ iff E G C A E G

Similarly for Vx $(x, ...). H Proof of the Generic Model Theorem

Having defined 9li[G], we shall now use the forcing theorem to prove that: (a) (b) (c) (d)

91[G] is a model of ZFC. 91 c 91[G] and G E 9l[G]. Ord"[cl = Ord'l". If 91 is a transitive model of Z F such that 91 E 9 and G

E

91, then

W[G] E 91. Throughout, G is a fixed 91-generic ultrafilter on B, a complete Boolean algebra in 91. We first dispose of (d). If 91 is a class in 9,then the assertion is trivial: The function iG is a class in 91 and thus W[G] is a class in 9.However, 91 need not be a class in 9;but the assertion is still true: For each a E O r P , 91: E 91 E 3, and thus, by induction, ic 191: E 9.Therefore i G ( x )E 9l for all x E WE and we have 91[G] E 9. Next we prove (b). To show that W E 91[G],we define canonical names for sets in the ground model. We define (in 91) an embedding of 91 in WE, by €-induction : (18.31) (i) 0= 0; (ii) for every x E W ,let K E 91' be the function whose domain is { j : y E x}, and for all y E x,

K(Y*)= 1 It is easy to see that for each x E 91, iG(K) = x

170

3.

FORCING A N D GENERIC MODELS

and so 2 is a name for x (for any generic G). Thus 911 G 91[G]. To show that G E 9I[G], we define a canonical name for G, the canonical generic ultrafilter: be as follows: Let C E dom(C) = {.’ : u E B},

(18.32)

C(.’) = u for every u E B

If G is any generic ultrafilter, then i,(G) = { i G ( x ): C(x) E G } = { i G ( i i ) : u

Thus G

E

E

G} = G

9 l [ G ] and we have proved (b).

Exercise 18.7. If x, y E 91, then

:1

E

1:

= jII = 1

if x

=y

and 0 if x # y. Similarly for

PII.

Exercise 18.8.

For every u

E

E , 116 E GI1 = u.

As for (c), this is easy. Since 91 E 91[G], we have Ordm G Ord’m’cl.On the other hand, note that for every x E %IB,

rank ic(x) I p ( x ) (by induction) and therefore all sets in 91[G] have rank equal to an ordinal in 911.

It remains to show that 9 l [ G ] is a model of ZFC: A. Extensionality, Regularity, Infinity. These axioms hold in 91[G] since 91[G] is transitive and 91[G]2 91. B. Separation. If cp is a formula (we suppress the parameter), 9ll[G] k VX 3Y V x ( x E Y t r x

Let X

E

%1[G] and let X

(18.33)

E

E

X ~cp(x))

91” be a name for X . We let Y

dom( Y ) = dom(X),

E

91Bbe as follows:

Y ( t )= X ( t ) . IIcp(t)ll

Now, if Y = iG( Y),it is clear that Y E X.Moreover, for every x E 91’we have IIX YII = Ilx E XI1 IIcp(x)ll X , 91[G] satisfies ’

and so for every x

E

X E

Y + + XE

x A C p ( X)

C . Pairing, Union, Power set. Before we verify these axioms in 91[G],let us make the following observation: Let S E ‘W be a set of names (i.e., S c !VIE). Then the set i G [ s ] = { i c ( t ): t E S } is in ‘1I1[G]. For, consider this X E !IllB: dom(X) = S , X ( t ) = 1 for all t E S . Clearly, i G ( X )= iG[S]. In particular, if x , y E %I”, we denote {x, y}” (the Boolean-valued pair), the z E 91P such that dom(z) = { x , y ) , and z(x) = z(y) = 1. Also, (x, y)” = {{x}”,{ x , y]”})”.Clearly, ic({x, y}”) = {iG(x),ic(y)). If 9i contains all ordinals, then !UI[G] is almost universal. Consequently, !UJI[G] is a model of ZF.

Exercise 18.9.

171

18. FORCINGAND BOOLEAN-VALUED MODELS

Since we are proving Theorem 39 even in the case when 9 l is only a set, let us continue with the proof. We have just shown that 91[G] satisfies the Pairing Axiom. To verify Union and Power set, it suffices to show that for every X E W[G]: (1 8.34)

uX

(18.35)

P(X) n 9ll[G]

Let X

E

E

Y for some Y c_

E

91[G];

Y for some Y E %1[G].

%1[G] and let X be a name for X . As for the union, let

S=

u

u

(dom(y):y

E

dom x)

and let Y = i,[S]. If z E X , then z E iCk) for some y E dom(X) and so z E Y ; hence (18.34) holds. As for the power set, let us first show that every subset of X has a name of a certain form. If 2 c X and if 2 is a name for 2, let Z’ be as follows: (18.36)

dom(Z’) = dom(X),

for all t

Z ‘ ( t )= X ( t ) . I(t E Zll

Clearly, i,(Z’) = X n iG(Z)= Z and so Z ‘ is a name for 2. Therefore, if we let S = { Z :dom(Z) = dom(X) and Z(t) I X ( t ) for all t )

then Y = ic[S] satisfies (18.35). D. Replacement. As shown in Section 9, the replacement schema follows from the collection principle. Thus we show that 91[G] satisfies

vx 3Y(Vu E X ) [ 3 U q ( u , o ) + (3o E Y ) q ( u , u ) ] (9.4) Let X E !UJI[G]and let X be a name for X . In 9l, there is for each u E dom(X) a set of names S, such that (18.37)

We let S = U {S,,: u E dom X),and Y = i,[S]. Now, if u E X is such that 9N[G] 1130(p(u, r), then 11 = ic(u) where X ( u )E G, and by (18.37), 911[G]satisfies 30 q ( u , u )

-+

(30 E Y ) q ( u , u )

Therefore 9ll[G] is a model of ZF. Note that so far we have not used the axiom of choice: Exercise 18.10. If N ! is a model of ZF, then 3)1[G] is a model of ZF.

To show that 911[G] satisfies the axiom of choice, we prove the following lemma : Lemma 18.10. k t i , he the interpretationfunction. I f S then iG 1 S E Yll[G]. Proqfi Let T = {(S, s ) :~s

E

S } . Clearly, i, 1 S = i,[T].

E

91 is a set ofnames,

172

3.

FORCING AND GENERIC MODELS

In fact, in the case that 91 is a class in 9UI[G], then ic is a class in YlJI[G]. In any case, if X E 9l[G], then let X be a name for X and let S = dom(X). By Lemma 18.10, ic IS is in 9I[G], and since S can be well-ordered, it follows that X,an image of a well-ordered set, can be well-ordered (in 9l[G]). Generic Models and Independence Proofs We shall conclude this section by providing answers to the following questions about generic models: Why can we assume that a generic filter exists? And how are generic models used to obtain an independence proof? We shall first prove a theorem about the Boolean-valued model V”.

Theorem 43 (a) V” is a Boolean-valued model of ZFC. Every axiom of ZFC has Boolean value 1. (b) Let 9 = P(B),and let G be the canonical name for a generic ultrujilter on B. Then

((c is a &generic ultrafilter on B’I( = I

(18.38)

ProoJ (a) I am very reluctant to give a detailed proof that for every axiom cp of ZFC, (IcpIJ= 1. The main reason is that the arguments are exactly the same as in the proof of the generic model theorem, only less lucid since we are dealing not with a transitive class but with a Boolean-valued model. Thus what follows is a sketch of the proof, and the incredulous reader is requested to work out the details. When calculating the Boolean value of a formula, one may find the following observation helpful (cf. Exercise 18.1):

(18.39)

IVY

x)(P(x)ll =

I P Y E Xk(Y)II=

y

c n

E

dom(x)

y E

dom(x)

MY).

IIdY)II)

M Y ) * IIdY)II)

Now, let us consider the axioms of ZFC one by one: Extensionality: Proved in Lemma 18.4. Pairing. We prove that for every x , y E V” there is z E V” such that IIXEZhyEZ(I= 1

Let z = { x , y}”. Separation. We prove that for every X

E

V” there is Y E V’ such that

and II(Vz E X ) (cp(z)wz E Y)II = 1 IIY E XI1 = 1 Let Y E V Bbe as follows (see (18.33)):

dom(Y) = dom(X),

Y(t) = X ( t ) * IIcp(t)ll

18.

173

FORCING A N D BOOLEAN-VALUED MODELS

Union. We prove that for every X E I/” there is Y II(VU

We let dom Y =

u {dom(u)

:u

E

V” such that

X)(Vv E u)(u E Y)II = 1

E

Y ( t ) = 1 for all t

dom(X)),

Power set. We prove that for every X (18.40)

E

JIVU(U

E

V” there is Y

E

E

I/”

E

dom( Y)

such that

x + u E Y)ll = 1

Here we let dom( Y) = { u E V B: dom(u) = dom(X) and u ( t ) I X(t) for all t} Y ( u )= 1 for all u

E

dom( Y)

The argument given in (18.36) shows that (18.40) holds. Replucernenf. As in the proof of Theorem 39, we consider the axioms of collection; we prove that for every X E V” there is Y E V” such that I p u E X)(+(P(U, u ) + (30 E Y)cP(u, u)ll = 1

Here we let dom( Y) =

{S, : u

E

Y ( t )= 1 for all t E dom( Y)

dom(X)},

where S , is defined by the formula (18.37), for each u E dom(X). Injnity. Here we recommend the reader to verify that

[I& is an inductive set 11 = 1 (See also Lemma 18.11.) Regularity. We prove that for every X (18.41)

I1X is nonempty

+ (3y E

E

VB,

X)(Vz E y)z

4 X(I = 1

If (18.41) is false, then JI3u(uE X ) A (Vy E X ) ( ~ EZ y ) z E XI1 = b

is nonzero. Let y E V B be of least p ( y ) such that Ily E XI[ . b # 0. Then Ily E XI1 . b I 11(3z E y ) z E XI[, so there exists z E dom(y) such that llz E XI1 . Ily E XI( . b # 0. Since p(z) < p(y), this is a contradiction. Choice. As in the proof of Theorem 39. For every S, we have

11s’ can be well ordered 11 = 1 (Either verify or wait for Lemma (18.1 1): “can be well-ordered ” is C Now, as in Lemma 18.10, we prove that for every X E V Bthere is S and there i s f e V B such that ( 18.42)

llfis a function on $and ran(f) 2 XI1 = 1

174

FORCING AND GENERIC MODELS

3.

Then it follows that the axiom of choice is valid in V”. We let S = dom(X) andfE V” as follows: dom(f) = {(Z, x)” : x

E

S},

f ( t ) = 1 for all t E dom(f)

These f and S satisfy (18.42). (b) Recall that (IGc Bll = 1 and 113 E GI1 = u for every u E B. Thus II~EGII=O I [ u~ ’ ~~ G ~ v ( - u ) ’E G I I = u + -u=l,andweshallprove

If X

E

fl X E G(I= 1

I((VXE 5 )if X G G, then

(18.43)

9, then

Ilx’ G GI( = II(VU E x ’ ) E~ GI1 =

fl (13

E

G(I =

ucx

and so

112 E G+

(nX)’

E

fl X = “(fl X)”

E

GI1

GI1 = 1

Now (18.43) follows. The Boolean-valued version of absoluteness of C,-formulas:

Lemma 18.11. Ifcp(xl, . . . , x,) is a Co-formula, then ( 1 c p ( L . . ., 1I),: Proof: By induction on the complexity of cp. H cP(x1,

iff

. . x,) *

7

Corollary. Ifcp is C,, then q ( x , ...) implies Ilcp(Z,

...)[I

=1

= 1.

The next lemma states that V and V” “have the same ordinals”:

Lemma 18.12. For eoery x E VB, IIx is an ordinal11 =

1

1Ix = oil1

a E Ord

Proof: Since “ x is an ordinal ” is E, , we have, by Lemma 18.11,

1

a E Ord

IIx = oil1 I IIx is an ordinal11

On the other hand, let IIx is an ordinal11 = u. We first observe that if ordinal, then 1Ix is an ordinal and x E f l l I IIx = oil1 Also, for every a, we have

11

is an

1

u II(x E oil1

+ (Ix = oil( + 1:

E XI[

However, there is only a set of a’s such that lloi E XI/ # 0 (because lldi E do,,,(x) (116 = t ( 1 . x ( t ) ) ) . Hence there is y such that

Et

uI IIX E

and we have u 5 la
jq

xII

=

18. FORCING A N D BOOLEAN-VALUED MODELS

175

As a consequence, since a H L , is a Z,-function, we have

Corollary 18.13. For every x E VB, IIx is constructible11 =

1 IIx = iill

4eL

Converting Forcing Arguments into Independence Proofs

I shall use the example from Section 16 to indicate how one uses the method of forcing to prove independence of set-theoretical statements. Let us show that the statement cof is countable

(18.44)

is consistent relative to ZFC. We can use three different ways to d o it. I. Consistency of a generic ultrafilter We let 91 = L be the ground model, and we consider the notion of forcing ( P , < ) defined as follows (see Section 16): (18.45) p

E

P

iff

p is a finite sequence ( a o , . . . , an-

of ordinals less than coi

iff P=4 Now, let us assume that there is G E P generic over L Then, as shown in G, thenfis a function of o onto cof and therefore Section 16, if we let J'= P<4

u

911[G] k cof is countable

Thus all we have to d o is to show that it is consistent that P has a filter generic over L. We let B = (r.0. P)" and consider the Boolean-valued model LB.If G is the canonical generic ultrafilter and if 9 = PL(B),then we have, by Theorem 43b, I ~ isC a $-generic ultrafilter on

ill = I

Now, since constructibility is absolute for Boolean-valued models (cf. Corollary 18.13), and so is the definition of (this particular) P and B,and since also

( 1 5 is the set of all constructible subsets of B'(I = 1 we conclude that the following sentence has Boolean value 1 in LB: (18.46) There exists a filter on the set P defined in (18.45), generic over the

constructible universe. Therefore, the sentence (18.46) is consistent and hence we may assume that an L-generic G c_ P exists, and construct 9lUI[G].

I76

3.

FORCING A N D GENERIC MODELS

11. The method of Boolean-valued models

We can prove the consistency of (18.44) directly, by showing that

lithe constructible o1is countable11 = 1

(18.47)

in LB,where B is defined as in I. To do that, we have to carry out the arguments not in L[G] but in LB.By manipulating the Boolean values, we can show that

(1

u

G is a function from 6 onto (of)' 11 = 1

Now, since constructibility is absolute for Boolean-valued models, we have ll(w4)" is the constructible (ol

1) = 1

and therefore we conclude that (18.47) holds. Hence the statement (18.44) is consistent because it holds in L!. 111. The method of countable transitive models

We can also use the following argument to show that a statement A is consistent with ZFC (in our Example A is "wf is countable"). We have defined a notion of forcing P and shown (18.48) (V model 91 of ZFC)(VG E P") [if G is generic over 91, then 'X)I[G] k A]

If we analyze our proof of [if G is generic, then 91[G] k A], we can see that although we use the fact that 91 and 9Jl[G] are models of ZFC, we in fact use only finitely many axioms of ZFC, say cp . . . , (P, , in either 911 or W[G]. Assume that 1 A is provable in ZFC. Then 1 A holds in every model of ZFC, in particular in every generic model 91[G]. Now, if we look at the proof of the generic model theorem, we see that it takes only finitely many axioms of ZFC in W, say I(/ 1, . . . , I ( / k , to show that 91[G] is a model of 1A, cp 1, . . . , (P,, (and that Pm exists). By the reflection principle, there is a countable transitive n dLemma 16.1,there modelM such that M t cpl A * * . A ~ ~ ,~, .A. .I~( I/ (~/ , , a by exists G E PM generic over M. Then M [ G ] satisfies 1A, and also cp A * * A cpn . The latter is then used, as in (18.48), to show that W[G] k A, a contradiction. Thus 1 A is not provable in ZFC. 19. INDEPENDENCE OF THE CONTINUUM HYPOTHESIS AND THE AXIOM OF CHOICE

In this section we apply the method of forcing to give independence proofs of the continuum hypothesis and of the axiom of choice.

Theorem 44 (Cohen). There is a model of ZFC in which 2N0 > N,. Thus the continuum hypothesis is independent of the axioms of ZFC.

19. INDEPENDENCE

OF THE CONTINUUM HYPOTHESIS A N D AC

177

We shall prove Theorem 44 by constructing a generic model of ZFC in which the continuum hypothesis is violated. The independence of the continuum hypothesis then follows by one of the arguments at the end of Section 18. Given a ground model 91, it is possible to extend it to a model 91[G] in which the continuum is as large as prescribed. The precise statement follows. Lemma 19.1. For any cardinal ti in 91,that satisjes (in 911) tiKo= K, there exists a notion qfforcing (P,< ) such that the generic extension !Ul[G] has these properties : (i) 91 and 91[G] have the same cardinals and same cojnalities; (ii) !U1[G] k 2'O = ti.

["Same cofinahties" means: cfg"'"(a) = cfm(a)for all limit a E 91.1 1 have more or less started the construction in Section 16; cf. (16.4)-(16.6). Adding

K

Generic Reals

Let P be the set of all functions p such that (19.1)

(i) dom(p) is a finite subset of (ii) ran(p) c {0, l},

ti

x w,

u

and let p be stronger than q iff p 3 q. Let G be a generic set of conditions and let f = G. By a genericity argumentJis a function from ti x w into (0, 1). For each a < t i , we letf, be the function on o defined by (19.2)

L(n) = f ( a , 4

and (19.3)

a, = {n E w :f,(n) = 1)

Each a , is a real (a subset of 0); and as shown in Section 16, a , 4 (m and if a # /I,then a , # a a . For instance, the latter holds because the set D = { p E P : p(a, n ) # p(b, n ) for some n } is dense in P (and is in (m) and hence DnG#@. Thus if we show that %1[G] has the same cardinals as 94 (that (P,<) preserues cardinals), then it follows that 2N0in (m[G] is at least K. This brings us to the general problem of preservation of cardinals and the role of the chain condition. We recall that P is I-saturated (or satisfies the I-chain condition) if there is no incompatible W s P of size I, and that P is I-saturated if and only if B = r.o.(P) is. Lemma 19.2. Let 1 be a regular uncountable cardinal (in (m) and assume that (P, < ) is I-saturated (in %I). Then A is a regular cardinal in %I[G].

3.

178

FORCING A N D GENERIC MODELS

Corollary 1. Every cardinal in 91 greater than or equal to sat(P) remains a cardinal in

9Wl.

2. If P satisfies the countable chain condition, then 9l1[G] and cardinals and same cofinalities.

(xn have

the same

Proof: Let B = r.0. P. Assume that cfm[''(I) < I . Then there exist v < A, a namefc %1* and a condition w E B such that w forces: (19.4)

(i) f i s a function; (ii) f: 3 --t 1 and the range offis cofinal in

1.

P < I , let W ( % P ) = w . llf(4= Till It follows from (i) that w(a, P ) . w(a, y ) = 0 whenever P # y ; from (ii) it follows that for every P c I there exist a < v and y > such that w(a, y ) # 0. Thus (19.5) I {P < A : 3a(w(a,P ) # 0 ) )I = 1

For each a < v and each

and since v < I , there is a. < v such that (19.6)

I {P < I

w(a0

9

P ) # 0 )I

=I

However, the collection in (19.6) is incompatible, contrary to the assumption that B is I-saturated. Now if v 2 sat(P), then if v is regular in 911, then v is regular in %1[G];and if v is a singular cardinal in 91,then v > sat(P) and is a limit of regular cardinals, thus a cardinal in 91[G]. In particular, if sat(P) = K,, then all cardinals and confinalities are preserved. We shall now show that the notion of forcing ( P , < ) defined in (19.1) satisfies the countable chain condition. The following lemma is a slightly more general statement: Lemma 19.3. Let S be an arbitrary set and let C he at most countable. Let P he the set of al1,functions p whose domain is afinite subset ofS, with values in C. Let P be partially ordered by inverse inclusion. Then every incompatible subset W of P is at most countable.

Proof: Let W be an incompatible subset of P; that is, if p and q are distinct elements of W ,then p ( x ) # q ( x )for some x E dom(p) n dom(q). We shall show that there is a countable subset A E S such that dom(p) E A for all p E W . Then it follows that I WI IK O . We construct a sequence A , E A C_ ... C A , c *.. of subsets of S , and a sequence W, c W, c ... c W, E ... of subsets of W as follows: First we pick

19.

INDEPENDENCE OF THE CONTINUUM HYPOTHESIS A N D AC

179

q E W and Ipt W, = (4)and A0 = dom(q). Given W, and A,, we choose for each p E P with dom(p) E A , , some q E W whose restriction to A, is p (if such 4 exists). Then we let W,,, be W, along with these chosen q's, and A , , = {dom(q): q E W,, Finally, we let A = uFz0A , . It is obvious (by induction) that each W, and each A, is at most countable. We complete the proof by showing that W = uFE0W,. If q E W , then there is n such that dom(q) n A = dom(q) n A,. Thus if p = q [ A , , there exists q' E W,, such that q' I A, = p . Since dom(q') 5 A , , I G A , it follows that q and 4' are compatible; however, both are elements of W and thus q = q'. Hence q E K+i.

,

u ,

Now the proof of Lemma 19.1 is almost complete. Since P is K,-saturated, it preserves cardinals, and 9I[G] has at least ti subsets of o (namely the generic reals a,, u < K). It remains to show that 931[G] 1 2N0I K . This is a consequence of the following general estimate of the number of subsets of a cardinal I in a generic extension.

Lemma 19.4. Let I he a cardinal in 9131. I f G is an 931-generic ultrajlter on B, then Proof: Every subset A c I in I1)1[G] has a name A E 91'; every such A determines a function a- lldi E ,411 from I into B. Different subsets correspond to different functions, and thus the number of all subsets of 1 in 91[G] is not greater than the number of all functions from I into B in 311.

In our case, 1 P I = ti; and since P satisfies the countable chain condition, it follows that I r.0. P I ItiNo (see also Exercise 17.22); this is because every u E r.0. P is the sum of countably many elements of P. Considering that KNo = K, we have I BIN0= ti, and so (2No)'mrG1 = ti. Exercise 19.1. If P is defined as in (19.1) and if K is arbitrary, then (2Ho)m[CI = (K'~)'~. [(2Xo)WlCl = ((2No)no)mlcl 2 (KNO)Wl"l 2 (KHO)W.]

Adding Subsets of Regular Cardinals The proof of independence of the continuum hypothesis (Lemma 19.1) consisted of adding a large number of generic subsets of o.This construction is easily generalized to arbitrary regular cardinals. (For what happens at singular cardinals, see Exercises 19.3 and 19.4.) Let ti be a regular cardinal (in the ground model 9 31)and assume that 2'" = ti. (We shall discuss the case 2'" > ti in the exercises.) Let ( P , < ) be defined as follows: P is the set of all functions p such that: (19.7) (i) dom(p) c ti and I dom(p) 1 < K ; (ii) ran(p) c (0, 1). A condition p is stronger than q iff p

3

q.

3.

180

FORCING AND GENERIC MODELS

Let G be a set of conditions generic over 91 and let f = f i s a function from K into (0,l}, and

X

(19.8)

=

u G.

As before,

{a < K :f(a) = l }

is a subset of K and X # 1' 11. We shall show that !lland '111[G]have the same cardinals (and cofinalities). To start, since [PI = K < =~ K, we have sat(P) I IPI+ = K', and by Lemma 19.2 all regular cardinals I > K remain regular cardinals in the extension. We shall now show that all the regular cardinals I 5 K are preserved. In fact, we shall show that for every a < K, every a-sequence of ordinal numbers in '111[G]is a set in 1 ' 11. (In particular, P'lRIG1(o!) = P"(a) for all a < K.) Then it clearly follows that if I I K is a regular cardinal in 1 ' 11, then A is a regular cardinal in !ll[G]. Let I be an infinite cardinal. We say that a notion of forcing P is I-closed if for every a I 1,if po 2 p1 2 ... 2 p c 2 *..

(< < a) is a descending sequence of conditions, then there is p E P stronger than all p e , 5 < a. Let us call P A-distributive if the intersection of I open dense subsets of P is open dense. (If P = B - {0},B a complete Boolean algebra, then by Lemma 17.7 this definition is equivalent to I-distributivity of B.) Note that if B = r.o.(P), then B is I-distributive if and only if P is I-distributive. Lemma 19.5. r f P is I-closed, then it is I-distributive. Proof. Let ( D , : a < I } be a collection of open dense sets. The intersection D= a D, is clearly open; to show that D is dense, let p E P be arbitrary. By induction on o! < I, we construct a descending I-sequence of conditions p 2 po 2 p1 2 . .. . We let p a be a condition stronger than all p c , < a, and such that p a E D , . Finally, we let q be a condition stronger than all para < I.Clearly, qED.

na<

<

Lemma 19.6. Let I be an injnite cardinal and assume that (P,< ) is I distributive (in 91).Then f f 9)1[G] ~ is afunctionfrom I into 91, thenfE 91. In particular, P~['~(I) = ~"(1). Proof: Let f: I + 91 and f E 91[G], let f be a name fort There is A condition po E G such that po forces

(19.9)

f i s a function from

E

9 3 1 and a

2 into A'

For each a < I, the set D,

=

{ p Ip o : ( 3 E~A ) p

Iff(.*)

= 21

is dense below p o . (It goes without saying that {D,:a < I ) E 91.) Thus D= D, is dense below p o and therefore there is some p E D n G. Now we argue in 91:for each a < I there is x, such that p I t f ( G ) = 1,;let g: L + A

19.

181

INDEPENDENCEOF THE CONTINUUM HYPOTHESIS AND AC

be the function defined by g(a) = x,. However, it is easy to see thatf(a) = x, = g(a), for every a < 1,and thusfe '91. The notion of forcing defined in (19.7) is clearly I-closed for all 1 < ti and therefore all regular cardinals I K remain regular cardinals. 911[G] has no new subsets of any 1 < ti, while X is a new subset of ti. Exercise 1 9 2 . Let P be as in (19.7) and let Q = {p E P : dom(p) is an initial segment of K;. Then Q is dense in P and hence r.o.(Q) = r.o.(P). Exercise 19.3. Let K be a singular cardinal and let (P, < ) be defined as in (19.7). Then P collapses K to cf(K): in the generic extension, there is a 1-1 function g from K into cf K. [Let K = Ku, and let X be the added subset of K,; cf. (19.8). For each a < K,, let g ( a ) = the least n such that the order type of X n (w, - w,) is (0,+ a. Show that for every a and every p E P there is q 2 p and some n such that dom(q) z w , + , - w, and that the set ([ E (I)"+ I - w , : q ( ; ) = 1) has order-type w, a. By genericity of G,the function g is defined for every a < Nu;it is clearly one-to-one.]

+

Exercise 19.4. Again let K be singular, and let P be the set of all 0, l-functions whose domains are a bounded subset of K ; as usual, P is ordered by inverse inclusion. Show that P collapses K to cf K.

If we want to add a large number of subsets of a regular cardinal employ a technique similar to that in Lemma 19.1:

ti,

we

Lemma 19.7. Lrt ti, A he cardinals (in 911) such that ti is regular and 2'" = ti, and such that A" = A. There exists a notion of jorcing ( P , < ) such that the generic exrension 911[G] hus these properties: (i) 911 and 911[G] haw the same cardinals and same cofinalities; (ii) 911[G] k 2" = A. Proof Let P be the set of all functions p such that: (19.10)

(i) dom(p) G I x ti and \dom(p)I < K ? (ii) ran(p) G to, 11,

and let p be stronger than 4 iff p 2 4. Let G be a generic set of conditions and let ,f= (19.11)

a,

=

u

G. For each a < I , we let

{< < ti : , / ' ( a ,i') = 1)

Each a , is a subset of ti and by what is by now already a standard argument, each a, $ 911 and u, # us whenever a # p. Since P is 2-closed for each 2 < ti, it follows that P preserves all cardinals and cofinalities below ti (including ti). We shall show in the lemma below that P is ti +-saturated,and thus P also preserves cardinals and cofinalities above ti. Clearly. ti has at least A subsets in 91l[G]. On the other hand, we have (in 911) [ P I = A<' = E., and since P is h-+-saturated, we have IBI I IP 1' = I (where B = r.0. P). and by Lemma 19.4, (2")'m["' I ( I B 1") = A.

182

3.

FORCING A N D GENERIC MODELS

It remains t o show that P is ti+-saturated. The following lemma is a straightforward generalization of Lemma 19.3.

Lemma 19.8. Let ti be a regular cardinal such that 2'" = ti. Let S be an arbitrary set and let IC I 5 K . Let P be the set of all functions p whose domains are subsets ofS ofsize < K, with values in C . Then sat(P, 3 ) I .'it Proof: The proof is as in Lemma 19.3. Let W E P be incompatible. We construct a sequence A. c A , E ... E A, E ... (a < ti) of subsets of S , and Wo G W, G ... G W, G .. . of subsets of W. If a is a limit ordinal, we let W, = UP<,W, and A, = A,. Given A,and W,, we choose for each p E P with dom(p) E A, some q E W (if there is one) such that p = q I A,. Then we let W,, = W, u {the chosen q's}, and A,, = {dom(q) : q E W,, ,}; finally, A = U a C K A,* As in Lemma 19.3 one shows that W = W,: If q E W , then there is a -= ti such that dom(q) n A = dom(q) n A,, etc. The proof is completed by showing that I A,[ I ti and I W,l I ti for each a < ti. This is proved by inducti, then I A,\ 5 K because A, = {dom(q): q E W,}. If a is tion on a. If 1 Wal I a limit ordinal and I W,l I ti for all p < a, then I W,l = I Up., W,l I ti. Thus ti and let us show that I W,, 1 I ti. The set W,, is let us assume that I W, I I obtained by adding to W, at most one q E W for each p E P with dom(p) E A,. There are at most ti" subsets X of A, of size < ti, and since ti is regular and 2'" = ti, we have ti<" = ti. On each X there are I C I I x l functions with values in C, and therefore there are at most ti elements of p with dom(p) E A,. Hence I W,, I Iti. Then it follows that I W I I ti. H

us<,

,

u uacK u

,

,

Collapsing Cardinals In Section 16 we introduced a notion of forcing which collapses 0,: The cardinal 07 becomes a countable ordinal in 'M[G].This simple method of collapsing a cardinal is described in the following lemma. We shall present a slightly more subtle way of collapsing in the next section (namely, how to make an inaccessible cardinal accessible).

Lemma 19.9. L.et ti be a regular cardinal and let 1 > ti be a cardinal. There is a notion of forcing ( P , < ) that collapses 1 onto ti, i.e., 1 has cardinality ti in the generic extension. Moreover, (a) every cardinal a I ti in 9 1 remains a cardinal in 9l[G]; and (b) $1'" = 1,then every cardinal u > 1 remains a cardinal in the extension. [The condition in (b) is satisfied, e.g., if G C H holds and cf 1 2 K.] Proof: Let P be the set of all functions p such that: (19.12) (i) dom(p) E ti and Idom(p)) < ti, (ii) ran(p) G 1, and let p c q iff p 3 4.

19.

183

INDEPENDENCE OF THE CONTINUUM HYPOTHESIS AND AC

u

Let G be a generic set of conditions and letf= G. Clearly,fis a function, and it maps ti onto 1;this is because for all a < K and all /3 < 1,the sets

D, = {p : a E dom(p)},

ED = { p : fl E ran(p)}

are dense. ( P , < ) is a-closed for every a < K and therefore all cardinals Iti are preserved. If A<" = 1,then I P 1 = 1 and hence sat(P) I1'; it follows that all cardinals 2 I + are preserved. H If 1 '" > 1 in Lemma 19.9, then not only 1 is collapsed but "'1 instance :

as well. For

Exercise 19.5. In (19.12), let K = K1 and I, = K,. Then in W [ G ] there is a one-to-one function g : K? + K,. [If X is a countable subset of K,, let g ( X ) = the least a such thatf[(a + w ) - a] = X (wheref= G is the collapsing function). Use the fact that X E 91.1

u

If K is singular in (19.12),then (as was the case in adding a subset of collapsed to cf K. One of many ways to verify it:

ti) K

is

Exercise 19.6. In (19.12), let K = K,. Then in !UI[G], there is a one-to-one function g from I, into w . [Let f = U G, and let g(a) = the least n such that the function f I (a,+ - w,) is eventually constantly equal to a.]

,

In the above examples of forcing, either when adding a subset of ti or collapsing a cardinal to ti, we assumed that 2'" = ti. This is because if 2'" > ti, then either notion of forcing collapses 2'" to K. For instance: Exercise 19.7. Let ( P , < ) be the notion of forcing adding a subset of w , via (19.7). and let (Q,< ) be the notion of forcing that adds a mapping of Sl onto 2"O via (19.12). Then r.o.(P)= r.o.(Q). [Let Q'= ( q E Q : dom(q) is an initial segment of w , ; Q'is dense in Q.Show that P has a dense set P' isomorphic to Q': use the fact that every p E P has 2O ' mutually incompatible extensions.] [Another way to show that (P. < ) from (19.7) adjoins a G, and for every g E "{O, l}, let F ( g ) = one-to-one mapping of 2"O into K,: Letf= least a such that f ( a + n) = g ( n ) for all n.]

u

,

Thus adding a subset of w via (19.7) makes the continuum hypothesis true, and we have a forcing proof of the consistency of 2N0= K,.

Independence of the Axiom of Choice If the ground model satisfies the axiom of choice, then so does the generic extension. However, we can still use the method of forcing to construct a model in which AC fails; namely, we find a suitable submodel of the generic model, a model 91 such that 91 c 91 c 91[G]. I shall describe a general method of construction of models without choice in Section 21. In the present section I shall only prove the basic result.

3.

184

FORCING AND GENERIC MODELS

Thorem 45 (Cohen). There is a model of ZF in which the real numbers cannot be well-ordered. Thus the axiom of choice is independent of the axioms of ZF.

Before we construct a model without choice, we shall prove an easy but useful lemma on automorphisms of Boolean-valued models. Let B be a complete Boolean algebra and let n be an automorphism of B. We define, by induction on p ( x ) , an automorphism of the Boolean-valued universe I/”, and denote it also by n: (19.13)

.(a)=

(i) 0; if n(y) has been defined for all y E dom(x), let (ii) dom(nx) = n[dom(x)], and (nx)(ny) = n(x(y)) for all n(y) E dom(nx).

Clearly, n is a one-to-one function of V” onto itself, and n ( i )= 2 for every x. Lemma 19.10. Let cp(x,, . . ., x,) be aformula. l f n is an automorphism ofB, then for all xl, ..., x, E V”,

---.

(19.14)

JJcp(nx1,

nxn)JJ= n(JJcp(x1,

xn)Jl)

Proof: (a) If cp is an atomic formula, (19.14) is proved by induction (as in the definition of IIx E y(J,( J x= ~ 1 1 ) . For instance, IInx

ZYII

= =

c c . .1 (1

t s dom(ny)

z E dom(y)

(Ibx = tll = KZII

. (ny)(t)) . (ny)(nz))

(b) For other formulas, the proof is by induction on the complexity of 4).

In practice, (19.14) is used as follows: Let (P, c )be a separative partially ordered set. If n is an automorphism of P , then n extends to a unique automorphism of the complete Boolean algebra B = r.0. P :

44 = c M P ) : P 5 4

(19.15)

Then (19.14) takes this form: (19.16)

where xl,

p

..., x,

~p(x1,- . * , xn)

1‘ E

iff

np

~ ( n x , ,. . ., nx,)

V”.

Exercise 19.8. If G is an !Ill-generic ultrafilter on B and n an automorphism of B (in !IN), then H = n[G] is !IN-generic and !IN[H] = !Ill[G].

19.

INDEPENDENCEOF THE CONTINUUM HYPOTHESIS AND AC

185

Now, let the ground model 91 be the constructible universe: 911 = L We first extend 911 by adding countably many Cohen generic reals: Let P be the set of all functions p such that: (19.17)

(i) dom(p) is a finite subset of w x w, (ii) ran(p) E (0, l},

and let p < q iff p 2 q. [Note that any one-to-one mapping between (0 x w and w induces an isomorphism between this notion of forcing and the one that adds one Cohen generic real (when the conditions are functions with dom(p) E o).] Let G be a generic set of conditions. For each i E w, let a, = {n E (0 : ( 3 p E G ) p(i, n ) = 1)

(19.18) and let (19.19) Let u, , i (19.20)

A = {ai : i E

w}

w, and A be the canonical names for aiand A :

dom(ui) = {.' : n a,(.') =

( 19.21)

E

E

w)

C { p E P : p(i, n ) = I}

(n E w )

dom(A) = {ui : i E w } A ( u i )= 1

Lemma 19.11. I f i

(i

E

w)

# j , then every p forces ui # u j : /(a,= U j l ( = 0

ProoJ For every p there exists q =) p and n

E

w such that q(i, n)= 1 and

q ( j , n) = 0.

From now on, let us work in 'X)I[G].Let 9 be the class of all sets hereditarily ordinal-definable over A : 9 = HOD(A) As we have shown in Section 15, 9 is a transitive model of ZF. Since the elements of A are sets of integers, it is clear that A E 9.We shall show that the set of all real numbers cannot be well-ordered in the model 9 (in fact that A cannot be well-ordered). If R" is well-orderable in 9,then there is a one-to-one mappingfE 9 of Pm(o)into the ordinals. The following lemma shows that this is not true: Lemma 19.12. (In "m[G]) There is no one-to-onefunctionf: A + Ord, ordinaldefinable over A.

Proof: Assume that f: A 3 Ord is one-to-one and ordinal-definable over A. Then there is a finite sequence s = (xo, . . . , x r > in A such that f is ordinal-

3.

186

FORCING AND GENERIC MODELS

definable from s and A. Sincefis one-to-one, it is easy to see that every a E A is ordinal-definable from s and A. In particular, pick some a E A that is not among the x i , i 5 k. Since a E OD[s, A] (in W[G]), there is a formula cp such that (19.22)

W[G] k a is the unique set such that q(a, al, .. ., a,,, s, A )

for some ordinals al, . . ., a,, . We shall show that (19.22) is impossible. Let a be the canonical name for a, and let xo, . . . ,xk be the canonical names for xo , . . . , xk (see 19.20); let s be the obvious canonical name for the sequence (xo , . . . , x k ) . We shall show the following: (19.23) For every po that forces q(u. Oi,, . . . ,Oi, , s,A ) there exists band q Ipo that forces a # b and cp(b,Oil, ..., din, s, A ) . Let po 11 q(a, Oil, ..., i,,, s, A). Let i, io, ..., i k be such that a = a , , xo = uior. . . , xk = u i k .Let us pick j E w such that j # i, and that for all m E w, ( j , m) # dom(p,). See Fig. 19.1.

‘0

i

‘k

J

FIGURE 19.1

Now let n be the permutation of w that interchanges i and j , and nx = x otherwise. This permutation induces an automorphism of P: for every p E P , (19.24)

dom(np) = {(nx, m ): (x, m) E dom(p)} (.p)(.x,

m ) = P ( X , m)

( x E dom(p))

In turn, n induces an automorphism of B, and of 91’.It is easy to see (cf. 19.20, 19.21) that (19.25)

n(a,)= a j ,

n(a,) = a,

n(uj)= a, n(A)= A

(all x # i, j )

n(s)= s

and

Since ( j , m ) # dom(po), for all m,it follows that (i, m ) # dom(np,), for all m,and thus po and npo are compatible. Let q = po u np,. Now, on the one hand we have PO 11~

( a ih,, 9

* * *

9

hn

9

S, A )

20.

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MORE GENERIC MODELS

and on the other hand, since nti = 4, ns = s and nA = A, we have np,

ti,, ..., G", s, A )

Ik q+j,

Hence q ~kcp(ai,...) and cp(aj,...)

and by Lemma 19.11,4 Ik ai # a j . Thus we have proved (19.23),which contradicts (19.22). W 20. MORE GENERIC MODELS

Products of Partially Ordered Sets Let (Pi, < ) and (P,, < ) be two partially ordered sets. The product ( P , <) of (P,, < ) and (P,, < ) is defined as follows: P = P, x P,

(20.1) (Pi, P 2 )

5

(41,

42)

iff

Pi

41 and

P2

5

42

Note that (pl, p , ) and ( q , , q , ) are compatible iff p 1 is compatible with q , , and p 2 is compatible with q r ; they are incompatible if either p 1 is incompatible with 4 , . or p , is incompatible with 4 2 . We shall investigate generic models obtained via products of notions of forcing. The first thing worth noticing is the following: Exercise 20.1. If

r . 0 . P I = r.0. Q1 and r.o.(Q1 x Q2) (up to isomorphism).

r.0.

Pz = r.0. Q z , then

r.o.(P1x P2)=

Thus one can define a corresponding operation 0 on complete Boolean algebras: If B , and B, are complete Boolean algebras, let

B , 0B ,

= r.o.(P, x

P2)

where P,, P, are partially ordered sets, dense in B , and B, . We call B , 0B , the direcf sum of B , and 8 , . (In fact B , 0 B , is the direct sum of Boolean algebras B , , B , , defined algebraically.) Exercise 20.2. The Stone space of B1 @ B2 is the topological product of the Stone spaces of El and B2.

While direct sum of complete Boolean algebras is important in the theory of forcing, its dual notion, the direct product of algebras is uninteresting. The direct product of B , and B, corresponds to the disjoint union of their Stone spaces (see also Figure 17.1). If B is the direct product of B,, B,, then there is a E B such that B, is isomorphic to B, and B , is isomorphic to B - ~Clearly, . if G is a generic ultrafilter on B, then either a E G and W[G] = W[G n B J or the same thing for - a .

3.

188

FORCING AND GENERIC MODELS

If G is a generic filter on P, x P, , let G , and G, be the projections of G:

(20.2)

G , = {pI E P I : ( p , , p , )

E

G for some p , E P 2 )

G2 = {p, E P , : ( p , , p,) E G for some p ,

E

P,}

It is fairly easy to show that G, is generic on P, and G 2 is generic on P , , and G = G I x G , (see the forthcoming lemma). On the other hand, to obtain a generic filter on the product, we cannot just take any two generic filters G and G, on P, and P , :

,

Exercise 20.3. Show that except in trivial cases, G x G is never generic on P x P .

The characterization of generic filters on products is given in the following lemma.

Lemma 20.1. (Product Lemma). Let P, and P2 be two notions ojforcing in 91. In order that G E P , x Pz be generic over 91, it is necessary and sufJicient that G = G1 x G , where G1 E P , is generic over 91 and G , E P , is generic over 91[G,]. Moreover, 91[G]= rU1[Gl][G,]. As a corollary, if G1 is generic over 91 and G , is generic over 9I[G

due

to

symmetry

GI

is

generic

~1[GlI[G21= 91[G,I[GlI*

rU1[G2], and

over

,

we

,I, then have

,

Proof: First let G be an 91-generic filter on P x P , . We let G and G 2 be the projections of G (20.2). Clearly G, and G , are filters, and G E G , x G , . If ( p , , p , ) E G1 x G , , then there are p’, E G , and p ; E G , such that (p’,, p,) E G and ( p , , p i ) E G. Since G is a filter, there exist q1 I p , , p i and q2 I p , , p i such that (q,, 4,) E G. Hence (p,, p , ) E G and we have G = G , x G , . It is easy to see that G, is generic over 9.R: If D1E 91 is dense in P,, then D, x P , is dense in P, x P,; and since (Dlx P , ) n G # 0, we have D, n G, # 0. To show that G2 is generic over 9.R[G1],let D, E 9.R[G1]be dense in P , . Let IF be the forcing relation corresponding to P,. Let D, be a name for 0,and let p1 E G I be such that p1 forces “ D , is dense in p,.” Let p 2 E G2 be arbitrary. For every q , 5 p i and every q2 I p 2 there exist r l Iq1 and r2 I4, such that rl IF E D,; thus

;,

D = {(rl, r , ) : rl 5 p , and rl

IF

i., E 0,)

,

is dense in P , x P2 below ( p , , p z ) and so there are r,, 1, such that r l E G and rl IF 3, E D , . Hence r2 E 0,n G,. On the other hand, let G1 E P , be 9.R-generic and let G , c P 2 be %JI[G,]-generic. We let G = G, x G2 . Clearly G is a filter on P x P , . To show that G is %%generic, let D E 9.R be dense in P, x P,. We let D2 = (PZ

: (pi, PZ) E

D for some P I E GI}

The set 0, is in 9J[G1]; we shall show that 0,is dense in P , and thus

D

n (G, x

G2)#

0.

20.

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MORE GENERIC MODELS

Let q ,

E

P , be arbitrary. Since D is dense in P x P,, it follows that the set

D , = {Pl ($2 I 42)(P1* Pz) E Dl is dense in PI. Hence there is p1 E G I n D1 and so D, is dense in P,. Since G , x G, E 'JJI[G,][G,],it is obvious that 911[Gl x G,] = '!Ul[Gl][G2].

Products and k--Products of a Collection of Partial Orders Let Pi,i E I , be a collection of partially ordered sets. We define the product of P i , i E I , as a generalization of the product of two partially ordered sets. The product of partially ordered sets is closely related to the product of topological spaces. To simplify the definition, we shall assume that all the Pi have a greatest P i ,we let element 1. For every p = (pi : i E I ) E

Hi

(20.3)

s ( p ) = { i E I : pi # 1)

(the supporr of p ) and we let P ={ pE (20.4)

H Pi : s ( p ) is finite) i6I

p Iq

iff

pI I g, for all i

In practice, we consider finite functions p with dom(p) G I , and assume that p ( i ) = 1 whenever i # dom(p). 20.4. Compare the product of P , , i E I , with the topological product of the spaces P, whose topology is given by the basic open sets U p ,= {q, E P, : q, 5 p , ) .

Exercise

20.5. If r.o.(P,) = r.o.(Q,) for each i are the respective products. Exercise

E

I , then r.o.(P) = r.o.(Q),where P and Q

Exercise 20.6. If P is the product of P , , i E 1. and if S is the topological product of the Stone spaces of r.o.(P,),i E I , then r.o.(P) is (isomorphic to) the algebra of regular open sets of S.

If G is a generic filter on the product of Pi, i E I , then the projections Gi, i E I , are generic filters on Pi. An example of a product of notions of forcing is the notion of forcing (19.I), adding K subsets of to. There is an obvious isomorphism between this and the product of k- copies of the notion of forcing that produces a single Cohengeneric real. The notion of forcing adding subsets of a regular cardinal k- can also be considered a product; however, since the conditions are not finite, we have to generalize the notion of product: We define the k--product of P I , i E I , as follows: (20.5) P ={pE P i :Is(p)l < k-J

fl

iel

p I q

iff

pi I qi for all i

3.

190

FORCING A N D GENERIC MODELS

Chain Condition in Products and K-Products Since we want to use products and K-products of notions of forcing, let us investigate the properties of products that are of significance in the theory of forcing. To start, let us make the following trivial observation: Lemma 20.2. Let K be a regular cardinal, and let P be the ti-product of P i , i Let a < ti be an injinite cardinal. If each Piis a-closed, then so is P .

E

I.

Proof: Let a < ti and let pi = (p' : i E I ) , 5 < a, be a descending a-sequence of conditions in P. If we let s = UC<, s(p<), then 1st < K, and since each P i is u-closed, it is easy to find p = (pi : i E I) such that s ( p ) = s and that pi I p: for each i E I and each ( < a. W It is more interesting to determine the chain condition of a product (or a ti-product). It turns out that sat(P) does not depend on the number of factors in the product; it depends (a) on ti, and (b) on sizes of the factors. The chain condition of a product (or a ti-product) can be evaluated by using the method of Lemmas 19.3 and 19.8.The following lemma gives an estimate of sat(P) in the most important cases of the ti-product.

Lemma 20.3. Let ti be a regular cardinal, and let P be the K-product of P i , i E I . Let W be an incompatible subset of P . (a) If each I PiI 5 ti, then I W I I 2'". (In particular, if ti = KO then W is at most countable.) (b) I f 1 2 t i a n d i f e a c h ( P i [ I 1 , t h e n IWI 51'". ( c ) If1 > ti is inaccessible and each I Pi I < 1,then I W I < 1. Proof: If p = ( p i : i E I) and q = (qi : i E I) are incompatible in P, then for some i E s ( p ) n s(q), pi and qi are incompatible in Pi, and in particular pi # q i . Thus we can regard elements of Was functions whose domain is a subset of I of size < ti, with values in the P i , and show that if W consists of pairwise incompatible functions then I W 1 has the required bound. This we do precisely as in Lemma 19.8 (or 19.3): We construct A,

A, c

... & A,

G

...

and

W,

E

u

W, E ... G W, E . . *

(a <

u,.,

ti)

where each W, is a subset of W and A, = (dom(p) : p E W,}.Construction of the W, is described in Lemma 19.8, and it is shown there that W = W,. The bound on I W 1 is verified by induction on a < ti. For example, let us prove (b); then (a) holds because ti'" = 2'" if K is regular, and (c) is proved similarly. If I W, I I A<", then obviously I A, I 5 ti .A'" = A'". If a < K is a limit and if I W,I I 1'" for each /3 < a, then 1 W,I s lal.A<" = Thus let us assume that I W,l I1'" and let us show that I W,, I I The set W,, is obtained by adding to W, at most one q for each p E P with dom(p) c_ A,. However, since

,

,

20.

191

MORE GENERIC MODELS

IA.1 I I < " ,there are at most " '1 functions p with dom(p) E A,, Idom(p))< K, and L possible values for each i E dom(p). Thus 1 W,, I I A'". It follows that IWI < I < " .

The relationship between the chain condition of a product and the chain conditions of the factors, even in the case of a product of two factors P x Q, is more complicated. The statement that the product of two C.C.C. partial orderings satisfies the C.C.C. is independent of ZFC; see Exercise 22.4 and Theorem 53 (Section 23). One can prove only that if P and Q satisfy the c.c.c., then sat(P x Q) I (2'O)' (see Exercise 29.2), and the bound is best possible.

The Livy Collapse In Lemma 19.9 we have described a notion of forcing that collapses a cardinal I onto a smaller regular cardinal K. We shall now present a technique (due to Levy) of collapsing all cardinals below an inaccessible cardinal L while preserving I, thus making I a successor cardinal in the generic extension. Lemma 20.4. Let K he a regular cardinal and let I > K be an inaccessible cardinal. There is a notion offorcing ( P , < ) such that: (a) every a such that K I a < I has cardinality K in 'XII[G];and (b) every cardinal I K and every cardinal 2 I remains a cardinal in W[G]. In particular, 'XII[G] 1 I = 'K Proof: For each a < I, let P , be the set of all functions pa such that

(20.6)

dom(p,) c

K

and

Idom(p,)l < K,

ran(p,) G a

and let p?, < q1 iff p 3 3 q?,. Let ( P , < ) be the K-product of P I , a < I. Note that we can regard the conditions p E P as functions on subsets of I x K such that (20.7)

(i) Idom(p)I < K ; (ii) p(a, 5 ) < a for each (a,5 ) E dom(p).

Let G be a generic set of conditions; and for each a < A, let G , be the projection of G on P, . Then G , is a generic filter on P a ; and as in Lemma 19.9, the setf, = G,is a function that maps K onto a. Thus W[G] k I a I I I K 1, for every a < I. The notion of forcing ( P , <) is y-closed for every y c K, and hence it preserves all cardinals and cofinalities I K. In particular, K is a cardinal in 9l[G]. By Lemma 20.3, ( P , < ) satisfies the I-chain condition. Hence I remains a cardinal in 9 I [ G ] ,and so do all cardinals greater than I. It follows that in 'XII[G], 1 is the cardinal successor of K.

u

3.

192

FORCING A N D GENERIC MODELS

The Continuum Function 2" for Regular Cardinals ti In Section 19 we have presented a method of adjoining a large number of subsets of a regular cardinal K, thus making 2" as large as prescribed. We shall now describe a construction that does this simultaneously for a collection of regular cardinals. This method completely solves the problem of the behavior of the continuum function at regular cardinals. We recall that for each K, cf(2") > K.

Theorem 46 (Easton). Let 'XI1 be a transitive model ofZFC and assume that the generalized continuum hypothesis holds in 91. Let F be a function (in 91)whose arguments are regular cardinals and whose values are cardinals, such that for all regular K and I : (20.8)

(i) F(K)> K ; (ii) F(K)IF(I) whenever (iii) cf F(K)> K.

ti

I I;

Then there is a generic extension 'XI1[G] of 91 such that 'XI1 and 'XI1[G]have the same cardinals and cofinalities, and for every regular K , W[G] 1 2" = F(K) Remark 1. We have to point out that the generic extension is obtained by forcing with a class of conditions. By Lemma 19.4, a notion of forcing can only increase the size of 2" for K < I r.o.(P) I ; thus we have to use a class of conditions. We shall describe the appropriate generalization of the forcing method.

Remark 2. Theorem 46 leaves open the problem of the behavior of 2" at singular cardinals. As I have remarked in Section 8, unless a certain large cardinal hypothesis holds, cardinal exponentiation is completely determined by the continuum function on regular cardinals (see Lemma 8.1). Besides, although some conditions on the continuum function are known (see Sections 6 and 8), n o "necessary and sufficient " description has been formulated. The problem of adding subsets of singular cardinals is considerably more difficult (and of necessity involves large cardinals), and although substantial progress has been made recently, there are still many open questions and no analog of Easton's theorem for singular cardinals seems to be within reach.

Since the proof of Easton's theorem involves forcing with a class of conditions, we shall first give a proof of the special case, when the "continuum function " F is prescribed for only a set of regular cardinals. Thus let us work in a ground model !XI that satisfies the GCH and let F be a function defined on a set A of regular cardinals and having the properties (20.8)(ik(iii). For each K E dom(F), let (P", 3 )be the notion of forcing that adjoins F ( K ) subsets of K (cf. (19.10)):

(20.9) dom(p,) G

K x

F ( K ) and Idom(p,)l < K,

ran(p,)

E (0, 1)

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MORE GENERIC MODELS

We let ( P , < ) be the Easton product of P,, ti E A : A condition p is a function p = ( p , : K E A ) E P , such that if we denote s ( p ) = {ti E A : p , # $3)(cf. 20.3), then

nKEA

(20.10)

1s(p) n y I < y

for every regular cardinal y,

We can regard the conditions as functions with values 0 and 1, whose domain consists of triples (ti, a, p) where K E A , a < ti, and p < F(ti), and such that for every regular cardinal 7,

I {(ti, a, p) E dom(p) : K I y} I < y

(20.11)

(and p is stronger than q iff p 3 q). Note that (20.1 1) implies that for each A, Idom(p,)( < ti. where p , is defined by

ti E

PA4

B) = P ( K

a,

a)

Let G be a generic set of conditions, and let for each K E A, G , be the projection of G on P ,. Each G , is a generic filter on P , and thus produces F ( K ) new subsets of ti: a; = { a < K : ( 3 p E G)p(ti, a, p) = 1)

(p < F ( t i ) )

We shall show that ( P , < ) preserves cardinals, and that each K E A has exactly F ( K )subsets in WI[G].The condition (20.11) is instrumental in the proof. Given a regular cardinal 1,we can break each condition p E P into two parts : (20.12)

p"

=

~I{(K,

a, 1): K 5

A},

p"

=

~ I { ( K , a, p ) : K > 1)

Clearly p = p" u p > a .We let p > a = {p'" * P E P ) p-' is the Easton product of P ,, K E A and K > 1. Moreover, P is (isomorphic to) the product Ps' x P>'. First we notice that P>' is 1-closed: If C I P'A consists of pairwise compatible conditions and ICI I 1, then p = U C is a condition in P"; (20.11) holds for all regular y > 1, and holds trivially for y II because if (ti, a, p) E dom(p), then K > 1. Furthermore, P' satisfies the L+-chaincondition: If W c P s a is incompatible, then I W I I1.The proof given in Lemma 19.8 works in this case as well because Idom(p)( < I for each p E P" (and because the GCH holds). Thus P = P>' x Ps' where P>' is bclosed and sat(PsA)
Lemma 20.5. Let G x H be an W-genericjilter on P x Q, where P is I-closed Then every function f: A + !IN in W[G x HI is in !IN[H]. I n particular, pWG x "](A) = p'WW(1) and sat(Q) I. '1

3.

194

FORCING AND GENERIC MODELS

Proof: Letfbe a name f o r k let us assume, without loss of generality, that for some A, [Ifis a function from 1 into A'" = 1. For each a < 1, let D, E P be defined (in 'xn) as follows: p E D, iff there is a maximal incompatible W G Q and a family {a',".b: q E W) such that for each q E W,

(20.14)

(p, 4 )

lv(q= a:;

We claim that each D, is open dense in P. Clearly, D, is open; thus let po E P be arbitrary and let us find p E D,such that p I po . There exist p I p o , q 1 E Q and a , E A such that ( p , , 4,)Ikf(G) = 6,. By induction on y < A+, we construct p y E P, qy E Q, and ay E A such that p o 2 p 2 . . * 2 p y 2 . . . ,that the qy are pairwise incompatible and that (pyrq,,)forcesf(6) = 6,. If { q s :5 < y} is not maximal incompatible, we can find such p , , , q y ,and a,, since P is I-closed. By the 1+-chain condition, there is p < I + such that W = { q y: y < p} is maximal incompatible; then we find p E P stronger than all p y , y < fl. Thus D, is open dense in P. Since P is 1-closed, it follows that a D,is open dense, and so there exists p E G such that p E D, for all a < 1.We pick (in 'm) for each a < 1 a maximal incompatible W, E Q and a family {a(,,\: q E W,} such that (20.14) holds for each q E W, . By genericity of H, for every a there is a unique q E W, such that q E H,and we have, for every a < 1,

,

(20.15)

f ( a ) = a::,

where q is the unique q

However, (20.15)defines the function f i n 9l[H].

E

W, n H

H

Now we can finish the proof of Easton's theorem, that is, at least in the case when F is defined on a set A of regular cardinals. Let K be a regular cardinal in 91;we shall show that K is a regular cardinal in 91[G]. If K fails to be a regular cardinal, then there exists a function f that maps some 1< K, regular in 91, cofinally into K. We consider P as a product: P = P>' x P". Then G = G>' x G" and 9 l [ G ] = 911[G>'][G5'] = 9Il[G"'][G'']. By Lemma 20.5, f is in 9ll[G"] and so K is not a regular 1 IK cardinal in 'J)l[G"']].However, this is a contradiction since sat(P") I ' and hence K is regular in 9l[G"]. It remains to prove that (2'rmG1= F(1), for each 1~ A. Again, we regard P as a product P>' x P" and G = G>' x Gs'. By Lemma 20.5, every subset of 1 in 9)1[G] is in 91[Gs'] and we have (2'rWc1= (2')mc'Al. However, an easy computation shows that IPS'( = F(1); and since sat(P") = A + , we have lr.o.(Psa)l = F(1) and hence (2'r)'mlGl5 F ( I ) . O n the other hand, we have exhibited F(1) subsets of 1for each 1E A, and so 9ll[G] k 2A= F(1). H

20.

1%

MORE GENERIC MODELS

Exercise 20.7. The singular cardinals hypothesis holds in Easton's model.

then

[If K is singular then every$ c f ( K ) -+ K is in 91 = 91[Gscf"],and so if F(cf K ) < K ( K ~ ~ =' (K" ~ ')'I ~ ~I;(2K)* ~ Ilr.o.(Ps cf ' ) I K = (F(cf K))" = K+.]

Forcing with a Class of Conditions We shall now show how to generalize the preceding construction to prove Theorem 46 in full generality, when the function F is defined for all regular cardinals (in 91).This generalization involves forcing with a proper class of conditions. Although it is possible to give a general method of forcing with a class, we shall concentrate only on the particular example. Thus let 911 be a transitive model of ZFC + GCH. Moreover, we assume that 91 has a well-ordering of the universe (e.g., if 91satisfies I/ = L). Let F be a function (in 91) defined on all regular cardinals and having the properties (20.8)(i)-(iii). We define a class P of forcing conditions as follows: P is the class of all functions p with values 0 and 1, whose domain consists of triples (K, a, fi) where K is a regular cardinal, a < K and fi < F(K),and such that for every regular cardinal y, (20.11) I {(K, a, f i ) E dom(p) : K 5 Y ) I < Y (and p is stronger than q iff p 2 4). As before, we define P" and P" for every regular cardinal I,. Note that P" is a set. To define the Boolean-valued model 91' and the forcing relation, we use the fact that P is the Easton product of P,, K a regular cardinal. For each regular A, we let B A = r.o.(P"). If A < p then the inclusion P" s P'" defines an obvious embedding of BAinto B,; thus we arrange the definition of the B A so that BAis a complete subalgebra of B, whenever I, < p. Then we let B= B A . B is a proper class; otherwise it has all the features of a complete Boolean algebra. In particular, X exists for every set X c B. Also, P is dense in B. T o define 911', we cannot quite use the inductive definition (18.8) since B is 91IBi;the formal definition of 91' not a set. However, we simply let 91' = does not present any problem. Similarly, todefine IIx E ylJ and Ilx = yll, we first and I 5 p, then IIx E yllB1= IIx E yllB, and so we let notice that if x, y E IJxE yll = JIxE ~ ~ where 1 1A is ~ such~ that x, y E %IBi. The same for IIx = y(1. As for the forcing relation in general, we cannot define llqll unless cp is X does not generally exist if X s B is a class. restricted; this is because However, we can still define p I1 q using the formulas from Lemma 18.8. Now, we call G c P generic over 1' 11 if (i) p c q and p E G implies q E G,(ii) p , q E G implies p u q E G,and (iii) if D is a class in 91 and D is dense in P, then D n G # 0. The question of existence of a generic filter can be settled in a more or less the same way as in case when P is a set. One possible way is to assume that 1 ' 11is

uA

1

uA

1

3.

1%

FORCING AND GENERIC MODELS

a countable transitive model. Then there are only countably many classes in W and G exists. Another possible way is to use the canonical generic ultrafilter. It is the class G in ’IllB defined by G($)= p for all p E P (here we need the assumption that 3ll is a class in %IB). Thus let G be an W-generic filter on P.For every regular 1,G A= G n P 5 is generic on P“. If x E W B A and 15 p, then iG1(x) = iG,(x),and so we define i c ( x ) = ic,(x) where 1 is such that x E WB1. Then we let W [ G ]= ic[Wm”];it follows that W[G] = W[G,]. Using genericity of G and properties of the forcing relation, we get

’

uA

Forcing Theorem :

(20.16)

9131[C] lk cp(xl,

where x . . . , x,

E

. . ., x,)

iff

9NB are names for x

(3p E G)p 1 cp(xl, . . ., x,)

. . ., x,

.

The formula (20.16) is proved first for atomic formulas and then by induction on cp; in the induction step involving the quantifiers, we use the fact that G intersects every dense class of W. We shall now show that W [ G ]is a model of ZFC. The proofs of all axioms of ZFC except power set and replacement go through as when we forced with a set. It is no surprise that the power set and replacement axioms present problems. It is easy to generalize the constructions of Section 19 either to get a class of forcing conditions adding a proper class of Cohen reals, or a class of conditions collapsing Ord onto o.The present proof of the power set and replacement axioms uses the fact that for every regular 1 (or at least for arbitrarily large regular A), P = P I x P where P, is 1-closed and P is a set and sat(P,) 5 1’. (Here P = P” x P “.) Power set. Let 1 be a regular cardinal. Lemma 20.5 remains true even when applied to P’’ x P“. It does not matter that each D, in (20.14) is a class. The “sequence” of classes (D, : a < A ) can be defined (e.g., as a class of pairs { ( p , a ) : p E D,)) and since P>’is 1-closed, the intersection D, is dense, and there exists p E G n P” such that p E D, for all a c 1.The rest of the proof of Lemma 20.5 remains unchanged, and thus we have proved that every subset of L in %l[G]is in 9lI[G,]. Since P 5’ is a set, it follows that the power set axiom holds in W[G]. Replacement. To show that the axioms of replacement hold in Yt[C], we combine the proof for ordinary generic extension with Lemma 20.5. It suffices to prove that if in YJ[G], q(a. u ) defines a function K : Ord -+9l[G], then K[A] is a set in !Ul[C]for every regular cardinal 1. Without loss of generality, let us assume that for every p E P,

nacA

(20.17) p

IF

for every a there is a unique u such that cp(a, v).

Let 1be a regular cardinal, and let us consider again P = P>’x P “, and G = (G n P>’)x G1. As in Lemma 20.5, let us define, for each a < 1,a class

21.

197

SYMMETRIC SUBMODELS OF GENERIC MODELS

D, c P>': p E D, iff there is a maximal incompatible W E P S a and a family {a',".;: q E W } such that for each q E W , (20.18) As in Lemma 20.5, each D,, a < I, is open dense; since P" is I-closed, D, is dense and there exists p E G n P >' such that p E D, for all a < I. We pick (in W )for each a < 1 a maximal incompatible W, E P s and a family {a',".;: q E W,} such that (20.18) holds for each q E W,. Now if we let S = {at; : a < I and q E W,}, then it follows that

nu<

K[1] s i,[S] However, i,[S] is a set in 9UI[G]: there is y such that S

c 9UIBy, and we have ic[S] = ic,[S] E 9UI[G,]. Thus 91[G] is a model of ZFC and it remains to show that 9UI[G] has the same cardinals and cofinalities as 91, and that in 9UI[G], 2" = F(K)for every regular cardinal k-. However, this is proved in exactly the same way as when we forced with a set of Easton conditions. H Exercise 20.8.* Let 91 be a transitive model of ZFC. There is a generic extension 9)1[G]

such that !Ut[C]satisfies the GCH. [For each a, let P , be the notion of forcing which collapses 1 = a,+ I onto K = (aa)' (see 19.12).P , is >,-closed and sat(P,) = 1'. Let P be an Easton product ofP,,a E Ord; namely, we require that 1 s ( p ) n y I < y for every inaccessible y = aa.Show that for each a, K = (3,)' is a cardinal in !UUI[G],K = K?!?],and 9I[C] 1 2N1= K, + Apply Lemma 20.5 in two ways: (a) for each a, consider Ps' x P": sat(Psu)= a:+, and P'* is 2=+,-closed; (b) if a is inaccessible and a = &,consider P < a x P r o :sat(P <') = 3; and P is a,-closed.]

,.

We conclude the section with a remark on the Bernays-Gbdel axiomatic set theory. At the end of Chapter 1,1 made a remark to the effect that if a sentence involving only set variables is provable in BGC = BG + Axiom E, then it is provable in BG + AC. This is a consequence of the following: If 911 is a transithat has the tive model of BG + AC, then there is a generic extension 91[C] same sets and has a choice function F defined for all nonempty sets. The forcing conditions p E P used in the proof are choice functions whose domain is a set of nonempty sets (and p < q iff p 3 q). The proof that !UUI[G]is a model of BG is rather easy since no new sets are added (P is k--closed for all k-). The generic filter on P defines a choice function F = G , and F is defined for all nonempty sets X E 911[G].

u

21. SYMMETRIC SUBMODELS OF GENERIC MODELS

In Section 19, we constructed a model of set theory in which the reals cannot be well-ordered, thus showing that the axiom of choice is independent

3.

198

FORCING AND GENERIC MODELS

of the axioms of ZF. What follows is a more systematic study of models in which the axiom of choice fails. I shall present a general method of construction of submodels of generic extensions. The construction uses symmetry arguments similar to those used in Theorem 45, and the models obtained are generally models of ZF and d o not satisfy the axiom of choice. This method has been widely used t o obtain a number of results about the relative strength of various weaker versions and consequences of the axiom of choice. The main idea of the construction of symmetric models is the use of automorphisms of the Boolean-valued model V Band the "symmetry lemma " 19.10. In fact, the idea of using automorphisms of the universe to show that the axiom of choice is unprovable dates back into the preforcing era of set theory. I shall describe this older construction first. To d o it, we have to modify somewhat the axioms of set theory.

Set 'Zheory with Atoms (ZFA) In this modified set theory we have not only sets, but also additional objects, atoms. These atoms d o not have any elements themselves but can be collected into sets. Obviously, we have to modify the axiom of extensionality, for any two atoms have the same elements-none. The language of ZFA has, in addition t o the predicate E, a constant A. The elements of A are called atoms; all other objects are sets. Axiom A.

If a E A, then there is no x such that x

E a.

The axiom of extensionality takes this form: (21.1) If two sets X and Y have the same elements, then X = Y. All the other axioms of ZF remain unchanged. In particular, the axiom of regularity states that every nonempty set has an €-minimal element. This minimal element may be an atom. The effect of atoms is that the universe is no longer obtained by iterated power set operation from the empty set. In ZFA, the universe is built up from atoms. Ordinal numbers are defined as usual except that one has to add that an ordinal does not contain any atom. For any set, let us define the following cummulative hierarchy: (2 1.2)

P O ( S ) = S,

P ( s )=

U P@(S)

if a is a limit

@
Pa+'(s)= P(S)u P(P"(S)) P"(S) =

u

P(S)

a E Ord

It follows that (21.3)

V = P"(A)

21.

199

SYMMETRIC SUBMODELS OF GENERIC MODELS

and that the kernel, the class POo(@)of “ hereditary ” sets, is a model of ZF. If A is empty, then we have just ZF; right now we are more interested in the case when A is infinite.

+

+

Exercise 21.1. The theory ZFA AC “ A is infinite” is consistent relative to ZFC. [Construct a model of ZFA. Let C be an infinite set of sets of the same rank (so that X $ TC( Y ) for any A’, Y E C). Consider one X o E C as the empty set, and all other X E C as atoms. Build up the model from C by iterating the operation P * W = P ( Z ) - (01.1

Permutation Models of ZFA

While in ZF, the universe does not admit nontrivial automorphisms, the important feature of ZFA is that every permutation of atoms induces an automorphism of V : If n is a one-to-one mapping of A onto A (a permutation of A), then we define for every x (by €-induction): K ( X ) = K[X] = { n ( t ): t E x) (21.4) Clearly, K is an E-automorphism, and we have K(X) = x for every x in the kernel P“ (0). We use these automorphisms to construct transitive models of ZFA. First we point out that the analog of Theorem 3 1 in Section 11 is true in ZFA: If W is a transitive, closed, almost universal class and if A E 9.Jl (or at least A n W E a),then 91 is a model of ZFA. Let Y be a group of permutations of a set S. A set 9of subgroups of Y is a normalfilter on Y,if for all subgroups H , K of Y: (21.5)

(i) 3 ’ E 9; (ii) if H E 9 and H E K,then K E 9; (iii) if H E 9 and K E 9, then H n K E 9; (iv) if n E Y and H E 9 then ~ H n - E l 9. For a given group of permutations Y of the set A of atoms and a normal filter 9 on 9,we say that x is symmetric if the group sym,(x) = {n E Y : K ( X ) = x} (21.6) belongs to 9. Let us further assume that (21.7)

sym,(a) E 9

for all a

E

A

that is, that all atoms are symmetric (w.r.t. Y and 9) and let all hereditarily symmetric objects: (21.8)

23 = {x : every z

E TC({x}) is

‘1) be

the class of

symmetric}

The class 23 (a permutation model) is a transitive class and obviously includes the kernel (because sym(x) = Y for all x E P m ( 0 ) ) ,moreover, all atoms are in SandAEB.

3. FORCING A N D GENERIC MODELS

200

Lemma 21.1. 9 is a transitioe model of ZFA. Proof: We have to show that '$ isIclosed under Godel operations and almost universal. It is easy to see that &(nx, ny) = n(&(x, y ) ) for i = 1, ..., 10, and therefore

(21.9)

sym(&(x, y ) ) 2 syrn(x) n sym(y)

(i = 1, . . . , 10)

It follows that if x and y are hereditarily symmetric, then so is &(x, y ) . To show that 9 is almost universal, it suffices to verify that for each a, 2) n P"(A) is symmetric. For all x and all n E Y we have rank(nx) = rank x. Also, (21.10)

sym(nx) = n . sym(x) . n-'

and hence, by property (iv) in (2t.5), if x is symmetric and x E Y,then n ( x )is symmetric. Thus for all n E Y we have n[% n P"(A)] = 2) n P " ( A )and therefore, sym(2) n P " ( A ) )= 9. In the following examples we construct permutation models as follows: For every finite E c A, we let (21.11) fix&) = {n E Y : nu = a for all a E E } and let 9be the filter on Y generated by {fix(E): E c A is finite}..Fis a normal filter since n . fix(E) . 11-l = fix(n[E]), and satisfies (21.7). Thus x is svmmetric if and only if there exists a finite set of atoms E , a support for x, such that n(x)= x whenever n E 9 and nu = a for all a E E. We shall now give two examples of permutation models. Example I. Let A be infinite, and let Y be the group of all permutations of A. Let 4 be generated by {fix(E): E c A is finite), and let 3 be the permutation model.

Lemma 21.2. In the model 3 from Example I , the set A , although infinite, has no countable subset. Hence the axiom of choice fails in 21. PrmJ Assume that there existsf'€ 3 which is a one-to-one mapping of ( 1 ) into A. Let E be a finite subset of A such that ?f=ffor every n E fix(E). Since E is finite, there exists a E A - E such that a = f ' ( n )for some n ; also, let b E A - E be arbitrary such that b # a. Now, let n be a permutation of A such that nu = b but AX = x for all x E E . Then ?f=.L and since n is in the kernel, we have nn = n. It follows that n ( . f ( n )= ) (?f)(nn)= . f ( n L however,.f(n) = a while n ( , f ( n )= ) nu # a. A contradiction. Example I/. Let A be a disjoint countable union of pairs: T I

(21.12)

P,.

A = n=O

P,= { a n ,h,]

H

21.

SYMMETRIC SUBMODELS OF GENERIC MODELS

201

and let Y be the group of all permutations of A such that n({a,, b,}) = {a,,,b,}, for all n. Let B be generated by {fix(E):E c A is finite}, and let % be the permutation model. Lemma 21.3. In the model %from Example 11, ( P , : n E o}is a countable set of

pairs and has no choice function. ProoJ: Each P, is a symmetric set since n(P,) = P, for all n E Y. For the same reason, R ( P , : n E w ) = n({(n, P,) : n E w } )= (P,: n E w ) , for all n E 9,and so (P, : n E o) E 2). Hence S = {P, : n E o}is a countable set in 9. We show that there is no function f E 9 such that dom(f ) = S and f (P,)E P, for all n. Assume thatfis such a function and let E be a support off: There exists n such that neither a,, nor b, is in E, and we let n E Y be such that nu, = b, but nx = x for all x E E. Then nf =L nP, = P, , and so

n ( f (Pn)) = (nf )(nPn) = f (pn) but n(f (P,))= b, while f (P,)= a,; a contradiction. W Exercise 21.2. We recall that a set is called D-finite if it has no countable subset. Show that in Example I1 A is D-finite. Exercise 21.3. Show that in either Example I or 11, A cannot be linearly ordered.

The method of permutation models gives numerous examples of violation of the axiom of choice. One usually uses the set of atoms to produce a counterexample (in the permutation model) to some consequence of the axiom of choice, thus showing the limitations of proofs not using the axiom of choice. (A typical example is a vector space that has no basis, a set that cannot be linearly ordered, etc.) However, these examples d o not give any information about the “true” sets, like real numbers, sets of real numbers, etc., since those sets are in the kernel. It is clear that a different method has to be used t o investigate the role of the axiom of choice in ZF. We shall now describe such a method and exploit the similarities between it and permutation models. Symmetric Models

Although in ZF, transitive classes have no nontrivial automorphisms, we can still use automorphisms (symmetries) to construct submodels of generic extensions. As shown in Section 19, every automorphism n of a complete Boolean algebra B induces an automorphism of the Boolean-valued model V B (see (19.13)). The important property of such an automorphism is the “symmetry lemma” 19.10: (19.14)

Il~(nx1,..., nxn)ll = n(ll~(x,,...( xn)ll)

for all names xl,. .., x,. Let Y be a group of automorphisms of B. A set 9 of subgroups of Y is a normalfilter on Y if it has the properties (21.5). For each name x E V Bwe define

3.

202

FORCING AND GENERIC MODELS

its symmetry group sym,(x) = {n E 9 : n(x) = x}

(21.13) If

A

is an automorphism of B, then

(21.14)

sym(nx) = n.sym(x).n-'

This is because a(nx)= nx if and only if (n-'an)(x)= x. Given a normal filter 9 on 9,we call x symmetric if sym(x) E 8.The class HS of hereditarily symmetric names is defined by induction on p(x): if dom(x) 5 HS and if x is symmetric, then x E HS. Note that n(3) = 3 for all x and all a, and so all 3 are in HS. If a name x is symmetric, and if n E 9,then by (21.14) and since 8 is normal, nx is also symmetric. It follows that nx E HS whenever x E HS and n E Ce. The class HS constitutes a submodel of the Boolean-valued model V E ,and satisfies all axioms of ZF. We shall prove this in the context of generic extensions. Thus let 91 be the ground model, let B be a complete Boolean algebra in 91, and let Ce and 9 be respectively (in W),a group of automorphisms of B and a normal filter on 9.Let G be an W-generic ultrafilter on B. We let 91 = { i c ( x ): x E HS}

(21.15)

be the class of all elements of W[G] that have a hereditarily symmetric name. 91 is called a symmetric submodel of W[G]. We will prove that 3 is a transitive model of ZF. Before we do so, we notice that HS is a Boolean-valued model Thus we can define IIq 1 HS for every (with the same IIx E yll and IIx = yll as WE). formula q. Note that (21.16) and that IlqllHs= 11q11 whenever q is a restricted formula. We also have a forcing theorem for the model 3 : (2 1.17)

3

q(x1,

..

9

x,)

ifl

IIq(X1, ...( XnllHs E G

where xi, . . . , x, E HS are names for xl, . . . , x,. Finally, since n[HS] = HS for all n E 9,we have the symmetry lemma for ( 1 I(Hs: If n E 3 ' and xl, . . . ,x, E HS, then (2 1.18)

IIq(nx1, ...- nxn)llW= n(IIq(xI9 * * * . xn)llHs) Lemma 21.4. A symmetric submodel 91 of W[9] is a transitive model of ZF, and E 3 E W[G]. Proof. Since 3 E HS for every x E 91,we have 9.R C 91. The heredity of HS implies that 91 is transitive. To verify that the axioms of ZF hold in 9,we

21.

SYMMETRIC SUBMODELS OF GENERIC MODELS

203

follow closely the proof of the generic model theorem. As there, we have t o show that certain sets exist in the model by exhibiting names for the sets; here we have to find such names in HS. A. Extensionality, Regularity, Injnity. These axioms hold in % since % is transitive and % 2 9.R. B. Separation. Let cp be a formula and let

Y = { x E x : % k cp(x, p ) }

(21.19) where X, p

E

91. Let X, p E HS be names for X, p . We let Y E 9.R'

as follows:

dam( Y)= dom(X),

Y ( t )= x(t) . IIcp(r, P)IIHS A routine argument shows that Y is a name for Y; it remains to show that Y is symmetric. We shall show that sym( Y) zsym(X) n symb). Thus let n E Y be such that .nX= X and np = p. For every t E dom(X) we have nt E dom(nx) = dom x and (21.20)

IIcp(nt7 P)II"S = nll44C P)IIHS N n t ) = (nx)(nt)= n(X(t)) and so Y(nr)= n( Y(r)).Therefore, nY = Y. C . Pairing, Union, Power set. Let X E % and let X E HS be a name for X. For the union, we let

S=

u {domk)

:y

E dom(X)}

If n E sym(X) then n[S] = S and so the set Y = iG[S]has a hereditarily symmetric name Y : Y ( t )= 1 for all t E S . Moreover, Y z U X. Pairing and Power set are handled similarly. D. Replacement. We show that if X E 91, then there exists Y E %such that for all u E X, % satisfies 30 c p ( 4 u )

+

(3u E Y) q ( u , u )

We proceed as in (18.37) except that (we deal with 1 1 Hs instead of 1 1 and that) we look for S c HS such that n[S]= S for all n E Y (for then Y = ic[S] has a name in HS). This is accomplished by taking for S the set HS n 9.R: for large enough a. Since every n preserves the rank and since each n E Y preserves HS, we have n[S] = S for all n E '3'. In general, the set G is not a member of %, and % does not satisfy the axiom of choice. We shall use symmetry arguments to show that AC fails in the forthcoming examples. The models in Examples I and I1 are due to Cohen. Example I. This example is an analog of the first example of a permutation model and in fact, describes the same model that was used in Chapter 19 in the proof of Theorem 45. We let W[G] be the generic extension adjoining countably many Cohen reals: P is the set of all finite 0-l-functions p with

3.

204

FORCING A N D GENERIC MODELS

dom(p) c w x w (cf. (19.18)). We define a,, n E w, and E w } , as well as their canonical names as in (19.18)-(19.21). Every permutation A of w induces an automorphism of P (and in turn an automorphism of B) via (19.24). We can view such permutations as permutations of the set {u, : n E w}. Let Y be the group of all such automorphisms of B that are induced by permutations of w. For every finite E c w, let

domain

A = {an: n

(2 1.21 )

fix(E) = {n E Y : nn = n for each n

E

EJ

Let 9 be the filter on Y generated by {fix@) : E c w is finite}; 9 is a normal filter. Now let HS be the class of all hereditarily symmetric names, and let % be the corresponding symmetric submodel of 9J[G]. It is easy t o see that all u, are in HS and so is A. Moreover, the a, are distinct subsets of w (see Lemma 19.1 I), and so A is an infinite set of reals in 9. We claim that (21.22)

% k A is D-finite

To prove (21.22), let us show that there is n o f E % that maps w onto A. Assuming that there is suchf, IetfE HS and let po E G be such that po Ikfmaps 6 onto A The contradiction is obtained as in Lemma 19.12. We let E be a support off, i.e., a finite subset of w such that sym(f) 2 fix(E). We pick i E o such that i # E, and find p Ipo and n E w such that p Ikf(ri) = ui Then we find a permutation n

E

S (see Fig. 19.1) such that:

(i) np and p are compatible; (ii) II E fix(E); (iii) ni = j # i. Then n f = J n(ri)= ti, and we have (21.23)

p u np ~ k f ( i i ) = ui a contradiction.

and

p u np ~ k f ( t i ) = uj

Exercise 21.4. Show that if there is an infinite, D-finite set of reals, then the statements in Exercises 5.5 and 5.6 are false.

The similarity between the model HOD(A) used in Theorem 45 and the above symmetric model % is no accident: If the ground model is I, then these models are the same. (If 9A is arbitrary, then '9 is (in 9A[G]),the class of all sets hereditarily 9R-definable over A.)

21.

205

SYMMETRIC SUBMODELS OF GENERIC MODELS

Example II. This example is an analog of the second example of a permutation model. We construct a symmetric model 'J1 such that

(21.24) 91 I=there exists a countable set { P , : n function.

E

o}of pairs without a choice

Unlike Example I, where we used generic reals to play the role of the atoms in the corresponding permutation model, here we use sets of reals to be the indistinguishable objects. The reason for this is that the natural ordering of real numbers implies existence of a choice function on any family of pairs of reals. We first construct a generic model %UI[G]that has countably many pairs

p , = {anrb,,}

(21.25)

(n E 0)

where each a, and each b,, is an infinite set of generic reals: a,, = {a,,, : rn

(21.26)

E

o}, b,,, = {b,,, : rn

E w}

Thus we let ( P , 2 )consist of all finite 0-1-functions p such that dom(p) E ({O, 1) x o x o)x o

(21.27)

Let G be a generic filter on P. For each n and each rn, let (21.28)

a,,, = {i E o : p ( 0 , n, rn; i ) = 1 for some p E G}

b,, = {i E o :p(1, n, rn; i ) = 1 for some p E G} Then we define a,, b,, and P,, for each n by (21.26) and (21.25). The sets a,,,,,, b,,, , a,,, b,,, P , have obvious canonical names a,,, , b,,, , a,,, b,,, P,,: a,,,,,({)=

C { p : p(0, n, rn; i) = 1)

a,,(a,,,) = 1

(all rn E o)

etc. See Fig. 21.1.

0 0

FIGURE 21.1

3.

206

FORCING A N D GENERIC MODELS

A familiar argument (see Lemma 19.11) shows that every condition p E P forces a,,, # b,, for any n, m,k , I, and a,,, # a,,,,,, b,,, # b,,,,, whenever (n,m) # (n’, m’).It follows that every condition forces

(21.29)

a,

# b, ,

(all n, m,n # n‘)

6, # b,.

a, # a,,,,

To construct the symmetric model, we want to consider automorphisms n which have the property (21.30)

n(P,,)= P,,

(all n

E

a)

Thus we consider all permutations n of {O, 1) x w x w such that if n ( ~n,, m) = ( E * , n*, m*),then: (21.31) (i) n* = n for all n ; (ii) for every n, either n ( ~n,, m) = ( E , n, m*)for all m or ?I(&, n, m) = ( 1 - E, n, m*) for all m. Every permutation of this kind induces an automorphism of P (and in turn, an automorphism of r.0.P): (21.32)

( n p ) ( n (n, ~ ,m);i ) = P ( E , n, m ; i )

Let Y be the group of all such automorphisms. Note that for every K E Y, we have (21.33)

n(anm) =

n(a,) =

or brim* a,, or 6, an,*

4Pn) = Pn

For every finite E c (0, l} x w x w, let (21.34) fix(E) = {n E Y : n ( ~n,, m )= ( E , n, m) for all

(E,

n, m) E E}

Let 9be the filter on Y generated by {fix@) : E c (0, 1) x w x w finite}; 9is a normal filter. Now let HS be the class of all hereditarily symmetric names, and let 9 be the symmetric submodel of %I[G].It is easy to see that all a,,, ,b,,, ,a, ,b,,,P,,are symmetric. Moreover, the set {P,: n E o}is countable in 91. We letfbe the following name (for the function$ n - P , ) (21.35)

d o m f = {(fi, P,,)”: n E ( u } , f ( ( f iP,,)”) , =1

(all n )

I t follows from (2 1.30)that n ( f )= .f for all n E 9 and s o f ~HS. Hence$ n - P,, is in 91. The following lemma completes the proof of (21.24):

Lemma 21.5. There is nojunctionfe 91 such t h a t f ( n ) E P,,for all n E w.

21.

207

SYMMETRIC SUBMODELS OF GENERIC MODELS

Proof: Assuming that there is suchf, letfE HS and let po

E

G be such that

po it for every n E w,f(n) E P,

We let E c (0, 1) x w x o be a support off sym(f) 2 fix@). Let n be such that ( E , n, m ) # E for any E , m. There exists p I po such that (for instance) p

We shall find n

(21.36)

E

It f ( A ) = a,

Y such that:

(i) I I and ~ p are compatible; (ii) n E f i x ( E ) ; (iii) ~ ( a ,=) b, .

Then n ( f )= L n ( i )= ti, and we shall have p u np Ikf(A) = a,,

p u np I + f ( A ) = b,

a contradiction. To find n with properties (21.36), let mo be large enough so that 4 dom(p) for any m 2 mo and any E, i. See Fig. 21.2.

(E,

n, m, i )

FIGURE 21.2

We let n be the permutation of (0, 1) x w x w defined as follows:

(21.37)

X(E,

if n' # n

n', m ) = (c. n', m)

+

\ ( I , n, m mo) n(0, n, m ) = '(1, n, m - mo) ((1, n , m )

if m < m o if mo Im < 2mo if m 2 2 m o

n(1. n, m ) . . . correspondingly The permutation x is in 92,and nu, = b,. Since n does not occur in E, we have n E fix(E). It follows from (21.37) that p and np are compatible. Thus II satisfies (2 1.36). Exercise 21.5. The set

Example 11.

u."=oP , cannot be

linearly ordered in the model 91 of

3.

208

FORCING A N D GENERIC MODELS

On the other hand, J i v y has shown that every set can be linearly ordered in the symmetric model of Example I. In fact, Halpern and Levy proved that the model even satisfies the prime ideal theorem, thus establishing the independence of the axiom of choice from the prime ideal theorem. We note that numerous consequences of the axiom of choice in mathematics can be proved using the prime ideal theorem instead-among others the Hahn-Banach theorem, compactification theorems, the completeness theorem, the Tychonoff theorem for Hausdorff spaces, etc.

The JechSochor Embedding Theorem The above examples suggest that there is a similarity between permutation models and the symmetric submodels of generic models. We shall present a general theorem which shows that any permutation model can be embedded in a symmetric model of ZF, " with a prescribed degree of accuracy."

Theorem 47 (Jech-Sochor Embedding Theorem). Let 9 be a permutation model, let A be its set ofatoms, and let a be an ordinal (in 9).There exists a symmetric model 91 of ZF and an embedding X H ~of 9 in 91 such that (P"(A))'"is E-isomorphic to (P"(A))". Proof: See Fig. 21.3. We work in the set theory with atoms, ZFA, plus the axiom of choice. We denote by A the set of all atoms, and let 91 be the kernel, '93 = P " ( 0 ) . We consider a group 59 of permutations of A, and a normal filter 9 on 59, and let 9 be the permutation model given by Y and 9. Let c1 be an ordinal number.

\ A

FIGURE 21.3

We shall construct a generic extension 9LII[G] of the kernel, and then the model 9 as a symmetric submodel of M[G]. We construct !U1[G] by adjoining to M a number of subsets of a regular cardinal K, K of them for each a E A. We use these to embed 9 in M[G]. Let K be a regular cardinal such that K > I P " ( A ) ( .The set P of forcing

21.

209

SYMMETRIC SUBMODELS OF GENERIC MODELS

conditions consists of functions p with values 0, 1 such that 1 dom(p) 1 < K and dom(p) c ( A x K ) x K ; as usual, p < q iff p =I q. Let G be an 9bgeneric filter on P. For each a E A and each < K, we let

<

(21.38)

xar = { q E K : p(a, <; q ) = 1 for some p E G}

Each xar has a canonical name xaC: (21.39)

xa<(ii)=

C {P E P : ~ ( a5;, I‘)

Then we define, for every a (21.40) and let names.

A’ = {ii : a E

E

(IIE K )

= 1)

A,

ii = { X a < : < < K ) A ) (see Fig. 21.4). The sets ii and A’have obvious canonical

FIGURE 21.4

Having defined ii for each a for each x by €-induction: (21.41)

.C = { j : y

E

A , we can define 2 (and its canonical name i)

E x},

We shall show that the function x - 2

i@) = 1 (all y

E

x)

is an E-isomorphism.

Lemma 21.6. For all x , y, xEy

-$I

i E j ,

x=y

iff

i = j

Proof: First we note (cf. Lemma 19.11) that llxa5= x . , ~11 = 0 whenever (a, <) # (a’, <’), and that IIxat= ill = 0 for all z E 91.Consequently, we have ii # gwhenever a # b are atoms. We claim that for all x , 2 # xatforany a, 5. If x E %I, then i = x and so i # xac.If x 4 %I, then i is of higher rank than a n y xat: xar is a subset of K, while the transitive closure of 2 contains some of the Xat

*

Now we can prove the lemma, simultaneously for E and =, by induction on rank : (a) If x E y, then i E 9 follows from the definition (21.41). If 2 E j,then y cannot be an atom because then we would have i = xacfor some a, 5, which is impossible. Hence 2 = Z for some z E y and we have x = z by the induction hypothesis; thus x E y. (b) If x = y , then 2 = j . Conversely, if x # y , then either both x and y are

3.

210

FORCING AND GENERIC MODELS

atoms and then 2 # j j ; or, e.g., x contains some z that is not in y, and then, by the induction hypothesis, Z E 2 and Z 4 j j ; thus 2 # jj. W Note that the proof of Lemma 21.6 does not depend on the particular G and

so in fact we have proved (21.42)

x =y

iff

I I i = 911 = 1

iff

I ) i = j I )# 0

and similarly for E. Now we shall construct a symmetric submodel 91 of 'lJI[G].We construct 91 so that for every x E 23,2 is in 9 and that (P"(A))*'is isomorphic t o (Pa@))¶. For every permutation a of A, let 8 be the group of all permutations n of A x K such that for all a, 5, (21.43)

for some 4'

n(a, <) = (oa, <')

(n permutes the blocks in Fig. 21.4 as prescribed by a). We let

A = {tf : a E H)

(21.44)

for every subgroup H of $9. Since every permutation n of A x automorphism of P (and hence of r.0.P) via (nP)(n(a?<), 'I)= P(a9

<7

'I)

K

induces an

(all 4 5, 'I)

we consider d as a group of automorphisms of B = r.0.P. For every finite E E A x K we let (21.45)

and we let (21.46)

fix@) = {n E 9 : n(a,5 ) = (a, t ) for all (a, t) E E }

9 be the filter on 3 generated by the set { A: H

E

9} u { f i x ( E ) :E

EA x K

finite}

It is easy to verify that is a normal filter. Let HS be the class of all hereditarily symmetric names and let 9 be the corresponding symmetric submodel of 'Jn[G].It is an immediate consequence of the definition of 9 (21.46) that all xatrall ti (a E A ) , and A are symmetric, and so A is in 91. The following two lemmas show that for any x, x' is in 91 if and only if x E $3.

Lemma 21.7. For all x, X E %

iff

~ E H S

Proof: It suffices to show that sym,+(x)E 9 iff syrn,+(i)E g . If (r E $9 and n E 5, then n i is the canonical name for (ax)-, and so sym,4(5)= sym,4(x); then sym , ( 5 )E 9. O n the other hand, if sym,q(5)E 9, thus if sym (x) E 9, then symT4(x)2 A n f i x ( E ) for some H E 9 and a finite E G A x K . If e = {a E A : (a, t) E E for some t}, then sym,@(x) 2 H n fix(e), and since fix(e) E 8 by (21.7), we have syrn,#(x) E 9.

21. SYMMETRIC SUBMODELS OF GENERIC MODELS

211

Lemma 21.8. For all x, X € %

ijj

i E 9 1

ProoJ: By Lemma 21.7, it suffices to show that if 2 E 3, then x E 9. Assume otherwise, and let x be of least rank such that iE (n and x 4 9. Thus x c 23, and since iE 3, there exists z E HS and p E G such that p IF z = i. Since sym,$(t)E .F, we have sym(r) 2 A n fix(E) for some H E 9 and a finite E s A x k-. We shall find (T E $9 and n E i? such that: (21.47)

(i) np and p are compatible; (ii) n E R n fix(E); (iii) ox # x.

Then we shall have nz = t by (ii), llni = ill = 0 by (iii) and (21.42); and since np IF nt = n i , we shall have (21.48) npuplFt=i npupllz=ni a contradiction. To find n, note that x is not symmetric, so that there is G E 9 such that ox # x and o E H n fix(e), where e = {u E A : (a, 5 ) E E for some 5 ) . Since IpI < k-, there exists 7 < K such that (a, 5 ) $ dom(p) for any a E A and 5 > y. Thus we define n E 5 as follows: (21.49) if u

E

e, then n(a, 5 ) = (a, 5 )

(all 5 )

n(u, 5 ) = (ou, i) It follows that n

E

H

if i > ; ” 2

n fix@) and that p and np are compatible.

1

We complete the proof of Theorem 47 by showing that ( ( P = ( A ) ) y -= (Pa(A))’n

The left-hand side is clearly included in the right-hand side; we prove the converse by induction. Thus let x E P ” ( A ) n 3 and let y E 91 be a subset of x; we shall show that = Ifor some Z E 21. Let y be a name for y. The notion of forcing that we are using here is k--closed; and since we have chosen k- large, it follows (sec Lemmas 19.5 and 19.6) that there is p E G that decides TE y for all r E X. Hence y = Z. where := {t E x : p IF i E y ) ,and by Lemma 21.8 we have 2 E

21.

As for applications of Theorem 47, consider a formula cp(X, 7 )such that the only quantifiers in cp are 31E P ; ’ ( X )and Vu E P’(X). Let 21 be a permutation model such that

( 2 1.SO)

21 1 3x q ( X , 1,)

212

3.

FORCING AND GENERIC MODELS

Let X E 9 be such that 9 I=cp(X, y); let c( be such that (in 9)P y ( X )C P”(A). By the embedding theorem, 2)can be embedded in a model 91 of ZF such that ( P “ ( A ) ) is isomorphic to (P”(2))”.Since the quantifiers in cp are restricted to P y ( X ) ,it follows that 91 I= q ( f ,y ) , and so 91 I= 3x q(X,Y)

(21.51)

Therefore, if we wish to prove consistency (with ZF) of an existential statement of the kind just described, it suffices to construct a permutation model (of ZFA). Note that X cannot be well-ordered,” ‘‘ X cannot be linearly ordered are formulas of the above type, and so is “ X is a countable set of pairs without a choice function.” “

”

Exercise 21.6. Show that the set A of atoms in the first example of a permutation model has the following properties, and by Theorem 47 it is consistent with ZF that such A

exists:

(i) A is infinite; (ii) A is not the union of two disjoint infinite sets; (iii) P ( A ) is D-finite.

The embedding theorem, in conjunction with the construction of permutation models, has interesting applications in algebra. One can construct various abstract counterexamples to theorems whose proofs use the axiom of choice. For example, one can construct a vector space that has no basis, etc. We shall give one example of an application of Theorem 47: Without the axiom of choice, the partial ordering by cardinality is quite arbitrary: Example 111. Let 91 be a transitive model of ZFC and let ( I , 5 )be a partially ordered set in u1.There exists a model 91 2 91 of ZFC that contains a family of sets { S i : i E I } such that

(21.52)

i5j

iff

Isi[Il S j l

for all i, j E I . It suffices to construct a permutation model 2) with kernel 91 such that (21.52) holds in 2‘3 for some {Si: i E I } . Then one can embed 9 in a model 91 of ZF such that 9 will satisfy (21.52). Since ( I , 5 ) is embedded in (P(I), G ) by the mapping iw { j E I :j 5 i}, it suffices to find in 3 a family { S , : p c I } such that

(21.53)

p z q

iff

% k IS,I I IS,]

Let A be the set of all atoms; let us assume that / A I = ( I ( . K O and let {ain: i E I , n E w } be an enumeration of A . For each p c I , let S , = {ai, : i E p , n E 0). We shall construct a permutation model 2l such that the mapping h: PI-+S,, is in %, and that (21.53) holds.

21. SYMMETRIC SUBMODELS OF GENERIC MODELS

2 13

Let Y be the group of all permutations n of A such that n&) = S,,, for every i E I ; i.e., n(ain)= a,, for some rn. Let 9 be the filter on Y generated by { f i x ( E ): E E A finite}, and let 23 be the permutation model. Clearly, each S , ( p G I ) a s well as the function h : pt+ S , are preserved by all permutations n E Y and thus belong to 23. Also, if p E q, then S , c S , and so (S,I I IS,\; it remains to show that if p $ q, then 23 k IS,\ $ IS,l. Let p $ q, let i E p - q and assume that g E 2 3 is a one-to-one function of S, into S,; let E be a finite support of g. Let n, m be such that a,,, a,, $ E and let A be a permutation of A such that n(ain)= a,,, n(aim)= a,,, and .(a) = a otherwise. Clearly n E fix(E) and so n(g)= g. Since i $ q, we have g(a,,) = ujkwhere j # i ; hence + j k ) = u j k ,and we have g ( a i n ) = ajk = n(ajk) = n(g(ai,)) = (ng)(nain)= g(%) Hence g is not one-to-one, a contradiction. H

Two More Examples of Symmetric Models We conclude this section by sketching two examples of models of Z F in which the axiom of choice fails. The first model was constructed by Feferman and Levy, the other by Feferman. Example IV. Let 91 be a transitive model of ZFC. There is a model 9 2 911 such that (K,)"= (K,)"; hence K, is singular in 91.

First we construct a generic extension 911[G] by adjoining collapsing maps f,: w w, , for all n E (0:We let (P, 2 )consist of finite functions with domain G w x w , such that p(n, i ) < w, for all (n,i ) E dom(p). If G is a generic filter on P, then.f= G is a function on o x w, and for every n, the functionf, defined on w byf,(i) = f ( n , i ) maps w onto w,. We shall construct a symmetric model 91 c !Ul[G] such that eachf, is in 91 but K, is a cardinal in %. Let Y be the group of all permutations n of w x w such that for every n, n(n, i) = (n,j ) , for some j. Every n induces an automorphism of P via

u

(21.54) dom(np) = {n(n,i ) : (n,i ) E dom p } ,

(np)(n(n,i ) ) = p(n(n, i ) ) Let 9 be the (normal) filter on Y generated by {H,, : n E w}, where H , consists of all II such that n(k, i ) = ( k , i ) for all k In, all i E w. Let HS be the class of all hereditarily symmetric names and let 9 be the symmetric model. It is easy to verify that for each n, the canonical namef, ofj, is symmetric and s o h E %. To show that K, remains a cardinal in %, we shall use the following lemma: Lemma 21.9. If sym(x) 2 H, and p Ik cp(x), then p : n Ik cp(x), where p :n is the restriction ofp to { ( k , i ) : k I n}. Proof: Let us assume that p : n does not force q(x) and let q 2 p : n be such that q Ik -I&). It is easy to find A E H,, such that ap and q are compatible; since np Ik ~ ( n x )and , AX = X, we get a contradiction.

3.

214

FORCING AND GENERIC MODELS

Now let us assume that g E 9 is a function of w onto K,, and let g be a symmetric name for g. Let p o E G be such that p o forces “ g is a function from cij onto K, .” Let n be such that p o : n = p o and that sym@) 2 H,.Since g takes K, values, it follows that for some k E w, there exists an incompatible set W of conditions p 2 po such that J W I 2 K,, and distinct ordinals a,.,, p E Wl such that for each p E W , p k g(k) = d,. By Lemma 21.9, we have p : n IF g(k) = G,,, for each p E W, which is a contradiction: O n the one hand, the conditions p : n, p E W , must be mutually incompatible, and on the other hand, the set { ( p :n ) : p E P } has size only K, .

,,

If the ground model 9ll in the above example satisfies GCH, then one can show that in 91,the set of all reals is the countable union of countable sets. To show that a successor cardinal K,, is regular, one uses the axiom of choice. It is an open problem to construct a model of ZF in which every infinite cardinal has cofinality w. Somewhat surprisingly, this problem is related to large cardinal axioms: It follows from a recent result of Jensen that if both K, and K2 are singular, then 0’ exists (see Exercise 30.7).

,

Example V. Let (xn be a transitive model of ZFC. There is a model 9 2 91 such that in 9,there is no nonprincipal ultrafilter over w.

The model 91 is obtained by adjoining t o 9Jfinfinitely many generic reals a,, n < o,without putting in 91 the set {on: n E w } (unlike in Example I where {a, : n E w } is in 91). First we construct 9l[G] as in Example I: ( P , 2 )is the set of all finite O-l-functions with domain E w x w. Let G be generic and let a, = { m : p(n, m ) = 1 for some p E G},for each n E w. Now let 91 be as follows. Every X G w x w induces a symmetry a x , an automorphism of P defined by if (n, m ) # X (oXp)(nm , ) = (p‘“’ m, (21.55) 1 - p(n, m ) if (n, m ) E X Let 9 be the group of all ox, X E w x w, and let 9 be the normal filter on Y generated by {fix(E): E c w finite}, where fix(E) = {ax : X n ( E x w ) = O}. Let Cfl be the symmetric model. Let D E 91 be an ultrafilter over w ; we shall show that D is principal. Let D E HS be a name for D and let p E G be such that p forces D is an ultrafilter over G.”Let E c w be finite, such that sym(D) 3 fix@), and let n 4 E. Then there is q 5 p , q E G , which decides u, E D (where u, is the canonical name for a,). For example, assume that q IF a, E D (the proof is similar if q IF a, $ D). Let rn, be such that for all m 2 m,, (n,m) $ dom(q), and let X = {(n,m ) : m 2 mo}. Let b, = aX(an), and b, = iG(b,).Since for each m 2 m,, 1 1 6 E b, 11 = - 1 1 6 E a, 1 , it follows that a, n b, is a finite set. However, oxqlkaxa,E a , D ; it is fairly obvious that axq = q and since ox E fix(E), we have axD = D . Thus q IF b, E D and hence a, n b, E D. Consequently, D is principal. “

21. SYMMETRIC SUBMODELS OF GENERIC MODELS

215

A modification of the construction in Example V gives a model 9 in which all ultrafilters are principal. It can be shown that Feferman’s model (Example V) and Cohen’s model (Example I), although different, have the same sets of ordinals. Compare this fact with Lemma 15.1. Finally, let me mention the following open problem. We have used the axiom of choice to show that there exist disjoint stationary subsets of o1(see Lemma 7.6). Is there a model of ZF which satisfies the following? (21.56) The closed unbounded filter over

K, is an ultrafilter.

It is known that (21.56) implies (assuming the countable axiom of choice) that there exists a model of ZFC that has a measurable cardinal; thus this problem is related to large cardinal axioms. The problem is particularly interesting because (21.56) is a consequence of the axiom of determinacy.

CHAPTER 4

SOME APPLICATIONS OF FORCING

22. SUSLIN’S PROBLEM

The real line is, up to isomorphism, the unique linearly ordered set that is dense, unbounded, complete, and separable. In 1920 Suslin raised the question as to whether “separable ” can be replaced by a weaker condition: Each collection of disjoint open intervals is at most countable (the countable chain condition, c.c.c.). Suslin’s Problem. Let S be a linearly ordered set such that: (i) S is dense and unbounded; (ii) S is complete; (iii) S satisfies the countable chain condition.

Is S isomorphic to the real line? This problem cannot be decided in the axiomatic set theory. In this and the following section, we shall show that Suslin’s problem is independent of the axioms of ZFC. A simple argument using Dedekind completions and embedding of countable ordered sets in the rationals shows that Suslin’s problem can be formulated as follows (see Section 4): Is there a dense linearly ordered set S that satisjes the countable chain condition and is not separable? We shall call such an ordered set S a Suslin line. Upon completion and removal of endpoints if necessary, a Suslin line provides a counterexample to the problem of Suslin. We shall therefore investigate whether Suslin lines exist. The first step toward solution of Suslin’s problem is its reformulation in terms of trees. Trees (which can be considered a generalization of ordinal numbers) are often useful in point set topology and infinitary combinatorics and make an interesting object of investigation in combinatorial set theory. A tree is a partially ordered set (T, < ) with the property that for each x E T, the set ( y : y < x} of all predecessors of x is well-ordered by <. See Fig. 216

22.

217

SUSLIN'S PROBLEM

22.1. The ath lervlof T consists ofallx E T such that {I,: < xi has order-type a. The height of T is the least a such that the ath level of T is empty; in other words, it is the length of the well-founded relation < : 19

(22.1)

o(x) = order-type of {y : y < x}

ath level = {x : o(x) = a) height(T) = sup{o(x)+ 1 : x

E

T}

A branch in T is a maximal linearly ordered subset of T. The length of a branch b is the order-type of b. An a-branch is a branch of length a.

FIGURE 22.1 Example of a (finite) tree. Exercise 22.1. I f T is a tree of height w such that each level of T is finite, then T has an infinite branch. [To construct a branch {xo, x l , ..., x,, ...} in T , pick xo at level 0 such that {y : y > xo} is infinite. Then pick xl, x 2 , . . . similarly.]

An antichain in T is a set A G T such that any two distinct elements x, y of A are incomparable, i.e., neither x < y nor y < x. Let us call a tree T a Suslin tree if:

(22.2)

(i) the height of T is w , ; (ii) every branch in T is at most countable; (iii) every antichain in T is at most countable.

Lemma 22.1. There exists a Suslin line if and only ifthere exists a Suslin tree. Proof. (a) Let S be a Suslin line. We shall construct a Suslin tree, The tree will consist of closed (nondegenerate) intervals on the Suslin line S. The partial ordering of T is by inverse inclusion : If I, J E T, then I 5 J if and only if I 3 J.

4. SOME APPLICATIONS OF

218

FORCING

The collection T of intervals is constructed by induction on a < w I . We let I, = [a,, b,] be arbitrary (such that a, < b,). Having constructed I,, /3 < a, we consider the countable set C = {(I, : /3 < a} u {b, : /3 < a} of endpoints of the intervals I,, /3 < a. Since S is a Suslin line, C is not dense in S and so there exists an interval [a, b] disjoint from C;we pick some such [a,, bJ = I,. The set T = {I, : a < q}is uncountable and partially ordered by 2. If a < /3, then either I, 3 I, or I, is disjoint from I,. It follows that for each a, {I E T : I 3 I,} is well-ordered by 2 and thus T is a tree. We shall show that T has no uncountable branches and no uncountable antichains. Then it is immediate that the height of T is at most 0 , ;and since every level is an antichain and T is uncountable, we have height(T) = wl. If I, J E Tare incomparable, then they are disjoint intervals of S; and since S satisfies the countable chain condition, every antichain in T is at most countable. To show that T has no uncountable branch, we note first that if b is a branch of length wlr then the left endpoints of the intervals I E b form an increasing sequence {x, : a < ol} of points of S. It is clear that the intervals (x,, x,+ a < wl, form a disjoint uncountable collection of open intervals in S, contrary to the assumption that S satisfies the C.C.C. (b) We shall now assume that there exists a Suslin tree and construct a Suslin line. The idea is to let S consist of branches of the tree. However, since the set S so constructed might not generally be dense, we shall first normalize the tree. Let a be an ordinal number, a I wl. A normal a-tree is a tree T with the following properties: (22.3)

(i) (ii) (iii) (iv)

height(T) = a ; T has a unique least point (the root); each level of T is at most countable; if x is not maximal in T, then there are infinitely many y > x at the next level (immediate successors of x); (v) for each x E T there is some y > x at each greater level (less than a);

(vi) if

/3 < a is a limit ordinal and

{z : z

< x} = {z : z < y}, then x

x, y are both at level

/3 and if

= y.

See the following exercise for justification of conditions (ii) and (vi): Exercise 223. If T is a normal a-tree, then T is isomorphic to a tree T whose elements are 8-sequences (0 < a), ordered by extension; i f f G sand s E T, then t E T,and the 8th level of T is the set {t E T : dom(t) = 8).

A normal Suslin tree is a Suslin tree that satisfies the normality conditions (22.3). We shall show that given a Suslin tree, we can obtain a normal Suslin tree. Let T be a Suslin tree. T has height wl, and each level of T is countable. We first discard all points x E T such that T, = {y E T : y 2 x} is at most count-

22.

SUSLIN'S PROBLEM

219

able, and let T, = {.x E T : T, is uncountable). Note that if x E TI and a > o(x), then I T.1= S1for some y > x at level a. Hence TI satisfies condition (v). Next, we consider only branching points of the tree, i.e., points that have at least two immediate successors. For each x E T,, there are uncountably many branching points z > x in 7''(otherwise, all branching points z > x would be below a certain level and then there would exist an uncountable branch). Thus T2 = {thebranching points of TI}is a Suslin tree with properties (i), (iii), (v), and (iv*), where (iv*) stands for: each x has at least two immediate successors. Next we satisfy property (vi): We consider all chains C in T2 with the property that C = { z : z < x} = { z : z < y) for some x # y. For every such C we add to T2 an extra node a, and stipulate that z < a, for all z E C , and a, < x for each .Y such that x > z for all z E C . The resulting tree T3 satisfies (i), (iii), (iv*), (v), and (vi). To get property (iv), let T4 consist of all x E T, at limit levels of T,. The tree T4 satisfies (i), (iii), (iv), and (v); and then T5 c T4 satisfying ( i i ) as well is easily obtained. Now we can finish the proof of Lemma 22.1. Let T be a normal Suslin tree. The line S will consist of all branches of T (which are all countable). Each x E T has countably many immediate successors, and we order these successors as rational numbers. Then we order the elements of S lexicographically: If a is the least level where two branches a, b E S differ, then a is a successor ordinal and the points a, E a and b, E b at level a are both successors of the same point at level a - 1 ; we let a < b or b < a according to whether a, < b, or b, < a,. It is easy to see that S is linearly ordered and dense. If (a, b ) is an interval in S, then one can find x E T such that I , c (a, b), where I , is the interval I , = {c E S : x E c). And if I , and I , are disjoint, then x and yare incomparable points of T . Thus every disjoint collection of open intervals of S must be at most countable, and so S satisfies the countable chain condition. The line S is not separable: If C is a countable set of branches of T, let a be a countable ordinal bigger than the length of any branch b E C . Then if x is any point at level a, the interval 1,does not contain any h E C , and so C is not dense in S. Before going further, let me make the following observation: If T is a normal tu,-tree and if T has a branch of length ol, then T has an uncountable antichain. For, if h is an uncountable branch, pick for each x E b a successor z , of x such that z , $ h. Then A = ( z , : x E b ) is an antichain. Thus in order that a normal (]),-tree T be Suslin, it suffices that T has no uncountable antichain. Suslin lines have often been used in point set topology, mostly to construct nonseparable examples of various types of spaces; it turns out that since existence of Suslin lines cannot be proved in ZFC, the same is true for many of these topological examples: Their existence cannot be proved in ZFC. If T is a tree, let P , denote the partially ordered set that is obtained from ( T , I ) by reversing the order. Note that any two elements of P I . are either

220

4.

SOME APPLICATIONS OF FORCING

comparable or incompatible; they are incompatible just in case they are incomparable as points of the tree T . Thus if T is a Suslin tree, then P T satisfies the countable chain condition. Lemma 22.2. If T is a normal Suslin tree, then P , is K,-distriburive. Corollary. If T is a normal Suslin tree, then B c.c.c., atomless, complete Boolean algebra.

= r.0.

P , is an KO-distributive,

ProoJ Let D , , n = 0, 1,2, ...,be open dense subsets of P , .We shall prove that (-)7."=,, D, is dense in PT , First I claim that if D G P , is open dense, then there is a < o,such that D contains all levels of T above a. To prove this, let A be a maximal incompatible subset of D. A is an antichain in T and hence countable. Thus pick a < w 1 such that all a E A are below level a. Now if x E T is at level >a, there is d E D such that a 5 , d (by density), and by maximality of A, d is comparable with some a E A . Hence a I d (draw a picture), and consequently a 5 , x <, d. Since D is open, we have x E D. End of claim. Now if D,,n = 0, 1, . . . , are open dense, we pick countable ordinals a, such that D, contains all levels above a,, and let a = sup, a,. The intersection D, contains all levels of T above a ; and since T is normal (cf. 22.3.v), this implies that D, is dense in P T . Finally, it is an easy consequence of normality of T that the algebra B = r.0. PT is atomless. H

n;=o

A complete Boolean algebra that is atomless, K,-distributive, and satisfies the countable chain condition is called a Suslin algebra. Thus if there exists a Suslin tree, there exists a Suslin algebra. Exercise 22.3. If there exists a Suslin algebra, then there exists a Suslin tree. [If ( B , 5 )is a Suslin algebra, construct a Suslin tree T c B - {0},ordered by 2 ,by

induction on levels. To get successors oft E T , pick an infinite partition oft. At limit levels take all possible nonzero Boolean products along branches of the tree constructed so far. By KO-distributivity,each level of T is nonempty (and is a partition of B).]

As I mentioned earlier, it is independent of ZFC whether the countable chain condition of partially ordered sets is preserved under products. A Suslin tree provides an easy counterexample. If T is a normal Suslin tree then P r x Pr is not C.C.C. [For each x E T , pick two immediate successors px and qx of x. The set { ( p x , q x ): x E T } c P , x P , is pairwise incompatible.] Exercise 22.4.

So far, we have reduced the original problem of Suslin to an equivalent problem on trees. It is worth mentioning-and an interesting result on its own-that if the definition of Suslin tree is somewhat weakened, then trees with the weaker properties exist. An Aronszajn tree is a tree of height o all of whose levels are at most countable and which has no uncountable branches.

22.

22 1

SUSLIN'S PROBLEM

Lemma 22.3. There exists an Aronszajn tree. Proof. We construct a tree T whose elements are some bounded increasing transfinite sequences of rational numbers. If x, y E T are two such sequences, then we let x I y just in case y extends x, i.e., x E y. Also, if y E T and x is an initial segment of y, then we let x E T ; thus the ath level of T will consist of all those x E T whose length is a. Whatever the construction is, it is clear that an uncountable branch in T yields an increasing (0,-sequence of rational numbers, which is impossible. Thus T will be an Aronszajn tree, provided we arrange that T has K 1levels, all of them at most countable. We construct T by induction on levels. For each a < w,, we construct a set U , of increasing a-sequences of rationals; U , will be the ath level of T . We construct the U , so that for each a, I U , I I KO,and that:

(22.4) For each p < a, each x E U o and each q > sup(x) there is y that x c y and q 2 sup(y).

E

U a ,such

Condition (22.4) enables us to continue the construction at limit steps. To start, we let U , = (125). The successor steps of the construction are also fairly easy. Given U,, we let U , , be the set of all extensions x n r of sequences in U , such that r > sup(x). It is clear that since U , satisfies the condition (22.4), U , , satisfies it also (for a + l), and it is equally clear that U , , is at most countable. Thus let a be a limit ordinal (a < ol) and assume that we have constructed all levels U , , y < a, of T below a, and that all the U , satisfy (22.4); we shall construct U,. The points x E T below level a form a (normal) tree T, of length a. I claim that T, has the following property:

,

,

,

=-

(22.5) For each x E T, and each q sup(x) there is an increasing a-sequence of rationals y such that x c y and q 2 sup(y) and that y 1 p E T, for all p < a. The last condition means that {y I p : p < a} is a branch in T,. To prove the claim, we let a,, n = 0, 1, . . ., be an increasing sequence of ordinals such that x E Ul0 and lim a,, = a, and let {q,),"=, be an increasing sequence of rational q. Using repeatedly condition numbers such that qo > sup(x) and lim q,, I (22.4), for all a, (n = 0, 1, ...), we can construct a sequence y o c y , c ... c yn c . . . such that yo = x, y, E U,", and sup(y,) I q, for each n. Then we let y = y,; clearly, y satisfies (22.5). Now we construct U , as follows: For each x E T, and each rational number q such that q > sup(x), we choose a branch y in T, that satisfies (22.5),and let U , consist of all these y: a + Q.The set U , so constructed is countable and satisfies the condition (22.4). Then T = I/, is an Aronszajn tree. H

u."=o

u,
We shall now show that existence of a Suslin line is consistent with ZFC.

4. SOME APPLICATIONS OF

222

FORCING

Theorem 48 (Tennenbaum, Jech). There is a model of ZFC in which a Suslin tree exists. Thus existence of Suslin trees is consistent with the axioms of ZFC. The forcing construction given below uses countable trees as forcing conditions. An alternative forcing construction, using finite trees, is described in Exercises 9 and 10. And Theorem 49 to be proved later in this section shows that Suslin trees exist in the constructible universe L. Let P be the collection of all T of the following form: There is a < w1 such that: (22.6) (i) each t E T is a function t : /3 w for some /3 < a ; (ii) if t E T and s is an initial segment of t (i.e., s = t I y for some y), then s E T; (iii) if /3 + 1 < a and t : /3 + w is in T, then t"n E T for all n E o ; (iv) if /3 < a and t : /3 + w is in T, then for every y such that /3 Iy < a there exists s: y -+ w in T such that t G s; (v) T n is at most countable for all B < a. Note that each T E P is a normal a-tree (for some a < w I ) under partial ordering by s,and in fact every normal a-tree is isomorphic to some T E P. We shall construct a generic model of ZFC in which there is a Suslin tree. Given a ground model W,we extend it to a model W[G] such that wFG1= w17 and W[G] contains a normal Suslin tree. Let P be the set of all countable normal trees defined in (22.6),and let Tl be stronger than T2 if Tl is an extension of T2, i.e., -+

(22.7) Tl < T2

iff

3cr < height(Tl) such that

T2 = { t la : t

u

E

TI}

(in other words, Tl is obtained from T2 by adding more levels on top of T2). Let G be an 9bgeneric set of conditions and let F = { T : T E G}. We shall show that in W[G], F is a normal Suslin tree. First we note that if Tl and T2 are two conditions, then either one is an extension of the other, or Tl and T2 are incompatible. Thus G consists of pairwise comparable trees and one can easily verify that .Fis a normal tree (of height Iw1). If To, Tl, . . ., T,,, . . . is a sequence of conditions such that for each n, T,, is an extension of T,,, then T. is a normal countable tree (and extends each T,,). Hence P is K,-closed, and consequently, the cardinal K 1is preserved (and 9JI[G] has the same countable sequences in W as 91). To show that the height of .F is ol, we verify that for every a < wl, G contains a condition T of height at least a. We show that the set {TE P : height(T) 2 a} is dense in P, for any a < ol. In other words, we show that for each To E P and each a < wl, there is an extension T E P of To, of height at least a. It suffices to show that each To E P has an extension T E P that has one more level; for then we can proceed by induction and take unions at limit steps.

u."=o

22.

223

SUSLIN’S PROBLEM

If height(To) is a successor ordinal, then an extension of To is easily obtained. If height(To) is a limit ordinal, then we first observe that for each t E To there exists a branch b of length’ a in To such that t E b : Using an increasing sequence a. < al < * * < a, < * * * with limit a, we use the normality condition (22.6.i~)(or 22.3.v) to obtain such a branch. Now we construct an extension T of T,, of height a + 1, as follows: For each t E To, we pick a branch b, of length a in To such that t E b, and let T = To u {s: s = b, for some t E T}; we “extend all the branches b,, t E To, of To.” Since To is countable, the added level is countable, and one can verify that T E P. It remains to show that F has no uncountable antichain. If T is a tree and A is an antichain in T, then A is called a maximal antichain if there is n o antichain A‘ in T such that A‘ =I A : Each t E T is comparable with some a E A. If A a maximal antichain in T and if T is an extension of T, then A is not necessarily maximal in T‘. Let us call a set S c T bounded in T if there is a < height(T) such that all elements of S are at levels
u

Lemma 22.4. If A is a maximal antichain in a normal tree T and if A is bounded in T (in particular, ifthe height of T is a successor ordinal), then A is maximal in every extension of T. Proof: Let T’ be an extension of T. Let a < height(T) be such that each a E A is at level
The key element of the proof is the following lemma: Lemma 22.5. Let a be a countable limit ordinal, let T E P be a normal a-tree and let A be a maximal antichain in T. Then there exists an extension T‘ E P of T of height a + 1 such that A is a maximal antichain in T’ (and hence A is a bounded maximal antichain in TI). Proof: For each t E T there exists a E A such that either t c a or a G t. In either case, there exists a branch b = b, of length a in T such that t E b and a E b. Let T’ be the extension of T obtained by extending the branches b,, for all t E T: T‘ = T u {U b,: t E T}. The tree T‘ is a normal (a + 1)-tree and extends T; moreover, since every s E T‘ is comparable with some a E A, A is maximal in T’. W

Now I can finish the proof of Theorem 48 by showing that (22.8)

W[G] t= every antichain in F is countable

Since every antichain can be extended to a maximal antichain, it suffices to show that every maximal antichain is countable. Thus let A be a maximal

4.

224

SOME APPLICATIONS OF FORCING

antichain in F . There is a name A for A and a condition T E G such that

(22.9)

T

~k A

is a maximal antichain in F

(here I use ,7rather than its canonical name). We intend to show that the following set of conditions is dense below T:

(22.10) D = {T' I T : there is a bounded maximal antichain A' in T' such that T'

Ik

A'

sA}

Then some T' E D is in G and there is a bounded maximal antichain A' in T' such that A' G A. However, ,Fis an extension of T', and by Lemma 22.4, A' is maximal in ,F.Consequently, A = A', and since A' is countable, we are done. To show that D is dense below T let To IT be arbitrary. We shall construct T' I To such that T' E D. Since To Ik (A is a maximal antichain in .F and .T is there exists for each s E To an extension Tb of To,and some an extension of fo), t , E Tb such that s and t, are comparable

(22.11 )

and

T;, IF r', E A

Since To is countable, we repeat this countably many times and obtain an extension To< To such that (22.11) holds for every s E To. Let T, = Tb. Then we proceed by induction and construct a sequence of trees To 2 T, 2 ... 2 T,, L ... such that for each n, T,,, extends T,, and

(22.12) (Vs E T,,) (3tS E T,, We let T,

(22.13)

=

u."=o T,,, and let

[s and

t,

are comparable and T,,

, It isE A ]

A' = { t , : s E T,}

By (22.12), A' is a maximal antichain in T, , and T, lk A' E A Now we apply Lemma 22.5 and get an extension T' of T such that A' is a bounded maximal antichain in T'. Clearly, T' IF A' E A, and hence T' E D. Exercise 225. The notion of forcing ( P . < ) used in the proof of Theorem 48 is isomorphic to the partially ordered set ( Q , < ) used in Exercise 19.7.Thus (a) the model W[G] is obtained by adjoining a generic subset of o, via (19.7), and (b) the continuum hypothesis holds in W[G].

If Tl and T2are trees, then an isomorphism n: Tl -,T2between T1and T2is a one-to-one mapping of Tl onto T2such that x < y if and only if n ( x ) < n(y).An isomorphism maps level a of T, onto level a of T2(for all a); and if b is a branch in T,, then n[b]is a branch in T 2 .An automorphism of T is an isomorphism of T, onto T2.A tree T is rigid if it has no nontrivial automorphism, i.e., the only automorphism of T is the identity mapping. T is homogeneous if for any x, y at the same level of T , there exists an automorphism n of T such that n ( x ) = y.

22.

225

SUSLIN'S PROBLEM

Exercise 22.6. If T is a normal a-tree where a < w 1 is a limit ordinal and if 7t is a nontrivial automorphism of T , then T has an extension T E P of height a + 1 such that 7t cannot be extended to an automorphism of T . [Construct T so that for some branch b in T, b is extended while n[b]is not.] Exercise 22.7.* The generic Suslin tree constructed above is rigid. [If T IF p is a nontrivial automorphism of .T,then the set {T' I T : 3automorphism A

T

of an initial segment of T that cannot be extended to an automorphism of T and 117t c pi is dense below T ; a contradiction.]

If s: a + o and t : a -,o,let s many 5 < a.

-

t just in case s(() = t ( ( ) for all but finitely

a homogeneous Suslin tree. [Let the forcing conditions be normal countable trees satisfying (22.6) with the additional properties: (vi) i f f E T and s t , then s E T ; and (vii) if s E T and t E T a r e at the same level, then s t . ] Exercise 22.8.* There is a generic model !)Jl[G] in which there exists

-

-

Similar, albeit somewhat more complicated constructions give examples of rigid or homogeneous Suslin lines. Suslin trees can also be obtained by forcing with finite conditions. The construction is described below ((22.14)and Exercises 22.9 and 22.10). In the proof one uses the following lemma that is often used in topology and combinatorics. The collection Z in Lemma 22.6 is called a A-system by combinatorists.

Lemma 22.6 (A-lemma). Let W be an uncountable collection offinite sets. Then there exists an uncountable Z c W , and a j n i t e set S such that X n Y = S for any distinct elements X . Y E Z. Proof. Since W is uncountable, it is clear that uncountably many X E W have the same size; thus we may assume that for some n, I X I = n for all X E W. We prove the lemma by induction on n. If n = 1, the lemma is trivial. Thus assume that the lemma holds for n, and let W be such that 1 X 1 = n 1 for all X E W . If there is some element a that belongs to uncountably many X E W, we apply the induction hypothesis to the collection { X - {a}: X E Wand a E X } , and obtain Z c W with the required properties. Otherwise, each a belongs to at most countably many X E W , and we construct a disjoint collection Z = {X,: a < ol} as follows, by induction on a. Given X,,5 < a, we find X = X , E W that is disjoint from all X,, < a.

+

Let (P, < ) be the following notion of forcing:

(22.14) A condition is a finite tree ( T , < ) such that T = wl, and if a < T /3, then a < /3; (Tl, < T , ) is stronger than (T2, < T 2 ) if Tl z T2 and


n (T2 x T2).

4. SOME APPLICATIONS

226

OF FORCING

u

If G is a generic set of conditions, then F = {T : T E G } is a Suslin tree. The crucial properties to verify are: (a) (P, <) satisfies the countable chain condition, and (b) .T has no uncountable antichain. Exercise 22.9.* ( P , < ) satisfies the C.C.C. [Given an uncountable set W ofconditions, use the A-lemma to find a n uncountable Z c W such that any X, Y E Z are compatible.] Exercise 22.10.* 9 has no uncountable antichain. [If To IF A is uncountable, we first find an uncountable set W of pairs (T, aT) such that T I To and T ~k ixT E A. By the A-lemma, find an uncountable Z c W with the property that if T,, T2E Z, then there is T stronger than both Tl and T2such that T I+ & T I is comparable with 6 T 2 .Then some T ITo forces that A is not an antichain.]

Since the notion of forcing (P, < ) satisfies the c.c.c., the cardinals of 91 remain cardinals in 9J[G].In particular, if the continuum hypothesis fails in 9J, then it fails in !UI[G].Thus existence of Suslin trees is consistent with 2'O > K1. We shall now turn our attention to the constructible universe.

Theorem 49 (Jensen). The axiom of constructibility implies that Suslin trees exist. We break the proof into two parts. First we formulate an important combinatorial principle and use it to construct a Suslin tree. Then we prove that the principle holds in L (22.15) Jensen's Principle (0).There exists a sequence of sets S , , a < wl, such that for every X G wlr the set { a < w 1 : X n a = S,} is a stationary subset of wl. Notice that the diamond principle implies the continuum hypothesis: If = S, for some a < 0,; thus 2N0= K,.

X c w, then X

Exercise 22.11. I f 0 holds, then there exists a family 9 ofstationary subsets o f w , such that 1 9 1 = 2H1and IS1 n Sz I I No whenever Sl and Sz are distinct elements of 9. [Let 9 = { S x : X c a,},where S x = {a : X n a = S.}.]

We shall prove that 0 implies existence of Suslin trees. First we make a useful observation. If T is a normal a,-tree, let T,, for each a < wl, consist of all x E T at levels below a : T, = { x E T : o ( x ) < a}.

Lemma 22.7. If A is a maximal antichain in T, then the set

C = {a : A

r\

T, is a maximal antichain in T,}

is closed unbounded.

Proof: It is easy to see that C is closed. To show that C is unbounded, let a. < w 1 be arbitrary. Since T,, is countable, there exists a countable ordinal

22.

SUSLIN’S PROBLEM

227

such that every t E T,, is comparable with some a E A n TI.Then there is a2 > a1 such that each t E is comparable with some a E A n etc. If a. < a l < a2 < ... a, < ... is constructed in this way and if a = limn a,, then A n T, is a maximal antichain in T,. W a1 > a.

x2,

x,

Now we are ready to construct a Suslin tree, using element (besides 0 )is Lemma 22.5. Lemma 22.8. If

0.Again,

the key

0 holds, then there exists a s u s m tree.

Pro08 We construct a normal Suslin tree ( T , < T ) by induction on levels. To facilitate the use of 0,we let points of T be countable ordinals, T = wl, and in fact each T, (the first a levels of T ) is an initial segment of wl. We construct T, , a < w l rsuch that each T, is a normal a-tree and such that 5 extends T, whenever p > a. TI consists of one point. If a is a limit ordinal, then (T,, < T ) is the union of the trees ( T p , < T ) , /? < a. If a is a successor < T ) obtained by adjoining ordinal, then (T,,,, < T ) is an extension of infinitely many immediate successors to each x at the top level of T,. It remains to describe the construction of T,, if a is a limit ordinal. By 0, we are given a sequence S,, a < wl, with property (22.15). IfS, happens to be a maximal antichain in ( T , , < T), then we use Lemma 22.5 and find an extension (T,, < T ) of T, such that S, is maximal in T,, 1. Otherwise, we let T,+ be any extension of T, that is a normal (a + 1)-tree. (In either case, we let the set T,+ be an initial segment of countable ordinals.) T, is a normal Suslin tree. It We shall now show that the tree T = suffices to verify that T has no uncountable antichain. If A c T (= w is a maximal antichain in T, then by Lemma 22.7, A n T, is a maximal antichain in T,, for a closed unbounded set of a’s. It follows easily from the construction that for a closed unbounded set of a’s, T, = a. Thus using the diamond principle, we find a limit ordinal a such that A n a = S, and A n a is a maximal antichain in T,. However, we constructed T,+l in such a way that A n a is maximal in T , + l , and therefore in T. It follows that A = A n a and so A is countable. W

(x,

u,
Consistency of 0 is easily obtained by forcing. In fact, if 9 l [ G ] is obtained by adjoining a generic subset of w 1 via (19.7), then W[G] k 0 : Let P consist of all countable sequences p = ( S , : < a ) (a < wl) where S , G 5 for all ( < a ; let p I q just in case p extends q. P is clearly K,-closed. Let G be P-generic. Then !U1[G] C 0. [If plt (C is a closed unbounded set and X G ol),find q Ip such that q = ( S , : I a ) and q It (a E C and X n a = S,).] Exercise 22.12.*

<

isomorphic to the partially ordered set (Q, < ) in Exercise 19.7. Thus !U1[G] is obtained by adjoining a generic subset of o1via (19.7).

Exercise 22.13. ( P , < ) is

Lemma 22.9. Assume that V = L Then

0 holds.

228

4. SOME APPLICATIONS OF FORCING

Proof: The proof uses an argument very similar to the proof of the continuum hypothesis (Lemma 13.1). I shall use the notation and terminology of Section 13. By induction on a < o we define a sequence of pairs ( S , , C,), a < o such that S, G a and C , is a closed unbounded subset of a (in the obvious meaning of “closed unbounded”). We let S o = Co = 125 and S,, = C,, = a 1 for all a. If a is a limit ordinal, we define:

,,

,,

+

(22.16) ( S , , C,) is the <,-least pair (in the canonical well ordering of L) such that S, E a, C , is a closed unbounded subset of a, and S, n 5 # S, for all 5 E C,; if no such pair exists, let S, = C, = a. We are going to show that the sequence S,, a < o,, defined above satisfies (22.15). Thus assume that the sequence does not satisfy (22.15); then for some X G o,, the set { a : X n a = S,} is thin, and hence there is a closed unbounded set C such that

X n a # S,

(22.17)

for all a E C

Let (X, C) be the <,-least pair such that X E ol, C is a closed unbounded subset of w , , and such that (22.17) holds. is a o,-sequence of pairs of subsets of w , , it Since ( ( S a , C,): a < 0,) belongs to t, , and moreover, it satisfies the same definition (22.16) in the model (L,, E) (since L,is a model of ZF-). Also, (X, C) E t,*, and (X, C) is, in (L,,, E), the <,-least pair such that X c_ m l r C is a closed unbounded subset of wl, and such that (22.17) holds. Let (N, E ) be an elementary submodel of (L2, E) such that:

(22.18)

(i) (ii) (iii) (iv)

N is countable; if l < q < a,and q E N, then E N ; if 5 E N, then ( S , , C,) E N; ((St, C,) : { < 0,) E N, and (X, C) E N .

In fact, since (N, E ) < (L,,, E), and ( ( S , , C,) : a < o,)and (X, C) are definable elements of L,, conditions (iii) and (iv) are satisfied automatically; and the argument in Exercise 13.15 shows that (ii) holds also automatically. There exists an isomorphism n of N onto a transitive set M = n [ N ] ;in fact, M = Lp for some p < w , . Because of condition (ii), if a is the least ordinal not in N, then a = n ( o l ) and n(5)= 5 for all 5 < a. Consequently, K(S C,) = (St, C,) for all 5 < a, and n((S,, C,): 5 < w , ) = ( ( S , , C , ) : 5 < ay. Also, n ( X )= X n a and n(C)= C n a, and by the definition of X and C , the model M satisfies: (22.19) ( X n a, C n a ) is the <,-least pair (Z, D )s.t. Z !G a, D c a is closed unbounded and Z n 5 # S, for all 5 E D.

By our choice of the model N, the definition (22.16) is absolute for M and therefore (22.19) holds (in the universe L); comparing (22.19) with (22.16), we

23.

229

MARTIN’S AXIOM AND ITERATED FORCING

have X n a = S, (and C n a = C,). Since C n a is unbounded in a, and C is closed, it follows that a E C. Thus we have X n a = S , for some a E C , contrary to the assumption (22.17). A purely combinatorial argument can be used to show that to the following statement:

0 is equivalent

(22.20) There exists a sequence of functions ha, a < ul, such that for every$ w , + w l , the set { a < w 1 :f l a = ha}

is stationary. Exercise 22.14. Show that the version (22.20) of Exercise 22.15.* Use (22.20) and

0 holds in L.

0 to construct a rigid

Suslin tree in L.

[Use Exercise 22.6.1

0 holds, then there exists a homogeneous Suslin tree. [Modify the construction in Lemma 22.8.. as in Exercise 22.8.1

Exercise 22.16.* I f

It should be noted that the construction of a Suslin tree in L given above is A , over Lu,: Exercise 22.1 7.* ( V = L) There is a function a H S , (a < wl), A I over L,,, such that the S , satisfy the diamond principle. Exercise 22.18.* ( V = L) There is a normal Suslin tree T such that the function a H T, (where T, = (x E T : o(x) < a ) ) is A I over Lo,.

I shall conclude this section with the statement of the following theorem, which will be proved in the next section. Together with Theorem 48, the theorem establishes the independence of Suslin’s problem on the axioms of ZFC: By Theorem 48, there are Suslin trees in some models; and by Theorem 50, there are none in other models. Theorem 50 (Solovay-Tennenbaum). There is a model of ZFC in which there are no Suslin trees. Thus existence of Suslin trees is independent of ZFC.

23. MARTIN’S AXIOM AND ITERATED FORCING

Martin’s Axiom Rather than giving a direct forcing proof of Theorem 50, we shall formulate an axiom that easily implies nonexistence of Suslin trees, and then give a consistency proof of the axiom. The axiom, known as Martin’s axiom, is an interesting alternative to the continuum hypothesis, and is an important tool in point set topology and combinatorics. In Section 16 I proved the following lemma (Lemma 16.1): If ( P , < ) is a

230

4. SOME APPLICATIONS OF FORCING

partially ordered set and 9 is a countable collection of dense subsets in P, then there exists a %generic filter on P. (Lemma 17.5 gives a Boolean algebraic version: If % is a countable family of subsets of a Boolean algebra B, then there exists an %-complete ultrafilter on B.) The generalization of Lemma 16.1 for uncountable 9 is false: Exercise 23.1. Let ( P , < ) be the notion of forcing that collapses w 1 (see (18.45)). There is an uncountable collection 9 of dense subsets of P such that there is no 9-generic filter on P. [For each a < o l rlet D, = {p E P : a E ran(p)}.]

Thus consider the following generalization of Lemma 16.1. Let K be an infinite cardinal: (23.1) MA,: If ( P , < ) is a partially ordered set that satisfies the countable chain condition, and if 9 is a collection of at most K dense subsets of P, then there exists a %generic filter on P. MA,, is just Lemma 16.1 and is provable in ZFC. Note that MA, implies that K < 2,O: Exercise 23.2. Let ( P , < ) be the notion of forcing used to obtain a Cohen-generic real. There is a collection 59 of size 2 ” O of dense subsets of P such that there is no %generic filter on P. [For each g: o + (0, l), let D, = {p E P : p $ g}.]

Martin’s Axiom states that MA, holds for every

K

< 2Ho:

Martin’s Axiom (MA). If (P, < ) is partially ordered set that satisfies the countable chain condition and $ 9 is a collection of less than 2’O dense subsets of P, then there exists a %generic filter on P. Since MA,, is always true, it follows that Martin’s axiom is a consequence of the continuum hypothesis. We shall see later that MA can replace the continuum hypothesis in numerous proofs that use CH; this is important since the axiom can hold even when CH fails. Martin’s axiom plus 2N0> K1settles Suslin’s problem, rather easily in fact:

Lemma 23.1. If MA,, holds, then there is no Suslin tree. Proof. Let us assume that T is a normal Suslin tree and let P T be the partially ordered set obtained from T by reversing the order. PTsatisfies the countable chain condition. For each a c a,,let D, be the union of all levels above a : D, = {x E T : o(x) > a}. Each D, is dense in P,;ifweIet 9 = {D,:a < wl} and if G is a %generic filter on P, then G is a branch in T of length al.A contradiction. H There are several, more or less natural, ways of reformulating Martin’s axiom. The most obvious reformulation is in terms of Boolean algebras (just as

23.

231

MARTIN'S AXIOM AND ITERATED FORCING

Lemma 17.5 is a reformulation of Lemma 16.1 and just as Lemma 17.4 relates the two definitions of genericity). The formulation (ii) below will be useful in the forthcoming consistency proof. Lemma 23.2. The following are equivalent, for any injinite cardinal

K:

(i) (ii) l f ( P , <) is a partially ordered set satisfying the C.C.C. such that I P 1 I K and if 9 is a collection of at most K dense subsets of K , then there is a %generic jilter on P . (iii) l f B is a complete, C.C.C. Boolean algebra and i f X is a family of at most K subsets of B, then there exists an X-complete ultrajilter on B. Proof: (i) + (iii):We follow the proof of Lemma 17.4b. Let us assume MA, and let B be a complete Boolean algebra satisfying the C.C.C. Let %be a collection of at most K subsets of B. Let P = B - {0}, and let for each X E %, D , = { u # 0 : either u I X,or for some x E X , u * x = 0). Each D , is dense and by MA, there is a filter F on B that meets each D,. It is easy to see that any ultrafilter G that extends F is an X-complete ultrafilter on B. (iii) + (ii): We follow the proof of Lemma 17.4a. Let us assume (iii) and let P be a C.C.C. partially ordered set of size at most K. Let B = r.0. P and let e: P + B be the canonical embedding. First we notice that if (iii) holds, then for any collection 9 of at most K dense subsets of B there is an ultrafilter on B that meets all D E 9 (for each D consider X, = { - u : u E D}).If 9 is a collection of dense sets in P, 19I I K, we consider the following dense sets in B: (a)all e[D], D E 9; and (b) all e [ E , J for p , q E P, where Ep,q.= {r E P : either r I p and r I q, or r is incompatible with p , or r is incompatible with q}. Since 1 P 1 I K , (iii) gives an ultrafilter G on B that meets all e[D] and all e [ E p , J ,and then G' = { p E P : e ( p ) E G } is a %generic filter on P. (ii) + (i): Let P be a C.C.C. partially ordered set and let us assume that (ii) holds. Let 9 be a family of at most K dense subsets of P. For each D E 9, we let W, be a maximal incompatible subset of D. Since each W, is countable, there and (b) if exists a set Q E P of size at most K such that (a) W, G Q for all D E 9, p , q E Q are compatible, then there exists r E Q such that r I p and r I q. (Condition (b) guarantees that p , q E Q are compatible in (Q, < ) if and only if they are compatible in (P, <).) Each WDis a maximal incompatible subset of Q; let E D = {q E Q : q w for some w E W,}. Each E D is dense in Q. The partially ordered set Q has size at most K and satisfies the C.C.C. By (ii), there is a filter G on Q that meets every E D . It is easily seen that G generates a %generic filter on P.

n

Let me state (without proof) another equivalent of Martin's axiom in terms of point set topology: In every compact Hausdorff C.C.C. space, the intersection of fewer than 2'O dense open sets is dense.

232

4. SOME APPLICATIONS OF FORCING

Theorem 51 (Solovay-Tennenbaum). There is a model of ZFC in which Martin’s axiom holds and 2’O > K1. By Lemma 23.1, there are no Suslin trees in this model. Hence Suslin’s problem is independent of the axioms of ZFC (Theorem 50). Since the continuum hypothesis fails in the model too, one could still ask whether existence of Suslin trees follows just from 2N0= K1.This has been resolved too: R. Jensen constructed a model that satisfies GCH and in which there are no Suslin trees. The proof of Theorem 51 consists in construction of the model in stages: One successively destroys counterexamples (or potential counterexamples) to Martin’s axiom. The method, known as iferated forcing, deserves a detailed exposition.

Iterated Forcing: Tw+Step Iteration Let P be a notion of forcing in the ground model 91111, and let G I c P be generic over 911. In ‘M[G,], let Q be a partially ordered set and let G 2 c Q be 311[G,]-generic. Then 91[G,][G,] is a model of ZFC extending 911 and the natural question to ask is whether 911[G,][G2] is a generic extension of 911, and what is the corresponding notion of forcing in 911. The answer to the first question is yes, and the two-step generic extension can be described uniformly for all generic filters on P. That is, if Q is a name for Q , then there is a notion of forcing P * Q in 911 that gives the same model 911[G,][G,] as the two step iteration of forcing: first P and then ic,(Q). We shall first describe the two-step iteration in terms of complete Boolean algebras. Let B be a complete Boolean algebra in 911, and let C E 9JB be such that (23.2) IIC is a complete Boolean algebra(lB= I We shall define a complete Boolean algebra D = B * C in 91 such that the (“constructed in D-valued model 911” represents” the C-valued model (91B)c 911B ”). = 1 and the equivLet us consider the class of all c E !IllB such that (Ic E C(( alence relation c I c2 defined by (Icl = c2 ( 1 = 1. There is a set D that contains exactly one element c in each --equivalence class; thus D is a maximal subset of ’91’ such that: “

-

(23.3)

(i) IIc E CJI= 1 for each c E D; (ii) llcl = c2 11 < I whenever cl, c2 E D and c , # c 2 . We make D a Boolean algebra as follows: If cl, c 2 E D, then there is a unique c E D such that ((c=c1 +cc2)) =

we let c = c1 + c 2 . The operations

s D

1

,

and - are defined similarly.

23.

233

MARTIN’S AXIOM A N D ITERATEDFORCING

I t is easily verified that D is a Boolean algebra; I leave this to the reader. Note that the partial ordering 5 of D induced by the Boolean operations relates naturally to the partial ordering I of C (in 91’):

,,

(23.4)

iff

~1 I D C ,

1 1 ~ 1< r ~ 2 ( I =

1

Also, we see that if c,, c2 E D - {OD} then c,, c 2 are incompatible (in D) if and only if

(23.5)

JIG, . c,

= 011 =

1

Let me show that D is a complete Boolean algebra. Let X be a subset of D; we show that X exists. Let X E 911’ be as follows: dom(X) = X, and X ( c ) = 1 for all c E X. We have IIX G CI( = 1, and since Cis a complete Boolean algebra in 911’ and since 911’ is full (Lemma 18.6), there exists c such that

ID

(23.6)

IIc = C r XI1 = 1 Therefore there is c E D that satisfies (23.6), and clearly c = For each b E B, there is a unique c E D such that

(23.7) Let c

IJc=

1,)1 = b

and

I(c = 0,I) =

I DX .

-b

= e(b).

Lemma 23.3. The rnapping e : B + D is an embedding ( a complete one-to-one homomorphism) of B in D . Proof. By easy calculation. Thus we can assume, for all practical purposes, that B is a complete subalgebra of the algebra D = B * C. Remark. If C, and C, are complete Boolean algebras in 91l’ and if IIC, is a complete subalgebra of C , IJB = 1, then B * C , is a complete subalgebra of B * C, . In fact, if h E 911’ is such that

(23.8)

Jlh:C, + C , is an embedding](, = 1

then there is an embedding k : B * C, + B Namely. if c I E B * C,, we let

(23.9)

k ( c , )= the unique c2 E B

* C,

* C,such

that k(b) = b for all b

such that llh(c,) = c2 1

=

E

B.

1

We recall that if G is a generic ultrafilter on B, then i c is the G-interpretation of 911’ (see (18.28)).

Lemma 23.4. Let B be a complete Boolean algebra in 911, let C E 911’ be such that IJCis a complete Boolean algebraII, = 1, and let D = B * C be such that B is a complete subalgebra of D.

234

4. SOME APPLICATIONS OF FORCING

(a) If G I is an 'Ill-generic ultrafilter on B, if C = iGI(C), and if G 2 is an W[G,]-generic ultrafilter on C, then there is an W-generic ultrafilter G on

B * C such that 'IllPI = ~ [ G l I [ G , I (b) If G is an 'Ill-generic ultrafilter on B * C,if G I= G n B and C = iG,(C), fhen there is an W[G,]-genericultrafilter G , on C such that 'IllUI[G1IEGzI= 9Wl ProoJ (a) Let D = B * C,and let G C a n d G 2be as assumed and let us define G G B * C as follows: for each c E D,

(23.10)

CE

iff

G

iG,(c)E G 2

I claim that G is an W-generic ultrafilter on D. I shall show that if A E W is a dense subset of D, then A n G # 0,and leave verification of the other properties to the reader. If A is dense in D and A E W,then

(23.11 )

A , = ic,[A] = {iG1(c): c E A} is a dense subset of C = iG,[q (here we use (23.4)),and A, E !l)JIIG1].Since G , is 91[G,]-generic, there is c E D such that iG,(c)E A , n G 2 . By (23.11 ) we may assume that c E A and hence c E A n G. The definition (23.10)of G can be done in 9UI[G,][G2]; thus G E 'UI[G,][G2] and 211[G] G 91[G,][G,]. On the other hand, it follows from (23.10) (and the embedding (23.7) of B in D)that G = G n B, and so G E YIl[G];furthermore, G, satisfies in 'Ill[G] the formula (23.10) and we have G 2 E 'Ill[G]. Thus 911[G,][G2] G 'UI[G]. (b) If G, G, and C are as assumed, then we define G2 G C by the formula (23.10). I claim that G 2 is 9J[G,]-generic, and again verify only the following property: If A, E 9R[G,] is dense in C, then A, n G , # 0. Let A , E 9Jfl be a name for A , and let b, E G, be such that

,

(23.12)

b,

~kA ,

,

is dense in C - (0,)

We let

(23.13)

A={cED-{OD}:CI

IICEAlJI}

and show that A is dense in D below b,. Thus let d E D - {OD} be arbitrary such that d I b,. There is some (nonzero) b Ib , such that b Ik d # O , , and by (23.12)we have

(23.14)

b Ik(3xE1)x<,d

,

,

Since 91' is full, there is x such that b IF (x E A and x 5 d). We may assume that x E D. Now we let c = x b. Note that by (23.7), b Ik b = 1, and so a D

23.

235

MARTIN'S AXIOM AND ITERATED FORCING

hence b Ik c E A,. Thus we have c I IIc E A , 1 ; i.e., c E A. Since we have I1c I dll = 1 and so c s Dd. We have proved that A is dense in D below bo. Therefore, there exists c E A such that c E G. By (23.13), IJcE A ( 1 is in G and hence in G, = G n B. Thus iG1(c)E A,, and by the definition of G2,iGl(c) also belongs to G,. b

Ik x = c ;

- b Ik c = 0,

,

Let B, C, D, G1, G2,and G have the same meaning as above. Let P tie a dense subset of B and let Q E !ME be such that IIQ is a dense subset of CI( = 1. Let us define a subset P * Q of D as follows:

*Q

(23.15) c E P Clearly, P '

41

SD

where p E P and 114 E Q(I = 1

c = p 'D q

* Q is a dense subset of B * C.

Lemma 23.5. If P1

iff

P2

'

42

p, . q1

ifand

and p2 . q 2 are ifpl 5 PZ and P1

elements of 5 C 42.

P * Q, then

I' 41

Proof: Routine.

If P E (xn is a notion of forcing and Q is a notion of forcing in (xnr.o.(P),then Lemma 23.5 gives an alternative definition of the two-step forcing P * Q. (If P is not separative, then the p's have to be replaced by their images under the natural embedding of P into r.0. P ; similarly for Q.) Then we have r.o.(P * Q) = (r.0. P ) * (r,o. Q)w'.o'p. The two-step iteration is a generalization of the product P x Q: Exercise 23.3. If P , Q E !Ill, then P x Q is densely embedded in P [Consider the elements p . 4 where (p, q ) E P x Q.]

* Q.

In this sense, Lemma 23.4 is a generalization of the product Lemma 20.1. Lemma 23.6. Let D = B * C. (a) If B satisfies the C.C.C. and if IIC satisfies the c.c.c.11 = 1, then D satisfies the C.C.C. (b) On the other hand, if0 satisfies the c.c.c., then B satisfies the C.C.C. and IIC satisfies the c.c.c.11 = 1. Proof: (a) Let B and C satisfy the C.C.C. and let W be a pairwise disjoint set of nonzero elements of D. Let W E (xnE be such that dom( W) = W and W(c)= 1 for all c E W; clearly, 11 W E CII = 1. It follows from (23.5) that

[Ifor any distinct x, y

E

W, x . y = 011 = 1

Since C satisfies the c.c.c., we have ( 1 W is countable11 = 1 Now we use the countable chain condition in B, in a manner similar to the proof of Lemma 19.2. There isfc !ME such that [Ifis a function of G onto WI(= 1

236

4. SOME APPLICATIONS

For every n, there is a partition U , of B and a set Z, that for every u E U , ,

IIf(4= cull

= {c, : u E

OF FORCING

U,}

G

W such

2

Since B satisfies the c.c.c., each U, is countable, and so there is a countable set Z G W such that for every c E W,

1 {llc = zll : t E Z } = 1

(23.16)

Now we use the fact that the elements of Ware pairwise disjoint in D. If c, d are distinct elements of W, then IIc. d = 011 = 1, and since IIc = dll I IIc . d = cII, we have (23.17)

IIC = dlJI 1Ic = 0 11 Thus if c E W - Z, we have by (23.16) and (23.17): IJc= 01) = 1. Since we assumed that c # 0, for each c E W, we conclude that W = Z. Thus W is countable. (b) Let D satisfy the C.C.C. Since B is a complete subalgebra of D, B also satisfies the C.C.C. T o show that IIC satisfies the c.c.c.11 = 1, let W E WE, and let b, E B - (0) be such that

(23.18)

b,

IN-W

is an uncountable set of nonzero elements of C

We shall show that there exists a nonzero b I bo such that (23.19)

b

IF

W is not pairwise incompatible

There is f E W Esuch that b, Ikfis a one-to-one function of GIinto W

-=

For every a wl, 60 IF (3x E W) x = f ( d i ) ; and since W Eis full, there exists c, E D such that b, IF (c, E W and c, =f(di)). Let d, = b, * c,. Since bo IF c, # cs, for all a # B, the set {d, : a < w I} is an uncountable subset of D. Since D satisfies the c.c.c., there exist a and /3 such that d, and d, are compatible. Hence there exists a nonzero d E D such that d I d,.d,; moreover, we can find dsuch that d = b * c, where0 # b I b,and b lk (c # Oand c Ic, . cp). Now (23.19) follows. H As a consequence of the above lemma, we observe that if P and Q are two notions of forcing, then P x Q satisfies the C.C.C.if and only if Q is C.C.C.in mr.0.P if and only if P is C.C.C. in !W.o.Q.

C.C.C.

Exercise 23.4. Let T be a Suslin tree in Wand let P be the notion of forcing that adjoins

generic reals (19.1). Let G c P be W-generic. Then T is a Suslin tree in W[G]. [Let PT be the notion of forcing associated with the Suslin tree T . P satisfies the C.C.C. in any model 9 2 tR, in particular in any WIGI] where G is a generic filter on PT . Thus P T x P is c.c.c.,and SO P T is C.C.C. in Zn[G].] K

23.

237

MARTIN’S AXIOM A N D ITERATED FORCING

The above argument gives another consistency proof for “there exists a Suslin tree and 2’O > K,.” Exercise 23.5. If P is 1-closed and IIQ is ,klosedII,,,,

=

1, then P * Q is I-closed.

One can define the inverse operation to the operation B * C.If B is a complete subalgebra of the complete Boolean algebra D (in W),let C E 331’ be the Boolean algebra that is (in 331’) the quotient of d modulo the filter generated by the canonical generic ultrafilter on B. In other words, C = t>/= where the (B-valued) equivalence relation = is defined as follows:

[Id, = d, 11 = C { b E B : b is disjoint from d , Ad,} (where dl Ad2 = ( d , - d , )

+ (d2 - d , ) . Let us denote C = D :B.

Exercise 23.6. I(D:B is a complete Boolean algebra”, = 1, and D is isomorphic to B * (D:B).

It follows that if (in 91) B is a complete subalgebra of a complete Boolean algebra D and if G is a generic ultrafilter on D,and G = G n B, then 9J[G] is a generic extension of 91[G,].

,

Exercise 23.7.* Let B, D,G, G I be as above and let P be a dense subset of D.There exists Q E ‘lI1[Gl], Q c P, such that G n Q = G n P is an WIGI]-generic filter on (Q, 2). [For each p E P, let h ( p ) = { b E B : b 2 p}. and let Q = {p E P : h ( p ) E GI}.]

Direct Limits of Sequences of Complete Boolean Algebras Let a be a limit ordinal number, and let for each i < a, B i be a complete Boolean algebra such that if i < j, then Bi is a complete subalgebra of B j . Then

C=

u

Bi

i
is a Boolean algebra (not necessarily complete), and we call the completion B of C the direct limit of {Bi : i c a } : (23.20)

B

= lim

dir Bi

i-a

The direct limit of { B i :i < a} is the unique (up to isomorphism) complete Boolean algebra B such that Bi is dense in B. More generally, if B i , i < a, are complete Boolean algebras and if for each i, j < a, eij is a complete embedding of Bi into Bj such that the eij commute (i.e., ejk 0 eij = en), then there is a unique (up to isomorphism) complete Boolean algebra B and complete embeddings eia of Bi into B such that the e, commute with the eij and that U i < a eia[Bi]is dense in B. [Even more generally, if I is a directed partially ordered set, i.e., for any i , , i2 E I there is j E I such that i, 2 j and i2 Ij, and if {Bi,e i j } i , j e ,is a directed system of complete Boolean algebras and complete embeddings, then there is a unique complete Boolean algebra B, and embeddings eim,i E I , etc.; B = lim dirie, Bi.]

ui
238

4. SOME APPLICATIONS OF FORCING

We shall prove that direct limits of sequences of complete Boolean algebras preserve the chain condition. Before giving the proof, let me state a few facts about canonical projections of complete Boolean algebras. Let B be a complete subalgebra of a complete Boolean algebra D. For each d E D, let us define (23.21)

h(d) =

n {b

E

B : d I b}

h(d) is the least b E B such that d 5 b. The function h maps D onto B and it is easy to see that h has the following properties: (23.22)

(i) (ii) (iii) (iv)

h(d) 2 d ; h(b) = b for all 6 E B; h(dl d 2 ) = h(d,) h(d2); h(dl d 2 ) 5 h ( d l )- h(d,).

+

+

Regarding (iv), we might have d , * d2 = 0 while h(dl) . h(d,) # 0. However, the following follows easily from the definition of h(d):

(2 3.23)

if b E B and 6 d = 0,

Lemma 23.7. Let

K

then b * h(d) = 0

be a regular uncountable cardinal. Let a be a limit ordinal,

and let

Bo 5 Bl 5

G Bi E

(i < a )

be a sequence of complete Boolean algebras such that for any i < j , Bi is a complete subalgebra of Bj and such that for each limit ordinal y < a, B, = lim diri
<

n

1 {he(w):w E W } = 1

Each B, is K-saturated and so there exists a set W, c W of size
1 {h,(w):w E W,}= 1

For each <, let q(<) be an ordinal q < K such that W,c B, (such an q exists because I W, 1 < K). Now, there exists a limit ordinal y < K such that for every 5 < y. q ( < )< y. I claim that W G B, and therefore I W I < K, contrary to our assumption.

23.

239

MARTIN'S AXIOM AND ITERATED FORCING

Let w E W be arbitrary and consider h,(w) E B,. Since Uc<,,B , is dense in B,, there exists < y and b E B , such that 0 # b Ih,(w). By (23.24), there exists u E W, such that b and h,(u) are compatible. By the choice of y, we have u E B, . By (23.23) (applied to hc), we have that b and u are compatible. Since b I h,(w), u and h,(w) are compatible. By (23.23) again (applied to h,), u and w are compatible. Since u and w are both in W, it follows that w = u, and we have w E B,, as claimed. Exercise 23.8. In Lemma 23.7 let a = K and assume that B , = lim diriey Bi for a stationary set of y's (rather than for all y < K ) . The conclusion of Lemma 23.7 remains valid.

Proof of Theorem 51

+

Let !UI be a transitive model of ZFC GCH. We shall construct (in %I) a complete Boolean algebra B such that if G E B is generic, then iUI[G] satisfies 2,O I K 2 and MA,,. Thus %{[GI will be a model of MA + 2,O = K,. The algebra B will satisfy the countable chain condition and so cardinals will be preserved. Also, we shall have IB I = K, and so (2Ho)WIG1 = K2 will follow by Lemma 19.4. We shall construct a sequence B,, a -= o,, of complete Boolean algebras such that: (23.25)

(i) B,is a complete subalgebra of B, whenever a < /?; (ii) if a is a limit ordinal, then B, = lim diri4, Bi; (iii) each B, satisfies the c.c.c.; (iv) I B, I I K2 for all a < w , .

Then we let B = lim diru+02B, . By Lemma 23.7, B satisfies the c.c.c., and since it has a dense subset B, of size I K2,it follows that I B I I K t o = K 2 . At successor stages, each B,, will be constructed by two-step forcing: is such that B,, = B, * C,where C E

uu<02

(23.26) IIC is a complete Boolean algebra, satisfies the C.C.C. and

has a dense subset of size I K1/Isa = 1 Then by Lemmas 23.6a and 23.7, each B, satisfies the countable chain condition. Let me show, by induction on a, that the condition (23.26) also guarantees that 1 B, 1 IK 2 ,for each a c 0,. If a is a limit ordinal, then B , has a dense subset of size IK, and so I B , I I K2.If 1 B , I I K2and Csatisfies (23.26), then let Q E %IBB. be such that 1 ) I Q I IK, and Q is dense in CII = 1. Then the following set is dense in the algebra B, * C: (23.27)

B,

* Q = { h . q : 0 # b E B,, q E B, * C,and

114 E QIJ = 1)

The number of B,-valued elements of Q is at most equal to the number of functions from K, to B,; since 1B.I I K2,this number is IN:' = K,. Thus

4. SOME APPLICATIONS OF FORCING

240

there are at most K2 q's in (23.27) and so the set B, * Q has size IN,.Since B,, satisfies the C.C.C. and has a dense subset of size
,

(23.28) b = IIR is a partial ordering of G I and (GI, R) satisfies the c.c.c.IIBa Let C E 9JlBnbe the (B,-valued) complete Boolean algebra defined in 9RBmas follows: (23.29)

IIC = r.o.(Gi, R)ll = b IIC is the trivial algebra {0, 1}11 = - b

Let B,,, = B, * C. It follows from the definition of C that C has property (23.26), and hence B,, satisfies the C.C.C. and has size at most K2. Finally, we let B = lim dir,+Ol B, . Let G be a generic ultrafilter on B. Since B satisfies the countable chain condition, IUI[G] has the same cardinals as 9.R. Since JBI = K,, we have 9l[G] k 2O ' I K,. We shall show that 9Jl[G] satisfies MA,,. By Exercise 23.2, MA,, implies 2,O 2 K,, and it follows that Martin's Axiom holds in m[G]. For each a < 02, let G, = G n B,.

,

Lemma 23.8. ff X

E

IDI[G] is a subset o f o , , then X E Im[Gdfor some a < 02.

Proof: Let X E !INB be a name for X.The set X is determined by the B-values b, = E XII, 5 < ol. Since B, is dense in B, and B satisfies the c.c.c., countably many b," E B,, n < o,such that there exist, for each t c o,, b, = b,". It follows that there is a < o2such that all the b, belong to B, , and so X has a &valued name. H

u,<,,,2

c."=o

u,<02

To prove that MA,, holds in W[G], let (P, < ) be a partially ordered set in IDI[G] such that IDI[G] t ( P , c ) satisfies the c.c.c., and let 9 E IDI[G] be a collec-

23.

24 1

MARTIN'S AXIOM A N D ITERATEDFORCING

tion of at most N dense subsets of P. By Lemma 23.2(ii), we may assume that I P I = N l; there is a binary relation R on w isomorphic to ( P , < ), and so we may as well assume that (P, < ) = (wI, R). Now we use Lemma 23.8, which of course applies also to each X G w 1 x wl. 9 is a set of at most K I subsets of o and as such can be encoded by a subset of w 1 x o1; therefore Lemma 23.8 applies to 9 as well. Thus there is p < w 2 such that both R and 9 belong to 911[Gp]. Let R E 9W . be a Bp-valued name for R. Because of the preparations we made before the construction, there is y < w 2 such that R is in fact the 7th Bp-valued relation on G I . Let tl 2 p be such that (p, y) = (pa,7,). Since 91i[G,] is a submodel of 9ll[G], and 9li[G] 1 ( w l , R ) satisfies the c.c.c., it follows that !Ul[G,] k (wl, R) satisfies the C.C.C. Thus

,

(23.30)

b = II(G1,R) satisfies the C.C.C.~)~, E G,

We have constructed B , , to be B, * C where 1)C= r.o.(Gl, RJI = b (see (23.29)). Hence (by Lemma 23.4b), 9J[G,+,] = 9i[G,][H] where H is an 9ll[G,]-generic filter on (w R). Now because 9 E 9i[GJ, we have H n D # 0 for each D E 9. Hence H is 9-generic. Therefore MA,, holds in 9J[G], and the proof of Theorem 51 is complete. Exercise 23.9.l If 91 is a model of ZFC + GCH, and if K is a regular uncountable cardinal in 91, then there is a generic extension 91I1[G] that satisfies MA and 2N0= K. [Iterate to K instead to w 2 and make sure that MA1 holds for all I < K . ]

More about Martin's Axiom I shall prove two more consequences of Martin's axiom. One is that 2" = 2,O for every K < 2N0,and the other is that if MAN,,then products of partially ordered sets (or topological spaces) preserve the countable chain condition. There will be more applications of Martin's axiom in Section 24, and more in Chapter 7 (concerning real numbers, particularly Lebesgue measure). Let me also mention that MA,, has a number of consequences in point set topology, usually involving separability and the countable chain condition. A typical theorem using MA,, is, e.g.: Every compact first countable C.C.C.Hausdorff space is separable (while a Suslin line can be used to produce a counterexample).

Theorem 52 (Martin-Solovay). Martin's axiom implies that 2" = 2'" for all K < 2". Corollary. 2O ' is a regular cardinal. = cf 2" > K, for all K < 2'O.I [Proof of Corollary: cf The proof uses almost disjoint subsets of w. If A,, A2 are subsets of w then we say that A, and A 2 are almost disjoint if A , n A2 is finite.

4. SOME APPLICATIONS OF

242

FORCING

Lemma 23.9. There exists a family of 2O ' pairwise almost disjoint subsets of w. Proof: Let S be the set of all finite (0,1)-sequences: S = uF=f=o "(0,1). For every f: w + {0, l}, let A, E S be the set A, = {s E S : s c f } = {f n : n E w } .Clearly, A, n A, is finite i f f # g; thus {A, :f E "(0, 1)) is a family of 2O ' almost disjoint subsets of the countable set S, and the lemma follows.

I

Let K < 2,O and let us assume that MA, holds. We fix a family {A, : a c K ) of almost disjoint subsets of o. Lemma 23.10. For every X c K there exists A (23.31)

a

E

X

iff

Cw

such that for all a < K ,

A, n A is infinite

Granted Lemma 23.10, the mapping that sends each X E K to some A E o such that (23.31) holds is one-to-one, and maps P ( K )into P(w).Hence 2" I 2'O. Proof. Let X c K. We let (PI c )be the following notion of forcing: A condition is a function p from a subset of w into (0, 1) such that:

(23.32)

(i) dom(p) n A, is finite for every a E X; (ii) { n : p ( n ) = 1) is finite.

The set P is partially ordered by inverse inclusion: p 5 q iff p extends q. We first show that P satisfies the countable chain condition. If p and q are incompatible, then { n : p ( n ) = 1) # { n : q(n) = 1) and since there are only countably many finite subsets of w, it follows that P satisfies the C.C.C. For each p E K - X, let D, = ( p E P : A, G dom(p)}. Any q E P can be extended to some p E D,: Simply let p ( n ) = 0 for all n E A, - dom(p). Since A, is almost disjoint from all A,, a E X,p has property (i) in (23.32) and hence is a condition. Thus each D, is dense. For each a E X and each k E w. let El,k = { p E P : { n E A , : p ( n ) = 1 ) has size at least k} It is easy to see that each Ea,k is dense in P. Let 9 be the collection of all D, (p E K - X )and all E,.,(a E X,k E w ) . By MA,, there exists a $generic filter G on P. Note thatf= G is a function on a subset of w. We let

u

A

= { n : f ( n )=

1) = { n : p(n) = 1 for some p

E

G)

If OL E X,then A n A, is infinite because for each k there is some p E G n E l . k , If p E K - X, then A n A, is finite because for some p E G, A, c dom(p) and { n : p ( n ) = 1 ) is finite. Theorem 53. MA,, implies that if P i , i ing the countable chain condition, then condition.

E

Hi

I, are partially ordered sets, all satigyE,

P i also satisjes the countable chain

23.

243

MARTIN’S AXIOM AND ITERATED FORCING

We first observe that it is enough to show that (assuming MA,,), the product of two partially ordered sets preserves the C.C.C.

ni

Lemma 23.1 1. fj t, Pi does not satisfy the c.c.c., then there is a finite S such that Pi does not satisjj the C.C.C.

ni

C

f

Proof: Let W be an uncountable pairwise incompatible subset of the product P . For each p E P , the support of p, s ( p ) = { i E I : p i # 1) is finite. By the A-lemma (Lemma 22.6), there exists an uncountable Z c W,and a finite set S c I such that s ( p ) n s(q) = S for any distinct elements p , q E Z. Then {p 1 S : p E Z} is an uncountable incompatible subset of Pi.

flies

If ( P , < ) is a partially ordered set, we say that P satisfies the strong countable chain condition if every uncountable W E P has an uncountable subset Z such that the elements of Z are pairwise compatible.

Lemma 23.12. If P and Q are partially ordered sets such that P satisfies the strong C.C.C. and Q satisfies the c.c.c., then P x Q satisfies the C.C.C. Proof: Let W be an uncountable subset of P x Q. If the set

W,

= {p E

P : ( p , q ) E W for some q E Q}

is at most countable, then there exists some p such that W, = ( 4 E

Q : (P,4 ) E

W)

is uncountable. Since Q satisfies the c.c.c., there are ql, q2 E Wpthat are compatible. Then (p, q l ) , (p, q 2 ) are compatible elements of W. If W, is uncountable, then since P satisfies the strong c.c.c., there exists an uncountable Z G W, such that any two elements of Z are compatible. For each p E Z pick q p E Q such that ( p , q p ) E W. If q p , = qp2for some pl, p 2 E Z, then ( p , , q p , ) ,( p 2 , q p , )are compatible elements of W. Otherwise, the set {qp: p E Z} is uncountable, and hence some q p , and qp2 are compatible. Then (pl, q p , ) , ( p 2 , q p 2 )are compatible. W Exercise 23.10. If P and Q satisfy the strong c.c.c., then P x Q satisfies the strong C.C.C. Exercise 23.1 1 . I f P the strong C.C.C.

=

nit,Pi and ifall Pi.i

E

I , satisfy the strong c.c.c., then P satisfies

In particular, if P is the notion of forcing that adjoins K generic reals ( 1 9 4 then P satisfies the strong C.C.C.(compare with Exercise 23.4). Theorem 53 will now follow from this lemma:

Lemma 23.13. MA,, implies that every partially ordered set that satisfies the satisfies the strong C.C.C.

C.C.C.

Proof: Let P be a partially ordered set that satisfies the C.C.C. and let W = {wa: a < wl} be an uncountable subset of P . We will use MA,, to find a filter G such that Z = G n W is uncountable.

4. SOME APPLICATIONS OF FORCING

244

First I claim that there is some p o E W such that every p I p o is compatible with uncountably many w , . Otherwise, for each a < o there is p > a and some u, I w, which is incompatible with all w , , y 2 /3; then we can construct an of pairwise incompatible elements. ol-sequence {u,, : i < ol} For each a < olrlet

D, = { p I po : p I w, for some y 2 a } By the above claim, each D, is dense below po . By MA,,. there exists a filter G on P such that p o E G and G n D, # 0 for all a < wl. It follows that G n W is uncountable.

H

24. SOME COMBINATORIAL PROBLEMS

In this section I shall discuss various combinatorial problems that are related to the topics covered in the two preceding sections. I shall prove several theorems dealing with trees, almost disjoint sets and functions, and ultrafilters, and state several related results without proof. Trees

First let us consider Aronszajn trees (see Section 22), that is, trees of height o1such that each level is at most countable. The Aronszajn tree constructed in

Lemma 22.3 has the following property: Its nodes (elements of T) are one-toone functions from some countable ordinal into a fixed countable set, and x I y just in case x c y . Let us call an Aronszajn tree that can be represented this way a special Aronszajn tree. [Special Aronszajn trees (and their generalization to greater heights) have application in model theory; this is because " there is a special Aronszajn tree" can be expressed as: " a certain sentence has a model of type (Kl, KO)."] Now let us consider a stronger property: An Aronszajn tree T is regular if there exists a function f: T -P o such that (24.1)

f ( x ) #f(y) whenever x and y are comparable

Exercise 24.1. A regular Aronszajn tree is special. [If x E T is at level a, represent x by the (a 1)-sequence ( f ( x o ) , j ( x , ) , j ( x e ) , . . . , j ( x ) ) , where xo < x, < ... < xc < ... are the predecessors of x.]

+

. . .,

Exercise 24.2. If T is a special Aronszajn tree, then the set of all successor nodes of T is a regular Aronszajn tree.

Lemma 24.1. r f T is an Aronszajn tree, then the following are equivalent: (i) T is regular (a) T is the union of countably many antichains.

24.

245

SOME COMBINATORIAL PROBLEMS

(iii) there exists a function R from T into the set Q of all rationals, such that x
(24.2)

implies

R(X)

<~ ( y )

Proof: (i) + (ii): Let $ T + o be a function that satisfies (24.1). For each natural number n, we let A, = { x E T :f ( x )= n}. Then clearly each A , is antichain and T = uFz0A , . (ii) + (iii): Let T = A , , where each A, is an antichain. We define K : T Q by induction on the A , . First we let R ( X ) = 0 for all x E A o . Having defined R ( X ) for all x E u:I: A , , such that R is order-preserving, we can easily define n ( x ) for all x E A,, such that R stays order-preserving. (iii) (i) is obvious. H

u:=o

-+

,

-+

Notice that by (ii), a regular Aronszajn tree is not a Suslin tree. Thus the following theorem is a strong refutation ofexistence of Suslin trees from MA,,. Theorem 54. If MA,, holds, then every Aronszajn tree is regular. Given an Aronszajn tree T, we shall use MA,, to construct a function ,f: T + w that will attest to the regularity of T. First we prove the following

lemma. Lemma 24.2. If T is an Aronszajn tree and W is an uncountable collection of finite pairwise disjoint subsets of T , then there exist S, S’ E W such that any x E S is incomparable with any y E S’. Proof: Since uncountably many elements of W have the same size, we may as well assume that there exists a natural number n such that IS1 = n for all S E W; furthermore, let us consider a fixed enumeration {zl, . . ., z,} of each set S E W. Let D be a uniform ultrafilter over W , i.e., an ultrafilter such that every X E D is uncountable. Let us assume that the lemma is false. For each x E T and each k = 1, . . ., n, let Yx,k be the set of all S E W such that x is comparable with the kth element of S. Since any S and S’ contain comparable elements, we have

uu k=n

XES

k=l

Yxk

=

for every S E W. Thus pick, for each S E W , an element x = xs of S and k = k s such that YX,kE D. Now, there is k I n such that the set 2 = {S E W : ks = k } is uncountable. We shall show that the elements xs, S E 2, are pairwise comparable; and that will be a contradiction since T has no uncountable branch. If S1, S , E 2 and x = x s , , y = xs2, then the set Y = Yxk n 5, is in the ultrafilter and thus uncountable. If S E Y,then the kth element of S is comparable with both x and y . Since Y is uncountable, there must exist S E Y such that the kth element of S is greafer than both x and y . But then it follows that x and y are comparable.

4. SOME APPLICATIONS OF FORCING

246

To prove Theorem 54 let T be an Aronszajn tree and let us consider the following notion of forcing (P, c):forcing conditions are functions p such that: (24.3) (i) dom(p) is a finite subset of T; (ii) ran(p) c o ; (iii) if x, y E dom(p) and x and y are comparable, then p ( x ) # p ( y ) . p is stronger than q just in case p extends q.

Lemma 24.3. ( P , < ) satisfies the countable chain condition. Proof: Let W be an uncountable subset of P. Note that the set {dom(p): p E W } is uncountable (there are only countably many functions from a finite set into 0). By the A-lemma (Lemma 22.6), there is an uncountable W , c W, and a finite set S c T such that dom(p) n dom(q) = S for any distinct elements p, q E W,. Then there is an uncountable W, c W, such that p I S = q IS for any p, q E W, . By Lemma 24.4, there exist p and q E W, such that any x E dom(p) - S is incomparable with any y E dom(q) - S. Then p u q is clearly a function that satisfies (24.3)and extends both p and q. Thus p and q are compatible elements of W and so (P, < ) satisfies the C.C.C. W For each x E T, let D, be the set of all p E P such that x E dom(p); clearly, each D, is dense in P. Let 9 = (0, : x E T}. It follows from MA,,, that ( P , c )has a %generic filter G. The elements of G are pairwise compatible and since G is %generic, every x E T is in the domain of the function f = G. The function f maps T into o and satisfies (24.1). Thus T is a regular Aronszajn tree and the proof of Theorem 54 is complete. A few more facts on Aronszajn trees:

u

Exercise 24.3.* The following are equivalent for any Aronszajn tree T: (i) T is special, (ii) the set ofall successor nodes of T is the union of countably many antichains, and (iii) there exists a function R from T into the reals such that x < y implies n ( x ) < ~ ( y ) .

In the next exercise, P , denotes the partially ordered set obtained from (T, < ) by reversing the order.

u."=,,

Exercise 24.4.* The following are equivalent for any normal Aronszajn tree T: (i) there A,, is cofinal in T; (ii) whenever G c P T is are antichains {A,,}:=o such that Wgeneric, then XF is countable in 9t[G]; and (iii) whenever G c P, is %I-generic, then 9)l[G] contains a subset w that is not in 91.

A last word on Aronszajn trees. A simple modification of the construction in Lemma 22.3 gives a homogeneous Aronszajn tree. A construction of a rigid Aronszajn tree is more difficult but possible (a result of Baumgartner). We shall consider a generalization of Aronszajn trees to higher cardinals later, in the chapter on large cardinal properties.

24.

247

SOME COMBINATORIALPROBLEMS

A tree of height w 1 whose levels are countable can have at most 2'' uncountable branches. Aronszajn (and Suslin) trees are examples of trees without uncountable branches. While it is easy to construct a tree with countable levels and at most K1 uncountable branches (see Exercise 24.5), existence of such trees with more than K 1branches turns out to be an interesting problem. Exercise 24.5. Construct a normal tree of height o,which has exactly K , uncountable branches. [Let the tree consist of functions t: a + o such that t ( ( ) = 0 except for finitely many

0.1 A tree (T, < ) is a Kurepa tree if:

(24.4)

(i) height (T) = w , ; (ii) each level of T is at most countable; (iii) T has at least K, uncountable branches.

Theorem 55. Existence of Kurepa trees is independent of the axioms of set theory: (a) There is a model of ZFC in which a Kurepa tree exists. (b) lf there is an inaccessible cardinal, then there is a model of ZFC in which there are no Kurepa trees. I shall give two examples of generic models; in one, Kurepa trees exist, and in the other, Kurepa trees do not exist. In fact, Kurepa trees exist in the constructible universe, and I shall sketch the construction afterward. The assumption of an inaccessible cardinal in (b) is necessary since one can show that if there is no Kurepa tree, then K, is inaccessible in L. First we construct a generic model of ZFC in which there is a Kurepa tree. Given a ground model 91 that satisfies 2H0= K1,we extend it to a model 9ll[G] such that 9ll[G] has the same cardinals as 91 and contains a Kurepa tree. To define the notion of forcing (P, <), let us consider the normal a-trees (a < w , ) , described in (22.6).A condition is a pair p = (T,f), where: (24.5) (i) T is a normal a-tree, for some a < w , ; (ii) f i s a one-to-one function whose domain is a countable subset of w 2 and such that for each i E dom(f),f(i) is an a-branch in T. We let (7",f')I ( T ,f)if: (24.6) (i) T' is an extension of T, i.e., T = ( t ) a : t E T'}; (ii) dom(f') 2 dom(f), and for every i

3c(

s height(7")

E dom(f),f'(i)

such that

zf(i).

Let G be an 91-generic set of conditions and let ,F=

u { T :( T , f )

E

G for some f }

We shall show that in 911[G] .F is a Kurepa tree. The same argument that I gave in the generic construction of a Suslin tree

4. SOME APPLICATIONS OF FORCING

248

shows that 9- is a normal tree of height w , . Since for every i E w 2 , the set of conditions ( T , f )such that i E dom(f) is dense, it follows that from the generic set G one obtains a one-to-one function F such that dom(F) = w 2and each F ( i ) is an w,-branch in 9. Thus 9- is a Kurepa tree, provided w 1 and w 2 are cardinals in !UI[G]. The notion of forcing (P, < ) is clearly Ko-closed and hence K, remains a cardinal. In order to show that all cardinals 2 K2are preserved, we prove that (P, < ) satisfies the K,-chain condition. This is proved using a generalization of the A-lemma (Lemma 22.6): Lemma 24.4. Assume 2O ' = K,. Let W be a collection of countable sets, such that I W 1 = K2. Then there exists Z c W such that 1 Z I = K2,and a set A, I A 1 IKO,such that X n Y = A for any distinct elements X , Y E Z .

ProoJ: It is not very difficult to give a direct generalization of the proof of Lemma 22.6. Let me give an alternative proof (which can also be used to prove the A-lemma). Since I W I = K,, we may as well assume that X c 0,for each X E W. Let us enumerate W: W = {Xu: a < 0,). Let So = { a < w2 : cf a = 0,). For each a E So there is B < a such that X u n a c B; call this B = f ( a ) . The function f is regressive on the stationary set S o , and so there is y < 0,and a stationary set S c So such that Xu n a c y for all a E S. By the continuum hypothesis, there are only K, countable subsets of y, and so there is a set U c S of size K, and A c y such that X u n a = A for every a E U . Then one can easily construct (by transfinite induction) a set Z c U of size K, such that Xu n X, = A for any distinct a, E Z. H Now it is easy to show that the notion of forcing (P, <) defined in (24.5) satisfies the K,-C.C. (provided 2'O = K1 holds in !UI): Let W be a subset of P, I W J = K,. The set {dom(f): ( T , f ) E W} has size K, by the continuum hypothesis; there are only KI normal trees of height
u

r)

As usual, p < q iff p

3 q.

24.

249

SOME COMBINATORIALPROBLEMS

( P , < ) is Xo-closed, and so 91 and 911[G] have the same countable sequences of elements of 91. Also, N F G 1= X l , and NyGl= I. The following lemma plays the key role in the proof:

Lemma 24.5. If P is an Ko-closed notion of forcing and T is an o,-tree in the ground model 9Jl such that every level of T is countable, then in W [ G ] T has no other branches than those that are in 91. Proof: Assume that T has a branch b E W [ G ]that is not in 9Jl; since W [ G ]has no new countable sets, b has length wl. There is a name b for b and a condition po E G such that

(24.8)

po

Ik

b # ii

for all a E 9Jl

We construct, by induction, conditions p s < p o and nodes x, E T for all finite sequences s of0's and 1's. Having constructed p s , we can find (because of (24.8)) two incomparable nodes xsoand xsl,both > x,, and two conditions pso and psl, both stronger than ps such that pso IF Zs0 E b and psl ~ki sE, b. Moreover, we can find such xsoand xsl at the same level of T. Let a < w I be such that all x, lie below level a in T. For each$ w -+ (0, l}, let p, be a condition stronger than all ps,", n E 0. Since p o IF b is uncountable, there exists q < p s and x, at the ath level of T such that q Ik if E 6. Now it is clear that x, # xg wheneverf and g are distinct 0, 1-functionson w. Thus the ath level of T has at least 2Noelements, contrary to our assumption. H It follows immediately from the lemma that in the model 'ilJI[G](given by (24.7)), no tree T E 9Jl whose levels are countable can be a Kurepa tree: since every branch of T in 'N[G] is in YJI, T has at most (2N1)9R branches, but < I = KFG1,and so T has (in 9Jl[G])less than K2 branches. (2N*)w A similar argument can be used for any tree in W [ G ] , with a slight modification. For each a < 1, let Pa denote the set of all conditions whose domain is a subset of a x w , ; similarly, let P" = {p E P : dom(p) c ( K - a ) x w , } .Clearly, P is (isomorphic to) the product Pa x P".Let X E W [ G ] be a subset of wl, and let X be a name of X ; since P satisfies the I-c.c., there exists for each 5 < w1 a set of conditions W, c P of size less than I such that E XI(= { p : p E W,}. There exists a < I such that W, c P a , for all 5 < wl. It follows that X E 9N[G n Pa]. Now let T E W [ G ] be an ol-tree with countable levels. Since T can be considered a binary relation on wl, there exists a < 1 such that T E W[C n Pa]. By the product lemma 20.1, G n P" is P"-generic over W[G n Pa] and W G ] = W[G n Pa][G n P"]. Since W[G n Pa] and W have the same countable sequences in roZ, it follows that P" is KO-closednot only in mZ, but also WG n Pa] k P" is KO-closed.Thus Lemma 24.5 applies and every nPd < I = KFG1,and branch of T in 9N[G] is in WG n PJ. However, (2N1)W[G so T is not a Kurepa tree in W G ] . This completes the proof of Theorem 55.

1

W

4. SOME APPLICATIONS OF FORCING

250

As I mentioned earlier, I/ = L implies that Kurepa trees exist, and I shall now sketch the construction in L But first let me give another formulation of the problem. Lemma 24.6. A Kurepa tree exists i f and only i f there exists a family 9 of subsets of w , such that: (24.9)

(i) 1 9 1 2 K,; (ii) for each a < w , , I {X n a : X

E

9) 1 I KO.

ProoJ: (a) Let (T, < T ) be a Kurepa tree. Since T has size K,,we may assume and moreover that a < j? whenever a c T p. If we let 9 be the that T = a,, family of all w,-branches of T, then 9 satisfies (24.9). (b) Let 9 be a family of subsets of w , such that (24.9) holds. For each X E 9, let f x be the function on w , defined by fx(a)= X n a

(a < q )

For each a < w , , let and let T = Uac,,U a .Then (T, c )is a tree, the Uaare the levels of Tand the functions f x correspond to branches of T. By (24.9)(ii), every U , is countable, and it follows that T is a Kurepa tree. Now assume V = L I shall construct a family 9 of subsets of w 1 that satisfies (24.9). I present this construction to give the reader a taste of the methods used in the theory of constructible sets. The theory (" the fine structure of L"),a mixture of model theory and recursion theory, is a powerful technique that is used to prove various combinatorial principles in L. (I shall introduce some of them below.) The culmination of the theory is Jensen's covering theorem, which will be discussed in Section 30. We shall use the facts on models (&, E ) mentioned in Section 13. In particular, we recall (Exercise 13.14) that if Lais a model of ZF-, then for every X c L, there is a least elementary submodel M < L a such that X G M. Also w , . We (see Exercise 13.16), if M < (L,,, E ) , then w , n M = a for some a I also use the fact (from model theory) that elementary submodels of a given model are closed under unions of chains. For each a < w , , there is a least elementary submodel M of (L,,, E ) such that a E M. Moreover (see Exercise 13.15), M = L,,for some y < w,, and we let Y =f(a): (24.10)

f (a) = least y such that

a

E

Ly

< (L,, E)

Let 9 be the following family of subsets of w1: (24.1 1)

9 = {X E w1 : X n a E Lfc,, for every a < wl)

24. SOME COMBINATORIAL PROBLEMS

251

It is immediately clear that {X n a : X E Y}is countable for each a < w , ; and hence if we show that 1 9I = K 2 ,9 will satisfy (24.9). Assume that I .FI IK1.Then 9 has an enumeration

c = (xc:5 < wl)

(24.12)

and any such enumeration is in L,, . If we let C be the <,-least such C in Lo,, then since the functionfis a definable element of L2 (by the definition (24.10)) and the X, satisfy (24.11) in (Lz, E), it follows that C is a definable element of

(L,, d.

Now, we construct an elementary chain of submodels of (L,,, N,
E):

(v<wl)

as follows: N o is the least elementary submodel of L,; N,+l is the least N 4 L,, such that N, c N and N , E N ; if q is a limit ordinal, then N,, = N,. Note that each N , is countable, and w1 n N, = a,, for some a, < wl.Moreover,

uv<,,

(24.13)

(a, : v < 0,) is a continuous increasing sequence of countable ordinals. Now, we let X = {a, : a, 4 X,}.Obviously, X # X rfor all 5 < wlr and we shall show that X satisfies the condition in (24.11), which will contradict the assumption that (24.12) is an enumeration of all elements of 9. We want to show that X n a E L,(,) for all a < wl. By induction on a, if a is not a limit point of the sequence (24.13), then let a, be the largest a, < a. Then either X n a = X n a, or X n a = (X n a,) u {a,};in either case, since X n a, E L,(,”)c I,,(,) (by the induction hypothesis), we have X n a E L,(,). Thus it suffices to show that X n a,, E L,(,,,, whenever q is a limit ordinal. We shall show that (24.14)

(i) (a, : v < I t )

E

L,(,,,);

(4 (x,n a,, 5 < a,,) E L,(,,,).

Since L,(,,,) is a model of ZF-, the set X n a,, has then the following definition in Lfc,,,): X n a = {a, : v < q and a, $ X, n a,,} For each v < w 1, let II,be the transitive collapse of N,. Each N , is isomorphic to some La(,),and since w 1 n N, = a,, we have n,(wl)= a,. Since C is a we have C E N, for all v and one can see that definable element of LZ, II,(C)= (x,n a, : 5 < a,). Note that a,, is uncountable in La(,,,,while it is countable in L,(all).It follows that S ( q )
4. SOME APPLICATIONS

252

OF FORCING

,

submodel of La,, N:, is the least N < 4, such that N v u {N,}c N, etc. It is not difficultto show, by induction on v < q, that for each v, N , is isomorphic to N,. Then if n: denotes the transitive collapse of N v ,we have n’JNV]= L,,,), and so (L,,,, : v < q ) E Lfca,).It follows that (a, : v < q ) E Lfca,),and we have (i). W It should be mentioned that a modification of the above construction proves that if V = LEA] where A C w , , then a Kurepa tree exists. This fact can be used to show that the assumption of an inaccessible cardinal in part (b) of Theorem 55 is necessary: If w z is not an inaccessible cardinal in L, then there exists A c w such that oft”] = w , and w!jtA1= w z (see Exercise 15.8). There exists a tree T E LEA] that is a Kurepa tree in LEA]. It easily follows that T is a Kurepa tree in the universe (it has height N countable levels, and at least K o ,-branches). Thus Kurepa trees exist unless K, is inaccessible in L Almost Disjoint Subsets of w , If A,, A2 are subsets of wl, then we say that A , and A z are almost disjoint if 5 KO.This definition generalizes the notion of almost disjoint subsets of w introduced in Section 23. In Lemma 23.9 we showed that there exists an almost disjoint family of infinite subsets of w, of size 2N0.The counterpart for w1 is true in case the continuum hypothesis holds:

I A , n A, I

Exercise 24.6. If 2Ko= HI,then there exists a family of 2K1pairwise almost disjoint uncountable subsets of 0 , . [As in Lemma 23.9, let S = “(0, 1); ( S I = K,.]

u.
If the continuum hypothesis fails, the problem becomes more interesting.

To start, we have

Lemma 24.7. There exists a family of K, pairwise almost disjoint uncountable subsets of w , . Proof. Let me prove a little stronger statement: There exists an almost disjoint family offunctionsf: w1 + w , , ofsize Kz.[we recall (cf. Section 8) thatf, g are almost disjoint iff (a) # g(a) for all but countably many a’s.] Since functionsf: w1 + w1 are uncountable subsets of w 1 x w 1 and since if 1; g are almost disjoint, then I f n g I I KO,the lemma will follow. It suffices to show that if given Hlalmost disjoint functions { f v : v < a,}, then there existsf w1 + o1almost disjoint from allf,, v < 0,; this we do by simple diagonalization: L.etf(a) # allf,(a), v < a. W

If T is a Kurepa tree, then the family of all a,-branches is an almost disjoint family of uncountable subsets of T. In fact, since the levels of Tare countable, we can identify the w,-branches with functions from w 1 into u and get the following result: There exists an almost disjoint family of K functionsf: w + w.

,

24.

253

SOME COMBINATORIALPROBLEMS

Exercise 24.7. Given a family 9of K 2almost disjoint functionsf: o1-+ o,there exists a collection 9'of K2 pairwise almost disjoint stationary subsets of wl. [Eachfe 9 is constant on a stationary set S , with value n,. There is Y c 9ofsize K2 such that nI is the same for allfE '3. Let Y = {S, : f Y}.] ~

I shall return to the problem of size of almost disjoint families in the section on saturation of ideals (Section 35). One of the results to be proved there is the following: if the ground model !Ul satisfies GCH and if !Ul[G] is obtained by adjoining any number of Cohen generic reals, then any almost disjoint family of uncountable subsets of w , in IUI[G] has size at most K2. Now let T be a tree with K 1nodes, of height w , ; we do not require that T has countable levels; let us ask whether Tcan have more than K 1a,-branches. As in Lemma 24.6, we have: Exercise 24.8. Let 1 2 K 2 .There is a tree T of size K1and height o1with 1 ol-branches if and only if there is a family .F of subsets of w1 such that 1 91 = I and

I{X n a : X

E

9 and a < ol}l= KI

Note that if 2'O = K,,then there is such a tree with 2"' a,-branches (let T = Ua<,,"(0, 1)). Note also that if there is a tree like that with L a,-branches, then there exists an almost disjoint family of L functionsf: w 1 + w1 (and hence an almost disjoint family of I uncountable subsets of w ,). However, there exists a generic model (using an inaccessible cardinal) in which 2'O = 2" = K2,and not only is there no Kurepa tree, but no tree of size K, has K2 w,-branches. Finally, if T is an a,-tree of size K,,let us ask how many branches, countable or uncountable, T can have. This problem is related to a problem on linearly ordered sets. I t can be shown that: There exists a tree of height w with K1 nodes and at least I branches if and only if there exists a dense linearly ordered set of size I with a dense subset of size 8,. Exercise 24.9. There is always a tree of height w l with K1 nodes and K2 branches. [Distinguish two cases: Either 2N0 = KI, or 2'O 2 Kz .]

But even in this case T does not necessarily have 2" branches. Let !Ul[G]be the generic model obtained from a model of GCH by adjoining K,, Cohen generic reals. In !Ul[G] any tree T of size K 1 and height K, has at most 2no = H,, countable branches. Moreover, T has at most K2a,-branches since by the result mentioned above, any family of uncountable almost disjoint subsets of a, in W[G] has size at most K2. However, 2" > K,, (because cf 2n1> Ec,), and so T has less than 2" branches. It follows that in !Ul[G]there is no dense linearly ordered set of size 2" with a dense subset of size K,. Some More Combinatorial Principles in L

We have so far encountered two combinatorial principles true in L: the diamond and the existence of a Kurepa tree. I shall state below several other

4. SOME APPLICATIONS OF FORCING

254

principles that have been used in combinatorial set theory, model theory, and algebra and that can be proved in L (although the proofs are in general more complicated than the proofs given above). Let me start with an easy generalization of 0 : (24.15) Given K 1stationary subsets E , , v < wl, of wl, there exists a sequence of sets S, , a < w l , such that for every X G w , , and every v < wl, the set

is stationary.

{a E E , : X n a = S,}

Exercise 24.10.+ If V = L,then (24.15) holds. [Modify the proof of Lemma 22.9. To define S,, take into account the sets E , , v < a.] Exercise 24.11. Assume that (24.15) holds. If E , , v < wl, are stationary subsets of w,. then there exist pairwise disjoint stationary sets F , , v < wI,such that F,. G E , for all v < w1. [Let X,, v < wl. be distinct subsets of w l . Let FI= { a E E , : X, n a = &I; the sets

F: are pairwise almost disjoint, so let F , = F: -

u,,<" 5.1

The following generalization of a Kurepa tree is quite useful: (24.16) Silver's Principle W. There exists a family 9 of subsets of w l r and a sequence W,, a < 04,of sets such that: (i) (ii) (iii) (iv)

(91 2 K2; for each a < w , , ((Xn a : X E 9) 1 IKO; for each a < ol, I W, I I KO; for every countable C c 9, there exists a. < w 1 such that for all a >a,, {X n a : X E C} E W,.

Clauses (i) and (ii) just assert that there exists a Kurepa tree (see (24.9)).Each W, is a countable collection of subsets of the ath level of the Kurepa tree. The proof of (24.16) in L is quite complicated. It is easier to obtain (24.16) by forcing: .Fthat satisfies (24.9), then there is a Xo-closed notion of forcing such that !Ul[G] satisfies Silver's principle W. [A forcing condition is a pair ( W , U),where I' = ( W, : [ < a ) and (6 is a countable set of countable subsets of 9 ;(W', V ) is stronger than (Ir, U ) if 1 " extends I., (8' z %, and for every { such that a 5 [ < a' and every C E '6. { X n 5 : X E C) E W'.] Exercise 24.12.* If U ! u1 has a family

An example of an application of (24.16) is the following: Using 0 and W, one can construct a Suslin algebra of cardinality K2.There are other applications in the partition calculus and in point set topology. Another theorem of L is the following. It has several applications in algebra. If a < w 2 is an ordinal of cofinality w then closed unbounded, stationary, and thin subsets of a are defined in the obvious way.

255

24. SOME COMBINATORIALPROBLEMS

(24.17) There exists a stationary subset S of w 2 such that for every CL < o2of cofinality ol, S n a is a thin subset of a. Exercise 24.13.j There exists a No-closed, Nl-distributive notion of forcing such that

\UI[G] satisfies (24.17). [A forcing condition is a function p: u -P (0,1) such that a < w 2 , and that {< < y : p ( < ) = 1) is thin in y. for every y I u of cofinality w , ; p’ is stronger than p iff p’ 2 p. To verify N,-distributivity, let S b e a name for a function on w 1 and let po be a condition. Construct an w,-chain of conditions p o c p , c ... c p , c ..., u < wl, such that each pat decides /.(a) and that for each limit ordinal u < 0,. if j’, = Iimt4%(dom p,), then p,(y.) = 0; then p = p.is a condition and decidesf(6) for all u < w , . ]

Finally, let us consider the following principle (also true in L) which has several applications, notably in model theory:

(24.18) The Box Principle (0): There exists a collection (C, : a < w 2 ,a a limit ordinal} such that for every limit a < w 2 , (i) C , is a closed unbounded subset of a ; (ii) if cf a = w , then 1 C , I IKO; (iii) if y is a limit point of C,, then C , = C , n y. (Note that by (ii) and (iii), C , has order-type w 1 whenever cf a = wl.) Exercise 24.14.j There exists an No-closed, N,-distributive notion of forcing such that

‘IJI[G]satisfies the box principle. [A forcing condition is a collection p = {C, : u 5 y}. where y < w 2 is a limit ordinal, y} is stronger than (C;: u I y’} if y 2 y’ and the C, satisfy (24.18). A condition (C,: a I and C, = C: for all u < y’. To verify N,-distributivity. letfbe a name for a function on w , and let po be a condition. Construct an wl-chain of conditions p o c p I c ... c p . c ..., u < w , . such that each p.+ decidesf(6) and that for each limit ordinal u < w , . if )I* = limt+, (dom pt), then y. E dom p a , and for each limit ordinal p < u, C,,is an initial segment of C?,. Then if 1) = lim,,,, y., let C, = C , and p = {C, : < y}; p is a condition and decides f ( i )for all u < w , . ]

,

<

Note that the box principle implies (24.17): For each limit ordinal 5 < ulr let S, = { a < w 2 : cf a = w and C , has order type There is 5 such that S = S , is stationary. If a < w , has cofinality w l r then S n a is not stationary in a because if C: denotes the set of all limit points of C,, then Ch is closed unbounded in a but S A C, has at most one element.

r}.

Ultrafilters over w Since every ultrafilter over w is a set of subsets ofw, there exist at most 2’”’ ultrafilters over w ; more generally, if K is an infinite cardinal number, then there

4. SOME APPLICATIONS OF

256

FORCING

are at most 2'" ultrafilters over K . The next theorem shows that the number of ultrafilters over K is exactly 22".To get a slightly stronger result, recall that an ultrafilter D over K is uniform if I X I = K for all X E D. Theorem 56 (PospiSil). For every infinite cardinal

ultrafilters over

K,

there exist 2" uniform

K.

We prove first the following lemma. Let us call a family %' of subsets of K uniformly independent if for any distinct sets XI,. . ., X,, Yl, . . ., Y, in %', the intersect ion (24.19)

has cardinality

X I n ... n X n n ( K

-

Yl) n ... n ( K - Y , )

K.

Lemma 24.8. There exists a uniformly independent family of subsets of I%[ = 2".

K

such

that

Proof: Let us consider the set P of all pairs (F, 9) where F is a finite subset of K and 9 is a finite set of finite subsets of K. Since I P I = K , it suffices to find a uniformly independent family V of subsets of P, of size 2". For each u G K, let

X u = { ( F , 9) E P :F

n u E 9}

and let %=

{xuU

K}

If u and u are distinct subsets of K, then Xu # Xu:For example, if a E u but 4 v, then let F = {a}, 9= IF}, and (F, 9) E X uwhile (F, 9) 4 X,. Hence I % I = 2". To show that % is uniformly independent, let ul, ..., u,, v l , ..., u , be distinct subsets of K . For each i I n and each j I m, let aijbe some element of K such that either aij E ui - v j or aij E uj - ui . Now let F be any finite subset of K such that F z {aij: i I n , j I m} (note that there are K such finite sets). Clearly, we have a

F n ui

+ F n vj

(any i I n, j I m )

Thus if we let 9 = {F n ui : i In}, we have ( F , 9) E XUi,

i In

( F , 9) $ Xu,, j I m Consequently, the intersection

Xu,n has cardinality

K.

H

n Xu"n ( K

- Xul)n ... n ( K

- XJ

24.

257

SOME COMBINATORIALPROBLEMS

Proof of Theorem 56. Let V? be a uniformly independent family of subsets of K. For every functionj W + (0, l), consider this family of subsets of K :

IK-

(24.20) G, = { X :

XI < K ) u { X : f ( X ) = 1)

u {K- X :f(X)=O)

By (24.19), the family G j has the finite intersection property, and so there exists an ultrafilter Dj such that D z G j . It follows from (24.20) that Dj is uniform. I f f # g, then for some X E C, f ( X ) # g ( X ) ; e.g., f ( X ) = 1 and g ( X ) = 0 and then X E D, while K - X E D,.Thus we obtain 2’“ distinct uniform ultrafilters over K. Let D be a nonprincipal ultrafilter over w . D is called a p-point if for every partition { A , : n E o}of w into KO pieces such that A,, q! D for all n, there exists X E D such that X n A,, is finite, for all n E w The notion of a p-point comes from point set topology, where the topological properties of p-points in the space PN of all ultrafilters over w are of some interest. First we notice that it is easy to find an ultrafilter that is not a p-point: Let {A, : n E w} be any partition of w into KO infinite pieces, and let F be the following filter over w :

(24.21) X E F iff except for finitely many n, F n A, contains all but finitely many elements of A,, .

If D is any ultrafilter extending F, then D is not a p-point. Theorem 57 below shows that existence of p-points follows from the continuum hypothesis. A recent result of Shelah states that there exists a model of ZFC in which there are no p-points. Let me first give a trivial reformulation of the definition of a p-point. Lemma 24.9. A nonprincipal ultrajilter D over w is a p-point i f and only i f it satisjies the following: If A . z A l 2 . ’ ’ 2 A,, z * . . is a decreasing sequence of elements of D, then there exists X E D such that for each n,

(24.22)

X - A,, i s j n i t e

(If (24.22) holds, we may say that “ X is included in A , modulo afinite set,” or: (24.23)

X

G

A, m0d.f.)

< ) is a countable linearly ordered set and if D is a p-point over P, then there exists X E D such that the order type of X is either w or w*. (X has order type w* iff X = {x.}.“=~and xo > x, > ... > x,, > ....) Exercise 24.15. If ( P ,

If 2N0= K,,then there exists a p-point. (b) More generally, if Martin’s axiom holds, then there exists a p-point.

Theorem 57 (a)

258

4. SOME APPLICATIONS OF FORCING

In the proof of Theorem 57 we shall use functions from o to o.Let us say that a function h: u -+ u majorizes a functionf: o -, w if h(n) > f ( n )for all n. A family of functions X majorizes all functions f: u -+ w if for every f there is h E X such that h majorizesf: An easy diagonalization shows that no countable family majorizes all functionsf: w -, o.We can use MA to extend this observation: Lemma 24.10. (a) If .#' is countable, then X does not majorize all functionsf: o o. (b) If Martin's Axiom holds and I X I < 2'O, then X does not majorize all functions. -+

Proof: (a) If cR= {h,};=o, let f be a function such that f ( n ) I h,(n) for all k = 0, ..., n. ( b ) Let P be the collection of all finite sequences p = (p(O), . . .,p(n - 1)) of natural numbers; p < q iff p extends 4. The notion of forcing (PI < ) satisfies the C.C.C.

For each n, let D , = { p E P : n E dom(p)}; each D , is dense in P . For each h E .R, let Eh = {p E P : p(n) 2 h(n) for some n ) ; again, each E h is dense in P . Since I H 1 < 2'O, there exists a filter G on P that intersects each D,and each E,,. Then f = G is a function from o to o that is not majorized by any h e x .

u

The key element of the proof of Theorem 57 is the following lemma: Lemma 24.11. Lpt Y be a family of subsets of o with the finite intersection property and let us assume that (a) Y is countable, or more generally, that (b) MA holds and ( Y I < 2K0.kt A. 2 A , 2 . * * 2 A, 2 be a decreasing sequence of elements of Y. Then there exists Z G w such that: (i) 9 u {Z}has thefinite intersection property; (ii) Z c A, mod.f., for all n E o. Proof: We may assume that if X,Y E Y,then X n Y E Y. For each X E 9, let h,: w -+ o be some function such that h d n ) E X n A,. B y Lemma 24.10, the family {h, : X E 9 }does not majorize all functions; and so there is a functionf: o o such that for every X E Y, -+

(24.24)

f ( n ) 2 h,(n)

for some n

Now we let

u {k~A,:kIf(n)} m

(24.25)

Z=

n=O

It is readily verified that Z - A,, is finite for each n, and by (24.24),Z n X # 0 for every X E 9.

24.

SOME COMBINATORIALPROBLEMS

259

Proof of Theorem 57. Let .d,, a < 2no, be an enumeration of all decreasing sequences {A,},“=, of subsets of w. We construct, by induction on a < 2N0,a chain of families Yo E . . . E Yuc .. * of subsets of w, such that each 9,is closed under finite intersections and 19, I < 2N0for all a. We let 9, = {X C w : o - X is finite}. If a is a limit ordinal, we let 9, = 9,. Having constructed 9,, we construct 9,+1 as follows: Let .d,= {A,},“=, be a decreasing sequence of subsets of w. If some A, is disjoint from some X E g U ,then we let Y,, = 9. Otherwise, the family 9 = teu u { A , : n E w } has the finite intersection property and by Lemma 24.1 1, there exists Z c w such that Z c A, m0d.f. for all n, and 9‘ = te u {Z}has the finite intersection property. Then we let 9,+ consist of all finite intersections X , n ... n X, of elements of 9“. Finally, we let Y = (9, : a < 2’O}, and let D be any ultrafilter such that D 2 3.We claim that D is a p-point: if A, 2 A , 2 ... 2 A, 2 .. . is any decreasing sequence ofelements of D, then {A&“=, = .d, for some a < 2”O and we have 2 E Y,, such that Z c A, m0d.f. for all n.

up<,

u

,

Incidentally, it is easy to obtain a p-point by forcing. Let P be the set of all infinite subsets of w and let us identify two elements of P if they differ by a finite set. (Strictly speaking, let P = B - {0}where B is the Boolean algebra P(w)/the ideal of finite sets.) Let X IY just in case X c Y m0d.f. Note that if X, 2 X , 2 ... 2 X, z... are infinite subsets of w, then there exists an infinite X E w such that X c X, m0d.f. for all n. Exercise 24.16. ( P 3 < ) is an &-closed notion of forcing and in W[G],G is a p-point. [If .cl = {A,}:=o is a partition of w, then .d E 91; consider the dense set { X E P : either X c A, for some n, or X n A, is finite for all n).] The same argument gives an alternative construction of a p-point if 2N0= K 1.. Let d , , a < w,, be all partitions of w ; we construct an w,-sequence X,, a < w , , of infinite subsets of w as follows: Given X,,let X,, c X ube such that either X , + E A for some A E . d , or that X u + ln A is finite for all A E d,. If a is a limit, let X ube such that X, G X, m0d.f. for all p c a. Then we let D = {X : X 2 Xu for some a < wI}. D is a p-point. If in the definition of a p-point, one replaces “ X n A, is finite” by “ X n A, has at most one element,” one gets a stronger notion, known as the Ramsey ulrrajiltrr. Again, Martin’s axiom implies that Ramsey ultrafilters exist. Note that in Exercise 24.16 G is in fact a Ramsey ultrafilter and that the construction of a p-point outlined above can be modified to give a Ramsey ultrafilter (if 2“O = X l ) . In Theorem 91 in Section 38 we shall construct a model of ZFC in which there are no Ramsey ultrafilters. Let me conclude this section with two problems that arise naturally in the preceding considerations. First, let us consider the following partial ordering on functionsf: w + o: (24.26) fxg iff 3no V n 2 no f ( n ) < g ( n )

,

4. SOME APPLICATIONS

260

If

K

OF FORCING

is a cardinal number, we say that S = {fa : a < K} is a K-scale if:

(24.27) (i) fa
E

S such that g
If Y c is countable then we can easily findf w w such that g
u

(24.28) (i) s c s', and E c E'; (ii) if k E dom(s') - dom(s), then s ( k ) > h(k) for all h E E. If (s,, E l ) and (s,, E , ) are conditions and s 1 = s,, then (sl, E l ) and (s,, E , ) are compatible. Hence (P, < ) satisfies the C.C.C.Let G c P be W-generic; we let f= {s : (s, E) E G for some E}. We claim that h h(k) for all k # dom(s). Thus in W[G], there is$ w -+ w such that h
u

25.

261

FORCING A N D COMPLETE BOOLEAN ALGEBRAS

W[G] has an K,-scale. If we start with a ground model u1 in which the continuum hypothesis fails, W[G] also satisfies 2N0> K1 and still has an K,-scale. The above notion of forcing can be also used with Martin’s axiom and we get:

Lemma 24.12. Martin’s axiom implies that i f % c *ohas size <2’O, then there exists f such that h
X 5Y

iff

X

-

Y is finite

[5is not a partial ordering since we can have both X IY and Y 5 X,but it is a partial ordering of the Boolean algebra P(o)/finite sets.] Every countable descending sequence X, 2 X 2 . .. 2 X, 2 * . * has a lower bound.

,

Exercise 24.1 7. Martin’s axiom implies that any descending sequence X o 2 X I2 . . ..2

X, 2 ..., a < K, of length K < 2u0 has a lower bound.

[A forcing condition is a pair (s, F ) where s is a finite subset of w and F is a finite subset of K ; (s’, F‘) < (s, F ) just in case s’ 2 s, F‘ z F, and s’ - s G X. for all OL E F . Consider the dense sets D , = {(s, F ) : Is I 2 n}, n < w, and Em = {(s, F ) : a E F } , a < K.] 25. FORCING AND COMPLETE BOOLEAN ALGEBRAS

An important feature of the method of forcing is that properties of the generic extension are determined by properties of the complete Boolean algebra that is used to construct the generic extension. For instance, we have used extensively the lemma which states that if B satisfies the K-chain condition (in the ground model), then K remains a cardinal in the generic extension. In this section, I shall look more closely into this correspondence between properties of B and properties of the B-valued model, and also give examples how, conversely, we may use Boolean-valued models to obtain results on complete Boolean algebras. The main theme of this section is the correspondence between a complete Boolean algebra B and the B-valued model VB.Since B-valued arguments are often cumbersome, I intend to use generic extensions whenever possible. For that reason, let 111 be a fixed ground model, a model of ZFC, and let us have an understanding that B, C,D, etc. are complete Boolean algebras in 9.l, and all properties of these complete Boolean algebras are their properties in the ground model. Also, unless expressly stated otherwise, a statement: (25.1)

“The generic extension W[G] has such and such properties.”

will mean that for all W-generic G G B, W[G] has such properties. In fact, this interpretation of (25.1) will sometimes need the tacit assumption that for every

262

4. SOME APPLICATIONS OF FORCING

nonzero u E B there exists G such that u E G (this is true if 9.N is countable; cf. Lemmas 16.1 and 17.5). The correct interpretation of (25.1) is: ]Ithe universe has such and such propertiesll, = 1 Most of the problems considered below deal with the relationship of the extension W[G] to the ground model 9.N. Although in general W may not be a class of W[G] (i.e., x E 9.N may not be expressible by a formula), we shall treat !Illas if it were a class of 9.N[G].The reason is that we can easily compute the Boolean value of x E 9.N: (25.2)

IIx E Wll = C { I I x =ill: a E 9.N}

(A logical purist would perhaps prefer to augment the language with a new unary predicate !Ill.) In the first part of this section, I shall present various variations on the main theme. Some of these are collected in Theorem 58 below. First, the relevant definitions. Let K and I be cardinals. A complete Boolean algebra B is (K, I)-distributive if (25.3)

Note that (25.3) is a special case of (17.27); B is K-distributive if and only if it is (~,‘I)-distributivefor all I . As in Lemma 17.7, we can reformulate ( K , I)distributivity as follows: Lemma 25.1. B is ( K , I)-distributive ifand only ifeuery collection of K partitions of B of size at most I has a common rejinement. W We recall that a complete subalgebra B of a complete Boolean algebra D is (completely)generated by a set X C _ D if B is the least complete subalgebra of D such that X G B. Let K be a cardinal. We say that a complete Boolean algebra B is K-generated if there exists X G B of size at most K such that the complete subalgebra of B generated by X is equal to B. A complete Boolean algebra is rigid if there is no automorphism of B except the identity function Ita = a. For every nonzero a E B, let B, denote the complete Boolean algebra B,={u~B:u
The properties of complete Boolean algebras defined above are “ hereditary,” i.e., if B has the property, then each B,, a # 0, has the property; more generally, if B, has the property and b 5 a, then Bb has the property. Given such a “hereditary ” property, we say that B has this property locally if the set of all a E B such that B, has the property is dense. Thus, for instance B is locally rigid if {u # 0 : B,, is rigid} is dense. We similarly define locally K-generated, etc.

25.

263

FORCING AND COMPLETE BOOLEAN ALGEBRAS

Note that B is locally rigid, (etc.) iff there is a partition W of B such that each B,, a E W, is rigid (etc.) iff B is the product of rigid algebras (cf. Fig. 17.1).

Theorem 58. Let B be a complete Boolean algebra (in Iu1). (a) (b) (c) (d)

B B B B

is atomless ifand only iflm[G] # Iu1. is (K, 1)-distributioe if and only if eoeryf K I in !W[G] is in Iu1. is locally K-generated if and only if Iu1[G] = IUI[A]for some A G K. is locally rigid ifand only if every set in !W[G] is Iu1-deJnuble (in -+

WC]).

The four cases of Theorem 58 show a typical correspondence between complete Boolean algebras and generic extensions. For instance, if there exists a (nontrivial) rigid complete Boolean algebra, then we can obtain a model 9 l [ G ] such that

Iu1 # (HIulD)SWG1 = Iu1[G]

(25.4)

On the other hand, if there is a generic model W[G] such that (25.4) holds, then there exists a rigid complete Boolean algebra. Note that clauses (a) and (b) are formulated without the localness assumption; this is because “ locally atomless ” implies “atomless,” and similarly for distributivity. Proof of Theorem 58a. If B is atomless and G is generic on B, then it is easy to see that G 4 91 (cf. Exercise 16.6). Conversely, if a E B is an atom, let G = {u E B : a Iu } . G is clearly a generic ultrafilter on B, and G E Iu1; hence !Ul[G] = 911. Proof of Theorem 58b. Let B be (K,l)-distributive, and let G be generic on B ; let us show that every$ K -+ 1 in 9l[G] is in 911. (Compare with Lemma 19.6.) Let $ K -+ I and let f be its name; there is b E G such that (25.5)

b ikfis a function from ii- into

/If@)

x

For each a < K and each p < 1, let uUB= b . = fill. For each a < K, the set W, = {uus: p < A} is a partition of b, of size at most 1. By distributivity, the partitions W,, a < K, have a common refinement W. There exists u E W such that u E G ; then for each a < K there is a unique p = p(a) such that u Iu , ~ . Clearly,f(a) = P(a) for all a < K, and s o f ~91. Conversely, assume that B is not ( K , I)-distributive. There exist partitions W, = {uas: p < A), a < K (some uaS may be zero), such that

c

(25.6)

g:K+A

U
Let b = -u, and letfbe a B-valued name for a function such that

/If(&)

=

=b

. uUB= oUB

(a < K ,

p c 1)

4. SOME APPLICATIONS OF FORCING

264

Note that b forces thatfis a function from {uap : f l < A} is a partition of b and

K

into 1. For each a -= K, the set

(25.7)

I claim that b forces thatf$ 9.R. Let h E 9.R be a function from K into I, and let p Ib. It follows from (25.7) that there is a such that for some p # y, p is and compatible with both uU8 and u,,. Then p * ua8 I k f ( d i ) = p - u,, ~ k f ( d i ) = jk Thus either p . ua8 or p . u,, forces that f ( a ) # h(a), and the claim follows. W

B

Exercise 25.1. If B is ( K , 2)-distributive, then B is ( K , K)-distributive. [Use Theorem 58b and work in 9I[G]. If there is a function$ K + K that is not in !)It then there is a subset of K that is not in 91.1 Exercise 25.2. Let I = 2". If B is ( K , 2)-distributive, then B is ( K , I)-distributive. [Workin W[G]. Assume that every subset of K is in W and show that every F: K + I is in W.Let F: K -+ P(K).Define X G K x K as follows: (a, b) E X iff fl E F(a). Since X E !Ill, and F is defined from X by F ( a ) = (/I: (a, /?) E X),we also have F E W.]

B is called weakly

(25.8)

(K,

2)-distributive, if

nc

=

c nc

uas

Exercise 25.3.* B is weakly ( K , I)-distributive if and only if every f: K + I in W[G] is majorized by some g : K -+Ithat is in W (i.e.,f(a) < g(a) for all a < K ) . Exercise 25.4. Let P be the notion of forcing that adjoins a single Cohen-generic real. The algebra B = r.o.(P)is not weakly (w, o)-distributive. [See the proof of Lemma 24.10b.l Exercise 25.5. B is weakly (w. w,)-distributive iff w , is a cardinal in WI[G].

Let us turn our attention to part (c) of Theorem 58. To start, the notation W[A] needs explanation. If A is a subset of then W [ A ] denotes the least model 'ill2 W of ZFC (if it exists) that has the same ordinals as 111 and contains A. In general (e.g., if W is a set), 9."A] need not exist. But if 9.R[G] is a generic then W[A] does exist, and in fact (as we prove extension of W and A E 9Jl[G], below) +9"A] is itself a generic extension of 9.R.

Let B be a complete Boolean algebra and let D be a complete subalgebra of B. If G is generic on B, then it is easy to see that G n D is generic on D,and so W[G n D] is a model of ZFC, and (25.9)

WGEGnD]ZW[G]

We shall prove that every model of ZFC between Wand '%l(G]is obtained this way, and that for every subset A of ID1 in WG] there is a complete subalgebra D of B such that %N[Gn D]= WA].

25. FORCING

265

A N D COMPLETE BOOLEAN ALGEBRAS

Lemma 25.2. Lat X be a subset of a complete Boolean algebra B such thnt B is completely generated by X. Then for every generic G on B, (25.10)

91[G] = 9J[X n GI

Proof: We want to show that 9ll[G] is the least model such that the set A = X n G is in 9JI[G].I t suffices to show that G can be defined in terms of A (in any model !Jl z 911 containing A). Since B is generated by X, every element of B can be obtained from elements of X by successive (transfinite) application of the operations - and Thus let X, be subsets of B defined inductively as follows:

1.

xo = x

(25.1 1)

X, = { - a : a E Xu} X, = la : a = 1 Z where 2 C_

I

u (X,u X,)) ,
Then B = Ua<@ X, for some 0 (in fact 0 I I B I+). If we denote G, = G n X u ,G, = G n X,,we have (25.12)

Go = A

G,

={-a:

a

E

X, - G,}

G , = {a E X, : a =

1 Z where Z contains at least one b in some G, or G,,

p < a}

and G = IJa<e G, . Thus given A, we define G,and G, inductively using (25.12), and let G = IJ,<e G,. Corollary 1. If B is K-generated, then 9J[G] = W [ A ] f o rsome A

C_ K .

Corollary 2. IfC is generic on B and A E 93[G] is a subset O ~ K then , there exists a K-generated complete subalgebra D o f B such that W[Dn GI = W [ A ] . Proof: Let A be a name for A. We let X = {u, : a < K}, where (25.13)

u, = 116 E All

Now we let D be the complete subalgebra completely generated by X ;by the lemma we have W[X n GI = W[Dn GI. It remains to show that 'M[X n GI = W [ A ] . On the one hand, we have A = ( a : u , E X n G}. O n the other hand, X nG={u,:ol~A}. Lemma 25.3. Let G be generic on B. If '31i s a model of ZFC such that '131 5 '3 G q G ] , then there exists a complete subalgebra D C _ B such that '3 = W[D n GI.

266

4. SOME APPLICATIONS OF

FORCING

Proof. We shall show that 91 = W[A], where A is a set of ordinals. Then the lemma follows by Corollary 2 of Lemma 25.1. First we note that since satisfies the axiom of choice, there exists for every X E % a set of ordinals A, E % such that X E W[A,]. We let Z = P ( B ) n %, and let A = A,; we shall show that % = W[A]. If X E 9,consider the set of ordinals A,; by Corollary 2 above, there exists a subalgebra D , c B such that W[A,] = W [ D , n GI. Hence D , n G E %, and we have D, n G E Z. Since Z E W[A], it follows that D, n G E 9UI[A] and hence X E W[A]. Thus a = W[A]. H

The following lemma gives a necessary and sufficient condition for a subalgebra to give the same Boolean valued model as the given algebra. Let B be a complete subalgebra of a complete Boolean algebra D. Let us recall the definition of canonical projection h of D onto B: h(x) = C {u E B : u 2 x }

(25.14)

Let a be a nonzero element of D. Even if a @ B, we may define (25.15)

B,={x.a:x~B}

Again, B, is a complete Boolean algebra. In accordance with the discussion preceding Theorem 58, we say that B is locally equal to D, if the set { a : B, = D,} is dense in D . Lemma 25.4. Let B be a complete subalgebra of D. Then B is locally equal to D if if 9UI[G] = W[G n B] (for all generic G G D).

and only

Proof. If B is locally equal to D, and if G is generic on D, then there exists a E G such that B, = D, . Since D, n G = B, n G = { a . x : x E B n G} it follows that D, n G E W[Bn GI. Clearly, G E W[D, n GI, and so W[G] = W[G n B]. Thus let B G D be such that B is not locally equal to D. There is d o E D such that for all nonzero a I d o , there is u I a such that u # B,. Let G be a generic ultrafilter on D such that do E G. We shall show that 9 t [ G n B] # 9lt[G]. Let G, = G n B. Let C E W[G,] be the algebra C = D / = , where (25.16)

d,

=d,

iff

h(d, Ad,) $ G ,

(C is the G,-interpretation of the B-valued algebra D : B from Exercise 23.6.) Note that if d , = d , and d , E G, then d , E G (if h(d1Ad2)@ G , , then d , Ad, E G). Let [d] denote the =-equivalence class of d. Let H = {[d] : d E G} By the preceding remark, we have G = {d: [d] E H}, and it follows that W G I = W[Gll[HI*

25.

267

FORCINGA N D COMPLETE BOOLEAN ALGEBRAS

H is an ultrafilter on C , and an argument similar to the proof of Lemma 23.4b shows that H is 91[Gl]-generic. We shall show that H is nonprincipal. Then it follows easily that H # 9XII[Gl] (see Exercise 16.6). Thus let [u] E H; we shall find u such that [u] E H and [u] < [u]. We have u E G, and we may assume that u I d o . Note that if 11 I a is such that u 4 B,, then h ( o ) . a> u ; also, a - u 4 B,, and hence h(a - u ) . u # 0. Now for every a I u, we have B, # D,, and hence the following set is dense below u : {w

# 0 : w = u . h(u - u ) for some u 5 u)

Hence there is u I u such that u . h(u - u ) E G. Thus h(u) E G, and h(u - u ) E G, and we have u f 0 and u - u f 0. In other words, we have found H 11 I u such that [u] E H and [u] # [u]. Proof of Theorem 58c. Let B be locally K-generated, and let G be generic on B. There is a E G such that B, is K-generated. By Corollary 1 of Lemma 25.2, 9l[G n B,] = 9ll[A] for some A 5 K. But 91[G n B,] = 91[G]. Conversely, assume that 9ll[G] = 9ll[A] for some A E K, for all generic G c B. Hence there is a name A E 911’ such that IIA G k and W[A] = 911[qlI

=

1

Let D be the subalgebra of B generated by 115 E All, a < K (see the proof of Corollary 2 of Lemma 25.2). Then 91[G] = 911[G n D]for all generic G; and by Lemma 25.4, D is locally equal to B. Since D is K-generated, it follows that B is locally K-generated (verify that if D, = B,, then B, is K-generated). H Exercise 25.6. IBl Ih - < ’ .

If a complete Boolean algebra is h--generated and I-saturated, then

Exercise 25.7.

Every infinite countably generated C.C.C. complete Boolean algebra has

size 2-.

If B and C are complete Boolean algebras and if n is an isomorphism between B and C , then it is clear that for any generic ultrafilter G on B, n[G]is a generic ultrafilter on C and W[G] = 9l[n[G]]. We shall now establish the “converse” of this statement.

Lemma 25.5. (a) Let B and C be complete Boolean algebras (in W). Let G be a generic ultrafilter on B and let H be a generic ultrajlter on C such that W[G] = W [ H ] . Then there exist b E G and c E H and an isomorphism n (in %I) between B b and Cb such that n[G n Bb] = H n c b . (b) Let G and H be generic ultrafilters on B, such that W[G] = 3n[H]. Then there is an automorphism n o f B (in 3n) such that n[G] = H .

4. SOME APPLICATIONS OF

268

Proof. (a) There is (25.17)

FORCING

H E W8such that iG(H)= H and

I(H is a generic ultrafilter on

c'

and W[Hl = 91[q11,,= b, E G

(where G is the canonical name for a generic ultrafilter on B). Without loss of generality, we shall assume that bo = 1 (otherwisewe replace the algebra B by the algebra Bbo)' Next I claim that there is c, E H such that for every nonzero a I c,, 114 E HI1 # 0. Let S be the set of all x E C such that 112 E HI1 = 0, and let s = S. Then

1

(25.18)

(IS E H(I = (((%E 3)x E HI( =

1 ((2E HI( = 0

XES

and it follows that s 4 H.Thus c,, = -s satisfies the claim. Again, let us assume that c, = 1, for otherwise we replace C by Cco. Now we define an embedding e of C into B as follows: (25.19)

e(a) = 114 E H(J It is easy to verify that e is a complete homomorphism and that for every a E C, (25.20)

a~

H

iff

e(a) E G

In fact, we have (25.21)

I14 E HI1 = Ile(4" E GI1 I claim that e is one-to-one. It suffices to show that e(a) # 0 whenever a # 0. However, this has been guaranteed by the choice of c, above. Let D = e [ C ] ;D is a complete subalgebra of B; and for every G generic on B, if H = iG(li), we have (25.22)

W[G n D ] = m[e[H]]

=

W [ H ]= W[G]

Thus by Lemma 25.4, D is locally equal to B.Therefore, there exists b E G such that Db = Bb. Now, let h be the canonical projection of B onto D, and let p be this mapping of Dh(b)Onto Db:

dd)=

(d Dh(b)) It is easily verified that p is an isomorphism between Dh(b)and D b .Thus if we let c = e-'(h(b)), the composition of e and p gives us an isomorphism n between C, and Bb . Moreover, it follows that for all x E C , , x E H iff nx E G. (b) Let G and H be generic on B such that W G ] = YJl[H]. If G = H,then the assertion holds trivially; if G # H,then there are disjoint u, u E B such that u E G and u E H,and by (a) there exist a Iu and b 5 u and an isomorphism n between B. and Bb such that a E G, b E H,and n[G n Ba]= x [ H n Bb]. We can now extend n to an automorphism p of B as follows (see Fig. 25.1): (25.23)

*

25.

FORClNG A N D COMPLETE BOOLEAN ALGEBRAS

269

FIGURE 25.1

For x 5 a, we let px = K X ; for x 5 b, let px = n - ' x . If x is disjoint from a b, we let px = x. In general, each x E B can be written uniquely as a disjoint sum

+

x=a*x+b.x+y

and we define p x = n(a - x ) + n- ' ( b * x ) + y

It is easily seen that p is an automorphism of B, and n[G]= H. Lemma 25.5 enables us to give the following characterization of Wdefinable sets in the generic extension W[G]: lheorem 59 (Vophka). If B is a complete Boolean algebra and G is a generic ultrafilter on B, then (HWD)mG1= W[G n B*] where

B*

={aE

B : na = a jor every automorphism n of B}

Proof: Let us work in W[G]. First we use Lemma 25.5 to show that W[G n B*] c HWD.If H is a generic ultrafilter on B such that UI[H] = W[G], then H = n[G]for some automorphism R of B. Since n is the identity on B*, we have H n B* = G n B*. Thus the set

{H n B* : H

G

B is generic and q H ] is the universe)

has exactly one element, G n B*. Hence G n B* is definable, and UI[G n B*] E HWD follows. In order to prove the converse, let us first prove that if X is an W-definable subset of %I, then X E 9I[G n B*]. Let X E W Ebe a name for X ;we shall find another name for X, a B*-valued name Z E!l)uIpThen . X = iG nB.(Z)and hence X E %l[G n B*]. Since X is W-definable, there is a formula q ( X , ...), with parameters in and some a E G such that (25.24)

a

IF

X is the unique

X such that q(X,.. .)

4. SOME APPLICATIONS OF FORCING

270

Let A E W be such that a ik X c A. We may assume that for each x E A, 115 E XI1 E a (otherwise we replace X by a name X' such that 112 E XI1 = 112 E XJI. a). Let Z be the name (for a subset of A ) defined as follows: (25.25) ((3E Z ( (=

(a(I(3E

XI(): a is an automorphism of B}

(x

E

A)

It follows from definition (25.25)that for any x E A and any automorphism p of B, ~(113E 211)= 11% E 211. Hence 2 is B*-valued. It remains t o show that X = Z, where Z = iG(Z)= iG nB.(Z).Since ((2E XI( I112 E 2 1 1 for every x, we have X c_ Z. O n the other hand, let x E Z. Since 112 E 2 1 1 E G,there exists by (25.25)an automorphism a of B such that ~(112E XI[) E G.Let u = a n(II2 E XI[)and let u = a-lu. Note that u E G.We shall show that u IF 2 E X (and hence x E X). On the one hand, u In(II2 E XI[) and so u Ik 2 E a(X)

(25.26)

O n the other hand, we have u I a (since u I 112 E XJJIa), and so u which gives us

11 q(X,

. . .),

au = u Ik q(nX, ...)

(25.27)

But also u Ia, and (25.24), together with (25.27), implies that u Ik a ( X ) =

x

Hence u ~k i E X and we have X = Z. Now we shall show that the model 91 = 9I[G n B*] includes HfUID. Let Y E HmD be of least rank such that Y # '31. Then Y G '31, and there is A E '31 such that Y G A. There is a one-to-one function J E '31 such that f [ A ] E W. Then X =f[Y] is an W-definable subset of 91 and hence X E '31. Therefore Y = f - ' [ X ] is in '31. Corollary 1. I f B is rigid, then !lIl[G]

Ik

euery set is 'JJ1-definable.

A complete Boolean algebra is weakly homogeneous if for any nonzero u and u there is an automorphism a such that n ( u ) . u # 0. It is easy to see that then no u # 0, 1 is fixed under all automorphisms, and we have Corollary 2.

If B is weakly homogeneous, then (H'lIID)SWG1 = '%I.

P be the notion of forcing that adjoins a single Cohen generic real, r.o.(P). Show that B is weakly homogeneous.

Exercise 25.8. Let

and let B

=

This exercise gives us consistency of the following:

L=HOD#V The above Corollary 2 can be proved directly, using:

25.

271

FORCING AND COMPLETE BOOLEAN ALGEBRAS

Exercise 25.9. If B is weakly homogeneous, if cp(xl, .. . ,x.) is any formula, and if xlr . .., E ! I l l , then the value lIcp(g1, ..* , ic.)ll is either 0 or 1.

x,

Proof of Theorem 58d. If B is locally rigid and G is generic on B, then there is a E G such that B, is rigid, and since Zn[G] = W[G n BJ, every set in WG] is Zndefinable in W[G], by Corollary 1 of Theorem 59. Conversely, assume that B is not locally rigid. Then there is u # 0 such that for every nonzero u I u, the algebra B, has a nontrivial automorphism. Let G C B be generic such that u E G; we shall show that G is not Zndefinable in 9w31. Otherwise, there is some q ( X , . . .), with parameters in W,and u I u such that (25.28)

u Ik G is the unique

X such that q ( X , . . .)

Let n be a nontrivial automorphism of B,; it is easy to find disjoint zlr z 2 I u such that nzl = z 2 . Then z1 ~k2, E G and z2 4 G, and z2 (= nzl) forces 2, E n(C),hence z 2 Ik G # n(G) This contradicts (25.28) since u

~k q(C)and

also u = nu

ik

q(zG). W

By Theorem 58d, there are two possible ways to obtain a rigid complete Boolean algebra. Besides the "direct " algebraic construction, one can also achieve the same result by constructing a notion of forcing such that in the generic extension every set is !JJI-definable.And indeed, the first construction of a rigid complete Boolean algebra (due to McAloon) was accomplished by the indirect method. I shall not present the construction here, but let me mention one simple idea, namely how one can obtain a set of integers, not in W but definable: Construct a two-step generic extension. In the first step, adjoin a Cohen-generic subset A of o,and then use Easton's method (Theorem 46) to change the continuum function so that in the resulting model we have 2'" = K,, if n E A, and 2Hn= Kn+2otherwise. In the model, the set A is definable by A = {n E 0 :2" = Kn+l}

The above discussion and Exercise 25.8 show that both L = HOD # V and L # HOD = V are consistent (and so is L # HOD # V). Moreover, there are models in which (HOD)"OD# HOD: The submodel HOD has fewer definable subsets than the given model. In fact, one can even construct, for any ordinal 0, a transfinite sequence of models 'WU,a < 0, such that W U + = (HOD)%

for every a < 0 (and !IllA=

nu<,!Nuif I is a limit).

4.

272

SOME APPLICATIONS OF FORCING

Thus the notion of ordinal definability is, unlike constructibility, not absolute. In the constructible universe, a rigid complete Boolean algebra can be constructed with much less effort, particularly if the reader is already familiar with the construction of a Suslin tree (Section 22). Below, I shall sketch a construction of a rigid complete Boolean algebra using 0 (or its version (22.20)). Construction of a Rigid Complete Boolean Algebra in L (a sketch) Any normal Suslin tree can be embedded upside down as a dense set in an atomless complete Boolean algebra: Let (T, 5 T) be a normal Suslin tree, let 5 = 2 and let BT = r.o.(T, I ). Let U , , for each a c o 1, denote the ath level of T. Since B satisfies the c.c.c., there is for each b E B some a and some u G U , such that b = { t : t E u } ; thus elements of B are represented by subsets of levels of T. In Exercise 22.15 we used 0 to construct a rigid Suslin tree. Now 1 shall show that a slight modification yields a tree T such that B , is a rigid complete Boolean algebra. If C is a closed unbounded subset of ol. let ?c denote the subtree of T consisting of elements at levels a E C:

F=

u {U,:a

E

C}= { t E T : o(t) E C}

Lemma 25.6. If n is an automorphism of B T , then there is a closed unbounded C G o1such that n I F is an automorphism of F . Proof. Let C be the set of all a such that for each t E U , , there is s E U,such that s = x ( t ) , and there is z E U , such that z = n-'(t). It is easy to verify that C is closed. To show that C is unbounded, let a. be arbitrary. Let A G B be the countable set {nt : t E Uuo}u {n-'t : t E Uuo}

There exists a1 > a. such that all elements of A are represented by subsets of Ua1.In a similar way we find az > a l r a3, ad, . . ., and then let a = a,,. I claim that a E C: If t E U , , let t,, n c o,be the predecessors oft in T at levels a,. For each n, let s, E LlUnbe such that t,+ E II- '(s,). See Fig. 25.2. The s, form a chain in T and it follows that there is s E U , that has the s, as predecessors; and s = n(t). We find z = n - ' ( t ) in U , similarly. I By Lemma 25.6, we have reduced the problem of construction of a rigid complete Boolean algebra to a construction of a normal Suslin tree T such that is rigid, for all closed unbounded C E ol. The key step in the construction is a modification of Exercise 22.6:

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FORCING A N D COMPLETE BOOLEAN ALGEBRAS

I

T

FIGURE 25.2

Lemma 25.7. I f T is a normal a-tree where a < w , is a limit ordinal, ifC is a closed unbounded subset of a, and i f n is a nontrivial automorphism of F,then T has an extension T' of height a 1 such that A cannot be extended to the ath level.

+

Now we use 0 and construct a normal Suslin tree as in Lemma 22.8, with the added twist that we also use the version (22.20) of 0,and whenever the h, of (22.20) is an automorphism of for some closed unbounded C G a, we make sure that h, does not extend to the ath level. This way we obtain a normal Suslin tree T such that is rigid for all closed unbounded C G wI. Then B , is a rigid complete Boolean algebra. H Simple Complete Boolean Algebras

Let us consider now the following concept, which is stronger than rigidity: An atomless complete Boolean algebra is simple if it has no proper atomless complete subalgebra. Lemma 25.8. If B is simple and G is a generic ultrajlter on B, then:

(i) every set in !N[G] is 211-dejnable in !IR[G]; (ii) there is no model 9 of ZFC such that !IR G 91 G !IR[G]. Proof: The assertion (ii) follows from Lemma 25.3; (i) follows from Corollary 1 of Theorem 59 and the following lemma: Lemma 25.9. Every simple complete Boolean algebra is rigid.

Proof. Let A be a nontrivial automorphism of B ; we shall construct a proper atomless subalgebra of B. There exist disjoint a and b such that na = b. Let us write each x E B uniquely as a disjoint sum x=a.x+b.x+y

(see Figure 25.1), and let D be the complete subalgebra of B consisting of all those x such that b . x = x(a . x). Then D # B (since, e.g., a 4 D),and is easily seen t o be atomless. H

4. SOME APPLICATIONS OF FORCING

274

It can be proved that simple complete Boolean algebras exist in the constructible universe. One possible way to construct such an algebra in L is to modify further the construction of a rigid complete Boolean algebra given above. Given a normal Suslin tree T one can characterize all complete subalgebras of B T (in fact, they can be identified with certain equivalence relations on T); and using 0, one can arrange the construction of a Suslin tree T such that 87. is simple. Thus existence of simple complete Boolean algebras is consistent; however, it is open whether one can prove their existence in ZFC. The construction of a simple complete Boolean algebra in L mentioned above has another interesting application. The Suslin tree T that yields the simple algebra B T can be constructed so that the function a- T,is A I over L,, (see Exercise 22.18). Then if G is an Lgeneric ultrafilter on B,, one can use the fact that BT is simple and define in qG] a X binary relation which well-orders the universe. Thus we obtain a model of ZFC other than the constructible universe, which has a El well-ordering.

Suslin Algebras In Section 22 we defined a Suslin algebra as a complete Boolean algebra that is atomless, KO-distributive,and satisfies the countable chain condition. We have also shown that if T is a normal Suslin tree, then B , is a Suslin algebra (Lemma 22.2). Conversely, it is shown in Exercise 22.3 that if B is a Suslin algebra, then there is a normal Suslin tree T c B - {0},such that s = 2 B . It is not difficult to see that then BT is a complete subalgebra of B. If T is a normal Suslin tree, then the Suslin algebra B T has cardinality 2'9 The following theorem shows that every Suslin algebra has cardinality at most 2". Using a notion of forcing that generalizes the forcing conditions in (22.6), we can obtain consistency of a Suslin algebra of arbitrary prescribed size K (satisfying K'O = K); also, using combinational principles that hold in L (in fact, 0 and W, see (24.16)), one can construct in L a Suslin algebra of size K 2 . lheorern 60. If a complete Boolean algebra B is a Suslin algebra, then IB( I 2 ' l . Proof. The proof presented here uses Boolean-valued models. I t follows closely the proof of Theorem 51 (consistency of Martin's axiom). Let K = 2"l. We shall assume that there is a Suslin algebra of size > K and reach a contradiction. Let B be a Suslin algebra such that 1 B 1 > K ; without loss of generality we may assume that I B, I > K for every u # 0. We shall construct a transfinite sequence (25.29)

Bo c_ B1 c

G

B, c ...,

a
of complete subalgebras of B, each of size SK.If D E B and ID I S K, then there are K~~ = K D-valued names for binary relations on w thus for each such

25.

275

FORCING A N D COMPLETE BOOLEAN ALGEBRAS

D let R:, y < K, be a fixed enumeration of all such names. Let a -+ (pa, y.) be the canonical mapping of K onto K x K ; we recall that Ia for all a. The sequence (25.29) is constructed as follows: We let B, = (0, 1); if a is a limit ordinal, then B, is the complete subalgebra of B generated by ,<, B,. If 1 B, 1 I K for each v < a, then 1 B, 1 < K. At successor steps, we construct B,, as follows: Let D = B,, and let R = RF,. If

/?.

(25.30)

II(G,, R) is a Suslin treelIBm= 1

if C E VB1is the Suslin algebra (in V B ncorresponding ) to the Suslin tree and if B, * C is (isomorphic to) a complete subalgebra of B, then we let B,, = B, * C.Otherwise, we let B,, = B,. In either case, if I B, I IK, then IB,+l

I

B,. Clearly, Now let B, be the complete subalgebra of B generated by Urn<, I B, 1 IK. Let A E V B Kbe the complete Boolean algebra B : B, (in VBK).Since both B and B, satisfy the c.c.c., we have, by Lemma 23.6, (IA satisfies the c.c.c.IIBX= 1

Similarly, since both B and B, are &distributive, we have (using for instance Theorem 58b) I(A is K,-distribctive(lBK= 1

We have assumed that I B , I > K for all u # 0, and we also have I B, I IK. Thus

IIIAl

>KIIB~=

and consequently IIA is not atomicllBs= 1

Now we work inside V B K Using : Exercise 22.3, we find T G A such that ( T, 2 A ) is Suslin tree; let B, G A be the Suslin algebra corresponding to T (B, is the complete subalgebra of A generated by T).Let R be a binary relation on GIisomorphic to T. The name R is B,-valued; and since B, satisfies the c.c.c., R involves at most K, elements of B, (compare with Lemma 23.8). Since cf K > K,, there exists /? < ti such that R E V B @furthermore, ; let y < K be such that R is the yth BB-valued binary relation on 6,. R = RY. Let a < ti be such that /? = /I, and y = y,. Since II(Gl, R) is a Suslin treellBK= 1

we also have (25.30): II(Gl, R) is a Suslin treell,

=

1

If C denotes the corresponding Suslin algebra in VBa,we have (cf. remark following Lemma 23.3) B,

* C G B, * B, c B, * A = B

4. SOME APPLICATIONS

276

OF FORCING

and it follows that B,, = 8, * C.However, as the reader knows, forcing with a Suslin tree destroys its Suslinity, and we have

R) is not a Suslin

=1

~~(&l,

Which of course is a contradiction. W Collapsing Algebras

In Section 19 we defined a notion of forcing that collapses a cardinal A onto a cardinal K IA. As defined in (19.12), the forcing conditions p E P are functions such that dom(p) G

ran(p) E A,

K,

and (dom(p)I < K

I shall refer to the corresponding complete Boolean algebra as the collapsing algebra, and use the notation

(K,

A)-

C o l ( ~A), = r.o.(P)

(25.31)

I shall present below several interesting theorems on the collapsing algebra Col(Ko, A). In particular, it will be shown that every complete Boolean algebra can be embedded in a countably generated complete Boolean algebra. If A is an infinite cardinal, we let P Adenote the set of all finite sequences

P(n - 1)) (n E w ) of ordinals less than A; we have Col(Ko,A) = r.0. P A .For a fixed A, we use u,, (n < w, a < A) to denote the following elements of Col(Ko,A): (25.32)

(25.33)

p

=

(P(O),

u,, =

* * * 1

C { p E PA : p ( n ) = a}.

Lemma 25.10. Every (KO,A)-collapsing algebra is Ho-generared.

Proof: Let A be an infinite cardinal and let B = Col(Ko, A). We shall find countably many elements b,, (n,m < w ) of B such that B is the least complete subalgebra of B that contains all b,, . First we note that the collection {u,, : n < w, a < A} generates B: Since P Ais dense in B, it suffices to show that each p E P Ais so generated. If p = (p(O), ..., p(n - I)), let u = uo,p(o). . * * * 4-I.p(n- 1) and let us show that p = u. On the one hand, p Iu ~ , ~for( all ~ , k < n, and so p I: u. On the other hand, let us show that every q E P A , q I u, is compatible with p. Otherwise, q(k) # p(k) for some k E dom(p) n dom(q), and then it fol= 0 ; this contradicts the lows from the definition (25.33) of uk,p(k)that q * uk,p(k) assumption q I u. We define, for every n and rn < w, (25.34)

b", =

c {P

E PA :

P ( 4 IP b ) I

25.

277

FORCING A N D COMPLETEBOOLEAN ALGEBRAS

and show that each u,, is in the complete subalgebra C of B generated by {b,,,, : n, m < w ) . This we prove by induction on a. Let us assume that all urn@ (m < w and /l< a ) are in C and let us show that u,, E C. Since u n S .u,, = 0 whenever /3 # 7, we have

and thus it suffices to show that We shall show that

Ips, unDbelongs to C .

(25.35) Let u denote the left-hand side of (25.35) and let u denote the right-hand side. Let us first show that u I u. Given m < w and p E Pa,p I u, it suffices to find q 5 p such that either q I b,, or q I umO.So let p I u ; we let q Ip be arbitrary such that both n, m E dom(q). Since q I u, it is clear that q(n) 5 a. Now-if q ( m ) 2 q(n), we have q 5 6,,; if q(m) < q(n), we have q ( m ) < u and so 4 2 C f i < a UmB. Now let us show that v I u. We shall show that every p I v is compatible with u. Otherwise, there is p I u such that n E dom(p) and p ( n ) > u. Then we find q stronger than p such that for some m, q(m) = a. But then q is incompatible with &<. u,,,,,; and also, since q ( n ) > q(m), q is incompatible with b,,,,. This contradicts the assumption that q I p I b,, urn@. Thus each u,, belongs to C and therefore C = B. Hence B is &-generated. W &
+ zB<,

Since I Col(N,, I)1 = I P ,

IKo

= I K o ,we obtain

Theorem 61 (Gaifman, Hales). There are arbifrarily large N,-generated complete Boolean algebras. W

The next theorem will give an even better result.

Lemma 25.11. Let (Q, < ) be a notion offorcing such that 1 Q 1 such that Q collapses I onto No,i.e..

=

A > K O and

Ix is countablell,,,,,Q,= 1 Then r.o.(Q) = Col(N,, A). Proof: Without loss of generality we may assume that (Q, < ) is a separative partial ordering. We shall find a dense subset of Q isomorphic to Pa. Let B = r.o.(Q), and let G be the canonical name for the generic filter on Q. Let YE V B be such that

IJf

maps 03 onto GI/,= 1

4. SOME APPLICATIONS OF

278

FORCING

For each p E P,, we shall construct q ( p ) E Q such that D = { q ( p ): p E PA}is dense in Q and that q ( p ) is an isomorphism of P , onto D.We construct q ( p ) by induction on the length of p. If p = ( p ( O ) ) , we construct q ( p ) as follows: Since Q collapses A, there exists an incompatible set W C _ Q of size A. Moreover, we may find such W of sue A with the additional property that each w E W decidesf(O), i.e., there is qw E Q such that w kf(d) = q w . Thus let W, be a maximal incompatible set of conditions with these properties, W , = {we : t A}, and for each p = ( p ( 0 ) ) E P , we let q ( p ) = w e , where 5 = p ( 0 ) . Having constructed q ( p ) , where p = (p(O), . . ., p ( n - l)), we construct q(pn<), < A, as follows: We let W, = {wt : < A} be a maximal incompatible set of conditions w < q ( p ) such that I W,l = A and that each w E W, decides f ( n ) .Then we let q(p"<) = w c , for all wc E W,. The set D = { 4 ( p ): p E P,} is clearly isomorphic to P,. Let us show that D is dense in Q. Let q E Q be arbitrary. Since q Ik 4 E G,and hence 4 It 4 E ran(f), there is r I q and n < w such that r It q =f(ri). Now there is p E P , of length n + 1 such that q ( p ) is compatible with r ; since q ( p )decidesf(n), we necessarily have q ( p ) Ik f(ri) = 4. Therefore, q ( p ) IF 4 E G.Since Q is separative, it follows that q ( p ) I q. This proves that D is dense in Q. W

-=

<

<

Theorem 62 (Kripke). Every complete Boolean algebra can be embedded into an KO-generated complete Boolean algebra; in fact, if I B I = A, then B can be embedded in Col(Ko,A). Proof: Let B be a complete Boolean algebra, let 1 B I = A. Then the notion of forcing Q = (B - (0)) x P, has cardinality A and collapses A. By Lemma 25.1 1, r.o.(Q) = Col(Ko,A). In other words, B 0 Col(Ko,A) is isomorphic to Col(Ko,A), and so B is isomorphic to a complete subalgebra of Col(K,, A). Since Col(Ko,A) is KO-generated,the theorem follows. W

We shall now prove a stronger version of Theorem 62. Lemma 25.12. Let B be a complete Boolean algebra, I B I = A. Let C be a complete subalgebra of B such that I C I < A, and let ho be an embedding of C in Col(Ko,A). Then there exists an embedding h of B in Col(Ko,A) such that h(c)= ho(c)forall c E C . Proof: See Fig. 25.3. Let D be the image of C under the embedding h , . Let Col be an abbreviation for Col(Ko,A); let Col' and COPdenote, respectively, the ( K O ,X)-collapsing algebra in the Boolean valued models If and VD. First, we find an embedding k of B in C * Col': Working in F, we observe that X is a cardinal and that B: C is complete Boolean algebra that collapses X onto Go. Also, since B:C is a quotient of B, B : C has cardinality X. Thus by Theorem 61 (we are still in V'), there is an embedding of B : C in Colr. By the remark following Lemma 23.3, there is an embedding k of C * (B: C) into C * Colc such that k ( c ) = c for all c E C (by Lemma 23.3, C is considered a

25.

279

FORCING A N D COMPLETE BOOLEAN ALGEBRAS

FIGURE 25.3

complete subalgebra of both those algebras). Since C

k: B + C

* (E: C) = C, we have

* Col‘

such that k ( c ) = c for all c E C. Next we find an isomorphism between Col and D * Col”: Working in VD, we observe that 1is a cardinal, and that Col:D collapses 1onto G o . Also, since the algebra Col“ has a dense subset P, of size 1,its quotient Col: D has a dense subset Q of size 1.Thus by Lemma 25.1 1 (applied in V”), there is an isomorphism between Col: D and Col”. By the same argument as above, we get an isomorphism a between C o l = D * (Col: D) and D * COP: a : Col+ D

* ColD

such that n(d)= d for all d E D. Since C and D are isomorphic, there exists an isomorphism 6 :C

* Colt+ D * Col”

such that o(c) = h,(c) for all c E C. Thus we define h: E + Col as follows: h(b)= a-’(o(k(b))),for all b E E (see Fig. 25.4): C.

k

B

,COl

FIGURE 25.4

Clearly, h is an embedding of E into Col, and h(c) = ho(c)for all c

E

C. rn

4. SOME APPLICATIONS OF

280

FORCING

Exercise 25.10.* Let G be a generic filter on PAand let X be a set of ordinals in !UI[G]. Then either !UJI[X] = !.UJI[G] or there exists an !Ut[X]-generic filter H on P , such that W[X][H] = W[G]. [If I, is uncountable in W[X], then !Ut[G] is a generic extension of !UJI[X] by C O ~ * ~ ( K ,, ,K), where K = I I, lax]. However, PI is isomorphic in !Ut[X] to P A , If L is countable in W[X] and W[X] # W[G], then W[G] is a generic extension of !Ut[X] by a countable atomless notion of forcing Q.There is only one atomless complete Boolean algebra with a countable dense subset (see Exercise 17.6), and so r.o.(Q)is isomorphic (in w[X]) to r.o.(P,).]

The Lctvy Algebra Let I be an inaccessible cardinal. We shall now consider the notion of forcing P introduced in (20.7), which collapses each a < I onto KO:The conditions are functions p on subsets of I x w such that each dom(p) is finite, and p(a, n) < a whenever (a, n ) E dom(p). Let B denote the corresponding complete Boolean algebra, the U v y algebra. We have shown in Section 20 that sat@) = I. Since I is inaccessible, we have 1B1 = 1. For each v such that o Iv < I, let P, denote the set of all p E P such that a < v whenever (a, n ) E dom(p). Similarly, let P" denote the set of all p E P such that a 2 v whenever (a, n) E dom(p). Clearly, P is isomorphic to P, x P'. Let B, = r.o.(PV)and B' = r.o.(P'). Since B is isomorphic to B,0 B',B , is isomorphic to a complete subalgebra of B having P, as a dense subset, and so we may identify B, with the complete subalgebra of B generated by P,. Thus B, is a complete subalgebra of B, whenever v c p c I. Moreover, if y > w is a limit B,, and so we have ordinal, B, is the completion of

uv<,

(25.36)

By = lim dir B, V d Y

If u E B, then because sat@) = I,there exists X c P of size less than I such that u E XX. Hence u E B, for some v < I, and it follows that B = B,. We also note that if K is an uncountable cardinal, then JP,, I = K and P , , collapses K onto KO, and so by Lemma 25.11, is (isomorphic to) Col(K0, K ) .

uvcl

If C is a complete Boolean algebra and I C I < I, then by Theorem 62 C can be embedded in some B,, v < I, and hence in B. Moreover, by Lemma 25.12 every embedding of a complete subalgebra of C in B can be extended to an embedding of C in B. We use this to prove the following theorem. Also, using Exercise 25.10, we obtain the followingfactor lemma for the U v y collapse; we shall use this factor lemma in the proof of Theorem 101 in Section 42: Exercise25.11.* Let G be a generic filter on the LCvy collapse P, and let X be a countable set of ordinals in W[G]. Then there exists an W[X]-generic filter H on P such that !JX[X][H]= W[G].

25.

28 1

FORCING A N D COMPLETEBOOLEAN ALGEBRAS

[Let 1' < 1 be such that X E 9I[G n B,]. Let K G P , be 'UI[X]-generic such that 9N[G n B,.] = ' U I [ X ] [ K ] Hence . 9lI[G] = 9 I [ X ] [ K ] [ Gn B'].]

Theorem 63. The L h y algebra B is homogeneous, in the following sense: If A and A' are isomorphic complete subalgebras of B such that I A I = 1 A' I < I B I and if no is an isomorphism between A and A', then there exists an automorphism n of B such thar .(a) = no(a)for all a E A. Proof: First we construct increasing sequences of complete subalgebras. A o S A , ~ . . . E A , , G . . A. ,b E A ; E . . . c _ A ~ G..., asfollows:

Ah

5

B,,

A; G

A;

E Bv, = A; &

FIGURE 25.5

We let A. = A and A; = A'. There is v , such that A; E B,,; we let A ; = B v , . The embedding no of Ah in B can be extended to an embedding n; of A'' in B, and we let A , = n;'[A',]. Then there is v , > v 1 such that A , s B v 2 ;we let A, = B v z . Then 11,: A , -+ B extends to some n 2 : Az-+ B, and we let A; = n 2 [ A z ] .We proceed in this manner ad infinitum. I?,", and n, is an automorClearly, A,, = uFz0A: = phism of this Boolean algebra. This automorphism extends t o a unique automorphism nu of the completion, which is B,, where v = v,. Now B = B, 0 B', and the automorphism n, of B, can be extended to an automorphism n of B, 0 B". (For instance, let us define K on the dense set (B, - (0)) x (B' - {O}) by n(u, 0 ) = (nuu, v ) . )

'

u:=o

u;=o

Corollary. I f u and 11 are elements of the Lpvy algebra B such that u # 0, 1 and v # 0, 1, then there exists an automorphism a of B such that n ( u ) = v.

',

It follows (see Exercise 25.9) that for any formula cp and all x . . ., x, E im, I(cp(x*,, ..., x*,,)llB is either 1 or 0. I will conclude this section with a stronger property of the Lkvy algebra (Lemma 25.14 below). Lemma 25.13. Let C be a complete subalgebra of a complete Boolean algebra B and let u E B be such that u # C. Then the complete subalgebra D of B generated by C u {u}has an automorphism n such that nu # u, and nu = a for all a E C.

4. SOME APPLICATIONS OF FORCING

282

Proof. First we note that

(25.37)

D={xEB : x=a*u+b-uforsomea,b~C}

We define an automorphism x of D as follows: Let u = { t E C ; t I u), and w = x { t ~ C : - ut }~ and let z = - ( u + w ) . Each X E D can be written uniquely as a disjoint sum

x=u'u+~-u+c where a, b, c E C , and c I u + w while a I z and b 5 z. It suffices to define a separately for a . u, b - u, and c. If c I u w, we let RC = c. If x = a u where a I z (see Fig. 25.6) we let

+

R X = ~ - U

~~

L

u

W

E

FIGURE 25.6

Similarly, if x = b - u where b I z, we let xx

=b

. u.

I t is not difficult to verify that n is an automorphism, and n(z * u) = z na = a for all a E C.

- u. Also,

Lemma 25.14. Let B be the Levy algebra (in the ground model W), and let C be a complete subalgebra of B such that 1 C I < I B I. For any formula cp(x), if x is a C-valued name, i.e., x E 91Ic, then IIdx)llB

Proof. Let b = II&)llB. If n is an automorphism of B such that nc = c for all c E C , then A ( X ) = x and it follows that n(b) = b. Thus it suffices to show that for any u # C, there exists an automorphism a of B such that nu # u, and n(c)= c for all c E C. By Lemma 25.13, there exists an automorphism A of a complete subalgebra D of B such that nu # u, and nc = c for all c E C; and I D I is still less than I B 1. By Theorem 63, we can extend such automorphism to an automorphism of

B.

26.

283

MORE APPLICATIONS OF FORCING

26. MORE APPLICATIONS OF FORCING

I shall conclude the chapter on forcing with a few more applications and several interesting results on generic models. Further applications of the forcing method will be presented in the chapters on large cardinals and real numbers. Alrnmt Disjoint Forcing

I shall briefly discuss a useful notion of forcing that has various applications in descriptive set theory. The forcing conditions are constructed with the use of almost disjoint sets; we have already used the method once in Section 23: The forcing conditions were defined in (23.32), in the proof of Theorem 52. Let .d = { A , : a < wl) be a family of almost disjoint subsets ofw, and let X be a subset of w,. We consider the following notion of forcing P = P d , x : A condition is a function p from a subset of w into (0, I } such that: (26.1)

(i) dom(p) n A , is finite for every a (ii) { n : p ( n ) = 1) is finite.

E

X;

The set P is partially ordered by inverse inclusion: p 5 4 iff p extends 4. If p and q are incompatible, then {n : p ( n ) = 1) # { r ~: 4(n)= 1); and it follows that P satisfies the countable chain condition. If G is a generic filter on P , then 91[G] = 9ll[A], where

(26.2)

A

= {n : p ( n ) =

1 for some p E G }

The argument given in Lemma 23.10 shows that

(26.3)

X

= {a < w1 : A n

A , is infinite}

A typical application of almost disjoint forcing is the following: Let W [ X ] be a generic extension of the ground model 911 where X G w?; furthermore, = w'f. Let .d = { A , : a < wl} be an almost disjoint family in assume that 91\ and let us consider the notion of forcing P = P d s x in 9 l [ X ] . If G c P is generic over 9ll[X], then 9 1 [ X ] [ G ]= 9 l l [ X ] [ A ] ,where A is defined by (26.2). Note that w?wxl[A1 = w'f. Now in 9 l l [ X ] [ A ] the , set X satisfies the condition (26.3);and since .d E !Ul, it follows that X E 9ll[A], and we have

9 l l [ X ] [ A ]= %l[A] Thus we have found a generic extension 9 l [ A ] of 91 such that A G w and E 9ll[A]. In other words, undcr certain conditions one can generically code a set X c w , by a set A c o.

x

Example I. Let 911 be a model of ZFC. There is a generic extension 9 l [ A ] of 9l, where A c w, such that w F A 1= w? and w? is collapsed.

4. SOME APPLICATIONS OF FORCING

284

Proof. First we add generically (in the usual way) a collapsing mapfof o 1onto o,,such that w 1 is preserved. There is in 91[f] a set X c w l such that %1[X] k lwFl = K,. Then we apply the method described above to find a generic extension q A ] of 911 such that A c_ w and X E 91[A], and that o?”’=wY. I

The method of almost disjoint forcing can be generalized to uncountable be an almost disjoint family of cardinals. For instance, let A = {A, : a < 02} subsets of o1(i.e., 1 A, n A , I I KO whenever a # p). If X is a subset ofw,, we can define by a straightforward generalization of (26.1),replacing w by w 1 and “finite” by “at most countable.” P d s x is Ko-closed; and assuming 2N0= K,,P satisfies the K,-chain condition and hence preserves cardinals. This generalization enables us, under certain conditions, to code generically a set X c 0,by a set A c w y (which in turn might be coded by A’ c w). Exercise 26.1. Let 91 be a model of ZFC + GCH. There is a generic extension 91[A] of W, where A c w, such that wf”]= my, w f ” ] = w 2 , and w y is collapsed.

Perfect Set Forcing

Several interesting results have been obtained by employing as forcing conditions perfect sets of reals, or a variant of such conditions. I shall present here two such results. Let W be a model of ZFC. We say that a set of ordinals A is minimal ouer 91 if A # 112 and if for every set of ordinals X E 9.n[A],we have either A E Im[X] or x E 9.n. Theorem 64 (Sacks). Let W be a model of AFC. There exists a generic extension W [ x ] ofW where x is a real, such that x is minimal ouer W. Exercise 26.2. A Cohen-generic real is not minimal over the ground model. [Show that P is isomorphic to P x P , and therefore W[x] = W[xI][x2], where x 1 is Cohen-generic over Zn and x2 is Cohen-generic over Z n [ x , ] . Consequently, x, # Zn and x

#

“x11.1

We note in passing that if B is a simple complete Boolean algebra, then the corresponding generic extension is minimal. In connection with the problem of existence of a simple complete Boolean algebra, it is known that the algebra used in the proof of Theorem 64 is not simple (it is not rigid). Let me recall that a set of reals is perfect if it is nonempty, closed, and has no isolated points. The following lemma is instrumental in the proof of Theorem

64.

u.“=o

Let Seq denote the set Seq = “(0,1) of all finite 0, l-sequences. Let {Ps: s E Seq) be a family of perfect sets such that: (26.4)

(i) s G t implies P , 2 P,; (ii) Pso n P,, = 0,for all s E Seq.

26. MORE APPLICATIONS OF FORCING

285

It is easy to see that x

i*

(26.5) Let us call the set (26.5) the fusion of {Ps: s E Seq}. Lemma 26.1 (Fusion Lemma). Let {Ps: s E Seq} be a family of perfect sets satisfying (26.4).Furthermore, assume that the diameter of P , tends to 0 with increasing length of s, i t . ,

(26.6)

lim (diameter PI,,) = 0

( a l l f E '"(0,1))

n- x

Then the fusion of { P s: s E Seq} i s a perfect set.

n;=,

P,,,; the fusion is the set Proof. For each f: w -+ {O, l}, let P, = P = { P , : . f "'~{O, li}. By (26.5), P is a closed set. By (26.6), each P , is nonempty and has exactly one element: P , = {x,}. We shall show that no x , is an isolated point of P. Let E > 0. There exists n such that the diameter of P,,, is < E . Let g : w + (0, I } be such that g I n =f In and g(n) # f (n).Then xg # x,and I x g - x,( < E . Thus x, is not isolated and it follows that P is perfect.

u

Let us recall the Cantor-Bendixson analysis of closed sets: Every uncountable closed set is the union of a perfect set and a countable set. One of its consequences is part (a) of the following lemma, which is a very weak instance of a general theorem to be proved in Chapter 7. Lemma 26.2.

(a) If P and Q are perfect sets, then either P n Q or P

-Q

contains a perfect

(b) For every perfect set P there exist perfect sets Q 1 c P, Q 2 c P such that Proof. (a) Let U be the complement of Q. The open set U is the union U= I, of countably many closed intervals (with rational endpoints). Each P n I, is either perfect or finite, and so P - Q = P n U = lJ;=o ( P n I,) either contains a perfect subset or is at most countable. In the latter case, P n Q is uncountable and hence contains a perfect subset.

u;=o

Proof of Theorem 64. The forcing conditions are the perfect sets of reals (in m) and P I Q just in case P 5 Q. Thus P and Q are incompatible if and only if P n Q does not contain a perfect subset. (Then by Lemma 26.2 there exists Q 1 < Q such that P n Q i = 0.)

286

4. SOME APPLICATIONS OF

FORCING

Let G be an W-generic set of conditions. (By Lemma 26.2b, G is not in 91.) We shall first show that there is a real xG E W[G] such that 91[G] = 91[xG]. Every condition P , being a closed set, is the complement of a certain union of open intervals with rational endpoints (in W).If we carry out the same construction in !M[G], we obtain a perfect set P* = PwG1;clearly, we have P = P* n W ,and moreover, P* is the closure of the set P (in !M[G]). Let us consider (in 91[G])the collection 9 = {P* : P E G } of perfect sets. 9 has the finite intersection property and, by genericity of G, contains sets of arbitrarily small diameter. Hence its intersection contains a single element x

= xc:

(26.7) The following lemma shows that 91[G]= 911[x].

Lemma 26.3. G E 911[x].

Proof. For each condition P , let P" = Pwxl; as before, P = P" n 9l,and P" = P* n W1[x]. We shall show that (26.8)

G ={P:x

E

P")

thus giving a definition of G in 9 1 [ x ] . On the one hand, if P E G, then x E P* and so x E P". O n the other hand, let P .$ G . By the genericity of G, if a condition is compatible with every element of G , then it is itself in G . Thus there is Q I E G such that P and Q I are incompatible. Now it follows by Lemma 26.2a (and by genericity again) that there is Q E G such that Q c Q , and P n Q = 0. We may also assume that Q is bounded, and then P and Q have positive distance; i.e., there is E > 0 such that any p E P and q E Q have distance 2 E . Thus in 91[G], P* and Q* have distance 2 E . This means that we have found Q E G such that P* n Q* = 0. Hence x E Q* and therefore x .$ P*. Thus we have x .$ P", and (26.7) is proved. We shall show that x is minimal over %I. Let X be a set of ordinals in 911[G]; we wish to show that either X E 911 or .x E 911[X]. Let X be a name for X. Let D ,and D, be the following sets of conditions:

D 1= ( P : P decides 6 E X for all a } D 2= { P : (VQ 5 P ) & such that Q does not decide di E X) Clearly D = D , u D 2is a dense set of conditions and so G n D # 0. If there is P , E G n D,,then X E 91l because x = { a : P, IF 6 E X } So let us assume that G n D , is empty, let P 2 E G n D and let us show that XG

E

91I[X].

26.

287

MORE APPLICATIONS OF FORCING

For every P G P , , we construct a family { P , : s E Seq} as follows: For s = 0, we let P , = P. Having chosen P, for all s E Seq of length n, we pick P,, and P,, so that: (26.9)

(i) (ii) (iii) (iv)

Pso c P, and PSIc P,; P,, n PSI= @; the diameter of both P,, and P,, is at most l/n; there exists a = a, such that ~b4, E

4, $ X Condition (iv) can be satisfied because P, G P , and P , E D,. The family ( P , : s E Seq} satisfies (26.4) and (26.6), and therefore its fusion QP is a perfect set. Since the set of conditions { Q P: P G Pz}is dense below P , and P , E G, there exists P 5 P , such that Q P E G. Let us consider such P, and the corresponding family { P , : s E Seq}. Pso

For every n E w, we have Qp

and

X

u

P,,

Ib

{ P , : s E “{O, I}}

and so there is a unique s of length n such that P , E G. Moreover, if P , E G, then by (26.9(iv)) Pso E G iff a, E X and P S IE G iff a, $ X Thus if we define in 911[X] a function$ w 0 1 we have for each s

(26.10)

E

(0, 1) as follows (by induction):

if aflnE X if afln$ X

Seq,

iff

P,EG

scf

We shall complete the proof by showing that x G E 91[X]: Let P x denote, for each condition P , the perfect set Proamin 9lI[X], and let us consider the intersection (in 911[X])

(26.1 1 )

ii pnt:

n=O

By (26.9)(iii),the set in (26.1 1 ) has exactly one element y. Similarly, the intersection (in W[G]) (26.12)

(where P* = PmG1)has also exactly one element; and since x G E PTlnfor all n (because P,,, E G), it follows that y = x G . This shows that x G E 9 I [ X ] , and so xG is minimal over 911. H

4.

288

SOME APPLICATIONS OF FORCING

An argument employing the fusion lemma is used below to show that forcing with perfect sets does not collapse N If, in addition, one assumes that 2K0= N 1 holds in 311, then there are only K 1 forcing conditions, and so all cardinals above (and including) K2 are preserved. Lemma 26.4. Iff€ 1)1[G] is afunctionfrom w into w?, rhen the range off is

bounded by some a < w?. Proof. Letfbe a name for a function from w into wl. It suffices to show that for every condition P there exists a stronger condition Q and an ordinal a < w 1 such that

Q

(26.13)

Ik

ran(f) c di

Let P be a condition. We construct a family { P , : s E Seq} as follows: For s = 0, we let P , = P. Having chosen P , for all s E Seq of length n, we pick P,, and PSIso that: (26.14)

(i) (ii) (iii) (iv)

Pso c P , and PSIc P,; Pso n P,, = 0; the diameter of both P,, and PSIis at most l/n; there exist countable ordinals a,, and ctS1 such that Pso ~ k f ( i i ) = tiso and PSI~ k f ( A ) = hsl

Then we let Q be the fusion of {Ps: s E Seq}, and let a be a countable ordinal greater than all a,, s E Seq. Let us show that Q and a satisfy (26.13).Otherwise, there is Q 1c Q, and there is n E o such that (26.15)

u

Q1

Ikf(A)

2 6.

However, since Q C { P s : s E "{0, l}}, there exists s E Seq of length n such that Q 1n P , contains a perfect subset Q2.Then by (26.14)(iv), Q 2 forces f($ = 6,, contrary to assumption (26.15). Hence Q forces that the range offis bounded by a, and the lemma follows. W Exercise 26.3. Every function8 w w in W[G] is majorized by some g: w -+ w, g E W. [Use fusion: For each s E "{O, I} consider m, such that P , ~ k f ( h )= 6 1 ~then ; let g(n) = max{m, : s E "{0,I}}.]

In the proof of Theorem 64 we could have used perfect sets in the Baire space, i.e., perfect trees T c Seq; cf. Section 4 (with the same results since NNis homeomorphic to the set of all irrationals). In that case, if G is a generic set of perfect trees, one has m[G] = !Wl[f], wherefis the unique infinite path through all T E G. Exercise 26.4. Formulate and prove a fusion lemma for perfect trees T G Seq.

The following example (due to Namba) uses a generalization of perfect trees.

26.

289

MORE APPLICATIONSOF FORCING

Example I/. Let 911 be a model of ZFC and 2K0= K l . There is a generic extension 911[G] of '31 such that wFG1= w'? and cfmG1(wy)= w.

What makes this example interesting is that co2 is collapsed, but its cofinality is changed not to w 1 (as is the case when the "ordinary" collapse is performed) but to w. It follows from Jensen's work on singular cardinals that in the absence of large cardinals, this example is the only possible nontrivial change of cofinality. For more, see Jensen's covering theorem in Section 30. Let S be the set of all finite sequences of ordinals less than w 2 : rx

S = (J "w2

A free is a set T G S such that (26.16) if t

E

n=O

T and s = f In for some n, then s

E

T.

A nonempty tree T E S is pe$ect if every t E T has K2extensions s 3 t in T. (Note that then every t E T has N2 incomparihle extensions in T.) In analogy with perfect sets in the Baire space, we have a Cantor-Bendixson analysis of trees T c S: Let

T'

= {r E

T : r has K2 extensions in T}

nP<=

and let To = T, T,, = T i , T, = T, if a is a limit. Let 19< w 3 be the least 0 such that Tb = T,. Then T, is either empty or perfect. If T has no perfect G T, then the above procedure leads to To= 125,and we can associate with each t E T an ordinal number

(26.17)

h T ( t )= the least a such that t $ T,,

If s c t , then hT(s) 2 h T ( f ) ,and for every t

(26.18)

E

T,

1 {s E T : t c s and hT(s)= h T ( f ) I} < K2

Now let me describe the notion of forcing that is used in this example. Let P be (in the ground model 9J) the set of all perfect trees T G S, partially ordered by inclusion. We shall show that in the generic extension, w 2 has cofinality w and o,is preserved. If G is a generic set of conditions, we define in W[G] a functionf: w + w? as follows : 3T E G 3 E T such that s(n) = a iff f(n) = a An easy argument using genericity of G shows thatf(n) is uniquely defined for each n and that the functionfmaps w cofinally into w?. Let me prove now that w1 is preserved in the extension by showing that every$ w -+ (0, 1) in W[G]is in the ground model. Thus let T be a condition, let f~ WE, and let T Ik f is a function from Ci, into {0, l}

4. SOME APPLICATIONS OF FORCING

290

We shall find a stronger condition that decides eachf(n'); that is, we shall find a function g: w + (0, 1) such that some condition stronger than T forces f(n') = g(n), for all n. We proceed as follows. By induction on length of s, we construct, for each s E S, conditions T, and numbers a, E (0, I} such that: (26.19)

(i) if s1 G s 2 , then T,, 2 T,, (ii) if length(s) = n, then T, I k f ( i i ) = Oi, (iii) for every n, the conditions T,, s E " w 2 ,are mutually incompatible, and moreover, there are mutually incompatible sequences such that for each s E "w2 t , E S, s E t,

E

T,

and

for all t E T,, either t

G

t, or t ,

G

t

The "moreover" clause in (iii) is stronger than incompatibility of the condiTsis compatible tions and implies that any condition stronger than USEnw2 with some T,, s E "a2. The construction of conditions satisfying (26.19)is straightforward: We let t , = 0 and T0 G T be any condition that decides f ( 0 ) :T0 1 k f ( 0 ) = a O . Having defined T,, t , , and a, for s E " w 2we first pick K 2 incompatible extensions t S i , i < w 2 , of t s in T,, and then find TSiE T, and a s i ,i < m 2 , such that T i ~ k f ( n + 1 ) = aSi and that each t E Ti is compatible with t s i . Note that if sI E s 2 , then r,, E t s 2 . For any function g: w + {O, l}, we define a tree T(g )C S (not necessarily a perfect tree) as follows: If 8 = (Po, .. ., p,) is a finite sequence of zeros and ones, we let (26.20) T ( p )=

u {T,:

s E "w2 and

( P o , .. ., fin) = ( a o , ..., a,,k, . . ., a,)}

and (26.21) Each T ( p )is a condition (a perfect tree) and by the remark following (26.19),we have (k = 0, . . ., n) T(B)I k f ( i ) = p k Thus, if we show that there is at least one g: w -+ (0, 1) such that the tree T ( g ) contains a perfect subtree, our proof will be complete. Lemma 26.5. There exists g : w + (0, 1) such that T(g) contains a perfect

subtree. Proof: Let us assume that no T ( g)has a perfect subtree. Then by (26.18) there exists, for each g: w - + (0, l}, a function h,: T ( g ) + w , such that h d s ) 2 hAt) whenever s E t , and that for each r E T(g),there are at most K , elements s S, t in T(g)such that h,(s) = h,(t).

26. MORE APPLICATIONS OF FORCING

291

By induction, we construct a sequence so c s1 c ... c s, c ... such that for each n, s, E "02. Having constructed s,, we consider the node t," of T,-. Since there are only K 1functions g: o + (0, l}, there exists i < w 2 such that hs(tSmi)< hg(tsn)for all g for which hg(tsni)is defined. We let s,+ = snni. Given the sequence s,, n = 0, 1, .. .,we consider the function g : o + (0, 1) defined by g(n) = a,",

n = 0, 1, ...

Now, by (26.20) and (26.21),each tSnbelongs to T(g),and so hg(tsJis defined for all n. However, this yields a descending sequence of ordinals hs(tso)> hs(rs,)> ..., a contradiction.

Degrees of Constructibility The relation between two sets of ordinals

(26.22) x is constructible from y , i.e., x

E

yy]

is a transitive relation (and reflexive). Thus one may be tempted to investigate the partial ordering of the corresponding equivalence classes, the degrees of consrructihilify. One might even expect a certain similarity between degrees of constructibility and degrees of recursive unsolvability (Turing degrees) investigated in recursion theory. Unlike the case for Turing degrees, there has been no systematic study of degrees of constructibility. There are a few facts known and a few results that I should mention. Theorem 64 provides an example of a minimal degree. Using a similar technique, models have been obtained where the degrees of constructibility form various finite patterns. Not much is known about, however, how, for instance, the initial segments of degrees look in general. (One result worth mentioning is a model in which the nonconstructible degrees have order-type w*, i.e., the order-type of negative integers.) Any two degrees have a least upper bound. (This is easy: If A and Bare sets of ordinals, let C = r[A x B] where r is the canonical mapping of Ord2 onto Ord; the degree of C is the 1.u.b. of the degrees of A and B.) The following theorem of Hajek, stated here without proof, shows that the same is not true for lower bounds: I t is consistent that there are two r e d s X , Y, Cohen-generic over L such that the constructibility degrees of X and of Y have no greatest lower bound. It follows that 4x1 n YY] is not a model of ZFC (and it is unclear whether it is even a model of ZF). The situation is similar for suprema and infima of countable sequences of degrees. For instance, if d o , d , , ..., d , , ... is an increasing (or decreasing) sequence of Ldegrees, then its supremum (infimum) may or may not exist.

4.

292

SOME APPLICATIONS OF FORCING

A related problem is the structure of transitive models of ZFC (containing all ordinals). Above, I mentioned an example of two models %Il and %I2 of ZFC such that (xn, n (xn2 is not a model of ZFC.Similarly, there is an example 3 Iu1, 3 . * . 2 91" 3 ... of models such that of a decreasing sequence 91, is not a model. Or consider this:

n;=o

Exercise 26.5. I f lllo c 111, c ... c 111, c ... is a strictly increasing sequence of models 111, is not a model of ZF. (containing all ordinals), then U = [If U n V, E U for all a, then there is n such that U n V, E 9.R" for arbitrarily large a ; hence U G Zn, .]

u."=o

A somewhat better picture can be obtained when generic extensions are used. For instance: Exercise 26.6. If Bo 2 B , 1 .. . z B, 2 ... is a sequence in 111 of complete Boolean algebras and if G c B is generic over !Ill, then m

W [ G n B,]

91 = n=O

is a model of ZF. [31 is transitive and closed under Godel operations. To show that for each a, V. n 9l E W, note that for every n, V. n 31 E 91[G n B,]:

n (V. n W[G, 00

V, n 9 l =

n

B,])

k=n

where G, = G n B, .]

Even when we have a sequence of models Iu1[G n BJ as in Exercise 26.6, the intersection need not satisfy the axiom of choice. I shall conclude this discussion with the following lemma, which shows that in some cases the intersection of a decreasing sequence of models of ZFC is a model of ZFC:

Lemma 26.6. Let K be a cardinal and let B be a K-distributive complete Boolean algebra. Let Bo 2 B 2 . * * 2 B, 2 * . .,a < K , be a descending K-sequence of comB, . lf G is an 9J-generic ultrajilter on plete subalgebras of B, and let B, = B, and G , = G n B, for all a I K , then

nu<,

n

m[GU]

=

m[GK]

(1x1

Proof. Clearly, G, E I1I1[GU]for all a < K, and so W[G,] is included in the intersection. To prove the opposite inclusion, it suffices to show that every set of ordinals A that is in each W[G,], a < K, belongs to WG,]. Let A be a set of ordinals, A E W[G J for all a < K. Working in q G ] , we assign to each a < K a B,-valued name A, for A. Since B is K-distributive, it follows that the sequence {A, : a < K} is in I1I1. We wish to define a B,-valued name A, for A. For each ordinal t, we shall define wt = IJtE A , (1 E B,, and then show that A is the G,-interpretation of A,; i.e., that wt E G if and only if { E A.

26.

293

MORE APPLICATIONSOF FORCING

Let

4 be an ordinal; I shall describe the construction of w = W~ . For each

11

a < K, let u, = E A , E B,. By K-distributivity, there exists a partition .(pi: i E I } of B such that for each a < K and each i E I, either p i I u, or p i . u, = 0. Hence for each a < K there exists 1 , I ~ such that u, = {pi : i E l a ] .We let

C

(26.23)

w=

1 {pi : 3,Va 2 a.

pi I u,}

We shall show that w E B,, i.e., w E B, for all a < K. Let a < K. It suffices to o I w. Thus let pi I w. show that for each pi I w there is I ) E B, such that pi I Then there is a. 2 a such that pi I u p for all p 2 a o . Let u = { u p : p 2 ao}. Clearly v E B,; and also pi I v. By the choice of the p i , there is J G I such that u = {pi :.j E J}. Now if p j Iu, then by (26.23), p j I w ; hence u I w, and we have pi Iu I w. I t follows that w E B, for all a < K. It remains to show that for each <,the element w = wr,is in G if and only if 5 E A. Let pi be the unique pi that is in G. Then w E G just in case pi I w. If 4 E A, then u, E G and pi I u, for each a, and hence pi I w. Conversely, if pi I w, then p i I u, for eventually all a ; hence u, E G and so 5 E A. H

fl

1

Genericity of Extensions At the conclusion of Part 11, 1 shall very briefly touch the question of ! and 9 are models of ZFC (with the genericity of extensions of models. If M same ordinals), the question is, When is the model 9 a generic extension of the model W? Or more specifically, Which transitive models of ZFC containing all ordinals are generic extensions of the constructible universe? It is known that in the presence of large cardinals, the universe is not a generic extension of L There are also several results that characterize generic extensions. The following theorem shows that the universe is a generic extension of the class of all hereditarily ordinal definable sets (at least if V = 4 A ] for some set A).

Theorem 65 (Vognka). Let V = LEA] where A is a set of ordinals. Then V is a generic extension of the model W = HOD. There is a Boolean algebra B E ‘JJZ, complete in !M, and there is an ultrafilter G c B, generic ouer 9JI such that V = 9JI[G]. Proof: Let

K

be such that A c K. We let

C = OD n P ( P ( K ) ) be the family of all ordinal definable sets of subsets of K. Let us consider the partial ordering (C, C-). First 1 claim that there is a hereditarily ordinal definable partially ordered set (B, 5 ) and an ordinal definable isomorphism n between (C, C ) and (B, I): There is a definable one-to-one mapping F of OD into the ordinals. The set C is an ordinal definable set of ordinal definable sets and so F 1 C is an

4.

294

SOME APPLICATIONS OF FORCING

OD one-to-one mapping of C onto F[C]. We let B = F[C], and define the partial ordering of B so that (B, I)is isomorphic to (C, 5 ) .Since c n C2is an OD relation, we have (B, I ) E HOD. Now (C, 5 ) is clearly a Boolean algebra. Moreover, if X c C is ordinal X is ordinal definable and so X = X. Hence the definable, then algebra C is OD-complete; and using the OD isomorphism x, we conclude that (B, I ) is a complete Boolean algebra in W = HOD. Now we let

u

u

H = {u E c : A E u} Clearly, H is an ultrafilter on C,and if X c H is OD, then

n

X E H. Hence G = x [ H ] is an %&genericultrafilter on B. It remains to show that V = W[G]. Let$ K + B be the function defined by f ( a ) = n({ZG K : a E 2)).Clearlyfis OD, and s o f ~W = HOD. Now we note that for every a < K, a

and therefore A

E

E

A

iff

f(a)

E

G

%)1[G]. Now it follows that V = q

A ] = 91[G].

PART I11

Large Sets

CHAPTER 5

MEASURABLE CARDINALS

27. THE MEASURE PROBLEM

Let S be an infinite set. A measure on S (a nontrivial a-additive measure, to be more precise) is a real-valued function p on P ( S ) such that: (27.1)

(i) 40) = 0 and p ( S ) = 1; (ii) if X C_ Y , then p ( X ) I p( Y ) ; (iii) p ( { a ) )= 0 for all a E S (nontriviality); (iv) if X , , n = 0, 1. 2, ... are pairwise disjoint, then

49

z,

=

1 p(X,)

(a-additivity)

“=O

I t follows from (ii) that p ( X ) , the nieasure o f X , is nonnegative for every X G S ; in a special case of (iv) we get p ( X u Y ) = p ( X ) + p( Y ) whenever X n Y = 0 (finite additivity). More generally, let .Sbe a a-complete field of sets. A measure on .Sis a real-valued function 11 on 9 satisfying ( i t ( i v ) . Thus a measure on S is in fact a measure on P ( S ) . An example of a measure on a a-complete field of sets is the Lebesgue measure on the field of all Lebesgue measurable subsets of the interval [0, 11 = {x : 0 I x I 1). The Lebesgue measure has, in addition to (ib(iv),the following property: (27.2) if X is congruent by translation to a measurable set Y , then X is measurable and p ( X ) = p( Y ) . It is well known that there exist sets of reals that are not Lebesgue measurable, and in fact that there is no measure on [0, 13 with the property (27.2) (translation inuariant measure): Exercise 27.1. Let M be a maximal (under C ) subset of [0, 11 with the property that - y is not a rational number, for any pair of distinct ?I, y E M . Show that M is not

x

Lebesgue measurable. 295

2%

5.

MEASURABLE CARDINALS

[Consider the sets M , = (x + q : x E M } where q is rational. They are pairwise {M, : q E Q n [ - 1, 11) [ - 1, 21.1 disjoint and [0, I ] G

u

The natural question to ask is whether the Lebesgue measure can be extended to some measure (not translation invariant) such that all subsets of [0, 13 are measurable, or whether there exists any measure on [0, 13. Or, whether there exists a measure on some set S. The investigation of this problem has lead to important discoveries in set theory, opening up a new field, the theory of large cardinal numbers, which has far-reaching consequences both in “abstract ” set theory and in descriptive set theory. A measure p on S is two-valued if p ( X ) is either 0 or 1 for all X C_ S. If 11 is a two-valued measure on S, let

u = {X E S : p(X) = 1)

(27.3)

It is easy to verify that U is an ultrafilter over S. For instance, let me show that if X E U and Y E U , then X n Y E U . If p ( X ) = p ( Y ) = 1, then X = ( X - Y ) u ( X n Y ) a n d Y = ( Y - X ) u ( X n Y ) . I f p ( X n Y)werenot 1, then p(X - Y) = p( Y - X) = 1, and we would have p ( X u Y) = 2. Next we note that the ultrafilter U is a-complete. This is so because p is a-additive, and an ultrafilter U over S is a-complete if and only if there is no X, such that partition of S into countably many disjoint parts S = X, # U , for all n. In fact:

u:=,

Exercise 27.2.

into less than

u,<./

An ultrafilter over S is K-complete if and only if there is no partition o f S pieces, S = X , , y < K, such that X , $ U for all a < y.

K

Thus if p is a two-valued measure on S, U is a a-complete ultrafilter over S. Conversely, if U is an a-complete ultrafilter over S, then the following function is a two-valued measure on S : (27.4) Let p be a measure on S . A set A G S is an atom of p if p ( A ) > 0 and if for every X c A, we have either p(X) = 0 or p(X) = p ( A ) . If p has an atom A, then U = {X C _ S : p ( X n A ) = p ( A ) } (27.5) is again a a-complete ultrafilter over S. A measure p on S is atomless if it has no atoms. Then every set X G S of positive measure can be split into two disjoint sets of positive measure: X = Y u 2, and p( Y) > 0, p ( 2 ) > 0. is an atomless measure on S , there exists Z c S such that p ( Z ) = 4. More generally, given Z o E S , there exists Z c Zo such that p ( 2 ) = fp(Z,). [Construct a sequence S = So 2 S I z ... z S , 2 ..., a < w , , such that p(S,) 2 i,

Exercise 27.3. If p

27.

297

THE MEASURE PROBLEM

and if p ( S , ) > i,then i 5 p(&+ ,) .= p ( S J ; ifa is a limit ordinal, let S, exists ct < w , such that p(S,) = 4.3

=

S,. There

We shall eventually prove various strong consequences of existence of a nontrivial a-additive measure and establish the relationship between the measure problem and large cardinals. Our starting point is the following theorem which shows that if a measure exists, then there exists at least a weakly inaccessible cardinal.

Theorem 66 (Ulam). Ifthere is a a-additive nontrivial measure on S , then either there exists a two-valued measure on S and I S I 2 the least (strongly)inaccessible cardinal, or there exists an atomless measure on 2'O and 2'O 2 the least weakly inaccessible cardinal. Theorem 66 will be proved in a sequence of lemmas, which will also provide additional information on the measure problem and introduce basic notions and methods of the theory of large cardinals. First we make the following observation. Let K be the least cardinal that carries a nontrivial a-additive two-valued measure. Clearly, K is uncountable and is also the least cardinal that has a nonprincipal countably complete ultrafilter. And we observe that such an ultrafilter is in fact K-complete:

Lemma 27.1. Let K be the least cardinal with the property that there is a nonprincipal a-complete ultrafilter over K , and let U be such an ultrafilter. Then U is K-complerr. Pro05 Let U be a a-complete ultrafilter over K, and let us assume that U is not ri-complete. Then there exists a partition {X, : a < y} of K such that y < K and X , 4 U for all a < y. We shall now use this partition to construct a nonprincipal a-complete ultrafilter over y. thus contradicting the choice of K as the least cardinal that carries such an ultrafilter. Let f be the mapping of K onto y defined as follows:

f ( x )= a

x E X,

iff

(x E

K)

Now, any mapping of K onto y induces a a-complete ultrafilter over y; we define D E P ( y ) thus: (27.6)

ZED

iff

~ - , ( Z ) UE

(We leave it to the reader to verify this.) Moreover, our mappingfis such that D is nonprincipal: Assume that { a ) E D for some a < y. Then X, E U , contrary to our assumption on X,. Thus y carries a a-complete nonprincipal ultrafilter. We are now ready to define the central notion of this chapter. An uncountable cardinal K is mra.surable if there exists a K-complete nonprincipal ultrafilter U over K .

5.

298

MEASURABLE CARDlNALS

By Lemma 27.1, the least cardinal that carries a nontrivial two-valued a-additive measure is measurable. Note that if U is a K-complete nonprincipal ultrafilter over K, then every set X E U has cardinality K because every set of smaller size is the union of less than K singletons. For similar reasons, K is a regular cardinal because if K is singular, then it is the union of less than K small sets. The next lemma gives a first link of the measure problem with large cardinals. Lemma 27.2. Every measurable cardinal is inaccessible.

ProoJ We have just given an argument why a measurable cardinal is regular. Let me show that measurable cardinals are strong limit cardinals. Let K be measurable, and let us assume that there exists A < K such that 2L 2 K ; we shall reach a contradiction. Let S be a set of functionsf: 1 + {0, 1) such that I S 1 = K , and let U be a K-complete nonprincipal ultrafilter over S. For each a < 1,let X,be that one of the two sets { f S ~: f ( a )= 0), { f S~: f ( a )= 1) which is in U,and let &,be0or 1 accordingly. Since U is K-complete, the set X = X,is in U . However, X has at most one element, viz. the functionf that has the valuesf(cr) = E , . A contradict ion.

n,<

Let us now turn our attention to measures that are not necessarily twovalued. Let p be a nontrivial a-additive measure on a set S. In analogy with (27.3) we consider the ideal of all null sets: (27.7)

I,

= {X

c S : p ( X ) = 0)

I, is a nonprincipal a-complete ideal over S. Moreover, it has these properties: (27.8)

(i) {x) E I for every x E S ; (ii) every family of pairwise disjoint sets X E S that are not in I is at most countable.

To see that (ii) holds, note that if W is a disjoint family of set of positive measure, then for each integer n > 0, there are only finitely many sets X E W of measure 2 l/n. A a-complete nonprincipal ideal I over S is called a-saturated if it satisfies (27.8). More generally, a K-complete nonprincipal ideal I over S is tc-saturated if it satisfies (27.8)(i) and if every disjoint family W c P ( S ) of sets not in I has size < K. Let me make a small digression: Let I be a K-complete nonprincipal ideal over S, and let B be the quotient Boolean algebra B = P ( S ) / I . Note that B is K-saturated (cf. (17.15)) just in case there is no W c P ( S ) of size K such that X 4 I for each X E W and X n Y E I whenever X and Y are distinct elements of w. Exercise 27.4. Let S, 1, and B be as above. Show that B is K-saturated if and only if I is K-saturated.

27.

299

THE MEASURE PROBLEM

[If W c P ( S ) - I has size K , and X n Y E I whenever X # Y E W , let W = { X , : tl < K ) , and let Z , = X, X , ; the family (2, : a < K } is disjoint, and Z , $ I for each tl.]

Up..

The following lemma is an analog of Lemma 27.1 : Lemma 27.3.

(a) Lut K he the least cardinal that carries a nontrivial a-additive measure and let ,u he such a measure on K . Then the ideal I , of null sets is K-complete. (b) Let K he the least cardinal with the property that there is a o-complete a-saturated ideal over K, and let I he such an ideal. Then I is K-complete. Proof. (a) Let us assume that I, is not K-complete. There exists a collection of nuli sets { X , : ~1 < $ such that y < K and that their union X has positive measure. We may assume without loss of generality that the sets X a , a < y, are pairwise disjoint; let m = p ( X ) . Let f be the following mapping of X onto y:

f(x)=cc

iff

XEX,

(XE

X)

The mapping f induces a measure v on y : (27.9) The measure v is o-additive and is nontrivial since v{a} = p ( X a ) = 0 for each a E y. This contradicts the choice of K as the least cardinal that carries a measure. (b) The proof is similar. We define an ideal J over y by: 2 E J ifff- l(Z) E I. The induced ideal J is a-complete and maturated. Let {ri : i (27.10)

E

I} be a collection of nonnegative real numbers. We define

1 ri = sup{ 1 ri : E is a finite subset of I } icl

isE

Note that if the sum (27.10) is finite (i.e., is not a), then at most countably many terms ri are nonzero. Let K be an uncountable cardinal. A measure p on S is called K-additive if for every y < K and for every disjoint collection X a , a < y, of subsets of S, we have (27.11) Exercise 27.5. Let p be a two-valued measure, and let U be the ultrafilter of all sets of measure one. The measure p is K-additive if and only if U is K-complete.

If p is a K-additive measure, then the ideal I, of null sets is K-complete. The converse is also true, and we get a better analog of Lemma 27.1 for real-valued measures :

5.

300

MEASURABLE CARDINALS

Lemma 27.4. Let p be a measure on S, and let I , be the ideal of null sets. If I , is K-complete, then p is u-additive.

Proof. Let y < K, and let X u ,a < y, be disjoint subsets of S. Since the X, are disjoint, at most countably many of them have positive measure. Thus let us write

{Xu: a < y } = (Yn : n = 0, 1, 2, . . .} u { Z , : a < y } where each Z , has measure 0. Then we have

Now first p is a-additive, and we have

and secondly I, is K-complete and

c.

P(Xd Thus PCU. Xu)= Corollary. Let K be the least cardinal that carries a nontrivial a-additive measure and let p be such a measure. Then p is K-additive.

An uncountable cardinal K is real-valued measurable if there exists a nontrivial u-additive measure p on K. By the corollary above, the least cardinal that carries a nontrivial a-additive measure is real-valued measurable. We shall show that if a real-valued measurable cardinal K is not measurable, then K I2'O. Note that if p is a nontrivial u-additive measure on K, then every set of size < K has measure 0, and moreover K cannot be the union of less than K sets of size < K. Thus a real-valued measurable cardinal is regular. We shall show that it is weakly inaccessible. We shall first prove the first claim made in the preceding paragraph. Lemma 27.5.

(a) If there exists an atomless nontrivial a-additive measure, then there exists a nontrivial a-additive measure on some K I2'O. (b) If I is a a-complete a-saturated ideal over S, then either there exists Z G S such that I ( Z = {X c Z : X E I } is a prime ideal, or there exists a acomplete maturated ideal over some K I2'O. Proof: (a) Let p be a measure on S. We construct a tree T of subsets of S, partially ordered by reverse inclusion. The 0th level of T is {S}.Each level of T

27.

30 1

THE MEASURE PROBLEM

consists of pairwise disjoint subsets of S of positive measure. Each X E T has two immediate successors: We choose two sets Y, Z of positive measure such that Y u Z = X and Y n Z = 0. If a is a limit ordinal, then the ath level consists of all intersections X = X, such that each X, is on the 5th level of T and such that X has positive measure. We observe that every branch of T has countable length: If {X, : 5 < a} is a branch in T , then the set { 6 : 4 < a}, where 6 = X, - X,+ is a disjoint collection of sets of positive measure. Consequently, T has height at most wl. Similarly, each level of T is at most countable, and it follows that T has at most O 2' branches. Let { b , : a < K}, K I 2'O, be an enumeration ofall branches b = { X , : 5 < y} such that X, is nonempty; for each a < K, let Z , = {X : X E b,}. The collection { Z , : a < K } is a partition of S into K sets of measure 0. Using what is by now a standard argument, we induce a measure v on K as follows: Letfbe the mapping of S onto K defined by

n,<,

nE
x E Z,

iff

f(x)=a

(x E S )

and let V

W

=

Af- l(Z))

for all Z s K. It follows that v is a nontrivial a-additive measure on K. (b) The proof is similar. We define a tree T as above and then induce an ideal J over K by letting Z E J ifff- l(Z) E I. W The proof of Lemma 27.5 shows that if p is atomless, then there is a partition of S into at most 2'O null sets; in other words, p is not (2'O)+-additive. Hence if K carries an atomless K-additive measure, then K 5 2O ' and we have: Corollary. ff K is a real-valued measurable cardinal, then either K is measurable or K I 2'O. Similarly, if^ carries a K-complete a-saturated ideal, then either K is measurable or K I 2O '.

The measure v obtained in (a) above is atomless; this follows from the fact ' and Lemma 27.2. that K I 2O Exercise 27.6.* Prove directly that the measure v defined in the proof of Lemma 27.5a is atomless. [Assume that Z is an atom of v, and let Y = j - l ( Z ) . If X E 7' is such that p ( Y n X ) # 0 and if XI,X 2 are the two immediate successors of X , then either p( Y n X = 0 or p( Y n X,) = 0. Prove by induction that on each level of T there is a unique X such that p( Y n X ) # 0, and that these X's constitute a branch in T of length w I; a contradiction.]

By Lemma 27.5a, if there exists an atomless 0-additive measure, then there is one on some K I 2'O. Clearly, such a measure can be extended to a measure

5.

302

MEASURABLE CARDINALS

on 2? For X c_ 2K0,we let p ( X ) = p ( X n K ) . Thus we conclude that there exists an atomless a-additive measure on the set R of all reals. It turns out that using the same assumption, we can obtain a a-additive measure on R that extends Lebesgue measure. This can be done by a slight modification of the proof of Lemma 27.5: Using Exercise 27.3, we construct for each finite 0-1 sequence s, a set X,G S such that X, = S, and for every s E Seq, X s o u X , , = X , , X s o n X , , = 0,and p ( X s 0 )= p ( X , , ) = i + p ( X , ) .Then we define a measure v, on "'2 by Vl(Z) = P ( U

{ X , : f e 4)

where X , = ();=o X,,, for each f e '"2. Using the mapping F: "2 -+ [0, 13 defined by

we obtain a nontrivial a-additive measure v on [0, 11. This measure agrees with Lebesgue measure on all intervals [k/2", (k + 1)/2"],and hence on all Borel sets. Every set of Lebesgue measure 0 is included in a Borel (in fact, Gd)set of Lebesgue measure 0 and hence has v-measure 0. Every Lebesgue measurable set X can be written as X = (B - N,) u N , , where N , and N, have Lebesgue measure 0, and hence the Lebesgue measure of X is equal to v ( X ) .Thus v agrees with Lebesgue measure on all Lebesgue measurable subsets of [0, 11. We shall now show that a real-valued measurable cardinal is weakly inaccessible. The proof is by a combinatorial argument, using matrices of sets. An Ulum matrix (more precisely, an Ulam (HI, KO)-matrix)is a collection : a < w , , n < o}of subsets of w , such that: (27.12) (i) if a # p, then Am.nn Afl,n= 0. for every n < o ; (ii) for each a, the set o,- ur=o Am,nis at most countable. See Fig. 27.1.

FIGURE 27.1. An (N,,N,)-matrix.

An Ulam matrix has K l rows and KOcolumns. Each column consists of pairwise disjoint sets, and the union of each row contains all but countably many elements of wl.

27.

303

THE MEASURE PROBLEM

Lemma 27.6. A n U lam matrix exists. ProoJ For each 5 < w l , let fs be a function on w such that 5 G ran(fc). Let us define A,, for a < w I and n < o by

5 E A,,

(27.13)

f s ( n )= a

iff

If n < w, then for each 5 E w 1 there is only one a such that ( E A,,, namely a = f s ( n ) ; and we have property (i) of (27.12). If a c ol, then for each 5 > a there is an n such thatf,(n) = a and hence (wl Am")G a ; that verifies

u."=o

property (ii). Using an Ulam matrix, we can show that there is no measure on w I :

Lemma 27.7. There is no nontrirlial a-additive measure on ol. More generally, there is no a-complete a-saturated ideal over wl. froo& Let [ A z , , : a < oil, n < m } be an Ulam matrix. Assuming that we have a measure on wl, there is for each a some n = n, such that A,, has positive measure (because of (27.12)ii). Hence there is an uncountable set W E w 1 and some n < oi such that n, = n for all a E W . Then {Am,: a E W } is an uncountable, pairwise disjoint (by (27.12)) family of sets of positive measure; a contradiction. A straightforward generalization of Lemmas 27.6 and 27.7 gives the result mentioned above:

Lemma 27.8. If K = A+, then there is no K-complete a-saturated ideal over K (not w e n a tc-saturuted irlral). Corollary. Ewrp rral-valued measurable cardinal i s weakly inaccessible.

Proofi For each 5 < A + , we letft; be a function on A such that 5 E ran(ft;). and let 5 E A,, iff f&) = a Then {A,,, : a < A+, q < A! is an Ulam subsets of A such that: (27.14)

(A+,A)-matrix, that

is a collection of

# p < A', and q < A; uvcl0whenever A for each < 1'.

(i) A,, n A,,, (ii) [ A + -

a

=

A,,I

I

a

The proof of Lemma 27.7 easily generalizes to show that no K-saturated complete ideal exists over K.

K-

Lemma 27.8 completes the proof of Theorem 66: If there is a a-additive nontrivial measure on S, then either the measure has an atom A and we can construct a two-valued measure on S via a a-complete nonprincipal ultrafilter over A, and then I S / 2 the least measurable cardinal, which is inaccessible; or

5.

304

MEASURABLE C A R D I N A L S

the measure on S is atomless and we construct, as in Lemma 27.5, an atomless measure on 2"0, and then 2"O 2 the least real-valued measurable cardinal, which is weakly inaccessible. W We conclude this section with an application of the concept of a scale that we introduced in Section 24. Historically this result is interesting since it preceded Theorem 66 by showing that the assumption of existence of a nontrivial a-additive measure violates the continuum hypothesis. We recall (cf. (24.27)) that a K-scale is a K-sequence of functionsf,: w + w, ct < K, such that fo
K

is not a real-valued measurable

Proof: Let fa, a < K , be a K-scale. We define an (KO, K,)-matrix of subsets of K as follows: For n, k < w, let

(27.15)

j$)=

iff

a E A,,

k

(a E K )

Since for each n and each a there is k such that a E Ank, we have

u m

=

k=O

for every n = 0, 1, 2, . . . . Now assume that p is a nontrivial K-additive measure on K. For each n, let k, be such that p(A.0 u

An1

u

u Ankn)2 1

-

(1/2"+*)

n;=o

and let B, = A,, u ... u A&. If we let B = B,, then we clearly have p( B) 2 Let g: w + w be the function g ( n ) = k , . If a E B, then by the definition of B and by (27.15), we have

4.

f.(n)5 s(n) for all n = 0, 1, 2, . . .; hence g +fa. However, since B has positive measure, B has size K, and therefore we have g $J, for cofinally many a < K. This contradicts the assumption that thef, form a scale. Corollary 1. (which also follows from Lemma 27.7). Ifthere is a measure on 2'O, then 2O ' > Kl. (Because an K,-scale exists.)

' is not a real-valued measurable Corollary 2. If Martin's axiom holds, then 2O cardinal.

28.

305

ULTRAPOWERS A N D ELEMENTARY EMBEDDINGS

28. ULTRAPOWERS AND ELEMENTARY EMBEDDINGS

In the preceding section we introduced measurable cardinals and showed that measurable cardinals are inaccessible. That result was obtained by a more or less straightforward application of the definition of measurability. To obtain much stronger results, we shall borrow from model theory a powerful method which is well suited for measurable cardinals-the method of ultrapowers In Section 7 (cf. (7.7)) we defined the reduced producr of a collection of sets modulo a filter (over the set of indexes). Reduced products, not just of sets but of models, are a useful tool in model theory, and become particularly helpful when the filter is an ultrafilter. Let S be a set and let U be an ultrafilter over S. For each x E S, let 91, be a the same language for eeach x E S. For clarity of model of some language 9, exposition, let me assume that 9 is the language of set theory, 9 = {E}, and that all formulas of 9 have only the binary predicates = and E. (The subsequent discussion of ultraproducts is however valid for arbitrary languages as well.) Thus for each x E S, we have a model of the language of set theory. Let us consider the Cartesian product P=

n

A,

X E S

and let us consider the following equivalence relation

=* over P:

As we have seen in Section 7, the relation = * is indeed an equivalence relation, due solely to the fact that V is a filter. For each f E P, let [f] be the equivalence class mod = *, and let A* be the collection of all equivalence classes of allfE P: A*

=

(n

A,)/=*

xeS

ForJ g (28.2)

E

P , we define

f ~ g*

iff

{x E S :f(x)E,g(x)}E U

and we note that =* is a congruence with respect to E*, i.e., iffi = * f z , g 1 = * g 2 , and f,E* g l , then f2 E* g 2 . Thus we can define a binary relation over A* (and by abuse of notation denote it E* as well) as follows: (28.3)

5.

306

MEASURABLE CARDINALS

Now, consider the model YI* = (A*, E*), which is called the ultraproduct of E S (mod V ) : YI* = Vlt"{YI, : x E sj (28.4) The importance of ultraproducts is due mainly to the following basic property of ultraproducts:

YI,, x

Lemma 28.1 (Fundamental Theorem of Ultraproducts). Let U be an ultrujilter over S and let YI* be the ultraproduct of PI,, x E S (mod V). (a) ~ f isqaformula, thenfor anyf,, ...,f. (28.5) PI* != ~ ~ [ [ f l .].,. , [f.]] (b) If o is a sentence, then

PI* b c

iff

iff

E HxeS

{X E

A,,

S : 91, k cp[f,(x), . . .,fn(x)])E U

{ x E S : 91, =! o} E U

Part (b) is a consequence of (a). Note that by the lemma, the satisfaction of cp at [ f , ] , . . ., [f.] does not depend on the choice of representativesf,, . . .,fn for the equivalence classes [ f J . . ., [fJ. Thus we may further abuse the notation and write %* k cp[f,, .* . ,f"] It will also be convenient to adopt a measure theoretic terminology. If { x E S : 91, 1 cp[f,(x), . . .,f.(x)]} E

u

we say that PI, satisfies cp(f,, . . .,fn)for almost all x , or that PI, b cp(fl, . . .,fn) holds almost euerywhere. In this terminology, the fundamental theorem states holds in the ultraproduct just in case for almost all x , that cp(f,, ...,f,) q ( f , , . . .,f,)holds in PI,. ProofofLemma 28.1. We shall prove (a) by induction on complexity of formulas. We shall prove that (28.5) holds for atomic formulas, and then prove the induction step for 1,A , and 3; then it follows that (28.5) is true for all formulas. Atomicformulas: First we consider the formula u = u, and we have

PI* !=

[ f l = [sl

f=*g

iff

{ x : f ( x )= g ( x ) } E V

iff

{ x : 91, b f ( x ) = g ( x ) } E U

iff iff

[fl E* [sl

iff

{ x : PI, k f ( x ) E g ( x ) } E U

For u E u, we have

PI* != [ f l E [sl

[ f l = [sl

iff iff

{ x : f ( x )E x g ( x ) } E U

28.

307

ULTRAPOWERSA N D ELEMENTARY EMBEDDINGS

Logical connectioes: First we assume that (28.5) holds for cp and show that it also holds for l c p (here we use that U is an ultrafilter):

PI* != 1 c p [ f ]

iff

not PI* != ~ [ f ]

iff

{ x : PI, != c p [ f ( x ) ] }6 U

iff

{ x : PI, f c p [ f ( x ) ] }E U

iff

{ x : 91, k l c p [ f ( x ) }E U

Similarly, if 28.5 is true for cp and $, we have

PI* 1 P A $

iff

PI*

iff

{ x : PI, != cp} E U and { x : PI, I= $} E U

iff

{ x : PI, != cp A tj)E U

!= cp and

PI* !=

i,b

(The last equivalence uses this fact: X E U and Y E U iff X n Y E U . ) Existential quantifier. We assume that (28.5) is true for ~ ( u o, I, . . ., on) and show that it remains true for the formula 3u cp. Let us assume first that

PI* != 3.4&-I,

(28.6) Then there is g E

. . . A3

nx A, such that PI* != cp[g,

fl,

...,f,], and therefore

{ x : PI, k cp[g(x),f&), . . .,f , ( X ) ] } E

(28.7)

u,

and it clearly follows that { x : 91, != 3~ ~ p [ ~ ? f l (*x* )*,, f n ( x )E] }u

(28.8)

Now let us assume that (28.8) holds. For each x E S , let u, E A, be such that != cp[u,, f l ( x ) ,. . .,f,(x)] if such as u, exists, and arbitrary otherwise. If we define g E HxES A , by g ( x ) = u,, then we have (28.7) and therefore

PI,

PI* k

cp[g,fl, . . * , f n ]

Now (28.6) follows. Exercise 28.1. Let U be a principal ultrafilter over S , such that {x} E CJ. Show that the ultraproduct YI* is isomorphic to PI,.

Let us consider now the special case of ultraproducts, when each PI, is the same model PI. Then the ultraproduct is called an ultrapower of 91:

(28.9)

91* = Ult"(P1)

An important corollary of the fundamental theorem is Lemma 28.2. An ultrapower of a model 91 is elementarily equivalent to PI, i.e.,for

any sentence

Q

we haw

PI* != o

iff

PI != Q

308

5.

MEASURABLE CARDINALS

Proof. By part (b) of Lemma 28.1 we have PI* k cr iff { x : PI t= cr} E U , but { x : PI k cr} is either S or empty, according to whether PI k cr or not. We shall now show that a model 91 is elementarily embeddable in its ultrapower. If U is an ultrafilter over S and YI* = Ult,(PI), we define the natural embedding j: YI -, PI* as follows: For each a E A, let c, be the constant function with value a : c,(x) = a

(28.10)

(for every x

E

S)

and let (28.1 1) = [cO1 Lemma 28.3. The natural embedding j : YI + PI* is an elementary embedding, i.e., P I k cp[a,, ..., a,,]

(28.12)

iff

Y t * b ( a l ) ,. . .,j(a,,)]

for any formula q. Proof. By the fundamental theorem: PI* k &(a)] iff PI* k cp[c,] iff PI t= q [ a ] for almost all x iff PI k q [ a ] . Exercise 28.2. If U is a principal ultrafilter, then j is an isomorphism between PI and

Ult(Y1).

In Chapter 2 we introduced models of set theory as (possibly proper) classes endowed with a binary relation, the interpretation of the membership predicate E. As is the case with many model-theoretical concepts and constructions, the method of ultraproducts can be used in the context of these models of set theory. For instance, let us consider the ultrapower of the universe ( V , E) modulo a given ultrafilter U over S. Let us consider the class of all functionsf with domain S ; the equivalence relation = * and the relation E* are defined as before: (28.13) f

=* g

iff

{x E S : f ( x )= g(x)) E U

(28.14)

f ~ g*

iff

{ x E S :f (x) E g(x)} E V

We encounter a slight difficulty when dealing with the equivalence classes [f ] since each equivalence class is a proper class. However, we can avoid all complications by using the trick introduced in (9.3): We simply define (28.15)

Now we let

[f]= [g :f=*g and V h ( h = * f + rank h 2 rank g ) } Ult = mu( V)

be the class of all [f]where f is a function on S, and consider the model Ult = (Ult, E*) (here we make the same abuse of notation as in (28.3) and furthermore use the symbol Ult to denote both the class and the model).

28.

309

ULTRAPOWERS AND ELEMENTARY EMBEDDINGS

A quick glance at the proof of the fundamental theorem of ultraproducts should convince the reader that the theorem is true also in the present context. That is, if cp(x,, . . . , x,) is a formula of set theory, then

If cr is a sentence, then Ulr k u

(28.17)

iff

0

and so we may say that the ultrapower is elementarily equivalent to the universe. Moreover, using the constants c, (defined in (28.10)), we obtain a function (which is a proper class definable from U ) j,: V + Ult

defined by

(28.1 1 ) and the function j,, the natural embedding of V in Ult, is an elementary embedding of the universe in the model U l t : We have

(28.18)

cp(a1, ..., an)

iff

Ult 1 cp(ja,, . . .,ja,,)

whenever cp(x,, ..., x,) is a formula of set theory. The most important applications of ultrapowers in set theory are those in which the (Ult, E*) is well founded. As we shall show presently, well-founded ultrapowers are very closely related to measurable cardinals. The model Ult,( V) is well-founded (cf. (9.8))just in case (i) every nonempty set X c Ulr has a €*-minimal element, and (ii) ext(f) is a set for everyf, where ext(f) = (191 9

E*f)

The second condition is clearly satisfied for any ultrafilter U : for every g ~ * f there is h = * g such that rank h I rank f. As for condition (i), this is satisfied if and only if there exists no infinite descending E* sequence f o 3*f,

3*

* ' *

3*fk 3* *..

(k

E 0)

of elements of the ultrapower. Exercise 28.3. Let U be a nonprincipal ultrafilter over w. Then Ultu(V) is not well founded. [For each k E w, leth be a function on w such thath(n) = n - k for all n 2 k . Then fo 3*fi 3 * f 2 3* . . . is a descending €*-sequence in Ult.]

Lemma 28.4. If U is a u-complete ultrafilter, then (Ult, E*) is a well-founded model. Proof. We shall show that there is no infinite descending €*-sequence in Ulr if U is a a-complete ultrafilter over S. Let us assume thatfo,fi, ...,fn, ... is such a

5.

310

MEASURABLE CARDINALS

descending sequence. Thus for each n, the set Xn = {X E S : f n + 1(x) Efn(X)J is in the ultrafilter. Since U is a-complete, the intersection X = Xnis also in U and hence nonempty; let x be an arbitrary element of X. Then we have

fob) .fl(x)

3 f h )3

an infinite descending €-sequence, which is a contradiction. In fact, a-completeness of U is a necessary and sufficient condition for well-roundedness of the ultrapower: Exercise 28.4. If U is not a-complete, then Ulr,( V ) is not well-founded. [There exists a countable partition {X, : n = 0, 1, 2, .. .} ofS such that X, $ U for all n. For each k, let X be a function on S such that X(x) = n - k for all x E X , .]

By the Mostowski collapsing theorem (Theorem 28), every well-founded model is isomorphic to a transitive model (and the collapsing isomorphism is unique). Thus if U is a-complete, there exists a one-to-one mapping II of Ult onto a transitive class such thatfE* g if and only if n ( [ f ] E) ~([g]).In order to simplify notation, we shall identify each [f]with its image n ( [ f ] )Thus . if U is a-complete, the symbol Ult denotes the transitive collapse of the ultrapower, and for each functionfon S, [f] is an element of the transitive class U l t ; we say the functionfrepresents [f]E Ult. Exercise 28.5. Every ordinal number a is represented by a function f: S + Ord.

If U is principal and xo E S is such that U

= {X : xo E

X}, then Ult = V and

[f] = f ( x o ) for each f: The natural embedding j,, is the identity mapping. The situation is more interesting when U is a nonprincipal a-complete ultrafilter. Let us consider (without loss of generality) a nonprincipal a-complete ultrafilter U over a cardinal number K. Let us denote the ultrapower by M, M = U/t,,(V), and let j = j , be the natural embedding: (28.19)

j: V-M

Since j is elementary, it is clear that if a is an ordinal, then j ( a ) is an ordinal; moreover, a < fi implies j ( a ) < j ( p ) . Thus we have a Ij ( a ) for every ordinal number a. Note that j ( a + 1) = j ( a ) + 1, and j ( n ) = n for all natural numbers n. It is also easy to see that j ( w ) = w : If [f]< w, then f ( a ) < Q for almost all a < K, and by a-completeness, there exists n < w such that/(a) = n for almost all a. By the same argument, if U is I-complete, then j ( y ) = y for all y < 2. We have proved (Lemma 27.1) that if there exists a nonprincipal acomplete ultrafilter, then there exists a cardinal K that carries a nonprincipal k--complete ultrafilter. Thus let us assume further that U is a nonprincipal K-complete ultrafilter over k-. (And hence j ( y ) = y for all y < K.) We shall show

28.

311

ULTRAPOWERS AND ELEMENTARY EMBEDDINGS

that j ( K ) > K. Let d (the diagonalfunction) be the function on

(28.20)

d ( a )= a

K

defined by

(a < K )

Since U is K-complete every bounded subset of K has measure 0 and so for every y < K, we have d(u) > y for almost all a . Hence [dl > y for all y < K and thus [ d ] 2 K. However, we clearly have [d] < j ( ~and ) it follows that j ( ~>) K. Exercise 28.6. Let U be a nonprincipal a-complete ultrafilter over S and let A be the largest cardinal such that U is I-complete. Then j ( A ) > A. [Let { X , : a < A} be a partition of S into sets of measure 0; letfbe a function on S such that /(x) = a if x E X , . Then [f]2 A.]

We have thus proved that if there is a measurable cardinal, then there is an elementary embedding j of the universe in a transitive model M such thatj is not the identity mapping; let us call j a nontrivial elementary embedding of the uniuerse. We shall use this to prove the following theorem : Theorem 67 (Scott).

If V = ,!I then there are no measurable cardinals.

Proof. Let us assume V = L and that measurable cardinals exist; let K be the least measurable cardinal. Let U be a nonprincipal K-complete ultrafilter over K and let j : V + M be the corresponding natural embedding. As we have shown, J(K) > K. Since V = L, the only transitive model containing all ordinals is the universe itself: V = M = L. Since j is an elementary embedding and K is the least measurable cardinal, we have M 1 j ( K ) is the least measurable cardinal; and hence, j ( K ) is the least measurable cardinal. This is impossible since j(K) > K. If there exists a measurable cardinal, then we obtain a nontrivial elementary embedding of the universe. Let me show that conversely, if j : V + M is a nontrivial elementary embedding then there exists a measurable cardinal. (A word of caution: “There exists a nontrivial elementary embedding of V ” or even “ jis an elementary embedding of V” is not a statement expressible in the language of ZF.) Lemma 28.5. l f j is a nontrivial elementary embedding of the universe, then there exists a measurable cardinal.

Proof. Let j : V -+ M be a nontrivial elementary embedding. Notice that there exists an ordinal a such that j ( a ) # a ; otherwise, we would have rank(jx) = rank(x) for all x, and then we could prove by induction on rank that j(x) = x for all x. Thus let K be the least ordinal number such that j ( K ) # K (and hence j ( ~>) K). It is clear that j ( n ) = n for all n andj(o) = o since 0, n + 1, and o are

5.

312

MEASURABLE CARDINALS

absolute notions and j is elementary. Hence K > o.We shall show that measurable cardinal. Let D be the collection of subsets of K defined as follows: (28.21 )

XED

K

is a

KE~(X) (XGK)

iff

Since K < j ( K ) , i.e., K E ~ ( K ) we , have K E D ; also 0 4 D because j ( 0 )= 0. Using the fact that j(X n Y) = j ( X ) n j ( Y) and that j ( X ) c j ( Y) whenever X c Y , we see that D is a filter: If K E j ( X ) and K E j ( Y), then K E j(X n Y);if X E Y and K E ~ ( X )then , K € j ( Y ) .Similarly,j ( K - X ) = j ( ~- )j ( X ) and thus D is an ultrafilter. D is a nonprincipal ultrafilter: For every a < K, we have j({a})= {j(a)}= {a}, and so K 4 j({a})and we have {a}4 D. We shall now show that D is K-complete. Let y < K and let X = (X,: a < y) be a sequence of subsets of K such that K E ~ ( X ,for ) each a < y. We shall show that X, E D.In M (and thus in V), j ( X ) is a sequence of lengthj(y) of subsets ofj(K); for each a < y, the j(a)th term of j ( X ) is j ( X , ) . Since j(a) = a for all a < y and j ( y ) = y, it follows that j ( X ) = ( j ( X , ) : a .c 7). Hence if X = Om<,X,, we have j ( X ) = j(X,). Now it is clear that K E j ( X ) and hence X E D.

na<,

na<,

The construction of a K-complete ultrafilter from an elementary embedding described in (28.21) gives rise to a neat commutative diagram of elementary embeddings:

Lemma 28.6. Let j : V + M be a nontrivial elementary embedding, let K be the least ordinal moved, and let D be the ultrafilter over K defined in (28.21). Let j,: V -,Ult be the natural embedding of V in the ultrapower U l t d V ) .Then there is an elementary embedding k of Ult in M such that k(j,(a)) = j ( a ) f o r all a :

FIGURE 28.1

Proof. For each [f]E Ult, let

(28.22)

k([fI)

= (J(f))(K)

(Heref is a function on K and J ( f ) is a function on j ( x ) . ) We shall first show that definition (28.22)does not depend on the choice off representing [f].Iff= g, then the set X = {a : f ( a )= g(o()}is in D and hence K

28.

ULTRAPOWERSA N D ELEMENTARY EMBEDDINGS

313

is in the set A X ) = {a 4

4 : ( d k )= (jg)(.)l

~ (is)(.). ) Therefore ( f l ) ( = Next we show that k is elementary. Let cp(x) be a formula and let Ult 1 c p ( [ f ] ) we ; shall show that M k cp(k([f])). The set X = { a : cp(f(a))}is in D and hence K belongs to : M 1 cp((d)(a))} A X ) = {a Since ( j f ) ( ~=) k ( [ f ] ) , we have M 1 cp(k([f])). Finally, we show that k(j,(a)) = j ( a ) for all a. Since j,(a) = [cJ, where c , is the constant function on K with value a, we have k ( j d a ) ) = (j(c,))(K).Nowj(c,) is the constant function on j ( K ) with value j ( a ) and hence ( ~ ( c , ) ) ( K=)j ( a ) . W

Before we continue our investigation of elementary embeddings, notice that ifj: V - + M , then the model M is determined by the function j : Exercise 28.7. If j is an elementary embedding of the universe in a transitive model M , then M = ord i( V.).

urn

If X is a class defined by a formula cp, then j ( X ) is the class of the model M , defined in M by the same formula cp. Note that j ( X ) = U a E O r d j ( n X V,). In particular, M = j ( V ) . We proved in Lemma 27.2 that every measurable cardinal is inaccessible. We shall now obtain some stronger results, as a consequence of the following simple lemma. Lemma 28.7. Let j be an elementary embedding of the universe and let K be the least ordinal moved (i.e.,j ( K ) > K ) . If C is a closed unbounded subset of K , then K E j(C). Proof. Since j ( a ) = a for all a < K, we have j ( C ) n K = C. Thus j ( C ) n K is unbounded in K ; and becausej(C) is closed (in j ( V ) and hence in the universe), we have K E j ( C ) . W Corollary. Every measurable cardinal K t h inaccessible card i na 1.

K

is a Mahlo cardinal, and hence K is the

Proof. Let K be a measurable cardinal; we shall show that the set X of all regular cardinals a < K is stationary in K. Let j : V + M be an elementary embedding such that K is the least ordinal moved (e.g., let j = j , where U is a nonprincipal K-complete ultrafilter). Since K is a regular cardinal in M, we have K E j ( X ) .Now if C is any closed unbounded subset of K, we havej(X n C) # 125 because K E j(X) n j ( C ) and hence X n C # @. Thus X is stationary.

The above argument can be pushed further: Let X denote the set of all Mahlo cardinals below K. Since K is Mahlo. it is a Mahlo cardinal in M and so

314

5.

MEASURABLE CARDINALS

E j(X). The same argument as above shows that X is stationary and thus K is the Kth Mahlo cardinal. One can formulate an abstract version of the preceding argument. We define the Mahlo operation on classes of ordinals: If X is a class, let K

(28.23)

M(X) = { a : X n a is stationary in a }

If X is the class of all inaccessible cardinals, then M(X) is the class of all Mahlo cardinals.) Exercise 28.8. Let j be an elementary embedding of the universe and let K be the least ordinal moved. If X is a class of ordinals such that K E j ( X ) , then K E M(X).

While measurable cardinals induce elementary embeddings of the universe in a transitive model, the following theorem shows that no nontrivial elementary embedding exists of V into itself.

Theorem 68 (Kunen). If j : V + M is a nontrivial elementary embedding, then M # V. The proof below makes use of the following combinatorial lemma:

Lemma 28.8. Let I be an injnite cardinal such that 2A = INo.There exists a function F : “ I + I such that whenever A is a subset of I of size I and y < I , there exists s E “‘A such that F(s) = y. (The lemma is in fact true for every infinite A.) Proof. Let { ( A , , y,) : a < 2’} be an enumeration of all pairs ( A , y) where y < L and A is a subset of I of size A. We define, by induction on a, a sequence s,, a < 2’, of elements of “I as follows: If a < I, then since INo = 2’ > la 1, there exists s, E ”A, such that s, # s g for all p < a. For each a < 2’, we define F(s,) = y, (and let F ( s )be arbitrary if s is not one of the s,). The function F has the required property: If A G A has size L and y < I, then ( A , y ) = ( A , , 7,) for some a, and then y, = F(s,). Proof of Theorem 68. Let us assume that j is a nontrivial elementary embedding of V in V . Let K = K~ be the least ordinal moved; K~ is measurable, and so are = j(Kn). Let 1 = hn,,+a K,. K , = j(K0), K~ = j(Kl), and every K,, where K , + Since j((K, : n < w ) ) = (j(K,) : n < w ) = ( K , + : n o), we have j ( A ) = limn.+mj ( x n )= A. Let G = j[I] = j j ( a ) : a < A}; we shall use the set G and Lemma 28.8 to obtain a contradiction. The cardinal I is the limit of a sequence of measurable cardinals and hence is a strong limit cardinal. Since cf I = w, we have 2’ = INo.By Lemma 28.8 there is a function F: “A + I such that F[”A] = I for all A C_ I of size A. Sincej is elementary, and jw = w and j l = I, the function j(F) has the same property. Thus, considering the set A = G, there exists s E “’G such that (jF)(s) = K. Now, s is a function, s: w+j[I], and hence there is t: w + I such that

-=

28.

315

ULTRAPOWERS A N D ELEMENTARY EMBEDDINGS

s(n) = j ( t ( n ) ) for all n < o. It follows that s = j ( t ) . Thus we have K = (jF)(jt) = j(F(f)); in other words, K = j(a) where a = F(t). However, this is impossible since j(a) = a for all a < K, and j(K) > K. Exercise 28.9. Ifj: V + M is a nontrivial elementary embedding, if K is the least ordinal moved, and if 1 = lim{K, j ( K ) , j(j(K)), . . .}, then there exists A c 1 such that A $ M. [Assuming that M contains all bounded subsets of 1, the above proof that G M goes through.]

Let me now consider ultrapowers and the corresponding natural embeddings j,: V + Ult. To introduce the following lemma, let us observe that if j: V -,M and if K is the least ordinal moved, thenj(x) = x for every x E V,, and j(X) n V, = X for every X G V,. Hence VF+ = V,, (and P ( K=)P ( K ) ) .

Lemma 28.9. Let U be a nonprincipal K-complete ultrafilter over M = Ult,(V) and let j = j, be the natural embedding of V in M .

K,

let

(a) " M c M , i.e., every K-sequence (a, : a < K ) of elements of M is itself a member of M . (b) u M . (c) 2" I ( 2 " ) M < j ( K ) < (2")+. (d) I f I i s a limit ordinal and ifcf I = K, then j(I)> lim,+Aj(a); ifcf 1 # K, then j ( A ) = lim,+Aj(a). (e) If I > K is a strong limit cardinal and cf I # K, then j ( I ) = I .

z

Proof. (a) Let (aC: ( < K ) be a K-sequence of elements of M. For each 5 < K, let ge be a function that represents a t , and let h be a function that represents K : [SCI = at.

[hl = K

We shall construct a function F such that [ F ] = (at : 5 < K). We let, for each a < K, F(a) = ( s t ( a ) : 5 < h ( 4 ) Since for each a, F(a) is an h(a)-sequence, [ F ] is a K-sequence. Let 5 < K ; we want to show that the 5th term of [F] is a t . Since [h] > 5, we have 5 < h(a) for almost all a ; and for each a such that ( < h(a), the 5th term of F(a) is gda). But [ce]= 5 and [ g C ]= a t , and we are done. (b) Assume that U E M, and let us consider the mapping e of " K onto&) defined by e( f )= [ f .Since " K E M and U E M, the mapping e is in M . It follows that M 1 J ~ ( KI) 5 2". This is a contradiction since K < j(K) a n d j ( ~is) inaccessible in M. (c) 2" I (2")M holds because P M ( ~=)P ( K )and M c V; (2")M is less than j ( K ) since j ( K ) is inaccessible in M; finally, we have Ij ( K ) I = 2" and hence j ( K ) < (2")+. (d) If cf I = K, let 1 = lim,.+KI, and letf(a) = 1, for all a c K . Then [f]>

5.

316

MEASURABLE CARDINALS

j(1,) for all a < K and [ f ] c j(1). If cf I > K, then for everyf:

K + 1 there exists c 1such that [f]<j(a). If cf I = y < K, let I = Iim,+, 1,;for everyf: K + I there exists (by K-completeness) v < y such that [f]< j ( I , ) . (e) For every a c 12, the ordinals below a are represented by functions f: K + a ; hence I j ( a ) I I I ' a 1 < 1;by (d) we have j(1)= lim,,+A j ( a ) = 1. H

a

Remark. Note that in (e) it suffices to assume that cf I # cardinals a < I.

K

and a' c I for all

Normal Measures We defined the notion of a normalfilter in (7.18); a filter is normal if it is closed under diagonal intersections. If D is a nonprincipal complete ultrafilter over K and if D is normal (as a filter), then we call D a normal (two-valued) measure on K. Lemma 28.10. Let D be a nonprincipal K-complete ultrajilter over following are equivalent :

K.

Then the

(i) D is normal. (ii) Iff is such that f(a) c a for almost all a, then there is y c K such that f ( a ) = yfor almost all a. (iii) I n the ultrapower UltD(V), K =

where d is the diagonal function. (iv) For every X C K , X E D if and only

[d]

if^ €jD(X).

Thus D is normal just in case the diagonal function is the least nonconstant function (i-e., the least function that is not constant almost everywhere). Proof: (i) + (ii): Let us recall that the diagonal intersection of {X, : y < K} is the set

A X , = { ~ < K : ~0EX,) Y
Let D be closed under diagonal intersections and letfbe a regressive function on a set Y E D.If we assume thatfis not constant almost everywhere, then for each y c K, the set X, = {a E Y : f ( a )# y } is in D. Let X = Ay<'X,. For each a E X, we havef(a) 2 a becausef(a) = y for all y c a. A contradiction. (ii) -,(iii): Obvious. (iii) + (iv): If X G K, then X E D if and only if d ( a ) E X for almost all a, that is if and only if [d] EJ,(X). If [dl = K, we get X E D iff K E ~ ( X ) . (iv) --* (i): Let us assume that X E D iff K € j D ( X )and let us show that D is closed under diagonal intersections. Let X,,y < K, be such that K E j(X,) for all y < K , let X = A,<*X, and let us show that K E j(X). Nowj(X) is (in U l t ) the

28.

317

ULTRAPOWERSA N D ELEMENTARY EMBEDDINGS

diagonal intersection ofj({X, : y < K}), and K E j(X) just in case K € j ( X y )for all y < K. Thus we have K E ~ ( Xand ) so X E D. The following lemma shows that normal measures on measurable cardinals exist and in fact can be obtained from a given measure in a canonical way:

Lemma 28.11. Eoery measurable cardinal carries a normal measure. If U is a nonprincipal K-complete ultrafilter over K , then there exists a functionf such that (28.24) D = { X G K : ~ - , ( D )UE} is a normal measure.

K

K

We denote the measure D obtained from U as in (28.24) by D =f,(U), and call D a prqjection of U (byf). Proof: Let f K + K be a function such that [f] = K in the ultrapower Ult,( V ) . It is easy to see that for any function f K + K , D = f * ( U ) is a K-complete ultrafilter (see also (27.6)). Since [f] > y for every y < K, we havef- , ( { y } ) # U , and so D is nonprincipal. In order to show that D is normal, let us consider a function h : K + K such that h ( l ) < 5 on a set X E D , and let us show that h is constant on some set in D. Let g be the function on K defined by gb) = h

(f(4 We have g(a) < / ( a ) for all a ~ f I (-X ) ;and since f is the least nonconstant function in U l t , , there is Y E U such that g is constant on Y. Then h is constant onf[Y] andf[Y] E D. W Exercise 28.10. Let j be an elementary embedding of the universe, let K be the least ordinal moved, and let D = { X c K : K E j ( X ) } (cf. 28.21). Then D is a normal measure on K. [Let f be a function regressive on some X E D.Then ( d ) (<~K )and show that if y =(d)(~ thenf(a) ) , = y almost everywhere.]

If D is a normal measure on K, then every closed unbounded subset of K has measure 1. (In other words, every X E D is stationary.) This follows either from the characterization of normality in Lemma 28.10(iv) or using the fact that if a set X G K is disjoint from a closed unbounded set C, then there exists a regressive function f on X that is unbounded and nondecreasing:

f ( a ) = sup(C n a ) (a E X ) (28.25) This and the fact that K is represented by the diagonal function are used in numerous applications of normal measures. For instance, we have:

Lemma 28.12. If D is a normal measure on K , then the set of all inaccessible cardinals below K has measure one. Similarly, almost all cardinals below K are Mahlo.

5.

318

MEASURABLE CARDINALS

Proof. Simply because U l f D 1 [d is inaccessible, and U l t D k [d] is Mahlo (compare with the corollary of Lemma 28.7).

One important reason the above proof works is that since K is inaccessible (in the universe), K is also inaccessible in the ultrapower; similarly, since K is Mahlo, K is Mahlo in U l f , etc. We shall use arguments like this for other properties of K ; however, measurability of K need not be preserved (and is not preserved if K is the least measurable cardinal). Exercise 28.11. If K is measurable, then there exists a normal measure D on K such that Ult,(V) 1 K is not measurable. [Let D be a normal measure such that j , ( ~ ) is the least possible ordinal; let

M = Ult,( V ) . If K is measurable in M,then there is a normal measure U on K such that U E M . Since P ( K )c M , the ultrapower U l t , is a class in M . By Lemma 2 8 . 9 ~(applied in M ) , we have jU(K) < ((2")+)'. Since ((2")+)'" < j D ( K ) . we get a contradiction.]

Note that in the exercise above almost all (mod D ) cardinals below K are nonmeasurable. If D is a normal measure on K, then D extends the closed unbounded filter on K. We shall see later that this property does not necessarily imply normality. Let f be a function on K ; let us say thatf is monotone iff(.) I f ( P ) whenever a < p. Exercise 28.12. Let U be a nonprincipal k--complete ultrafilter over K. Then U extends the closed unbounded filter if and only if the diagonal function is the least nonconstant monotone function in U l t , . [If U extends the closed unbounded filter and iff is monotone and regressive on some X E U,then since X is stationary,fis constant on an unbounded set and hence constant almost everywhere. If U does not extend the closed unbounded filter, then use (28.25) to get a nonconstant monotone function regressive on X E V . ]

Let us consider the relation D I U (the Rudin-Keisler order) among nonprincipal K-complete ultrafilters over K, defined as follows: (28.26)

D IU

iff

D=h,(U)

for some h :

K+K

The relation I is transitive and induces a partial ordering among the equivD. Let D < U denote alence classes, where D E U means D I U and U I DI U and U D.

<

Exercise 28.13. If D = h,(U), then the function k : Ult,( V ) + Ult,( V ) defined by

k([f],)

= [f

h ] , is an elementary embedding.

It follows that if j' is an ordinal function, then [ , f ] ,~ [ A],, f and in particular [d], I [ h ] " . Another consequence is that if D I U , thenj,(h-) 5 ,j[,(h-). If the function h in (28.26) is one-to-one, or at least one-to-one on some set X E U , then D = U because U = g,(D) where g = /I-'. Moreover, [.(ID = [ f A], for every1; and O h D ( V) = Ulr,( V ) . Also, [ d ] , = [AIL).

28.

319

ULTRAPOWERSAND ELEMENTARY EMBEDDINGS

If U is a nonprincipal K-complete ultrafilter over K , let 6(V) = [ f l u wheref is the least ordinal function that is one-to-one on a set X E U. Exercise 28.14. If D < U , then 6 ( D ) < 6 ( U ) .Consequently, the Rudin-Keisler order is

well-founded. [Let D < U . In view of the discussion above, we may assume that the diagonal function is the least one-to-one function in both Ulr, and U l t , . Let D = h,(U). Claim: h ( a ) 5 a almost everywhere. (Otherwise let f ( [ ) be the least such that h ( a ) = ( ; f i s one-to-one since 5 = h(f(r)), andf(<) < 6 almost everywhere mod D.) Since D < U we have [ h ] , < [ d ] , , and so 6 ( D ) = [ d ] , [h]" < [d]" = 6 ( U ) . ]

r

Measurable Cardinals and Cardinal Arithmetic One of the remarkable consequences of existence of measurable cardinals is that if the generalized continuum hypothesis holds below a measurable cardinal K , then 2" = K + . In fact, as we shall see in the next chapter, the failure of the GCH at a measurable cardinal has rather strong consequences. For now, we shall just use the ultrapower to show that 2" depends on the behavior of 2" below ti.

Lemma 28.13. LA

K

be a measurable cardinal. If 2" > t i + , then the set

{ a < K : 2" > a + } has measure one, for every normal measure on K . Consequently, if2" = a+ for all cardinals a < K , then 2" = K + .

-= K : tl is a cardinal} is closed unbounded and

(As for the notation, note that { a hence has a normal measure one.)

Proof: Let D be a normal measure on K , and let M almost all a, then, since [d], = K, we have

M k

=

Ult,( V ) . If 2" = a+ for

2'=K+

In other words, there is a one-to-one mapping in M between P"(K) and ( K + ) ~ = K + because ~ K + (in fact, (K')" However, P"(K) = P(ti) and ( K + ) 5 P M ( t i )= P(ti)),and so 2" = t i + . Similarly, we have: Exercise 28.15. If D is a normal measure on K and { a : 2" I a + + } E D,then 2" 5 K + + . More generally, if D < K and {K,: 2NmI K.+B} E D, then 2" 5 K.+B. [Iffis such thatf(K,) = K.+P for all a c K , then [ f I D = (K,+jc,#' I KK+B.] Exercise 28.16. I f D is a normal measure on 2" < K K + K . [If{(.) = %+,. then [/I = ( K K + K ) M . ]

K

and { a : 2"< K,+,}

E

D, then

A similar method can be used for singular cardinals of measurable cofinality. The following results (which have been partly superseded by the more general theorems of Silver and Galvin-Hajnal; see Section 8) illustrate the method.

.

5.

320

MEASURABLE CARDINALS

Lemma 28.14. Let K be a measurable cardinal, let D be a normal measure on K and let j : V -+ M be the corresponding elementary embedding. Let I > K be a strong limit cardinal of cojnality K . Then 2A< j ( A ) . Proof. We recall that by Lemma 28.9(d), since cf A = K, we have j ( A ) > A. We shall show that (28.27)

2' = I" I

(A")"

I( A j ' K ) ) M c j ( A )

The first equality in (28.27) holds because I is a strong limit. We have I" I (I")" because ("I)" = "I.As for the last inequality, we have

M C j ( A ) is a strong limit cardinal and since I < j ( I ) and j ( K ) < j(A), we have M k

A]'") < j ( I ) .

W

For applications of the lemma, see Exercises 28.18-28.20 below. K = cf 1 is measurable. If 2" = a' for all a < 1,then 2" = 1+. [Let D be a normal measure on K ; let 1= [ e l D .For almost all a, e(a). I (e(a))+,and SO Ult C 1"2 1 '. Thus 2" = 1" 5 (1")"2 (A')" 2 A+.]

Exercise 28.1 7. Let 1 be a singular cardinal such that

The result of this exercise is a special case of Theorem 23. In fact, Silver's original proof was motivated by measurable cardinals and is similar to the present method. I shall present the ultrapower proof in Chapter 6. Exercise 28.18. Let K be measurable and let

2" < K ( 2 . ) + . [Use Lemma 28.14: j(1) = (N~(,+.,)" 2

1 = K,+, be a strong limit. Then

Nj(K)+j,K);

j ( K ) + j ( ~<) (2")+.]

Exercise 28.1 9. Let K be measurable, let 1be a strong limit, cf 1= K, such that 1< N". Then 2" c K".(Compare with Theorem 24.) [Use Lemma 28.14: If 1= N., thenj(1) = (N,(.))" I Nj("), and j ( a ) < (aK)+< 2.1

Exercise 28.20. Let @(a)denote the ath fixed point of N,i.e., the ath ordinal 5 such that Ke = 5. Let K be measurable and let 1= @ ( K K ) be a strong limit. Then 2' < @((2")+).

+

[Use Lemma 28.14 as in Exercise 28.18, and use the fact that (@(a))" I @(a)for all a.1

The above applications of the ultrapower modulo a normal measure on K can be expressed in a general abstract form (see also Lemma 32.2):

Lemma 28.15. Let K be a measurable cardinal and let D be a normal measure on Then the model ( K + ] , E) is isomorphic to the ultraproduct Ult,{( V,, E) : a < K } . Thefunction

K.

n ( x ) = [(x n V , : a < K)] is an isomorphism. Proof. Exercise.

(x E

v,+

29.

321

INFINITARYCOMBINATORICS

Corollary. L.et cp be a formula. Under the assumptions of the lemma, ifx S G V , , we hatie (28.28)

V,,

k q ( x , S)

iff

{ a : V,,

I=cp(x, S n

E

V, and

V.)}E D

29. INFINITARY COMBINATORICS

In this section, I shall introduce a new method, which in conjunction with the method of ultrapowers will enable us to obtain various consequences of large cardinal axioms. The method, which uses combinatorial arguments, has wide applications in model theory, and its most remarkable application in set theory (Section 30) is a mixture of combinatorial and model-theoretic techniques. Partition Properties

Let us consider the following argument (the pigeonhole principle): If seven pigeons occupy three pigeonholes, then at least one pigeonhole is occupied by three pigeons. More generally: If an infinite set is partitioned into finitely many pieces, then at least one piece is infinite. Recall that a partition of a set S is a pairwise disjoint family P = { X i : i E I } such that X i = S . With the partition P we can associate a function F : S + 1 such that F ( x ) = F ( y ) if and only if x and y are in the same X E P. Conversely, any function F: S + I determines a partition of S. (I shall sometimes say that F is a partition of S.) For any set A and any natural number n > 0,

uiE,

(29.1)

[ A ] " = { X G A : 1x1 = n )

is the set of all subsets of A that have exactly n elements. It is sometimes convenient, when A is a set of ordinals, to identify [A]" with the set of all sequences (alr . .., a,,) in A such that a l -= c a,, . We shall consider partitions of sets [A]" for various infinite sets A and natural numbers n. Our starting point is the theorem of Ramsey dealing with finite partitions of [o]". If { X i : i E I } is a partition of [A]",then a set H c A is homogeneous for the partition if for some i, [HI"is included in X i ; that is, if all the n-element subsets of H are in the same piece of the partition. Lemma 29.1 (Ramsey's Theorem). Let n and k be natural numbers. Every partition { X , , . . ., X , } of [orinto k pieces has an infinite homogeneous set. Equivalently, for every F : [o]" + { 1, .. ., k } there exists an infinite H C o such that F is constant on [HI". Proof: By induction on n. If n = 1, the lemma is trivial, so we assume that it holds for. n and prove it for n + 1. Let F be a function from into

322

5.

MEASURABLE CARDINALS

11, . . ., k}. For each a E o,let F, be the function on [w - {a}]" defined as follows:

W )= F ( { 4 " X )

By the induction hypothesis, there exists for each a E o an infinite set H , G w - {a} such that F , is constant on [H,]". We construct an infinite sequence {ai : i = 0, 1, 2, ...}: We let a. = 0, and ai+,

- the least element of Hai bigger than ai

It is clear that for each i E o,the function Faiis constant on [ { a , : rn > i}]"; let G(ai)be its value. Now there is an infinite subset H E {ai : i E o}such that G is constant on H. It follows that F is constant on [HI""; this is because for x1 < . " < x , + ~ i n H w e h a v e F ( x l ,..., x,+l)=F,,(xz , . . . , x , + 1). To facilitate our investigation of generalizations of Ramsey's theorem, I shall now introduce the arrow notation. Let K and A be infinite cardinal numbers, let n be a natural number and let in be a (finite or infinite) cardinal. The symbol (29.2) K --* (A)", (read: K arrows A) denotes the following partition property: Every partition of into rn pieces has a homogeneous set of size 1. In other words, every F: [K]" + rn is constant on [HI" for some H 5 K such that I H I = A. Using the arrow notation, Ramsey's theorem is expressed as follows :

[KT

No (No):

(n, k E a) The subscript rn (in 29.2) is usually deleted when rn = 2, and so (29.3)

-+

K+

(A).

is the same as K .+ (A);. The relation K + stays true if K is made larger or if A or rn are made smaller. A moment's reflection is sufficient to see that the relation also remains true when n is made smaller. Obviously, the relation (29.2) makes any sense only if K L 1 and K > m ; if rn = K, then it is clearly false. Thus we always assume 2 I rn < K and 1 I K . If n = 1, then (29.2) holds just in case either K > A, or K = 1 and cf K > rn. We shall concentrate on the nontrivial case: n 2 2. We start with two negative partition relations

(AL

Lemma 29.2. For all K and 1, 2" % (4: I n other words, there is a partition of 2" into K pieces that does not have an injnite homogeneous set.

Proof. In fact, our partition has no homogeneous set of size 2 3. Let S = "(0, 1) and let F: [S]' + K be defined by F ( { J g } ) = the least a < K such

323

29. INFINITARY COMBINATORICS

that f ( a ) # g(a). I f J g, h are distinct elements of S, it is impossible to have F ( { J sf) = W J 4 )= F ( { g , h}). Lemma 29.3. For every K, (Thus the obvious generalization of Ramsey’s theorem, namely K 1-, (K1)i, is false.) To construct a partition of [2“12that violates the partition property, let us consider the linearly ordered set ( P , < ) where P = “(0,l}, a n d f < g if and only if f ( a ) < g(a) where a is the least a such that f ( a ) # g(a) (the lexicographic ordering of P). Lemma 29.4. The lexicographically ordered set “(0, 1) has no increasing or decreasing K -sequence. +

Proof: Assume that W = {f,: a < K + } c “2 is such thatf,
fu(cm)

r,

ra

Proof of Lemma 29.3. Let 2“ = I and let { f , : a c A} be an enumeration of the set P = “(0, I}. Let 4 be a linear ordering of I induced by the lexicographic ordering of P: a < p iff, < f s . Now we define a partition F: [A]’ -+ (0, 1) by letting F({a, p}) = 1 when the ordering < of {a, p} agrees with the natural ordering; and letting F({a, P } ) = 0 otherwise. If H c_ I is a homogeneous set of order type K + , then {f,:a E H} constitutes an increasing or decreasing K+-sequence in (P, <); a contradiction.

if

K

By Lemma 29.3, the relation K + (Nl)2is false if K I 2”’. On the other hand, > 2’O, then K -+ (N,)’is true, as follows from this more general theorem:

Theorem 69 (Erdos-Rado Partition Theorem).

3,’

+

I n particular, (2’0)+

-+

(Wd

(K1)2,, .

Proof: We shall first prove the case n = 1 since the induction step parallels closely this case. Thus let K = (2’O)’ and let F: [K]’ -+ o be a partition of [K]’ into No pieces. We want to find a homogeneous H c K of size N l . For each a E K, let F, be a function on K - { a } defined by F d x ) = F({a, x } ) .

5.

324

MEASURABLE CARDINALS

We shall first prove the following claim: There exists a set A c ti such that I A I = ZNOand such that for every countable C c A and every u E K - C there exists u E A - C such that F , agrees with F , on C. To prove the claim, we construct an o ,-sequence A. C A G . .. G A, C . . ., a < a,, of subsets of K, each of size 2"0, as follows: Let A. be arbitrary, and for each limit a, let A, = U,<, A,. Given A,, there exists a set A,, 2 A, of size 2N0such that for each countable C c A, and every u E ti - C there exists u E A,+, - C such that F , agrees with F , on C (because the number of such A,, and clearly A has the required functions is I 2N0).Then we let A = property. Next we choose some a E ti - A, and construct a sequence {xu: a < o ,}in A as follows: Let xo be arbitrary, and given {x, : p c a } = C,let xu be some u E A - C such that F , agrees with F, on C. Let X = {xu: a < wl). Now we consider the function G:X -,o defined by G(x) = F A X ) .It is clear that if a c 0, then F({x,, x,}) = F,,(x,) = F,(x,) = G(x,). Since the range of G is countable, there exists H G X of size N, such that G is constant on If. It follows that F is constant on [HI'. Thus we have proved the theorem for n = 1. The general case is proved by induction. Let us assume that &f + (N1)"Noand let F: [K]"" -,w, where K = For each a E K , let F,: [K - {a}]"-+w be defined by F,,(x) = F ( x u {a}).As in the case n = 1, there exists a set A c K of size >,,such that for every C c A of size I C I I 3,,- and every u E K - C there exists u E A - C such that F , agrees with F, on [C]". Next we choose a E K - A and construct a set X = {xu:a c 3 : - ,}c A such that for each a, F,. agrees with F, on [{x, : p < a}T. Then we consider G: [XI" + o where G(x) = F,(x). As before, if a , c ... c a,+l, then F({xUI, ..., xu"+]})= G({x,,, ..., x.3). By the induction hypothesis, there exists H , G X of size such that G is constant on [HI". It follows that F is constant on W

u,
x.

,

It can be proved that for each n, the partition property 3 : + (Nl)nN+olis best possible. The property also generalizes easily to larger cardinals. For each K, let exp0(K)= K and exp,, ,(K) = 2expn(K). Exercise 29.1. For every K , (exp,(K))+ -+

(,+):+I

In particular, we have (2")' + (K')'. Both Ramsey's theorem and the Erdos-Rado theorem have numerous applications in model theory. The following easy application of the ErdosRado theorem in point set topology is of some interest to us. We recall that if (P, <) and (Q, <) are two partially ordered sets such that sat(P) = sat(Q) = N, (that is, both satisfy the countable chain condition) then sat(P x Q) is not necessarily N,;see Exercise 22.4,and also Theorem 53 and the discussion preceding Lemma 20.3.

29.

325

INFINITAR Y COMBIN ATORICS

Exercise 29.2. If P and Q are partially ordered sets such that sat(P) = sat(Q) = N l , then sat(P x Q) I( 2 w ) t . [Let K = (2&)’ and let {x. : a < K} be an incompatible subset of P x Q. For all a, fl < K, let F ( { a , fl}) = 0 or 1 according to whether the first coordinates of x, and xg are incompatible or compatible. Apply the Erdos-Rado theorem and get either N Iincompatible elements of P or Nl incompatible elements of Q.]

A natural generalization of the partition property (29.2) is when we allow I to be a limit ordinal, not just a cardinal. Let K, n and rn be as in (29.2) and let a > 0 be a limit ordinal. The symbol

(29.4) stands for: For every F: [K]” + rn there exists H G K of order-type a such that F is constant on [HI”. There are various results about the partition relation (29.4).For instance, it has been proved that K , + (a)’ for all a < a,.The analogous case for K, is unclear: If 2N0= K1,then K2 + (ol)’ (by Erdos-Rado). However, it is consistent (with 2N0= K , ) that K 2 f , (a,+ a)’(either by forcing or in L, using the principle W in (24.16)). Moreover, it is not known whether K2+ (a1+ is consistent . In the positive results given by the Erdos-Rado theorem, the size of the homogeneous set is smaller than the size of the set being partitioned. A natural question arises, whether the relation K + (K)’ can hold for cardinals other than K

= 0.

A cardinal K is weakly compact if it is uncountable and satisfies the partition property K + (K)’. The reason for the name “weakly compact is that these cardinals satisfy a certain compactness theorem for infinitary languages; we shall investigate weakly compact cardinals in some detail in Section 32. ”

Lemma 29.5. Every weakly compact cardinal is inaccessible.

u

Proof. Let K be a weakly compact cardinal. To show that K is regular, let us assume that K is the disjoint union { A , : y < A} such that I c K and ( A , I < K for each y < 1. We define a partition F: [K]’ + (0, 1) as follows: F({a, /3}) = 0 just in case a and /3 are in the same A , . Obviously, this partition does not have a homogeneous set H C K of size K. That K is a strong limit cardinal follows from Lemma 29.3: If K 2 2 ’ for some I c K, then because 2’+(I+)’, we have ~ f r ( A + ) ’ and hence K

% (K)’.

We shall prove in Section 32 that every weakly compact cardinal K is the Kth inaccessible cardinal. I shall now give an equivalent formulation of weak compactness. Recall that an Aronszajn tree is a tree of height a 1all of whose levels are at most countable and which has no uncountable branches (see Lemma 22.3). For any

5. MEASURABLE CARDINALS

326

regular uncountable cardinal K, let us call an Aronszajn K-tree a tree of height K with levels of sue < K and with no branches of size K. I shall look into this generalization of Aronszajn trees later in this section; for now let us just say that K has the tree property if there is no Aronszajn K-tree, that is, if every tree of height K with small levels has a branch of size K. Lemma 29.6.

(a) If K is weakly compact, then K has the tree property. (b) I ~ isK inaccessible and has the tree property, then K is weakly compact, and in fact K + (.):for eoery m < K . Proof: (a) Let K be weakly compact and let (T, < T) be a tree of height K such that each level of T has size < K. Since K is inaccessible, I TI = K and we may assume that T = K. We extend the partial ordering < of K to a linear ordering i : If a < /I, then we let a < p; if a and p are incomparable and if 5 is the first level where the predecessors a t , pc of a and p are distinct, we let a < p if and only if a, < p,. Let F: [K]' + (0, I } be the partition defined by F({a, p)) = 1just in case i agrees with < on {a, p}. By weak compactness, let H C K be homogeneous for F , If!( = K . We now consider the set B G K of all x E K such that {a E H : x < a } has size K. Since every level has size < K, it is clear that at each level there is at least one x in B. Thus if we show that any two elements of B are < Tcomparable, we shall have proved that B is a branch in T of size K. Thus assume that x, y are incomparable elements of B ; let x < y. Since both x and y have K successors in H, there exist a < p < y in H such that x < ., a, y < p, and x < y. By the definition of <, we have a < p and y < p. Thus F({a, p}) = 1 and F ( { y , p } ) = 0, contrary to the homogeneity of H. (b) Let K be an inaccessible cardinal with the tree property, and let F: [K]' + I be a partition such that 11 I < K. We shall find a homogeneous H G K of size K. We construct a tree (T, G ) whose elements are some functions t : y + I , y < K. We construct T by induction: At step a < K, we put into T one more Having constructed t o , .. .,f a , .. .,p < a, let element t , calling it t,. Let t o = 0. us construct t , as follows, by induction on 5. Having constructed t , I (, we look first whether r, I is among the t,, fl < a (note that for 5 = 0 we have r, 10 = t o ) . If not, then we consider t , constructed: t , = t a l 5. If t,1< = t o for some p < a, then we let fa(<) = i, where i = F ( { p , a } ) . (T, G ) is a tree of size K ; and since K is inaccessible, each level of T has size < K and the height of T is K. It follows from the construction that if t , c t , , then p < a and F({P, a } ) = t,(length(t,)). By the assumption, T has a branch B of size K. If we now let, for each i E I,

<

(29.5)

H i = {a :t,

E

B and r,"i

E

B)

29.

327

INFINITARYCOMBINATORICS

then each Hi is clearly homogeneous for the partition F, and at least one H i has size K.

It should be mentioned that an argument similar to the one above, only more complicated, shows that if K is inaccessible and has the tree property, then K+ for all n E w , m < K . We have proved that every measurable cardinal K is inaccessible and is in fact the Kth inaccessible cardinal. We shall now improve on this as follows:

(KK

Lemma 29.7. Every measurable cardinal K is weakly compact. Moreover, i f D is a normal measure on K , then the set { a < K : u is weakly compact} is in D. Pro05 Let K be a measurable cardinal. To show that K is weakly compact, it sufficesto prove the tree property. Let (T, < ) be a tree of height K with levels of size < K. We consider a nonprincipal K-complete ultrafilter U over T. Let B be the set of all x E T such that the set of all successors of x is in CJ. It is clear that B is a branch in T and it is easy to verify that each level of T has one element in B; thus B is a branch of size K. Let D be a normal measure on K and consider the elementary embedding jD: V + M where M = Ult,( V). Since P ' ( ~ ) = P ( K ) ,it follows that K is weakly compact in M , and since [ d ] , = K, we have { a : a is weakly compact} E D. We shall consider a generalization of the partition properties introduced so far. But first let us consider this: If A is an infinite set of ordinals and a an ordinal, let [A]" denote the set of all increasing a-sequences in A. The symbol (29.6) K -b (1). denotes: For every partition F : [ K ] " + (0, 1) of [ K ] " into two pieces, there exists a set H of order type I such that F is constant on [HI". Exercise 29.3. For all infinite cardinals K , K $, (w)"'. [For s, I E [K]"' let s = t iff ( n : s(n) # I(.)) is finite. Pick a representative in each equivalence class. Let F ( s ) = 0 if s differs from the representative of its class at an even number of places; let F ( s ) = 1 otherwise. F has no infinite homogeneous set.]

Let me introduce now the following partition property: Let K be an infinite cardinal, let a be an infinite limit ordinal, a I K, and let m be a cardinal, 2 5 m < K. The symbol (29.7) K + (a),iw

u:=o

denotes the property that for every partition F of the set [ K ] < " = [K]" into rn pieces, there exists a set H s K of order type a such that for each n E o,F is constant on [HI". (Again,the subscript m is deleted when m = 2.) We say that H is homogeneous for F . Exercise 2 9.4. (0 jt ((0)< '". [For x E [w]<'".let F ( x ) = 1 if I x 1

pick n

E

E x, and F ( x ) = 0 otherwise. I f H H and show that F is not constant on [HI".]

Gw

is infinite,

320

5.

MEASURABLE C A R D I N A L S

We shall see later that the least K such that K f , (w)<'* is inaccessible, and greater than the least weakly compact cardinal. But first let us investigate the stronger property. (29.8)

K + (K)<"

The cardinals that satisfy the partition relation (29.8) are called Ramsey cardinals. We shall show that every measurable cardinal is Ramsey and use this property to improve considerably on Scott's theorem (Theorem 67). Note that since Xo is not Ramsey, every Ramsey cardinal is weakly compact and hence inaccessible.

Theorem 70. Let K be a measurable cardinal, let D be a normal measure on K , and let F be a partition of [K]<'* into less than K pieces. Then there exists a set H E D homogeneous for F . Herice every measurable cardinal is Ramsey. Proof. Let D be a normal measure on K, and let F be a partition of [ K ] < " into less than K pieces. It suffices to show that for each n = I , 2, . . . ,there is H , E D such that F is constant on [ H J ; then H = H , is homogeneous for F. We prove, by induction on n, that every partition of [K]" into less than K pieces is constant on [HI" for some H E D. The assertion is trivial for n = 1, so we assume that it is true for n and prove that it holds also for n + 1. Let F : [K]"" + I , where I I / < K . For each a < K , we define F , on [K - {a}]" by F,(x) = F({a} u x ) . By the induction hypothesis, there exists for each a < K a set X uE D such that F , is constant on [X,]"; let i , be its constant value. Let X be the diagonal intersection

n:=

X=(a
n x,}

Y
We have X

E

D since D is normal; also, if y < a l < . * * < a,, are in X , then

and so F ( ( y , a l , .. ., a,}) = F y ( { a l ,..., a,}) = i , . Now, there is i E I and H C X in D such that i , = i for all y E H . It follows that F ( x ) = i for all x E [ H I n + ' . (a1,..., a,,} E [ X , Y

Exercise 29.5. If D is a normal measure on K, then the set of all Ramsey cardinals

below

K

has D-measure 1.

[As in Lemma 29.7.1

Exercise 29.6. Let D be a normal measure on K and let f: [ K ] ' " --t K be such that f ( x ) = 0 o r f ( x ) < min(x) for all x E [ K ] ' " (min(x) is the least element ofx).Then there is H E D such that for each n,fis constant on [HI". [By induction, as in Theorem 70. Given f on [K]"", let f&)= f ( { a } u s) for a < min(s);fu is constant on [X,]" with value y. c. a. Let X be the diagonal intersection of X., a < K, and let y and H c X be such that H E D and y, = y for all tl E H.]

29.

329

INFINITARY COMBINATORICS

Let U be a K-complete nonprincipal ultrafilter over K. For each n = 1,2, . . ., we define an ultrafilter U , over [K]" as follows: If X G [ K ] " + ' , let (29.9) We let U (29.10)

X , 2 )= {x E

[K]"

: a < min(x) and { a } u x

E

X}

, = U , and define by induction X U,+ , iff { a : X ( o l ) U,} E U E

E

(Thus X E U , iff for almost all a,, for almost all a z , ..., for almost all a,,, . .., a,,} E X.)It is easy to verify that each U , is a nonprincipal K-complete ultrafilter over [K]". Moreover, if Z E U , then [Z]" E U,.

{a,,

Exercise 29.7. If U is normal and X E U , , then there is H E U such that [HI" G X . [Either X or [K]" - X contains [HI" for some H E U , by Theorem 70. It must be X

since [HI" E U , . ]

Partitions and Models Partition theorems play an important role in model theory, and I shall have more opportunity to present some applications in the next section. For now, let us consider a model

YI = ( A , P", ..., F", ..., c", ...)

(29.11)

of a (not necessarily countable) language Y Y = { P , ..., F,..., C, ...}

(29.12)

Let K be an infinite cardinal and let us assume that the universe A of the model P I contains all ordinals a < K: A 2 K. A set I c K is a sef ofindiscernibles for the model PI if for every n E w, and every formula cp(u,, . . ., u,,), iff PI b cp[a,, . . ., a,,] PI b cp[P,, . . ., p n ] whenever a , < . . < a, and fl , < . . . < p, are two increasing sequences of elements of I.

(29.13)

*

Exercise 29.8. Let F : [ K ] ' " + {O, l}, and let F , = F I [ K ] " , for each n. Let / be a set of indiscernibles for the model ( K , <, F1, F 2 ,F 3 , ...). Then / is homogeneous for the partition F. [Consider the formula F,(al, .. ., a,)= F , ( P , , .. ., B,,).]

Thus the partition properties are, in a way, a special case of existence of indiscernibles. We shall now show that on the other hand, partition properties imply existence of indiscernibles in models:

Lemma 29.8. Let

K

be an injinite cardinal and assume that K

+

(a);Am

330

5. MEASURABLE CARDINALS

where a is a limit ordinal and I is an infinite cardinal. Let Y be a language of size I I and let P I be a model of 9 such that A 2 K . Then P I has a set of indiscernibles of order type a. Proof. Let 0 be the set of all formulas of the language 9. We consider the ~ P(@) defined as follows: If x E [K]" and x = (alr .. . , a,,} function F : [ K ] < 4 where a l < . * .< a,,, then F ( x ) = {cp(ol, ..., u,,) E 0 : P I k cp[a,, . . ., a,,]} The function F is a partition into at most 2A pieces and thus has a homogeneous set 1 c K of order-type a. It is now easy to verify that I is a set of indiscernibles for P[. We shall see in Section 32 that for a given limit ordinal a, the least K that satisfies K + (a)<" is inaccessible and satisfies K + (a);'" for all I < K . Now we shall prove this for Ramsey cardinals.

Lemma 29.9. lf

K

4

(K)<"'

and if 1 i s a cardinal less than

K,

then

K + (K):'".

Corollary. If K is a Ramsey cardinal and $PI 2 K is a model of a language of size < ti, then PI has a set of indiscernibles of size K . Proof. Let F : [ K ] < " + I be a partition into I < K pieces. We consider the following partition G of [K]'" into two pieces: If a I < ... < ak < ak+ I < ... < t l 2 k are elements of K and if F({a,, . . ., MI,}) = F({ak+ . . ., a,,}), then we let G({a,, ..., C(2k))= 1; for all other x E [K]'", we let G(x) = 0. Now, let H C_ K be a homogeneous set for G, I H 1 = ti. We claim that for each k and each x E [HIZk,G(x) = 1: This is because I H I = K > A, and therefore we can find a 1 < ... C(k < a k + ]< ... < C(2k in H such that F ( ( a l , ..., a,})= F({ak+l, a2k}). It follows that H is homogeneous for F : If a1 < ... < a,, and PI < ... < /I,, are two sequences in H, we choose a sequence y 1 < < y,,in H such that both x,, < y1 and p,, < yl. Then

-=

.-.?

The combinatorial methods introduced in this section will now be employed to obtain a result on measurable cardinals considerably stronger than Scott's theorem in the preceding section. Let me first make a few observations about models with definable Skolem function. Let 91 be a model of a language Y such that 91 2 K and let I G K be a set of indiscernibles for PI. Let us assume that the model VI has definable Skolem

29.

331

INFINITARY COMBINATORICS

functions; i.e., for every formula* q ( u . u l , . . .,0), there exists an n-ary function h, in PI such that: (i) lg, is dejnable in PI, i.e., there is a formula $ such that

iff

y = h(xl, ..., x,)

PI 1 $[y, xl,

..., x,,]

for all y , xl, . . ., x,, E A ; and (ii) lg, is a Skolemfunctionfor cp (see Section 10). Let B E PI be the closure of I under all functions in Y and all the functions h,,,, cp a formula. B is an elementary submodel of PI, and in fact is the least elemen-

tary submodel of PI that includes the set I; let us call B the Skolem hull of I and let us say that I generates '8. We augment the language of PI by adding function symbols for all the Skolem functions and call Skolem terms the terms built up from variables and constant symbols (which are 0-ary functions) by applications of functions in Y and the Skolem functions. Since B is an elementary submodel of PI, the interpretation of each Skolem term t is the same in B as in PI. For every element x E B there is a Skolem term r and indiscernibles y 1 < ... < y,, (elements of I) such that (29.14) x = t'"[yl, Y,] = t " [ y l , ..., Y,] a * . ,

Now if $ is a formula of the augmented language, i.e., if $ also contains the Skolem terms, it still does not distinguish between the indiscernibles: If a l < ... < a, and p1 < ... < p, are two sequences in I, then $ ( a l , ..., a,,) holds (either in '21 or in B) if and only if $(/II, . . ., p,) holds.

Theorem 71 (Rowbottom). If K is a Ramsey cardinal, then the set of all constructible reals is countable. More generally, i f A is an injnite cardinal less than K , then I P"(A)l = A. ProoJ Let K be a Ramsey cardinal and let A < K. Since K is inaccessible, we have PL(A)c L, . Consider the model (29.15)

PI = ( L , E, P"(A),

PI is a model of the language V = {E, Q,c a } z 5 A where Q is a one-place predicate (interpreted in PI as P(A) n L) and c,, a I A, are constant symbols (interpreted as ordinals less than or equal to A). Since ti is Ramsey, there exists a set I of size ti of indiscernibles for PI. The model PI has definable Skolem functions: Since K is inaccessible, LKis a model of ZFC + V = L and therefore has a definable well-ordering (see also Exercise 13.14). Thus let B G L, be the elementary submodel of PI generated by the set I. Every element x E B is expressible as x = t ( y l , ..., y,,) where t is a Skolem term and y 1 < ... < y, are elements of I.

5.

332

MEASURABLE CARDINALS

We shall now show that the set S = P L ( I ) n 9 has at most I elements. Since S is the interpretation in 9 of the one-place predicate Q, it suffices to show that there are at most I elements x E b such that b C Q(x). Let t be a Skolem term. Let us consider the truth value of the formula (29.16)

t ( a l , ...,a,)=

[(Ply..., Pn)

for a sequence of indiscernibles a l < ... < a , < PI -= ... < P,. The formula (29.16) is either true for all increasing sequences in I or false for all increasing sequences in I. If (29.16) is true, then it is true for any two sequences al
a: <

9

.

.

< a: < a ! <

< a,! < . * .< a: <

< af; <

(< < K )

<

in I and then t(ai, . . ., af;)# t(a7, . .., a:) whenever # g. In conclusion, the set (29.17)

{ [ ( a l ,..., a,): a I < * . -< a , are in I}

has either one or K elements. Now we apply this to evaluate the size of the set S. We know that IS I < K because S C P L ( I )G P ( I ) and K is inaccessible. Iff is a Skolem term for which the set (29.17) has size K, then t(a . . ., a,) is not in S, for any a < . * . < a , in I; by indiscernibility, Q ( t ( a l ,.. .,a,)) is true or false simultaneously for all increasing sequences in I. Thus if t(a . . ., a,) E S, the set (29.17) has only one element. However, since 1 9I I I,there are at most I Skolem terms. And since every x E b has the form t(a . . ., a,) for some Skolem term and a < . . . < a , in I, it follows that IS I 5 I . Thus we have proved that S = Q" = P L ( I )n 93 has at most I elements. Now b < L.and 193 I = K ; hence the transitive collapse of b is L, and we have an isomorphism

,,

n:92:L,

Since each a I I has a name in PI, we have I u {I} c 23 and so z ( X ) = X for each X G I in 9. In particular n ( X )= X for all X E S; and since QnI'] = n[S]= S, we have S = P ( I ) n n [ S ]= P L ( I )n & = P(1) This completes the proof: On the one hand, we proved that I S I I I; and on the other hand, IPL(I)I 2 I; thus IPL(I)I = I. If K is Ramsey, then h', is inaccessible in L. [Show that PLixl(w) is countable for every x C _ w.]

Exercise 29.9.

Exercise 29.10. If K is Ramsey, and A < K, then every set definable in ( L w ,E, a ) , s i has size either K or 2 1.

29.

333

INFINITARY COMBINATORICS

The proof of Theorem 71 is an important application of the Ramsey property of K and can be used in a more general situation. Below we give an abstract version of the result just proved.

Lemma 29.10. Let K be a Ramsey cardinal, and let 1 be an infinite cardinal less than K . Let PI = ( A , . . .) be a model of a language Y such that Y I IA, and let A 2 K . If P G A is such that (PI < K then A has an elementary submodel B = ( B , ...) such thar ( E l = K and IP n BI 5 A. Moreooer, i f X E A is of size at most A, then we canjnd B such that X E B. Moreover, if^ is a measurable cardinal and D is a normal measure on K , then we c a n j n d 9 such that B n K E D.

I

Proof: First we add to the language Y one unary predicate whose interpretation is the set P ; we also add constant symbols for all x E X.Next we find some Skolem functions h, (in ( A , . . . , P, x ) x )~for~every formula cp, and extend the language further by adding function symbols for the functions h, . Next we find a set of indiscernibles I E K, of size K, for the expanded model PI'; if K is measurable and D is a normal measure, we find I E D. We let B be the elementary submodel of PI' generated by I . As in the proof of Theorem 71, one proves that if I P n BI < K then IP n BI I 1.

Let us consider now models of a countable language Y with a designated one-place predicate Q. A model PI = ( A , Q", . . .)of Y is said to have type (K, 1) if I A I = K and I Q" I = 1.One of the favorite subjects of model theorists is investigation of possible types of models of a given theory-so-called twocardinal problems. Many of these problems are either independent of ZFC or lead to large cardinals. Let us call a cardinal K > N, a Rowbottom cardinal if for every uncountable A < K, every model of type (K,1)has an elementary submodel of type (K, No). By Lemma 29.10, every Ramsey cardinal is Rowbottom. Using the argument from Exercise 29.9, one can see that if there is a Rowbottom cardinal, then N, is inaccessible in L. Let us call an infinite cardinal K a Jdnsson cardinal if every model PI (of a countable language) of size K has a proper elementary submodel B of size K (i.e., Z 3 G PI and B # 9I). Clearly every Rowbottom cardinal is Jonsson. Both these properties can be formulated in terms of partitions. For instance : Exercise 29.11.* An infinite cardinal K is a Jonsson cardinal if and only if for every F : [ K ] < " ' + K there exists H E K of size K such that the image of [ H I < " under F is not the whole set K. [To show that the condition is necessary, consider the model (K, <, F1, F 2 , ...) where F , = F 1 [K]". To show that the condition is sufficient, let 91 = ( K , . . .) be a model. Let {h, : n < w } be a set of Skolem functions for PI, closed under composition and arranged so that each h, is n-ary. For each x E [ K ] " , let F ( x ) = h,(x). If H E K , the image of [HI<"' under F is an elementary submodel of 9 1 . 1

334

5.

MEASURABLE C A R D I N A L S

We shall see in Section 32 that existence of Jonsson cardinals has the same strong consequences we stated in Theorem 71; in particular, there are no Jonsson cardinals if I/ = L. Exercise 29.12. No is not a Jonsson cardinal. [Let dl = ( ( u , f ) wheref(n) = n - 1 for all n > 0.1

It has been proved among other things that no K,, n < w, is a Jonsson cardinal and that the least Jonsson cardinal is either weakly inaccessible or has cofinality w . While it is consistent (relative to ZFC + there exists a measurable cardinal) that all Jonsson cardinals are Ramsey cardinals, there is also a model in which a Rowbottom (and hence Jonsson) cardinal exists and has cofinality w. One of the still open questions is whether K,,can be J6nsson. Somewhat more is known about Rowbottom cardinals: Exercise 29.13. I f

cf

K

ti

is a Rowbottom cardinal, then either

ti

is weakly inaccessible or

= Q.

[To show that K = 1’ is not Rowbottom, letf, be a one-to-one mapping ofa onto 1, for each a, such that 1 2 a < ti. Let PI = (ti, I , <, R ) where R(a, p, y) ifff,(p) = y. If ( B , B n 1, < , R n B 3 ) < A and (BI = ti, let a be the 1th element of B ; then f z [ B n a ] G B n 1 and hence ( B n 11 = 1 > No. To show that ti is not Rowbottom if K > cf K = 1 > No,let f be a nondecreasing function of ti onto I and use f to produce a counterexample.]

A related problem is Chang’s conjecture: Every model of type (N2,K,)has an elementary submodel of type (N,,No). Chang’s conjecture is consistent relative to the theory ZFC + “there exists a Ramsey cardinal.” We shall show in Section 32 that Chang’s conjecture has the strong consequences stated in Theorem 71. For the following exercise, we recall that functionsfand g on w 1 are almost disjoinr i f f @ ) # g(a) for all a 2 some a. . Exercise 29.24.* If there is a family 9 of K 2 almost disjoint functionsf: w , + w then Chang’s conjecture fails. [Consider a model 91 with the universe 9 u w 1 and the designated predicate wl. If (9 u E, B, ...) <9I and lY(= h’,, (BI =No.then B s a for some a < w I .Show that f ( a ) # g(u) for allJ g E 9,a contradiction.]

Trees We dealt with trees of height w , quite extensively in Sections 22 and 24. We shall now look at trees of larger height. In Lemma 29.6 we proved that for an inaccessible cardinal K, nonexistence of an Aronszajn rc-tree is equivalent to weak compactness. There is a close relationship between partitions and trees; the following equivalence can be extracted from the proof of Lemma 29.6: Let us call a tree T binary if each x E T has at most two immediate successors. K be a cardinal and 1 a regular cardinal. Then K only if every binary tree of size K has a branch of length 1.

Exercise 29.15.* Let

--t

holds if and

29.

IN FIN ITARY COMBINATORICS

335

We have called a tree of height K an Aronszajn tree if all its levels have size < K and if it has no branch of length K ; K has the tree property if there is no Aronszajn tree of height ti. The tree property of cardinals has been studied in some detail; let me consider the case K = K,. This case is representative of the general case when K is the successor of a regular cardinal. In Lemma 22.3 I presented a construction of an Aronszajn tree of height o This construction can be generalized to K, provided 2’O = K , (or any K = A + such that = A). Instead of the set Q of all rational numbers, we consider a linearly ordered set Q of size K such that every a < 0,can be embedded in any interval of Q (e.g., let Q be the lexicographically ordered set of all finite sequences of countable ordinals). We construct a tree T whose elements are certain bounded increasing sequences in Q of length < w,. We proceed very much as in Lemma 22.3: We try to obey the same condition (22.4).The difference is at limit steps of cofinality w ; if cf a = w, then we simply extend all a-branches represented by bounded a-sequences in Q. By the assumption 2”O = K, their number is only K,,and so the ath level has still size K,.At limit steps of cofinality 0 ,we do more or less the same as in Lemma 23.3. We extend only some branches so that the size of the limit level is K , and that the condition (22.4) is satisfied. The Aronszajn w,-tree constructed thus under the assumption 2’O = K , is special (see Section 24): Its nodes are one-to-one functions from some a < 0, into a fixed set of size K,,and the ordering of the tree is by inclusion. Special Aronszajn trees are of interest to model theorists: There is a sentence that has a model of type (I?, K ) just in case there exists a special Aronszajn K+-tree. There is a model of set theory in which 2“O = K, and no special Aronszajn w,-trees exist. This consistency result therefore establishes consistency of the statement: “There exists a model of type (HI, KO)that has no elementarily equivalent model of type (K,,K,)” (this statement is known to be false if the continuum hypothesis holds). The above-mentioned consistency result is relative to the consistency of there exists a Mahlo cardinal.” This assumption is necessary since it has also been proved that there is no special Aronszajn w,-tree” implies that K, is a Mahlo cardinal in L. (It is easy to see that K 2 is inaccessible in L: If K, is a = 0 ,and successor cardinal in L, then there exists A G w,, such that w4IA1= o,-see Exercise 15.8. In YA],2N0= K , holds and hence there exists a special Aronszajn o,-tree T. It is obvious that T is a special Aronszajn 0,-tree in the universe.) Similarly, ifs,has the tree property (that is, if there are no Aronszajn trees of height w,, special or otherwise), then K 2 is weakly compact in L; and the statement “K, has the tree property and 2O ’ = K,” is consistent relative to the existence of weakly compact cardinals. Let me finally consider a generalization of Suslin trees. If K is a regular cardinal, we call a Suslin K-tree a tree whose height is K and that has no branches of length K and no antichains of size K.

,.

,

“

“

5.

336

MEASURABLE CARDINALS

If W is a model of ZFC + GCH and if K = 1 ' is a successor of a regular cardinal (in W),then a straightforward generalization of the forcing notion (22.6) yields a generic extension !Ul[G] in which there is a Suslin K-tree: The forcing conditions are normal a-trees, a < K , whose nodes are functions t : /I + 1.The generalization consists in adding one more requirement on the trees T E P: If a c height(T) is a limit ordinal of cofinality < K , then b: a + I that represents a branch in the tree T, = {t E T : dom(t) < a} belongs to the ath level of T. Since A<' = 1,it is possible to satisfy this requirement and still have I TI 5 1. The proof that W[G] has a Suslin K-tree follows closely the case K =

K,.

The construction of a Suslin tree in L also generalizes to higher cardinals. As before, let me concentrate on the simplest generalization, the case K = K2. Let us consider the analog of the diamond principle (22.15)for K,; since we will distinguish between ordinals of cofinality w and cofinality w , , let me be a little more general. Let E be a stationary subset of w , . (29.18)

O(E)

There exists a collection of sets {S, : a X G 0,, the set

E

E } such that for every

{ a €E : X n a = & } is a stationary subset of 0,.

The principle O ( E ) is proved in Lexactly as Lemma 22.9. The easiest construction of a Suslin o,-tree in L follows closely the proof of Lemma 22.8. We let E = {a : cf a = w l } . We construct a tree by induction on levels. We do different things at limit levels of cofinality w and at limit levels of cofinality w , . If cf a = o,then we construct the ath level by extending all branches in T,; since 2'O = K,,the ath level has still size K,.If cf a = w , , then we use O ( E ) to destroy potential antichains of size K2.Note that since all branches are extended at cofinality w levels, the tree T, for cf a = w , has a plenty of a-branches: Each x E T, has an a-branch going through x. The proof that the resulting tree is a Suslin w,-tree follows closely Lemma 22.8. There is another, more sophisticated construction of a Suslin w,-tree in L (which, unlike the construction above, generalizes to successors of singular cardinals). Let me recall the box principle 0 (cf. 24.18) which is true in L. The construction uses the box principle and the principle O(E)for every stationary set of ordinals of cofinality w. Roughly speaking, the 0 principle is used to destroy antichains, while 0 enables us to keep going at limit levels ofcofinality ol. In this construction, not all branches are extended at cofinality w levels. Now a modification of O ( E ) for E = {a : cf a = w } has been proved from GCH alone. Since it is also known that 0is true unless K, is a Mahlo cardinal in L, it follows that if GCH holds and K2 is not Mahlo in L, then there exists a Suslin w,-tree. There is a model of ZFC in which 2"O = K, and there are no Suslin w,-trees.

30.

337

SILVER INDISCERNIBLES

The construction uses a measurable cardinal, and the model satisfies 2H' = K3. It is still open whether nonexistence of Suslin 02-trees is consistent with the GCH. 30. SILVER INDISCERNIBLES

We have shown in the previous two sections that existence of measurable cardinals contradicts the axiom of constructibility, and in fact the incompatibility of measurable cardinals with V = L is quite substantial: If there is a measurable cardinal, then there are only countable many constructible reals and Sl is inaccessible in L. We shall now use more model theoretical methods to obtain quite striking consequences of existence of measurable cardinals. Before stating the main theorem of this section, let me define (or recall) the and let notions involved. Let PI = (A, . ..) be a model (of a language 9) X G A. An element a E A is definable from X if there exists a formula cp(u, u l r ..., 0,) of 9, and some xl, ..., x, E X such that (30.1) PI k a is the unique u such that cp(u, xlr ..., x,). Let I be a limit ordinal and let I c I be a set of ordinals in LA.The set I is a set of indiscernibles for LAif for every formula cp(v,, . . ., 0,) of the language of set theory (30.2) whenever a

LA k cp[al, . . ., a,] iff LA k ~ p [ P 1 , pn] < . . . < a, and p1< * * . < p, are two increasing sequences in I.

Theorem 12. I f there exists a measurable cardinal, then: (A) I f K and I are uncountable cardinals and K < I , then (L, E) is an elementary submodel of ( L A ,E). (B) There is a unique class S of ordinals containing all uncountable cardinals such that for each uncountable cardinal K : (a) S n K has order type K , and if^ is regular, then S n K is closed and unbounded in K ; (b) S n K is a set of indiscernibles for (L, E); (c) every a E L, is definable in L,from S n K . Elements of the class S are the Silver indiscernibles for L The proof of Theorem 72 will constitute about half of this section. Thus I shall first state some consequences of the conclusion of Theorem 72. By the reflection principle, if cp is a formula, then there exists an uncountable cardinal K such that L k cp(xl, ..., xn) for all xlr

iff

L k cp(x,, ..., x,)

. .., x , E L.By (A), we have then L k cp(xl, ..., x,)

iff

LA k cp(x,, ..., x,)

338

5.

MEASURABLE C A R D I N A L S

for all cardinals A 2 K and all x,, . . ., x, . In view of this, we can define satisfaction in L for all formulas cp E Form: If cp(v . . ., v,) is a formula of the language L = {E} and if (al, . . ., a,) is an n-termed sequence in L, we define (30.3)

L i=

~[a,,

ss.9

an]

as follows: For every uncountable cardinal

L, i=

ti

such that a,, ..., a , L,, ~

cp[a,, . . . a,].

Note that this gives us a truth definition for the constructible universe (see 14.17): For each sentence a, let #a be its Godel number and let T = {#a: L,, i= a} Then given any actual (metamathematical) sentence D of set theory, we have (30.4)

#GET

iff

Lka

(Note that the set T is constructible but not definable in L: Otherwise, T would be a truth definition in L. Hence the (real) cardinal K l is not definable in L.) As a consequence of (A) we have (30.5)

(L,,E)<(LE)

for every uncountable cardinal K. (We can unabashedly write (30.5) since we have defined satisfaction for L.)

Corollary 1. Every constructible set definable in L is countable. Proof. If x E L is definable in L by a formula cp, then by (30.5) the same formula defines x in L,, and hence x E LN,.

In particular, every ordinal number definable in L is countable. It follows from (B)and (30.5) that the Silver indiscernibles are indeed indiscernibles for L: If cp(v,, . . ., on) is a formula, then (30.6)

-=

whenever a , c * - .c u,, and p1< ... B, are increasing sequences in S . Moreover, every constructible set is definable from S in the following sense: If a E L,there exists an increasing sequence (y . . . , 7), of Silver indiscernibles and a formula cp such that

,,

L i= a is the unique x such that cp(x, yl,

. . ., y.)

Corollary 2. Every uncountable cardinal is inaccessible in L

,

Proof. (For each a, K, denotes the real K, ,not K;.) Since L k K is regular, we have

L i=K, is regular

30.

339

SILVER INDISCERNIBLES

for every

CY

2

1. Similarly, L k K,,, is a limit cardinal, and hence L k K, is a limit cardinal

for every CY 2 1. Thus every uncountable cardinal (and in fact every 7 E S) is an inaccessible cardinal in L. W Corollary 3. Ewry uncountable cardinal is a Mahlo cardinal in L Proqf. By Corollary 2, every Silver indiscernible is an inaccessible cardinal in L. Since S n w , is closed unbounded in w,, K , is a Mahlo cardinal in L. Corollary 4. For everj’ a 2 w, I V , n L I I I ci 1. In particular, the set of all constructible reals is countublr. Proqf: The set V , n L is definable in L from a. By (30.5),V, n L is also definable from a in where K is the least cardinal > a. Hence V, n L c L, for some /?such that ( a (= (/?I.However, 141= I p I . H

By (30.6), every formula (p(ol, ..., v,) is either true or false in L for any increasing sequence (y,, .. ., 7,) of Silver indiscernibles; moreover, the truth value coincides with the truth value of (30.7) L&”i= q“1, . . K,] since LU,*< L and K,,. .., K, are Silver indiscernibles. Thus let us define . 1

(30.8)

0‘ = {

~ :p Lu0,

+ ~ [ h ’ , ., . * ,

Kn]}

(zero-sharp).Later in this section we shall give another definition of the set O”, not depending on the conclusion of Theorem 72. We shall show that the (new) definition of 0’ is absolute for transitive models of set theory containing all ordinals and that a set 0’ satisfying the definition exists if and only if the conclusion of Theorem 72 holds (and then 0’ is as in 30.8). With this equivalence in mind, I shall henceforth use the statement

“0’ exists” to denote the conclusion of Theorem 72; i.e., (A) and (B). The axiom “0‘ exists admits a considerably simpler formulation. We shall prove : ”

Theorem 73 (Kunen). The following are equivalent:

(i) 0’ exists. (ii) For some limit ordinal 1, the model ( L A ,E) has an uncountable set of indiscernibles. (iii) There exists a nontrivial elementary embedding of L into L. It will also follow from the proof of Theorem 72 that the conclusion “0’ exists” follows not only from existence of a measurable cardinal but in fact from weaker assumptions, like “ there exists a Ramsey cardinal.”

5.

340

MEASURABLE CARDINALS

Review of L

Constructible sets were introduced in Section 12, and we proved in Sections 12 and 13 that constructibility and the corresponding well-ordering < are absolute for models of sufficiently many axioms of ZF. In particular, we have defined the function a w L , (in 12.2) and the canonical well-ordering of L (Section 13) and showed that both are absolute for all adequate transitive sets. We shall need a somewhat finer property of constructibility, namely that both the function a- L, and the well-ordering < are absolute for every L A , where 1is a limit ordinal. This does not necessarily follow from our definition of L, in (12.2) and so we have to amend slightly the defining formula as to obtain this absoluteness result. (However, the sets L, remain the same.) First we weaken the definition (13.9) of adequacy: Let us call a transitive set M adequate if (a) M is closed, (b) if U E M, then (5(U)E M where @ ( U ) = {Si(x, y ) : x, y E U , i

=

1,

..., lo}

and (c) if a E M, then (L, : /3 < a ) E M . Then we prove the analog of Lemma 13.2: We show that the function a- L, is absolute for every adequate transitive M. The Godel operations, as well as the operation (5 are absolute and defined in M. Neither cl(X) nor def(X) is necessarily defined in M, but since the function W(U,n) defined in (13.13) is absolute, we easily see that the formula x E def(U) is absolute: xEdef(U)

-

x

c I! and 3 n x W(U ~ u {U),n)

And hence def'(U) = def(U) for each U E M. Now, the standard definition of L, by induction (13.14) is absolute for M, because for each a the function f in (13.14) is unique,f= ( L p : /3 < a), and def(f(t)) = L 6 + 1E M for each 5 < a. The second part of Lemma 13.2 also goes through: If M is an adequate transitive set and satisfies the axiom of constructibility, then M = LAfor some limit ordinal 1. The definition (13.7)-( 13.8) of the canonical well-ordering < is easily seen to be absolute for adequate transitive sets. Next we show that each L, where A is a limit, is adequate. If x, y E La,then for each i = 1, . . ., 10, Six, y) c La+*and hence E L a + 3 .Thus LA is closed. If U E L,, then G(U ) G L,+ and is definable over La+ 3 , hence E L, + . Thus if U E LAthen G(U ) E LA.Finally, we prove by induction on 1that if a E LA,then ( 4 : < a ) E L A .If a < A is a limit ordinal, then La is adequate by the induction hypothesis and so (L, : /3 c a ) is definable in La; hence (L, : /3 < a ) E I&+, G La. And clearly, if (L, : /? c a ) E La then (L, : /? < a 1) E LA. Finally, as in Lemma 13.3, adequacy is a syntactical property of a transitive set: There is a sentence 0 such that a transitive set M is adequate if and only if M k 0. Thus we have:

.,

+

30.

34 1

SILVER INDISCERNIBLES

Lemma 30.1. (i) The function a HL,, and the canonical well ordering of L are absolute for euery La, where A is a limit ordinal. (ii) If M is transitive and ( M , E ) is elementarily equivalent to some La, rl a limit, then M = L,, for some a. In particular, i f M < (La, E) where I is a limit, then M is isomorphic t o some L,,, a 51. H

We shall now use the canonical well-ordering of L to endow the models LA with definable Skolem functions. For each formula ~ ( uu,l , ..., u,), let hq be the n-ary function defined as follows: (30.9)

h,,(uI, . . ., 0 , ) =

the <,-least u such that cp(u, u l , otherwise

. . ., u,)

We call h,,, cp E Form, the canonical Skolem functions. For each limit ordinal 1, @ A is an n-ary function on La, the &-interpretation of h,,, and is definable in (La, E). When dealing with models (La, E), we shall freely use terms and formulas involving the h,, since they as definable functions can be eliminated and the formulas can be replaced by €-formulas. For each limit ordinal A, the functions %*, cp E Form, are Skolem functions for (La, E) and so a set M E LA is an elementary submodel of ( 4 ,E ) if and only if M is closed under the % I . If X E La,then the closure of X under the hkAis the smallest elementary submodel M < La such that X E M ,and is the collection of all elements of Ladefinable in La from X. The fact that the well-ordering c a of La is definable in La uniformly for all limit ordinals I (by the same formula) now brings the following benefit:

Lemma 30.2. If a and are limit ordinals and i f j : L,, + Lo is an elementary embedding of (L,,, E ) in (Lo, E ) , then for eachformula cp and all x . . ., x, E La, (30.10)

bLW1), . ..,A%))= j ( % h , .. .

x,))

Hence j remains elementary with respect t o the enriched language L?* = {E} u {b: cp E Form}.

Proofi The mapping j is elementary with respect to formulas involving
Models with Indiscernibles The method of this section is based on a theorem of Ehrenfeucht and Mostowski in model theory, stating that every infinite model is elementarily equivalent to a model that has a set of indiscernibles of prescribed order-type. We shall deal only with models (La, E) (and models elementarily equivalent to these); Lemma 30.3 is a special case of the Ehrenfeucht-Mostowski theorem.

342

5.

MEASURABLE CARDINALS

Let I be a limit ordinal, and let PI = ( A , E ) be a model elementarily equivalent to ( L A ,E). The set O r 8 of all ordinal numbers of the model VI is linearly ordered by E ; let us use x < y rather than x E y for x, y E O r p . A set I s Ord" is a set ofindiscernibles for PI if for every formula cp, (30.11)

91 I= cp[x,, ..., x,]

iff

91 k cp[y,,

..., y,]

whenever x1 < ... < x, and y , < ... < y, are elements of 1. Let h," denote the PI-interpretation of the canonical Skolem functions (30.9). Given a set X G A, let us denote X ' ( X ) the closure of X under all h,", cp E Form. The set X'"(X)is the Skolem hull of X and is an elementary submodel of PI. If I is a set of indiscernibles for PI, let Z(%, I) be the set of all formulas cp(o,, . . . , u,) true in PI for increasing sequences of elements of I: (30.12) cp(ol,

..., 0,)

E

C(PI, I)-PI

k cp[x,,

..., x,]

for some x,, ..., x,

E 1

such that x, < . * . < x, A set of formulas C is called an E.M. set (Ehrenfeucht-Mostowski) if there exists a model PI elementarily equivalent to some L A ,I a limit ordinal, and an infinite set I of indiscernibles for PI such that C = C(PI, I ) . Lemma 30.3. If C is an E.M. set and a an injnite ordinal number, then there exists a model '41 and a set of indiscernibles I for PI such that:

(a) E = E(PI, I ) ; (b) the order-type of I is a ; (c) PI = P ( 1 ) . Moreover, the pair (PI, I ) is unique up to isomorphism. ProoJ We prove uniqueness first. Let (PI, I) and (8,J ) be two pairs, each satisfying (a), (b), (c). Since both I and J have order-type a, let A be the isomorphism between I and J. We shall extend a to an isomorphism between PI and 8. Since I! is the Skolem hull of I, there is for each a E PI a Skolem term ~(o,, ..., v,) (a combination of the Skolem functions h,) such that a = tVL [x,, , , . , x,] for some x < . * * < x, in I; similarly for '23, J . Thus we define

(30.13)

a(t"[x,,

..., x,])

= r'H[n(xJ,

..., n(x,)]

for each Skolem term t and all x,, ..., X,E I such that x 1 < C(PI. I ) = C ( 8 , J), we have

< x,. Since

(30.14) t'y[xl,

..., x,]

= t'g[yl$* * . , y,]

iff

t'f[ax,,

..., ax,]

= t;[ay,,

..., ny,]

30.

343

SILVER INDISCERNIBLES

,,

for any terms t , , t 2 and indiscernibles x , y: Let z . . ., z,+, be the enumeration of the set { x , , ..., x , , y , , ..., y;} in increasing order. Then the equality in (30.14) holds (simultaneously in 91 and 9)just in case cp(ol, ..., on+,) E C where cp(zl, ..., z,+,) is the formula that says that t , [ x , , ..., x d = t2[y1, ..., y,]. Hence IC is well-defined by (30.13) and is an isomorphism between VI and 9 extending the order-isomorphism of I and J . To prove existence of a model with indiscernibles with properties (a), (b), (c), we use the compactness theorem. Since E is an E.M. set, there exists (YI,, I,) such that C = C(PIo, I,). Let us extend the language {E} by adding a constant symbols ci, t < a. Let A be the following set of sentences:

(30.15)

ci is an ordinal

(all

c< < c,

(all

cp(ci,,

. . . , CJ

5 < a)

5. 9 such that 5 < q < a ) (all cp E Z and all 5 , < ... < 5.

< a)

We shall show that every finite subset of A has a model. Let D c A be finite. There exist t < . . . < ( k such that ql,. . .,cgkare the only constants mentioned in D. Let o(ccl,. . ., cJ be the sentence that is the conjunction of all sentences in D. Since I is infinite, there are i . . ., i, E I,, such that i < * .. < i k . Let us take the model V I , and expand it by interpreting the constant symbols ctl, . . ., cCtas i,, . . ., i k . Since X = C(Vlo, I,) and D c A, it is clear that YI, 1 o[i,,. . ., i J and hence the expansion (VI,, i,, . . ., i k ) is a model of o, hence of D. By the compactness theorem, the set A has a model 91 = ( M , E, c?)~,,. Let I = (c," : ( < a). I is a set of ordinals of 9 x 1 1and has order-type a. It is clear that if cp(u,, ..., I>,,) is an E-formula and 5 , < ... < t,,, then (M, E) k cp[cr, . . . , cE] if and only if cp E C.Thus I is a set of indiscernibles for (M, E). Now we let A be the Skolem hull of I in ( M , E ) . Since 91 = (A, E) is an elementary submodel of ( M , E), it follows that I is a set of indiscernibles for 91, Z(%, I) = C, and that f " " ( I ) = . R ( M . E )=( IA). Hence (VI, I) satisfies (a), (b), (c). For each E.M. set I: and each ordinal a, let us call the (C, a)-model the unique pair (VI, I) given by Lemma 30.3. The uniqueness proof of Lemma 30.3 easily extends to give the following:

,

,,

Lemma 30.4. Lat C be an E.M. set, let a Ip, and let j : a + p be orderpreserijiny. Then ,j can be extended to an elementary embedding of the (X,a)-model into the (1, ,!+model.

Proof. Extend j as in (30.13).

We shall eventually show that a measurable cardinal implies existence of an E.M. set Z having a certain syntactical property (remarkabilitp) and such that a)-model is well-founded. Let us investigate well-roundedness first. every (1,

5.

344

MEASURABLE CARDINALS

Lemma 30.5. The following are equivalent, for any given E.M. set Z: (i) For every ordinal a, the (C, a)-model is well-founded. (ii) For some ordinal a L ol, the (C, a)-model is well-founded. (iii) For every ordinal a < ul, the (C, a)-model is well-founded. ProoJ: (i) + (ii) is trivial. (ii) + (iii): If (PI, I) is the (C,a)-model and if /? Ia, let J be the initial segment of the first /? elements of I; let 9 = M " ( J ) . Clearly, (23,J ) is the (Z, /?)-model. Since a submodel of a well-founded model is well-founded, it follows that if p 5 a and the (Z, a)-model is well-founded, then the (X,j?)-model is also well-founded, and thus (ii) implies (iii). (iii) + (i): Let us assume that there is a limit ordinal a such that the (X,a)-model is not well-founded; let (PI, I) be the model. There is an infinite sequence a, ,a I, a 2 , . . . in PI such that a E a , , a E a etc. Each a,,is definable from I; that is, for each n there is a Skolem term t,, such that a, = t:[x l , .. ., x,J for some xl, , . . , x k nE I. Therefore there is a countable subset I , of I such that a,, E .#'"(I,) for all n E o.The order-type of I , is a countable ordinal p and (.#"(Io), I,) is the (C, /?)-model.This model is clearly non-well-founded since it contains all the a,,. Hence for some countable /?,the (C,/?)-model is not well-founded.

I shall now define remarkability. To avoid unpleasant surprises in the middle of a proof, let me from now on consider only (C,a)-models where a is an infinite limit ordinal. Let us say that a (Z, a)-model (PI, I) is unbounded if the set I is unbounded in the ordinals of 91, that is, if for every x E Ord" there is y E I such that x < y. Lemma 30.6. The following are equivalent, for any given E.M. set Z: (i) For all a, rhe (C, a)-model i s unbounded. (ii) For some a, the (C,a)-model is unbounded. (iii) For every Skolem term t ( v I ,..., on) the set Z contains the formula ( p , ( v l , ..., u,,+ ,) staring (30.16) i f t ( u l , . . ., u,,) is an ordinal, then

t(o,,

. . ., v,,) < u,+

ProoJ: (i) + (ii) is trivial. (ii) + (iii): Let (PI, I) be a (Z, a)-model, where a is a limit ordinal, and assume that I is unbounded in Or#'. To prove (iii), it suffices to show that for any t , (30.16) is true in PI for some increasing sequence x 1 < ... < x , + in I . Let t be a Skolem term. Let us choose x < * * . < x,, E I and let y = t"[x 1, . . ., x,J. If y # Orrf', then (30.16) is vacuously true; if y E Or#', then there exists x,+ E I such that y < x,+ and we have PI t= t [ x , , ..., x n ] < x,,+ (iii)+ (i): Let (PI, I) be a (Z, a)-model, where a is a limit ordinal, and assume (iii). To prove that I is unbounded in Or&', let y E O r 6 . There is a

30.

345

SILVER INDISCERNIBLES

Skolem term rand x 1 < ... < x, E I such that y = t " [ x , , . . ., xn]. Now ifx,, any element of I greater than x,, (iii) implies that y < x,+ 1.

is

Thus we say that Z is unbounded if it contains the formulas (30.16) for all Skolem terms t. Let a be a limit ordinal, a > w, and let (PI, I )be the (C, a)-model. For each 5 < a, let i, denote the 5th element of I. We say that (PI, I) is remarkable i f it is unbounded and if every ordinal x of PI less than i,,, is in .f'({i, : n E o}).

Lemma 30.7. The following are equivalent for any given unbounded E.M. set C : (i) For all a > w, the (C, a)-model is remurkable. (ii) For some a > w, the (C, a)-model is remarkable. (iii) For every Skolem term t(x,, . . ., x,, y,, . . ., y,), the set I; contains the formula cp(x,, ..., x,, y , , . ... y,,, z , , ..., z,) stating (30.17) ift(xl, . . . , x,,

. . ., y,) is an ordinal and

J'~,

t(x1, . . . . X , , J ' ~ ?

...,yn)

<

then t(X1,

..., x r n , y 1 ,...,.Y,)=

t(x,,...,x,,z1,

...A).

Moreover, if(PI, I ) is a remarkable (X,a)-model and y < a is a limit ordinal, then eoery ordinal x of PI less than i, is in .#"({it : 5 < y}). Proof: (i) (ii) is trivial. (ii) -+ (iii): Let a > o be a limit ordinal and let (PI, I ) be a remarkable (Z, a)-model. To prove (iii), it suffices to show that for any t, (30.17) is true in PI for some increasing sequence x , < * . . < x, < y , < ... < y,, < z , < ... < z , in I . Let t be a Skolem term. We let x , < ... < x,< y 1 < ... < y n < z I < ... < z, E I be such that x,, ..., x, are the first m members of I and that y1 is the wth member of I, y1 = i,,,. Now if a = t"(xl, . . ., v,) is an ordinal of 91 and less than y,, we have, by remarkability of (PI, I),a E X " ( { i , : n < 0)).Hence there is k < Q, k 2 m, and a term s such that -+

(30.18)

PI l= t[XI, . . ., x,, y,, . . ., y,] = $ 0 , ..., ik]

In other words (30.18) says that PI satisfies a certain formula cp[io, ..., i,, y,, . . ., y,]. By indiscernibility, 91 also satisfies cp[io, . . ., i,, zl, . . ., z,], i.e., (30.19)

91 k

t[x,,

. . ., x,, z , , . . ., z,]

.

= s[io, . . , ik]

Therefore t Q [ x , , ..., x,, )I,,..., y,] = f [XI. .... Xmr z1, ..., 2.1. (iii) -+ (i) and "moreover": Let (91, I ) be a (X,a)-model, where a > w is a limit ordinal, and assume (iii). Let y 2 w be a limit ordinal and let x E Ord' be less than i;,, the yth element of I. We shall show that x E -R"({i, : < y } ) . Since 9 I = .#"'(I), there is a Skolem term t and x , < ... < x, < y l < < y, E I such that y, = i;, and x = ?'[XI, ..., x,, y,, . . ., y,]. Let us choose w,, . .., w, Q

5.

346

MEASURABLE C A R D I N A L S

and zl, . . . , z , in I such that XI

< ... < x, < w1 < ... < w, < y1 < ... < J'" < z1 < ... < z,

Now since x < y l r it follows from (30.17) that 91 I=

f[X1,

..., x,

y1,

... y,] (

= t [ x , , ..., x,,

Z,,

..., 2.1

Z1'

...) Z,]

However, by indiscernibility, this implies that 91 I= t[Xl. ...) x,,

and hence x

= f"[x1,

w1,

..., w,] = t[x,, ...) x,

. . ., x,, wl,.. ., w,]. Therefore x

E

H'*(i, : 5 < y).

Thus we say that C is remarkable if it is unbounded and contains the formulas (30.17) for all Skolem terms f . The important consequence of remarkability is the following: Let (91, I ) be a remarkable (C,a)-model and let y < a be a limit ordinal. Let J = {it;: < y} and let 6 = .Y"(J).Then (B, J ) is the (C, 7)-model, and the ordinals of B form an initial segment of the ordinals of 91. Another consequence of remarkability is that the indiscernibles form a closed unbounded subset of the ordinals. Let (91, I ) be the (X,a)-model. We say that the set I is closed in O r P if for every limit y < a, i , is the least upper bound (in the linearly ordered set O r P ) of the set {is : 5 < y}.

<

Lemma 30.8. If (91, I ) i s remarkable, then I is closed in Ord". Proof: Let y < a be a limit ordinal. If x is an ordinal of 91 less than i , , then by remarkability, x is in the (C,y)-model 6 = X ( { i <: 5 < y}). However, since C is unbounded, 6 is an unbounded (Z, 7)-model and hence x < i , for some 5 < y. Hence i, is the least upper bound of { i t : < 7).

Proof of Theorem 72 Let us call an E.M. set Z well-founded if every (Z, a)-model is well founded, and let us consider the statement (30.20) There exists a well-founded remarkable E.M. set. We shall prove Theorem 72 in two steps: First we shall show that both (A) and (B) are consequences of the assumption that there exists a well-founded remarkable E.M. set, and then we shall show that if there exists a measurable cardinal, then (30.20) holds. (Note that by Lemma 30.5 it suffices to find a well-founded remarkable model with uncountably many indiscernibles.) Thus let us assume that there is a well-founded remarkable E.M. set and let C be such set. For every limit ordinal a, the (Z, a)-model is a well-founded model elementarily equivalent to some L A , and so by Lemma 30.1 is (isomorphic to) some 4 .

30.

347

SILVER INDISCERNIBLES

Lemma 30.9. I f K i s an uncountable cardinal, then the unioerse of the (C,K)-rnoilel is L,,..

Proqf. The ( C , k-)-model is (L,,, I ) for some p; since I 11 = K , we clearly have fi 2 K . To prove that p = K , assume that p > K . Since I is unbounded in p and

has order-type K , there is a limit ordinal 1' < ti such that K < i y . By remarkability, all ordinals less than i , are in the (C,7)-model PI = X({ir: 5 < y}). This is a contradiction since on the one hand we have ti G PI, and on the other hand 1911 = I y J


For each uncountable cardinal K , let I , be the unique subset of K such that

(L,,, I , ) is the (1,K)-model. By Lemma 30.8, I , is closed and unbounded in Lemma 30.10. If

.W(I,)

=

r,.

K

< I are Uncountable cardinals, then I , n K

=

K.

I , , and

Proof. Let J be the set consisting of the first K members of I , and let P I = .W"(J). Then (PI, J ) is a (C,k-)-modeland the ordinals of PI are an initial segment of A, say Ord" = p. Since (PI, J ) is isomorphic to (L,, I,), it is clear that p = K and J = I , . Hence I , n K = I , . Now since PI < La, PI is closed under the definable function F(a) = the ath set in the well-ordering < L , and since Ordu = K , we have A = F [ K ] = L,.

Using this lemma, we can now prove both (A) and (B) of Theorem 72, except for the uniqueness of S. We let (30.21)

S=

u {I,

: K an uncountable cardinal}

For each uncountable cardinal K , S n K = I , is a closed and unbounded set of order type K , and is a set of indiscernibles for ( L , , E); moreover, by Lemma 30.3~.every a E r, is definable in L, from S n K . Let K c I be uncountable cardinals. Since I , is closed in La and I , n K = I , , it follows that K E I , ; hence S contains all uncountable cardinals. Also, since L, = .#'OLA(IK), we have r,
Proof. Assuming that there is one such C,we define the class S as in (30.21). Now since LN,ois the (C,K,)-model and K, E S for each n 2 1, we have

(30.22)

cp(v,,

. . ., u,)

E

C

iff

LNfocp[KI, ..., K,]

which proves that C is unique. We therefore define 0" (zero-sharp): (30.23) 0' is the unique well-founded remarkable E.M. set if it exists.

348

5.

MEASURABLE CARDINALS

Uniqueness of S now follows from:

Lemma 30.12. For every regular uncountable cardinal K there is at most one closed unbounded set of indiscernibles I for L, such that L, = X'LK(f). Proof. Let X = Z(L,, I). Since I is closed unbounded, it follows that X is remarkable; since K is uncountable, Z is well-founded (by Lemma 30.5). Hence Z = O* and since (L,,I ) is the (X,K)-model, we have I = S n K . Thus we have proved (A) and (B) of Theorem 72 under the assumption that 0' exists (i.e., (30.20)).O n the other hand, if (B) holds, then O x exists because, e.g., (L,,, S n ol) is a remarkable well-founded model with K , indiscernibles. To complete the proof of Theorem 72, it remains to show that if there is a measurable cardinal, then O* exists. That will follow from:

Lemma 30.13. Lat K be an uncountable cardinal. If there exists a limit ordinal I such that ( L , , E ) has a set of indiscernibles of order-type K , then there exists y and I G y of order type K such that (4,I ) is remarkable. Then if ti is a measurable cardinal, K is Ramsey; and by the corollary of Lemma 29.9 (L,,E) has a set of indiscernibles of order-type K. By Lemma 30.13, there exists a remarkable model ( 4 ,I ) where I has order-type K. By Lemma 30.5, X ( 4 , I) is well-founded and remarkable and hence 0' exists. Note that Lemma 30.13 also gives a proof of ( i ) H (ii) in Theorem 73. Proof. Let I be the least limit ordinal such that (La, E) has a set of indiscernibles of order-type ti. We shall show first that there is a set of indiscernibles I G I for La, of order-type K, such that . @ ^ ( I ) = L,. Let J be any set of indiscernibles for La of order type K, and let 91 = M L A ( J )Then . 91 I),,] is a set of indiscernibles for (b, E). Ifcp(c,, ..., O k ) is a formula, then for any i l < ... < i, E J , L, satisfies ' p [ i , , . . ., ik] if and only if r, satisfies the formula (30.24) L,k c p [ i , , ..., i k ] The formula (30.24)is a formula about a, i , , . . ., i,, and since a = tLA"y,, , . ., yn], there is a formula $ ( u l , ..., u,,. ~ 1 ..., ~ . i l k ) such that L, satisfies (30.24) if and only if (30.25) L, k $[rl, .. ., y n , i , , .. . , ik] By indiscernibility of I , the truth of (30.25) is independent of the choice of i, < ... < i k in I provided y,, ii , . Hence the truth of (30.24) is independent of the choice of i , < < ik in J . Hence J is a set of indiscernibles for L,, and this contradicts the minimality of I since a < I and the order type of J is ti.

349

30. SILVER INDISCERNIBLES

Finally, let I be a set of indiscernibles for I of order-type K such that the wth element of I is least possible. I shall show that ( L A ,I) is remarkable. , is not remarkable. Then there is a Skolem term Let us assume that ( L A I) t ( x l , ..., x,, y,, ..., y n ) such that the following holds in L, for any x1 < ... < x, < y , < ... < yn < z1 < ... < z,: X‘’((I)= L A ,and that i,

(30.26)

and (30.27)

...) Xnr

[(XI,

..., x,, y1,

[(XI,

yn)

~ 1 * ., * )

yn)
# t ( x 1 , ..., x,,

zlr

***,

zn)

We now consider the following increasing n-termed sequences in I: Let uo be the sequence of first n indiscernibles after x,, let u , be the first n indiscernibles after uo ,etc.; for each a < K, let u, be the first n indiscernibles after all u s , p < a. For each a < K, let y. == t ( x 1 , x m , urn) By indiscernibility, applied to the formula (30.27), we have y. # ys whenever # p. In fact, in (30.27) we have either < or > (in place of # ); but > is impossible since that would mean that y,, > ys whenever a < j. Thus (7. : a < K ) is an increasing sequence of ordinals. I claim that J = {yrn: a < K} is a set of indiscernibles for L,. This is so because for any formula cp, the truth value of cp(y,,, . . ., y a k ) in L, does not depend on the choice of y z , < ... < y., in J because by the definition of the urn, the truth value of

a

cp(r(x1,

..., x,,

u,,),

..., t ( x , , ..., X,,

Urnk))

does not depend on the choice of a1 < . . * < a,, . Hence {y. : a < K } is a set of indiscernibles for L , . Since i,, is the first member of u,, , it follows by (30.26) that y,, < i,, . Now if A = H ( J ) and n is the transitive collapse of A, then, as we proved in the first paragraph, n[A] = L, and K = n[J] is a set of indiscernibles for L, of order-type K such that . f L i ( K ) = LI . However, n(yJ I y,,> < i,, , and so the wth member of K is smaller than i,,, contrary to our assumption. Hence (L,, l) is remarkable. This completes the proof of Theorem 72. Absoluteness of 0‘

The set 0’ is not only unique if it exists, but is the unique well-founded remarkable E.M. set in every transitive model of ZF that contains it. And conversely, if 91 is a transitive model of ZF containing all ordinals and X E 911 is such that 911 k (Cis a well-founded remarkable E.M. set), then X = O x . This follows from the fact that C = 0’ can be expressed as a ll property and from the Levy-Shoenfield absoluteness lemma.

,

350

5.

MEASURABLE CARDINALS

Lemma 30.14. There is a ll, formula @(C) such that @(C) holds i f and only if C is a well-founded remarkable E.M. set. Corollary. The property ’‘ C is a well-founded remarkable E.M. set ” is absolute for euery transitive model u1 of Z F containing all ordinals. Hence YJI k (0* exists) ifand only if0” E “,in which case (O‘)IIR = 0”. The corollary is a consequence of Theorem 36’ (the parametric version of the Levy-Shoenfield absoluteness lemma). Although the theorem is stated for Z E: w, it is applicable here since a set of formulas Z can be replaced by a subset of o,the set of all Godel numbers of cp E C.

Proof: A set of formulas C is a well-founded remarkable E.M. set if: (30.28) (a) C is an E.M. set; (b) X is remarkable; (c) for every a, the (C,a)-model is well-founded. Since (a) does not readily admit a nice formulation, we shall first replace it by another condition. Let 2 be the language {E, cI, c 2 , . . ., c,, . . .} where c,. n < w, are constant symbols. For every formula cp(ul, . . ., u,) let @ be the sentence cp(cI,. . ., c,) of 2.For each set of formulas Z, let 2 be the set containing (i) all @ for cp E C,(ii) the sentence “c, is an ordinal and c , < c2,” and (iii) the sentence “cp(ci,, ..., ci,)++q(cj,, ..., cj,)” for every cp E C and any i , < ... < i n , j , < ... < j,, (iv) all axioms of ZFC + I/ = L. Let (30.28) (a’) 2 is consistent. Clearly if Z is an E.M.set, then 2 is consistent for we simply interpret the constants c,, n < o,as some Silver indiscernibles. Conversely, if 2 is consistent, then 2 has a model and that model provides us with a (C,a)-model (with indiscernibles c, , n < w ) and the proof of Lemma 30.3 goes through. Therefore if C satisfies (a’), (b), and (c), we proceed as in the proof of Theorem 72 and prove that C = 0’. Thus C is a well-founded remarkable E.M. set if and only if it satisfies (30.28)(af),(b), (c). Now (a’) and (b) are syntactical conditions, and so the conjunction of (a‘)and (b) is a Ao-property over HF (= Vw),the collection of all hereditarily finite sets. We shall now show that (c) can be written as a ll, property of C. If C satisfies (a’) and (b), then for every limit ordinal a there is a unique (up to isomorphism) (C,a)-model and we can find one ((A, E), I) such that I = a and that <“ (i.e., E) agrees with < on a. If ((A, E), a) is such, we say that ((A, E), a) is a (Z, a)-model.’’This last property is a A. property of C,(A, 15) and a (with parameter F,,). Then Z satisfies (c) if and only i f “

(30.28) (c’)Va V ( A , E) (if ( ( A , E), a) is a (X,a)-model, then (A, E ) is well-founded).

30.

351

SILVER INDISCERNIBLES

Now ‘ ‘ ( A , E ) is well-founded” is A 1 (see Section 14) and hence the conjunction of (a’), (b), and (c’) is ll,. 30.f.The property “ Z is a well-founded remarkable E.M. set ” is ll over HC. [Note that “ V a the (Z, a)-model is well-founded” can be replaced by Va E HC the (Z, a)-model is well-founded.] Exercise

Exercise 302. Let 91 be a model of ZFC, let B E 91 be a complete Boolean algebra in %If and let G be an 91-generic ultrafilter on B. If 311 k 0 ” does not exist, then W[G] k O x does not exist. [All cardinals 2 1 B + 1 remain cardinals in 9 I [ G ] . Let y I < y2 < ... < y. < ... < yo be an increasing sequence of cardinals in 91 such that y1 2 1 B+ I. If 0” exists in 9Jl[G], then O x = (cp : Lyv, k ‘ p [ y , , .. ., y.]} and hence 0” E 91.1

Elementary Embeddings of L In Section 28 we proved that a well-founded ultrapower of the universe induces an elementary embedding j,: I/ + U l f ,and conversely, ifj: I/ + M is a nontrivial elementary embedding, then (28.21)defines a normal measure on the least ordinal moved by j . Let j be a nontrivial elementary embedding of the universe, and let M be a transitive model of ZFC, containing all ordinals. Let N = j ( M ) = UaEOrdj(M n V,). Then N is a transitive model of ZF and j : M + N is elementary: (30.29)

M k ~ ( a 1 ...) , an)

iff

N 1 q(j(a,), ...,J(an))

((30.29)is proved by induction on the complexity of cp). In particular, if M = L then j ( V ) b ( N is the constructible universe), and so N = L and j l L is an elementary embedding of L in L. Note that by Scott’s theorem, the functionj I L is not a class in L; thus if there is an elementary embedding of L (into L), then v # L. If 0’ exists, then there are nontrivial elementary embeddings of L. In fact, letj be any order-preserving function from the class S of all Silver indiscernibles into S. Then j can be extended to an elementary embedding of L; we simply let (30.30)

for every Skolem term t and any Silver indiscernibles y < . . . < y n . We shall prove that the converse is true, that if there is a nontrivial elementary embedding of L then 0’ exists. But let us first look into the duality of elementary embeddings and ultrapowers, in the case when the model being embedded is not the whole universe. Let M be a transitive model of ZFC containing all ordinals, and let j : M + N be an elementary embedding; let y be the least ordinal moved byj. Let D be the collection of subsets X of y. X E M , defined as follows: (30.31)

X E D

iff

y E~(x)

352

5.

MEASURABLE CARDINALS

We have D E PM(y)but D is not necessarily in M. D clearly has the attributes of an ultrafilter: (30.32)

y E D and 125 # D ; if X E D and Y E D,then X n Y E D ; if X E D and Y E M is such that X G Y, then Y E D ; for every X G y such that X E M, either X or y - X is in D.

(i) (ii) (iii) (iv)

We shall call every D satisfying (30.32) an M-ultrafilter. The M-ultrafilter D defined in (30.31) is also M-y-complete and M-normal: If a < y and {X,: 5 < a} E M is such that X, E D for all { < a, then X , E D ;and if f~ M is a regressive function on X E D,thenfis constant on some Y E D.The proofs are the same as in Lemma 28.5. If D is an M-ultrafilter over y, one can construct the ultrapower of M mod D as follows: Consider, in M (!), the class of all functionsfwith domain y (that is {fE M : dom(f) = y}). Using D,define an equivalence relation = * and E* as usual;

n,<.

f=*g

(30.33)

{a < y : f ( a )= g(a)}E D

iff

(note that the set on the right-hand side of (30.33) is in M, and D decides whetherf= * g orf # * 9). Then define the equivalence classes and the model Ult

=

UltD(M)

The fundamental theorem of ultraproducts is easily verified: Ult b

v([fll,..., [f.])

If for each a

E

iff

{a < Y: M b v ( f l ( a ) ,..., L ( a ) ) }E D

M, ca denotes the constant function with value a, then

defines an elementary embedding of M in Ult (the natural embedding). The ultrapower of M modulo an M-ultrafilter D is not necessarily wellfounded, even if D is M-countably-complete. However, if D is obtained from an elementary embedding by (30.3l), then UltD(M) is well-founded because we have the commutative diagram (as in Lemma 28.6):

FIGURE 30.1

30.

353

SILVER INDISCERNIBLES

(If [f,]3* [fl] 3* ... were a descending sequence in Ult, then k[f,,] k [ f l ] 3 ... would be a descending sequence in N.)

3

Proof of Theorem 73 If 0“ exists, then there are nontrivial elementary embeddings of L; each order-preserving j : S -+ S generates one. It was proved in Lemma 30.13 that if some LA has uncountably many indiscernibles, then there exists a remarkable (&, I) with uncountably many indiscernibles, and by Lemma 30.5, 0” exists (and 0” = C(L.,,I)). We shall now show that if there exists an elementary embedding of L, then there is LAwith K1 indiscernibles and hence 0” exists. Let j: L + L be an elementary embedding. We shall first replace j by a more manageable embedding. We let D be the Lultrafilter defined by X E D

iff

y ~ j ( X )

where y is the least ordinal moved by j . The ultrapower Ult ,(L) is well-founded (see Fig. 30.1) and so we identify Ulr with its transitive collapse L; let j, be the associated natural embedding, j,: L-+ L. The least ordinal moved by j, is y (because D is Ly-complete).

Lemma 30.15. I ~ isK Q limir cardinal such that cf

K

> y, then j D ( K ) = K .

Proof Every constructible function f: y -+ K is bounded by some a < K and hence [f]< [cm](where c, is the constant function with value a). Thus jdK) = j,(a). Now if a < K, then Ij,(a) I I I (‘a)” 1, hence jdct) < K. It follows thatj,(K) = K. Let me drop the subscript D and simply assume that j: L -+ L is an elementary embedding, that y is the least ordinal moved, and that j ( K ) = K for every limit cardinal K such that cf K > y. Let U, be the class of all limit cardinals K with cf K > y; by transfinite induction we define a sequence of classes Uo 3 U 1 3 . . * 3 U, ... as follows: (30.34)

U z + l = { ~ ~ U) u , : , n ~= J K} U, = (? U,

(2 a limit)

,< A

(That is, consists of fixed points of the increasing enumeration of U z . ) Each U, is nonempty, and in fact a proper class. To see this, verify, by induction on a, that each U, is a proper class and is 6-closed, for each 6 with cf 6 > y ; that is, whenever ( K :~ 5 < 6) is an increasing sequence in U,, then lims+a K; E U,. Hence each U, is nonempty, and we choose a cardinal K E U,,,,. Thus K is such that cf K > y and K is the Kth element ofeach U,, a < ol. We E). shall find a set of K 1 indiscernibles for (k, Since j : L + L is elementary and j(K) = K, it is clear that the mapping i = j 1 L, is an elementary embedding of (LK, E) into ( L K ,E). We shall use i

354

5.

MEASURABLE CARDINALS

and the sets U, n K, a < w l , to produce indiscernibles y,, u < o l ,for L,. Let X, = U, n K for each u < ol, and recall that y is the least ordinal moved by i . For each a < w l , we let M, = .#"*'(y u X,) (30.35) M, is an elementary submodel of L,. If n, is the transitive collapse of M,, then because IX,I = K, we have nu[M,] = L, . Thus if we denote i , = n; then i, is an elementary embedding of L, in L,.Let (30.36)

Yo

= &(Y)

Lemma 30.16. (a) The ordinal y, is the least ordinal > y in M , . (b) l f a < p and x E M , , then i,(x) = x . In particular, i,(y,) (c) l f a < It then y, < 7,.

= y,.

Proof. (a) Since y c M,, i,(y) is the least ordinal in M,greater than or equal to y ; thus it suffices to show that y $ M, . If x E M,, then x = t[q . . ., q,,l where t is a Skolem term and the q's are either < y or elements of X,. For all such q, i ( q ) = q and hence i ( x ) = i ( t ( q l , . . . , q,,)) = t ( i ( q l ) ,. . . , i(q,,))= x. However, i ( Y ) # 7 and so Y $ M , * (b) Each x E M , is of the form t[q 1 , . . ., q,,l where the q's are either < y or in X,. If q < y, then clearly i,(q) = q. If q E X,, then because a < /3, we have IX, n q I = q and hence n,(q) = q ; in other words, i,(q) = q. Therefore i,(x) = x . (c) If u < j,then M a 2 M, and hence yu I y s . To see that y, # y,, note that because y , > y, we have i,(ya) > i , ( y ) = y , , while io(ya) = y, .

Lemma 30.17. If u < p, then there is an elementary embedding i,: L, -+ L, such that for every t; that is either < a or > we have i u a ( y e )= y c , and i,,(y,) = y,. Proof. Let M u , = S L K ( y au X,),and let i,, = nibS'where napis the transitive collapse of M,, . The mapping i,, is an elementary embedding of L. in L, . If q < y,, then clearly iUp(q)= q ; in particular ia,(ys) = ye if < a. If x E M , , 1, then x = t(ql, . . ., q,,) where the q's are either y or E X,, If q E X,, 1 , then 1 X, n q 1 = q and therefore ims(q)= q. Hence iua(x)= x for every x E M , , I , and in particular i+(ye) = yc if 5 > fl. Now we shall show that i,,(y,) = y , . Since M u , 2 M,, we have y , E Mu,; and since y . c Ma,, iu,(ya) is the least ordinal in M a , greater than or equal to y.; hence we have y . I iaa(ya) I y , . Thus it suffices to show that there is no ordinal 6 E Mu, such that y, I 6 < y s . Otherwise there is some 6 = t(< . . ., <",ql, . . ., q k ) such that the 5's are < y . and the q's are in X, (and t is a Skolem term) and that y o I6 < y , . Thus we have:

-=

,,

(30.37)

(L, E) b Y < y .

such that ya I t(t,q) < y s .

<

30.

355

SILVER INDISCERNIBLES

The formula in (30.37) is a formula cp about jt1, q, and y,,. At this point, we apply the elementary embedding i,: L, + L, backward. That is, yZ. the q’s and y, are all in the range of i,: y, = i3(y),q = i,(q), and y, = i,(y,); and since L, 1 cp[i,(j!), il(q)?iz(y,,)],we conclude that L, b cp[y. q, y,] namely

(30.38) (L,,E ) k 35 < y such that 1’ I t(5, q ) < yo. Thus pick some 5’s < 7 such that j’ I t ( < , q ) < y,, . Since 5 E y and q E X,,we have r ( ( , q ) E M,,, which means that t(<, q ) is an ordinal in M , between y and y,, and that contradicts Lemma 30.16a. The proof of Theorem 73 will be complete when we show:

Lemma 30.18. The set (7, : a < (uI j i s a set of indiscernibles for (L,, Pro@ Let cp be a formula and let a1 < ... < a, and quences of countable ordinals. We wish to show that

E).

< ... < Pn be two se-

(30.39) L, b c p [ Y Z , , ...*r3,I iff L, 1 d Y , P . * . .Y/?J Let us pick 6, < ... < 6, such that a, < 6, and p, < 6,. First we apply the elementary embedding iln.d, and get with = yarn,and preserves the other 7’s. Then we apply iZnbecause izn.am(ymn) we get a similar effect, and by a successive application of i, n - 2 , a n - 2 , . . .,

L, b cp[Y,,? . .

. 1

Y3J

iff

L, b cp[Ya,, * .

.?

YdJ

Then we do the same for the p’s and 6’s as we did for the a’s and 6’s, and (30.39) follows. This completes the proof of Theorem 7 3 . Exercise 30.3. Let ti be an uncountable regular cardinal. I f 0 ” exists, then for every constructible set X & h-, either X or ti - X contains a closed unbounded subset. [Let X = ((al, . . . , 2,. PI, ..., pm) where a l < . . . < a, < < ... < pm are Silver Show that either X or K - X contains all Silver indiscernibles such that a, < ti 5 /I1. indiscernibles 1’ such that a, < y < K : The truth value of y E r(a,p) is the same for all such y.] Exercise 30.4. Let us assume that for some uncountable regular cardinal K, every constructible X G K either contains or is disjoint from a closed unbounded set. Then O* exists. [Let D be the collection of all constructible subsets of K containing a closed unbounded subset. D is an Lultrafilter and every intersection of less than K elements of D is nonempty; hence the ultrapower UIr,(L) is well-founded and gives an elementary embedding of L in L.] Exercise 30.5. Assume that 0” exists. I f A G o 1is such that A n u E Lfor every a < w l r then A E L [For every a E S n w 1 there is r. such that A n a = rm($, . .., yico,, u,& ..., a;(=)).

356

5.

MEASURABLE CARDINALS

Clearly, A n LY = tm($, . ... a, N2... ., Nk,,,+I ) . Since there are only countably many Skolern terms, and by the Fodor theorem, there is a stationary subset X of S n w , and r, y I , ..., y,, such that for all a E X , A n 01 = t(yl, .... y,,. a, K 2 , .... Nk+I ) . Show that A = r ( r l > f . . , y n ? N,,N2.

...?

Nk+l).]

The Sharps

All results about 0’ and Silver indiscernibles for L proved in the present section can be relativized to obtain similar results for the models a x ] , where X G

K“.

In particular, under the assumption of existence of measurable (or just Ramsey) cardinals, there is for every x E a unique class S, containing all uncountable cardinals such that for each uncountable cardinal K, S, n K is a set of indiscernibles for the model (L,[x], E, x) and all elements of LK[x]are definable in the model from S, n K. Here x is considered a one-place predicate. Also, for each regular uncountable cardinal K, S, n K is closed unbounded in K. The proof of the relativization of Theorem 72 uses models with indiscernibles (PI, I) where 91 is elementary equivalent to some (L,[x], E, x) where L > Q is a limit ordinal (note that x E L,[x]). If K is a Ramsey cardinal, then (&[XI, E, x) has a set of indiscernibles of size K, and the theorem follows. We define x’ as the unique set Z = C((L,[x], E, x), I) that is well-founded and remarkable. If x ” exists, then we have x‘ = {‘p : (L&

€9

x)

‘p[K* .

., K,]}

Here ‘p is a formula of the language {E, P} where P is a one-place predicate symbol (interpreted as x). Note that x is definable in the model (Ldx], E, x ) (by the formula P(o)). The property (of Z) ‘‘Eis x’ is Il,with a parameter x. Thus x’ is absolute for all transitive models or ZF containing all ordinals such that x’ E 9A. Also, “x’ exists” is equivalent to the existence of a nontrivial elementary embedding j : 4x1 + a x ] . ”

Jensen’s Covering Theorem

I shall state without proof an important theorem which decides many problems about singular cardinals if 0” does not exist. In particular, the theorem implies that unless 0” exists, the singular cardinals hypothesis is true and hence all cardinal exponentiation is determined by the continuum function on regular cardinals (see Section 8). It follows that in order to establish independence of the singular cardinals hypothesis (or of related problems), one needs large cardinals, and that such independence results cannot be accomplished by generic extensions of the constructible universe. (By more recent results, in order to violate the singular cardinal hypothesis, one needs to assume existence of measurable cardinals.)

30.

357

SILVER INDISCERNIBLES

The proof of Theorem 74 is rather long and uses recursion theoretic methods (the fine structure of 15). Theorem 74 (Jensen's Covering Theorem). If 0' does not exist, then for every uncountable set X of ordinals there exists a constructible set Y 2 X such that

P I = 1x1. The conclusion of Theorem 74 is in fact equivalent to nonexistence ofOx : If exists, it is easy to find a counterexample. For instance, let X = {K,: u < w , } ; since K,, is a regular cardinal in I., every constructible Y 2 X has size at least K,, while I X 1 = H I . The assumption that X is uncountable cannot be dropped :

Ox

I

Exercise 30.6. I t is consistent that O x does not exist and cf(w4) = w while I w i = N, (then there is a countable X c w4 that cannot be covered by a countable constructible

set). [Example I 1 in Section 26 gives a generic extension of L with these properties; O* does not exist in a generic extension of L (Exercise 30.2).]

However, note that the theorem implies that a countable X can be covered by constructible Y of size Nl. I shall prove some consequences of the covering theorem. The most important is Corollary 1 which states that unless 0' exists, the singular cardinals hypothesis is true. It follows that if O x does not exist, then cardinal exponentiation is determined by the continuum function on regular cardinals, and K' always has the least possible value (see Lemma 8.1); in particular, 2" = K + for all strong limit singular cardinals.

Corollary. Assume that 0' does not exist. Then:

if^ is a strong limit singular cardinal, then 2" = .'K ( 2 ) If K is a singular cardinal and if there exists a nonconstructible subset of K , then some u < K has a nonconstructible subset. ( 3 ) If 1 is a regular cardinal in I., then either I I 5 K 1or cf I = I I I n particular, I is not a singular cardinal. (4) If K is a singular cardinal, then ( K + ) = ~ K+. (1) For every singular cardinal K , if2cr " < K , then K"" = K + . In particular,

I

I.

Before giving a proof of the corollary, let me make a few comments. Result (2) suggests that in the absence of O', singular cardinals behave like large cardinals: We shall show in Section 32 that every weakly compact cardinal has this property. Results (3) and (4) shed light on the difficulties with collapsing cardinals and changing cofinalities by forcing: The standard method of collapsing K + onto K does not work if K is singular; and the only method we have dealt with so far of changing the cofinality of a regular cardinal I such that cf I < 11 1 in the generic extension is Example I1 in Section 26, where cf I = w and 111 = K,.

5.

358

Proofofthe Corollary. (1) Let

K

MEASURABLE CARDINALS

be such that 2c'K< K, and let A

=[K]"~

be the set of all subsets of K of size cf K. We shall show that I A I I K + . By the covering theorem, for every X E A there exists a constructible Y G K such that X C Yand IY1 = I w h e r e A = S , . c f ~ . T h u s (30.40)

As

u {[YlCfK Y :

E

C}

where C = { Y S K : IYI = L a n d Y E L }

it follows from (30.40) that I A 1 5 K + . (2) Let ti be a singular cardinal and assume that each a < K has only constructible subsets; we shall show that every subset of K is constructible. I t suffices to show that each subset of K of size cf K is constructible: If A c K, let {av : v < cf K } be such that lim,, a,, = ti; then .d = { A n a, : v < cf K } is a subset of L , of size I cf K and hence constructible. I t follows that A is constructible. Let X G K be such that X I I cf(K). By the covering theorem, there exists a constructible set of ordinals Y z X such that I Y I < K. Let n be the isomorphism between Y and its order-type a ; the function n is constructible and one-to-one. Since 101 I = 1 Y I < K, we have a < ti. Let Z = n [ X ] .Then Z C a is constructible by the assumption, and hence X = n- '[Z] is also constructible. (3) Let 1 be a limit ordinal such that A 2 w 2 and that 2 is a regular cardinal in L. Let X be an unbounded subset of I such that I X 1 = cf I . By the covering theorem, there exists a constructible set Y such that X c Y G I and that I Y I = I X I . N,.Since Y is unbounded in I and 1 is a regular cardinal in L we have 1 Y I = [ I 1. This gives us IA I = X I . cf A and since we assume that 111 2 N 2 , we have [A1 = c f L (4) Let K be a singular cardinal and let I be the successor cardinal of K in L; we want to show that A = ti+. If not, then IA I = K ; and since K is singular, we have cf I < K. However, this means that cf I < IAI which contradicts (3).

I

I shall conclude with the following application of Jensen's covering theorem. In Section 21, Example IV, we have seen a model of ZF without the axiom of choice in which N I is singular. A still unsolved problem is to construct a model in which all uncountable alephs are singular. The following exercise

31. THE MODEL L[U]

359

shows that some large cardinal assumption is necessary for such consistency result: Exercise 30.7. If both w , and w2 are singular, then 0’ exists. [Let K = (oI and let 1 be the successor cardinal of K in L. Since cf K = cf 1 = o,there are sets X G K and Y G 1, both of order type w such that sup X = K and sup Y = 1. Let M = YX, Y ] ; M is a model of ZFC and in M , K is a singular cardinal, and 1 is not a cardinal. By Corollary 4, O x exists in M . ] 31. THE MODEL t [ V ]

The results of preceding sections show that measurable cardinals are inconsistent with the axiom of constructibility, and in fact existence of measurable cardinals implies that very few sets are constructible. This eliminates the constructible universe L with its nice structural properties as a tool for consistency results; thus if we want to show consistency of, say, the continuum hypothesis relative to existence of measurable cardinals, we have to look for models with nonconstructible sets. I t turns out that the theory ZFC + “there is a measurable cardinal” has a canonical model that shares with L many of its useful properties. If K is a measurable cardinal, let U be any ti-complete nonprincipal ultrafilter over K and let us consider the model a U ] of all sets constructible from U . We shall prove that the model q U ] is unique; it is the same model regardless of our choice of U . Moreover, in q U ] K is the only measurable cardinal and K has only one normal measure; and L [ U ] is the smallest model in which ti is measurable. Every set in Y U ] is the least model in which K is measurable. Every set in q U ] is constructible from the unique normal measure and hence all sets in a U ] are ordinal definable and q U ] has a definable well-ordering of the universe. We shall also prove that if U‘ is a K’-completenonprincipal ultrafilter over some K’ # K, then the models q U ] and U U ’ ] are elementarily equivalent. Besides uniqueness of the model q U 3 , its most important property is that it satisfies the generalized continuum hypothesis; in that respect, q U ] behaves like L. In fact, the model Y U ] has a “fine structure” analogous to the “fine structure” of L,and so Y U ] satisfies many combinatorial principles that hold in L. We shall first investigate the model YO] where D is some normal measure. In Theorem 75 we prove that Y D ]satisfies GCH. The proof is not very difficult and is very much like the proof of GCH in L. They key step is the use of Rowbottom’s theorem (Lemma 29.10). The uniqueness theorem for Y V ] and further results on L [ U ] will require a new technique, the method of iterated ultrapowers. Let me first recall some facts about relative constructibility (Section 15). If A is a set, the hierarchy L J A ] is defined in analogy with the L;hierarchy, except that L+,[A1 = def,(k[Al)

360

5.

MEASURABLE CARDINALS

where def,(M) is the Giidel closure of the set M u { M } u { A n M}. Equivalently, def,(M) is the collection of all X E M definable over the model (M, E, A n M ) (where A n M is taken as a one-place predicate). If we denote A = A n LEA], then a A ] = 44, and the model Y A ] satisfies " V = 44." By Lemma 15.6, there is a sentence @* such that if M is a transitive model of O* and A is a set in M, then the function a H L , J A ] is absolute for M . The sentence O* holds in L,[A] for every infinite cardinal I and any A. It follows that if M is an elementary submodel of (LJA], E) and A E M, then M is isomorphic to some LJa(A)] where a is the transitive collapse of M. Also, if M is a transitive model of @* and if A is such that A n M E M , then for every a E M , (31.1)

u 4 1 = &[A n MI = (&,[A n MIIM

This is proved in Lemma 15.4. In Section 15 we dealt mostly with LEA] where A is a set of ordinals; among others, we have shown (Lemma 15.7) that if V = q A ] and if I is an infinite cardinal such that A C_ I, then 2 A= I f . In the present situation we want to investigate Y U ] where U is an ultrafilter, thus a set of sets of ordinals. The following lemma is a straightforward generalization of Lemma 15.7: Lemma 31.1. If V = q A ] , and if A C P(w,), then 2Nn= K,, Proof: Let X be a subset of 0,. Let I be a cardinal such that A E L,[A] and X E L,[A]. Let M be an elementary submodel of (L,[A], E) such that w,c M , A E M , X E M , and IM I = K,. Let x be the transitive collapse of M , and let N = a [ M ] . Since w, c M, we have a ( Z ) = Z for every Z c w, that is in M and in particular n ( X ) = X ; also, n(A) = n[A n MI = A n N . Now N = L,[A n N ] for some y, and hence, by (31.1), N = L,[A]. Since IN I = K,, we have y < w,+ and hence X E Lua+,[A].It follows that every subset of w, is in &,+,[A] and therefore 2'*= H a + , .

Let K be a measurable cardinal and let U be a K-complete nonprincipal ultrafilter over K. Let us consider the model LEU]. To start with, qLr]= 4U], where U = U n a U ] . Lemma 31.2. In Y U ] , fl is a K-complete nonprincipal ultrajilter over Moreover, if U is normal, then Y U ] I=U is normal.

K.

Proof: A straightforward verification. For instance if U is normal andfE U U ] is a regressive function on K, then for some y < K, the set X = {a :f(ac)= y} is in U ; since x E LEU], Yu] I f is constant on some x E U .

Thus the model LEU] satisfies that K is a measurable cardinal and that V = qfl]for some K-complete nonprincipal ultrafilter over K. The following theorem shows that if D is normal, then 401 satisfies the GCH.

Theorem 75 (Silver). If V = 4 D ] where D is a normal measure on a measurable cardinal K , then the generalized continuum hypothesis holds. Proof. If 1 2 K, then D E P ( I ) and hence 2' = I + by Lemma 31.1. Thus it suffices to show that 2' = 1 ' for any infinite cardinal I < K. Let 3, < K and let us assume that there are more than A+ subsets of I;we shall reach a contradiction. If 2" I + ,then there exists a set X G I that is the I'th subset of I in the canonical well-ordering a and that D E h [ D ] ,and consider the model 91 = (A, E) where A = LAD].We have K E A, and we consider the set P = P ( I ) n A. Since 2' < K, we have I P I < K. By Lemma 29.10, there is an elementary submodel 9 < 91 such that I u {D, X,a } C B, K n B E D and IP n 6 I I I. Let a be the transitive collapse of B onto a transitive set M ; we have M = 4 [ a ( D ) ]for some y. Using normality of D, we show that n(D) = D n M . Clearly, A(K) = K ben 23 I = K . The function a is one-to-one, and for every { < K , cause a ( < )I {. Since D is normal, there is a set Z E D such that A ( ( ) = { for all 5 E Z. Hence if Y E 6 is a set in D, then a( Y ) 2 a( Y n Z ) = Y n Z , and so A( Y ) is also in D ; similarly, if Y E B and a ( Y ) E D, then Y E D. It follows that n(D)= A[D n %] = D n M. Hence by (31.1), M = LJD]. Since I G B, a maps every subset of I onto itself, and so P ( I ) n M = P ( I ) n 6.In particular, we have n ( X )= X and so X E L J D ] . By the minimality assumption on a, we have a I y, and this is a contradiction since on the one hand ( P ( I )n &[D] 1 2 I + ,and on the other hand 1 P ( I ) n L,[D] 1 < I .

IK

One proves rather easily that the model Y O ] has only one measurable cardinal:

Lemma 31.3. f f V = U D ] and D is a normal measure on measurable cardinal.

K,

then

K

is the only

Proof: Let us assume that there is a measurable cardinal I # K and let us consider the elementary embeddingj,: V -+ M where U is some nonprincipal I-complete ultrafilter over I.We shall prove that M = L[D] = V thus getting a contradiction since U 4 M by Lemma 28.9b. Since j is elementary, it is clear that M = 1!&(0)]. If I > K, then j ( 0 )= D and so M = Y O ] .Thus assume that I < K. Since K is measurable, the set Z = { a < K : a is inaccessible and a > I} belongs to D. By Lemma 28.9e,j(~) = K and j ( a ) = a for all a E Z. I shall show that j ( D ) = D n M .It suffices to show that j ( D ) E D n M since j ( D ) is (in M ) an ultrafilter. Let X E j ( D ) be represented byfi I + D. Let Y = n,<J(t); we

5.

362

MEASURABLE C A R D I N A L S

have Y E D, and clearly j ( Y ) G X . Now if a E Y n Z, then j ( a ) = a and so X 2 j ( Y ) 2 j [ Y n Z ] = Y n Z E D and hence X E D. Thus j ( D ) = D n M , and we have (by Lemma 15.4) M = q j ( D ) ]= q D n MI = q D ] . W I shall now introduce the method of iterated ultrapowers and use it to obtain further results on the model q U ] . But first I shall make a few remarks on the model LEU] related to the results above. The proof of Silver's theorem, in particular the use of Lemma 29.10, gives more information on the construction of q U ] . One can show (using normality of D ) that the set of indiscernibles I E D for LAD] has the property that 7413 is a closed unbounded subset of K. One can deduce from this that every subset of K in L J D ] contains or is disjoint from a closed unbounded set. This fact can be used to show that the canonical well-ordering of P ( o ) n LED] does not depend on D and that it is in fact Zz over HC. Another application of the method of proof of Theorem 75 is the proof of Jensen's principle 0 in q D ] . The proof of 0 in L combined with the proof of Silver's theorem, shows that q D ] satisfies 0. As a last remark, if j is an elementary embedding of the universe, there is a least model M , containing all ordinals, such that j I M is a class in M and is an elementary embedding. (This has to be formulated in Godel-Bernays theory of sets and classes rather than in ZF.) This model, denoted L(j), is obtained by constructing a hierarchy L J j ), using Godel operations and in addition closing under the function j . The model L ( j ) is not necessarily equal to the model YU], and it is not known whether it has some nice properties like GCH.

Iterated Ultrapowers Let K be a measurable cardinal and let U be a K-complete nonprincipal ultrafilter over K. Using U , we construct an ultrapower of V modulo U ; and since the ultrapower is well-founded, we identify the ultrapower with its transitive collapse, a transitive model M = Ulr,( V). Let us denote this transitive model Ul#)( V )or just Ult"). Let j = j , be the natural embedding of V in Ult('), and let K(') = j ( K ) and U") = j ( U ) . In the model Ult"), the ordinal ~ ( lis) a measurable cardinal and U ( ' ) is a K(')-complete nonprincipal ultrafilter over K ( ~ ) Thus, . working inside Ult"), we can construct an ultrapower modulo U"): Ultv,,,(Ult(')).Let us denote this ultrapower Ult"), and let j c l )be the natural embedding of Ult'" in Ult") given by this ultrapower. Let K(') = j ( ' ) ( K ( ' ) ) and U ( 2 )= j'')(U(')). We can continue this procedure and obtain transitive models Ult'", U P , . . ., Ult'"', . . . (n < 0) [That we can indeed construct such a sequence of classes follows from the observation that for each a, the initial segment V, n Ult(")of each ultrapower

363

31. THE MODEL L [ U ]

in the sequence is defined from an initial segment V, of the universe (where /l is something like K + a + l).] Thus we get a sequence of models U P ) ,n < 01. For any n < m, we have an elementary embedding inm: UIP" -+ UIP" which is just the composition of the embeddings j("),j ( " +'), . . . , ,j("- I ) :

(31.2)

For n

i n m ( X ) = . / . ( m - I ).I . ( m - 2)

...j y x )

( x E UI+"))

let in, be the identity mapping on Ult'"). denote the universe V . These embeddings form a commutative system ; that is, = in,

Also, let

(31.3)

lmk

ink

l,,,

(m

<0
We also let K(")= i O n ( ~ and ) , U(")= io,,(U). Note that K(') < ~ .... and U l t ' " 3 Ult'l' 2 ... 2 Ult'") =I .... See Fig. 31.1.

< . . . < ti(") <

( l )

Ulr'"'

/ FIGURE 31.1

We shall now define a direct limit of the system

(3 1.4)

{(Ult'"', E);

inm}n.m<w

a model (M, E), together with elementary embeddings inw: Ulr'")-+M which commute with all the embeddings in,,,. [Recall the construction of the direct limit of a sequence of complete Boolean algebras. The underlying idea in the present situation is the same (and can be formulated algebraically): If the embeddings in,were inclusion mappings, i.e., i n m ( x )= x for all x , then we would take M to be the union of the Ult'"),n < 0.3 Let us call a thread a function r whose domain is the set {k, k + 1, . . ., n, . . . ) k 5 n < o , and such that r(k) E Ult(k)and for all n 2 k, t(n) = it,,(@)). By abuse of notation, let us use (31.5)

(xk,

xk+l,

>

...

where x = r(k), etc., to denote the thread r. Let us introduce an equivalence on threads: Two threads (xk+ xk+ I , . . .) and (y,, y l + ...) (where k I 1) are equioalent if x , = y,. Note that in each equivalence class of threads there exists one maximal thread; moreover, a thread (31.5) is maximal just in case there is no X E Ult'k- ') such that X k = i k - l , k ( X ) .

5.

364

MEASURABLE CARDINALS

Let M be the class of all maximal threads. We define a relation E over M as follows: If t and s are maximal threads, then

(31.6)

rE

s

for some n, r(n) E s(n)

iff

It should be clear that (31.6) holds just in case t ( n ) E s ( n ) for all n 2 k where k is the least element of dom(r) n dom(s). For each n c o,we define a mapping in, from U P ) into M as follows: For each x E Ult'"', consider the thread (x,,, x,+ . . .) where for each k 2 n, x, = ink(x).Let i,,Jx) be the maximal thread that extends (x,,, x,+ . . .). See Fig. 3 1.2.

Ulr'"'

FIGURE 31.2

We shall show that for each n, the mapping in,: U P ) + (M, E) is an elementary embedding. We prove this by induction on complexity of formulas, simultaneously for all n. Clearly, in, preserves both = and E: (31.7)

i,,(x) = i,,Jy)

iff

x =y

ino(x) E ino(y)

iff

x

E

Y

The induction step is easy to prove for propositional connectives. Thus consider a formula 32 q ( z , x , ...). Let n < w,let x, .. . E U P , and assume that (31.8)

M k 32 q ( z , i,,,(x), ...)

We want to show that (31.9)

Ult'") k 3 q ( z , x ,

. ..)

Let c be some maximal thread such that M C q(t, ...). If n E dom(t), then we simply let z = t(n), and have, by the induction hypothesis, Ult'")C q ( z , . . .). Otherwise, we consider some rn > n such that rn E dom(c). Since i,,(x) = imW(im(x)),we have U P ) C 3z q ( z , i,,,,(x)) and since i,,,,,: U l P + U l P ) is elementary, (31.9) follows. Thus (M, E) is a model of ZFC, elementarily equivalent t o each Ult(') (hence to V ) ; the mappings in, are elementary and commute with the i,,,,,.

31. THE

MODEL

365

L[U]

Moreover, for each x E M there exists n < o and x, E UlP" such that x = ino(x,). We call (M, E) the direct limit of the system (31.4):

(31.10)

(M, E) = lim dir { Ult'"); i,,,,} n-#

I t is clear that the above construction works in general, for any system of models and elementary embeddings

{(Mn En);inrn}n.m < 3

(0

Moreover, there is no reason to restrict ourselves to an o-sequence of models. Let us assume that we have a system {(Mi Ea); i i , } a , , 7

< i.

of models and elementary embeddings i,,: M , M , where rl is a limit ordinal and the i,, commute. Then we define the direct limit --f

(31.11)

( M A ,E,) = lim dir { M a ;i,,} ,-a

analogously: Threads are now functions t with domain { a : a. Ia c A} and such that t , E M a for each a E dom(r) and t ( p ) = i,,(t(a)). Again, we obtain elementary embeddings i,,: M l + M, that commute with the i, and each x E M A is in the range of some i,, . There is no reason to expect that the direct limit of well-founded models is a well-founded model. However, as we shall see shortly, the direct limit of Ult(")is well-founded. Exercise 31.1. Let K be a measurable cardinal and j : V + M be the corresponding elementary embedding. Let M o = V , M I = M , and for each n < w, M , , I = j ( M , ) and i,,,, I = j ( M , . The direct limit of {Mn;in,,,} is not well founded. [iOlo(K), i l c u ( ~ ).,. . , ~,,JK), . . . is a descending sequence of ordinals in the model.]

Let us return to iterated ultrapowers. We started with a K-complete nonprincipal ultrafilter U over K and defined the models U P and elementary embeddings i,, . Let us denote by ( U P ) ,E'")) (or just Ult'")) the direct limit

(31.12)

( U P ) ,E'")) = lim dir {( Ult'"),E); i,,} n-w

Let K'") = iO&) and U(") = iou( U). Since Ult'") satisfies that U('") is a P)-complete nonprincipal ultrafilter over K("), we can construct, working inside the model U P ' ) , the ultrapower of Ult@')modulo U'")and the corresponding natural embedding j'"). [This is easy to understand if Ult'") is a transitive model. If ( U P ) ,E'")) is not well-founded, then the ultrapower and the mapping j(") are classes in the sense of the non-well-founded model. In general, note that a " class " C in a nonstandard model (M,E) is just a subclass of M ,namely {x E M : M b x belongs to C},and similarly, a " function " F in (M, E) is a function from M to M :{(x, y ) E M x M : M =! F(x) = y}.]

366

5.

MEASURABLE CARDINALS

Let us denote by ( U l t ( w + l )E'"')) , the ultrapower of (Ult'"), E'")) modulo U(,) and let i,,,+ be the corresponding natural embedding. For each n < o, let in+,,+ = i,,,, o in,,,. [Another way of looking at Ult'w+": Let E be the set algebra {extEI,,,(X) : U P ) k X E K ( ~ ) } and let U * be the ultrafilter { e x t ( X ): X E@') U(w)}.For each fsuch that Ult'") k f is a function on dW), let f * = { ( x , y ) : U P ) k f ( x ) = y } . Then Ult'"' is (isomorphic to) the ultrapower of Ult'") using U* and only the functions f *.] This procedure can be continued, and so we define the iterated ultrapower of V modulo U as follows: (3 1.13)

( U P ' , E ' O ' ) = ( V , E) (Ulf''+1), E ( " + 1 ) )= Ult"l.,(Ult'a',E'"')

(Ult'"', E'")

=

lim dir {( Ult'"),E'")); iua)

(A a limit)

"-11

where U(') = ion(U ) , for each a. We do not know yet that all the models Ult(') are well-founded; but we make a convention that if Ult'") is well-founded, then we identify it with its transitive collapse. The first lemma about iterated ultrapowers is the Factor Lemma below. If M is a transitive model of set theory and U is (in M ) a K-complete nonprincipal ultrafilter over K, we can construct, within M , the iterated ultrapowers. Let us denote by U l t t ' ( M ) the ath iterated ultrapower, constructed in M . (Thus Lilt'"' = Ult:.'( V)).

Lemma 31.4 (Factor Lemma). Let us assume that Ulr'"' is well-founded. Thenfor each fi, the iterated ultrapower Ultf,$(Ult('))taken in Ult'") is isomorphic to the iterated ultrapower Ult''+ 0). Moreover, there is for each /?an isomorphism e r ) such that iffor all 5 and q, iyi denotes the elementary embedding of Ult$/,l(Ult'"))into Ultg),,(Ult'')), then the following diagram (Fig. 3 1.3) commutes :

FIGURE 31.3

Proof: Although the lemma sounds complicated, it is actually quite simple, and also easy to prove. The proof is by induction on p. If /I = 0, then the 0th iterated ultrapower in U l P ) is Ult'");and we let eg) be the identity mapping. If Ult',fLI and Ult:"+P'are isomorphic and e;) is the isomorphism, then UltjJI: and

367

31. THE MODEL LEU]

U f t g + f l +l ) are ultrapowers of Uftf,&and Ult',"+8), respectively, modulo itJ(Va)) and i,,, +8( U ) , respectively; and since io.a+fl(U ) = eF)(i$J(Uca))),the isomorphism e f ) induces an isomorphism efi between Ultg,: and U l t g + B + l ) . If I is a limit ordinal, then Ult$!, is the direct limit (in Ult'")) of {UlrVJ,;if,!}B,y
Lemma 31.5. Let U be a K-complete nonprincipaf ultrafilter over every a, the ath iterated ultrapower Ult(a)is well-Jounded.

K.

Then for

Proof. Clearly, if Ult(a)is well-founded, then Ult'"' is well-founded. Thus if y is the least y such that Ult(Y)is not well-founded, then y is a limit ordinal. The ordinals of the model Ult(Y)are not well-ordered; let be the least ordinal such that the ordinals of Ulr(y)below ioy(<)are not well-ordered; see Fig. 31.4.

<

V

0

z

s

I

FIGURE 31.4

Let xo , x,, x 2 , . . . be a descending sequence of ordinals in the model Ult(Y) such that xo is less than ioy(<). Since U l P ) is the direct limit of Uft'"),a < y, there is a < y and an ordinal v (less than iom(<)) such that x, = i,,(v). Let /3 be such that a + /3 = y. By our assumptions, the following is true (in V ) : (31.14)

(Vy' I y)(Vt' < t) the ordinals below joy,(<') in U l P a r e well-ordered.

When we apply the elementary embedding i,, to (31.14), we get: (31.15)

Ult'")k (Vy' Ii,,(y))(Vt' < i,,(t)) the ordinals below u w well-ordered.

it;,(<') in Ulf(UY;!)

5. MEASURABLE

368

Now p Iy Iioz(y), and v < ioa(<). Hence if we let y' = p and 5' we get

=

CARDINALS

v in ( 3 1-15),

(31.16) Ult(")k the ordinals below @ ( v ) in Ulrgk, are well-ordered.

By the factor lemma, Ultf?,,is (isomorphic to) Ult(a+8),and iFh(v) is ia,.+@(v). Since a + p = y and iay(v)= x o , and since being well-ordered 1s absolute (for the transitive model Lilt'")), we have: (3 1.17) The ordinals below xo in Ult(Y)are well-ordered. But this is a contradiction since x l , x 2 , x 3 , ... is a descending sequence of ordinals below xo in U l P . H Thus for any given K-complete nonprincipal ultrafilter U over K we have a transfinite sequence of transitive models, the iterated ultrapowers Ultg)(V ) , and the elementary embeddings iaB: Ult'")+ Ult(B).Let K(') = i o a ( K ) for each a ; we shall show that the sequence d"),a E Ord, is a normal sequence, and consequently obtain a characterization of the iterations Ult(') in the case V = YO].

Lemma 31.6. (a) l f y c K('), then ias(y) = y for all fl 2 a. (b) If X G K(') and X E Ult'"), then X G ia8(X) for all p 2 a ; in fact X = K(') n ia8(X). Prooj: By the factor lemma, it suffices to give the proof for a = 0. (a) As we know, iol(y) = y for all y < K. By induction on p, if i o a ( y ) = y, then i0,@+l(y) = is,B+l(y) = y because y K @ ) ; if 1is a limit and ios(<) = 5 for all 5 I y and fl c 1,then ioA(y) = y. (b) Follows from (a). H

-=

Lemma 31.7. The sequence continuous.

(K(')

:a

E

O r d ) is normal; i.e., increasing and

Proof. For each a, K ( ' + ~ ) = i a , a + l ( ~ ( a> ) ) K('). To show that the sequence is continuous, let A be a limit ordinal; we want to show that K(') = lim a+* I&'). If y < K('), then y = i,@) for some a < 1 and 6 < K('). Hence y = 6 and so y < K").

Lemma 31.8. Let D be a normal measure on K , and let for each a, Ult(') be the ath iterated ultrapower mod D, K(') = i o a ( K ) , and D(')= ioa(D).Let 1 be an infinite limit ordinal. Then for each X (31.18)

X E D(')

iff

E

Ult('), X

c K(*),

(hr < l ) X 3 {dY) : a I y < A}

Proof. Since for no X can both X and its complement contain a final segment of the sequence ( K ( ~ :) y < A), it suffices to show that if X E D('),then there is a such that K ( ~ E) X for all y 2 a.

369

31. THE MODEL L [ U ]

There exists a < A such that X = iaA(Y) for some Y E Da).Let me show that X for all y, a I y < 1. Let y 2 a and let Z = &,( Y). Then Z E DtY)and since D(Y) is a normal measure on K ( ~ in ) U / T ( ~we ) , have K ( , ) E i,,,, l(Z). However, i,.,, l(Z)G i,, l,d(iy,y+ l(Z))= X and hence K ( , ) E X . ~ ( 7E )

Representation of Iterated Ultrapowers We shall now give an alternative description of each of the models Ult") by means of a single ultrapower of the universe modulo an ultrafilter on a certain Boolean algebra of subsets of ' K . This will enable us to obtain more precise information about the embeddings ion: V + Ult(') and also to generalize the iteration to the case when U is an M-ultrafilter for some model M. We shall deal first with the finite iterations. Let U be a K-complete nonprincipal ultrafilter over K. Let us use the symbol V*a for "for almost all a < K " : (31.19)

If X

V*a q ( a )

{ a < K : cp(a)}E U

iff

< K , let

E " K and a

~ ( ~ ) = { ( a ~ , . . . , a , - ~ ) :, .( .a.,, O aL ~,-~)EX}

(31.20)

We define ultrafilters U , over ,K, by induction on n:

u, = u

(31.21)

X

E

U,+l

iff

V*a X t u )E

U,

Each U , is a nonprincipal K-complete ultrafilter over "Z E U , . It is easy to see that for all X C "K,

x E U,

iff

and if Z E U , then

'K,

V*a, V*a1 ... V*anP1( a o , ..., a,- ,) E x

Although the present definition varies slightly from definition (29.10), it is practically the same because for each n, U , concentrates on increasing n-sequences: { ( a o , .. ., a,-

(31.22)

I) E "K:

a. < ... < a,-

1}

(because Vao(Va, > a ) ... (Van- > a,-z)[ao < ... < a,-

E

U,

I]).

Lemma 31.9. For every n,

mu"( V ) = Ult'"'( V ) and jum= ion.

[Here j u mis the natural embedding j : V + Ulfu,(V).] Proof: By induction on n. The case n = 1 is trivial. Let us assume that the lemma is true for n and let us consider Ult,,, ,. Letfbe a function on ' K . For each t = (ao, . .., a,E "K, let As be the function on K defined by A,(() = f ( a o , ..., a,- (), and let F be a function on "K such that F ( t ) =f(?,for all "

,,

+

370

5.

MEASURABLE CARDINALS

In UlrUn= Ult'"), the function F represents a function on j u . ( K ) = ti("): Let .f= [Flu". This way we assign to each function .f on ""K a function . f U~ P on ti("). Conversely, if h E UIt(")is a function on d"),there is f on 'K such that h =$ There exists F on "K such thatf= [F],, and that for each I E "K, F ( r ) is a at a,, . function on K ; thus we let f ( a o , .. ., a,) be the value of F ( a o , . . ., a,We shall show that the correspondence [ f l u . +I w [flu,., is an isomorphism U P ) ) .We have between Ult,,, I and UlP" = Ulf,,,")( I E "ti.

" +

[flu,,1

=

[slu.+ I

V*ao ... V*a,,-, V*5

iff

f ( a o , . . . , a n - 1, 5 ) = g(ao, ..., a,,-', 5 ) iff

v * a
iff

U11," k {5 < i u . ( K ) :f(5) = at)}E J ' U . ( U ) U P k {< < K(") :f(5) = E u(")

: A t ) ( < )=

g(,)(olE U

a(<.)}

iff

[a],,.,

iff [f]Ul", = and similarly for E in place of =. Thus Ultun+I = Ult("+').To show that j u m += iO,n+ let f = c, be the constant function on I K with value x . It follows thatfis the constant function on d")with value ion(x),and therefore "+

h.+, ( x ) = [cxlu,+

1

= in,,+ I(ion(x))= i O . n + i ( x )

Exercise 31.2. Show that if m In, then for eachfon '"K, imn([flu,) function on " K defined by g(ao, ..., a,- ,) = f ( a o , . .., a,,,-

= [glumwhere g

is the

The infinite iterations are described with the help of ultrafilters U Eover where E ranges over finite sets of ordinal numbers. If E is a finite set of ordinals, then the order isomorphism II between n = I E I and E induces, in a natural way, an ultrafilter U E corresponding to U,: (31.23)

U E= { x [ x ] : x

5 "K}

where a ( ( a o , . .. , a,,- ,))= I E E~ where t ( n ( k ) )= ak for all k = 0, . . ., n - 1. If S is any set of ordinals and E 5 S is a finite set, we define a mapping inE.s (an inclusion map) of P ( E ~into ) P(%) as follows:

(3 1.24)

inE+&) = {I E ' K : I I E E X }

(all X

5 'K)

Lemma 31.10. lf E E F a r e j n i t e sets of ordinals, then for each X G

x E UE

fl

E ~ ,

inE,F(X) E UF

Proof. By induction on (m,n ) where m = 1 E I and n = I F I. Let E 5 F be finite sets of ordinals. Let a be the least element of F , and let us assume that a E E (if a 4 E, then the proof is similar). Let E' = E - {a} and F' = F - { a } .

31. THE MODEL L [ U ]

37 I

I f X c ' K , let us define for each z < K , the set X ( l )G " K as follows: XI,, = { I 1 E' : I E X and [ ( a )= a ) ; for Z c F ~ let , us define Z(,) G " K similarly (for all a < K ) . I t should be clear that (31.25) X

E UE

iff

V*a X l l ) E UE,;

Z

E

iff

UF

V*a Z(,) E U F ,

Now we observe that if Z = inE,F(X), then Z ( , )= inE,,F,(X,,,), and the lemma for E, F follows from (31.25) and the induction hypothesis. W Let us now consider an ordinal number a. I f E that a set Z L ' K has support E if Z

=

C_

a is a finite set, let

us say

inE.,(X)

for some X G ' K . Note that if Z has support E and E c F , then Z also has support F . Let B, denote the collection of all subsets of 'K that have finite support. ( B , , G ) is a Boolean algebra. Let U , be the following ultrafilter on B,: For each Z E B , , if Z = inE,,(X) where X G E ~ let , (3 1.26)

Z EU,

iff

X

E

UE

By Lemma 31.10, the definition of Uldoes not depend on the choice ofsupport E of Z . We shall now construct an ultrapower modulo U,. Iffis a function on 'K, let us say thatfhas a j n i t e support E G a iff(t) = f ( s ) whenever t and s E 'K are such that f I E = s I E . (In other words, there is g on E~ such thatf(r) = g(t I E ) for all t E ' K . Let us consider only functionsf on 'K with finite support and define

(31.27)

9

iff

f El 9

iff

f = 1

:f(t)= 9 0 ) ) E u s { t :f(t)E s(t)lE u, {f

The sets on the right-hand side of (31.27) have finite support, namely E w F where E and F are respectively supports offand 9. Let (Ult,,(V), E,) be the model whose elements are equivalence classes mod =' of functions on ' K with finite support. Exercise 31.3.

Prove the fundamental theorem:

(Ulf",, E , )

d[fIl,

..., [.f"]) iff

{ f : cp(fl(f).

....f " ( t ) ) urn

We are now in a position to state the main lemma.

Lemma31.11 (Representation Lemma). For every a, the model (Ultua(V),E,) i s (isomorphic to) the ath iterated ultrapower Ultt'( V ) , and the natural embedding j u x : V + UltuS is equal to io,. Moreover, if a I p and [flu, E Ult'"), then i , a ( [ f ] v , )= [g]", where g is dejned by g ( t ) = f ( t 1a)for all t

EB ~ .

5.

372

MEASURABLE CARDINALS

Proof. By induction on a. The induction step from a to a + 1 follows closely the proof of Lemma 31.9; thus let me describe only how to assign to [flu,+, the corresponding [TIu,,,in Ult('+').Let f be a function on " ' K with support E u {a} where E c_ a. For each t E ' K letf;,,(<) = f ( t " < ) for all 5 < K , and let F be a function on ' K (with support E) such that F ( t ) = f , for all t E ' K . Let?= [Flu,; f i s in Ult'")and is a function on K('). The induction step is rather simple for a limit ordinal A. Let us assume that Ultue = Ult'"' for all a < I. Iffis a function on * K with a finite support E, we assign tofa thread in the direct limit of {Ult(")},
One advantage of this description of iterated ultrapowers is that we can define Ultt'(M) even if M is a model of ZFC and U is an M-ultrafilter over y, provided U satisfies the following condition: Let us call U an iterable Multrafilter if, in addition to (30.32), U has the following property: (3 1.28)

if (X, : a < y )

E

then {a : X, E U } E M

M,

[Note that the M-ultrafilter defined in (30.31) from j : M + N is iterable, provided P N ( ~=)P ' ( ~ ) . ] Let A4 be a transitive model of ZFC containing all ordinals. Given an iterable M-ultrafilter U over y. we can define M-ultrafilters U , , n < o,over " y by induction: (31.29)

X E U,,+

{ a < y : X,,,} E U

iff

where X(,)= { ( a 1 , ..., a,) : (a, al, ..., a,) E X}. The definition (31.19) works because U, is iterable, which is proved by induction on n. Then one defines M-ultrafilters U , as in (31.26), and the ultrapowers Ult,,,(M) as in (31.27), using functionsf€ M with finite support. The ultrapowers Ult,, are not necessarily well-founded. Exercise 31.4. Each U , is an iterable M-ultrafilter on the algebra (in M ) of subsets of " y with finite support. Exercise 31.5 (Fundamental Theorem). Prove that

Uh",(M) c

cP([fIl)7

'.. *[Ll)

iff

It : M I=cp(fi(t).

...,L(t)))E u.

3 1. THE MODEL

373

V]

Exercise 31.6. I f I is a limit ordinal, then U l r u A ( M ) is the direct limit of ( ~ l r u z ( M i'&.B<.I. );

If the model Ult,,(M) is well-founded, let us define U @ )= {[f]", : { t : f ( t ) E V }E V,} Exercise 31.7.

If N, = U l t u , ( M ) is well-founded, then U(') is an iterable N,-ultrafilter

over io,(y).

One can also prove the factor lemma for Ult,,(M) if this model is well-founded. Exercise 31.8.* I f Ulr,,(M) is well-founded for all a < wI.then U I f u J M ) is wellfounded for all a. [Assume that U l t u , ( M ) is not well-founded and let fo. f l , ..., j n ,... constitute a counterexample. Eachf, has a finite support E n . Let be the order-type of En;we < w l . Produce a counterexample in Ulru,(M).] have /l

u.'=o

An important sufficient condition for well-foundedness of iterated ultrapowers : I f arbitrary countable intersections ofelements of U are nonempty, then U l t u , ( M ) is well-founded for all a. [Letfo.fl, . .. , f n , . . . be a counterexample, let X , , = { r E ' y :f,(f) 3 f n + I(t)}. To reach a contradiction, find r E X , . Construct I by induction such that for each v < a if a = v + q, then f I v has the property that for all n, { s E " y : ( r I v ) " s E X,} E U , : Given f I v, there is r ( v ) such that the condition is satisfied for I 1 ( v 1). Then r E n:=o X , .] Exercise 31.9.*

n;=o

+

As an application, consider the following: Exercise 31.10. Assume that every constructible subset of w 1 either contains or is disjoint from a closed unbounded set. Let F be the closed unbounded filter over wI.Then D = F n L is an iterable Lultrafilter and UIt,,(L) is well-founded (and hence equal to L ) for all a.

Uniqueness of the Model Y O ] Theorem 76 (Kunen). (a) If V = q D ] and D is a normal measure on K, then K is the only measurable cardinal and D is the only normal measure on K. (b) For every ordinal K , there is at most one D G P ( K )such that D E Y D ] and Y D ] k D is a normal measure on K

(c) If

K,

< K~ are ordinals and

if D , , D , , are such that

q D i ] k Diis a normal measure on

K~

( i = 1, 2 )

then Y D 2 ] is an iterated ultrapower of q D , ] ; i.e., there is a such that q D , ] = LrlrF!(L[D,]),and D2 = ioa(D1).

5.

374

MEASURABLE CARDINALS

The proof of Theorem 76 uses iterated ultrapowers. The following lemma uses the representation of iterated ultrapowers (Lemma 31.1 1) (compare with Lemma 28.9). Lemma 31.12. Let U be a ti-complete nonprincipal ultrajilter over K and let, for each a, iOa: V - , Ult'"' be the natural embedding of V in its ath iterated ultrapower. (a) If a is a cardinal and a > 2", then i o , ( ~ = ) a. (b) If 1 is a strong limit cardinal, Iz > a, and ifcf 1 > ti, then io&)

= 1.

Proof. I t follows from the representation lemma that for all [, q, the ordinals below iot(q)are represented by functions with finite support from 5~ into q and hence I iot(rt)I 5 I I . I rt 1". (a) We have i O b ( K ) = lim+ iO&), and for each 5 < a, l i o t ( K ) I I 15 1 . 2" < a. Hence iO$) = a. (b) Since cf A > ti, every function8 "ti -,A with finite support is bounded below 1: There exists y < A such that f ( t )< y for all t E 'ti. Hence io,(A) = limy+Aiom(y). Since 1 is a strong limit, we have lio,(y)I < A for all y < A and hence io,(Iz) = 2.

r

Remark. It is clear from the proof that in (b) it is enough to assume that y" < A for all cardinals y < A, instead of that 2 is a strong limit. Let U c P(ti). If 0 is a cardinal and U E Le[U], then by absoluteness of relative constructibility, every elementary submodel of (&,[V], E) that contains U and all ordinals < ti, is isomorphic to M = L J U ] for some a. (If A is the transitive collapse of the submodel, then A ( U ) = U n M E M , and M = L J U ] follows by (31.1).) Let 8 be a cardinal such that U E L B [ U ]and let us consider the model ( & , [ I / ] ,E, U ) where U is regarded as a constant. This model has a definable well-ordering, hence definable Skolem functions, and so we can talk about Skolem hulls of subsets of &[U].

Lemma 31.13. Let V = LED] where D is a normal measure on ti. Let A be a set of ordinals of size at least ti+ and let 8 be a cardinal such that D E Le[D] and A E Le[D].Let

M

(Le[D], D > ~9

be the Skolem hull of K u A. Then M contains all subsets of K. Hence every X s K is dejnable in ,!,[Dl from K u A u { D } ;in other words, [here is a Skolem term t such that for some a l , . . ., a, < K and yl, . .., ym E A,

b [ D ] k X = t [ a l ,..., a,. yl,

..., ymrD ]

Proof. Let K be the transitive collapse of M . We have A [ M ] = L,[D] for some u, and since A c M, we have necessarily a 2 K ' . By Lemma 3 1.1, every X c_ K is in L,+[D],and since K is the identity on K , we have X E M for all X C K. H

31.

L[U]

THE MODEL

375

The following is the key lemma in the proof of uniqueness of Y D ]

Lemma 31.14. h t D

G

P ( K )be such that D

E

Y D ] and

U D ] b D is a normal measure on

K

For each CL, let U l t ~ ) ( ~ D denote ] ) the ath iterated ultrapower, constructed inside 401,modulo D. Let iOa he the corresponding natural embedding. Let 1 be a regular cardinal greater than K + , and let F be the closed unbounded filter over A. Then:

(a) ioA(D)= F n U l t g ) ( Y D ] ) ; (b) U l t g ’ ( q D ] )= Y F ] . ~ J Proof. First, we have i o i ( r i ) = 3, by Lemma 31.12a because 3, > K + 2 ( K + ) ~ [ = (2K)LID1. Let D‘” = ioA(D)and let M = U l r g ) ( Y D ] ) If . X E D“), then by (31.18), X contains a closed unbounded subset and hence X E F. Since D(’) is an ultrafilter in M and F is a filter, it follows that D(’) = F n M . As for (b), we have

M

=

Ult(A)(YD= ] ) Y D ‘ ” ] = 4 F n MI = q F ]

by Lemma 15.4. W We shall now prove statements (a) and (b) of Theorem 76. We already know by Lemma 3 1.3 that in Y D ] K is the only measurable cardinal. Thus (a) and (b) follow from this lemma:

Lemma 31.15. Let D , , D, G P ( K )be such that D , 4 D i ] b Di i s a normal measure on

ti

E

Y D , ] ,D ,

E

q D 2 ] and

(i = 1, 2)

Then D , = D , . Pro05 Let D,,D , G P ( K )be such that 4 D i ] b Di is a normal measure on K , for i = 1, 2 ; we want to show that D , = D , . By reason of symmetry, it suffices to show that if X E K is in D,,then X E D , . Let 1 be a regular cardinal greater than K + and let F be the closed unbounded filter over 1. Let us consider the Ith iterated ultrapowers M i = U l t g , ) ( ~ D , ]()i = 1, 2), and the corresponding natural embeddings .1

.2

‘01,101.

By Lemma 31.14, M I = M , = q _F ]_ , and iAA(D)= i&(D2) = F n q F ] . Let nYF]. Let A be a set of ordinals, I A I = K + , such that all y E A are bigger than I and that i & ( y ) = i & ( y ) = y for all y E A ; such a set exists by Lemma 31.12b. Let 0 be a cardinal bigger than all y E A such that &(O) = i&(O) = 0. Now let X be a subset of K such that X E D 1 . By Lemma 31.13, X is in the Thus there is a Skolem term t such Skolem hull of K u A in ( L , [ D , ] , E, 0,).

G =F

5.

376

MEASURABLE CARDINALS

that for some a,,

. . ., a,, < K and yl, . .., ym E A,

(3 1.30)

&[D,] b X = t [ a i , ...,a,,, YI, ..., Ymr

0 1 3

Let Y E &[D2] be such that (31.31)

&[D2]

b Y = t[al,

a,,, ~

..-,Y m r

1 .

D2]

We shall show that Y E D , and Y = X, hence X E D , . First we observe that iAA(X)= i&( Y): Let 2 , = iA,(X)and Z, = i&( Y). We have &:'(a) = a, i'J(y) = y, iVj(8) = 0, and i;,(D,) = G ; and thus when we apply iA, to (31.30), we get

&[GI b 2 , = [[a,, ..., a,,

(3 1.32)

..., Y n i G ]

~ 1 ,

Similarly, when we apply i ; , to (31.30), we get (31.32) with 2 , instead of 2,. Thus 2 1 = 2 2 . Now, by Lemma 31.6b, we have X = 2 , n K and Y = 2, n K. Hence X = Y. Finally, since i&( Y) E F , it follows that ii,( Y) E i i A ( D 2and ) hence Y E D , . Thus X E D2 and this completes the proof of D1 = D , . The key lemma in the proof of the clause (c) in Theorem 76 is the following:

Lemma 31.16. Let K , D be such that 4 D ] k D is a normal measure on K , and let y be an ordinal such that K < y < iol(K), where iol is the natural embedding of YO] in U l t , ( y D ] ) . Then there is no U S P(y) such that 4 U ] != U is a normal measure on y.

Pro05 Let us assume that on the contrary there is such a U . Let j be the natural embedding of 4 U - j in U l t u ( 4 U ] ) .Let I be a regular cardinal greater than I y I+, and let F be the closed unbounded filter over I . Let G = F n YF]. Since 4 U ] != GCH, we have j(I)= I (see the remark following Lemma 28.9). In YU],G is the Ith iterate of U , and in a j ( U ) ] ,G is the j(rl)th iterate of j(U ) ; hence j ( G ) = G. Let f: K + K be a function in a D ] such that f represents y in U l t D ( 4 D ] ) . Since D is normal, the diagonal d(a) = a represents K, and thus we have (iol(f))(K) = y. Let io, be the natural embedding of L[D] in Ultg)(L[D]) = L[G]. It is clear that (ioa(f ) ) ( K ) = y. Now let A be a set of ordinals such that IA1 = ,'K that all E A are greater than I, and that ioA(<)= and j(<)= for all E A. Let 8 be a cardinal greater than all E A, such that iO#) = B and j(0) = 8. By Lemma 31.13, the functionfis definable in b [ D ] from A u K u {D}; thus i o a ( f )is definable in L,[G] from A u K u {G}.Hence y is definable in &[GI from A u K u {G}u {K}, and so there is a Skolem term t such that

<

<

(31.33)

&[GI

<

y = (a,,

for some a,, ..., a, < K and

..., a,,

tl, ..., 5,

<

<


E A.

..., 5 m ,

G, K]

31.

THE MODEL

L[U]

377

Now we apply the elementary embedding j to (31.33); and sincej(0) = 0, j ( G ) = G , j ( 5 ) = 5 for 5 E A, and j ( a ) = a for all a < y (hence j ( t i ) = t i ) , we have

b [ G ] j ( y ) = t[a,, ..., an, T I , ..., which is a contradiction because j ( y ) > y.

5.7

G , ti]

Proofof Theorern 76c. Let ti1 < ti2 and let D , , D , be such that L[Di] k D is a normal measure on tii ( i = 1, 2). Let i,, denote the natural embedding of 40 ,] in U l r g ; ( q D , ] )and let a be the unique a such that i o z ( K l ) I K~ < i o . z + I(til). By Lemma 3 1.16 (if we let K = i O a ( t i l ) D , = i 0 @ ,), and y = K ~ ) it, is necessary that ti2

= i0,(KI).

Now the statement follows from the uniqueness of i o , ( D l ) (Theorem 76b). Thus we have proved that the model I/ = L [ D ] (where D is a normal measure on t i ) is unique, has only one measurable cardinal and only one normal measure on ti, and satisfies the generalized continuum hypothesis. The next lemma completes the characterization of L[D] by showing that for any Kcomplete nonprincipal ultrafilter U over ti, YCJ] is equal to YO]. Lemma 31.17, Let U be a nonprincipal K-complete ultraJilter over U U ] = q D ] where D is the normal measure on ti in q D ] .

ti.

Then

Pro05 By the absoluteness of Y D ] ,we have YO] c 4 U ] because LEU] satisfies that K is measurable. Thus it suffices to prove that U n 401 E UD].Let j = j , be the natural embedding j: V + Ult,(V), and let y = j(ti). Let d(a) = a be the diagonal function and let 6 be the ordinal represented in CJlru(V) by d; thus (31.34)

X E U

iff

SEj(X)

for all X E K . Since q j ( D ) ] kj(D) is a normal measure on y, there exists a such that y = i o a ( ~ )j, ( D ) = io,(D), and Y j ( D ) ] = U l t g ) ( q D ] ) .We shall show that for every X G K, X E LED], (31.35)

=

idx)

This, together with (3 1.34), gives (3 1.36)

U n LED] = ( X

E

L[D] : X E K and S E ioa(X)}

and therefore CJ n Y O ] E LED]. The proof of (31.35) again uses Lemma 31.13. We let A be aset ofsize K + of ordinals greater than a such that ioa(r) = j ( [ ) = [ for all 5 E A, and let 8 be a cardinal greater than all 5 E A, such that ioA(8)= j(0) = 8. If X E K is in 4 D ] , then there is a Skolem term t such that 4 [ D ] k X = t[a,, -..,an, tl, ..., tm, D]

378

5.

MEASURABLE CARDINALS

,,

for some a l , . . ., a,, < ti and 5 . . ., 5 , E A. Since io2and j agree on {(I!, and ion(D)= j ( D ) , it follows that i o 2 ( X )= j ( X ) . W

ti

u A u

The proof of Lemma 3 1.17 gives additional information about K-complete ultrafilters in 401. Let us assume that V = 401 and let U be a nonprincipal ti-complete ultrafilter over K. By (31.36), we have U ={X

(31.37)

c ti : 6 E i o 2 ( X ) }

where a is such that j ( ~=) iOz(ti),and 6 < j ( ~ )Note . that for any 0 2 a, we also have U = { X : 6 E i o p ( X ) ) .Now a simple observation gives the following characterization of ti-complete ultrafilters over ti in q D ] :

q D ] . Then if U i s a nonprincipal ti-complete there exists 6 < iO&) such that

Lemma 31.18. Assume V =

ultrafilter o w

ti,

(31.38)

U

= {X

c

K

: 6 E i0,,(X)}

Proof. Let j = j , be the natural embedding of V = L[D] in U l t , . We have j ( i i ) = ion(ii)for some a. We shall show that a is a finite number; then the

lemma follows by (31.37). First we note that because V = q D 3 = Y U ] , we have Ultg' = U l t , = q i o n ( D ) ]= q j (U ) ] . Now if a 2 w, then in Ultg), i O c o ( K ) is an inaccessible cardinal (because it is measurable in Ult(,o)),while in Ult,, ioo(.) has cofinality w (because it has cofinality w in V and U l t , contains all w-sequences of ordinals). Hence a < w. W

Corollary. If V = Y O ] ,there are exactly K + nonprincipal K-complete ultrafilters ouer ti. Proof: If ti is measurable, then it is easy to obtain 2" nonprincipal K-complete ultrafilters over K (because there are 2" subsets of K of size K such that I X n Y I < ti for any two of them). By the lemma, if V = 401,there are at most I iO&) I = K + of them.

Recall (Exercise 28.12) that a K-complete ultrafilter over K extends the closed unbounded filter just in case the diagonal function d is the least (mod U ) nonconstant monotone function. If V = 401, then the characterization (31.38) shows that D is the only K-complete ultrafilter over K with this property; see Exercise 31.12. Let D, be the measure on "K defined from a normal measure D by (31.21). Let K ( , ) = iOn(K)(where ion: V + Ultg)). Exercise 31.11. The ordinal

K("-')

is represented in

Ulr,

by the function

dn(al.. . ., a,) = a,.

Exercise 31.12. Assume V = YO].If U is a K-complete nonprincipal ultrafilter over K and if U # D, then there is a monotone function f: K + K such that K I [flu < [dIu. (Hence U does not extend the closed unbounded filter.) [ U satisfies (31.38) for some 6; if 6 = K ' ~ )for some n, then U = D. Let n be such that

379

31. THE MODEL L[U]

< 6 < tiin). , let y: "ti + h- represent 6 in U l f D n .Let f ( 5 ) = least a such that g ( a l . . . . . 2 " - z ) 2 5 for some a , < . . . < a,- < a. The function f is monotone. To show that [/It, < [ t l ] , . we argue as follows: For almost all (mod D,) a I . ..., an, g(o7) > 2,; hence for almost all z I . . .., r n ,./(g(c?)) < g(o7). Hence ( j D n ( f ) ) ( i 5< ) 6. and hence for almost all 5 (mod U ) . f ( [ ) < 5. Thus [flLI < [d]" .] ,$n-

I)

,

,.

Some Generalizations The uniqueness theorem for r_ID]can be generalized to the situation involving several measurable cardinals. If A,, a < 0, is a sequence of sets, let us define the model L(A, : a < 0)

(31.39)

as the model

UA] where A

= {(a, X):X E

A,)

Under this definition, L(A, : a < 0 ) = y(B,: a < 0 ) ] , where B, = A, n L(A, : a < O ) for all a < 8. If K,, a < 8, is a sequence of measurable cardinals, and for each a, U , is a K,-complete nonprincipal ultrafilter over K,, then in L( U , : a < 0 ) , each U , n L( U , : a < e) is again a K,-compIete nonprincipal ultrafilter over K, . The results of the present section can be generalized as follows: If (K, : a < O ) is an increasing sequence of measurable cardinals, and if O < K ~ then for any sequence ( D m: a < 0) of normal measures on the K ~the , model M = L(D, : a < 0) has the property that the K, are the only measurable cardinals and for each a, D, n M is the only normal measure on K,. The following exercise gives a canonical model with two normal measures, under the assumption of a normal measure concentrating on measurable cardinals. It can be proved that in M there are only two normal measures on K. Let D be a normal measure on K such that X = { a < K :a is measurable} E D . For each a < K, let CJmbe a normal measure on a, and let U be the measure on K in U l t , represented by the function a- U , ; hence Z E U iff { a : Z n a E U , ) E D. Let M = L ( U , , U , D : a E X). M , U n M and D n M are two distinct normal measures on K . [ Z E U n M iff { a : Z n a E U , n M ) E D n M . Hence U n M is in the ultrapower U l t , n M ( M ) ;while D n M is not.] Exercise 31.13. In

Indiscernibles for YO] If there exist two measurable cardinals, K < A, then it is possible to prove analogous theorems for the model UD] as we did in Section 30 for L under the assumption of one measurable cardinal. More specifically, one can prove existence of a closed unbounded set I c K and a closed unbounded class J of ordinals bigger than K, such that I u J contains all uncountable cardinals

,

380

5.

MEASURABLE C A R D I N A L S

except K, that every X E YO] is definable in D from I u J , and that the elements of I u J are indiscernibles for YO] in the following sense: The truth value of

.... Pm] is independent of the choice of a1 < ... < a,, E I and PI < . * . < p,,, E J. In analLED] k q [ a I ,..., a,,,

PI,

ogy with Silver indiscernibles, the above situation can be described by means of a certain set of formulas q ( x . . ., x,,, y . . ., y,), which is usually called 0' (zero-dagger). If 0' exists, then one can prove consistency of the theory ZFC + "there exists a measurable cardinal ";and hence one cannot prove the relative consistency of "0' exists" with ZFC + " there exists a measurable cardinal." I shall not give details of the theory of indiscernibles for YO]. Instead, let me give a short argument showing that if there are two measurable cardinals, K < 1,then there is a proper class of cardinals that are inaccessible in 4 D ] . Let U be a normal measure on 1 and let for each a, io. be the natural embedding of V in Ultg); let i,,: U P + U P ) . Let K be the class of all cardinals a such that cf a > 1and y' < a for all y < a. By Lemma 31.12, if a E K, then ion@) = a and i0@) = p for all p E K greater than a. Hence if a, fl E K, then &,(a) = p and i,,(y) = y for all y E K that are greater than p or less than a. Now if D is a normal measure on K, then because K < 1,we have i,,(D) = D for all a, p E K. Thus each i,, (a, p E K), restricted to YD], is an elementary embedding of YO] in YO] such that i,,(a) = p andi,,(y) for all y E K below a or above p. Using these embeddings i,, (as in the proof of Lemma 30.19), one shows that the elements of K are indiscernibles for the model 401. Since some elements of K are regular cardinals, and some are limit cardinals, it follows that all elements of K are inaccessible cardinals in 4 0 1 . In the above argument, it was not necessary that K be a measurable cardinal, only that K be measurable in YO]. Thus we have proved :

Lemma 31.19. Let (31.40)

K

be a measurable cardinal, and assume that.

For some y < K , there is D G P ( y ) such that YO] k D is a normal measure on y.

Then there are arbitrarily large successor cardinals that are inaccessible in 4 0 1 .

We have proved in Lemma 31.18 that if U is a nonprincipal K-complete ultrafilter over K, then j&) < io&), where ioo is the natural embedding of LED] in Ult',")(YD]). We can now prove a stronger statement:

Lemma 31.20. If there is a K-complete nonprincipal ultrafilter U over K such that 2 i0&), then (31.40) holds.

j&)

Proof. Let us work in the model M = Ult,(V). The cardinal&) is measurable . F be the collection of all while ioo(tc) has cofinality w, and so i0&) < j ( ~ )Let

32.

381

LARGE CARDINALS BELOW A MEASURABLE CARDINAL

subsets X of i0&) such that X 2 { i O n ( K ) : n 2 no} for some n o . Using Lemma 31.8, we proceed as in the proof of Lemma 31.14 to show that

YF]1 F n q F 3 is a normal measure on ioO(.) Thus (31.40) holds in M for j ( K ) . Sincej is an elementary embedding, (31.40) holds in V for K. Corollary. ff K is a measurable cardinal and 2" > K + , then (31.40) holds. Consequently, it is impossible to prove consistency of 2" > K + relative t o ZFC + '' K is a measurable cardinal." Proof. On the one hand, I i0&) 1 = ( K + ) " ~ I IK + ; on the other hand, if LI is any K-complete nonprincipal ultrafilter over K , we have jU(K) > 2" > K + . W

We shall return to the problem of 2" > K + at a measurable cardinal in Section 36. 32. LARGE CARDINALS BELOW A MEASURABLE CARDINAL

Weakly Compact Cardinals In Section 29 we defined weakly compact cardinals as the uncountable cardinals satisfying the partition property K -,( K) '. We have shown that weakly compact cardinals are inaccessible, and in fact, exactly those inaccessible cardinals that have the tree property, that is, every tree of height K whose levels have size less than K has a branch of length K. We have shown that every measurable cardinal is weakly compact and moreover if D is a normal measure on K, then the set of all weakly compact cardinals below K is in D. We shall now investigate weakly compact cardinals in some detail. First we give a characterization of weakly compact cardinals which explains the name "weakly compact." This aspect of weakly compact cardinals has, as many other large cardinal properties, motivation in model theory. We shall consider infinitary languages which are generalizations of the ordinary first order language. Let K be an infinite cardinal number. The language YK*,, consists of: (a) K variables; (b) various relation, function, and constant symbols; (c) logical connectives and the infinitary connectives a < K (infinite disjunction and conjunction); (d) quantifiers 30, Vv.

v5<.q S ,AS<.

'pc for

The language Y K , is K like Y K Sexcept u that it also contains infinitary quantifiers: (e)

+<.

vc', Vt<. vc for a < K.

5.

382

MEASURABLE CARDINALS

The interpretation of the infinitary symbols of Y K ,isKthe obvious generalization of the finitary case where cpci is c p o ~ ~ ~ ~ ~ c 3<<,, p n - ul crstands for' 3u0 ... u , , - ~ ,etc.). The language Y,,,uis just the language of the first order predicate calculus. The finitary language Yu,usatisfies the compactness theorem: If C is a set of sentences such that every finite S E C has a model, then C has a model. Let us (or PK,J satisfies the weak compactness theorem if say that the language .YK,K whenever Z is a set of sentences of 2'K,K ( Y K .such J that IZ I 2 K and that every S G C with IS I < K has a model, then C has a model. Clearly, if Y K , K satisfies the weak compactness theorem, then so does YK,"because

v<<,,

Y K . 0 ,

E

9 K . K

-

Lemma 32.1.

(a) If K is a weakly compact cardinal, then the language 2'K,K satis$es the weak compactness theorem. (b) If K is an inaccessible cardinal and if PK,+, satisfies the weak compactness theorem, then K is weakly compact. Proof: (a) The proof of the weak compactness theorem for Y K is s very K much Let C be a set of like the standard proof of the compactness theorem for 9u,u. of size K such that if S c C and I S I < K, then S has a model. sentences of 9K,K Let us assume that the language 2' = Y K ,has K only the symbols that occur in C;thus 1 2 ' 1 = K. First we extend the language as follows: For each formula cp with free variables q ,l < CL, we introduce new constant symbols cr, < a (Skolem constants); let 9') be the extended language. Then we d o the same for each and obtain LF2)1 9'). We d o this for each n < o,and then formula cp of 9") P"). Since K is inaccessible, it follows that I Y * I = K. Y * has let Y * = the property that for each formula cp with free variables o, 5 < a, there are in 9* constant symbols cr, < a (which d o not occur in cp). For each q ( q , ...)<., let a,,, be the sentence (a Skolern sentence)

<

u;=

<

(32.1)

3

U < cpp(U<,

...) < < , - , c p ( C f , ...)<<,

and let C* = C u {a, : cp a formula of Y * } . Note that if S c C* and IS I < K, then S has a model: Take a model for S n C (for 9) and then expand it to a model for Y * by interpreting the Skolem constants so that each sentence (32.1) is true. Let {a, : a < K}be an enumeration of all the sentences in Y * .Let (T, G ) be the binary K-tree consisting of all t : y (0,l}, y < K, such that:

(32.2) there exists a model 21 of C n {a, : a E dom(t)} such that Va E dom(t) t ( a )= 1

iff

PI I=a,

32.

383

LARGE CARDINALS BELOW A MEASURABLE CARDINAL

Since K has the tree property, there exists a branch B in T of length A

= {a, : [ ( a )=

= 72

Let

N,

be the

1 for some t E B}

Clearly, X* c A. Let A, be the set of all constant terms of Y * ,and let equivalence relation over A, defined by 7,

K.

iff

(zl

= T ~ ) EA

and let A = A , / = . We make A into a model PI for Y * as follows: and similarly for function and constant symbols. The proof is then completed by showing that PI is a model for A (and hence for X). The proof of (32.3)

81ba

iff

~

E

A

is done by induction on the number of quantifier blocks in a: If a = 3,,, vg q ( q , ...), then by induction hypothesis we have PI 1 ...)<<, iff a(c7, . . . ) e < m E A and (32.3) follows. (b) Let K be inaccessible and assume that the language Y K S satisfies u the weak compactness theorem. We shall show that K has the tree property. Let (T, < ) be a tree of height K such that each level of T has size < K. Let us consider the YK.u language with one unary predicate B and constant symbols cx, for all x E T. Let X be the following set of sentences:

(32.4) 1(B(c,)A B(c,))

v

B(c,)

for all x, y

E

T that are incomparable

for all a < K, where U , is the ath level of T

XEU.

(X says that B is branch in T of length K ) . If S c X and IS 1 < K, then we get a model for S by taking a sufficiently large initial segment of T and some branch Z has a model, in this segment. By the weak compactness theorem for =YK,@, which obviously yields a branch of length K. W The weak compactness property for 2'K,K alone does not imply that inaccessible; but it does imply that K is at least weakly inaccessible:

K

is

K = I + is a successor cardinal, then the weak compactness theorem for false. [Consider constants c., a I K, a binary relation < and a ternary relation R . Consider the sentences saying that (a) < is a linear ordering; (b) c. < c,9; and (c)eachf, is a function, where f x ( y ) = z stands for R(x, y, z). Let Z consist of these sentences, the sentence z < x -P 3y R ( x , y , z ) (saying that ran(fx) 2 { z : z < x}), and the infinitary sentence R ( x , y , z ) -+ Vc<*(y = cc) (saying that dom(fx) G {cc: ( < 1). Show that each S c Z, IS1 < 1, has a model, but Z does not.]

Exercise 32.1. If gPK,o is

384

5.

MEASURABLE CARDINALS

Exercise 323. If K is a singular cardinal, then the weak compactness theorem for YK,,is false. [Let A c K be a cofinal subset of size < K. Consider constants c., a 5 K, and a linear ordering <. There is Z that says on the one hand that {c. : a E A} is cofinal in the universe, and on the other hand that c. > all c, , a < K ; and each S c Z, I S I < K, has a model.]

We have given three different characterizations of weak compactness of inaccessible cardinals: the partition property K -,( K ) ’ , the tree property, and K. 32.3 and 32.4 give yet another equivthe weak compactness of Y K sExercises alence; Exercise 32.5 is just a formulation of what was proved in the course of proof of Lemma 29.6. Exercise 32.3: If K is weakly compact and if (23,G ) is a K-complete algebra of subsets of K such that [.I[ = K, then every K-complete filter F on 1 can be extended to a K-complete ultrafilter on 1. [Consider constants cx for all X E A?, and a unary predicate U.Let Z be the following set of L,,,-sentences: 1U(c,), U ( c x )v U(c,- *) for all X E .a, U ( C , ) + U ( c , ) for all X Y in .a, U ( c x )for all X E F, and Ax, U ( c x )+ U ( c n d ) for all .d c .a such that I .d I < K. Show that Z has a model.] Exercise32.4. If K is inaccessible and if every K-complete filter on any K-complete algebra .4of subsets of K such that 19I = K can be extended to a K-complete ultrafilter, then K is weakly compact. [Proceed as in the proof of Lemma 29.7.1 Exercise 32.5. If ( P , < ) is a linearly ordered set of size K, and K is weakly compact, then there is a subset W c P of size K that is either well-ordered or conversely well-ordered by <.

The characterization of weak compactness from Exercise 32.3 can be used to obtain the following interesting result : Exercise 32.6.* If K is weakly compact and if 1 ( K ’ ) ~I = K, then O x exists. [Let .d be the least K-complete algebra of subsets of K closed under inverses of

constructible functionsf: K .+ K ; we have 1.11 = K . Let U be a K-complete ultrafilter on .d containing all final segments {a : a. 5 a < K } . U n L is a nonprincipal Lultrafilter, and U l f , nL(L)is well-founded. Thus there is a nontrivial elementary embedding of L in L.]

In Theorem 77 below, we prove equivalence of weak compactness with lli-indescribability, which will enable us to obtain further interesting results on weakly compact cardinals. Let n > 0 be a natural number and let us consider the nth order predicate calculus. There are variables of orders 1, 2, ..., n, and the quantifiers are applied to variables of all orders. An nth order formula contains, in addition to first order symbols and higher order quantifiers, predicates X ( z ) where X and z are variables of order k + 1 and k respectively (for any k c n). Satisfaction for an nth order formula in a model P I = ( A , P, . . . , J ..., c, ...) is defined as

32.

385

LARGE CARDINALS BELOW A MEASURABLE CARDINAL

follows: Variables of first order are interpreted as elements of the set A, variables of second order as elements of P ( A ) (as subsets of A). etc.; variables of order n are interpreted as elements of P " - ' ( A ) . The predicate X(z)is interpreted as z E X.A ll;forrnula is a formula of order n + 1 of the form

vx 3Y ... $

(32.5)

- 7 -

111

quantifiers

where X, Y, . . . are ( n + 1)th order variables and I) is such that all quantified variables are of order at most n. Similarly, a Z; formula is as in (32.5), but with 3 and V interchanged. We shall often exhibit a sentence cr and claim that it is ll; (or Z); although it is only equioalenr to a l: (or Xi) sentence, in the following sense: We are considering a specific type of models in which r~ is interpreted (e.g., the models ( V , , E)) and there is a ll; (or Z): sentence d such that the equivalence 0 - 6 holds in all these models. Note that every first order formula is equivalent to some I I g formula (and also to some xi formula). A cardinal K is llz-indescribable if whenever U c_ V, and cr is a ll; sentence such that ( V , , E, U ) k 6,then for somea < K, (K, E, U n V , ) k cr. Here U is considered a unary predicate. Lemma 32.2. Every measurable cardinal is lli-indescribable.

Proofi Let K be a measurable cardinal, let U c V, and let cr be a ll: sentence of the (third order) language {E, U}. Let us assume that (V,, E, U ) k cr. We have cr = VX cp(X)where X is a third order variable and q(X)contains only second and first order quantifiers. Thus (32.6)

vxc_VK/,,l

( V K + 1 7 E 7 x ? K 9 U ) ' $

where $ is the (first order) sentence obtained from cp by replacing the first order quantifiers by the restricted quantifiers Vx E V, and 3x E V,. Now let D be a normal measure on K and let M = U l r J V ) . Since (V,, l)M = V,, we know that (32.6) holds also in M.Using the fact that V , is represented in the ultrapower by the function a w V , , V , , by a w V,, and U by a- U n V , , we conclude that for almost all a,

(32.7)

v x c v,,, ( K + ~ , EX, , K, I/ n K)

k$

Then, translating (32.7) back into the third order language, we obtain

(K,

€7

U n K)

for almost all, and hence for some, a < K.

0

H

Exercise 32.7. The least measurable cardinal is Xi-describable.

[Consider the sentence 3U ( U is a K-complete nonprincipal ultrafilter over

K).]

5.

386

MEASURABLE C A R D I N A L S

Lemma 32.3. I f K is not inaccessible, then it is describable by a first order sentence, i.e., lli-describable .for some m. Pro05 Let K be a singular cardinal, and letfbe a function with dom(f) = A < K and ran(f) cofinal in K. Let U 1 =,f and U , = {A), and let o be the first order sentence saying that U 2 is nonempty and that the unique element of U , is the domain of U , . Clearly, K is describable in the sense that ( V , , E, U 1 , U,) k 0 and there is no a < K such that (V,, E, U 1 n V,, U 2 n V,) k 0 . It is routine to find a single U G V, and an (E, U)-sentence d attesting to describability of K. If K I 2 A for some 1 < K, there is a functionfthat maps P(A) onto K. We let U 1 =fand U , = {P(A)};then K is described by the same sentence as above. Finally, K = w is describable as follows: ( V , , E) k Vx 3y x E y .

The converse of Lemma 32.3 is also true: Exercise 32.8. Every inaccessible cardinal is lT:-indescribable, for all m. [Let U G V , . The model (V', E. U ) has a countable elementary submodel M o . Let a. Y be such that M o G V&. For each n, let M , + I be an elementary submodel of ( V , , E, U ) such that V.," M,, and let M,, G V,"+ Let a = limn,,,, a n ;then V, is an elementary submodel of ( V , , E, V).]

-=

,.

Theorem 77 (Hanf-Scott). A cardinal weakly compact.

K

is ll:-indescribable fi and only if i t is

Corollary. Every weakly cornpact cardinal Mahlo cardinals below K is stationary.

K

is a Mahlo cardinal, and the set of

Proofofcorollary. Let C G K be a closed unbounded set. Since K is inaccessible, (V,, E, C) satisfies the following Il: sentence

1 3 F ( F is a function from some A < K cofinally into K ) and C is unbounded in K.

By ll:-indescribability, there exists a regular a < K such that C n a is unbounded in a ; hence a E C . Thus K is Mahlo. Being Mahlo is also expressible by a Il: sentence: VX(if X is closed unbounded, then 3 regular a in X) and so the same argument as above shows that there is a stationary set of Mahlo cardinals below K. The proof that a weakly compact cardinal is Hi-indescribable uses the following property of weakly compact cardinals.

Lemma 32.4. lf K is a weakly compact cardinal, thenfor every U G V,, the model ( V , , E, U ) has a transitive elementary extension ( M , E, U')such that K E M . [It will follow from the proof of Theorem 77 that the property in Lemma 32.4 is yet another characterization of weakly compact cardinals.]

32.

387

LARGE CARDINALS BELOW A MEASURABLE CARDINAL

Proof: Let X be the set ofall L,.,sentences true in the model (V,, plus the sentences c' is an ordinal

E,

U,x

)

~

(all ci < K )

c>a

Clearly 1 X I = K, and if S c X is such that I S I < K, then S has a model (namely V,, where the constant c can be interpreted as some ordinal greater than all the a's mentioned in S). Hence X has a model 91 = (A, E, U", we may assume that AzV,, En(V,xV,)=E, U"nV,=U and x " = x for all X E V , . Moreover, V , < (A, E , U') because 91 satisfies all formulas true in V, of all x E V,. If we show that the model ( A , E ) is well-founded, then the lemma follows. Here we make use of the expressive power of the infinitary language L,,: We consider the sentence

(32.8)

l l L 1 o 3L'i

"'

30,

.'*

A

(U,+1

E U,)

n E u

The sentence (32.8) holds in a model 91 = (A, E) if and only if 91 is well founded. Since X contains the sentence (32.8), every model of I: is well founded. W Exercise 32.9. If K is weakly compact and if A G K is not constructible, then for some a < K, A n a is not constructible. [Let A G K be such that A n a E L for all a < K . By Lemma 32.4 there is transitive (M, E, A ' ) > ( V , , E, A ) such that K E M . Consider the sentence Va 3x(x is constructible and x = A n a).]

Proof of Theorem 7 7 . First we show that every l7:-indescribable cardinal is weakly compact. If K is Il;-indescribable, then by Lemma 32.3, K is inaccessible, and it suffices to show that K has the tree property. In fact, by the proof of Lemma 29.6 (or by the proof of Lemma 32.1), it suffices to show that every binary K-tree has a branch of length K . Let ( T , G ) be a binary K-tree, consisting of sequences t: y -, (0, l), y < K. sentence Clearly, for every a < K, the model ( V , , E, T n V , ) satisfies the

(32.9)

3E(E c T and E is a branch of unbounded length)

Namely, let E = {t I tJ : ( < a } where t is any t E T with domain a. Since K is l l i indescribable, the sentence (32.9) holds in (V,, E, T ) and hence T has a branch of length K. To show that a weakly compact cardinal is lli-indescribable, we use Lemma 32.4. Let K be weakly compact, let [IG V, and let CJ be a Il: sentence true in ( V , , E, U). We have 0 = VX q(X)where X is a second order variable and cp has only first order quantifiers.

~

5.

388

K E

MEASURABLE CARDINALS

Let (M, E, V ’ ) be a transitive elementary extension of (V,, M. Since

(vxc K)

?K‘(

€9

‘)

E,

V ) such that

p(’)

and V y = VK,we have

( M , E, V ‘ ) b [(VX s

K) ( K , E, u’

K )

v(x)l

Therefore, ( M , E,

v‘)b [ ~ ( vsxK) ( K , E, U’ n a ) b dx)]

and so

(L V ) b [*(VX s K) (K, €3

6

u na) b d X ) l

Hence for some a < K, ( v , , E , W b

ll:-indescribability can be used to obtain various results on weakly compact cardinals. For instance: Exercise 32.10. If K is weakly compact and if S E K is stationary, then there is a regular uncountable 1< K such that S n 1is stationary in 1.(Compare this with (24.17).) [LLK is regular” is expressible by a ll: sentence in (V,, E) and so is “K is uncountable.” “S is stationary” is ni in (V,, E, S): For every C if C is closed unbounded, then

s n C # 0.1 We recall that a complete Boolean algebra B is countably generated if there is a countable set A c B such that B is the least complete subalgebra containing A. Exercise 32.11. If K is weakly compact, then there is no countably generated complete Boolean algebra B such that I B I = K. [Assume B is such. Note that sat(B) = K. We may assume that B = (K, +, ., - ); let A c K be a countable set of generators. Let U , be the set of all pairs (u. x) such that u E K, x c K, 1x1 < K, and u = x, let U 2= {A}. Let CJ be the conjunction of these sentences: (a) B is a Boolean algebra and B z A (first order), (b) Vx 3u (if x E K, then u = x) (first order), and (c) VX[if X c K and X is a partition of B, then 3x(x = X)] (here x is a first order variable; the sentence (c) is ll;).Since (V,,E, U I , U , ) satisfies CJ, there is a < K such that ( V , , E, U , n V,, U 2 n V , ) C CJ. Then (a, +, ., -) is a complete subalgebra of B containing A.]

A similar argument can be used to prove that if K is a weakly compact cardinal, then there is no simple complete Boolean algebra of size K. Unlike measurable (or Ramsey)cardinals, weakly compact cardinals do not contradict the axiom of constructibility (assuming that their existence is consistent with ZFC). In fact, we shall show that every weakly compact cardinal is weakly compact in L.

32.

LARGE CARDINALS BELOW A MEASURABLE CARDINAL

389

Lemma 32.5. Let K be a regular cardinal and assume that L k K is inaccessible. If has the tree property, then for every u < K ' , there exists an elementary embedding j : L+,+ L, vor some /I) such that j ( K ) > K and K is the least ordinal moved by i[It can be proved that the assumption that K is inaccessible in L is redundant and follows from the other assumptions.] K

Proof: Let u < K ' . We may assume that ( L a ,E) is a model of ZF- (by taking a larger a' that has this property). Let ( X , : 5 < K ) be an enumeration of P(K) n L * For each t : y {O, l}, y < K, let -+

x f = n Y= 5
where

6 = X , if t ( 5 ) = 1, and

(32.10)

T = {t E L : t

E

=K

- X, if t ( l )= 0. Let

y{O, 1) for some y < K and IX,I = K}

(T, c-) is a binary tree, and since for every y < K, 12yJ L < K, the tree T has height K and every level of T has size < K. By the tree property, T has a branch B of length K ; let F: K -+ {O, 1) be the function F = { t : t E B}. Let U be the filter on P ( K )n L, generated by {X<: F ( < ) = 1) u { K - X, : F ( < ) = 0). Clearly, U is a nonprincipal b-ultrafilter, is L,-K-complete, and moreover, the intersection of less than K elements of U is nonempty. Now we consider the ultrapower Ult,(L,); that is, we use functions f: K + L, such t h a t f e L, . The ultrapower is well-founded and thus isomorphic to some L,. The natural embedding j,: L, -,Ult,(L) has the required properties. H

u

Lemma 32.6. Let K be an inaccessible cardinal in L and assume that for every u < ( K ' ) ~ , there exists an elementary embeddingj : La+ L, vor some p) such that K is the least ordinal moved by j . Then K is weakly compact in L. Corollary 1. If

K

is weakly compact, then

K

is weakly compact in L

Proof: This is an immediate consequence of Lemma 32.5 and 32.6. H

Corollary 1 can be proved directly, using Lemma 32.4: Exercise 32.12. If K is weakly compact, then K has the tree property in L (and hence L k K is weakly compact). [In L let T = (K. < T ) be a tree of height K such that each level of T has size less than K . T has a branch B (in the universe). By Exercise 32.9, B is constructible.]

Cwollary 2. If there is no Aronszajn w,-tree, then K, is weakly compact in L. Proof: We have shown in Section 29 that if K, has the tree property then K, is inaccessible in L Now. the statement follows from Lemmas 32.5 and 32.6. H

390

5.

MEASURABLE CARDINALS

Corollary 3. Every Silver indiscernible is a weakly compact cardinal in L Proof: It suffices to show that i,, the o t h Silver indiscernible, is weakly compact in L. Let j be an order-preserving map from the class of Silver indiscernibles into itself such that j(in)= in for all n < o,and j(i,) > i,, and let us extend j to an elementary embedding of L into L. By remarkability, i , is the least ordinal moved. By Lemma 32.6, i, is weakly compact in L Proof of Lemma 32.6. We shall prove that in Levery binary tree of height K has a branch of length K . In L let us consider a tree (T, G )of height K whose elements on level y < K are functions t : y + (0, I}. If t E T, then t I y E T for each y E dom(t). Since K is inaccessible in I, we have t E L, for all t E Tand so T c L;let a < ( K + ) be ~ such that T E La, that Lais a model of ZF- and that L, k (T, G ) is a tree of height K . Let j be an elementary embedding of L,,into some L, such that K is the least ordinal moved. Then j ( x ) = x for all x E L,, and L, k j ( T ) is a tree of height greater than K . Thus let f E j ( T ) be such that dom(f )= K (an element of the K t h level of j ( T ) ) . We shall show that f I y E T for all y < IC, and hence B = { f 1 y : y < K } is a branch in T of length K (and clearly B is constructible). If y < K , let t = f I y ; clearly t E ~ ( T )Now . because t E L, we have j ( t ) = t ; hence j ( t ) E j ( T ) and it follows that t E T. 1

We remark that if K is weakly compact, then although K is weakly compact in L, it is not necessarily true that K is weakly compact in any transitive model M of ZFC containing all ordinals (a counterexample is known). One way to prove that a measurable cardinal K is weakly compact is to use a normal measure D on K to prove K + ( K ) ~ to ; show that every partition f [K]’ -,(0, 1) has a homogeneous set H of D-measure one. The homogeneous set H has not only size K , but because H E D, H is stationary. Let us call an uncountable cardinal K inegable if every partitionf [ K ] ’ + {0, 1) has a stationary homogeneous set H. Every measurable cardinal is ineffable, and every ineffable cardinal is weakly compact. The following characterization of ineffable cardinals is useful: Lemma 32.7. A cardinal K i s ineffable ifand only ifit has thefollowing property: If ( A , : c1 < K ) is such that A, G a for all a < K , then there is A G K such that {a E K : A, = A n a ) is stationary in K . Proofi (a) Let K be ineffable, and let A,, a < K , be such that A, G a for all a. Let < be the lexicographic ordering of subsets of K , i.e., X < Y whenever the least element of the symmetric difference XAY lies in Y. L e t t [ K ] ’ + (0, 1) be as follows:

32.

391

LARGE CARDINALS BELOW A MEASURABLE CARDINAL

By ineffability,f has a stationary homogeneous set H.Let us consider the case when A, 5 A, for all a < b in H ;the other case is handled similarly. We show, by induction on t < K, that for every t there is q ( t ) such that 5 E A, for all a 2 q ( < )in H or t 4 A, for all a 2 q ( t )in H.If this is true for all v < 5, then for all a 2 sup{q(v): v < t}in H,A, n t is the same; then (because A, 5 A, for all a < p in H) once is in A, for one such a, then is in A, for all greater a E H.Hence we find q(<)that has the claimed property. Now let C be the closed unbounded set

<

<

and let S be the stationary set H n C. If a < p are in S, then A, = A, n a, and we let A = (J {A, : a E S). (b) Assume that the condition holds and let$ [K]’ + {O, 1) be a partition of [K]’. Let us define sets A, 5 a, a < K, as follows: A, = {< < a :f(t,a) = 1)

Let A G K be such that the set S = {a : A, = A n a}is stationary. Either S n A or S - A is a stationary set; let us consider the case when S n A is stationary, and let H = S n A. The other case is handled similarly. If a < p are in H, then because p E S and a E A, we have a E A, and thereforef(a, p) = 1. Thus H is homogeneous for$ Corollary 1. I f

K

is ineffable, then

K

is ineffable in L.

Proo$ Let ( A , : a < K ) be a constructible sequence such that A , G a for all a. Since K is ineffable, there is A C_ K such that S = {a : A, = A n a}is stationary. For each a < K, A n a is constructible, and so, by Exercise 32.9, A is constructible. Hence S is also constructible; and because S is stationary, clearly L k S is stationary. Thus K is ineffable in L Corollary 2. Every Silver indiscernible is ineffable in L. Proof: It suffices to prove it for one Silver indiscernible. Let K be an uncountable cardinal (or K = i J and let j : L L be an elementary embedding such that K is the least ordinal moved. Let (A, : a < K ) E L be such that A, 5 a for all a. For each a < K. we have j ( A , ) = A,, and so j ( ( A , : a < K)) = ( A , : a < j ( K ) ) for some A,, K I a < j ( ~ )such , that A, G a for all a < j ( ~ )Let . A = A,; we shall prove that S = {a < K : A, = A n a} is stationary in L. Let C E L be a closed unbounded subset of K. Since C = j ( C ) n K and j ( C ) is closed unbounded in j ( ~ )we , have K E j ( C ) . We also have A = j ( A ) n K and hence K E j ( S ) = {a < j ( K ) : A, = j ( A ) n a}. Thus j ( C ) n j ( S ) # 0 and hence C n S is nonempty. Therefore S is stationary in L.

As for the relative strength of ineffable and Ramsey cardinals: Ramsey cardinals imply that 0‘ exists, while ineffable cardinals are consistent with

392

5.

MEASURABLE CARDINALS

V = L One can show that if Ramsey cardinals exist, then the least ineffable cardinal is smaller than the least Ramsey cardinal. However, it is not true that every Ramsey cardinal is ineffable: Using Lemma 32.7, one can show that ineffable cardinals are ll$indescribable; but the least Ramsey cardinal is llidescribable.

Partition Cardinals If X is a set of ordinals and n < w, [XI" denotes the set of all increasing n-sequences in X ;[XI'" is the set uF'l [XI". In (29.7) we have introduced the partition property K + (a);'"

where K is an infinite cardinal number, m is a cardinal number 2 2, and a is a limit ordinal: Every function f: [K]'" + m has a homogeneous set H E K of order type a ; that is, for every n < o,f is constant on [HI". Let a be a limit ordinal number and let us define (32.1 1 )

q, = the least

such that

K

K

-P

(a)'"

Note that a is Ramsey just in case q, = a.

Theorem 78. (a) Each qDlis an inaccessible cardinal, and every model PI 2 q , whose language has less than q, symbols has a set of indiscernibles of order type a. (b) .lf a c p, then q, < q a . (c) lf qw exists, then there exists a weakly compact cardinal K < qu. (d) If qw, exists, then 0" exists. (e) If K + (a)'",then L k K --+ (a)'". Our first step toward the proof of Theorem 78 is the following easy lemma:

Lemma 32.8. If

K + (a)<'", then K

-P

(a);&.

Note that by Lemma 29.8, every model PI 2 indiscernibles.

K

of a countable language

Pro05 Let f be a partition,f: [K]'" + "(0, 1). For each n and for each k < w, let h.k:[K]"+ (0,l} be as follows:

h.k(C(1,. . ., a,)

= h(k),

-= w, let fn =f

where h =h(ai,... an)

Let II be a one-to-one correspondence between w and o x w such that if n(m) = (n, k), then m 2 n ; for each m, let g,: [K]" + (0,1) be the partition defined by gm(air.-.,am)

where (n, k ) = z ( m ) .

=fn.k(al,

* - - )

an)

32.

LARGE CARDINALS BELOW A MEASURABLE CARDINAL

393

By the assumption, there exists H G K of order type a which is homogeneous for all g m . We claim that H is homogeneous for 1: If not, then fn(al, . . ., a,,) #fn(pl, . . ., p,,) for some a's and p's in H. Then for some k , f n k ( a l , .. ., a,,) #fnk(p1, ..., p,,). Contrary to the assumption that H is homogeneous for g m , where n(m)= (n,k ) . The following lemma is analogous to (but more difficult to prove than) Lemma 29.9 (which is the special case for qa = a): Lemma 32.9. For eoery

< qa, qX-+ (a): ".

K

An immediate consequence of this lemma is that qa is a strong limit cardinal: If K < qz, then 2" f , (a): by Lemma 29.2, and so 2" < qa . Proof. Let K < q a , and letf [qa]"" + K. We wish to find a homogeneous set for of order type a. Since K < q X , there exists g: [~]<"'+{0, I} that has no homogeneous set of order-type a. For each n, let f,,= f I [q.]" and g,, = g I [K]", and let P I be the model

fi

91 = (b,?c

fn,

gfl),,=0,1....

By Lemma 32.8, and by Lemma 29.8, the model P I has a set of indiscernibles H of order-type a. We shall show that H is homogeneous forf. It suffices to show that for each n, the formula (32.12)

fn(al,

a e - 9

an) = f n ( P 1 7

. . * 3

Pn)

holds in P I for any increasing sequence a l < . . . < a,, < PI < . . . < p,, of indiscernibles: Then if a1 < . .. < a,, and a; < ... < a: are arbitrary in H , we choose p1
Ye = f ( u , ) for all 5 < a, and let G = {ys : < a}. By indiscernibility, and since y o > y 1 > ..' > ye > ... is impossible, we have yo < y1 < ... < yc < .... We shall reach a contradiction by showing that G is homogeneous for g. For each k, consider the formula

<

(32.13)

. . %f(U<J)

.J-(u,,)) By indiscernibility, either (32.13) or its negation holds for all increasing sequences t1< ... <
= S(f(U",)?

* '

5.

394

MEASURABLE CARDINALS

Proof of Theorem 78a. We know already that qu is a strong limit cardinal. It suffices to prove that qu is regular since the second part of the assertion follows from Lemma 29.8. Let us assume that q, is singular and that K = cf(q,); let q, = Iim,-+,A,. For each v < K, let f":[IY]'" + (0, 1) be such that f" has no homogeneous set of order type a. For each n, let fi = f' I [A,]"; let PI be the model

"= ( & a ?

€9

' V ~ f ~ ) V < K , f l = ~ , 1 ,

...

Since q, is a strong limit and K < qm,the model PI has a set of indiscernibles H of order type a. Let v be such that I, is greater than the least element of H. Then by indiscernibility, all elements of H are smaller than A,. Since the functionf' takes only two values, it follows that for each n, it is the equality that holds for all increasing sequences at < ... < a, < c ... < /?,, in H, and not its negation. Hence H is homogeneous forf", contrary to the assumption onf'. H Proof of Theorem 78b. Let a < P be limit ordinals, and let us assume that q, = q p . For each 5 < q,, there exists a functionf;: [t]'" (0, l } that has no homogeneous subset of of order-type ct. Let us define 9 : [qp]<" 4(0, 1) by

<

g(t1, ...) t n )

=&.(t1,

t n - 1)

<

Now if H is homogeneous for g, then for each 5 E H,H n is homogeneous for fs. Hence the order-type of each H n is less than a, and therefore the order-type of H is at most a, which is less than P. A contradiction. H

<

Before giving the proof of (c), let me formulate a lemma that is an abstract version of something we proved earlier: Lemma 32.10. Let M and A! be transitiue models of ZFC and let j : M 4 N be a nontrivial elementary embedding; let K be the least ordinal moved. If P ' ( ~ ) = P N ( ~then ) , K i s an ineffable cardinal in M . Proof: [Note that P M ( ~G)P N ( t i )holds always, because if X G K is in M, then X = . j ( X ) n K and hence X E N.] The proof is virtually given in Corollary 2 of Lemma 32.7. Let ( A , : a < K ) E M be such that A, G a for all a. We have j ( A , ) = A, for all a < K . and hencej((A, : a < K ) ) = ( A , : a c j ( K ) ) (for some A,, K 5 a < j ( ~ ) ) . The set A, is in M and witnesses ineffability of K in M. Proof of Theorem 78c. Let h, , cp E Form, be Skolem functions for the language of set theory, and let us consider the model

{E}

PI = (v,,,

€9

h:),€For,,,

32.

LARGE CARDINALS BELOW A MEASURABLE CARDINAL

3%

where for each q,h: is a Skolem function for cp in (V,,", E). The model '21 has a set of indiscernibles l of order type o.Let B be the closure of l under the Skolem functions h f . Let us consider some nontrivial order-preserving mapping of H into H. Using the Skolem functions, we extend this mapping (in the unique way) to a nontrivial elementary embedding of B into B. Let M be the transitive set isomorphic to B and let j: M + M be the corresponding nontrivial elementary embedding. is a model of ZFC and thus M is a transitive Since qu is inaccessible, model of ZFC. Using Lemma 32.10, we conclude that M b there exists an ineffable cardinal; and therefore V,_ 1 there exists an ineffable cardinal. Hence there is a weakly compact (even ineffable) cardinal below q w . Exercise 32.13. If M is a transitive model of ZFC and if j: M 4 M is a nontrivial elementary embedding, then the least ordinal K moved by j is nz-indescribable in M , for all n and m. [If U c V,", then U = j ( U ) n V,". If CT is a l: sentence and M k [(V.. E, U ) C 61. then M C [(3a < j ( K ) ) ( K , E, j ( U ) n V.) k u].]

Thus the proof of (c) shows also that if qluexists, then there is K < qlusuch that K is nL-indescribable for all n and m. Another consequence of Exercise 32.13 is that every Silver indiscernible is n;-indescribable in L. Exercise 32.14. The cardinal qcois not weakly compact. [q," is n:-describable.]

Proof of Theorem 89d. I f K + ( w ~ )
being a definable set in indiscernible.)

L

+ (a)'".(And (v")~, is countable and in fact smaller than the least Silver

Proof. Let K be any uncountable cardinal. Iff: [K]'" --* (0, 1) is in L, then there is a Skolem term t and Silver indiscernibles a < - . . < a,, < K Ip I < . . . < p, such that f = [(a,, . .., a,, PI, , .., p,). I leave it to the reader to verify that the set H of all Silver indiscernibles between a, and K is homogeneous f o r t Hence L k K+(W)'". Proof of Lemma 32.11. Let T be the set of all finite increasing sequences t = (a,,, ..., a , , - l ) in K such that for every k s n , f is constant on

5.

3%

MEASURABLE CARDINALS

[ { a o , .. . , an-I}]k, and let us consider the tree (T, 3 ); clearly, T is constructible. We note that an infinite homogeneous set forfexists just in case (T, 2 )is not well-founded. However, being well-founded is an absolute property for models of ZFC; and so if the tree is not well-founded, then it is not well-founded in L, and the lemma follows. W

We conclude the discussion on partition cardinals with the following observation: Exercise 32.15. If K is a Ramsey cardinal, then K is the Kth weakly compact cardinal. [Given a c K, find a weakly compact cardinal L such that a I1 < K, as follows.

Consider a Skolem expansion YI of the model (V,, E, and proceed as in the proof of Theorem 78c to show that V, C 3 ineffable cardinal 1 2 a.]

A Few Remarks on Jhsson Cardinals

Jonsson cardinals were defined in Section 29; a cardinal K is Jonsson if every model (of countable language) of size K has a proper elementary submodel of size K. Here is what is known about Jonsson cardinals: Na Hn,n < w, can be a Jonsson cardinal (for KO,see Exercise 29.12). If GCH holds, then no successor cardinal is Jonsson. The least Jonsson cardinal is either weakly inaccessible or has cofinality w. Every Ramsey cardinal is a Jonsson cardinal (by Lemma 29.10). Conversely, the method of Section 31 has been used to show that in LEU], every Jonsson cardinal is Ramsey. In Chapter 6, we shall present a model that has a Jonsson cardinal of cofinality o. It is not known whether &can be a Jonsson cardinal; however, if it is, then K, is measurable in some transitive model M. We shall prove only one, rather easy fact about Jonsson cardinals, namely that they imply that 0” exists:

Lemma 32.12. b t j : I!, ordinal moved b y j . If y

-,Lp be an elementary embedding and let y be the least

-= la 1,

then 0“ exists.

Corollary 1. If there is a Jonsson cardinal, then O* exists. Proof. Let K be a Jonsson cardinal and let us consider the model ( L , , E). Let A be an elementary submodel, of size K, such that A # L,. Let n be the transitive collapse of A ; clearly, n[A] = L. Thus j = n-I is a nontrivial elementary embedding of L, in L, . Since K is a cardinal, Lemma 32.12 applies and hence 0” exists. Let us recall that Chang’s conjecture is the statement that every model of type (K2, K,) has an elementary submodel of type (Kl, No).

Corollary 2. Chang’s conjecture implies that 0‘ exists. Proof. Consider the model (L,, o,, E), and let 41 = (A, w l n A, E) be its elementary submodel such that I A I = K 1and 10, n A I = KO.Let n be the

32.

397

LARGE CARDINALS BELOW A MEASURABLE CARDINAL

transitive collapse of A ; clearly, K[A] = L, for some a such that o 1I a < 02. Also, n[wl n A] is a countable ordinal, and hence K ( W J < wl. Thenj = II- is an elementary embedding of La in Lw2,and the least ordinal moved is countable. By Lemma 32.12, 0’ exists. W Proof of Lemma 32.12. We define an Lultrafilter D over y as follows

X E D iff y ~ j ( X ) ( X s y , X E L )

(32.14)

Note that j(X) is defined for all constructible X c y because every such X is in L,. Let us consider the ultrapower Ult &). If the ultrapower is well-founded, then we are done because then the natural embedding j D : L + UltdL) is a nontrivial elementary embedding of L in L and the lemma follows by Theorem 73. Thus we shall prove that UlfD(L) is well-founded. ...,f,,... is a counterexample to wellLet us assume that jo,fl, foundedness of the ultrapower. Each f, is a constructible function on y and (5 < y :fn+ ~f,(l)} E D for all n < a.Let 0 be a limit ordinal such that f, E I+ for all n and let S be an elementary submodel of (I+, E ) such that ( S I = I y 1, y G S, and f, E S for all n. Let K be the transitive collapse of S, n[S]= 4 ,and let gn = n(f,),for all n. E Since ~ ( 5=) 5 for all 5 < y, we see that for each 5 < y and all n, gn+ g,,(r) just in casef,. 1([) €fn(t), and hence go, gl, . . ., g, . . . is also a counterexample to well-foundedness of the ultrapower. However, since each g,, is in L and 18 I = 1 y 1 < I a 1, we have g, E L, for all n. Thus j(g,,) is defined for all n, and we have, by (32.14),

{t<

: gn +

(5) E gn(5)) E D

iff

+ 1))(Y) E (j(g.))(y)

Now we reached a contradiction because ( j g o ) ( y ) 3 (jgl)(y) 3 descending sequence. H

- a *

would be a

CHAP'T'ER 6

OTHER LARGE CARDINALS

33. COMPACT CARDINALS

In Section 32 we studied weakly compact cardinals, which are inaccessible cardinals satisfying the weak compactness theorem for the infinitary language Y,, o. If we remove the restriction on the size of sets of sentences in the model theoretic characterization of weakly compact cardinals, we obtain a considerably stronger notion of large cardinals. This notion, (strong) compactness, turns out to be much stronger than measurability. Compact cardinals can be characterized in several different ways. Let me use, as a definition, the property that is a natural generalization of the ultrafilter theorem : An uncountable regular cardinal K is compact (or, strongly compact) if for any set S, every K-complete filter over S can be extended to a K-complete ultrafilter over S. Obviously, every compact cardinal K is measurable, for any ultrafilter over K that extends the filter {X : I K - X I < K } is nonprincipal. Let us say that the language Y K . o (or 2',J satisfies the compactness theorem if whenever I: is a set ofsentences of YK. (Y,,,,) such that every S s I: with S < K has a model, then C has a model. Let A be a set of size at least K. By P,(A), we denote the set (33.1)

P , ( A ) = { P c A : /PI < K }

For each P E P K ( A ) ,let (33.2)

P = {Q E P , ( A ) : P c Q}

and let us consider the filter over P , ( A ) generated by the sets P for all P E P , ( A ) ; that is, the filter (33.3)

{X E P , ( A ) : x

2

P for some P E P , ( A ) }

If K is a regular cardinal, then the filter (33.3) is K-complete. We call U a j n e measure on P , ( A ) if U is a K-complete ultrafilter over P , ( A ) that extends the filter (33.3); i.e., if P E u for aII P E P,(A). 3 98

33.

399

COMPACT CARDINALS

Lemma 33.1. The following are equivalent, for any regular cardinal

K:

(i) For any set S , every K-complete filter over S can be extended to a complete ultrajilter over S . (ii) For any A such that I A 1 2 K, there exists a fine measure on P,(A). (iii) The language Y K ,satisfies the compactness theorem.

K-

Proofi (i) -+ (ii) is clear. (ii) + (iii): Let C be a set of sentences of Y K, ,and assume that every S c C of size less than K has a model, say ‘us.Let I/ be a fine measure on PK(C),and let us consider the ultraproduct 91 = UltU(91s:S E P K ( C ) ) .It is routine to verify that the fundamental theorem on ultraproducts holds for the language YK,, provided the ultrafilter is K-complete; in order to prove the induction step for the infinitary connective A,<. ‘p,, one uses the K-completeness of U . Thus we have, for any sentence n of Y Ko. ,

(33.4)

{S: YIs k cr} E U

iff

91 l=n

Now if r7 E C, then {O}^EU and since Vls l= n whenever S 3 r7, (33.4) implies that cr holds in 91. Hence 91 is a model of C. (iii)+ (i): Let S be a set and let F be a K-complete filter over S. Let us consider the Y K,-language + which has a unary predicate symbol X for each X E S, and a constant symbol c. Let X be the set of Y K ,sentences consisting of: (a) all sentences true in ( S , X ) x 5 s , (b) X ( c ) for all X E F. Every set of less than K sentences in C has a model: Take S as the universe, interpret each X as X and let c be some element of S that lies in every X whose name is mentioned in the given set of sentences; since F is K-complete, such c exists. Hence X has a model 91 = (A, X’, c) X E S . Let us define U c P ( S ) as follows:

(33.5)

X

E

U

iff

91 b X ( c )

It is easy to verify that U is a K-complete ultrafilter and that U 2 F: For instance, U is K-complete because if a < K and X = X,,then 91 satisfies the sentence A,. X,(c) -+ X(c). H

nt<.

The proof of (ii) + (iii) gives a stronger result: 33.f. I f K is compact, then Y K ,satisfies the compactness theorem. [Prove the fundamental theorem on ultraproducts modulo a K-complete ultrafilter for Y K ,.] Exercise

Every compact cardinal is measurable, but not every measurable cardinal is necessarily compact (although it is consistent that there is exactly one measurable cardinal which is also compact). We shall show that existence of compact

400

6.

OTHER LARGE CARDINALS

cardinals is a much stronger assumption than existence of measurable cardinals. We start with the following theorem: 'Ibeorem 79 (Vophka-Hrbatek). is no set A such that V = Y A ] .

If there exists a compact cardinal, then there

Proof: Let us assume that V = YA] for some set A. Since there is a set of ordinals A' such that YA]= QA'], we may as well assume that A is a set of ordinals. Let K be a compact cardinal, and let A 2 K be a cardinal such that A c_ A. Since K is compact, there exists a complete ultrafilter U over A+ such that IX = A+ for every X E U (let U extend the filter (X: \A+ - X I A}). Since U is tc-complete, the ultrapower Ult,( V) is well-founded, and thus can be identified with a transitive model M. As usual, iffis a function on A+, then [f]denotes the element of M represented byf. Let j = j , be the natural elementary embedding of V into M given by U . Let us now consider another version of ultrapower. Let us consider only those functions on ' 1 that assume at most A values. For these functions, we still define f = * g (mod V ) and f ~ g* (mod U ) in the usual way, and therefore obtain a model of the language of set theory, which we denote U l t ; ( V ) . The fundamental theorem holds for this version of ultrapower too: Iff, . . . are functions on A+ with Iranf 5 A, then

I

I

I

(33.6)

Ult- k q(f,...)

iff

{a : q ( f ( a ) ,...)} E U

(Check the induction step for 3.) Hence Ult- is a model of ZFC, elementarily equivalent to V . Also, since U is complete, U l t - is well-founded and thus is isomorphic to a transitive model N. Every element of N is represented by a functionfon A+ such that 1 ranf I IA. We denote by [fl- the element of N represented by 1: We also define an elementary embedding i : V + N by i ( x ) = [c,]- where c, is the constant function on A+ with value x. For every functionfon ' 1 with Iranf I I 1, we let (33.7)

It is easy to see that the definition (33.7) of k ( [ f ] - ) does not depend on the choice offrepresenting [ f l - in N, and that k is an elementary embedding of N in M. In fact, we have the commutative diagram in Fig. 33.1. If y c I + , then every function from A+ into y has at most A values, and hence [f]-= [f]for all f: A+ + y. Iff: 1' -P 1' has at most 1 values, then f:A+ 4 y for some y < 1'; it follows that i ( A + ) = limy*A+i ( y ) , and we have k(5) = 5 for all 5 < i ( A + ) . Similarly, i ( A ) = j ( A ) , and we have M = L[J(A)]= L[i(A)]= N . Now we reach a contradiction by observing that j ( A + ) =- i ( A + ) : Since the diagonal function d(a) = a represents in M an ordinal greater than each j ( y ) , y < A+, we have@+) > limy-A+j ( y ) . While N thinks that i ( A + ) is the successor of i(A), M thinks that j ( A + ) is the successor of j ( A ) (and j ( A ) = i(A)). Thus M # N, a contradiction.

33.

401

COMPACT CARDINALS

FIGURE 33.1

The proof of Theorem 79 provides this additional information: If K is compact, 1 2 K , and A G 1, then 1 ' is an ineffable cardinal in q A ] . [Let U be as above, let M = Ult,(L[A]), N = U l t ; ( L [ A ] ) , let j : L[A] -+ M , i : L[A]+ N,and let k : N + M be as above. Again, M = N, and i(A') is the least ordinal moved. By Lemma 32.10, N thinks that i(A') is ineffable; hence 'A is ineffable in L[A].] Exercise 33.2.

We shall now state a theorem which shows that, consistencywise, strong compactness is much stronger than measurability. Theorem 80 (Kunen). If there exists a compact cardinal, then for every ordinal number 8 there exists a transitive model M of ZFC with at least 8 measurable cardinals (i.e., the set of measurable cardinals in M has order type at least 8).

I shall not give a detailed proof of this theorem. I shall give the proof only for the case 8 = 2, which is fairly representative of the method, and then make some remarks on the general version of the theorem. The key lemma, on which the proof of the theorem is based, is the following: Lemma 33.2. Let K be a compact cardinal. For every 6 < (2")+, there exists a K-complete ultrafilter U over K such that j u ( u )> 6. [We know that j u ( K ) < (2")+.]

The proof of Lemma 33.2 uses the following combinatorial result: Lemma 33.33 Let K be an inaccessible cardinal. There exists a family Y of functions g: K + K such that 13 ' I = 2", and whenever 2 c Y is a subfamily of size < K , and {PQ:g E 2 1 is any collection of ordinals < K , then there exists a such that g(a) = p, for all g E X . Proof: Let .d be a family of almost disjoint subsets of K (i.e., I A I = K for each A E .d and 1 A n B I < K for any distinct A, B E d ) ,such that I .d I = 2". For each A E d , let f A be a mapping of A onto K such that for each p < K, the set ( a E A : f A ( a )= p} has size K. Let s,, a < K, enumerate all subsets s c K of size

< K. For each A

E

d , let g A : K

+K

be defined as follows:

if s, n A = {x}, gA(a)= 0

Let Y = {g" : A

E

.d}.

then gA(a)=f A ( x )

otherwise

6.

402

OTHER LARGE CARDINALS

If A # B E d , then it is easy to find s, such that gA(a) # 0 and g B ( a ) = 0; hence 1% I = 2". If .X c .d has size < K and if { P A : A E X } are given, then for each A E X we choose x A E A such that x A 4 B for any other B E X and that f A ( x A ) = b A .Then if a is such that s, = { x A : A E X } , we have gA(a) = PA for every A E .R. W

Proofof Lemmu 33.2. Let 6 < (2")+.Let Y be a family of functions g: K 4K of size 16) with the property stated in Lemma 33.3; let us enumerate Y = {g2: a I 6). For any a < p I6, let X,, = {( : g,(() < g,(()}. Using the property of Y stated in Lemma 33.3, we can see that any collection of less than K of the X,, has a nonempty intersection and hence F = { X : X zX , , for some a < P I 6) is a K-complete filter over K. Since K is compact, there exists a K-complete ultrafilter U extending F . It is clear that if a < p I 6, then g, < g, mod U , and hence j " ( ~ >) 6. W Before I start the proof of the theorem, let us go back to iterated ultrapowers. Let D be a normal measure on K, and let ion denote, for each a, the elementary embedding io,: V + Ult'");let K(') = i O , ( K ) and D@)= i@). First recall (31.18): If I is a limit ordinal, then X E U/?(')belongs to D(')just in case X 2 {K(,): a I y < I ) for some o! < I . Let (33.8)

K

= {v

: v is a strong limit cardinal, v > 2", and cf v > K }

By Lemma 31.12, if v E K , then K(') = v, and io,(v) = v for all a < v. Thus if yo < y1 < ... < yn < . . * are elements of the class K , and if I = limn-,x y n , then K(') = I , and X E Ulr@)belongs to D(') just in case X 2 {y,,: no I n} for some no. If A is a set of ordinals of order type w, A = { Y , , ) ; = ~ ,we define a filter F ( A ) over A = sup A as follows (33.9)

X

E

F(A)

3noVn 2 no(?, E X )

iff

The above discussion leads us to this: If A c K has order type w , and if I = sup A, then for every X E U P (33.10)

X

E

D(')

iff

X

E

F(A)

In other words, (33.1 1 )

D(') = F ( A ) n Ulf(')

Hence F ( A ) n U l P ) E Ult"); and so by Lemma 15.4, L [ F ( A ) ]= L[D('"].Thus F ( A ) n L [ F ( A ) ]= D") n L[D'')],and we have (33.12) L [ F ( A ) ] k F ( A ) n L [ F ( A ) ] is a normal measure on 1. The only assumption needed to derive (33.12) is that K is measurable and A is a subset of the class K . We shall now use Lemma 33.2 and a similar construction to obtain a model with two measurable cardinals.

33.

403

COMPACT CARDINALS

Suppose that A = (y,},“,o is as above, and that A’ = {yb},“=o is another subset of K of order type w, such that 1 ~>’ ~1 = sup A ; let 1’= sup A’. Let F = F ( A ) and F’ = F ( A ’ ) . Our intention is to choose A and A’ such that the model L[F, F’] has two measurable cardinals, namely I and A’, and that F n L[F, F’] and F’ n L[F, F‘] are normal measures on 1 and A’, respectively. The argument leading to (33.12) can again be used to show that F n L[F, F‘] is a normal measure on 1 in L[F, F‘]. This is because we have again

D(L= ) F n lJ1$A) moreover, i o i ( y i )

=

y: for each n, and hence ioi(A’)= A’ and we have ioA(F‘)= F‘ n U I P )

(33.1 3)

Therefore

L[F,F’] = L[D‘”’,F‘]= L[D‘”, ioA(F’)] and

F n L[F, F‘] =

(33.14)

n L[D‘”’, ioA(F‘)]

which gives (33.15)

L[F, F’] k F n L[F, F’] is a normal measure on 1.

In order to find A, A’ so that F‘ also gives a normal measure in L[F, F‘], let us make the following observation: Let us think for a moment that A c K and A’ c K . Then i o i , ( A ) = A and D‘”)= F’ n Ult(”), and the same argument as above shows that (33.16)

L[F, F‘] k F‘ n L[F, F‘] is a normal measure on 1’.

We shall use this observation below. Let us define the following classes of cardinals (compare with (30.34)): KO = K

(33.17)

K 3 1 + , = { v ~ K ZI K: , n v ( = v )

K,

=

0K ,

(y a limit)

l
Each K, is nonempty; in fact each K , is unbounded and &closed for all 6 of cofinality > K. Now we let yn = the least element of K ,

(33.18)

A = {y,,};=o,

1 = lim y, n-m

and let A‘

= {Y;},”=~

be a subset of K , ,

6.

404

OTHER LARGE CARDINALS

Let us consider the model L[A, A'], and let for each n I w (33.19) M, = the Skolem hull of K , in = the

class of all x

L[A, A']

4 A , A']

E L [ A , A']

x = t[v,,

***,

such that

vk, y o ,

***,

&, A, A'] K,.

**.,

yk,

where r is a Skolem term and vlr ..., vk

E

(Let us not worry about the problem whether (33.19) is expressible in the language of set theory; it can be shown that it is, similarly as in the case of ordinal definable sets. Alternatively, we can consider the model L,[A, A'] where 8 is some large enough cardinal in K,, ,.) Each M , is an elementary submodel of L[A, A ' ] ; let IT,be the transitive collapse of M , ; then nn[Mn]= L[n,(A), 7rn(A')]and j , = IT;' is an elementary embedding (33.20)

j n : L[n,(A), xn(A')] + L[A, A']

Lemma 33.4. For each n < w, n,,(y,,)< (2")' Proof. By induction on n. First let n = 0. Let a < yo be in M,. Then a = t(v,, . ..,vk, A, A ' ) for some Skolem term t and some vtr . . .,vk E K O .Let ioa be the natural embedding into Ultg) for some U over K . Since y o is the least element of K O ,we have a < v for all v E K and hence ioa(v ) = v for all v E K o . Hence also iou(A)= A and iou(A')= A' and it follows that iou(a)= a. Now ioa(a)= a is possible only if a < K. Hence each a < yo in M, is less than K and therefore ITo(Y0) 5 K < (2")' and let us show that IT,+ ,(yn+ I ) < NOWlet us assume that IT,,(~,,) (2")'. By Lemma 33.2 there exists U such that j U ( K ) > n,(y,). We shall show that n,(a) <jU(.) for all a < yn+, in M,,,; since IT,, l(a) I n,(a) it follows that IT,+ 1(Yn+ 1) = SuP{nn+,(a): a < Y n + 1 and a E Mn+ 1) s j u ( K ) < (2")'. First notice that it follows from the definition of K,+' in (33.17) that n,(v) = v for all v E K,, Note also that y,,, E K,+ for all rn 2 n 1, and A' c K,, Let a < Y , , + ~ be in M , + l . Then (in L[A, A']),

,

,.

a=t(yO,

Ynr

v1, ..., vki A, A ' )

where t is some Skolem term and v l , . . ., vk E K,, nn(a) = f(nn(rO), *

-

- 9

+

nn(yn),

vlr

Hence (in L[n,(A),nn(A')]) vk, nn(A)? A')

Now we argue inside the model U f t , ( V ) (which contains both zn(A)and A'): Consider the ath iterated ultrapower (modulo some measure on j u ( K ) ) . Since n,(yo),,.., n,(y,) are all less than j U ( K ) , we have ioa(nn(yi))= zn(yi)for all i = 0, . .., n. We also have iod(v) = v for each v E K,, (because a < v for each

33.

405

COMPACT C A R D I N A L S

,

,

v E K , , and K , , c K ) . It follows that ioz(a,,(a))= n,,(a). Now (because a,,(a) I a ) this is only possible if x,(a) < j r , ( ~ ) . W

We can now complete the proof of Theorem 80 for 6 = 2. Let us consider the model M,, the Skolem hull in 4 A , A'] of K , . Let a, be the transitive collapse of M, and B = n,(A). Since A' c K,, we have n,(A') = A', and j , = a; is an elementary embedding

,,

j,:

YE, A'] + 4 A , A']

By Lemma 33.4, n,(y,) I n,,(yn)< (2")' for all n, and hence .,(A) < (2")'. Let U be a ti-complete ultrafilter over ti such that j&) > a,@) (by Lemma 33.2). In U l t , , B is a subset of j , ( t i ) and A' is a subset of the class K . Thus we can apply (33.16) and get Ult, 1 ( 4 F ( B ) ,F(A')] =! F ( A ' ) n 4 F ( B ) , F(A')] is a normal measure on A')

Hence L[B, A'] k

(L[F(B),F ( A ' ) ] 1 F ( A ' ) n L[F(B),F ( A ' ) ] is a normal measure on A') and applying j,, we get

44 A']

1

( L [ F ( A ) ,F ( A ' ) ] 1 F ( A ' ) n L [ F ( A ) ,F ( A ' ) ] is a normal measure on A') Therefore F' n q F , F'] is (in q F , F ' ] ) a normal measure on A'. This completes the proof of the theorem for 6 = 2. The proof of the general case is completely analogous for 6 < K . Instead of A, A', one uses a sequence ( A , : < 6 ) , where $, the nth element of A,, is defined as the least element of K , , , , , . One uses the models M,,.,,, to prove an analog of Lemma 33.4: ~,,,,+,,(y!,~)) < (2")'. The assumption 8 < ti is important because when one uses the ultrapower U l t , , one needs to know that ( A S: 5 < 6 ) E Ult, . Once the theorem is proved for all 6 < ti, then the general version is derived as follows: If 8 is arbitrary, let us consider U P ( V ) for some a > 8. Then the model Ult'")(V )has a strongly compact cardinal greater than 8 and therefore has a transitive model with 8 measurable cardinals. W

<

The following theorem is a further evidence of the impact of large cardinals on cardinal arithmetic. Theorem 81. (Solovay). If there exists a compact cardinal, then the singular cardinals hypothesis holds above the compact cardinal. That is, i f ti is strongly compact and A > K is a singular cardinal, then 2"' A < A implies Acf A = A'. (Consequently, i f A > ti is a singular strong limit cardinal, then 2l = A'.) We shall prove the theorem in a sequence of lemmas. An ultrafilter over a cardinal A is uniform if every set in the ultrafilter has size A.

406

6.

OTHER LARGE CARDINALS

Lemma 33.5. ff K is a compact cardinal and I > K is a regular cardinal, then there exists a K-complete uniform ultrajilter D over I with the property that almost all (mod D )ordinals a < I have cojinality less than K . ProoJ: Let U be a fine measure on P,(I). Since U is fine, every a < I belongs to almost all (mod V ) P E P K ( I ) Let . us consider the ultrapower Ult,( V ) and letf be the least ordinal function in U l t , greater than all the constant functions c , , y


(33.2 1 ) We note first thatf(P) < A for almost all P : Let g: P , ( I ) + I be the function g ( P ) = sup P . If y < I , then y Ig ( P ) for almost all P and hencej(y) I [ g ] ; thus

[ f l I [sl <j(4*

Let D be the ultrafilter over I defined as follows:

(33.22)

XED

iff

f-l(X)~U (XGI)

It is clear that D is K-complete, and sincefis greater than the constant functions, D is nonprincipal. For the same reason, the diagonal function d ( a ) = a is greater (in Ult,) than all the constant functions c , , y < I , and since I is regular, D is uniform. In order to show that almost all (mod D)a < I have cofinality < K , it suffices, by (33.22), to show that cf(f(P)) < K for almost all P (mod V ) . That will follow immediately once we show that for almost all P (mod U ) ,

f(P)= sup{a E P : a < f(P))

(33.23)

We clearly have 2 in (33.23). To prove I , consider the function h ( P ) = sup{a E P : a < f(P)}. For each y < I , y is in almost every P and hence y I h ( P ) almost everywhere. Thus [h] 2 j,(y) for all y < A and s o f ( P ) 5 h(P) almost everywhere. Lemma 33.6. ff K is compact and I > K is a regular cardinal, then there exists a K-complete nonprincipal ultrajilter D over I , and a collection { M , : a < I ) such that

(33.24) (a) I M , I < K for all a < I , (b) for euery y

-= I , y belongs to M,for almost all a (mod D)

[An ultrafilter D that has a family { M , : a < I } with property (33.24) is called (K, I)-regular.] Proof: Let D be the ultrafilter over I from Lemma 33.5. For almost all a (mod D), there exists A, c_ a of size less than K and cofinal in a. If cf a 2 K, let A, = 0. Let A be set of ordinals represented in Ult,(V) by the function ( A , : a < I ) . The set A is cofinal in the ordinal represented by the diagonal

33.

407

COMPACT CARDINALS

function d ; and since [dl > for all 7 < A, it follows that for each q < A there is q' > q such that A n {< :,jD(q)I < j D ( q ' ) } is nonempty. We construct a sequence ( q y : y < A) of ordinals < A as follows: let qo = 0 and q, = lima*y qa if y is a limit; let q, + be some ordinal such that there exists E A such that j&,) I t <j&+ l). In other words, if we denote I , the interval (5 : q, I5 < q y + then for every y , the interval I , has nonempty intersection with almost every A,. Thus if we let j,()l)

<

,

<

M, = { y < A : I , n A,

# @}

for each a < I , then {M, : a < A} has property (33.24b). To see that M , has property (a) as well, notice that I A , 1 < K for all a and that since the I, are mutually disjoint, each A, intersects less than K of them. Lemma 33.7. If K is compact and A > K is a regular cardinal, then there exists a collection {Ma: a < A) c P K ( l )such that (33.25) Consequently, A < K

=A

Proof. Let { M , : a < A) be as in Lemma 33.6. If P is a subset of A of size less than K , then by (33.24b) and by h--completeness of D,P c M a for almost all a. Hence P E P ( M , ) for some a < A. This proves (33.25); since K is inaccessible, it follows that IPK(A)I= A. W Proof of Theorem 81. Let K be a compact cardinal. If A > K is an arbitrary cardinal, then we have, by Lemma 33.7, A < K

5

(A+)-

= A+

In particular, we have AN" I 1 ' for every A > K. By Theorem 23 (or rather by Lemma 8.3), this implies that the singular cardinals hypothesis holds for every A > K. Supercompact Cardinals We proved in Lemma 33.1 that a compact cardinal K is characterized by the property that every PK(A)has a fine measure. If we require the fine measure to satisfy a normality condition, then we obtain a stronger notion-a supercornpact cardinal. Ultrapowers modulo normal measures on PK(A)induce elementary embeddings that can be used to derive strong consequences of supercompact cardinals. For instance, Theorems 79 and 80 become almost trivial if existence of a strongly compact cardinal is replaced by existence of a

6.

408

OTHER LARGE CARDINALS

supercompact cardinal. It is consistent to assume that a compact cardinal is not supercompact, or that every compact cardinal is supercompact, but it is not known whether supercompact cardinals are consistent relative to compact cardinals. A fine measure U on P , ( A ) is normal if wheneverf P , ( A ) + A is such that f ( P ) E P for almost all P , then f is constant on a set in U.A cardinal K is supercompact if for every A such that I A 1 2 K, there exists a normal measure on P,(A). Exercise 33.3. If

K

is measurable, then there is a normal measure on P r ( ~ ) .

Normality of filters over P , ( A ) is a generalization of normality of filters over K . Many results on normality generalize from K to P,(A) (but not the existence of a normal measure on a measurable cardinal). For instance, let us call C G PK(A)closed unbounded if (a) for every P in P,(A) there is Q E C such that P c_ Q, and (b) if D s C is a nonempty directed (that is, V P 1 , P , E D 3Q E D such that Q 2 P I u P 2 )set of size < K , then D E C. [It is not difficult to show that (b) can be replaced by: closed under unions of chains of length < K . ] The closed unboundedfilter over P , ( A ) is the collection of all subsets of P,(A) that have a closed unbounded subset. Stationary sets are defined as sets that intersect every closed unbounded set.

u

Exercise 33.4.

(a) Each = {Q : P E Q} is closed unbounded. (b) The closed unbounded filter is K-complete. (c) Iff(P) E P on a stationary set, thenfis constant on a stationary set. [For (c), prove that the closed unbounded filter is closed under the diagonal intersections A a G C, A = { P : P E C, for all a E P}.]

Let A 2 K be a cardinal and let us consider the ultrapower Ult,( V ) by a normal measure U over PK(A);let j = j , be the corresponding elementary embedding. Clearly, a set X G PK(A)belongs to U just in case [ d ] E j ( X ) , where d, the “diagonal function” is the function d ( P ) = P .

Lemma 33.8. If U is a normal measure on PK(A),then [ d ] = { j ( y ): y < A} and hencefor eoery X 5 PK(A), (33.26)

X

E

U

iff

{ j ( y ) :y < A} E j ( X )

Proof. On the one hand, if y < A, then y E P for almost all P and hence j ( y ) E [ d ] . On the other hand, if [ f ] E [ d ] , then f ( P ) E P for almost all P and by normality, there is y < 1 such that [f]= j ( y ) . H Let G = { j ( y ): y < A}. It follows from (33.26)that iffand g are functions on PK(A),then

I![

=

[sl

iff

( d ) ( G )= ( j g ) ( G )

33.

409

COMPACT CARDINALS

and

[fl E [sl

iff

( X ) ( G )E ( M ( G )

Consequently,

(33.27) [f 1 = ( X ) ( G ) for every function f on P I @ ) . For each P E P K ( l ) ,let us denote K , = P ~ K

1,

= the

order type of P

Note that if we let G = ( j ( y ) : y < A}, then the order type of G is 1and hence by (33.27), 1 is represented in the ultrapower by the function P - A p . Also, since 1, < K for all P, we have j ( K ) > 1.By K-completeness of U , we havej(y) = y for all y < K ; and since K is moved byj, it follows that G n j ( K ) = K and therefore K is represented by the function Pt-+ K ~ . This gives the following characterization of supercompact cardinals:

Lemma 33.9. Let 1 2 K . A normal measure on P K ( 2 exists ) ifand only ifthere i s an elementary embedding j : V M such that -+

(33.28)

(i) j ( y ) = y for all y < K ; (ii) j ( ~>)A; (iii) ' M G M; i.e., every sequence ( a , : a < 1 ) of elements of M is a member of M .

Proof: (a) Let U be a normal measure on P K ( l )We . let M = Ult,( V) and let j be the natural embeddingj: V + Ult. We have already proved (i) and (ii). To prove (iii), it suffices to show that whenever ( a , : a < 1)is such that a , E M for all a < A, then the set {a, : a < 1) belongs to M. Let h , a < 1, be functions representing elements of M: [fa] E M. We consider the function f on P J 1 ) defined as follows: f ( P ) = { j J P ) : a E P } ; I claim that [f ] = {aa: a < A}. On the one hand, if a < 1,then a E P for almost all P and hence [fa] E [f]. On the other hand, if [g] E [f], then for almost all P, g ( P ) = f , ( P ) for some a E P . By normality, there exists y < 1such that g ( P ) = f y ( P ) for almost all P , and hence [g] = a ? . (b) Let j : V -+ M be an elementary embedding that satisfies (i), (ii), and (iii). By (iii), the set { j ( y ) : y < A} belongs to M and so the following defines an ultrafilter over P K ( l ) :

(33.29)

XEU

iff

( j ( y ) : y < 1)E j ( X )

A standard argument shows that U is a K-complete ultrafilter. U is a fine measure because for every a E 1,{ P : a E P } is in U . Finally, U is normal: If

6.

410

OTHER LARGE CARDINALS

f ( P ) E P for almost all P, then ( j f ) ( G )E G where G = { j ( y ) : y < I ) . Hence ( j f ) ( G )= j ( y ) for some y < I , and s o f ( P ) = y for almost all P . Exercise 33.5. I f U is a normal measure on PK(2).then every closed unbounded subset of PK(2)is in U . [If C G PI(2) is closed unbounded, then D = ( j ( P ): P E Cl is a directed subset of . j ( C )and ID1 = 2'" < j ( K ) . Hence D E j ( C ) , and since D = { j ( y ) : y < 2 ) . we have

cE

u

U.]

u

We have seen several examples how large cardinals restrict the behavior of the continuum function (e.g., if K is measurable and 2" > K + , then 2' > a+ for cofinally many a < K ) . This is more so for supercompact cardinals: Exercise 33.6. If 12 K and if U is a normal measure on PK(2),then the ultraproduct Ult,( (V,, , E): P E P,(2)) is isomorphic to (V,, E).

As a typical application, if K is supercompact and if there is a regular cardinal a such that 2" = a + , then there is a regular cardinal a < K such that 2' = a + . The following two results illustrate the technique of elementary embeddings.

Lemma 33.10. If K is supercompact, then there exists a normal measure D on K such that almost every a < K (mod D ) is measurable. Proof. Let I = 2" and let j : V + M be an elementary embedding with properties (33.28). Let D be defined by D = { X : K E j ( X ) } ,and let j , : V + UltD be the corresponding elementary embedding. Let k : Ult, + M be the elementary embedding defined in Lemma 28.6: k([flD)

= (jf)(K)

Note that k ( ~ =) K . Now, P ( K )c M and every subset of M of size iis in M ; hence every U G P ( K )is in M and it follows that in M , K is a measurable cardinal. Since k is elementary and k ( ~=) K , we have UltD k K is a measurable cardinal, and the lemma follows. Corollary. If

K

is supercompact, then K is the Kth measurable cardinal.

Incidentally, it is consistent that there exists a compact cardinal with no measurable cardinals below it. Lemma 33.11. If K is supercompact, then for every W normal measure D on K such that W E ultD(v): Proof. Assume on the contrary that 3W

G

(33.30) For every normal measure U on

G

P ( K ) there exists a

P(K)@(K,W ) ,where

K,

W

# Ult, .

41 1

33. COMPACT CARDINALS

Let II

= 2", let j : V + M be an elementary embedding with the properties (33.28). let D = { X : K € j ( X ) } ,j,: V + UftD, and let k : UltD+ M be as in Lemma 28.6. By (33.28), every subset of P ( K ) is in M, and hence 9 4 ' M ~ ; we have

M !=

3W- c P(K)@(KW , )

and by elementarity of k (since k ( ~=) ti), UlfD != 36" c

P ( k . ) @ ( K , $k-)

Let E U l t D be such that U l t D =! @(K, #.'). Since (applying k ) M k @(ti, W ' ) # ' I

&($k"') =

W', we have

and because every U c P(ti) is in M, it follows that @(K, %*') holds (in V ) . In other words, W' satisfies (33.30), contrary to the fact that W' E U f t , . W Corollary. If

K

is supercompact, then there are 2'' normal measures on

ti.

Pro06 If D is a normal measure on K and 9 4 . c P(ti) is in UftD, then W is represented by a function f on K such that f ( a ) G P ( a ) for all a < K. Since the number of such functions is 2", it follows that U l t Dcontains only 2" subsets of P(.). However, by the lemma, each W G P(ti) is contained in some ultrapower Ult, where D is a normal measure on ti, and therefore there must exist 22K normal measures on K. H

It is not known whether the same conclusion follows if K is only compact. The following lemma indicates that a strongly compact cardinal is not necessarily supercompact. Lemma 33.12. Let ti be a measurable cardinal such that there are cardinals below K . Then K is compact.

K

compact

Proof. Let F be a nonprincipal ti-complete ultrafilter over K such that C E F where C = { a < K : a is compact}. Let A be such that I A I 2 K ; we shall show that there is a fine measure on P , ( A ) . For each a E C , let U , be a fine measure on P,(A), and let us define U G P , ( A ) as follows:

X

E

U

iff

{a E C : X n Pa@)

E

U,} E F

It is easy to verify that U is a fine measure on P,(A). Corollary. lf there exists a measurable cardinal that is a limit of compact cardinals, then the least such cardinal is compact but not supercompact.

412

6.

OTHER LARGE CARDINALS

Proof. Let K be the least measurable limit of compact cardinals. By the lemma, K is compact. Let us assume that K is supercompact. Let 1 = 2" and let j: V + M be an elementary embedding such that K is the least ordinal moved, and that 'M c_ M.If a < K is compact, then M t j ( a ) is compact, but j ( a ) = a and therefore M t K is a limit of compact cardinals. Since every U G P ( K )is in M, K is measurable in M and hence M t K is a measurable limit of compact cardinals. This is a contradiction because M k j ( K ) is the least measurable limit of compact cardinals.

The assumption of the corollary holds if there are extendible cardinals (see below). There is also a consistency proof that uses forcing showing that not every compact cardinal is supercompact. (And another consistency proof gives a model that has exactly one compact cardinal and the cardinal is supercompact.) Let K be a supercompact cardinal. Let 12 K, and let D be a normal measure on P,(I). Let G = { j D ( a :) a < I},where j , is the elementary embedding of V in Ult,(V). Recall that K, I are represented by K , , 1, where K , = P n K and I, = the order-type of P. If I is a cardinal, note that I, = IP I almost everywhere. In Section 28 I stated without proof the following combinatorial result (see the remark following Lemma 28.8): If X is an infinite set, then there exists F: "X + X such that whenever Y G X is such that 1 Y I = IX 1 then F["Y] = X. Let us call F an E-H function on X (Erdos-Hajnal). K , and let D be a normal measure on P$). There is X E D such that for any P, Q E X , if P c Q, then A p < & . [In Ult,, consider an E-H function F on G = jD[I].For each P E P,(I), let F P be the restriction of F to " P . F is represented in Ult by ( F P : P E P K ( I ) ) ,and there is X E D such that F P is an E-H function on P for all P E X. If P , Q E X such that P c_ Q and [ P I = IQI, then Q = Fp["Q] = Fp["Q] = P.]

Exercise 33.7. Let A be a cardinal, I 2

Exercise 33.8. Let I be a regular cardinal 2 K, let D be a normal measure on P K ( I ) . There is X E D such that the function Pwsup(P) is one-to-one on X. [Let F be, in Lllt,, an E-H function on G, and let F P be as above. There is X E D such that for each P E X, F P is an E-H function on P and P has the property that if s G P is countable, then sup(s) E P (because G has this property). Now if P , Q E X and s u p P = s u p Q = y , s h o w t h a t P n Q i s c o f i n a l i n y . H e n c e I P n Q I = JPI a n d a s i n Exercise 33.7 it follows that P n Q = P . Similarly, P n Q = Q, and hence P = Q.]

Note that Exercise 33.8 gives an alternative proof of Theorem 81 if K is supercompact: We have found X E D such that I X I I1,and since PK(l)= P(Q), we obtain " '1 = I. Let K be supercompact and let I 2 K be a cardinal. A normal measure D on P K ( l is ) strongly normal if there exists X E D such that for every functionfon X , if for each P E X,f(P) is in X andf(P) c P (andf(P) # P), thenfis constant on some Y E D.

u

33.

413

COMPACT CARDINALS

Exercise 33.9. The following are equivalent

(i) D is strongly normal. (ii) There is X E D such that if { Z p :P E X } c D, then A p E X Z p E D, where APEX Z p = ( Q : Q E Z p for each P c Q such that P E X}. (iii) D has this partition property: If F: [P,(L)]' + {0, 1) is a partition, then there is X E D such that F is constant on { { P , Q} E [ X I 2 : P c Q or Q c P}. (iv) There is X E D such that if P, Q E X and P c Q, then Lp < K Q . [(i) (ii): Let X E D be a witness to strong normality. Prove by contradiction that D is closed under A P E ZXp . -+

(ii) (iii): Let F: [PK(L)]' (0, 1); for each P, let F p : I' (0, l} be F P ( Q ) = F ( P , Q). For each P there is Z p G I',Z p E D,such that F P is constant on Z p .Let X E D be as in (ii) and such that the constant value of F P is the same for all P E X . Then X n APEX Z p is homogeneous for F in the sense of (iii). (iii) -+ (iv): Note that if X E D, then there exist P, Q E X such that P c Q and Lp < KQ. (iv) + (i): Let X E D be as in (iv) and let$ X -+ X be such thatf(P) c P for all P . In U / r D , if P E j , ( X ) and P c G (where G = j [ L ] ) , then [PI < G n K = K and hence P = J ( Q ) for some Q E P.(L). Hence ( i f ) ( G )= j ( Q ) for some Q and so f(P)= Q for almost all P.] -+

-+

-+

L = K', then every normal measure on P,(L) is strongly normal. , if P, Q E X , [There is X E D such that for all P E X , K~ is inaccessible, L, = ~ p fand P c Q, then AP < 1,. Hence if P , Q E X and P c Q, then K~ = K Q is impossible and so K~ < K~ and consequently Lp = K; < K Q .] Exercise 33.10. I f

It has been proved that if K is supercompact, then every PK(A)has a strongly normal measure; however, not every normal measure is necessarily strongly normal: Exercise 33.11. I f 1> K is measurable, then there is a normal measure U on P,(L) that is

not strongly normal. [Let j : V + M be elementary, K least moved, j ( ~>) L, and ' M G M. Let D be a normal measure on 1.Let us define a normal measure U on P K ( L )as follows: X E U iff { a < L :j [ a ] E j ( X ) ]E D. I f X E U , then there are a < p such that j [ a ] and j [ p ] are in j ( X ) ; hence Ult181 3P. Q E j ( X ) such that P is an initial segment of Q. Thus 3P, Q E X such that P is an initial segment of Q, and so l p2 K~ .]

Beyond Supercompactness

I shall conclude this section with several large cardinal notions that are even stronger than supercompactness and in fact are bordering on inconsistency. A cardinal K is iJxtrndihlr if for every 5~ > K there is /I and an elementary embeddingj: V, + V, such that K is the least ordinal moved by j. We shall show that extensibility is stronger than supercompactness. If there is a normal measure on PK(l),let us say that K is 1-.wpercompacr,and recall the equivalence (33.28).

414

6. OTHER LARGE

Lemma 33.13.

Lpt

I

2K

be a regular cardinal a d let

K

CARDINALS

be I-supercompact. Let

a < K . l f a is y-supercompact for all y < K , then a is I-supercompact.

Proof. Let U be a normal measure on P I ( I ) ,and let us consider jv: V -, UIt". Since j ( a ) = a, we have lilt k (a is y-supercompact for all y < j ( ~ ) )in; particular, Ult k a is I-supercompact. Hence there is D such that l i l t k D is a normal measure on P,(I). Now, I P&) I = A and 'Ult G Ult, and hence every subset of f,(I) is in Ult. It follows that D is a normal measure on P&). H

Lemma 33.14.

(a) l f is~ extendible, then K is supercompact. (b) l f K is extendible, then there is a normal measure D on { a < K : a is supercompact) E D .

K

such that

Proof. (a) Let a =- K be a limit cardinal with the property that if V, k ( K is I-supercompact for all I),then K is supercompact. (Such an a exists by the reflection principle.) Thus it suffices to show that K is A-supercompact for all regular I < a. Let j : V , -+ V, be such that K is the least ordinal moved. Consider the sequence K~ = K. K~ = j ( ~ )..,.,K, + = ~ ( I c , ).,.., as long asj(K,) is defined. First we note that as in Theorem 68 (see Exercise 28.9) either there is n such that K" < a < j ( ~ , or ) a= K,. Therefore, it is sufficient to prove, by induction on n, that K is I-supercompact for each regular A < K, (if A < a). Clearly, K is A-supercompact for each I < K ~ Thus . let n 2 1 and let us assume that K is I-supercompact for all I < K,. Applyingj, we get: Vs k ( j ( ~is) I-supercompact for all regular I < K,+ Now we also have V, k (K is ysupercompact for all y < j ( K ) ) , and we can apply Lemma 33.13 (in V,) to conclude that V, k (K is A-supercompact for all regular il< K,+ I). This completes the induction step. (b) Let a be some limit ordinal greater than K and let j: V, -,V, be such that K is the least moved. Let D = {X G K : K E j ( X ) } .By (a), K is supercompact, and so V, k ( K is y-supercompact for all y < j ( K ) ) . Hence A E D, where A = { a < K : a is y-supercompact for all y < K}. By Lemma 33.13, every a E A is supercompact.

,

Let me consider now the following axiom schema called Vopenka's principle (VP): Let C be a proper class of models of the same language. Then there exist two members PI, 'B of the class C such that 91 can be elementarily embedded in '8.

Lemma 33.15. If Vophka's principle holds, then there exists an extendible cardinal. Proof: Let A be the class of all limit ordinals a such that cf a = o and that for every K a, if V , k (K is extendible), then K is extendible. Using the reflection

-=

33.

415

COMPACT CARDINALS

principle, we see that A is a proper class. Let C consist of the models ( V , , E), for u E A. By Vophka's principle, there exist a, p E A and an elementary embedding j: V ,+ V, + . Since j ( a ) = fi, j moves some ordinal; the least ordinal moved is clearly measurable and so it is not a (which has cofinnality 0). Let K be the least ordinal moved. Now V , k ( K is extendible) because for every y < a, j ( K, is a witness to it. By the definition of A, K is extendible.

-

,

A similar argument shows that Vopgnka's principle implies existence of arbitrarily large extendible cardinals. Let us call a cardinal K huge if there is an elementary embeddingj: V --* M such that K is the least ordinal moved and that j(')M c M .

Lemma 33.16, ff K is a huge cardinal, then Vopznka's principle is consistenr: in fact ( V , , E) is a model of VP.

Proof. We shall show that if C is a set of models and rank ( C ) = K , then there exist two members 41, 8 E C and an elementary embedding h : 41 + 8. Let j: V M such that K is the least ordinal moved and that j(loM G M . Since rank(C) = K, there exists PIo E j ( C ) such that PIo 4 C . It follows that j(41,) # 41,. Let eo = j 191,; it is easy to see that e, is an elementary embedding of PI0 in j(4io), and since 19101 < j ( ~ )we , have eo E M . Hence

-

M

k there exists PI

Ej ( C ) ,

41 # j(PIO), and there exists an elementary

e : YI +j(PI0);

and so there exists PI E C,PI # PI,, and there exists an elementary e : PI Let PI, e be such; clearly,

-

Pio.

M i. e is an elementary embedding of PI in PI, and because rank(4I) < K, we have 4I = j(4I), and hence PI E j ( C ) , and so

M k there exist distinct YI, 6 E ~ ( C and ) there exists an elementary h: 91-23 It follows that there exist distinct 91, 6 E C , and an elementary embedding h : 9 I + 23. Exercise 33.12. A cardinal K is huge, with j : V + M and j ( K ) = A, if and only if there is a normal K-complete ultrafilter U over [A]. = { X c 1 : IX 1 = K ) . [ X E U iff j [ A ] E j ( X ) . ] Exercise 33.13. Let K be the least huge cardinal and let p be the least supercompact cardinal. Then K < p. [If K = p then by Lemmas 33.16, 33.15, 33.14, and 33.13 we get: V, C V P ,

6.

416

OTHER LARGE CARDINALS

V, b (3 supercompact a), there is a supercompact a < p, a contradiction. If K > p, let j : V -+ M with I = j ( K ) and * M G M. Since p is supercompact, let i: V + N be such that i(p) > I and V A + 2G N. If Lr is a normal measure on [A]., then U E N, and hence N t (3huge cardinal below i(p)).Thus there exists a huge cardinal below p, a contradiction.]

With the definition of huge cardinals, we are approaching the brink of inconsistency. By the proof of Theorem 68 (see Exercise 28.9), if j : V -, M is elementaryand~is theleast ordinalmoved,and i f l = limn+m{ K , j ( K ) , j ( j ( K ) ) , . . .}, then P ( I ) $! M. 34. REAL-VALUED MEASURABLE CARDINALS

In this section we prove the following theorems: Theorem 82 (Solovay). (a) If K is a real-valued measurable cardinal, then there is a transitive model of set theory in which K is measurable. (b) If K is a measurable cardinal, then there exists a generic extension in which K = 2N0and K is real-valued measurable. Theorem 83 (Prikry). If 2O ' is real-valued measurable, then 2A= 2"O for all infinite l < 2H0. We defined real-valued measurable cardinals in Section 27. If p is a K additive real-valued measure on K, then the ideal I, of all sets of measure O is a a-saturated K-complete ideal over K. We have proved that if an uncountable cardinal K carries a maturated K-complete ideal, then K is weakly inaccessible. We shall prove (a) in Theorem 82 and Theorem 83 for this generalization of real-valued measurability, namely under the assumption that K is uncountable and carries a maturated K-complete ideal. Thus let K be an uncountable cardinal and let I be a maturated K-complete ideal over K. Let us adopt the following terminology, motivated by real-valued measurability: We say that X G K has measure zero if X E I, and that X has positive measure if X q! I. We also say that X has measure one if K - X has measure zero. Let us call A E K an atom if A has positive measure and is not the union of two disjoint sets of positive measure. I is atomless if it has no atoms. What we proved in Lemma 27.5b can be formulated as follows: If I is atomless, then K s 2O ' . It follows that if 2O ' < K, then every set X of positive measure contains an atom A G X,and hence there exists an at most countable disjoint collection W of atoms such that K = { A : A E W}. We start with the following analog of Lemma 28.11. We recall that a K complete ideal over K is normal if every function$ S --* K regressive on a set S G K of positive measure is constant on some T c_ S of positive measure. A real-valued measure p is normal if I, is normal.

u

34.

417

REAL-VALUED MEASURABLE C A R D I N A L S

Lemma 34. I, (a) If I is a a-saturated K-complete ideal over an uncountable cardinal there exists a function$ K 3 K such that

J

= f,(l) =

{x

K

:f-

K,

then

~ ( xE )I }

is a normal a-saturated K-complete ideal over K . (b) I f p is a K-additive real-valued measure on K , then there exists a function f: K -+ K such that v =fJp) defined by

V ( X )= P(f- I ( W ) (X cK) is a normal K-additive real-valued measure on K . Proof: We shall prove (a) and leave the completely analogous proof of (b) to the reader. Let us say that a function g: S -+ K is unbounded on the set S of positive measure if there is no y < K and no T c S of positive measure such that g(a) < y for all y E T. Let us consider the family .S of all functions g into K defined on a set of positive measure and unbounded on its domain. Let us define g < h if dom(g) G dom(h) and if g(a) < h(a) everywhere on dom(g). Let us also define g 5 h if dom(g) c dom(h) and if g(a) Ih(a) everywhere on dom(g). (Note that g 5 h does not mean g < h or g = h.) Let us call g E 3 minimal if there is no h E 3 such that h < g . We shall first show that there exists a minimal g E 9. Otherwise, for every g E .S there is h E .S such that h < g. Thus let g E F be arbitrary. Let W be a maximal collection of elements of 3 such that h < g for each h E W, and that dom(h,) n dom(h2) = Qj whenever h , and h2 are distinct elements of W. Since I is o-saturated, W is at most countable and by our assumption, the set dom(g) - {dom(h):h E W} hasmeasurezero.Thus ifweletf = { h :h E W}, we have dom(g) - dom(f) E I, and f < g. Since g was arbitrary, we can construct a countable sequence go > g1 > ... > g,, > ... such that dom(g,,) dom(g,,+ E I for each n. It follows that (7;=o dom(g,,) has positive measure dom(g,,) we would have and we get a contradiction since for any a E SO(4 g1b) .... The same argument shows that for every h E 9 there exists a minimal g E 9 such that g I h. Thus if W is a maximal family of minimal functions g E 9 such that dom(g,) n dom(g,) = 0 whenever g1 and g2 are distinct elements of W, W is at most countable and U{dom(g): g E W} has measure {g: g E W}, then dom(f)has measure one and f is a one. Thus if we letf= least unbounded function: On the one hand, if y < K, then there is no S E K of positive measure such thatf(a) < y everywhere on S; on the other hand, if S is a set of positive measure and g is a function on S such that g(a) < f ( a ) everywhere on S, then g is constant on some T E S of positive measure. We can clearly assume that dom(f) = K .

u

u

n:=o

’ ’

u

6.

418

OTHER LARGE CARDINALS

L e t t K -+ K be a least unbounded function; we shall show that J =f,(l) is a normal a-saturated K-complete ideal over K . It is obvious that J is a K-complete ideal. For every y E K , fhas measure zero and hence {y} E J. If X t$ J , thenf- l(X) t$ I , and if X n Y = 0, thenf- l(X) n f- Y ) = 0, and hence J is a-saturated because I is a-saturated. To show that J is normal, let S t$ I , and let g(a) < LX for all a E S. Then g(f(l))
6-

Lemma 34.2. Let I be a normal a-saturated K-complete ideal over K . If S is a set of positive measure and f : S + K is regressive on S , then f is bounded almost everywhere on S ; that is, there exists y < K such that {a E S : f ( a )2 y } E I . Proof: For every X c S of positive measure there exists Y G X of positive measure such thatfis constant on Y . Thus let W be a maximal disjoint family of sets X c S of positive measure such that f is constant on X. Let T = {X : X E W } .The family W is at most countable and hence there is y such that f(a) < y for all a E T . Clearly, S - T has measure 0. W

u

Corollary. If K is real-valued measurable (or ideal), then K is a weakly Mahlo cardinal.

if^ carries a a-saturated K-complete

Proof: Let I be a normal a-saturated rc-complete ideal over K . Since I is normal, every closed unbounded set has I-measure one (see Lemma 7.7). Because K is weakly inaccessible, it suffices to show that the set of all regular cardinals a < K has measure one. Let us assume that the set S of all limit ordinals a < K such that cf a < a has positive measure. Considering the regressive function at+ cf(a), we find a set T of positive measure and some 1< K such that cf a = A for all a E T. For each a E T, let (a, : v < A) be an increasing A-sequence with the limit a. For each v < 1,the function a w a , is regressive on T and so, by Lemma 34.2 there is y, such that a, < y, for almost all a E T. Let y = sup{y,: v A). Since 1< K , we conclude, by K-completeness of I , that for almost all a E T, a, < y for all v < 1.But this means that for almost all a E T, a = lim, a, is I y. This is a contradiction since T is unbounded. W

-=

Since every closed unbounded set has measure one (if I is a normal asaturated K-complete ideal over K ) , every set of positive measure is stationary. It can even be proved that if S has positive measure, then S n a is stationary in a for almost all a. Then it follows that K is the Kth weakly Mahlo cardinal, K t h cardinal which is a limit of weakly Mahlo cardinals, etc. I shall return to this subject in the next section.

34.

419

REAL-VALUED MEASURABLE CARDINALS

Exercise 34.2. I f K carries a a-saturated K-complete ideal, then K has the tree property. [As in Lemma 29.7, let 7" be the set of all x E T such that the set of all successors of x has positive measure. T is a tree of height K and each level of 7" is at most countable; show that 7" has a b branch.]

We shall now show that every real-valued measurable cardinal is a Rowbottom cardinal; we shall show that the statement of Lemma 29.10 for a measurable cardinal holds under the weaker assumption that K carries a a-saturated u-complete ideal.

Lemma 34.3. Let I be a normal a-saturated u-complete ideal over K , and let I be an injnite cardinal less than K . Let PI = ( A , . . .)be a model of a language 2'such 2 'I 21,and let A 2 K . If P c A i s such rhat [ P I < K , then PI has an that 1 elementary submodel '23 = ( B , . . .) such that B n K has measure one and IP n B I I A. Moreover, i f X c A is of size at most I , then we canjnd '23 such that X c B. As in the proof of Lemma 29.10 (or Theorem 71), the result follows from this combinatorial lemma:

Lemma 34.4. Let I be as in Lemma 34.3, let y < K and let$ [K]'" + y be a partition. Then there exists H G K of measure one such that the image of [HI'" under f is at most countable. Proof. We proceed as in the proof of Theorem 70. It suffices to show that for each n = 1, 2, ... there is H, of measure one such that f[[H]"] is at most countable; then we take H = fin"= H,. We prove, by induction on n, that for every partition of [K]" into less than K pieces there is H G K of measure one such that f [[HI"] is at most countable. For n = 1, let ,f: K + y and y < K ; let W be a maximal pairwise disjoint family of subsets X G K such that X has positive measure and f is constant on X . Let H = { X : X E W ) .Since I WI I No,we have [HI[ 5 No,and since y < u and I is K-complete, we clearly have K - H E I . Let us assume that the assertion is true for n and let us prove that it holds also for n + 1. Let f: [K]"' + y where y < K. For each a < K , we define f, on [ K - {a)]" by ,f,(x) = J ( { a } u x). By the induction hypothesis, there exists for each a < K a set X , of measure one such thatL[[X,]"] is at most countable; let A , be the image of [ X , ] " u n d e r h . Let X be the diagonal intersection

u

If

'

The set X has measure one since I is normal; also if a < a < . . * < a, are in X , then {al, .. .. a,). E [X,]" and so f ( { a ,a l , . .., a,,}) = fa({al, ..., a,}) E A,. For each a E X , let A, = {a,," : n < w } . For each n, consider the function gn: X + y defined by g,(a) = a,.". There exists a set H, G X of measure one

6.

420

OTHER LARGE CARDINALS

such that g,[H,] is at most countable. Thus let H = H,; the set H has g,[H] is at most countable. measure one, and moreover {A, : a E H} = It follows thatf[[H]”+’] is at most countable. W

u

u:==o

We can now proceed as in Theorem 75 and prove that if I is a normal a-saturated K-complete ideal over K and V = LEI], then GCH holds. In fact, if D denotes the filter dual to I, that is, the filter of all sets of I-measure one, then the proof of Theorem 75 goes through verbatim in the present context. Now we recall the results of Section 27: If K carries a maturated K complete ideal then either K I 2N0or K is measurable. Thus we conclude: If I is a normal o-saturated K-complete ideal over K and V = L[I], then K is measurable. froofof Theorem 82a. Let K be real-valued measurable. Then there is a normal ic-additive measure p on K by Lemma 34.1. Let I be the ideal of sets of measure zero, and let us consider the model L[I]. Let J = I n L[I].We have L[J] = L[I],and it is easy to verify that in L[I],J is a normal a-saturated K-complete ideal over K. Thus in L[I], K is a measurable cardinal. W

Let me now prove the generalization of Theorem 82a for maturated ideals. Since I am not able to show directly that if I is countably saturated, then I n L[I] is countably saturated in L[I], let me consider a somewhat more general situation. Let v be a regular uncountable cardinal less than K, and let us consider v-saturated K-complete ideals over K. (Hence c-saturated is the special case when v = N,.) Lemmas 34.1 and 34.2 hold again; in Lemma 34.4, we have to replace “at most countable” by “of size less than v.” Lemma 34.3 holds for all i, 2 v and the analog of Theorem 75 is: If V = L[I] and I is normal, then 2‘‘ = v and 2A= L+ for all i, 2 v. Lemma 27.5 can also be generalized, and we get: If I is atomless, then K 5 2 < ” .Hence if V = L[I], every set of positive measure contains a subset that is an atom, and therefore K is the union of a disjoint family W of atoms such that ( W ( c v. Lemma 34.5. Let v < K be a regular uncountable cardinal, and let I be a normal v-saturated K-complete ideal over K ; let F be the dual filter. Then in L[F], F n L[F] is a normal measure on K (and L[F] is rhe model L[D] of Section 31). Proof: To start, it is easy to verify that L[F] = L[I], and that in L[F], I n L[I]is a normal v-saturated K-complete ideal over K. Thus we may assume that V = L[F]; we want to show that F is an ultrafilter. We know that K is the union of a disjoint family W of atoms. (What we want to show is that W has only one element.).For A E W, let FA = {X E K : X n A has positive measure)

Since A is an atom, F A is a filter, and it is easy to see that F Ais in fact a normal

34. REAL-VALUED

42 1

MEASURABLE C A R D I N A L S

measure on K. Hence FA n L [ F A ] is the unique normal measure D in L[FA], and L[FA] is rhe model L [ D ] . We shall now show that F n L[D] = D . Let X E L [ D ] be a subset of K. If X E F, then X E F A for all A E W and hence X E F A n L [ F A ]= D. If X 4 F, then there is A E W such that X 4 FA and hence X 4 FA n L [ F A ] = D . I t follows that F n L [ D ] = D and so F n L[D] E L[D]. Consequently, L[F] = L [ D ] ;since we assumed that V = L [ F ] and because F n L[D] = D,we have F = D. A Model in Which 2'O

is Real-Valued Measurable

Let K be a measurable cardinal, and let I 2 K be a cardinal such that ,INo= 1. We shall construct a generic extension in which 2N0= I and K is real-valued measurable. Let F be a a-complete field of sets and let p be a measure on F (cf. Section 27). Let I G F be the ideal of sets of measure 0 and let us consider the Boolean algebra B = F / I . That is, the members of B are equivalence classes [XI where X E F and where X = Y just in case p ( X A Y ) = 0. Since both F and I are countably complete, it follows easily that B is X , ] . Since p is a measure, I is countably complete and C:=o [X.] = countably saturated and so B satisfies the countable chain condition. Now a Boolean algebra that is both a-complete and a-saturated is complete, and so B is a complete Boolean algebra. For [XIE B, let us define m ( [ X ] )= p ( X ) . Clearly, the definition of m does not depend on the particular choice of X , and furthermore, rn has the following properties:

(34.1) (i) m is a real-valued function on B ; (ii) m(0) = 0, m(a) > 0 if a # 0, and m(1) = 1; (iii) if a I h, then m(a) I m(b); (iv ) if a n , n = 0, 1, ..., are pairwise disjoint, then

A Boolean algebra B with a measure m (satisfying (34.1))is called a measure algebra; a set S with a field of sets 9 and a measure p on .Fis called a measure space. We need from measure theory the following basic fact about products of measure spaces. Let I be a set (of indices), and for each i E I let (Si,Ti,pi) be a measure space. Let us consider the Cartesian product S = E , S i , and let us consider the following a-complete field of subsets of S: Let E be a finite subset of I, and for each i E E, let Z i E F i .Let Z E S be as follows: If t E Si , then

ni

ni

(34.2)

t

E

Z

iff

t ( i ) E Zi for all i

E

E

422

6.

OTHER LARGE CARDINALS

Let 9 be the least a-complete field of subsets of S such that 9 contains every Z G S of the form (34.2), for any finite E c I and any Zi E .Fi,i E E. There exists a unique measure p on F (the product measure) such that for every Z of the form (34.2), p ( 2 ) is the product of pi(Zi), i E E. (In case of the product S = S , x S , , the measure of a “rectangle” 2, x Z 2 is equal to P I (Z,1 . P2(Z*).) We shall use the following simple example of a product measure space. Let I be an infinite set, and for each i E I let us consider the space (0, 1) of two elements. We give measure $ to both (0} and (1): (34.3) si = (0, l}, 9 i = P ( S i )

f,

14121)= 0, P({Q 11) = 1 pi({O)) = pi(11))= (The measures pi do not satisfy the condition (iii) in 27.1.)Let S =

nis

Si,and let p be the product measure on 9,the least a-complete field of subsets of S containing the sets {r E ‘{O, 1) : t ( i ) = 0}, for all i E I. Let 9Jl be a transitive model of ZFC (the ground model). In 9 3 3 1 ,let I be an p ) be the product measure space infinite cardinal such that IN”= 1.Let (S, 9, ‘{O, 1) defined above, where I = I x w. Let B be the corresponding measure algebra .F/the ideal of sets of measure 0. Let G be an 9331-generic ultrafilter on B. Since B satisfies the countable chain condition, the generic extension 9331[G] preserves cardinals. We shall show that in 9UI[G], 2K0= 1. On the one hand, an easy computation gives ( JI = I (because AN” = I) and since B satisfies the c.c.c., we get I B I = I. Therefore =1 (2No)*G1I ( I B and we have (2N0r*G1I 1. On the other hand, we shall exhibit I distinct subsets of w in %331[G].For each a < I and each n < w, let ua+ = [ Uu.,,I, where U a , E -7is as follows:

(344)

Ua, = { t E

“(0, 1) : r(a, n) = I}

For a < I, let xu be the B-valued subset of w such that (34.5) IIA E x u 1 = u a n (n < W ) Let x, be the G-interpretation of x u . We shall show that x u # x p whenever a # 8, and in fact that IJx,= x p 11 = 0. Let k be any natural number. Then l7

k’ = x p

l7

= “ z . p. k ]

where N a . p ,= k { t : [(a, n ) = t(P, n) for all n < k }

It is easy to verify that for each k, CI“zBk)

=

lPk

34.

423

REAL-VALUED MEASURABLE CARDINALS

[n&

But ))XU= Xo 1) = flp=i)" a s k ] = NqJk]= 0 since p ( n & This completes the proof that 2N0= A in !W[C].

N@k) = 0.

Exercise 34.3. No x, is in the ground model. I f a E w , then IIx, = 611 = 0. = 1/2k.] [Show that for each k, IIx. n k' = 6 n k'll = [D..*]where

Now let us assume that K is a measurable cardinal in the ground model, and let A 2 K be such that AHo = A. We construct a generic extension 9R[G] of !Dl, using the measure algebra described above. In !DI[G],we have 2'O = A, and all we have to do is to show that K is real-valued measurable in W [ G ] .This follows from this general lemma:

Lemma 34.6. Let K be a measurable cardinal in the ground model 9l?, let B be (in '91a)measure algebra, and let G be an Wgeneric ultrafilter on B. Then in 91[G], there exists a nontrivial K-additive measure on K . Proof. Let U be a K-complete nonprincipal ultrafilter over K. Let B be a complete Boolean algebra and let m be a measure on B. We shall define a B-valued name p and show that if G is a generic ultrafilter, then the G-interpretation of I( is a nontrivial K-additive measure on K. Let a be a nonzero element of B, and let A E 91P be a B-valued name such that a ~k A E <. For each a < K, we let (34.6)

Since U is K-complete, there is a unique real number r such thatf,(A, a ) = r for almost all a (mod U ) . Thus let (34.7) p,(A) = the unique r such thatf,(A, a ) = r almost everywhere(mod U) A = A', then p,(A) = p,(A'). Also, if a Ik A , c A , , then c K is in 9J1, thenf,(X, a ) = 1 for all a E X and = 0 for all a X. Hence p , ( X ) = 1 if X E U and p , ( i ) = 0 if X $ U . Let y < K and let A,, < y, be such that a Ik A, c 2 for all 5 < y, and that a 11A, n A, = $3whenever # q. Let A be such that a 11 A = Ut;.yAg.Then f,(A, a ) = !,(A;, a ) for all a < K , and hence (because U is K-complete),

Note that if a

IF

p,(A,) I p,(A2). IrX

+

r

Ec
(34.8)

<

&(A) =

1

t
Let r be a real number, 0 I r I 1, and let {a,,}~=,-,be a partition of a E B. If p , J A ) < r for all n, then for almost all a, m(a, . E ,411) < r . m(a,), and it follows that for almost all a, m(a . (/ai E A ( ( )< r . m(a);hence p,(A) < r .

[I&

As a consequence, we obtain: (34.9) If for every nonzero 6 I a there is a nonzero c I b such that p c ( A )< r, then p,(A) < r.

424

6.

OTHER LARGE CARDINALS

(And a similar statement holds when < is replaced by Now if b Ik A c_ C, we define

I,

> or

2.)

(34.10)

Again, if b ~k A , G A , , then p t ( A ,) I p t ( A z ) ,and if X E 91t then p t ( i ) = 1 if X E U and = 0 if X E U . However, p t is not additive and we only have (using (34.8)), for y < K : (34.11)

under the assumption that b If A , n A , = 0 whenever 5 # q, and that b If A = UC.?A , . Note that if b , I b,, then C ( b , ( A ) 2 pb2(A). Now we are ready to define p. Let G be a generic ultrafilter; in WI[G],we define p : P ( K )-, [0, 11 as follows: P ( A ) = SUP Pb*(A)

(34.12)

bEC

where A is a name for A. Let p be the canonical name for p (defined in 91P by (34.12) using the canonical G). It is clear that p does not depend on the name A for A, that A G A implies p ( A , ) I p ( A 2 ) and that if X E rUl, then p ( X ) = 1 if X E U and p ( X ) = 0 if X # U . It remains to show that p is K-additive. I shall now explain the meaning of p t . Let r be a real number (in 91)such that 0 I r 5 1. I claim that

,

(34.13)

p:(A)>r

iff

b 1fp(A)2i

If p f ( A ) 2 r, then for every generic G such that b E G, p ( A ) 2 r, and hence b IF p ( A ) 2 i. Thus assume that b ~kp ( A ) 2 i. Then

b

If

vq < i 3d E

G pd*(A) 2 q

that is, V q < r Vc I b 3d I c p:(A) 2

(34.14)

Let q < r; I claim that p ; ( A ) 2 q. If a I b, then Vc Ia E Ic such that p:(A) 2 q and hence (by a variant of (34.9)), pa(A)2 q. Thus p;t(A) 2 q. Since this holds for any q < r, we have p $ ( A ) 2 r. Next we show that p is finitely additive. Let A, A , , and A , be such that every condition forces that A is the disjoint union of A and A ,. If r and r 2 are real numbers and if b ~k ( ! ( A 1 ) 2 i , and p ( A z ) 2 i,),then by (34.13) and (34.11), b ~kp ( A ) 2 ‘I, i z ;hence p ( A ) 2 p ( A , ) p ( A 2 ) . Conversely, let us assume that p ( A ) > p ( A , ) p(A,). There are reals r , , r2 E 91 and b E G such

,

+

+

+

that b

Since b

ik

~kp ( A l ) <

p ( A , ) < i l , p ( A 2 )< i,,and p ( A ) 2 ‘Il

+ i2

f1,there is for each c Ib some d I c such that pAA ,) < r , ;

34.

425

REAL-VALUED MEASURABLE CARDINALS

hence by (34.9), p b ( A l ) < r l . Similarly, &(A2) < r 2 , and so & ( A ) I &(A) < r l + rz . This is a contradiction. Now when we know that p is finitely additive, it suffices to show that p ( u , < u A,) I C, p(A,). Then there is r E 91 and b E G such that

utcy

b

1p(A,) < i and p ( A ) > i

IF

,
be an arbitrary finite set, let A E = UCEE A,. Since Ilp(AE)I If p ( A E )< i. By (34.9), we get C(b(AE)< r. Since p b ( A E )< r for all finite E c y, it follows from (34.8) that pb(A) I r. Hence p(b*(A)I r, a contradiction. This completes the proof that in 9I[G] ,u is a nontrivial K-additive measure on K. W Let E

G -y

C S Ep(A,)ll E = 1, we have b

Another Model in Which 2'O Carries a a-Saturated Ideal be a measurable cardinal, and let 1 2 K be a cardinal such that shall construct a generic extension that satisfies 2'O = 1 and such that there is a a-saturated K-complete ideal over K . Let P be the notion of forcing that adjoins I Cohen reals; i.e., a condition is a finite 0-1 function p with dom(p) c 1. If G is a generic filter on P, then 911[G] l=2K0= 1,and all cardinals are preserved in 91[G]because P satisfies the countable chain condition. That K carries in M[G] a maturated ideal follows from this lemma: Let

1'O

K

= 1.We

Lemma 34.7. Let K be a measurable cardinal (in '331) and let 1 be a nonprincipal

K-complete prime ideal over K . Let P be a notion of forcing that satisjes the countable chain condition. Then in 911[G], the ideal J generated by 1 is a asaturated K-complete ideal over K. Pro05 Let J be the ideal in 91[G] defined as follows: X

E

J

iff

X

G

Y for some Y E 1

First we show that J is K-complete. Let S = {X,: 5 < y} be a family of less than K elements of J ; let .lr be a name for S and let p o E G be such that po Ik vy < r' x,E J . For each 5 < y and each p I p o , there exist q I p and some Y E 1 such that q ~fX, G f. Let W, be a maximal incompatible family of q I p o for which there is Y,, such that q 11X, G ?<,p. Since P satisfies the countable chain condition, each W, is countable, and hence Y = { 6, : c y and q E Ws}belongs to 1. Now it is easy to verify that po If

and hence

0X E J.

u u x,

S
<

G

f

426

6.

To prove that J

OTHER LARGE CARDINALS

is countably saturated, let us assume that

x^ = {X,: tJ < oI} is a family of pairwise disjoint sets of positive J-measure. Let

T be a name for 3 and let p E G be such that p It X, 4 J, for each tJ < ol, and p It T , n T,,= 0 for any [ # q. For each 5 < olr let 5 = {a < K : some q I p forces 2 E X < } . Clearly p It X,E $, and so % # I . By the K-compactness of I, we have Y= Y, $ I. Thus Y # 0, and let a be some element of Y. For each < w , , let q, Ip be such that q, IF ai E T,. Since P satisfies the countable chain condition, there are 5, q such that q, and q,, are compatible. Let q be stronger than both q, and q,,; then q IF di E 3, n T,,,a contradiction. W

nccwl

<

Exercise 34.4. Let I be a a-saturated K-complete ideal over K, and let { Y< : ( < w , } be a family of sets of positive measure. Then there is an uncountable W E w 1 such that Ys is nonernpty. [Assume that { Y < :{ < w I } is a counterexample. For each v < w , . let Z, = (ti - Ye). Show that Z, E Z , E ... G Z, c_ ... and that Uv<w, Z, = K. Hence there is y < w , such that if Z = Ue
ns

n;>,

Exercise34.5.* I f I is a o-saturated K-complete ideal over K (in 911) and P is a a saturated notion of forcing then in !Ul[G], the ideal generated by I is a 0-saturated ti-complete ideal. [Proceed as in Lemma 34.7 and use Exercise 34.4 to show that J is a-saturated.]

K

If we adjoin 3, 2 K Cohen reals to a model 'Min which ti is measurable, then carries a a-saturated K-complete ideal but is not real-valued measurable:

Exercise 34.6.* Show that in 9ll[G] there are functionsf,: w + w, a: < I, such that whenever g : w -+ (0, then for at most countably many a's we havef,(n) < g ( n ) for all n. [!U1[G] is also obtained by forcing with the product of I copies of the notion of forcing that adjoins a generic function f: w -+ w, thus !UI[G] = !Ul[(f, : a: < A)]. Show that if g : (0 -+ o,then there is a countable A c I such that g E 911[(f9 : CL E A ) ] ; if fi $ A, use the genericity offp over !Ui[(L: a: E A)] to show thatfp(n) > g ( n ) for some n.] Exercise 34.7. In !Ul[G],ti is not real-valued measurable. [Use Exercise 34.6 and the proof of Lemma 27.9.1

Proof of Theorem 83

We shall prove that if 2"O carries a maturated 2Ko-completeideal then 2L = 2'0 for all 3, < 2'0. Let 3, be a regular cardinal; two functionsJT g on 3, are almost disjoint if there is y such that f ( a ) # g(a) for all a 2 y. Lemma 34.8. Let K carry a a-saturated K-complete ideal, and let I < K be a regular cardinal. If J is a family of almost disjoint functions f: 1 -+ K , then 1.9I K.

35.

SATURATION OF IDEALS A N D GENERIC ULTRAPOWERS

427

Proof: If 1 . 9I > K, then because every f: I -, K is bounded by some /? < K, there exists Y E 3,and some /3 < K such that I Y 1 = K and every f~ Y is bounded by /3 (i.e.,f E ‘/I). Let F: [Y]’+ /3 be the following partition: F ( { j 9)) = some y such that f ( a ) # g(a) for all a 2 y. By Lemma 34.4, there exists X’ s Y of size K such that the image A of [XI’ under F is at most countable. Let a > sup(A). Then f ( a ) # g(a) whenever 1; g E X , which is a contradiction since I -W I = K and ! ( a ) < j for all!€ .W.

Now we are ready to prove the theorem. Let K = 2 ’ O . We prove 2* = K by induction on A < K. If I is a singular cardinal and 2’ = K for all v < A, then 2‘ = K by Theorem 18. Thus let I < K be regular and let us assume that 2’ = K for all v < I. Foreach X r I , l e t , f , = ( X n a : a < A ) . I f X # Y , thenf,andf,are almost disjoint. For each a < I, the set { f x ( a ): X G A} has size K, and hence { f x : X E I}yields a family .Pof 2A almost disjoint functions from 2 into K. By Lemma 34.8 we get 1.FI 5 K and therefore 2 A = K. 35. SATURATION OF IDEALS AND GENERIC ULTRAPOWERS

In Section 34 we investigated countably saturated ideals over an uncountable regular cardinal K. In particular, we have shown that if such an ideal exists, then K is at least weakly Mahlo, and we have also shown that existence of such ideals is equiconsistent with existence of measurable cardinals. The present section deals with a related concept. We define the notion of a I-saturated ideal over K and investigate these ideals in the cases when I = K or I = K + (or generally when I I2“). Aside from the general question of existence of 2saturated ideals for various values of I, we are particularly interested in saturation of the two “natural” ideals over K : The ideal of small (i.e., of size < K ) subsets of K and the ideal of thin subsets of K. In the first case, the problem is the already familiar problem of the number of almost disjoint subsets; while in the second case, we obtain some results on stationary sets. Many of the results in this section are proved using the method of generic ultrapowers, which generalizes the method of ultrapowers used in Chapter 5. Let K be a regular uncountable cardinal. Throughout this section, let I be a k--complete ideal over k- containing all singletons; thus X E I whenever X c K is such that 1X I < K. As in Section 34, I shall be using the following terminology: X has measure zero if X E I, measure one if K - X E I, and positive measure if X 4 I ; the phrase almost all a means that the set of all contrary a’s has measure 0. There are two natural examples of such ideals: the ideal of small sets (X c K : 1 X I < k-) (the least possible such ideal) and the ideal of thin sets

6.

428

OTHER LARGE CARDINALS

{X c K : K - X contains a closed unbounded set},which is (see Lemma 7.7) the least normal K-complete ideal over K. Let us consider the Boolean algebra B = P ( K ) / I . Recall that if I is a cardinal, then B is called 1-saturated if every pairwise disjoint family of elements of B has size less than 1;sat@) is the least 1such that B is A-saturated. Let us say that I is I-saturated if B is I-saturated and let sat(I) = sat@) In other words, I is I-saturated just in case there exists no collection W of size 1 of subsets of K such that X .$ I for all X E Wand X n Y E I whenever X and Y are distinct members of W. If sat(I) is finite, then K is the union of finitely many atoms of I; if sat(1) is infinite, then it is uncountable and regular, by Lemma 17.6. If 1I K, then 1 is A-saturated if and only if there is no disjoint collection W of size A of subsets X of K such that X .$ I (see Exercise 27.4). Clearly, every I over K is (2'()+-saturated.Thus if I is atomless, then sat(I) is a regular cardinal and

K1 Isat(I) I(2")+ Since I is K-complete, it follows that B = P(K)/I is a K-complete Boolean algebra (Exercise 17.16). Exercise 35.1. If I is K+-saturated, then P(K)/Iis a complete Boolean algebra. [By Exercise 17.23 it suffices to show that B is complete. Show that W exists in B for every incompatible W c B. Extend W to a partition Z of B ; Z = {[X.] : a < K}. Let Y. = X, X,, and Y = { X : [Xu]E W}. Show that [Y] = W in E.]

uB<,

1

u

1

Exercise 352. If the GCH holds and B = P(K)/I is complete, then I is K+-saturated. [If B is not ~'-saturated,let W be an incompatible subset of B of size K + . For each X E W let ux = X.It follows that IBI 2 2"', but clearly IBI I 2"; a contradiction.]

1

Exercise 35.3. If / is normal, then P(K)// is sa complete. [Let X , , a < K, be disjoint subsets of K such that X , $ I for all a. For each a < K let Y, be X, without the least element of X,; let Y = Y,. On the one hand, [Y] 2 [X,] for all a ; on the other hand, if Z c Y and Z $ I, letfbe the function on Z defined such that for all x E Y, , f ( x ) = the least element of Xu.Sincefis regressive, and I is normal, f is constant on some S $ I, and hence Z n Y, 4 I for some a. Thus [Y] = [X,].]

u,<"

z<,

We proved in Section 27 that if there exists a K-saturated ideal I over K, then is weakly inaccessible. Furthermore, the proof of Lemma 27.5 shows that if sat(I) = I < K, then either K is measurable or K is not strongly inaccessible because K I 2<'. In the last section we proved that existence of a countably saturated ideal over 2'" is equiconsistent with existence of a measurable cardinal. Our next observation is that Lemmas 34.1 and 34.2 remain valid if 0saturated is replaced by K-saturated; the proof is exactly the same. Thus we have: K

35.

SATURATION OF IDEALS AND GENERIC ULTRAPOWERS

429

Lemma 35.1. (a) If there exists a K-saturated K-complete ideal over an uncountable cardinal K , then there exists a normal K-saturated K-complete ideal over K . (b) Let I be a normal K-saturated K-complete ideal over K . If S flI and if$ S + K is regressive on S, then there is y < K such that f ( a ) < y for almost all aES. H A consequence of (b) is that K is a weakly Mahlo cardinal. (I shall return to this later in the present section.) Exercise 35.4. If I is a K-complete ideal over K with the property that every regressive function is bomded almost everywhere (i.e., iff(a) < GI for almost all a, then there is y < K such thatf(a) < y for almost all a), then I is K-saturated (and normal). [Otherwise, let X,, a < K , be a partition of K into disjoint sets of positive measure. For a > 0, let Y, = X, - (a,} where a, = min X,, and let Yo = X o u {a, : a > 0). The function f that has value a. on each Y, is regressive almost everywhere but is not bounded almost everywhere.]

The proof I presented in Section 34 that if sat(1) < K, then K is measurable in a transitive model does not go through if sat(1) = K. It is nevertheless true that K is measurable in a transitive model, and later in this section I shall give a proof of a more general theorem. In Section 34 we gave an example of a generic model in which K = 2'O carries an ideal I with sat(1) < K. A different forcing construction gives a generic extension (assuming that K is measurable in the ground model) in which K = 2 " O has an ideal I with sat(1) = K. If sat(1) < K and K is inaccessible, then I cannot be atomless (and K is measurable). However, it is consistent (relative to ZFC + there is a measurable cardinal) that an inaccessible cardinal K carries an ideal I with sat(1) = K. There is a restriction on K though: Exercise 35.5. I f I is an atomless h--complete K-saturated ideal over an inaccessible cardinal h-, then h- is not weakly compact. [Show that h- does not have the tree property. Use I to construct a tree ( T , 3 )whose elements are sets of positive measure. At successor steps, split each X on the top level into two disjoint sets of positive measure. At limit steps, take all those intersections along branches that have positive measure. Since I is K-saturated, each level has size < K ; each levcl a < h- is nonempty because K is inaccessible and I is K-complete. Then use sat(1) IK to show that T has no branch of length K . ]

(As a consequence, no weakly compact K carries an ideal I such that sat(1) = K.)

Now, let us consider the question of whether a regular uncountable cardinal can carry a K-complete ideal I such that sat(1) = K ' . I shall prove later in this section that if that is the case, then K is measurable in a transitive model. Conversely. if K is measurable, then there exists a generic extension in which ii is inaccessible and carries an ideal I with sat(1) = K + . K

430

6.

OTHER LARGE CARDINALS

Existence of a K' -saturated ideal over K does not necessarily imply that K is a limit cardinal. Thus let us consider the problem of existence of an N,-saturated ideal over wl. It has been shown that this assumption implies that for every a there exists a transitive model with a measurable cardinals. And conversely, if K is a huge cardinal (cf. Section 33), then there is a generic extension in which K = N Iand K , carries an K,-saturated ideal. In the particular case when I is the ideal of small subsets of K, then I is not K+-saturated (see Lemma 24.7) and so K' I sat(I) I (2")'. If I is the ideal of thin subsets of K, then I is not K-saturated, as we shall prove presently, and so K + I sat(]) I (2")'. It is not known whether the thin idealcan be K+-saturated and the assumption that sat(l)= K + seems to be quite strong (if not inconsistent). If 2" > K + then a new possibility arises: an ideal I over K such that K' I sat(1) I2". As we shall see, this assumption does not entail large cardinal properties; we shall show that there is a generic extension of L in which the ideal of countable subsets of w 1 is K,-saturated (while 2'l is large). However, there is a relation between saturation of ideals and cardinal exponentiation, and I shall start with a theorem that deals with this relation. +

+

Almost Disjoint Families of Subsets of K Let I be the ideal of small subsets of K : I = {X E K : I X I < K}. Thus sets of positive measure are just the subsets of K of size K, and "almost everywhere" means everywhere except for a set of size < K. To say that I is I-saturated means that there exists no family W of subsets of K such that I W 1 = Iz and that each X E W has size K and IX n Y I < K whenever X and Y are distinct elements of W. Let us call such a W an almost disjoint family of subsets of K (and let us call X, Y almost disjoint if I X n Y I < K. The proof of Lemma 24.7 shows that there exists an almost disjoint family of size K + ,and hence the ideal of small sets is not K+-saturated. Also, Exercise 24.6shows that if2'" = K, then there isa familyof2" almost disjoint subsets of K. We shall now state a more general theorem and then show that the ideal of small sets does not necessarily have the property that sat(!) = (2")'. The theorem establishes a relation between saturation of the ideal of small sets (and, in part (b), between saturation of arbitrary ideals) and the continuum function. Although the following results can be formulated for an arbitrary uncountable regular cardinal K, we shall consider only the representative case K = w Theorem 84. Let us assume that 2N0< 2"I and also that O 2' < N,,. Then: (a) there exists an almost disjoint family of size 2'l of uncountable subsets of w (b) if1 i s a o-complete ideal over wl, then sat(1) 2 2'l.

Part (b) can be also formulated as follows: If I is a o-complete ideal over w 1 and if 2"O < N,,, then 2H1I 2O ' . sat(]) (35.1)

35.

431

SATURATION OF IDEALS AND GENERIC ULTRAPOWERS

If there exists an K,-saturated ideal over ol, then we get

(35.2) and (35.3)

then 2 K 1= K2

if 2"O = K,, if K l < 2H0< KUI,

then

2"O = 2''

I shall now prove only part (a) of the theorem. Part (b) can be proved in a similar manner, but I will postpone the proof until later in this section and then give a proof using generic ultrapowers. Lemma 35.2. Assume 2" < K,, . If K is a regular cardinal such that 2O ' < K 5 2'l, then there exists a familji of K almost disjoint functions from w , into ol.

Proof. Let 9 be a family of almost disjoint functions on 0,; we call 9 a branching farniljj if whenever 1; g E .% and a is such that f ( a ) = g(a), then f ( 5 ) = g(5) for all 5 I a. For each X c wl, let f x = (X n a : a < wl). The family .F = {fx : X E P(ol)} is a branching family of functions on o,,1 8I = 2''; and for each a < wl, the functions in ."2 take values in P(a).Thus there exists a branching family of 2" functions from w1 into 2N0. Let K be a regular cardinal such that 2Nn< K 5 2'l. We shall show that for every K, such that K 1 < K, I 2N0,if there is a branching family of K functions from coy into w y ,then there is a branching family of K functions from 0, into some (0,< 0,. Then the lemma clearly follows. First let K, = N,, where K 1IKd,and let 3 be a branching family of K functions from Q, into w , + , . Each f E .F is bounded below a d +(because , w,+ > w,), and because K is regular and K > w,+ there exists a < wd+ such that ran(f) c a for K functions in .P.Thus there exists a branching family of K functions from (ul into a ; and since la I I K,, there is also a branching family of K functions from (oI into wd. If K, is a limit cardinal, then cf(w,) = w because K, < K,, . Let Y be a branching family of K functions from w into w , . For each f E .P there exists an ordinal qr < wi such that f ( a ) < qr for uncountably many a's. Since K is a regular cardinal and K > K,, there exists K, such that K , I K, < K,, and a family C i c .F of size K such that for every f E 9,f ( a ) < w, for uncountably many a's. For each a < (ol, let S, = { f ( a ) :f E 9}. Since Y is a branching family and (4 c S , , it suffices to show that 1S,I IK, for all a < ol. Thus let a < (ol. We define a function t : S , + w 1 x w, as follows: For each x E S , , we first pick someJe B such that x = f ( a ) . Then there exists some 5 > a such that , f ( ( )< U J ~ and , we let

,

,,

,

,

nl<,l

We shall now complete the proof by showing that the function t is one-toone, and hence I S, I I K, . Let x, y E S, be such that t ( x ) = t(y). Let 5 > a and

432

6.

OTHER LARGE CARDINALS

J g E 9 be such that x = f ( a ) , y = g(a), and t ( x ) = t ( y ) = (t,f(t)) = (t,g(5)). Since Y is a branching family andf(<) = g(<),we necessarily havef(a) = g(a), and hence x

= y.

Part (a) of Theorem 84 now follows easily from Lemma 35.2. Every function from w I into wI is an uncountable subset of w lx w I and almost disjoint functions are almost disjoint subsets of w 1 x wl. Thus for every regular K 5 2'I, there exists a family of K almost disjoint uncountable subsets of wl. If 2" is regular, we are done; if K is singular, then it is easy to construct an almost I) disjoint sets, disjoint family of 2H1subsets of w1 by taking first ~ f ( 2 ~almost etc. (or simply invoke Lemma 17.6 saying that sat(I) is regular and hence sat(]) > 2"). Part (b) of Theorem 84 can be proved along the same lines. If I is a o-complete ideal over w l r let us call an I-function on w 1 a function whose domain is a subset of w 1 of positive measure. Two I-functionsfand g are almost disjoint if the set of all a such that f ( a ) = g ( a ) is at most countable. If we assume that 1= sat(I) < 2", then by Lemma 35.2, there is a family of A+ almost disjoint functions from w 1 into wl. It is not difficult then to obtain a family of 1 almost disjoint I-functions from w l into w. This is, however, a contradiction : Exercise 35.6. Let I, 2 K z be a regular cardinal. If there are I, almost disjoint [-functions from w 1 into w, then there exists an almost disjoint family of I, subsets of w , ofpositive measure. [As in Exercise 24.7.1

The following example shows that the assumption 2H0< 2'I in Theorem 84 is necessary.

Example I. A model in which the ideal of countable subsets of w 1 is K,-saturated while 2" is large. Let 91 be a transitive model of ZFC + GCH and let W[G] be the generic extension of 91 which adjoins to !IN K Cohen reals (see (19.1)). We shall show that in W[G], every almost disjoint family of uncountable subsets of w has size at most K 2 .In fact, we shall use only the facts that 911 satisfies the GCH and that the notion of forcing satisfies the countable chain condition. Let us assume that 9 E m[G] is an almost disjoint family of uncountable subsets of w l r9= {Ai : i < us}, and let 9 be a name for 9. We may as well assume that every condition forces that 9 be an almost disjoint family, etc. Let i # j < w , . For every condition p there is 4 5 p and y < w 1 such that (35.4)

q

~fAi n

A j ' 7~

Thus there is a maximal incompatible set W of conditions, and for every 4 E W there is y = yq with the property (35.4). Since W is at most countable, we can take y = sup{yq: q E W } and y is a countable ordinal with the property that (35.4) holds for every condition q. Let us denote y =f(i, j ) .

35.

SATURATION OF IDEALS A N D GENERIC ULTRAPOWERS

433

The function f is a partition of [a3]'into X, pieces. By the Erdos-Rado theorem (Exercise 29.l), and by the GCH, we have

and therefore there exists H c w 3 of size N 2 , and y < w 1 such that (35.4)holds for every condition q and every pair i, j E H . Now we argue in 'X)I[G]:For every pair i, j E H we have A i n A j G y, and so { A , - y : i E HJ is a disjoint collection of K 2 nonempty subsets of 0,;a contradiction.

The Ideal of Thin Sets The ideal of thin subsets of a regular uncountable cardinal K is another natural example of a K-complete ideal. Moreover, it is the least normal K complete ideal over K containing all singletons. We shall now investigate saturation of the thin ideal. We proved in Lemma 7.6 that K can be partitioned into K stationary sets. We shall now prove a stronger theorem: Every stationary subset of K can be partitioned into K stationary sets. Thus n o restriction I I S = {X G S : x

E

I}

(where S is stationary) is a K-saturated ideal. As a digression, let me point out that one consequence is that not every subset of w 1 has the property that either itself or its complement has a closed unbounded subset. Let us consider the following problem: Does every subset of w2 have the property that either itself or its complement (in 02) has a subset X of order-type w , that is w-closed (i.e., limn-,ua,, E X for every sequence {a,,}n
Theorem 85 (Solovay). Let K be a regular uncountable cardinal. Then every stationary subset of K is the disjoint union of K stationary subsets. We shall first prove the following lemma. Recall that if a is a limit ordinal of uncountable cofinality, then the notion of closed unbounded and stationary subsets of a is defined in the natural way, and the collection of closed unbounded subsets of a generates a filter over a (and the filter is cf a-complete).

Lemma 35.3. Let S be a stationary subset of T = (a E S : either cf a

= o,or

K, and

let

cf a > w and S n a is not stationary in a}

Then T is a stationary set.

Recall the Mahlo operation defined in (28.23):

M(X) = { a < K : X n a is a stationary subset of a}

6.

434

Thus Lemma 35.3 states that for every stationary S stationary.

OTHER LARGE C A R D I N A L S

C_ K ,

the set S - M(S) is

Proofi We prove that T intersects every closed unbounded subset of K . Let C be closed unbounded. The set C' of all limit points of C is also closed unbounded, Let a be the least element of S n C'. If cf a = w, then and hence S n C" # 0. a E T ; and since a E C' s C , we are done. If cf a > w, then C' n a is a closed unbounded subset of a and is disjoint from S n a, and so S n a is not a stationary subset of a. Hence a E T , and again C n T is nonempty. Proof of Theorem 85. I shall try to follow the proof of Lemma 7.6 as much as possible. Let A be a stationary subset of K . We let W be the set of all a E A such that either cf a = w, or cf a > w and A n a is not stationary. Thus there exists for each a E W a continuous increasing sequence (at : 5 < cf a ) such that a: cj W , for all a and <, and a = lim6+cfz a:. First we show that there is 5 such that for all q < K , the set

(35.5)

{a E W : a; 2 q}

<

is stationary. Otherwise, there is for each some q(5) and a closed unbounded set C , such that a: < q(<) for all a E C, n W if a! is defined. Let C be the diagonal intersection of the C , . Thus if a E C n W, then u:

< q(<)

for all 5 < cf a

Now let D be the closed unbounded set of all y E C such that q(5) < y for all 5 < 1'. Since W is stationary, W n D is also stationary; let y < a be two ordinals in W n D. Now if 5 < y (and 5 < cf a), then a: < q(5) < y and it follows that a; = y. This is a contradiction since y E W and a; 4 W . Once we have found 5 such that (35.5) is stationary for all q < K, we proceed as in Lemma 7.6. Letfbe the function on W defined by

f (a) = a; The functionfis regressive; and so for every q < K, we find by Fodor's theorem a stationary subset S,, of (35.5) and y,, 2 q such thatf(a) = y,on S , . If y, # y,,,, then S,, n S,, = 0;and since K is regular, we have / { S , : q < K } I = [ { y , : q < K } I = K. Since the thin ideal is not K-saturated, the question is whether it can be K+-saturated, or in general, 2"-saturated. While the first question is still open; the following example shows that we can have sat(J) I 2" for the thin ideal over K.

Example ZZ. A model in which the ideal of thin subsets of w 1 is K,-saturated while 2" is large.

35.

435

SATURATION OF IDEALS A N D GENERIC ULTRAPOWERS

(A refinement of the following argument shows that the thin ideal in this model is in fact S3-saturated.) As in Example I, let 91i be a transitive model of ZFC + GCH and let 9lJ[G] be the generic extension that adjoins K Cohen reals. We shall show that in 9lJ[G], if .Fis a family of stationary subsets of w1 such that the intersection of any two distinct elements of 9 is thin, then I.F( I S,. Again, we shall use only the fact that 9 3 1 satisfies the GCH and that the notion of forcing satisfies the C.C.C. First we need a lemma:

Lemma 35.4. Let 9I[G] be a generic extension of911 by a C.C.C. notion offorcing. Then every closed unbounded C E W , in 9UI[G] has a closed unbounded subset D such that D E 911. Consequently, ifS E 911 is stationary (in 911), then 9i[G] k S is stat ionary. Proof: Let C be a name for C such that every condition forces that Cis a closed unbounded subset of wl. Let

Clearly, D is a subset of C and is closed; we have to prove only that D is unbounded. Let oro < w 1;we wish to find a > a. such that every condition forces cr' E C. For every p, there is q 5 p and some p > a. such that q 11 E C. Thus there is a maximal incompatible set W of conditions, and for each q E W an ordinal p = j, such that q 11$ E C. Since W is at most countable, we let aI= sup{p,, : q E W); the ordinal al is countable, and we have

6

p

Ik

( 3E C) $0 < p I oil

for all conditions p. Similarly, we find a 1 < a 2 < or, < ... such that for every n and every condition p , p

1 1 (3p E

C)cr', < p I i , + l

If we let a = limn or,, it is clear that )Id E C)1 = 1. H Now let us assume that 9 E %1[G] is a family of stationary subsets of w , , .F = {Ai : i < w,,} such that Ai n A, is thin whenever i # j, and let .Bbe a

name for 9. Let i # j < 04.By Lemma 35.4, Ai n A, is included in a thin set that is in 91. Thus for every p there is q I p and a thin set X E W such that

(35.6)

q

11

n A, E

X

Thus there is a maximal incompatible set W of conditions, and for every q E W there is X = X, with the property (35.6). Since I W I IKO,we can take

6.

436

X

OTHER LARGE CARDINALS

u

= {X,: q E W}; and X is a thin set such that (35.6) holds for every condition q. Let us denote X = f ( i , j ) . The function f is a partition of [a4]'into 2" pieces. By the Erdos-Rado theorem and by the GCH, we have

and therefore there is H c w4 of size K3and a thin set X such that (35.6) holds for every q and every pair i, j E H . Now arguing in 9IUI[G], we have A i n A j G X for every i, j E H , and so { A i - X : i E H } is a disjoint collection of U3 stationary subsets of 0,; a contradiction. Exercise 35.7. Let 1 be a normal K-complete ideal over K. If 1 is not K+-saturated, then there exists an almost disjoint family of K + sets of positive measure. [Let X i , i < K ' , be sets of positive measure such that X i n X jhas measure zero. For each i < K'. enumerate {Xi:j < i} by { Z , : a < K } , and let yi be the diagonal intersection of { X i- Z , : a < K } . Now yi contains almost all elements of X i , and yi n Z. E a + 1 for every a c K. Thus any x, yj are almost disjoint.] Exercise 35.8. If the ideal of thin subsets of K is K+-saturated, then every normal K complete ideal over K containing all singletons is K+-saturated. [Use Exercise 35.7 and the fact that every set of positive measure is stationary.]

There are several results describing situations when the thin ideal is not K+-saturated. One such result is contained in Theorem 84b; for instance, if 2"O = K 1 and 2" > K,, then the thin ideal over o1 is not K,-saturated. By another result (cf. Exercise 22.1 1) Jensen's 0 implies that there are 2'' almost disjoint stationary sets. (And both these results generalize to other regular uncountable cardinals.) The following lemma shows that for some large cardinals K, the thin ideal over K is not K+-SatUrated. Lemma 35.5. Assume that the thin ideal over K is K+-saturated and let I be a normal K-complete ideal oveer K containing all singletons. Then there is a set A G K of measure one (i.e.,K - A E I ) such that every stationary subset of A has positive measure.

[Since every set of positive measure is stationary, this means that X E I} and lo is the thin ideal.]

I I A = I , I A where I I A = {X E A :

Proof: Let .P be a maximal almost disjoint (i.e., I S n TI < K ) family of sta9 I < K. If 1.FI = K , .F= { S , : a < K}. tionary sets of measure zero; we have 1 let A be the diagonal intersection of {K - S,: a < K}; if [.PI< K, let A = { K - S : S E ~ In } .any case, A has measure one, and [ A n SI < K for every S E 9 (because A n S, E a 1).

n

+

35.

437

SATURATION OF IDEALS AND GENERIC ULTRAPOWERS

Now if S E A is stationary, then S has positive measure, for otherwise 9 could not be maximal.

Corollary 1. I f

K

is measurable, then the thin ideal over

K

is not K+-saturated.

Proof: Let D be a normal measure on K and let I be the dual ideal. Every A can be divided into two disjoint stationary sets.

E

D

More generally, we have:

Corollary 2. I f K carries a K-saturated K-complete ideal, then the thin ideal over K is not K +-saturated. Proof. Apply Lemma 35.4 to a normal K-saturated K-complete ideal I over K, and use Theorem 85 to get a contradiction. The lemma can be further used to show, e.g., that if K is weakly compact, then the thin ideal is not K+-saturated.

Generic U ltrapowers We shall now introduce a powerful method for dealing with ideals over regular uncountable cardinals. The method is very similar to the method of ultrapowers that we used in Section 28. Let K be a regular uncountable cardinal and let I be an ideal over K. Let us view the universe as a ground model, let us denote this ground model 91, and let us consider the generic extension of 911 given by the completion of the Boolean algebra P ( K ) / I .In other words, consider the notion of forcing ( P , G), where P is the collection of all subsets of K of positive measure:

(35.7) X E P iff X c K and X # I ; X is stronger than Y iff X E Y. Let G be a generic filter on P.

Lemma 35.6. (i) G is an 911-ulrrajlter over K extending t h e j l t e r dual to I . (ii) If I i s h--completr (in %I), then G is 911-~-complete. (iii) f j f i s normal, then G is ‘IJLnormal.

P r m f (i) If X G K has measure one, then { Y E P : Y E X} is dense in P and hence X E G. That G is an 911-ultrafilter is obvious. (ii) If {X, : a < y}, y < h-, is (in 911) a partition of K, then by the Kcompleteness of I, the set { Y E P : Y c some X,} is dense in P and hence X, E G for some a. (iii) I f X E G and i f f € 911 is a regressive function on X,then { Y c_ X :,fis constant on Y ) is dense below X, and hencefis constant on some Y E G. H

6.

438

OTHER LARGE CARDINALS

From now on assume that I is a K-complete ideal over K containing all singletons. Then G is a nonprincipal 91-K-complete 911-ultrafilter over K. Note that if I is atomless, then G # 911 (but otherwise G may be in 911; e.g., if I is prime, then G is the dual of I). Let us consider (in 911[G]) the ultrapower u i f G ( 9 1 ) ; let us call this ultrapower a generic ultrapower. The generic ultrapower is a model of ZFC, but is not necessarily well-founded. We have of course the fundamental theorem, in this form (35.8) u l f G ( 9 1 ) k cp([fl],

. . ., [ f n ] )

iff

{ a : 91 k cp(fl(a),. . ., f.(a)} E G

wheneverf,, . . ., f n E 911 are functions defined on a set X E G. In particular, we have an elementary embedding, the narural embedding j , : 911 -,UlrG(Yll), defined by where c, is the constant function on K with value x, and [cJ is its equivalence class in the ultrapower. Let us denote the generic ultrapower by 91, andj, = j. The ordinal numbers of the model 91 form a linearly ordered class, not necessarily well-ordered, but we shall show that (because I is K-complete), Ord' has an initial segment of order-type K. If x E Ord', let us call the order-type of x the order-type of the set { y E Ord' : y -= ?x}. If the order-type of x is an ordinal number, let us take the liberty of identifying x with this ordinal. Lemma 35.7.

(i) For every y < K , j ( y ) = y ; hence Ord' has an initial segment of order-type K . (ii) j ( K ) # K. (iii) I f I is normal, then there exists x E Ord' such that x = K ; infact [d] = K , where d is the diagonal function d ( a ) = a. Proof. Routine. G is 91-K-complete, nonprincipal, and in case (iii) 911-normal. H

Let me mention again the fact that was already mentioned and that is fairly easy to verify: If P is the notion of forcing defined above (35.7),then r.0. P = r.o.(P(K)/I); the mapping X H[XI gives the natural correspondence. This means that sat(P) = sat(I), the fact that we are going to use in the applications. An important question is: When is the generic ultrapower well-founded? We shall now formulate a necessary and sufficient condition on I for this to be the case. Let S be a set of positive measure. An I-partition of S is a maximal family W of subsets of S of positive measure such that X n Y E I for any distinct X , Y E W. An I-partition W, of S is a refinement of an I-partition W2 of S, W, I W2, if every X E W, is a subset of some Y E W 2 .A functional on S is a

35.

SATURATION OF IDEALS AND GENERIC ULTRAPOWERS

439

collection F of functions such that W, = {dom(f) : f F} ~ is an I-partition of S and dom(f) # dom(g) wheneverf# g E F. We define F < G for two functionals on S to mean that: (1) eachfE F u G is a function into the ordinals; (ii) W, IW,; and (iii) i f f € F and g E G are such that dom(f) c dom(g), thenf(a) < g(a) for all a E dom(f).

The reason we define functionals is that they represent functions in the Boolean-valued model tul’ (and so are canonical representatives for elements of UltG(911)):Let f~ !JIBbe such that (35.9)

S Ikfis a function with dom(f) E C andfE tul

Then there is clearly an I-partition W of S , and for each X E W a functionf, on X such that for all X E W, X IF f 12 = Jx. Thus the functional { f x : X E W} “represents” the Boolean-valued f on S. Conversely, if F is functional on S, then there is f E 91’ such that (35.9) holds; and for eachfE F, if X = dom(f), then X IFf 12 = f Note also that if F < G are functionals on S andf; g are the corresponding Boolean-valued elements, then (35.10) g E !Ul and dom(f) E dom(g) and f(a) < g(a) for all a E dom(f) S IF Conversely, iff and g satisfy (35.lo), then there are functionals F and G that represent f and g, and F < G. Let I be a K-complete ideal over K containing all singletons. The ideal I is precipitous if whenever S is a set of positive measure and {W,: n < W ) are I-partitions of S such that wo2 w,2 ‘.‘ 2 w,2 * * . then there exists a sequence of sets

xo 2 X I2 ... 2 x,2 ... such that X, E W, for each n, and

n:=oX , is nonempty.

Lemma 35.8. The following are equivalent: (i) I is precipitous. (ii) For no set S of positive measure is there a sequence offunctionals on S such that Fo > F1 > ... > F > .... (iii) If G is a genericfilter on P (cf: 35.7), then the generic ultrapower Illt,(W) is well-founded.

Proof: In view of the preceding discussion on functionals, (ii) is equivalent to (iii): If Fo > F 1 > ... are functionals on S, andf, ,fl, . . ., the corresponding ele-

6.

440

OTHER LARGE CARDINALS

ments of !UP, then S forces that [f,],[f,], ... is a descending sequence of ordinals in the generic ultrapower. Conversely, if S forces that Ult,('Jn) has a descending sequence of ordinals, we construct F , , F , , ... on S such that F , > F , > .... The implication (i) + (ii) is easy. If F, > F , > * * . are functionals on S, then the partitions WFo,WFlr... constitute a counterexample to precipitousness: If X o 2 X , 2 ... are elements of W,, W,, . .., let fo E F , be the function with X, were nonempty, we domain X , , f , E F , with domain X,,etc; now if would getf,(a) >fl(a) > for a E X,. To show that (ii) + (i), let us assume that I is not precipitous and let W, 2 W, 2 ... be partitions of some S 4 I that attest to the fact that I is not precipitous. We shall construct functionals on S such that Fo > F1 > ... . Without loss of generality, let us assume that if X E W,, Y E W,,, and X G Y , then X # Y. Let T = W,,; note that the partially ordered set ( T , c )is an upside-down tree (of height w ) . For each z E S, let us consider the set T, = { X E T : z E X}.Since every descending sequence X, 1 X , 1 ... in T has empty intersection, it follows that for every z, T, has no infinite descending sequence X, 2 X I3 ..*; hence the relation c on T, is well-founded. Thus there is, for each z, an ordinal function pz on T (the rank function, cf. Section 9) such that p,(X) < p,( Y) when X c Y. It is clear that if X E W,, Y E W,, and z E X c Y , then p , ( X ) < p , ( Y ) . Thus we define, for each X E T , a functionf, on X as follows:

n,"=,,

n:=,

,,

u;=,

,,

f x b ) = PAX)

(all

E

X)

Now it is clear that if we let F, = {f,: X E W,} for each n, then F , , F , , . . . are functionals on S and F , > F , > . . . . W The next lemma shows that the ideal of small sets is not precipitous. Then we show that a K+-saturated ideal over K is precipitous. Later (in Theorem 86a)) we show that if there exists a precipitous ideal, then there is a transitive model with a measurable cardinal. Lemma 35.9. L.et I ={X E K : X

K

be

a regular

uncountable cardinal. The ideal

1 I < K } is nor precipitous. Proof. Let I = {X E K : I X I < K}. A set X G K has positive measure just in case I X I = K. For each such X,letf, be the unique order-preserving function

from X onto K. For each set X of positive measure there exists a set Y c X of positive measure such thatfy(a) < f x ( a )for all a E Y; namely if we let Y = {a E X : f,(a) is a successor ordinal}, thenf,(a) =fu(a) + 1 for all a E Y. Thus for each X $ I there is an I-partition W, of X such that for all Y E W x , f y ( a < ) f x ( a ) on Y. Now we construct I-partitions W, 2 W, 2 ... as follows: We let W, = {K}, and for each n, we let W,, = { W, : X E W,}. For each n, we let F, be the

,

u

35.

SATURATION OF IDEALS AND GENERIC ULTRAPOWERS

441

functional F , = {fx : X E W,}. It is clear that F , > F , > ... > F, > ..., and therefore I is not precipitous. H

Lemma 35.10. Let K be a regular uncountable cardinal. Every K+-saturated Kcomplete ideal over K is precipitous. Prooj: Let I be a K+-saturated K-complete ideal over K. Let S be a set of positive measure and let W, 2 W, 2 be I-partitions of S. We shall find X, 3 X , z... in W,, W,, ... such that X, is nonempty. We shall first modify each W, to obtain a new I-partition W ; that is almost like W, but is disjoint. We proceed by induction on n. Since I W,) IK, let W, = {X,: a < 8) where 8 IK, and for each a < 8, let Xi = X, X,; then we let W, = {X': X E W}. Since I is K-complete, we have X - X' E I for all X E W, and thus W , is an I-partition of S ; moreover, W, is disjoint, and is a partition of So = W, and S - So E I. Having constructed W ; ,we enumerate W,+, = {X,:ac 8) where 8 IK, and for each a < 8, let Xh = (X,- UB<,X,) n Z where Z is the unique Z E W, that is almost all of the unique Y E W, such that X, E Y. We let W,+ = {X': X E W,+,}; W, is a partition of S,+ = W,+ S - S,+ I E I , and X - X' E I for all X E W,+ 1. Since each S , is almost all of S , the set ():=, S , is nonempty; let z be an element of this intersection. For each n there is a unique Y, E W, such that z E Y,; let X, be the unique X, E W, such that Y, E X,. It is clear that x, 2 x12 ... 2 X, 2 ..., and x # 0.

u,<,

u

,

u

n:=,

In the following exercises, " minimal unbounded function '* and " least unbounded function" have the meaning defined in the proof of Lemma 34.1. Exercise 35.9. If I is a precipitous ideal over K, then there exists a minimal unbounded function. [There is a set X of positive measure and a functionfon X such that X forces thatf represents K in the generic ultrapower.] Exercise 35.10. If there exists a K+-saturated ideal over K, then there exists a normal k-+-saturated ideal over K. [Let g be a minimal unbounded function. Show that g , ( I ) is ~+-saturated.]

If I is K+-saturated, then the Boolean-valued names for functions on K in 91 can be represented not by functionals but by ordinary functions: Let F be a functional (on K). Let W = {dom(f) : f F}; ~ since I is K+-saturated, W can be replaced by a disjoint W' such that for each X E W there X' E W' such that X' G X and X - X' E I. If we replace eachf E F by its restriction to the corresponding X' E W', we get a functional F' whose elements have disjoint domains. Thenf = F' is a function, and iff E 9IBis the name corresponding t o F, then [If = 1 1 1= 1.

u

Exercise 35.11. If I is a K+-saturated ideal over function.

K,

then there exists a least unbounded

6.

442

OTHER LARGE CARDINALS

[Let f E 91' be such that Ilfrepresents 2 in the generic ultrapower11 = 1, and let $ K + K be such that = 1 . Show thatfis the least unbounded function.]

llf=rll

Applications of Generic Ultrapowers 1 . Proof of Theorem 23 Let me consider this typical special case: Let K be a singular cardinal of cofinality K1,and assume that 2l = I + for all I < K. We shall show that 2" = K + , using a generic ultrapower. Let I be the ideal of thin sets over w I, let P be the corresponding notion of forcing (i.e., forcing conditions are stationary sets) and let G be a generic filter on P. Note that since I P I = 2" < K (in !Ill), all cardinals 2 K remain cardinals in 331[G]. Let us work in 331[G]. G is an 331-normal 33l-countably-complete 91ultrafilter over a?. Let '37= UhG(911)be the generic ultrapower and let j : 331 + 91 be the corresponding elementary embedding. 91 is not necessarily well-founded. Let (K, : a < w l ) be (in 911) an increasing continuous sequence of cardinals converging to K. Let e be the cardinal number in 91 represented by the function .(a) = K, . Let e+ denote the successor cardinal of e in '37. For each X E K in 91 let f x be the function on w'f defined by f,(a) = X n K,. Clearly, eachf, represents in 91 a subset of e. Moreover, if X # Y, thenf, andf, are almost disjoint and hence represent distinct subsets of e. It follows that IP"(K)~I IP"(e)J, where Pa(,) denotes the collection of all subsets of e in 9. Now 9 k 2' = e + (because 91 k 2"' = K: for all a), which means that in the model '37 there is a one-to-one correspondence between the power set of e and e+. It follows that there is a one-to-one correspondence between Pm(e)and the lext(e+)l. set ext(e+)= {x E O r P : x <'e+}. Thus we have so far IP"(K)I I Next we observe that e = S U P { ~ ( K ? ) : y < a?}.This is because iffrepresents an ordinal less than e, then there is a set of limit ordinals X E G such that f ( a ) < K, for all a E X ;thusf(a) < K ~ ( , ,for some y(a) < a, and by normality of G, there is y such that [f] < " ' K ~ . Now for each y < w?, lext(j(KY))II I ( K : ' ) ~I < K, and therefore lext(e)I IK. If x < "e+, then there is in 91 a one-to-one mapping of x into e, and (ext(e)lI K. Thus ext(e+)is a linearly ordered set whose therefore, lext(x)l I each initial segment has size at most K. Therefore lext(e+)l5 K + , and we have IP*(K)I


We have argued so far in Yl[G]; in other words, we have proved that I P ( KImcl) < ( K + ) ~ ' ] . But since all cardinals 2 K in 91 remain cardinals in 331[G], it is necessary that I P ( K I* )I(K')"; in other words we have proved that 2" = K + (in 91).

35.

SATURATION OF IDEALS A N D GENERIC ULTRAPOWERS

443

2. Proof of Theorem 846 Let 2O ' = K, < K,, and let I be a a-complete A-saturated ideal over wl. We shall show that 2" I K, 1. Let P be the notion of forcing corresponding to I, and let G be generic on P. Since sat(P) = sat(1) 5 I, all cardinals 2 1 in 91 are cardinals in 91[G]. Let us work in 'lR[G],and let 9 = Ult,(!Vl) and j = j,: !Vl -+ 9 be as before. For each X c w1 in 91 letfx be the function on w? defined byfx(a) = X n a. Eachf, represents in 9 a subset of the countable ordinal d represented by the function d(a) = a ; moreover, if X = Y, thenf, andf, are almost disjoint and hence [ f x ] # [ f y ] . It follows that IP"(w?)I IIP"(d)I. Let e be the cardinal number in 9 such that 9 t= 2N0= e (we recall that d = w ) . Since 9 1 IP(d)l = e, we have IP"(d)l = lext(e)) and so IP"(w?)l I lext(e)l Next we shall compute the size of ext(e). Since 911 t= 2'O = K, and j : lm+ 'iJl is elementary, we have e = j(w,). We shall now prove by induction on y < that (35.12) Iext(i(w3) I 5 I * Ia? I (35.1 1)

Let me denote j(w7) = e, for all y < w?. By Lemma 35.7, the ordinals of 'iJl have an initial segment of order-type w?; thus the injnite cardinals of ! also l have an initial segment of order-type w?, namely {e, : y < w?}. If y = 0, then e, = w and (35.12) is true. If y is a limit ordinal, then e, = sup{e, : 6 < y} and (35.12) is again true provided it is true for all 6 < y. If y = 1, then ext(e,) is a linearly ordered set whose each initial segment is countable, and hence lext(el)l I Kl.Since I is a cardinal (now we are in 91[G]), we have K, I 1,and (35.12) holds. Let us assume that (35.12) holds for y and let us show that it also holds for y + 1. Every function f: w1 -,my+ in 91 is bounded by some constant function, and therefore j(wy+,) = sup{j(<): < o,+ ,}. Hence the linearly ordered set ext(e,+ ,) has a cofinal set of order type wy+ and each its initial segment has size 1 1 . Iwyl (becauseif<<w,+,,then lext(j(<))l I (exte,( 51. I w y I ) . It follows that lext(e,+ I II lw? 1. Now we put (35.11) and (35.12) together and get I P"(w7) I I 1 . K,"

<

,

,

This we proved in 91[G]; but since all cardinals 2 I in lm remain cardinals in 91[G], the same must be true in lm. Hence (in lm) 2" I k K,.

rn

3. The CCH at a Cardinal K that Carries a u+-saturated Ideal Let K be a regular uncountable cardinal and let I be a K+-saturated ideal over K. We shall prove

(35.13)

if 2A= 1+ for all I < K ,

then 2" = K +

6. OTHER LARGE CARDINALS

444

(We proved this for a measurable K in Lemma 28.13.) Let P be the notion of forcing corresponding to I , let G be generic on P, and let '92 = Ult,(%l). Since I is K+-saturated, 9 is well-founded and hence we identify it with a transitive model 'J1 s 9R[G]. Let j : 91 + (n be the natural embedding. We have j ( y ) = y for all y < K, and j ( ~ ) >K. If X E K and X E 91, then X E 'J1 because X = j ( X ) n K. Thus P w ( ~G) P'(K). We assume that 91 k (2' = ' 1 for all 1< K) and hence 9 k (2A= 1' for all L <j(K)), and in particular, '92 k IP(K)I = K', where a+ denotes the least cardinal greater than a. Now (K')" I ( K ' ) ~ ' ] ; and because sat(P) ='K (in !"), (K')" is a cardinal in 9)1[G] and we have also ( K ' ) ~= ~(K')". ~ Thus we have, in 9R[G], I Pw(K)I 5 (K')" and since all cardinals above K + in 91 are preserved, the last formula is also true in 91, and we have 2" = K'. W 4. Some Consequences of an K,-saturated Ideal Over o1 Let 1 be an K,-saturated ideal over ol. Then (35.14) (a) If 2N0= N1,then 2" = K,. (b) If Kl< 2N0< K,,, then 2'j = 2N0, (c) If 2N0= KW I ' then 2N1I Uu,. (d) If N,, is a strong limit, then 2N10i < K,,

.

Let @(a)denote the ath member of the class {K : K, = K } . (e) If @(oI) is a strong limit, then 2*(,') < @(w,). (a) and (b) are consequences of Theorem 84 (cf. (35.2) and (35.3)) (and (a) is also a consequence of (35.13)). (d) gives a better bound on 2n1*1 than that in Theorem 24, which gives 2Hwl< K, where y = (2")'. (e) gives a bound for 2@(01), which is an open question in ZFC. As before, let G be a generic ultrafilter over wlr let 91 = Lllrc(!U1) and let j : Iu1+ 9. W is a transitive model, 9 G %1[G]. k t us denote K = my. We havej(y) = y for all y < K, andj(K) > K. Thus K is a countable ordinal in W.Moreover, everyJ K + Ord in 91 belongs to 9 (as in Lemma 28.9a), and so every y < o? is countable in 91. Since sat(1) = K,, o? is a cardinal in %1[G], hence in W, and so j(wF)=

We shall now prove (c), (d), and (e). To prove (c), let us assume that C 2N0= K,, . Since W C 107 I = KO, and j is elementary, we have 9l k IP"(K)~= 2'O = NjK.Now NjKINFG1;and sincej(K) = a?,KFG1 is the o F t h cardinal in 9R[G]. However, all cardinals 2 K? are preserved and hence u y = KE2. Thus

" "

lp

5%

35. SATURATION OF IDEALS A N D GENERIC ULTRAPOWERS

445

holds in 91, therefore in 911[G];and because cardinals above K 2are preserved, this also holds in 911. To prove (d) or (e), note that if K,, (or @(wl)) is a strong limit, then 2% = K;: (2"'("'1)= (0(wl))NI).Let A denote K,, in (d) and @(q) in (e). I t is easy to see that j ( A ) > A. Now '9 k Va < j ( A ) I " a I < j ( A ) , and because ("A)" G ("A)", we have 91 k I("A)wl < j ( A ) . In case (d) we have j ( A ) I Kz: as in (c); and in case (e) we obtain similarly j(A) I @(j(K)) I @(a2). The rest of the proof of either (d) or (e) is as before. Exercise 35.12. I f 1 is a precipitous ideal over w Iand if @(wl) is a strong limit, then 2@"" < @ ( ( 2 N l ) + ) . [Proceed as in (e) above and show that j ( @ ( w l )5) @ ( j ( w l ) )and j ( w l ) < (2")+.]

As I mentioned, the bound on 2',1 in (d) is better than that in Theorem 24. The same bound K,, can be obtained directly from Lemma 8.7 if there is no function cp: w 1 + w , such that llcpll = w 2 (see (8.7) for the definition of II(p(I). For every y < w 2 there is cp,: w 1 + w , such that llcpy 11 = y. In fact, let us define functions q;,, y < w 2 , as follows: cpo(a)= 0, ~ ? + ~ = ( acp,(a) ) + 1; if y = limn y n , let cp,(a) = supn {(pym(a)};if y = lim,,,, y s , let cpy(a)= sup{cp,,(a) : 5 < a}. Then cpy < cpb on a closed unbounded set whenever y < 6, and ()qY= y for all y < w 2 and all stationary S. It follows that if IIcpII = w 2 , then for every y < w 2 ,q,,(a) < cp(a) on a closed unbounded set.

(Is

Exercise 35.13. I f there is cp: w I -+ w I such that llcpll = w z . then there is a family of K2 functions from w , into w such that any two differ on a closed unbounded set. [Consider ($, : y < w 2 ) G n,,,,(cp(a) + I), wheref,(a) = min{cp,(a), cp(a)}.]

Let me mention that Chang's conjecture implies that there is no family of K, functions from w1 into w such that any two differ on a closed unbounded set; the proof is analogous to the one in Exercise 29.14. Thus Chang's conjecture implies that if KUl is a strong limit then 2K1ul < KcO2. (This is because by Lemma 8.7,2"0~1< N@ where 8 is the norm of the constant function cU1on wl.)

5. h--Saturated Ideals Over K Let h- be a regular uncountable cardinal and assume that saturated ideal. We shall prove:

K

carries a

K-

(35.15) (a) K is weakly Mahlo; (b) ( a < K : a is weakly Mahlo) is stationary; (c) if X G K has measure one then X n M(X) has measure one, where M(X) = { a < K : cf a < Q and X n a is stationary in a } If there exists a K-saturated ideal over K , then h- is weakly inaccessible by Lemma 27.8, and there exists a normal h--saturated ideal over K (by Lemma 35.1). Let I be a normal h--saturated ideal over K. We first prove:

6.

446

Lemma 35.11. If S ary in a.

GK

OTHER LARGE CARDINALS

is stationary, then for almost all a < K , S n a is station-

Proof: If not, then there is a set X of positive measure such that S n a is not stationary in a (or cf a = w ) for all a E X. Let G be a generic ultrafilter over K (corresponding to I ) such that X E G. Let ! I= UltG(9Jl). 91 is a transitive model. Since I is normal, K is represented in '3 by the function d(a) = a. Since S = j ( S ) n K , we have ! It= S is not stationary. However, the notion of forcing is K-saturated and hence K is a regular cardinal in %1[G], and by Lemma 35.4 (with w l replaced by K ) 9Jl[G] 1 S is stationary. Now ! IG %I[G]and so '3 b S is stationary. A contradiction. W Now we prove (35.15). Since I is normal, every set of positive measure is stationary. Thus (c) follows since if X has measure one then M(X) has measure one by Lemma 35.11, and so does X n M(X). To prove (a), it suffices to show that almost all a < K are regular cardinals. Otherwise, let X be a set of positive measure such that all a E X are singular. Let G 3 X be generic and let '3 = LTltG(911).Then 91 1 K is singular, contrary to the fact that K is regular in 9l1[G] and 91 G 911[G]. Now (b) follows by an application of (c): Let X = { a < K : a is regular}, then X n M(X) = { a < K : a is weakly Mahlo}. H 6. Proof of Theorem 85 Let K be a regular uncountable cardinal and let S be a stationary subset of K . We shall give a quick proof that S is the disjoint union of K stationary subsets. Otherwise, the ideal I = {X G ti: X n S is thin} is a normal K-saturated ideal. By Lemma 35.1 1, S - M(S) has measure zero and hence is thin, which contradicts Lemma 35.3. W 7. Changing Cofinalities of Cardinals

Let K and 1 be regular cardinals, 1< K . Let us assume that the GCH holds and that the thin ideal over K is K+-saturated. Then there exists a generic extension in which all cardinals less than K remain cardinals, and cf K = 1. Compare this with Corollary 3 of Jensen's covering theorem (Theorem 74). For instance, if K = N3 and I = N l , then %I[G] 1 ( I K ~ = S2 and cf K = q ) if; K is inaccessible, then it becomes a singular cardinal of cofinality 1. Let I be the ideal of thin subsets of K , and let ( P , G ) be the corresponding notion of forcing. Since I is K+-saturated, the algebra B = P ( K ) / I is complete, and B = r.0. P . Every b E B is represented by a stationary set: h = [ S ] ; moreover, if {b, : < y } is a partition of B, then we can find disjoint S , such that b, = [S,] for all < y. Let E = { a < K : cf a = I } We shall show that in the generic extension, all cardinals less than ti are preserved, and that

<

(35.16)

E kcfZ=I

35. SATURATION OF IDEALS AND GENERIC ULTRAPOWERS

447

To show that the cardinals below K are preserved, it suffices to prove that if v is = P"(v). For this, it is enough to show a cardinal and v t < K, then PmG1(v) that B is (v. v)-distributive. Let W, = {b! : 5 < v}, q < v, be v partitions of B into v pieces. Each b2can be represented by a stationary set Stl, such that each Z , = {Stl,: 5 < v}, q c v, is a partition of K. For eachf v + v, consider S , = S'j[,);since the number off's is 2' = v + < K, we conclude by K-completeness of I that those S , that are stationary constitute a partition refining all the W,. Thus B is (v, v)-distributive. Now let me show that E forces that there is an increasing function on I converging to K. For each a E E, letf. be an increasing function on I with limit a. Let us define a B-valued nameffor a function from I to K as follows:

0,

IIfce) = rill = [ { a E E :fa(<) = dl

(5 < 1, rl < K )

I t is clear thatfis, with B-value 1, a function with domain c I and range c K. I t is easy to prove that f is increasing. Let me show that E IF domf= 1 ( E IF (ran f is unbounded in 2) is verified in a similar way). Let c I and let S c E be stationary; we want to find a stationary T 5 S and q < K such that T IF = 6. For a E S , let g(a) = f.(<). Since g is regressive, there exists T c S stationary such that g is constant on T. Obviously, this T will do. I t remains to show that cf K is exactly I (in a generic extension %l[G]such that E E G).We have in fact shown that cf K = cf I in %l[G];thus if cf K < A, then there is v < I , a cardinal in 91l, such that cf I = v in %JI[G].But this is impossible since an argument analogous to the one above shows that B is (v. I)-distributive. W

<

f(e)

Precipitous Ideals

I shall conclude this section with the following theorem, showing that existence of precipitous ideals, even over a successor cardinal, is equiconsistent with existence of measurable cardinals. It has also been shown that it is consistent, relative to existence of measurable cardinals, that the ideal of thin subsets of o, is precipitous. Theorem 86. (a) If K is a regular uncountable cardinal that carries a precipitous ideal, then K is measurable in some transitive model of ZFC. (b) If K is a measurable cardinal, then there exists a generic extension in which K = h',, and K carries a precipitous ideal. The proof of (a) uses the technique of iterated ultrapowers from Section 31 (see also the proof of Theorem 80, in particular (33.8k(33.12)). Let K be a regular uncountable cardinal, and let 1 be a precipitous ideal over K. Let K be the class of all strong limit cardinals v > 2" such that cf v > K . Let yo < y , < ... < yn < ... ( n < w ) , be elements of K such that Iy,, n K 1 = y,,, let A = { y n : n = 0, 1, . . .} and let I = sup(A).

448

6.

OTHER LARGE CARDINALS

Lemma 35.12. There exists an 4A]-ultraJilter W over K such that W is nonprincipal, qA]-~-complete, qA]-normal, iterable, and every iterated ultrapower Ult'@(q A 3 ) is well-founded. Proof. Since I is precipitous, the generic ultrapower is well-founded, and so the diagonal function d(a) = a represents some ordinal number in Ulr,(!M). Thus there is a set S of positive measure, and an ordinal y such that

s IF d represents 7 in

(35.17)

U/tG(W)

We shall first show that for every X E YA], X measure 0, and so

(35.18)

G K,

either S n X or S - X has

U = {X E P ( K )n q A ] : X n S has positive measure}

is an qA]-ultrafilter. Let H = X L t Au1 K ( ~u {A}) be the class of all sets definable in LEA] from elements of K u K u {A} (this is expressible in ZF similarly to the way in which ordinal definability is ;or we can use Lo[A] for some large 8 as in Lemma 3 1.13). Since IK n ynl = 1'" for each n, it follows that if n is the transitive collapse of H, then n ( A ) = A and n[H] = UA]. Now if X G K and X E YA],then because n is the identity on K , we have X = n ( X ) and therefore X E H. Thus for every X E P ( K )n Y A ] there is a formula cp and a finite set E c K u K such that

(35.19)

X

= {<

< K : q A ] 1 V(<, 6 A ) )

We shall now show that (35.18) defines an qA]-ultrafilter. Recall that for any generic ultrafilter G over ~ , j is, the identity on K, and moreover,j G ( v ) = v for all v E K (this follows from the definition of K). If X E U A ] and X n S has positive measure, then because X IF 2 E Gand because (35.17) holds, we have

x

n

s IF ~ E ~ ~ ( X I )

Now using (35.19), and the fact that j,(A) = A and j,(E) = E, we have

(35.20)

x n s (YA] I= ~ ( 76, A))

But the formula forced by X n S in (35.20) is about !Ul, and thus true. Hence

117 E j G ( 2 ) l l

=

and by (35.17),

S IFZEG This, however, means that S - X has measure 0. Since I is K-complete, it is clear that U is ~A]-~-complete, and moreover the intersection of any countable family of elements of U is nonempty. It is less clear that U is iterable: Let (X,: a < K ) E YA]; it suffices to show that

35.

SATURATION OF lDEALS A N D GENERIC ULTRAPOWERS

449

lk ( { a : X, E U } E UA]). If G is generic such { a < K : X , E U } = {a < K : y E j G ( X , ) } , but this is j G ( ( X m: a < K)) E a G ( A ) ] and j G ( A )= A.

that S E G, then in UA] because

S

By Exercise 31.9, Ult,(UA]) is well-founded; let f: K + K be the function that represents K in Ulr,(YA]). Let W = f * ( U ) . It is easy to verify that W is an YA]-normal, qA]-~-complete,iterable YA]-ultrafilter over K, and that the intersection of any countable family of elements of W is nonempty. By Exercise 31.9, every iterated ultrapower Ultg)(YA]) is well-founded.

Proof of Theorem 86u. Let A = {y. : n = 0, 1, . ..} be as above, let 1 = sup(A), and let W be an YA]-ultrafilter as in Lemma 35.12. Let us define in U A ]

(35.2 1 )

F = {X E I : 3no V n 2 no y,

E

X}

(compare with (33.9)). I claim that D = F n YF] is a normal measure on 1 in 4DI. For each a, let iou be the elementary embedding iOa: Y A ] + Ulr';;'(YA]). It follows from the definition of the class K that:

(i) if a < y,. then io,(y,) = y,; (ii) i0&) = 7;, (iii) if a < A, then io,(I) = A. Hence for all a < I, i O , ( Y A ] )= YA], iou(F)= F, and iou(D)= D . We shall now prove that D is an ultrafilter in YO]. Otherwise, let X E 1 be the least X (in the canonical well-ordering of YO]) such that X $ D and I - X $ D . Since iou(D)= D for all a < 1,we have i o u ( X )= X for all a < I; in particular, i o y m ( X=) X for all n. Now for any n, if 7, E X , then i o y , ( K ) E ioy.(X) and hence K E X,and vice versa. Hence either all yn are in X or none, and so either X E F or I - X E F, a contradiction. The proof that D is K-complete (in 401)and normal is similar and is left to the reader. Thus we have proved that the model Y A ] satisfies the following formula

d4: (35.22)

cp(A): 30 such that QD] k D is a normal measure on 1

The proof of Theorem 86a will be complete if we find a transitive model M and an elementary embedding i : M + UA] such that i ( ~ =) 1.Then M k ( P ( K ) and hence K is measurable in some transitive model. Let me recall that for each a, i,, is the elementary embedding iOa: YA]+ Ultg)(qA]). As we have seen, ifa I, then io,(l) = 1 and io,(LIA]) = YA]. Let C be a proper class of ordinals greater than 1 such that ioA(v)= v for all v E C. Let H = . f L tu A {A)1u(C) ~be the class of all sets definable in Y A ] from elements of K u {A) u C . (As before, the problem of expressibility of H in ZF

-=

450

6. OTHER LARGE

CARDINALS

can be overcome by replacing YA]by a suitable large segment LQA].) H is an elementary submodel of YA]. If a < I, then ioz(v)= v for all v E K u {I}u C ; it follows that ioz(x)= x for all x E H. Observing that for every v such that K 5 v < 1 there exists a < 1such that iOz(v)> v, we conclude that H contains no ordinal v such that K I v < 1. Hence if R is the transitive collapse of H, and M = R [ H ] , then n(1) = K ; thus i = R- ' is an elementary embedding ofsome transitive model M into q A ] , and i ( t i ) = I. W The proof of Theorem 86b uses the notion of forcing which collapses all a < K onto o and makes K = K I (the Lkvy collapsing algebra of Section 20). The proof that if K is measurable in 91, then in %1[G] it carries a precipitous ideal'uses some methods that we shall be dealing with in the next section. For this reason, and because this section is already quite long, I postpone the proof of Theorem 86b until Section 37. 36. MEASURABLE CARDINALS AND THE GENERALIZED CONTINUUM HYPOTHESIS

The model L [ U ] that we discussed in Section 31 establishes consistency of the generalized continuum hypothesis with the existence of measurable cardinals. The obvious question is whether we can use the method of forcing to obtain models in which measurable cardinals exist and the continuum function 2'l behaves in some prescribed manner. We are already aware of one limitation on the behavior of the continuum function if K is a measurable cardinal: The value of 2" depends on the values 2" for a < K ; for instance, if 2" = u+ for all a < K, then 2" = K'. We shall show below (Theorem 87) that this is not the only restriction on 2". It turns out that in order that 2" > K ' , the universe must have very strong large cardinal properties; the theorem shows that if 2" > K', then for every 0 there is a model with 6 measurable cardinals. As one may guess, the behavior of the continuum function above K (on regular cardinals) is not influenced by measurability of K ; we shall also show that for any I < ti, the values of 2" can be arbitrarily prescribed for all regular u < 1.Then we introduce a method of forcing that will enable us (under a strong large cardinal assumption) to construct models in which the generalized continuum hypothesis fails at a measurable cardinal. The two main theorems of this section are the following: Theorem 87 (Kunen). If there exists a measurable cardinal K such that 2" > K', then for every ordinal number 6 there exists a transitive model M of ZFC with at least 0 measurable cardinals. Theorem 88 (Silver). If there exists a supercompact cardinal K , then there is a generic extension in which K is a measurable cardinal and 2" > K + .

36.

MEASURABLE CARDINALSA N D THE GCH

45 1

The proof of Theorem 87 is very much like the proof of Theorem 80, and I shall closely follow that proof. In the proof of Theorem 80, the key step was provided by Lemma 33.2. The following lemma plays a similar role in the present proof: Lemma 36.1. Let K be a measurable cardinal and let us assume that 2" > K + .Let U be Q k--complete nonprincipal ultrafilter ooer K. I f X is a set of ordinals of size < K, and i f 6 is an ordinal, 6 < K + + , then there is a transitive model N of ZFC such that X E N, that W = U n N E N and if we denote j , the natural embedding j,: N + Ult,(N), then

6 < j , ( K ) < K+'

(36.1)

Proof. Let me first state explicitly what is implicit in the statement of the lemma: If W E N, then in N, W is a K-complete nonprincipal ultrafilter, and one can construct, within N, the ultrapower Ult,(N). Let X be a set of ordinals, I X I < ti, and let 6 < K + +.Let j,: I/ + Ult,( V ) . Since j u ( K ) > 2" and 2" 2 K + +.there are functions fs: K + K, 5 I6, such that & 6. The standard proof shows that N b 2" = K + (see, e.g., Lemma 31.1 and Exercise 15.3). Since N C. j & ) < (2")+, we have j , ( K ) < K + +.

Proof of Theorem 87 Let me follow closely the proof of Theorem 80. Again, I shall prove only the case 8 = 2. Let K be the class of ordinals defined in (33.8), let K , be as in (33.17) and let A = {y,};=o and A' = {Y;}:=~ be the sets defined in y:, and let F = F ( A ) ,F' = F ( A ' ) be y,, and I' = (33.18). Let I = the filters over 1 and 1' defined in (33.9). We claim that ,4F, F'] has two measurable cardinals, A and A', and that F n 4 F , F'] and F' n 4 F , F'] are normal measures. As in the proof of Theorem 80, one verifies that F n L[F, F'] is a normal measure on 1 in 4 F , F'], and additional work is required to show that F' n q F , F'] is a normal measure on 1'.For each n, let M , < aA,A'] be the Skolem hull of K, in YA, A ' ] ; cf. (33.19); let n, be the transitive collapse of M, and let j,: 4 n , ( A ) , n,(A')] + 4 A , A'] be the elementary embedding j , = n, The next step in the proof is to prove that

(36.2)

nn(yn)


for all n < w

As in the proof of Lemma 33.4, this is done by induction on n. For n = 0, we have n,(yo) I K exactly as in the proof of Lemma 33.4. Thus let us assume that n,(y,,) c K + and let us show that n,+ I(yn+ < K + +.By Lemma 36.1 there exists a transitive model N ofZFC such that n,,(A) E Nand A' E N, and W E N such that if j,: N -, Ult,(N), then n,(y,) < j & K ) < K + + . It suffices to show +

6.

452

OTHER LARGE CARDINALS

n,(a) c j w ( K ) for all a c Y , , + ~ in M,+l, since then Yn+ 1 and a E M,+ 11 s j & ) c K + +n n + I(?,+ 1 ) = sup{nn+i(a): a Let a c yn+ be in M,+ I . To show that n,(a) < j&), we proceed exactly as in the proof of Lemma 33.4, except that we argue inside the model Ult d N ) (instead of Ultu(V)as there); here we use the fact that U l t d N )contains both n,( A ) and A'. We complete the proof of the theorem (for 8 = 2) as in the proof of Theorem 80. We consider the model M , < L [ A , A'], the Skolem hull of K,; MI,> = L[B, A'], where B = nl,,(A),A' = nlo(A'),and

that

j,:

YB, A'] + 4 A , A']

is the transitive collapse of M , . where j, = n; and II, We use the fact that n,(A) c K + + (by 36.2)). Let N be a transitive model of ZFC such that B E N and A' E N,and let W E N be such that n,(A) < ~ W ( K ) , where j,: N + Ult,(N). In Ult,(N), B is a subset of j w ( K ) and A' is a subset of the class K . Thus we apply (33.16) and get UIt,(N) k (L[F(B),F(A')] k F(A') n L[F(B),F(A')] is a normal measure)

and upon application of j, to U F ( B ) , F ( A ' ) ] , we obtain, as in the proof of Theorem 80,

4F, F'] k F' n 4 F , F'] is a normal measure on A' This completes the proof of Theorem 87 for 0 = 2. For the general case, see the concluding remarks in the proof of Theorem 80. Preservation of Measurable Cardinals in Generic Extensions

Our goal now is to construct a model of ZFC in which there is a measurable cardinal K with the property 2" > K + . By Theorem 87, we have to start with a model of something stronger than just a measurable cardinal, and we also have to make sure that 2" > a+ holds for many cardinals a below K in the resulting model. We shall proceed gradually, introducing more and more sophisticated methods of getting generic extensions in which a measurable cardinal remains measurable. ' As a starting point, we prove: Lemma 36.2. Let K be a measurable cardinal in the ground niorlel 911. Let ( P , < ) he a notion of forcing such that I PI < K . Then ti i s riieasurable in the generic extension.

Proof. I shall give a proof using elementary embeddings since similar arguments will be used in subsequent constructions; for a direct proof, see Exercise 36.1. Let B = r.0. P ; since I B I c K, we may as well assume that B E V,. We can

36.

453

MEASURABLE CARDINALS A N D THE GCH

also assume that P is a dense subset of B. Let G be an 911-generic ultrafilter on B ; let us work in 911[G]. Since K is measurable in 91, there is an elementary embedding j : 91+ 91 (where 91 is the ultrapower of 911 by some U over K ) such that K is the least ordinal moved. We shall extend j to an elementary embedding (denoted also j ) of 91[G] into 91[G], thus showing that ti is measurable in 911[G]. Since B E V,, we have j ( B ) = B, and B is a complete Boolean algebra in 91. Since 91 G 911 and G is generic over 911, G is also generic over 91. Let i? denote the C-interpretation of %1’, i?: 331’- %1[G], and let iz denote the Ginterpretation of 91’, iz: 91’ -, 91[C]. We define j ( x ) for x E 911[G] as follows: Let x E 91’ be a name for x , x = i F ( x ) . Let (36.3) A x ) = i%(x)) Since x E 9J’, we have j ( x ) E 91’ and so i z ( j ( x ) )E 91[G]. However, we have to show that the definition (36.3)does not depend on which name for x we choose. Let y be another B-valued name and let p E G be such that (36.4)

p Ikx=y

That is, (36.4) holds in 911; we applyj to (36.4), and have (in 91) j(p)

~kA x ) = j d v )

But j ( p ) = p E C and therefore iz(j x ) = ig(j y ) .

Finally we show that j : 911[G] such that

-, 91[G]

is elementary. Let cp be a formula

911[G] 1 cp(x,

Let x, . .. be such that i&)

=

(36.5)

x,

. . .)

. . . . There is some p

E

G such that

cp(x, ...)

p

Applying j to (36.5), we get (in 91) p

I+

cp(jx, . . .)

(becausej ( p ) = p ) . Hence 91[G] k cp(j(.x), ...)

and since cp was arbitrary, j is elementary. Exercise 36.1. Let 91t, h-. P be as in Lemma 36.2; let U be (in W )a K-complete ultrafilter over K . Then in 911[G],the filter

W = ( X E h- : X

is a h--complete ultrafilter over

h-.

2

Y for some Y E Ul

454

6.

OTHER LARGE CARDINALS

[For instance, to show that W is an ultrafilter, consider X E 9J1' such that RI( E G. The function a++ Ilfi E XI1 is a partition of K into I BI < K pieces, and by K-completeness of U there is Y E U such that 115 E XI1 is the same for all 01 E Y. NOW either X E W or K - X E W according to whether this B-value is in G or not.] IIX

G

It is an immediate consequence of Lemma 36.2 that if K is a measurable cardinal, I < K, and F is a function on regular cardinals below 1 such that (i) F ( a ) 5 F ( P ) if a I P, (ii) cf(F(a)) > a, and (iii) F ( a ) < K for all a in its domain, then there is a model in which K is measurable and 2" = F ( a ) for all a E dom(F). One can also prescribe the values of the continuum function above a measurable cardinal (that is, for regular a < K); this follows from this easy observation: K is a measurable cardinal and P is a K-closed notion of forcing (or just K-distributive), then K is measurable in the generic extension.

Exercise 362. If

K be a measurable cardinal. There is a generic extension in which measurable and such that 2& = K,, for all K, 2 K. [Use Exercise 20.8. The notion of forcing is closed.]

Exercise 36.3. Let

K

is

In the following lemma we introduce a method that preserves measurability of K while adding subsets to an unbounded set of cardinals below K. It is vital however that the set A c_ K has a normal measure 0. Lemma 36.3. Let 91 be a transitive model of ZFC + GCH and let K be a measurable cardinal in 911. Let (in 911) D be a normal measure on K and let A be a set of regular cardinals below K such that A 5 D ; let F be a function on A such that F(a) < K for all a E A, and: (i) cf F ( a ) > a ; (ii) F ( a , ) I F ( a 2 )whenever a1 I a 2 .

Then there is a generic extension 911[G] of911 with the same cardinals, such that K i s measurable in 91[G],and for every a E A,

!Ul[G] k 2"

(36.6)

= F(a)

Proof. If we did not care about measurability of K in 9ll[G], we would of course use the Easton product as in the proof of Theorem 46 (see (20.11)). The actual construction is a modification of that method. Let j : (xn + 9 be the elementary embedding given by the ultrapower UltD(911).As we assume that A 5 D, we have K $ j ( A ) . Let ( P , < ) be the Easton product of P,, a E A, where each Pa is the notion of forcing that adjoins F ( a ) subsets of a ; cf. (19.10), (20.10), and (20.11). Thus conditions are 0, 1-functions whose domain consists of triples (a, <,q ) where a E A, < a, and q < F(a), and such that for every regular cardinal y,

<

It(a9 L q ) E dom(p) : a I

Y) I < Y

36.

455

MEASURABLE C A R D I N A L S AND THE GCH

In particular, I p I < K for all p E P , hence P c V, and s o j ( p ) = p for each p E P . We shall however use not P butj(P) as our notion of forcing. Thus j ( P )is, in 91, the Easton product of Pa for u E j ( A ) . Note that P c j ( P ) and that j ( P ) is isomorphic to P x Q where (in 91) P = (jP)'" and Q = (jP)'". Let G be an 911-generic filter on j ( P ) . We claim that 9I[G] has the same cardinals and cofinalities as 91 and satisfies (36.6) and that K is a measurable cardinal in 911[G]. Let G1 = G n P ; since j ( P ) is isomorphic to P x Q, there is an WI[G,]-generic filter G 2 on Q such that 9l[G] = 91[G, x G2]. As we have noted before, K # j ( A ) , and so Q = (jP)'" is in fact = (jP)'", and hence is, in 91, K-closed. But since "91 G 91, Q is K-closed in 91l. Moreover, we have / P I = K and l j P l = I jK I = K + . Thus for each regular I 5 K , we can break j ( P ) into a product of two notions of forcing, one that satisfies the I+-chain condition and one that is I-closed, and hence all cardinals I K + are preserved. Since I j P I = K', all cardinals > K + are also preserved. We prove (36.6) similarly to Theorem 46: For u E A, we regard j ( P ) as a product (jP)'" x (jP)'"; and since (jP)'" = P" and (jP)'" is a-closed, we conclude that (2")wG1= F (a 1. Thus it remains to show that K is a measurable cardinal in 911[G]. We shall first extend j : 911 + 91 to an elementary embedding (36.7)

j : 911[Gl] -, !N[G]

Let B = r.0. P and let j ( B ) = (r.o.(jP)rn;let i?, denote the G,-interpretation of 911' and let iz denote the G-interpretation of 91"'). We definej(x)for x E 9J[G ,] as follows: Let x E 91i' be such that x = iFi(x). Let (36.8) Ax) = i%(x)) As in the proof of Lemma 36.7, we show that (36.8) does not depend on the choice of x. Since .j(p) = p for all p E P and because G c G , it follows that if some p E G , forces x = y , then (in 91) p If j x = j y and therefore i z ( x ) = iz(y). The same reasoning shows that j : 911[G,] -, 91[G] is elementary. Using (36.7), we define an 91[G1]-ultrafilter U over K as follows: XEU

for all X E

K

iff

K E ~ ( X )

in 911[G,]. A standard argument shows that U is

!Ul[G,]- complete; and since j extends the original j = j D , U is nonprincipal.

Now we use again the fact that j ( P ) is isomorphic to P x Q, where I P 1 = K and Q is K-closed.Thus every subset of K is in 9J1[G ,] and therefore U is in 911[G] a K-complete nonprincipal ultrafilter over K . The method of Lemma 36.3 can be used to prove consistency of the GCH with a measurable cardinal by forcing. K be a measurable cardinal. There is a generic extension in which is measurable and which satisfies the GCH.

Exercise 36.4.* Let

K

6.

456

OTHER LARGE CARDINALS

[First get the GCH above K by Exercise 36.3. Then usej(P), where P is the Easton product from Exercise 20.8 making the GCH true below K . ]

Inverse Limits of Sequences of Complete Boolean Algebras

In Section 23 we defined direct limits of sequences of complete Boolean algebras. We shall now define a notion that subsumes the notion of a direct limit and will enable us to perform a new kind of iteration of forcing. Let a be a limit ordinal number and let for each i < a, Bi be a complete Boolean algebra such that if i < j , then Bi is a complete Boolean subalgebra of B j . We defined the direct limit lim dir Bi i-i

uiCI

as the completion of the Boolean algebra B i . More generally, let B i , i < a, be complete Boolean algebras and let {eij: i I j < a ) be a commutative system of complete embeddings, eij:Bi -, B j . Then

lim dir { B i , eijIi.j < 3 i-+i

is the unique complete Boolean algebra such that there exist complete embedeil[Bi]is dense in B. dings e,: Bi -, B commuting with the eij and that Let me recall the definition (23.21) of the canonical projecrion: If B is a complete Boolean subalgebra of D, then for d E D,

uiCl

h(d) = n { b E B : d I b} Let us generalize this to the case when e : B -,D is a fixed complete embedding of B into D ;then we let

h(d)= n { bE B : d I e(b)}

(36.9)

Now let B i , i < a, be complete Boolean algebras such that B i is a complete subalgebra of B j whenever i < j < a. Let hij be, for any i s j , the canonical projection of Bj onto B i . Let J be a collection of functions (hi : i < a ) E (Bi - ( 0 ) )such that

ni<,

(36.10) (i) hi = hi,@,) (i ~ j ) ; (ii) F contains all eventually constant functions c b for all nonzero hE B i , where

uicz

(36.11)

cb =

(hoi(b), h l i ( h ) ,..., b, b, h, .. .)

( b E Bi)

We partially order .F componentwise, i.e., (hi : i < a ) I(ci : i < a ) if hi I c i for all i . Let B = r.o.(.F, ), we call B an intlerse limit of B i , i < a, (36.12)

B = lim inv‘ i+z

’

Bi

36.

457

MEASURABLE CARDINALSAND THE GCH

[It should be clear how to define lim'inv'.'){B,, eidi.j < n i-+a

in the general case when the embeddings eij are not just identity functions.] For each i < a, there is a complete embedding e , of Bi into the inverse limit (36.12), namely (36.13)

eia(6)

= cb

It is easy to verify that (36.13) is indeed a complete embedding of Bi in B. Moreover, if i I j , then e , and ejaagree on Bi (in the general case, e , commutes with the eij.s).Thus we can identify each 6 E C = Bi with cb E B ; then each Bi is a complete Boolean subalgebra of B (and so C is a subalgebra of B, but the completion of C may not be embeddable in B). If .P is the collection of all eventually constant functions, then (36.12) is the direct limit of B i , i < a ; otherwise, various other choices of 3 may give various inverse limits of the sequence B i , i < a.

ui<.

Let us again assume that B i , i < a, are complete Boolean algebras and that Bi is a complete subalgebra of B j whenever i I j a ; and let for each i < a, P i be a dense subset of Bi such that for any i Ij ,

-=

(36.14)

(i) P i G Pj; (ii) if p E P j , then hij(p)E P i .

Let P be the set of all functions ( p i : i < a ) such that (36.15) Pi = hijbj) for all i 5 j . We partially order P componentwise, and call it the inoerse limit of Pi,i < a: P = lim inv Pi 1'1

(and r.0. P is an inverse limit of B i , i < a). Iterated Forcing

Iteration of generic extensions was introduced in Section .73: We defined two-step iterations and introduced direct limits of sequences of complete Boolean algebras. We shall now generalize the technique by considering other than direct limits at limit steps of the iteration. Recall the two-step iteration. (To simplify matters, let us consider only notions of forcing that are separative and have a greatest element 1.) Let P be a notion of forcing and let B = r.0. P. In WE,let Q be a notion of forcing, and let Q be a set of representatives of the class { X E %IB : ((xE Qll = I) in the equivalence relation JIx= yII = 1. Let us denote (36.16)

P*Q

6.

458

OTHER LARGE CARDINALS

the set of all pairs ( p , q ) E P x Q, partially ordered by (36.17)

(PI, 41)

I ( ~ ~ $ 4 2 iff )

pi I PZ and PI

‘k 41

I 42

(In P * Q,we identify any two pairs ( p , q,), ( p , q 2 )such that p IF q 1 = q 2 . )As we have shown in Section 23, the algebra B is embedded in the algebra D = r.o.(P * Q)by the embedding e: B + D defined in (23.7). It is obvious that for each p E P, (36.18) e(P) = 1) (where 1 denotes the greatest element of Q).Similarly, if h denotes the canonical projection of D onto B, then ( P 9

(36.19) h(P, 4 ) = P We shall now define the general notion of iterated forcing. Let us assume that in each Boolean-valued universe %IB we are given a fixed B-valued notion of forcing Q E 91’.By ireratedforcing, we understand a transfinite sequence of forcing notions P,, together with a commutative system of complete embedB, + B, where B, = r.o.(Pa) for each a such that: dings eUB: (36.20)

(i) P , + = P , * Q,,where Qa is the given notion of forcing in %IBa, and ea,,+l is the embedding satisfying (36.18); (ii) if a is a limit ordinal, then Pa is either the direct limit lim diri,a { P i ; eij} or the inverse limit lim inv,,, { P i ;eij}(and ei, are the corresponding embeddings).

There is a more intuitive representation of iterated forcing. The verification of the following facts may not be immediate but is routine: Elements of each P , can be identified with functions p = ( p i : i < a ) such that: (36.21)

(i) for every y < a, the restriction p 1 y = ( p i : i < y) belongs to P,; (ii) for every y < a, if Q, denotes the given By-valued notion of forcing, then lip, E Q, = 1; (iii) the partial ordering 5 a of P , is defined as follows: p I ,q iff for every y < a,

[ley

and PI? “7 PY < Q y % where IF7 denotes the forcing relation associated with ( P , , I,). P l y SYqly

If a is a successor ordinal, then P , consists of all functions that satisfy (36.21). If a is a limit ordinal and Pa is the inverse limit, then Pa consists of all functions that satisfy (36.21). If a is a limit ordinal and Pa is the direct limit, then Pa consists of all functions p = ( p , : i < a ) that satisfy (36.21) and such that there is some y < a such that pi = 1 for all i 2 y.

36.

459

MEASURABLE CARDINALS A N D THE GCH

The embeddings e,,: B,

+

B, satisfy

e38 (p ) = p n l n l n l n . . - ( P E Pa) (36.22) and similarly, the canonical projections h,, satisfy

hap(p)= P I a ( P E Pa) (36.23) [Instead of the verification of (36.21)-(36.23), the reader can take them for the definition of iterated forcing, and verify that {em,:a I B} is a commutative system of complete embeddings, that each h,, is the canonical projection, that each P, + is isomorphic to Pa * Q,, and that for a limit a, P, is isomorphic either to lim diri+, { P i ;eij}or lim inv,,, { P i ,eij}.] We proved in Section 23 that the two-step iteration preserves both the chain condition and the closedness of forcing (cf. Lemma 23.6 and Exercise 23.5). Also, if the only limits are direct limits, then iteration of forcing preserves chain conditions (cf. Lemma 23.7; see also Exercise 23.8). We shall now generalize these results. A notion of forcing (P, < ) is strongly I-closed if whenever D c P ( D for directed) is such that ID I I I and for any d d , E D there is e E D such that e Id , and e I d,, then there is p E P such that p I d for all d E D.

,

Lemma 36.4. (a) If P is strongly I-closed, and if IIQ is strongly X-closed)I,,,,p, then P * Q is strongly I-closed. (b) If P = lim dir,,, P i , ifeach Pi is strongly I-closed, and ifcf a > I , then P is strongly I-closed. (c) Let us consider iterated forcing:for each a, let P a ,B,, Q, be as above. Let I be a cardinal and assume that for each a, Q, is a strongly %closed notion of forcing in 91PS.Let us further assume that i f a is a limit ordinal and cf a I 1, then P, is the inverse limit of P,, fl < a. Then each Pa is strongly I-closed. Proof. (a) Let D

=

{ ( p , q , ) : a < I } be a directed subset of P

* Q.

Clearly,

D , = { p a : tl < I}is a directed subset of P and hence there is p E P stronger than all p, , a < I. Since for any a, B < I there is y < I such that ( p y q y )5 (paq,)and (p, q y )I (pa q,), it is clear that p ~f(qvIq, and q y 5 qa) and thus p forces that {q, : a < I } is a directed subset of Q. Hence

p

IF

3q E Q stronger than all the q,

and therefore there is q such that 114 E Q(I = 1 and p

I+

q I q, for all a

<X

It follows that (pq) I (p.4,) for all a < I . (b) Let D be a directed subset of P, ID 1 II . For each d E D there is y d c a Since 1 < cf a, there is such that if d = ( p i : i < a), then pi = 1 for all i 2 y < a such that each d E D is as follows: d

=

(d ( y ) n l n l n l n. * *

460

6.

OTHER LARGE CARDINALS

Now D, = {(dI y ) : d E D} is a directed subset of P,; and since P, is strongly I-closed, D, has a lower bound p E P, . Then p"l"l"l^ ... is a lower bound for D in P. (c) By induction on a. It follows from (a) and (b) that the assertion is true if a is a successor or if P, is the direct limit. Thus assume that P , = lim invi_.,Pi. Let D = (p' : v < I} be a directed subset of P,; for each v let p' = (py : i < a). We shall construct, by induction on i < a, a function p = (pi : i < a ) E P, stronger than all p', v < I . We construct p such that for each i c a, (pi:j c i ) is in P i and is (in P i ) stronger than all p' I i, v c 1. Having constructed (pi :j c i ) , we let pi be such that Moreover, if p; = 1 for all v < I , we let pi = 1 too. If y Ia is a limit ordinal, we have to show that ( p i : i c y ) E P ,. If P,is the inverse limit, then there is nothing to prove, so let us assume that P,is the direct limit. By the assumption, we have cf y > L and therefore there is 6 < y such that for all v c I , p; = 1 for all i such that 6 Ii c y. Hence we have pi = I for all i , such that 6 Ii c y, and so (pi : i < A) E P, . Thus we have p = (pi : i < a ) E P,, and it is clear from the construction that p I p" for all v < 2. Iterated forcing does not generally preserve chain conditions, but there is one important instance when it does (see Exercise 23.8): Lemma 36.5. Let K be a regular uncountable cardinal and let P be the direct limit of P, ,a c K . Assume that there is a stationary set S E K such that for all a E S, P , is the direct limit of P i , i < a. If each P , satisfies the k--chain condition, then P also satisjies the K-chain condition. Proof. The same proof as in Lemma 23.7. The most important property of iterated forcing is the following factor lemma. Although what the lemma states is very simple, it is somewhat complicated to formulate. Let us consider an iteration of forcing. For each tl, let P, be the ath iterate, let B, = r.0. P , , and let Q,E !UiBzbe the B,-valued notion of forcing that gives the ( a 1)th step of the iteration. Let us fix an ordinal number a, and let us do the following in the Booleanvalued model !UPz: Let us consider the following iteration of forcing:

+

(36.24)

(i) fg)= { 1); (ii) Pi!,= Pi)* Q l + * ; (iii) if p is a limit ordinal, then ff)is either the direct limit or the i < /l, according to whether in W , P , + , is inverse limit of the direct or inverse limit.

el,

36.

461

MEASURABLE CARDINALS AND THE GCH

Clause (ii) needs an explanation. Q,+Bis an element of the Boolean-valued model 911B1+@. Since B, is embedded in Em+,,,the model 91Pa+p can be con(by the algebra B’;) = B a + B :B,) and sidered a Boolean-valued extension of 9IB* so Q, + p can be considered a B’$)-valuedobject in !IllB*. Since the factor lemma shows (by induction) that P , + Bis isomorphic to P, * P,”), it follows that in 91Bx, B‘;) = r.o.(P;)),and hence the definition (ii) is legitimate. Lemma 36.6. (Factor Lemma). Let us consider iterated forcing and let the notation be as above. Let u be an ordinal and let Pi) be as in (36.24). Let us assume that for any limit ordinal P, ifcf P I 1 P , 1, then P , + B is the inverse limit. Thenfor each /?,

P,+B

is isomorphic to

P,

* Pi)

Proof: By induction on p. Let P be an ordinal number; we shall construct an and P , + p . isomorphism IC between P , * P,”’ If P = 0. then P, * = { ( p , 1) : p E P,} and we let a(p, 1) = p . Thus let p > 0. A typical element of P, * PF) is a pair ( p , q ) where p E P , and q is an element of91Bisuchthat in 9lBZ, 9 is a &sequence and satisfies the conditions on iterated forcing; in particular, for each i < $, q ( i is in pi”) and the ith term of q is in Q a + i . We shall define a P-sequence ( p , + : i < P ) and let n(p, q ) = p ( p , ,p , + . . ., P % + ~. . , .). This mapping K will be an isomorphism between P, * P;)and For i < P, let qiE 91Bz be such that with B,-value 1, qi is the ith term ofq. Hence (recall the remark following the definition (36.24) of P;))

c)

A

91lB1 1 (qi is @)-valued and IIqi E Q,+i11 gi,ll = 1)

e’,

By the induction hypothesis, B , + i is isomorphic to B, * and so 9P”’ is isomorphic to (911Ba)gi”’.Let p Z t i E 91P1+ibe the element corresponding to qi under the isomorphism between (911B1)Bi’x’ and 911B”i. Clearly, I ( P z + i E Q x + i I ~ B , + , = 1. Let n(p, 4)= p ( p , , p , + . . . , p I + i ,. . .). All we have to d o now is to show that R is an isomorphism between P, * P’)and P , + , . We shall show that for each ( p , q ) E P, * Pi),n(p, q ) is in P I + pand leave the rest to the reader, namely to show that

,,

(P4) 5 (P’4’)

iff

4P4)

4p’q’)

We want to show that for each y 5 p, p “ ( ~ : ~i <+y)~ is an element of P Z f y . Since IlpI+i E Q , + i = 1 for each i < P, we need only show that i f y is a limit ordinal and P, ti’ is the direct limit, then there exists i , < y such that p z + i = 1 for all i, i, I i < 7. Thus let 7 5 p be a limit ordinal such that P , + , is the direct limit of P l + i , i < ;’. Hence in 911B1,Pj1) is the direct limit, and therefore

(IBlt,

(36.25)

11(3io < j$Vi 2 io) the ith term of q is lJIB,= 1

462

6.

OTHER LARGE CARDINALS

Now we have made an assumption that if P,+yis the direct limit, then cf y > I P I I. I t is easy to see that because I P , I < cf y, (36.25)implies that there exists io < y such that for all i 2 io 11qi =

lllB,

=1

Thus for all i 2 io, ) ( p i + i=

lIIBl+ i

=1

and hence p I t i = 1 for all i 2 i o . Proof of Theorem88 Let 91 be a transitive model of ZFC and assume that K is a supercompact cardinal in 911 and that 2" = K + (otherwise there is nothing to prove). We shall construct a generic extension 91[G] of 911 in which K is measurable and 2" = K' '. We use iterated forcing, and successively adjoin to each inaccessible cardinal a < K, a t subsets of a. At limit steps of the iteration we use direct limits when the ordinal is a regular cardinal and inverse limits otherwise. Let us define, by induction on a, the ath iterate P , (and the corresponding forcing relation ki and the algebra B, = r.o.(Pi))and the B,-valued notion of forcing Q,: +

(36.26) (a) If a is an inaccessible cardinal, let Q, be the notion of forcing in 91UIBZthat adjoins a + +subsets of a ; that is, we let Q, be, in 9NBi, the set of all functions p whose domain is a subset of size < a of a x a + (and Q, is ordered by 2 ) .Further, we let 6,be a set of representatives for +

{X E

91" : JIx

E

Q, 1

=

l}

in the equivalence relation IIx = y ( J= 1. If a is not an inaccessible cardinal, let 0, = (1) (as usual, 1 denotes the greatest element of each notion of forcing). (b) P , is the set of all a-sequences (pi : i < a) satisfying the following: (1) foreveryyca, ( p i : i < y ) E P y a n d p , E b y ; (2) if a is a regular cardinal, then 3ioVi 2 i o p i = 1.

(c) If p , q E P , , then p < , q

I

1

iff

I

(Vy < a) [p y 5 q y and p y

ky p y

is stronger than qy].

Finally, let P = P,+ and let B = r.0. P . Let G be a generic filter on P and let 9UI[G] be the generic extension of 9J by G. We shall prove that K is a measurable cardinal in 9UI[G] and that 9UI[G] k 2" = K + +.To begin, P is isomorphic to the two-step iteration P , * Q,. Thus 9l[G] = 9UI[G,][H,], where G, is 9Lgeneric on P , and H , is 9lqG,]-generic on Q, = iGM(QK). Now P , is the direct limit of P , , a < K ; and since K is a Mahlo cardinal, there is a stationary set of a < K such that P , is also

36.

MEASURABLE CARDINALS AND THE GCH

463

a direct limit. Since clearly 1 P, I < K for all a < K, it follows by Lemma 36.5 that P, satisfies the h--chain condition and hence K is a regular cardinal in 911[G,]. Also, 1 P, I = K, and hence 911[GK]satisfies (Va < ~ ) 2 'I K. In 911[G,], Q K is the standard notion of forcing that adjoins K + + subsets of K and preserves all cardinals. Thus 91[G] k ( K is a regular cardinal and 2" = K++). I t remains to prove that K is a measurable cardinal in 91[G].The basic idea is as in Lemma 36.2 or Lemma 36.3: to find an elementary embedding of 911[G]; the actual construction is somewhat more complicated. Let 1 = K + + . Since K is supercompact in 911, there is an elementary embedding j : 911 + 91 such that K is the least ordinal moved, j ( K ) > I and "1

(36.27)

A

911 E 91

It is easy to verify that I P 1 = A, that (by (36.27)) P E 91, and moreover, P is defined in 91 by the same definition (36.26) as in 911. Also, (r.0. P)" = B. Since P E 91 and 91 E 911, G is also an !%generic filter on P, and we can consider the model 91[G]. Let me prove now the following formula which will be used toward the end of the proof: (36.28)

'991[G] n 911[G] E %[GI

I t suffices to show that i f f € 91[G] is a function from 1 into ordinals, then f~ 91[G]. Letfbe a name forfand let p o E G be a condition that forces thatfis a function from 1into the ordinals. For each a < 1, let A, = { p I p o : 3p p

IF.@)

=

B)

Each A, is dense below p o (and hence A , n G # fa). For each o! < I and each E A,, let us denote by g(a, p ) the unique p such that p I k f ( d ) = Since I P I = 1,we have I g 1 = 1 and hence g E 91. Now it is easy to see thatfe 91[G] because it is defined in 9l[G] as follows:f(a) = the unique p such that for some p E G , g(a, p ) = /IQ.E.D. . (36.28).

B.

p

Let us now consider j ( P ) . In 91, j P is a notion of forcing obtained by iteration (i.e., via (36.26)) up to j ( K ) + 1. I claim that in '31 we can apply the factor lemma to j P at a = K + 1. Indeed, the first nontrivial step above a in the iteration occurs at the least inaccessible cardinal (in 91) above K, thus the first nontrivial direct limit is taken far above 4, and then only at regular cardinals. Since I P,, I = I , the assumption of Lemma 36.6 is satisfied. Hence j P is isomorphic to a two-step iteration (in 91)

,

(36.29) (j%+ * (jp)g'+:) Now the first factor of (36.29) is equal to P , + = P. Let us denote Q the second factor of (36.29). By the factor lemma, Q is, in !JIB, a notion of forcing obtained by iteration (36.26) from K + 1 to j~ + 1. At each i > K, the iteration uses a that is either trivial or adjoins i + + subsets of i (if i is notion of forcing in !Pi,

,

464

6.

OTHER LARGE CARDINALS

inaccessible in 91);in either case, the notion of forcing is strongly I-closed in Pi.By Lemma 36.4c, Q is strongly 1-closed in %’. Thus we can write

jP=P*Q

(36.30)

where Q E 9’ is a strongly 1-closed notion of forcing. Let i: be the G-interpretation of %,’ and let Q = iz(Q).Thus Q is a strongly I-closed notion of forcing in %[GI. Let p E P. Then by (36.30),j p is (represented by) a pair (s, q ) where s E P and q~ %’ is such that 11qE QII = 1. By the definition of P, p = (pi : i < K 1) and there is io < K such that pi = 1 for all i, io I i < K . Thus j ( p ) = (pi : i < j ( K ) 1 ) and pi = 1 for all i, io I i < j ( K ) . In particular, p ; = 1; and since pi = pi for all i < K, and s = j ( p ) \ ( K + l), we have s = ( p l ~cI) 1. This implies that if p E G and j p = (s, q), then s E G. Now let (36.31) D = { q E Q : for some p E G, q = i&) where j ( p ) = (s, q)} Since j I P E !Illand has size I, we have j I P E 91 and therefore D E 91[G]. It is easy to see that D is directed, i.e., if ql, q2 E D,then there is q E D such that q I q1 and q I q2 (this is because G G P is directed). We have (in %[GI), ID I I I G I IIP I = I ; and because Q is strongly I-closed, there exists a E Q (a master condition) such that a I q for all q E D. We shall now consider (as an auxiliary model), a generic extension of 911[G]. Let H be an 911[G]-generic filter on Q such that H contains the master condition a. Since H is also 91[G]-generic, and j P = P * Q, there is an %-generic filter K on j P such that %[K] = %[G][H]; in fact

+

(36.32)

+

K

= {(s, q ) : s E

G and iE(q) E H}

Everything is ready now to extend the elementary embedding j : 91 + 91 to an embedding of 911[G] in 91[K]. We work in 91[G][H] and define, for x E 9ll[G], (36.33) where x E %IB is some name for x , and i t is the K-interpretation of 91’’. We have to show that the definition (36.33) does not depend on the choice of the name x; the verification of elementarity ofj is then straightforward. Here we use the master condition a. If p E G, then j p = (s, q ) where s E P and 114 E QII = 1 . We have shown that s E G,and if q = iz(q), then, because p E G, we have q 2 a and therefore q E H. Thus (s, q ) E K, and we have (36.34)

if

PE

G,

then j ( p ) E K

Now if p E G forces x = y , then j p E K forces j x = jy and hence i%(jx)= iz(jy). Thus we have (in W[G][H]) an elementary embedding

465

37. APPLICATIONS OF FORCING IN LARGE CARDINALS

and we can define, in the usual way, an 91[G]-ultrafilter over

K

u = {xE K : K E i(x)} U is nonprincipal and 9l[G]-~-complete.It suffices to show that U E 91[G];for then 9ll[G] satisfies that K is a measurable cardinal. By (36.28), Q is I-closed not only in 91[G], but also in 9 l [ G ] . Thus the generic extension 9l[G][H] of 9l[G] does not add any new I-sequences in 91[G], and because 1 U I = I we have U E 91[G]. W 37. SOME APPLICATIONS OF FORCING IN THE THEORY OF LARGE CARDINALS

In the preceding sections, in particular in Sections 34 and 36 we used the method of forcing to obtain consistency results concerning large cardinals. I shall now present several other applications of forcing and discuss various consistency results involving large cardinals.

Large Cardinals in Generic Extensions As we have already seen, various ingenious methods of forcing have been devised to ensure that a given large cardinal remains a large cardinal in the generic extension. There are in fact several quite general results saying that if the notion of forcing used has such and such properties and if K is a large cardinal of such and such kind, then in the generic extension K is still a large cardinal of that kind. Conversely, there are results that say that certain notions of forcing cannot make a given cardinal large. The two lemmas that are presented below are typical examples of such results. We proved (in Lemma 36.2) that if K is measurable and I P I < K, then K remains measurable in 331[G].It turns out that practically every large cardinal is preserved under "mild " extensions, i.e., such that I P I c K, and conversely K does not become large in a mild extension if it is not large already. The other lemma is included to show the relation between saturation of Boolean algebras and saturation of ideals in the corresponding Boolean-valued models.

Lemma 37.1. Let K be an infinite cardinal in 'm, and let ( P , c )be a notion of forcing such that I P I c K.Let G be an 91-generic filter on 91.Then K is inaccessible (Mahlo, weakly compact, Ramsey, measurable, compact) in Iul if and only if it is inaccessible (Mahlo, weakly compact, Ramsey, measurable, compact) in 311[GI. Proof: If K is inaccessible in Iu1, then firstly K is regular in 9I[G] because all cardinals and cofinalities above I P I are preserved. Secondly, if a c K, then (2")mG1 5 I r.0. P < K and hence K is inaccessible in W[G]. Conversely, if K is inaccessible in Iu1[G], then K is inaccessible in Iu1 because 331 c 91[G].

466

6.

OTHER LARGE CARDINALS

If K is Mahlo in 91, then firstly K is inaccessible in 91[G]. Secondly, every a > I P I is a regular cardinal in 91[G] if and only if it is a regular cardinal in 911 and so the set S = { a < K : I P I < a and a is regular in 91[G]} is stationary in 91.By Lemma 35.4, S is also stationary in 91[G] and hence K is Mahlo in 91[G]. Conversely, if K is Mahlo in 91[G], then K is Mahlo in 91 because 91 G 311[G]. If K is weakly compact in 91, let F : [K]' -, (0,1) be a partition of [K]' in 91[G]. Let F E !U1* be its name, such that [IF:[k]' + (0,1}11 = 1. For each a p E K, let G(a, p) = ~ [ ~ ( d i , = 0'11 ; G is (in 91) a partition of [K]' into I B I pieces (where B = r.0. P). Since I B I < K, there is H E K of size K homogeneous for G,and it is easy to see that H is homogeneous for F. Conversely, if K is weakly compact in 91[G], let F : [K]' --* (0, 1) be a partition of [K]' in 91. There is, in 91[G], a set K c K of size K, homogeneous for F. An easy forcing argument shows that K has a subset H E 91 of size K (note that K = { K , : p E G} where K, = {a : p Ik di E K},and that some K , must be unbounded in K); hence F has in 91 a homogeneous set of size K. The argument for Ramsey cardinals is exactly the same as for weakly compact cardinals. If K is measurable in 311, then it is measurable in 311[G] by Lemma 36.2. Conversely, if K is measurable in 91[G], let U E 9ll[G] be a K-complete nonprincipal ultrafilter over K, let J be the dual prime ideal, and let J E 91' be its name. (Without loss of generality we assume that JIJis a k-complete nonprincipal prime ideal11 = 1.) Let I = {X E K : I(? E J(I = 1). It is easy to verify that I is a K-complete ideal containing all singletons. We claim that I is I P -saturated: If p IF 2 $ J and p IF f I$ J, then p IF ? n f $ J (because J is prime). Hence if X and Y are such that X $ I, Y $ I, and X n Y E I, then 112 $ Jll . 1 f $ J(I = 0, and it follows that I is I P I+-saturated.However, since I is v-saturated for some v < K and K is inaccessible, we conclude by (a generalization of) Lemma 27.5-see also Section 34-that K is measurable. If K is compact, let 1 2 K and let us show that in 9l1[G], there is a fine measure on PK(A).Let U be a fine measure on PK(A)in %I, and let j = j , be the natural elementary embedding j,,: 91 + Ult,,(91). The standard argument shows that X E U if and only if H € j ( X ) , where H is the set in Ultu(91) represented by the function d ( Z ) = Z on PK(a);also, H z { j ( a ): a < A} (and is equal to it just in case U is normal). Similarly, as in the proof of Lemma 36.2, we extend j to !M[G] as follows:

+

u

It

j ( x ) = i:"(jx)

where x is a name for x; the definition is legitimate because we assume, without loss of generality, that P E V , and hence j ( p ) = p for all p E P, and j ( P ) = P. Now we define, in !M[G], an ultrafilter W over PK(A)as follows:

X EW

iff

H€j(X)

37.

467

APPLICATIONS OF FORCING IN LARGE CARDINALS

I t is routine to verify that W is a fine measure on P,,(A); for instance, if Z o E P$), then {Z E P @ ) : Z 2 Z,} E W becausej(2,) = { j ( a ): a E Z,} c H. Conversely, if K is compact in 911[G], let S be a set in 91 and let F be a K-complete filter over S (in 91);let us show that there is a K-complete ultrafilter in 91 extending F. Every set X c F of size < K in 91[G] is included in some Y c F of size < K such that Y E 91 (this is because 1 PI K). Hence F generates a K-complete filter in 91[G] and that in turn is included in a K-complete ultrafilter on U . Let J be the dual prime ideal. As in the proof for measurable 12 E Jll = 1) is K-complete and cardinals above, the ideal I = {X G S : 1 1 P /+-saturated,and clearly X E F implies S - X E I. Since K is inaccessible and I is v-saturated for some v < K, I has an atom A. If X E F, then X n A # I and so { X G S : X n A # I} is a K-complete ultrafilter extending F.

-=

There are of course numerous examples when K ceases to be large when we use a notion of forcing of size 2 K (a good example is the Lkvy collapse). The example given below is quite interesting since obviously inaccessibility of K is preserved by any notion of forcing that is a-distributive for all a < K. Note that if K is a Mahlo cardinal, then K has no unbounded subset A such that I A n y ( < y for all regular y < K (since the set of all regular y < K is stationary, an application of Fodor’s theorem would give I A I < K ) . Exercise 37.1. Let K be an inaccessible cardinal. There is a notion of forcing (P. < ) such that I P I = K and P is a-distributive for all a < K, and such that K is not a Mahlo cardinal in the generic extension. [Forcing conditions are sets p E K such that I p n y I < y for every regular y IK ; p < q iff p is an end-extension of q, i.e., if q = p n a for some a. To show that for any a < K. P does not add any new a-sequence, observe that Vp 3q Ip such that P , = {r E P : r < q ) is a-closed.]

The following lemma shows that there is a relation between saturation of ideals in generic extensions and saturation of the notion of forcing. Recall that in Exercise 34.5we proved that if K carries a a-saturated K-complete ideal, then so it does in the generic extension if the notion of forcing is maturated. The lemma below is a generalization. The proof of (a) uses generic ultrapowers (see Section 35). One particular property needed is the following: Exercise 37.2. Let I be a ri+-saturated h--complete ideal over K. Let G be a corresponding generic ultrafilter and let ‘91 = UltG(911)be the generic ultrapower. Then every s : K + 91 in 9I[G] is in ‘91. [Let s be a name for s; for each a < K, let s, be a name such that Ils(i) = s,II = 1. By the remark preceding Exercise 35.11, each s, is represented by a function 1; E 91 on K. Let h: K + K be the least unbounded function (that is, h represents K in 91). Let f be the function on K defined b y f ( a ) = ( f P ( a ): /3 < h(a)). Show that f represents s in the generic ultrapower.]

468

6.

OTHER LARGE CARDINALS

Lemma 37.2. Let B be a complete Boolean algebra in 91, let G be an 91-generic ultrafilter and let K be an uncountable regular cardinal in 91. (a) Let I I K + and assume that sat B II and sat B < K. If in 91, K carries a I-saturated K-complete ideal, then the same is true in 91[G]. (b) I f sat B I I and if in 91[G],K carries a I-saturated K-complete ideal, then the same is true in 91.

Proof: We shall prove (b) first since the proof is somewhat easier. Let J E 91’ be such that J1J. is a k-complete A-saturated ideal over 2)) = 1. We let I = { X E K : JIX E JIJ= I}. It is easy to see that I is a K-complete ideal containing all singletons. To show that I is I-saturated, let us consider the Boolean algebra D = P(K)/I and let us show that D is I-saturated. In 91’, let C be the Boolean algebra P(K‘)/J. Since B is I-saturated and 9P t C is 1-saturated, the algebra B * C is I-saturated (see Lemma 23.6). Thus it suffices to find a homomorphism h : D + B * C such that h(d) # 0 for every d # 0. If d E D, then d is the equivalence class (mod I) d = [XI for some X G K. Let h(d) = c where c is in 91’ the equivalence class of 2 mod J. It is routine to verify that the definition of h does not depend on the choice of X , that h is a homomorphism from D to B * C,and that h(d) # 0 whenever d # 0. I shall now sketch the proof of (a). Let I be, in 91,a K-complete I-saturated ideal over K. Let J E 9 l [ G ] be the ideal generated by I. Since sat B IK, it is routine to verify that J is K-complete. Let J E 9 31’be the canonical name for J , and let C E 91’ be the Boolean algebra C = P(k)/J. We want to show that 91’ t C is 1-saturated; by Lemma 23.6 it suffices to show that B * C is I-saturated because B is I-saturated. Let D = P ( K ) / I .We shall find in WID a Boolean algebra E such that 91Dt E is 1-saturated, and such that D * E is isomorphic to B * C . Since D is I-saturated, it will follow that D * E is I-saturated and we shall be done. In !WID, consider the generic ultrapower 91 = UltG(91),where G is the canonical ultrafilter on fi. Letj: 91 + 91 be the corresponding elementary embedding. Let E = j ( B ) . Let sat@) = v < K. By elementarity ofj, we have 91 k sat(@) = j ( $ ) ; and since j ( $ )= $ and by Exercise 37.2 all $-sequences in 9IDare in 91, we have 9IDt sat(E) = $. Thus 91DI= E is 1-saturated. It remains to show that B * C and D * E are isomorphic. Let c E B * C. Then c E 91’ and IIc E CI), = 1. Thus there is X E 91’ such that IIX c CllB = 1 and that

X mod J(JB =1 Let f: K + B be the function f (a) = I(di E XI/.Since f (a) E B for all a c K , f J(cis the equivalence class of

represents in 9l = U l t , ( i ) G WID an element e E j ( @ = E; and let h(c) = e.

eED

* E. We

37.

APPLICATIONS OF FORCING IN LARGE CARDINALS

469

We leave it to the reader to verify that the definition of h(c)does not depend on the choice of X and that h: B * C + D * E is an isomorphism. W Large Cardinals and the Generalized Continuum Hypothesis Considering the influence of large cardinals on cardinal arithmetic, it is important to know what behavior of the continuum function 2Nais compatible with various kinds of large cardinals. The most is known for measurable cardinals: First, there is a transitive model of ZFC + "there is a measurable cardinal," namely the model 4 U ] , which satisfies the GCH. Secondly, 2" > ti+ for a measurable cardinal implies existence of models with many measurable cardinals, thus showing that consistency of 2" > ti+ at a measurable cardinal requires some stronger assumption than just measurability. And thirdly, Theorem 88 provides a model for 2" > ti+ at a measurable cardinal. For some, "small," large cardinals, consistency with the GCH can be established by means of a transitive model. For instance, weakly compact cardinals are consistent with the GCH because every weakly compact cardinal is weakly compact in L. Or, if K is a Ramsey cardinal, it is known that there exists a transitive model (the core model K ) that satisfies the GCH and in which ti is Ramsey. On the other hand, no construction of a transitive model is known for compact (or supercompact) cardinals. However, compact and supercompact cardinals are consistent with the GCH, but the consistency proof uses a generic extension by iterated forcing, using the techniques and ideas presented in Section 36. Iterated forcing as in Section 36 can also be used to establish the consistency of 2" > K + and " K is weakly compact (Ramsey, supercompact)" relative to ZFC + " K is weakly compact (Ramsey, supercompact)." Changing Cofinality of Cardinals Let us address ourselves to the following problem: Can one construct a generic extension in which all cardinals are preserved but the cofinality of some cardinals is different from their cofinality in the ground model? Obviously, in order to achieve this we have to change some (weakly) inaccessible cardinal into a singular cardinal. Now Corollary 3 of Jensen's covering theorem tells us ' in the ground model. Thus we that for this, it is necessary to assume at least 0 formulate the problem as follows: Let ti be some large cardinal and let I < ti be a regular cardinal. Find a generic extension which preserves cardinals and in which cf ti = 1. We have seen one method of doing this in Section 35 (application 7 of generic ultrapowers); however, the assumptions we have used there are rather strong and perhaps not as plausible as better known large cardinal assumptions. Theorem 89 below provides an answer to the problem when ti is measurable

470

6.

OTHER LARGE CARDINALS

and I = K O .A generalization of the construction in Theorem 89 gives the solution for any regular I < K assuming that K is supercompact.

Theorem 89 (Prikry). Lat K be a measurable cardinal. There is a generic extension in which cf K = o and no cardinals are collapsed. Moreover, every bounded subset of. in 91[G] is in the ground model. Proof. Let K be a measurable cardinal and let D be a normal measure on K. Let ( P , < ) be the following notion of forcing. A forcing condition is a pair p = (s, A ) where s E [K]'", i.e., s is a finite subset of K,and A E D. A condition (s, A ) is stronger than a condition (t, B) if:

(37.1) (a) t is an initial segment of s, i.e., t = s n a for some a ; (b) A G B ; (c) s - t C B.

We immediately note that any two conditions (s, A), (s, B) with the same first coordinate are compatible, and hence any incompatible W G P has size at most K. We also note that if (s, A ) and ( t , B) are compatible, then either s is an initial segment o f t , or t is an initial segment of s. Let G be a generic filter on P. We shall show that in 91[G], K has cofinality o,that every bounded subset of K is in 91,and that all cardinals and cofinalities above K are preserved. The last statement is immediate since P satisfies the K+-chain condition. I t is also easy to show that cf K = w in 9 I [ G ] : If (s, A ) and ( t , B) are in G, then either s is an initial segment of t or vice versa; hence S = {s : (s, A ) E G for some A} is a subset of K of order type w. By genericity of G, S is clearly an unbounded subset of K, and hence cf K = w. I t remains to show that if X E 91[G] is such that X G I for some I < K,then X E 91. For this, we need the property of P stated in Lemma 37.3 below (and Lemma 34.4). The proof uses Theorem 70 which states that every partition of [ K ] ' " into less than K pieces has a homogeneous set H E D.

u

Lemma 37.3. Let o be a sentence of the forcing language. There exists A E D such that the condition (0, A ) decides o,i.e., either (0, A ) 11c,or (0, A ) tk lo. Proof. Let S + be the set of all s E [K]'" such that (s, X ) Ik o for some X and let S - = {s : 3X(s, X ) Ik lc}.Let T = [K]'" - ( S + u S - ) . By Theorem 70, there is a set A E D such that for every n, either [A]" c S + or [A]" G S - or [A]" c T. We shall prove that (0, A ) decides 6. If not, then there are conditions (s, X )and ( t , Y),both stronger than (0, A) such that one forces o and the other forces l o . We may assume that 1 s 1 = I t 1, say 1s I = It I = n. Since (s, X ) I(0, A), we have s E [A]"; and similarly, t E [A]".But s E S f and t E S - , which is a contradiction since S + and S - are disjoint and therefore cannot both have a nonempty intersection with [A]". W

37.

47 1

APPLICATIONS OF FORCING IN LARGE CARDINALS

Lemma 37.4. Let

be a sentence of the forcing language and let (so, A,) be a condition. Then there exists A G A , such that A E D and such that the condition (so, A ) decides (T. (T

Proof. A slight modification of the preceding proof; we may assume that min(A,)> max(s,). Let S + be the set of all S E [A,]'" such that (so u s, X ) ~kCT for some X c_ A, and let S- be defined similarly. As before, there exists A E A , in D such that for no n, [A]" intersects both S + and S - . It follows that (so, A ) decides 6.

Now let I < K and let X c 2; we will show that X E 91.Let X be a name for E G be a condition such that p lk X 5 2. It suffices to show that po there is q 5 p and a set Z G I such that q Ik X = 2. for each p I Let p I p , , p = (s, A ) . For each tl < I, there is, by Lemma 37.4, a set A, c A in D such that (s, A,) decides the sentence di E X.Let B = A,; we have B E D and q = (s, B) decides each Oi E X, a < I. Thus if Z = {a < I : q l k d i ~X), we have q IF X = 2. This completes the proof of Theorem 89.

X,and let p ,

on<,

We proved in Section 29 that every measurable cardinal is a Rowbottom cardinal and that a Rowbottom cardinal is either weakly inaccessible or has cofinality w . The model %(GI of Theorem 89 provides an example of a singular Rowbottom cardinal. The exercise below shows that K has in %1[G] the combinatorial property equivalent to being a Rowbottom cardinal. Exercise 37.3.* In '.X)I[G],if F : [ K ] ' " -+ A is a partition into 1 < K pieces, there exists E K of size K such that F takes at most Ho values on [HI'".

H

[Let F be a name for F and let (so, A o ) be a condition (such that max(so) < min(A,)). Let g be a partition of [K]'" x [ K] '" into 1 pieces, defined as follows: If s E [A,]'" and for some X E A , , (so u s, X ) IF F(r)= 2, then let g(s, t ) = a ; otherwise, let g(s, t ) = 0. Show that there is A G A, in D and a countable S E 1 such that g[[A]'"' x [ A ] ' " ] L S. Then (so, A ) I+ F[[A]'"'] E f.]

Singular Cardinals Problems It follows from Jensen's covering theorem that if the singular cardinals hypothesis fails, then O x exists; more recent results then show that, moreover, there is a model with a measurable cardinal. Thus to construct models violating the singular cardinals hypothesis one needs large cardinals in the ground model. And even then we cannot prescribe the behavior of the continuum function at singular cardinals arbitrarily: As we have seen in Section 8, the least singular cardinal at which the SCH fails has cofinality w, and there are also bounds on 2" in some cases. The last two theorems establish independence of the singular cardinals hypothesis.

6.

472

OTHER LARGE CARDINALS

Corollary (to Theorems 88 and 89). I t is consistent (relatioe to existence o f a supercompact cardinal) that there is a strong limit singular cardinal ti such that 2" > .'K Proof. Let K be a supercompact cardinal. First we construct a generic extension in which K is measurable and 2" > K'. Then we extend the model further to make K a singular cardinal. The new model still satisfies 2" > K', and K is a strong limit cardinal.

There are several forcing constructions giving answers to various singular cardinals problems. For instance, there is a model in which 2Nn= K m + for all n, and 2No= Nu+,. It is an open problem to construct a model in which 2ffi> K,, for all a.

,

2c" < K and K < K" < 2". [Extend the model in the corollary by adjoining a large number of subsets of wl.]

Exercise 37.4. It is consistent that

+

More Applications of Forcing

We shall now use the method introduced in Section 36 to show that it is consistent with a measurable cardinal ti to assume that there is more than one normal measure on ti. Then we use the method to prove Theorem 86b, giving a model with a precipitous ideal over ol. Example. Let 91 be a transitive model of ZFC + GCH + there is a measurable cardinal K. Let A > K be a regular cardinal. There is a generic extension in which 2" = K', 2"' = I, and there are I normal measures on ti. We use the method introduced in Lemma 36.3 (which will be followed closely). For each regular cardinal a, let P ( a ) be the notion of forcing that adjoins a generic subset of a ; let P be the Easton product of the P(a)'s for all successor cardinals a < ti. More precisely, P consists of 0, l-functions p such that dom(p) G ti x ti, and:

(37.2)

(i) if (a, p) E dom(p), then a is a successor cardinal < K, and fl < a ; (ii) for every regular y, I {(a, p) E dom p : a I y } I < y.

For each v, let P , consist of all p E P such that a < v whenever (a, p) E dom p ; similarly P' is the set of all p such that a 2 v for all (a, p) E dom p . For each v, P is isomorphic to P , x P'. Let U be a normal measure on K, let 9 = Ult,(%1), and let j = j , be the corresponding elementary embedding. Let us consider the notion of forcing j ( P ) in %. As in Lemma 36.3, we have (jP)" = P ; let Q = (jP)". Clearly, (K,p) !$ dom p for any p < K and hence Q is K-closed. If G x H is a generic filter on P x Q, then %I[G x HI has a normal measure on K defined in a similar way as in Lemma 36.3: (37.3)

X E D

iff

K E

iFX,,(j(X))

37.

APPLICATIONS OF FORCING IN LARGE CARDINALS

473

As before, every X E K in 'IlUI[G x HI is in 'IlUI[G];Xcan be any r.0. P-valued name for X , and its choice is irrelevant because p E G implies j p E G x H. Let D be the canonical name for D ;thus if p E P, q E Q, and if X E 'Illr.o.p, then

p uq

(37.4)

Ik

(XED*K*Ej(X))

Let p E P and q E Q; we have p = j ( p ) , and q is represented in the ultrapower 91 by a function ( q . : a < K ) where q= E P" for each a c K. From (37.4) we get (37.5)

p uq

X ED

11

iff

for almost all a, p u qa ik o i X~

Now we are ready to construct the model. Let R be the product of 1 copies of Q, i.e., the set of all functions f E'Q such that [{i < I : f ( i ) # 0 } (IK. We consider the notion of forcing P x R. Let G x H be an 'Ill-generic filter on P x R. We claim that the model 'IlUI[G x HI has the required properties. Since R is K-closed, all subsets of K are contributed by G ; hence 2" = K + holds in 'IlI[G x HI.Standard arguments show that cardinals are preserved and 2"' IL in 'IlUI[G x HI. For each i < 1,let Qi denote the ith copy of Q and let Di be the canonical name for a normal measure using Q iinstead of Q in (37.4). It suffices to show that for any i # k < 1,Diand D, denote different measures in 'Ill[G x HI. The last assertion follows by a standard argument using genericity, and we leave its proof to the reader: Given i # k and a condition (p, r ) in P x R, use (37.5) to find a stronger condition (p', r') and some r.0. P-valued name X such that (p', r') forces X E Dibut X # D k . rn The model Y V ] has the property that the only K-complete ultrafilter over K that extends the closed unbounded filter is the unique normal measure (see Exercise 3 1.12). Exercise 37.6 shows that if there are K distinct normal measures on K , then there exist nonnormal K-complete ultrafilters extending the closed unbounded filter. Exercise 37.5. Let U., u < K , be K-complete nonprincipal ultrafilters over K and let D be a normal measure on K , D # Urnfor all u. Then there exists X E D such that X $ U , for all u. [For each u < K let X , E D be such that X, $ U . . Let X be the diagonal intersection A,,, X , . ] Exercise 37.6. Assume that there exist K distinct normal measures D., u < K , on K . Then there exists a K-complete ultrafilter U over K such that U is not normal and such that C E U for every closed unbounded set C. [Let {A, : u < K } be a pairwise disjoint collection of subsets of K such that A, E D, for all u < K ; such a collection can be easily constructed using Exercise 37.5. Let W be any K-complete nonprincipal ultrafilter over K . We define U as follows: Let X E U iff (a : X n A, E DI}E W. U is a K-complete nonprincipal ultrafilter over K . If C is closed

6.

474

OTHER LARGE CARDINALS

unbounded, then C n A, E D , for all a, and hence C E U.For each a let a, = min A , , and let f be a function on K that has value a, on each A , . Then for each y < K, ( 5 : f ( t )> y ) E U and {( :f(() < l }E U ;hence U is not normal.]

Proof of Theorem 866. We shall now use the method of Lemma 36.3 to show that it is consistent, relative to a measurable cardinal, that K, carries a precipitous ideal. Let 9 1 1 be a transitive model of ZFC and let K be a measurable cardinal. We shall show that if 91[G] is the generic extension of 91 obtained by the Levy collapse such that K becomes K1,then 91[G] has a precipitous ideal over K1. Let P be the set of all functions p such that dom(p) is a finite subset of K x o and such that p(a, n) < a for all (a, n) E dom(p); p is stronger than q if p 2 q (see (20.7)). Let G be an 91-generic filter on P . In 91[C], K is K,. Let D be, in %I, a normal measure on K, let % = Ult,(91) and let j: 911 -,91 be the elementary embedding j = j , . In 91[G], let I be the ideal over K generated by the dual of D ; ie., (37.6)

XEI

XnY=@

iff

forsome

YED

A routine argument (using sat P = K ) shows that I is in 91[G] a countably

complete ideal containing all singletons. It can be proved that I is precipitous; instead, we shall prove a weaker (but sufficient) property, namely that there exists S E K l , S $ I, such that Z(S= {X E K l : X n S E I} is a precipitous ideal. For that, it suffices to show that there is S 3 I such that (when forcing with sets X $ I) S forces that the generic ultrapower is well-founded. In turn, it suffices to construct an extension of 91[G] in which there exists an 91[G]ultrafilter W over K, generic over 9I[G] (with respect to forcing with sets X # I) such that the generic ultrapower Ult,(9l[G]) is well-founded. Let us go back to Iu1 and consider again the notion of forcing P . For every v, let P , be the set of all p E P such that a < v whenever (a, n) E dom(p), and let P' = {p E P : a 2 v for all (a, n ) E dom(p)}; P is isomorphic to the product P' x P'. Let us consider the notion of forcing j ( P ) .Clearly, ( j P ) K= P , and thus j P is isomorphic to P x Q where Q = (jP)".Every q E Q is represented in the ultrapower % by a function (4. : a < K) such that q,, E P" for all a < K. Let H be an W[G]-generic filter on Q; thus G x H is 91-generic on P x Q. As in Lemma 36.3, we define in W[G x HI an 9l[G]-ultrafilter W over K as follows:

X

(37.7)

E

W

iff

K E

izxH(jX)

The definition (37.7) does not depend on the choice of the name X because E G implies j ( p ) E G x H.Let W be the canonical name for W. As in (37.5), we have for any p E P, q E Q, p

(37.8)

(p, q)

Ik

XE W

iff

for almost all a, p u qm

IF

dEX

37.

475

APPLICATIONS OF FORCING IN LARGE CARDINALS

(here X is an r.0. P-valued name and ( q l : a < K) represents q in 91; "almost all" refers to the normal measure D). First we observe that the ultrapower Uh,(%I[G]) is well-founded. This is because the following commutative diagram holds: 91[G x HI

9lI[G]

In the diagram, j is the extension o f j : 91 + 91 defined by j ( x ) = ic" &)

and k is defined as follows: I f f € %I[G] is a function on Ultw(911[G]),then

K

representing [f] in

k"1) = (jf)(K) Both j and k are elementary and the diagram commutes. It remains to show that W is 911[G]-generic with respect to forcing with sets X # I. It suffices to show that if .T = { X i :i < 8) is an I-partition of K, then Xi E W for some i. Let T E 91r.o.p be a name for 3 and let Xi, i < 8, be names for the X i . Let us assume that there are conditions p E G and q E H such that p

ik

X is an /-partition of 2

and for each i < 0, (P, 4 )

1'

Xi

# W

We shall derive a contradiction. Let q be represented in 91 by (4. : a < K ) . By (37.8) there is for each i a set A i E D such that for all a E A i , p u qa

(37.9)

Ik

oi #

xi

Let us define (in 9lI[G]), (37.10)

T = {a < K : q1 E G}

We shall prove that T # I and that T n X i E I for all i < 8, thus reaching a contradiction since . iis an I-partition. For each i < 0, if a E T and a E A i , then p u qQE G and hence, by (37.9), a # X i .I t follows that T n Xi n Ai = 0, and so by (37.6), T n Xi E I. Let us finally show that T 4 I. It suffices to show that T n Z # $3whenever Z E D. Thus let Z E D, and let us prove that qa E G for some ct E Z. Let

A = { r E P : r I qa for some a E Z} It is easy to see that A is dense in P because Z is unbounded and qa E P ' for each a < K. Thus A n G # 0 and hence T n Z # 0.

476

6.

OTHER LARGE CARDINALS

Large Cardinals and Set Theory Without the Axiom of Choice Since the methods of the theory of large cardinals use heavily the axiom of choice, most of the problems concerning large cardinals are not very interesting if the axiom of choice is not assumed. An exception is measurability. In ZF, a measurable cardinal is defined in the usual way: An uncountable aleph K is measurable if there exists a K-complete nonprincipal ultrafilter over K. One of the first properties of measurable cardinals we proved in ZFC in Section 27 is that every measurable cardinal is inaccessible.The same proof in ZF gives only the following: If K is measurable, then K is regular, and 2A 2 K for any 1 < K. In particular, it is not excluded that K1 might be measurable. And indeed, we prove in Theorem 90 below that ZF + ‘‘N,is measurable” is consistent. The consistency is relative to ZFC + “there is a measurable cardinal.” This is necessary because if K is measurable (in ZF) and if U is a K-complete nonprincipal ultrafilter over K, then YU] is a model of ZFC + “ K is measurable.”

Theorem 90 (Jech). Let 91 be a transitive model of ZFC + “there is a measurable cardinal.” There is a symmetric model 91 2 91 of ZF such that 91 1 K, is measurable. We shall construct a symmetric extension of 911. Recall ihe theory of symmetric models from Section 21. We consider a complete Boolean algebra B, a group 8 of automorphisms of B, and a normal filter 9 on Y (see (21.5)). For every x E 91’ we let sym(x) be the symmetry group of x, sym(x)= (nE 9 : nx = x}, and call x symmetric if sym(x)E 9. We denote by HS the class of all hereditarily symmetric names (21.14). If G is an 9bgeneric ultrafilter on B, we let % be the G-interpretation of the class HS; 91 is a model of ZF and 91 2 91. Let us call a subset A G B symmetric if (37.11)

{n E Y : na = a for all a E A } E ,F

Lemma 37.5. Let K be measurable in 331, and let 91 be a symmetric extension of 9.N (via B, G, 9,B). If every symmetric subset of B has size < K, then K is measurable in 3. Proof. Let U be, in 91,a K-complete nonprincipal ultrafilter over K. We show that U generates a K-complete nonprincipal ultrafilter over K in 91. It suffices to show that if y < K and {X,: a < y } is a partition of K in 91, then for some a < y, X, includes some Y E U . I shall give the proof for y = 2 since the general case is analogous. Let X E % be a subset of K, and let X E HS be a symmetric name for X. Let A = (118 E XI1 :c1 < K}. If n E 8 is such that n ( X ) = X, then (because nd = d for all a) na = a for all a E A ; thus A is a symmetric subset of B. Hence 1 A I < K. For each a E A, let Y, = {a : 116 E XI/ = a } . Clearly,

37.

APPLICATIONS OF FORCING IN LARGE CARDINALS

477

{ Y, : a E A ) is a partition of K into less than K pieces, and hence one Y = Y, is in I/. Now if a E G, then we have Y E X, and if a $ G, then Y E K - X. Hence either X or K - K has a subset that is in U . Proof of Theorem 90. Let K be a measurable cardinal in 911. Let P be the set of of ordinals less than K ; p all one-to-one finite sequences p = ( a o , al, . . ., a,is stronger than 4 if p extends q (this notion of forcing collapses K and makes it countable). Let '3 be the set of all permutations of K ; every a E '3 induces an automorphism of (P, < ) as follows:

n() = (n(ao),. . ., n(an- ,)) and, in turn, an automorphism of B = r.0. P. Thus we identify 3 ' with the group of automorphisms so induced. For each y < K, let H , = {n E 9 : na = a for all a < y} and let 3 be the normal filter on Y generated by {H, : y < K}. Thus x E 911' is symmetric just in case there is y < K such that nx = x whenever na = a for all a < y. Let G be an 911-generic ultrafilter on B and let 91 be the symmetric model given by B, G, 3 , Y . We shall show that K = (N,)' and that K is measurable in 91. If y < K, then y is countable in 91: Let f b e the name such that IIf(n') = $11 = C { p E P : p ( n ) = a }

for all n < w and a < y. Clearly, f is symmetric because n ( f )= f for every n E H , , and hence f E HS. The interpretation off is a one-to-one function of o onto y. It remains to show that K is measurable in 91. By Lemma 37.5 it suffices to show that every symmetric A c B has size < K . Let A c B be symmetric. There exists y < K such that nu = a for all a E A and all n E H,. For every a E A, let S, = {p E P : p Ia). If x E H,and p E S,, then xp E S, because np I nu = a. Let T, = { p E S,: p ( n ) < y + o for all n E dom(p)}. If n E H, and p E T,, then np E S,. Conversely, if p E S,, there is n E H , that maps all a E ran(p) greater than y into y + w ; since n E H,and p E S,, we have np E S, and hence np E T,. Thus S, = {n(p): p E T, and n E H,}, and consequently (37.12)

if a # b

E

A,

then

T,# &

However, each T, is a set of finite sequences in y sible, (37.12) implies that I A I < K .

+ o;and since K is inacces-

In connection with measurability of Nl,we recall the following open question: Is it consistent that the closed unbounded filter over N l is a countably complete ultrafilter? It has been proved by the method of Theorem 80 that if that

478

6.

OTHER LARGE CARDINALS

is the case, then for every 6' there is a transitive model with 6' measurable cardinals. Let me also mention another open problem. In Section 21 (Example IV) we presented a model of ZF in which K is a singular cardinal. By Jensen's covering theorem (see Exercise 30.7), if we want both K1and K2 singular, then we have to use large cardinals. Indeed, a model is known (using K < 1 such that K is supercompact and I is measurable) in which cf w 2 = cf w 1 = o.The problem of whether cf a = w for all limit a is consistent is still open. 38. MORE ON ULTRAFILTERS

We have used ultrafilters extensively in our study of large cardinals, particularly ultrafilters that are countably complete. In this last section of Part 111, we shall investigate ultrafilters that are not countably complete. First we shall consider ultrafilters over w that have a partition property related to Ramsey's theorem (Lemma 29.1), and then we shall investigate a property of ultrafilters called regularity.

Ramsey Ultrafilters Over w In Section 24, we introduced p-points and Ramsey ultrafilters. Recall that an ultrafilter D over w is a p-point if for every partition .d of w such that A # D for all A E d ,there exists X E D such that X n A is finite for each A E .d. D is a Ramsey ultrafilter if for every partition d of w such that A # D for all A E .d there exists X E D such that for every A E d , X n A has one element. If 2K0= K1,then Ramsey ultrafilters exist: Let d,,a < wl, enumerate all partitions of w and let us construct an w,-sequence of infinite subsets of w as E A for some follows: Given Xu,let Xu+ E X, be such that either A E d,, or that 1 X u + n A 1 5 1 for all A E d,. If a is a limit, let X, be such that X, E X, m0d.f. for all f l < a. Then D = {X : X 2 X,for some a < wl} is a Ramsey ultrafilter. We shall show below (Theorem 91) that existence of Ramsey ultrafilters is not provable in ZFC; we shall exhibit a model of ZFC in which there are no Ramsey ultrafilters. The following lemma explains the name of Ramsey ultrafilters:

Lemma 38.1. Let D be a nonprincipal ultrafilter over w. D is Ramsey ifand only if for all natural numbers n and k, every partition F : [w]" + { 1, .. ., k } has a homogeneous set H E D. ProoJ: First assume that D has the partition property stated in the lemma. Let .d be a partition of w such that A # D for all A E d ;we shall find X E D such that 1 X n A I I 1 for all A E .d.Let F: [a]'+ (0, I}be as follows: F(x, y ) = 0 if x and y are in the same A E .d, and F ( x , y ) = 1 if x and y are in different members of .d.If H E D is homogeneous for F, then clearly H has at most one element common with each A E .d.

38.

479

MORE ON ULTRAFILTERS

Now let us assume that D is a Ramsey ultrafilter. We shall first prove that D has the following property: (38.1) if X , 3 X I3 X, 2 . . . are sets in D,then there is a sequence a, < a , < a, < ... such that {a,}:=, E D and a , + , E Xan for all n.

Thus let X, 2 X , 2 ... be sets in D. Since D is a p-point, there exists Y E D such that each Y - X, is finite. Let us define a sequence y o < y, -= . - -in Y as follows: yo = least yo E Y such that {y E Y : y > yo} G X, y , = least y ,

y,,

E

Y such that y, > yo and {y E Y : y > y , }

, = least y,, ,

E

Y such that y,,

,> y, and {y

E

G

X,,

Y : y > y,,

c XYn

For each n, let A, = {y E Y : y , < y I y,, l}. Since D is Ramsey, there exists a set {z,,},"=, E D such that z, E A, for all n. We observe that for each n, z , , , E Xrn: Since z,+, > y,+,, we have z , + , E X y n + land , sincey,+l 2 z,, we have Xym+, G XZnandhencez,., E X,.. . D or Thus if we let a, = z,, and b, = z,,+ for all n, then either { a , } ~ = E {b,},"=, E D ;and in either case we get a sequence that satisfies (38.1). Now we use the property (38.1) to prove the partition property; we proceed by induction on n and follow closely the proof of Lemma 29.1. Let F be a function from [w]"' into { 1, . . ., k}. For each a E w, let F, be the function on [w - {a}]"defined by F,(x) = F ( x u {a}).By the induction hypothesis, there exists for each a E w a set Ha E D such that F , is constant on [HA".There exists X E D such that the constant value of F , is the same for all a E X ;say F,(x) = r for all a E X and all x E [H,]". For each n, let X, = X n H , n H , n ... n H,. By (38.1) there exists a sequence a, < a , < a, < ... such that a , + , E X," for each n, and that {a,,}~=,, E D. Let H = { u , , } ~ =It~ . is clear that for each i E w, a, E X and {a, : m > i} G H a i . Hence F J x ) = r for all x E [{a,: m > i}T, and it follows that F is constant on If D is an ultrafilter over w, let us call a function$ w w unbounded mod D if {n : f ( n )> k} E D for every k E 0 ;let us call f one-to-one mod D if there is X E D such that f is one-to-one on X.The following exercise is a straightforward application of the definition of Ramsey ultrafilter: Exercise 38.1. A n ultrafilter D over o is a

Ramsey ultrafilter ifand only ifevery function

unbounded mod D is one-to-one mod D.

If D and E are ultrafilters over w, then D I E means that for some function f:w+o,

D =f,(E)

= {X :f-,(X)E

El

480

6.

Exercise 38.2.* If D =f,(D), [Let X = { n : f ( n ) < n}.

OTHER LARGE CARDINALS

then { n : f ( n ) = n } E D.

Y = { n :f(n) > n}. For each n E X, let I(n) be the length of the maximal sequence such that n z f ( n ) > f ( f ( n ) ) > .... Let X o = { n E X : I(n) is even} and X , = { n E X : / ( n ) is odd). Neither X I nor X 2 can be in D since, e.g., X o n f- l(Xo)= 0.The set Y is handled similarly, except that it remains to show that

the set Z ofall n such that the sequence n
D = E means that there is a one-to-one function of w onto w such that E =f,(D).= is an equivalence relation (and if Dl = D 5 E = E',then D' I E'). Exercise 38.3. If D 2 E and E 2 D, then D

= E.

[Use Exercise 38.2.1

Thus 5 is a partial ordering of ultrafilters over w (or rather their equivalence classes mod =). A nonprincipal ultrafilter D over w is minimal if there is no nonprincipal E such that E I D and E f D. Exercise 38.4. An ultrafilter D over w is minimal if and only if it is Ramsey.

[If D is Ramsey and E =f*(D) is nonprincipal, thenfis unbounded mod D, hence one-to-one mod D and consequently, E = D. If D is minimal andfis unbounded mod D, then D I f , ( D ) and hence D = g , ( f * ( D ) ) for some g. It follows, by Exercise 38.2, thatfis one-to-one mod D.]

We shall now outline the proof of the following theorem which shows that it is not provable in ZFC that Ramsey ultrafilters exist.

Theorem 91 (Kunen). Let 911 be a transitive model ofZFC plus GCH. There is a generic extension of 91 in which there are no Ramsey ultrajlters. We start with the following observation. Let S be an infinite set. Ajnitely additive measure on S is a real-valued function p on P ( S ) such that: (38.2)

(i) p ( @ ) = 0 and p(S) = 1; (ii) if X c Y, then p ( X ) I p( Y); (iii) if X and Y are disjoint, then p ( X u Y) = p ( X ) + p( Y ) ;

(compare with (27.1)).A finitely additive measure p on S is atomless if for every

X G S of positive measure there is Y s X such that 0 < p( Y ) < p ( X ) .

Lemma 38.2. l f p is an atomless finitely additive measure on w, then there is no p-point extending thefilter { X G w : p ( X ) = 1). Proof. Assume that D is a p-point such that X E D whenever p ( X ) = 1. Hence every X E D has positive measure. We construct an a,-sequence Xu, a < a,, of sets in D as follows: Given Xu,let Xu+ c Xu be such that Xu+ E D and p(Xu+,) < p ( X J If a is a limit, let XuE D be such that X uE X, m0d.f. for all

38.

481

MORE ON ULTRAFILTERS

fl < a (see Lemma 24.9). Then p( X,) > p ( X ,) > . . > p ( X , ) > . . . is a descend*

ing w,-sequence of real numbers; a contradiction. Now let 911 be the ground model and assume that 91 satisfies the GCH. Let = w 2 and let us consider the measure space "(0,1) defined in (34.3); let B be the corresponding measure algebra. For every a < K, let B, be the measure K

algebra of "{O, l}, and B" the measure algebra of '-,{O, 1). We have B = B, 0 B";and if cf u = wl, then B, = UB<,B , . Also, B = Uu<.B , . Let G be an 9bgeneric ultrafilter on B. We shall show that in 91[G] there is no Ramsey ultrafilter over w. Let D E 9t[G] be an ultrafilter over o and let D be a name for D. For each a < K we have I B, I I K1 and so there are K 1 B,-valued names for subsets of w. Moreover, if cf a = w 1 and X is a Bu-valued name for a subset of w, then X is BB-valued for some p < a. For each a < w there is y < w such that whenever X is a B,valued name for a subset of w then IIX E Dll E B y .Consequently, there exists a (of cofinality 0,) such that for all X E 91B,I(XE 0511 E B, implies (IXEDI( E B, and hence if we denote G, = G n B, and U = D n 911[G,], then U E W[Ca]and U is an ultrafilter. A somewhat more complicated argument (which we leave out) shows that if D is a Ramsey ultrafilter in !Ut[G],then a can be found such that U is a Ramsey ultrafilter in %1[G,]. Now W[G] is a generic extension of 91[Gu],namely W[G] = W[GJ[G"] where G" is 9l[G,]-generic on B".We state without proof that the completion of B" in 91[G,] is a measure algebra (in 9l[G,]). Hence if we apply the following lemma in 91[G,], we obtain a contradiction, thus showing that W[G] has no Ramsey ultrafilters.

,

Lemma 38.3. I f U is a Ramsey ultrafilter (in 91)and if B is an atomless measure algebra, then in 91[G] there is no Ramsey ultrajlter extending U .

Proof: Let m be a measure on B. We shall use U and m to define a finitely additive measure on w in W[G] in very much the same way as we did in Lemma 34.6. If {a,}:=o is a bounded sequence of real numbers and if U is an ultrafilter over w, there is a unique real number a such that for every E > 0, the set { n : la, - a ( < E } is in U . We denote this unique a a = lim a, U

Now let me follow (34.6)-(34.12). If a E B is nonzero and if a let

As before, we further let

IF

A E 6,we

482

6.

OTHER LARGE CARDINALS

and finally All we said about p in the proofof Lemma 34.6 is again true except that p is finitely additive, not K-additive. In particular, if X E 91 is such that X E U , then p ( X ) = 1 . Also, since B is atomless, it can be shown that is atomless. We want to apply Lemma 38.2 and show that there is no Ramsey ultrafilter extending U . For this reason, we shall show that the filter { X : p ( X ) = 1) coincides with the filter generated by U . Thus it suffices to show that if p ( X ) = 0, then there is Y E U such that X n Y = 0. Let us assume, on the contrary, that there is X E 911' such that IIp(X) = 011 = 1, and (38.3)

I (VYE

C)Xn

~#011 = ~ Z O

Let E = m(b).To get a contradiction with (38.3), it suffices to find Y E U such 1 1= E XI1 : n E Y}, it that m(llX n f # 12/11) < E. Since IIX n f # 0 suffices to find Y E U such that

x{IIi

(38.4)

If Ilp(X) = 011 = 1, then we note that by (34.9), pa(A) = 0 for all a E B. Hence

lim llii

E

XI(= 0

0

and (38.4) follows from this lemma:

Lemma 38.4. If U is a Ramsey ultrafilter and is a sequence of nonnegative real numbers such that lim, a,, = 0, then for every E > 0 there is Y E U such a,, < E. that

I,,

ProoJ: For each k = 1, 2, 3, ... let X i = { n :~ / 2 ~ 5 + 'a,, < ~ / 2 ~ }If. {n : a,, = 0) E U , then there is nothing to prove; otherwise, Xk E U , and since lim, a,, = 0, we have xkcj U for all k. Let Y E U be such that I Y n xk I s 1 for all k. Clearly, CnEa,, < E.

urz1

This completes the proof of Theorem 91.

Regularity of Ultrafilters Let U be a nonprincipal ultrafilter over S. We say that U is regular if there exists a family { M x : x E S} of subsets of S such that (38.5) (a) each M , is a finite set; (b) for every a E S, {x : a E M,} E U .

[More precisely, we say that U is (No,ISI)-regular and, in general, call U (K, I)-regular if there exists a family {M,: x E S} of subsets of I such that

38.

483

MORE ON ULTRAFILTERS

I

I M , < K for all x, and {x : y E 'M,) E U for all y < 1.(Compare with (33.24).) Many of the results presented below can be generalized.] Let me recall that an ultrafilter U over S is uniform if I X I = 1 S I for every X E U . It is evident from the definition that every regular ultrafilter is uniform (if X E U and (XI< IS 1, then {M, : x E X } # S ) .

u

Exercise 38.5. U is regular if and only if there exists a collection { X , : a E S} E U such that for any infinite Z c S , X , = 0. [Let x E X , iffa E M,.]

naEZ

It follows that if U is a countably complete ultrafilter, then U is not regular. Furthermore, it is easy to see that every nonprincipal ultrafilter over a countable set is regular.

Lemma 38.5. For every infinite cardinal regular ultrafilter U over S .

K

there exists a set S of size

K

and a

Proof. Let S be the set of all finite subsets of K . For each a E S, let 2 a } . Let U be an ultrafilter such that ii E U for all a E S. For each x E S, let M , = P ( x ) . For every a E S , we have a E M , if and only if x E ii. Hence U is regular.

ii = {x E S : x

Regular ultrafilters have been studied in model theory, in connection with the cardinality problem for ultrapowers. If A is an infinite set, then clearly [ AI I I Ult,(A)I 5 IA llSl for any ultrafilter over S. If U is [ A [+-complete, then I Ult,(A)I = [ A 1, and the next lemma states that if U is regular, then I Ult,(A) I = I A [Is1.There are various results on the size (or order-type) of the ultrapower, and we present some of them below; most of the natural questions about the size of ultrapowers remain open however. I f U is a uniform ultrafilter over S, then I U/r,(S) I > IS I. [There is a family F of functions f: S -+S such that IF1 = IS[+ and that I{x : f ( x ) = g ( x ) }I < IS I whenever f # g E F ; compare with Lemma 24.7.1

Exercise 38.6.

Lemma 38.6. Let U be a regular ultrafilter over S and let A be an infinite set. Then I Ult,(A) I = I A Proof. We may assume that S = K is an infinite cardinal. Let B be the set of all finite sequences in A ; since A is infinite, we have I B I = I A I. We will show that

(38.6)

I A IK I I Ult,(B) I

-=

Let M,,5 < K, be a family of finite subsets of K such that each u K belongs to almost all (mod U ) M,.For each g E ' A , let f = F ( g ) be the following function from K into B: f(5)= (9("0). g(u,), d u r - 1)) . ' . 1

484

6.

OTHER LARGE CARDINALS

where a,, c a , c < a k - , enumerates M , . Now if g # h, then there is a < K such that g(a) # h(a), and for almost all 5, a is in M cand hence F ( g ) and F(h) differ almost everywhere. Thus F gives a one-to-one mapping of “A into Ult,(B), proving (38.6). I One of the results on size of ultrapowers is the following: Exercise 38.7. If U is not countably complete, if A is an infinite set, and if 1 = 1 U l t u ( A ) ( ,then I N 0= 1.

[Let B be the set of all finite sequences in A ; we show that I Ulr A Ixo I Ult B 1. Let {X,: n < w } be a partition of S such that X, U, for all n. Iff = (f, : n < w ) is a sequence of functionsf, : S + A, we let F ( 3 ) be the function h: S + B defined as follows: IfxEX,,leth(x)= (fo(x),fi(x),...,f,( x)).Iff= ( f , : n < w ) a n d i = ( g , : n < w ) a r e such that for some n,f, # g, mod U,then F ( f ) # F ( j ) mod U.]

+

If a is a limit ordinal, then Ult,(a, < ) is a linear ordering without a greatest element, and we may ask what its cofinality is. The following two exercises deal with this problem for Ult, o. Exercise 38.8. If U is not countably complete, then every countable subset of U/tu(w,< ) is bounded. Hence the cofinality of U h w is at least w , . [Let {X, : n < w } be a partition of S such that X, $ U for all n. Iffk, k < w, are functions f k : S -+ w, let J S + w be as follows: If x E X,, let f ( x ) = max(fo(x), . . ., f,(x)) 1. Clearly,f>& mod U,for all k.]

+

If U is a regular ultrafilter over S , then every subset of Ultv(w, < ) of size IS I is bounded. Hence the cofinality of Ulr w is at least IS[+.

Exercise 38.9.

[Let M,,x E S , be a family of finite subsets of S such that each a E S is in almost all M,. Iff., a E S, are functionsf.: S + w, let$ S + w be as follows: If x E S, letf(x) = max{f.(x) : a E M,} + 1. Clearly,f>f, mod U,for all a E S.]

Let us consider now a uniform ultrafilter U over an infinite cardinal K and the ultrapower Ult, K . The constant functions are the functions c , , y < K , such that cy(a)= y for all a < K . A function$ K -+ K is unbounded iff > c, mod U for all y -= K; f is monotone iff ( a ) If (p) whenever a < p. The diagonal function is the function d(a) = a. Lemma 38.7. If U is a regular ultrajilter over K , then there is no least unbounded function in the ultrapower; i.e., i f f K -,K is unbounded, then there is an unbounded g such that g c f mod U . Proof. Let M u ,a < K, be a family of finite subsets of K such that each y E K is in almost all M u ;let f: K -+ K be an unbounded function. Let g be defined as follows: Let g(a) be the greatest element of M uless than f (a),if such an element exists. Sincef is unbounded, g(a) is defined almost everywhere. It is clear that g is unbounded because g < c, mod U would imply f c c y + mod U .

The following lemma plays an important role in the investigation of regularity of ultrafilters:

38. MORE ON

485

ULTRAFILTERS

Lemma 38.8. Let K be a regular cardinal and let U be a nonregular uniform ultrafilter over K . Then the ultrapower has a least unbounded monotone function; i.e., there exists an unbounded monotone$ K + K such that there is no unbounded monotone g with g
Proof: Let U be a uniform ultrafilter over K and let us assume that the ultrapower has no least unbounded monotone function. We shall show that U is regular. We shall construct a K-sequence of unbounded monotone functionsf,, f,,. . .,fy,.. ., y < K, such that if y < 6 < K, thenf,(a) >fa(.) for almost all a < K, andf,(a) 2fa(a) for all a. Then it follows that U is regular, as witnessed by the family { M a : a < K}, where

Clearly, each y is in almost all M,, and each M , is finite.
,

<

Corollary. Let ultrajlter over

K K,

be a regular cardinal. If there exists a nonregular uniform then there exists one that extends the closed unboundedjilter.

Proof: Let U be a uniform nonregular ultrafilter over K and let$ K -+ K be the least unbounded monotone function in U l t , . Let D = f * ( U ) = {X :f-I(x)E U } . D is clearly a uniform ultrafilter over K, and it is easy to see that the diagonal function is the least unbounded monotone function in U l t D .Now if C is a closed unbounded subset of K, then C E D because otherwise consider the function g ( a ) = sup(C n a). The function g is unbounded and monotone, and g(a) < a whenever 0 # a @ C. Hence D extends the closed unbounded filter. (Compare with Exercise 28.12.) It remains to show that D is not regular. Thus assume, on the contrary, that M,, a < K, is a family of finite sets such that for every y < K, the set X, = {a : y E M , ] is in D.We shall show that U is regular. For each a < K, let N , = MI(,). Each N, is finite, and if y < K, then { a : y E N,} =f- I(X,) is in

u.

As for existence of nonregular ultrafilters, it is not very difficult to construct one over a measurable cardinal. The problem is to show that a nonregular ultrafilter can exist over a small cardinal. The best result to date in this direction is a model with a nonregular ultrafilter over 02.

6.

486

OTHER LARGE CARDINALS

Nonregular Ultrafilters Over w1 We shall now turn our attention to this particular problem: Can there be a nonregular uniform ultrafilter over w , ? Although the question is open, some consequences of existence of such ultrafilters are known; most significant is Theorem 92 below, which shows that existence of a nonregular ultrafilter over w 1 implies that 0’ exists. We start with a combinatorial lemma. Let us recall that two functionsf and g on w1 are almost disjoint if { a :f ( a ) = g(a)} is at most countable.

Lemma 38.9. Let U be a uniform ultrafilter over w 1. Each of the following conditions is suSJicientfor U t o be regular:

(a) There exist K1 almost disjoint functions fs, 5 < w , , from w1 into w, and a function f: w + w such that f,
,

r

Proof. (a) For each a < w , , let

Ma = {Y :&(a)
for all 5 < Y}

Since for each a, f (a) is a natural number, it is easy to see that each M a is finite. Also, every y belongs to almost all M a because the set {a :&(a) = f d a ) for some 5 < y} is countable and thus @ U . (b) Let 9 be a family of K, almost disjoint functions from w 1 to w. Distinct functions in 9 represent distinct ordinals in the ultrapower Ult,; these ordinals thus form a linearly ordered set of size K, . Therefore there existsf E 9 such that for uncountably many g E 9 we have g
r

,

,.

38. MORE ON ULTRAFILTERS

487

Note that (a) implies that the cofinality of Ult,(w, <) is w1 if U is a nonregular uniform ultrafilter'over wl. Note also that by (b), if there exists a Kurepa tree, then every uniform ultrafilter over w is regular (and hence if there is a nonregular uniform ultrafilter over w , , then K 2 is an inaccessible cardinal in L). Corollary. Assume that there exists a nonregular uniform ultrafilter over w Then :

,.

(a) if2"O = K,,then 2"l = K2; (b) ifN1 < 2"O < K , , , then 2"l = 2"O.

,

Proof: (a) We recall that there exist 2"' almost disjoint functions from w into 2N0.If 2N0= K, and 2" > K2,then we have K 3 almost disjoint functions from w 1 into w1 and hence U is regular. (b) Let us assume that K , < 2"" < K,, and that 2"O < 2"'. Then 2"' 2 K3 and by Lemma 35.2 there exists a family of K3 almost disjoint functions from w 1 into w l , and hence U is regular. W In the proof of Lemma 38.9 we used almost disjointness of the functions ft only to make sure that ( l ) f t # fqmod U if 5 # q, and (2) {a :f,(a) =,fs(a)for some < y} 4 U (if y is countable). Thus the lemma can be generalized:

Lemma 38.10. Let D be an ultrajilter over w 1 extending the closed unbounded filter. Throughout ( a t ( d ) of Lemma 38.9, replace "almost disjoint" by " { a : f t ( a )=&(a)} is thin." Then each ofthe modijied conditions (a)-(d) implies that D is regular. W Corollary. Assume that there exists a nonregular uniform ultrafilter over w lf K,, is a strong limit limit cardinal, then 2"oi < Ku2. Proof: By Theorem 24 (see Lemma 8.7), 2"wl < No where 0 is the norm of the constant function c,, on w 1 (or, in other words, 8 is the length of the wellfounded relationfx g ( J g : w1 + wl) defined b y f < g iff { a : f ( a )< g(a)}contains a closed unbounded set). Thus it suffices to show that there is no function f: w 1 -+ w 1 such that llfll = w , . By exercise 35.13, if there isf: w1 + w 1 that has norm w 2 , then there exist K, functionsf,: w1 + w such that any two differ on a closed unbounded set. By Lemma 38.10b, any ultrafilter over w 1 extending the closed unbounded filter must be regular. Hence by the corollary of Lemma 38.8, every uniform ultrafilter over w I is regular. W The following lemma is not necessary for the forthcoming proof of Theorem 92, but shows that a nonregular ultrafilter over w1 can be "normalized ":

Lemma 38.11. Let U be a uniform nonregular ultrafilter over w 1 and let f: w 1 + w I be an unbounded function in Ult,. Then there exists a monotone unbounded function h such that h I f mod U .

488

6.

OTHER LARGE CARDINALS

Proof. For every p < ul,let P , = { a : f ( a ) = p}; for every Po < u1we have { P , : p < Po} 4 U . We claim that there exists a function g : u1+ u1such that g( (a)) 2 a for almost all a. To prove the claim, let {gs : .= u2} be a family of almost disjoint functions let fs= ge; of. We note from o,into u,(see Lemma 24.7). For each < u 2 , thatfg #f,mod V whenever # q, and that if y < u 2and S G u 2is countable such that y # S, then

uf

<

r

(38.7)

{ a :f,(a) =&(a) for some

<

<

r

E

S} $ U

Let us assume that for all < 02, ft < d mod V (where d is the diagonal function). Now the property (38.7) is sufficient for the arguments in Lemma 38.9(c), (b), and (a) (see the remark preceding Lemma 38.10); and hence the existence of functions f t , 5 < u t ,with property (38.7) and bounded by d implies that U is regular. Hence there is { such that g t f 2 d mod U , and the claim is proved. Now we let A = { a : g ( f ( a ) ) 2 a } ; we have A E V . We shall show that for every p < u,,the set A n P , is at most countable. Iff(.) = 0 and a E A, then g ( p ) 2 a ; hence A n P, c g ( p ) . Thus the function f ( A has the property that each value p is taken at most countably many times. It is now easy to construct an unbounded monotone function h: u,+ u1 such that h(a) Sf( a ) for all aEA. H 0

,,

Corollary. If U is a uniform nonregular ultrajlter over u then the ultrapower has a least unbounded function. Consequenrly, i f there is a nonregular uniform then there is one that is normal in the following sense: Every ultrajlter over u,, regressive function is bounded in the ultrapower by some constant function c , , y < a,. Proof. By Lemma 38.8, there is a least monotone unbounded functionf; and by Lemma 38.1 1, f is the least unbounded function. Now D = f,( U ) is a nonregular ultrafilter and has the property that the diagonal is the least unbounded function.

Theorem 92 (Ketonen). Ifthere is a uniform nonregular ultrafilter o v e r u , , then 0" exists. Proof. Let us assume that there exists a uniform nonregular ultrafilter over u,. By the corollary of Lemma 38.8, there exists a nonregular ultrafilter over u , that extends the closed unbounded filter. Let D be such ultrafilter. To show that 0" exists it suffices to produce a nontrivial elementary embedding of L. We shall show that there is a nonprincipal Lultrafilter U over a countable ordinal a such that Vlr,(L) is well-founded; then the natural embeddingj, is a nontrivial embedding of L into L. We start the proof by constructing functions f,:u1+ ul,for K, ordinals y. Let r be the set of all limit ordinals y such that w , < y < u 2and that L, is an

38.

489

MORE ON ULTRAFILTERS

elementary submodel of (Lu2,E). Let y E r. We consider a relation E , on w , such that the model ( w l r E , ) is isomorphic to ( L , , E). Let K,: L, + w 1 be the unique isomorphism. Let C , be the set of all a < w 1 such that:

(38.8)

(i) (a, E , ) is an elementary submodel of (a,,E,) and n Y ( w lE) a ; (ii) for all < a, x,(<) E a ; (iii) for all < wl, if a 5 <, then K,(<) r$ a.

< <

,.

It is rather easy to verify that C , is a closed unbounded subset ofw Moreover, if a E C,, then (a, E , ) is isomorphic to some (unique) L,; letf,(a) be such that (a, E , ) is isomorphic to (LJY(=,, E ) , and let p;: L,y(a) --+ a be the unique isomorphism. Let j: be the elementary embedding

<

defined byj:(x) = K; ‘(~”yx)).If < a, then (38.8) (ii) implies that J ( < )= <. By (iii),j;(a) 2 w l ; and since nY(wl)is in (a, E,), we have];@) = wl. Thus a is the least ordinal moved by j; , and jy(a) = w , . We have defined, for each y E r, a functionf, from C , into w l . Lemma 38.12. If y, 6 E r and y < 6, there exists a closed unbounded set C,,, such rhatf,(a) <&(.)for all a E C,, .

Proof. Let y < 6 E r. Let i : w , 4 w1 be the embedding of (a,,EY) into (wlr E,) corresponding to the inclusion of Ly in L, (see the diagram below):

We let

(38.9)

C,, = {a E C , n C, : n,(Ly) E a and (V<

E

a)i(<)E a)

Clearly, C,, is closed unbounded. Let a E C,,; we shall show thatf,(a)
490

6. OTHER LARGE CARDINALS

subset of E,-order-type

h(.) 'P =fy(a). w

P. So

we let z = (p:)-'n,(L,). Now it follows that

Thusf,, y E r, are K2 functions from o1into w 1 such that any two differ on a closed unbounded set. Since D is a nonregular ultrafilter extending the closed unbounded filter, it is necessary, by Lemma 38.10~that for every function f : o1+ o1there is y E r such thatf
Proof. Assume that wI is not inaccessible in L. Then there is a. such that for all < olra is not a cardinal in L. For each a 2 cto, let f ( a ) be a countable ordinal such that L,,,, k a is not a cardinal. There exists y E r such that f < f, mod D. Hence for almost all a(mod D), LfY,,)k a is not a cardinal. However, for almost all a there is an elementary embedding :J of Lfv,a)into L, such that $(a) = ol, and L, is an elementary submodel of L,, . This is a contradiction because L,, k o1is a cardinal. a such that a. I a

Now consider the functionf(a) = (a')L; since w1 is inaccessible in L, f is a function from w1 into wl. Thus there exists yo E r such thatf (a')". If y E I' is such that y > y o , then for every a E A n C,,,, f,(a) > (a')"; hence LJ7(=) contains all constructible subsets of a. Therefore, the elementary + L, defines an Lultrafilter U ; over a : embedding]:: LJY(.)

iff

L)

(38.10)

x E U;

Lemma 38.14.

If y < 6 are in r and yo < y, then for almost all a, U ; = U: .

a

(x za,

EJ;(X)

x

E

Proof. Let a E A n C,,, n C,,, n Cya;we shall show that U ; = U : . Let us use again the diagram in Fig. 38.1. Let x E a be in L. We have x

E

U;

iff

a E];(x),

x

E

U:

iff

aE~:(x)

Thus it suffices to show that j ; ( x ) = j i ( x ) for all x E a, x E L. We follow the diagram in Fig. 38.1 from the upper left corner to the lower left corner. Let x E a, x E L and let z = j",x). We denote u = p",x) = n,(z),and ti = i ( u ) = x&). By (38.9), we have u E a, and therefore u = p:(y) for some Hence z = j:(y). y E Now both j ; and j: have the property that a is the least ordinal moved. Therefore j",x) n a = x , and A ( y ) n a = y. Thus we have y = z n a = x , and hence j:(x) = j : ( x ) . We are now ready to finish the proof of Theorem 92. It suffices to show that for some y and a for which U! is defined, the ultrapower Ult,:(L) is wellfounded. Thus let us assume that the ultrapower is not well-founded for any y and a.

38. MORE

ON ULTRAFILTERS

491

Let y > yo be a fixed ordinal in r. For each a E A n C,,, there is a sequence of constructible functions go, 'gl, ..., g,,, ... on a such that for each n, {< < a : g,,(<) > gnt 1(<)} E U ; . As in the proofof Lemma 32.12, we observe that if there are such functions, then there are such functions with values < ulrand hence we fix for each a a descending (mod U;) sequence {gi: n < w } of functions from a into wI. For each a there exists a countable ordinal /? such that g: E 4 ,for every n. Letf(a)denote such /?.The functionfis defined on A n C,,, with values y such that f < f a mod D. Hence for almost all a, g; E for all n. By Lemma 38.14, U ; = U ; for almost all a ; and therefore there exists a such that U: is defined, that g: E Ld(a) for all a, and that g: > g:+ mod U: for all n. Now we are reaching a contradiction. For each n, gt is in Lf,(,, and hence J(g:) is defined, and moreover, gt > git mod U ; means that This gives a descending sequence of ordinals

This Page Intentionally Left Blank

PART IV

Sets of R e d s

CHAPTER 7

DESCRIPTIVE SET THEORY

39. BOREL AND ANALYTIC SETS

Descriptive set theory deals with sets of reals that are described in some simple way: sets that have a simple topological structure (e.g., continuous images of closed sets) or are definable in a simple way. The main theme is that questions that are difficult to answer if asked for arbitrary sets of reals, become much easier when asked for sets that have a simple description. An example of that is the Cantor-Bendixson theorem (Theorem 14): Every closed set of reals is either at most countable or has size 2‘O; compare this with the independence of the continuum hypothesis. Since properties of definable sets can usually be established effectively, without use of the axiom of choice, we shall work in set theory without the axiom choice. When some statement depends on the axiom of choice, I shall explicitly say so. However, we shall assume a weak form of the axiom of choice. The reason is that in descriptive set theory one frequently considers unions and intersections of countably many sets of reals, and we shall often use facts like “the union of countably many countable sets is countable.” Thus we shall work, throughout this chapter, in set theory ZF + DC, where DC is the principle of dependent choices (cf. (5.4)): (39.1) If p is a binary relation over a set A and if for every a E A there exists b E A such that b p a, then there is a sequence ao, al, . .., a,, .. . in A such that a,, p a, for all n E N.

Recall some consequences of the principle of dependent choices: (39.2) (a) The countable axiom of choice (Exercise 5.7). (b) Every infinite set has a countable subset (Exercise 5.4). (c) The union of countably many countable sets is countable. (d) w1 is a regular cardinal. 493

7.

494

DESCRIPTIVE SET THEORY

(e) A linear ordering without infinite descending chains is a wellordering, and a binary relation without infinite descending chains is well-founded (Lemma 5.2). Most of the arguments in descriptive set theory apply not only to subsets of R (the real line), but to a number of other topological spaces with a similar structure; the n-dimensional Euclidean space R", the closed unit interval [0, 13, the Cantor space C,or the Baire space J1' = ". A topological space X is a Polish space if it is homeomorphic to a complete separable metric space. [A set D G X is dense if it intersects every nonempty open set; 3 is separable if it has a countable dense subset. If d(x, y ) is a metric on X,then {a,,}:=o is a Cauchy sequence if lim,,,+m d(a,, a,) = 0; a metric space X is complete if every Cauchy sequence in 3 converges to some point a E X.] The spaces mentioned above are Polish spaces. For instance, the Baire space Y can be endowed with the following metric: If a, b E NNand a # b, let (39.3)

d(a, b ) =

1 2" ~

where n is the least n such that a(.) # b(n)

As a matter of fact, the Baire space is somewhat easier to work with than the real line, and we shall therefore, starting in the next section, work almost exclusively with sets in the Baire space. Exercise 39.1. For each n, the product space Y" is homeomorphic to Y . The product t' is homeomorphic to Y .

space

Note that the Baire category theorem (Theorem 13) holds for any Polish space 3:Replace the rational intervals by a countable basis for the topology and use completeness of the metric (and the countable axiom of choice). The facts on compact sets given in Exercise 4.5 hold for any topological space. [A compacr set is closed set C with the property that whenever C is covered by a collection of open sets, C C _ I G i , then there are finitely many i l , . . . , ik E I such that C E Gi, u u Gir.]

uiE

Exercise 39.2. A closed set F E 4- is compact if and only if the tree TFhas the following property: Every node s E TFhas only finitely many immediate successors snnl, . . . , snnk in TF.

Bore1 Sets A a-algebra of sets is an algebra of sets (see Section 17) closed under countable unions and intersections: .a E P ( S ) is a o-algebra (of subsets of S) if:

(39.4)

(i) S E .d; (ii) if X E 49, then S - X E J ; (iii) if X , E .d,n = 0, 1, . . ., then

ut=oX , E .d and n:=oX, E .&.

39.

BOREL AND ANALYTIC SETS

4%

Let 9 be a Polish space. A set A G 3 is a Bore1 set if it belongs to the smallest a-algebra of subsets of S containing all closed sets. Let me first give a more explicit description of Borel sets. For each a < ol, let us define the collections Cp and n,"of subsets of S:

Xy = the collection of all open sets

(39.5)

l l y = the collection of all closed sets

u W

Z," = the collection of all sets A =

A,, where each A,

n=O

belongs to

n:

for some fi < a

n,"= the collection of all complements of sets in Z:

n A,, where each A, m

= the

collection of all sets A =

n=O

belongs to Z: for some fl < a It is clear (by induction on a ) that the elements of each Et and each ng are Borel sets. Since every open set is the union of countably many closed sets, we have Ey E E; , and consequently if a < p ,

then Z , " E E ~ ,E , " E H ~ , H,"c_n,", l l , " ~ Z ;

Hence

u z,"= u n,"

(39.6)

a
Z<W,

and it is easy to verify that the collection (39.6) is a a-algebra (here we use the countable axiom of choice). Hence every Borel set is in some E :, a c 0,. Note that each E,"(and each n,")is closed under finite unions, finite intersections, and inverse images by continuous functions (i.e., if A E Et in 9,then 1-, ( A ) E E: in X whenever$ T! + ?2( is a continuous function). The sets in 72; are customarily called F,-sets; the elements of Il: are the G,-se t s. Exercise 39.3. Let f be a real-valued function on R. Show that the set of all points at whichfis continuous is a G,-set.

If the Polish space .T is countable, then ofcourse every A E 3 is a Borel set, in fact an F,-set. Uncountable Polish spaces are however more interesting: Not all sets are Borel, and the collections Z: form a hierarchy. We shall show that for each a, E: $ n,",and hence

zy

c

E; c ... c g

(where c denotes proper inclusion).

c

...

4%

7.

DESCRIPTIVE SET THEORY

The following lemma holds for any uncountable Polish space; I shall however prove it only for the Baire space .Y.

Lemma 39.1. For each a 2 1 there exists a set U G . Y z such that U is C8 (in E,"set A in . Y there exists a E . Y such that

. Y'),and that for every

A = {x : ( x , a ) E U }

(39.7)

(U is a universal C,"set.) See Fig. 39.1.

a I I I

I I

I

A

FIGURE 39.1

Proof: By induction on a. To construct a universal open set in . V"', let G . . ., G , , . . . be an enumeration of all basic open sets in . I/, and let Go = 0. Let

(39.8)

( x , y) E U

iff

u."=o u."=o

x E G,,,,

for some n

Since U = H, where each H, = {(x, y ) : x E G,,,,} is an open set in . V.', we see that U is open. Now if G is an open set in. +<, we let a E NNbe such that G= G,,,,;then G = {x : ( x , a ) E U}. Next let U be a universal C,"set, and let us construct a universal C;+ set V . Let us consider some continuous mapping of. Y onto the product space Y V ; for each a E . Y and each n, let a,,, be the nth coordinate of the image of a. [For instance, let us define a,,) as follows: a,,,(k) = a ( r ( n , k)), where r is the canonical one-to-one correspondence r: N x N -+ N.]Now let (39.9) ( x , y) E V iff for some n, ( x , y,,,) # U Since V = H, where each H, = { ( x , y ) : (x, y,,,) q! U } is a flf-set, we see that V is Z,"+ If A is a Z;8+ set in ,Af, then A = A , where each A, is fit. For each n, let a , be such that ./lr - A, = { x : ( x , a,) E U } ,and let a be such that a,,, = a, for all n. Then A = { x : (x, a ) E V } . Finally, let a be a limit ordinal, and let U,, 1 5 p < a, be universal Cog-sets. Let 1 I a,, < a < * * . < a, < . . . be an increasing sequence of ordinals such that a, = a. We notice that if A is a E,"-set, then there exist sets A,, n < o, such that A = uF=oA, and that A, E Z,""for all n. Thus if we let

u."=o

(39.10)

(x,y ) E U

u:=o

iff

for Some n, ( x , y,,,) # U,,

the same argument as above shows that U is a universal Z," set. H

39.

BOREL AND ANALYTIC SETS

497

Corollary. For every a 2 1, there is a set A E Y- that is 72: but not I

ProoJ: Let U

G . .tm2be

n:.

a universal 72:-set. Let us consider the set

(39.11)

A = {x : (x, x)

E

U}

Clearly, A is a E;-set. If A were ng, then its complement would be Zt and there would be some a such that A = {x : (x, a ) # U }

But this contradicts (39.11): Simply let x = a. W While the collection of Borel sets of reals is closed under Boolean operations, and countable unions and intersections, it is not closed under continuous images: As we shall learn presently, the image of a Borel set by a continuous function need not be a Borel set. We shall now investigate the continuous images of Borel sets. A subset of A of a Polish space S is called analytic if there exists a continuous functionf . C’ + T such that A = f [. “1. The projection of a set S c S x b (into 5)is the set (39.12)

P

= {x E

x:3y (x, y ) E S }

See Fig. 39.2.

9-

P FIGURE 39.2

Lemma 39.2. The following are equivalent,for any set A in a Polish space S:

(a) A (b) A (c) A (d) A

is analytic. is the continuous image of a Borel set B (in some Polish space g). is the projection of a Borel set in S x b,for some Y. is the projection of a closed set in S x . V.

Proof: We shall prove that: (1) every closed set (in any Polish space) is analytic; and (2) every Borel set is the projection of a closed set in 5? x .h‘.

Then the lemma follows: Since the projection map II: X x tY + 5? defined by A ( X , y) = x is continuous, it follows that every Borel set is analytic and that the

498

7.

DESCRIPTIVE SET THEORY

continuous image of a Borel set is analytic. Conversely, if A c X is an analytic set, A = f [ $"I, then A is the projection of the set { ( f ( x ) *4 : x

E

Y}

which is a closed set in S x 4'. In order to prove (l), note that every closed set in a Polish space is itself a Polish space, and thus it sufficesto show that every Polish space is a continuous image of the Baire space Y. Let 3 be a complete separable metric space; we construct a mappingfof VP onto S as follows: It is easy to construct, by induction on the length ofs E Seq, a collection {C,: s E Seq} (where Seq = uF=o "N) of closed balls such that C, = F and (39.13)

(i) diameter (C,) I l/n where n = length(s); (ii) C , E Ur=oC,, (all s E Seq).

n

For each a E . Y', let f(a) be the unique point in {C, : s c a } ; it is easily checked thatfis continuous and that X =f[ Y ] . In order to prove that every Borel set in 3 is the projection of a closed set in S x Y , it suffices to show that the family P of all subsets of .Xthat are such projections contains all closed sets, all open sets, and is closed under countable unions and intersections. Clearly, the family P contains all closed sets. Moreover, every open set is a countable union of closed sets; thus it suffices to show that P is closed under U."=oand Recall the continuous mapping a - (a,,, : n E N) of 1.' onto 1 . from Lemma 39.1, and also recall that the inverse image of a closed set under a continuous function is closed. Let A,, n < o,be projections of closed sets in f x Y'; we shall show that u2=o A, and A, are projections of closed sets. For each n, let F, c 3 x 1." be a closed set such that

n:=o.

n;=o

A, = {X : h (x, a ) E F,}

Thus

u UI

xE

A,

iff

3n 3a (x, a ) E F ,

iff

h 3b (x, a ) E F,,,,

iff

3c

n=O

and

(x9

C ( 0 ) ) E FC,,,(O)

00

x

A,

E

iff

V n h (x, U ) E F,

iff

3c V n (x, q,))E F ,

n=O

00

iff

IC (x, C) E

n{(x,

n=O

C)

: (x, c(,))

E

F,)

39.

BOREL A N D ANALYTIC SETS

Hence and

Ul=, A,

499

is the projection of the closed set

n;=oA, is the projection of an intersection of closed sets. The Suslin Operation .d

First some notation: We recall that Seq is the set of all finite sequences of "N.If a E NN and n E N, let a(n) denote the finite integers: Seq = sequence

uz==o

a(n)= (a(O), . . . , a(n - 1 ) )

(39.14)

(hence a(n)= a l n E Seq). For each s E Seq, U s is the basic open set {a E . 1 ' : a(n) = s} of 1', where n = length(s); U sis also closed. A denotes the closure of A G 2". Let { A , : s E Seq} be a collection of sets indexed by elements of Seq. We define

(39.15)

.d{A, : s

E

Seq} =

u

LIE

Exercise 39.4. The operations

Note that if {B, : s

U 0

E

CX

A,(,,

NN n=O

u.'=oand n:=oare special cases of the operation .d

Seq} is arbitrary, then

Bticn)

=

U

(B,,o, n

n

... n B,(n,)

a € NN n = O

a E N N n=O

and hence .d{B,: s E Seq} = ..'{A, : s E Seq} where the sets A, are finite intersections of the sets B, and satisfy the following normality condition:

if s

(39.16)

G

t,

then

A, 2 A,

Thus we shall restrict our use of .a' to families that satisfy the condition (39.16).The operation &' is called the Suslin operation. Lemma 39.3. A set A G .'ris analytic ifand only i f A i s the result of the operation .d applied to a family of closed sets. Proof: First we show that if F,, s E Seq, are closed sets in 3, then A = .d{F,: s E Seq} is analytic. We have

n F,,,) m

x

E

A

iff

h E . Y'

xE

n=O m

iff

h (x, a ) E

B, n=O

where B, = {(x, a) : x hence A is analytic.

E

F,,,,}. Clearly, each €3, is a Bore1 set in 3 x .4"and

500

7.

DESCRIPTIVE SET THEORY

Conversely. let A C _ ,Tbe analytic. There is a continuous function$ . 1 . + .f such that A =f[."1. Notice that for every a E . i P , cu

n f [ U i d n f[Uci(n,I

(39.17)

00-

=

n=O

n=O

= If(a)l

(this is easy to verify; here 8 denotes the closure of a set X).Thus A

=f[.~1 = U

n f[Ui,nJ

a € . + n=O

and hence A is the result of the operation .$ applied to the closed sets f [ U J (which satisfy the normality condition (39.16)). W Exercise 39.5. Let A,, s E Seq, be Borel sets satisfying (39.16) and the additional condition: For each s E Seq and all n # rn. A,, n A,, = 0.Then . d { A , : s E Seq} is a Borel set.

[u

f)Ad,",= f)U {A, : length(s) =

a € .t n=O

n=O

It follows from the preceding lemmas that the collection of all analytic sets in a Polish space is closed under countable unions and intersections, continuous images, and inverse images, and the Suslin operation (the last statement is proved just like the first part of Lemma 39.3). I t is however not the case that the complement of an analytic set is analytic (if 3 is an uncountable Polish space). In the next lemma, we establish exactly that; we show that there exists an analytic set (in .)'6 whose complement is not analytic. The Hierarchy of Projective Sets

For each n 2 1, we define the collections Xi, Hi,and A: of subsets of a Polish space 9 as follows: (39.18)

E: = the collection of all analytic sets

Xi+

= the

complements of analytic sets

= the

collection of the projections of all sets in T x .V

ll:

= the

complements of the Xi sets in 3

A: = Zi n lli

The sets belonging to one of the collections EA or lli are called projective sets. It is easily seen that the following inclusions hold:

39.

50I

BOREL A N D ANALYTIC SETS

FIGURE 39.3

We shall show that for each n there is a Z,' set i n . Y that is not I l i ; thus the inclusions in Fig. 39.3 are proper inclusions.

Lemma 39.4, For each n 2 1, there exists a uniuersal Z,' set in . Y 2 ; i.e., a set U c . 1. 2 such that U is and that for every Z i set A in . Y there exists v E .4;" such that A = {x: (x, u ) E U ) Proof. Let h be a homeomorphism o f . t' x .1." onto . V. If n = 1, let V be a universal Zy set; if n > 1, let V be, by the induction hypothesis, a universal Zip set. Let (x,y) E U

(39.19)

h

iff

E . 9 " (h(x,a), y ) $ V

Since the set { ( x ,y, a ) ; (h(x,a), y ) $ V } is closed (if n = 1) or ni- (if n > l), U is Xi. If A c _ , 1 is Z,', there is a closed (or H,'- set B such that '

(39.20)

X E

iff

A

3 a E .V ( x , a ) E B

If we let C = h[.V 2 - B], we have (39.21)

(x,a) E B

iff

h(x, a ) $ C

The set C is open (or Z,'- ,) in . Y and since V is universal, there exists u such that c = {u : (u, u ) E V }

By (39.21), we have (x, a ) E B

iff

(h(x,Y).

0)

$ V

then (39.20)gives x

E

A

iff

31E .N(h(x,y ) , u ) $ V

and by (39.19), A = (x : (x, u ) E U ) Hence U is a universal Z; set.

Corollary. For each n 2 1, there is a set A E .Y that is C,' but not

njG

502

7.

DESCRIPTIVE SET THEORY

Proof: Let U G . Y 2be a universal Zi set and let A = {x : (x, x)

E

U}

rn

The collection of all A i sets in a Polish space is a a-algebra and contains all Borel sets. It turns out that A: is exactly the collection of all Borel sets. Theorem 93 (Suslin). Every analytic set whose complement is also analytic is a

Borel set. Thus A; is the collection of all Borel sets.

Let S be a Polish space and let A and B be two disjoint analytic sets in X. We say that A and B are separable by a Borel set if there exists a Borel set D such that A E D and B G X - D. Lemma 39.5. Any two disjoint analytic sets are separable by a Borel set.

This lemma is often called " the Z:-separation principle." It clearly implies Suslin's theorem since if A is an analytic set such that B = S - A is also analytic, A and B are separable by a Borel set D and we clearly have D = A. Proof: First we make the following observation: If A = u F = oA, and B= Emare such that for all n and m, A, and B, are separable, then A and B are separable. This is proved as follows: For each n and each m, let D,, be a Borel set such that A, E D, G X - Em.Then A and B are separable by D,,,. the Borel set D = Let A and B be two disjoint analytic sets in A?. Letfand g be continuous functions such that A =f[.Y] and B = g[.Y]. For each s E Seq, let A, =f[LrJ and B, = g[U,]; the sets A, and B , are all analytic sets. For each s we have B,",,,. If a E ", then A, = ASnnand B, =

urn"=,

u;=o n:=o

u:=o

m

OD

( f ( 4=) nf7 f[Uci(n)I = n Ati(n) =O n=O and similarly for the sets B,. Let a, b E NN be arbitrary. Since f[.4-3 and g[.V ] are disjoint, we have f ( a ) # g(b). Let G, and G , be two disjoint open neighborhoods off(a) and g(b), respectively. By the continuity offand g there exists n E Nsuch that A,(,,) G G, and Been) c G, . It follows that the sets A,(,, and B6(,)are separable by a Borel set. We shall now show, by contradiction, that the sets A and Bare separable by a Borel set. If A and B are not separable, then because A = A,,, and B= A(,,,). there exist no and mo such that the sets A(no>and B(,o> are not separable. Then similarly there exist nl and ml such that the sets A(noll) and B ~ m o m are l ) not separable, and so on. In other words, there exist a = (no n , n2 ..*) and b = (mo m , m2 . ..) such that for every k, the sets A(no...n ~ and ) B(,o ... ,,) are not separable. This is a contradiction since in the preceding paragraph we proved exactly the opposite: There is k such that &(k) and & ( k ) are separable. rn

u:=o

u:=o

39.

503

BOREL AND ANALYTIC SETS

Exercise 39.6. Let A,, n = 0, 1, 2, .. . , be pairwise disjoint analytic sets. Then there exist pairwise disjoint Borel sets D, such that A, c D, for all n. [Modify the proof of Lemma 39.5.1 Exercise 39.7. If B E

(. '

is a Borel set and ifjis a one-to-one continuous function, then

A =f[B]is a Borel set. [For each s E Seq, let B, = B n U sand A, =f[B,].The sets A, are analytic and for each n, the sets A, such that length(s) = n are pairwise disjoint. Let D,be Borel sets such that for each n, {D,: length(s) = n} are pairwise disjoint and such that A, E D, E A,. By (39.17), A = d { A , : s E Seq} = .d{A,: s E Seq}. and hence A = . d { D , : s E Seq}. By Exercise 39.5, A is Borel.]

Measure and Category

We shall now turn our attention to two properties that are often used as a measure of good behavior of sets of real numbers: Lebesgue measurability and the property of Baire. Our goal is to show that all analytic sets of reals are Lebesgue measurable and have the Baire property. Since most readers are probably more familiar with Lebesgue measurability, let me first give a review of the theory of Lebesgue measure (in the n-dimensional Euclidean space R"). The Lebesgue Measure

The standard way of defining Lebesgue measure is to define first the outer measure p * ( X ) of a set X E R" as the infimum of all possible sums c { o ( I k ) : k E N) where ( 1 , : k E N) is a collection of n-dimensional intervals such that X c I,, and u ( I ) denotes the volume of I . For each X , p * ( X ) 2 0 and possibly = 00. A set X is null if p * ( X ) = 0. A set A G R" is Lebesgue measurable if for each X G R",

uF=o

p * ( X )= p*(X n A ) + p*(X

- A)

For a measurable set A, we write p ( A ) instead of p * ( A ) and call p ( A ) the Lebesgue measure of A. The standard development of the theory of Lebesgue measure gives the following facts: Every interval is Lebesgue measurable, and its measure is equal to its volume. The Lebesgue measurable sets form a o-algebra; hence every Borel set is measurable. p is o-additive: If A,, n < o,are pairwise disjoint and measurable, then

504

7. DESCRIPTIVE SET THEORY

u;=,,

(d) p is a-finite: If A is measurable, then there exist measurable sets A,, n c o,such that A = A,, and p(A,) -= co for each n. (e) Every null set is measurable. The null sets form a a-ideal and contain all singletons. From the other basic facts on Lebesgue measure let me mention these two:

(f) If A is measurable, then p ( A ) = sup{p(K) : K E A is compact}

and

(g) If A is measurable, then there is an F,-set F and a Gsset G such that F E A G G and that G - F is null.

This last property gives this characterization of Lebesgue measurable sets: A set A E R" is measurable if and only if there is a Borel set B such that the symmetric difference A A B = ( A - B ) u (B - A ) is null. One consequence of this is that if we denote by .Bthe a-algebra of Borel sets and by A the a-algebra of measurable sets, and if I, is the ideal of all null sets, then B/IM= . M / l , . The Boolean algebra A?/l, is a-complete; and since a familiar argument shows that I, is (as an ideal in . X ) a-saturated, we conclude that B/I, is a complete Boolean algebra. We shall return to it later. Exercise 39.8. If X is a null set and a. 2 a, 2 ... 2 a, 2 ... is a sequence of positive numbers with lim a, = 0, then there exists a sequence G,, n = 0, 1, . .. ,of finite unions of open intervals such that X G u2=o G, and p(G,) < a, for each n. Moreover, the intervals can be required to have rational endpoints. [First find an open G a X of measure I a. and express G as a disjoint union of intervals.]

Assuming the axiom of choice one can show that there exists a set of reals that is not Lebesgue measurable. One such example is the Vitali set in Exercise 27.1. As another example there exists a set X G R" such that neither X nor its complement has a perfect subset (see Exercise 5.1 for a construction of such a set). The set X is not measurable: Otherwise, e.g., p ( X ) > 0 and by (39.22f) there is a closed K c X such that p ( K ) > 0; thus K is uncountable and hence contains a perfect subset, a contradiction. However, we shall show in Section 42 that it is consistent (with ZF + DC) that all sets of reals are Lebesgue measurable. I shall conclude this review of Lebesgue measurability with two lemmas. One is the well-known Fubini theorem, and I state it here, without proof, for the sake of completeness (we shall actually use it in Section 41). The other lemma will be used in the proof of Theorem 94. If A is a subset of the plane R2 and x E R, let A , denote the set {Y : (x, Y ) E

4.

39. BOREL A N D

505

ANALYTIC SETS

Lemma 39.6. Lat A c R2 be a measurable set. Then A is null ifand only iffor almost all x, A, is null (i.e., the set {x : A, is not null} is null). Lemma 39.7. For any set X E R" there exists a measurable set A 2 X with the property that whenever Z E A - X is measurable, then Z is null. Proof: If p * ( X ) < co, then because p * ( X ) = inf{p(A): A is measurable and A 2 X},there is a measurable A 2 X such that p(A) = p * ( X ) ;clearly such an A will do. If p * ( X ) = co, there exist pairwise disjoint X, such that X= X, and that for each n, p * ( X , ) < 00. Let A, 2 X , , n < o,be measurable sets such that p(A,) = p * ( X , ) , and let A = A,.

u."=o

u;=o

It should be mentioned that the main results of descriptive set theory on Lebesgue measure (e.g., the forthcoming Theorem 94) can be proved in a more general context, namely for reasonable a-additive measures in Polish spaces. An example of such a measure is the product measure in the Cantor space "{O, 1) (see (34.3) for details).

The Property of Baire In Section 14, we have proved the Baire category theorem (Theorem 13): The intersection of countably many dense open sets of reals is nonempty. It is fairly easy to see that the proof works not only for the real line R but for any Polish space. Let us work in an arbitrary Polish space T. Let us call a set A E A? rare (or nowhere dense) if the complement of A contains a dense open set. Note that A is rare just in case for every nonempty open set G, there is a nonempty open set H E G such that A n H = 125.A set A is rare if and only if its closure A is rare. A set A G X is meager (or offirst category) if A is the union of countably many rare sets. A nonmeager set is called a set of second category. The Baire category theorem states in effect that in a Polish space every nonempty open set is of second category. The meager sets form a a-ideal. Moreover, in case of R", . V , or the Cantor space, every singleton {x} is rare and so the ideal of meager sets contains all countable sets. A set A has the Baire property if there exists an open set G such that A A G is meager. Clearly, every meager set has the Baire property. Note that if G is open, then G - G is rare. Hence if A A G is meager then (T - A ) A (% - G ) = A A G is meager, and it follows that the complement of a set with the Baire property also has the Bake property. It is also easy to see that the union of countably many sets with the Baire property has the Baire property and we have: (39.23) (a) The sets having the Baire property form a a-algebra; hence every Bore1 set has the Baire property.

506

7.

DESCRIPTIVE SET THEORY

Exercise 39.9. For every set A with the Baire property, there is a Gd-setG and an F,-set F such that G E A E F and such that F - G is meager.

[Note that every meager set is included in a meager F,-set.]

If .M denotes again the a-algebra of Borel sets, and if we denote by 9? the a-algebra of sets with the Baire property, and if I is the a-ideal of meager sets, we have .&/I = % / I . Note that the algebra .&/I is a-saturated: Let Y be a countable topology base for 3. For each nonmeager set X with the Baire property there exists G E Y such that G - X is meager. Thus the set D = {[GI : G E 9)of equivalence classes is a dense set in .&/I. Hence .&/I is a-saturated and is a complete Boolean algebra. Exercise 39.10. For every set A with the Baire property, there exists a unique regular open set U such that A A U is meager. The algebra # / I is isomorphic to the algebra of regular open sets (with the operations defined in (17.5)).

As is the case with Lebesgue measurability, the axiom of choice implies that there exists a set of reals without the Baire property. Exercise 39.11. The Vitali set M from Exercise 27.1 does not have the Bake property.

[" Meager" and '' Baire property" are invariant under translation. If M has the Baire property, then there is an interval (a, b) such that (a, b) - M is meager. Then (a, b) n M, is meager for all rational q # 0, hence each M n (a - q. b - q ) is meager, hence M is meager, hence each M , is meager; a contradiction since R = U g E M Q,.]

If X G R" is such that neither X nor its complement has a perfect subset, then X does not have the Baire property: Otherwise, e.g., X is of second category and hence X contains a G,-subset G of second category. Now G is uncountable, and this is a contradiction since as we shall prove in Theorem 94, every uncountable Borel set (even analytic) has a perfect subset. The following two lemmas are analogs of Lemmas 39.6 and 39.7. The first one, although not very difficult to prove, is again stated without proof. Lemma 39.8. Let A E R2 have the property of Baire. Then A is meager only if A, is meager for all x except a meager set.

if and

Lemma 39.9. For any set X in a Polish space .T, there exists a set A 2 X that has the Baire property and such that whenever Z E A - X has the Baire property, then Z is meager. Proof: Let us consider a fixed countable topology basis Y for X.Let X E 3. Let

D ( X )= {x E I for : every U

E

9 such that x E U , U n X is not meager)

Note that the complement of D ( X ) is the union of open sets and hence open; thus D ( X ) is closed.

39.

507

BOREL AND ANALYTIC SETS

The set X - D ( X ) is the union of all X n U where U meager; since G is countable, X - D ( X ) is meager. Let A=

E

Ce and

X

n U is

x u D(X)

Since A = (X - D ( X ) ) u D ( X ) is the union of a meager and a closed set, A has the Baire property. Let Z G A - X have the Baire property; we shall show that Z is meager. Otherwise there is U E 9 such that U - Z is meager; hence U n X is meager. and hence Since U n Z # fzl and Z G D ( X ) , there is x E U such that x E D(X), U n X is not meager, a contradiction. Although both “ n u l l ” and “meager” mean in a sense “negligible,” the following exercise shows that the real line can be decomposed into a null set and a meager set: Exercise 39.12. There is a null set of reals whose complement is meager. [Let 4,. q 2 , . , . be an enumeration of the rationals. For each n 2 1 and k 2 1, let Ink be the open interval with center qn and length l/(k . 2”). Let Dk= I,,, and A = I D, . Each D, is open and dense, and p ( D k )5 l/k. Hence A is null and R - A is meager.]

u;=,

I shall conclude this section with the following theorem. Theorem 94.

(a) Every analytic set 01reals i s Lebesgue measurable. (b) Every analytic set has the Baire property. (c) Every uncountable analytic set contains a perfect subset. Corollary 1. Eriery propert J’.

set

ofreals i s Lebesgue ineasurable and has the Baire

Corollary 2. Every analytic (and in particular every Borel) set i s either at most countable or has cartlinality 2 “ O .

We prove (b) and (c) for an arbitrary Polish space. The proof of (a) is general enough to work for other measures (in Polish spaces) than just Lebesgue measure in R“. Proq-J:The proof of (a) and (b) is exactly the same and uses either Lemma 39.7 or Lemma 39.9 (and basic facts on Lebesgue measure and the Baire property). Let me give the proof of (a) and leave (b) to the reader. Let A be an analytic set of reals (or a subset of R”). Let ,fi 1 . + R be a continuous function such that A = f [ 1 ’I. For each s E Seq. let A , =.f[Us].We have

(39.24)

A = .d{A,: s E Seq] = .d{A, : s E Seq)

7.

508

DESCRIPTIVE SET THEORY

and for every s E Seq, (39.25)

By Lemma 39.7, there exists for each s E Seq a measurable set B, 2 Assuch that every measurable 2 E B, - A, is null. Since A, is measurable, we may actually find B, such that A, E B, C A,. Let B = B,. Since B is measurable, it suffices to show that B - A is a null set. Notice that because A, E Bs c A,, and because (39.24) holds, we have A = .d{B, : s E Seq}

Thus

I claim that m

(39.26)

B-

m

u n Bacn) u ( B , - u

a€

E

NN n = O

s

€

b k )

k=O

Seq

To prove (39.26), assume that x E B is such that x is not a member of the right-hand side. Then for every s, if x E B, ,then x E B,nk for some k. Hence there is ko such that x E B,,,,; then there is k l such that x E B
n2=o

B- A E

u (B , - u B,nk 1

s E Seq

k 1 0

u;=

Since seq is a countable set, it suffices to show that each B, Let s E Seq, and let Z = B, B,nk. We have Z = B, -

m

m

k=O

k=O

U BsnkS B, - U A,nk

uc=oBSnkis null.

= Bs - A,

Now because Z E B, - A, and because 2 is measurable, 2 must be null. (c) The proof is a variant of the Cantor-Bendixson argument for closed sets in the Baire space. Recall that every closed set F in. 1’ is of the form F = [TI = {a : Vn iS(n) E T}, where T is a tree, T G Seq. For each tree T E Seq and each s E Seq, let Tsdenote the tree { t E T : t E s or s E t}; note that [T’] = [TI n U s . Let A be an analytic set (in a Polish space T), and let f be a continuous function such that A = f [ . Y ] .For each tree T E Seq, we define

T ‘ = {s E T :f[[T]] is uncountable}

40. Ej AND

n;

SETS AND RELATIONS IN THE BAIRE SPACE

509

(f[[TJ] is the image of the set [TJ by the function f ) . For each a c ol, we define F') as follows:

Po' = Seq T(z+ 1 )

P

=

(py To)

=

if a is a limit

P
Let a c w I be the least ordinal such that F a + ' )= T@)and let T = P . If T = 0,then A=

u {f[[T;')]]

: s E T"'

- T @ +" f

B
and hence A is countable. Thus if A is uncountable, T is nonempty and for ] uncountable. In this case, we shall find a perfect subset of every s E T , f [ [ T J is A. Let s E T be arbitrary. Since f[[TJ] has at least two elements, there exist so 3 s and s 1 2 s (in T ) such that f[[T,,]] andf[[T,,]] are disjoint. Then there are soo 3 so and sol 3 s o , and s l 0 =) sl, s I 13 s 1 such that the four setsf[[T,J], i , j = 0. I , are pairwise disjoint. In this fashion we construct s, E T for each finite 0-1-sequence t. These elements s, generate a subtree U = {s : s C_ s, for some t} of T such that (1) U is perfect, (2) every s has at most two immediate successors in U (hence [ U ] is a compact set in . U ) , and (3)fis one-to-one on PI. Let P be the image of [U] under the function5 Since [U] is compact andfis continuous, P is also compact, and hence closed. Moreover, P has no isolated points because [ U ] is perfect and f is continuous. Thus P is a perfect subset of A. W 40. Z/ AND

n:

SETS AND RELATIONS IN THE BAIRE SPACE

In Section 39 we introduced analytic subsets of a Polish space and showed that a set A E X is analytic if and only if there is a closed set F c 3 x I/' such that (40.1)

-

X E A

h ( x , a ) ~ F

Let us consider the product space X = V r ,for an arbitrary integer r 2 1. As in the case r = 1, the closed subsets o f . +", can be represented by trees: Let Seq, denote the set of all r-tuples ( s l , .. ., s,) E S e d such that length(s,) = ... = length(s,). A set T c_ Seq, is an (r-dimensional sequential) tree if for every (sI,..., s,) E T and each n I length(s,), (sl In, ..., s , l n ) is also in T. Let (40.2)

[ T ]= {(al, . . ., a,)

E

tpr: Vn(n,(n), ..., n,(n)) E T }

510

7.

DESCRIPTIVE SET THEORY

The set [TI is closed, and every closed set in . Y r has the form (40.2),for some tree T. I t follows that a set A G .Y’ is analytic if and only if there is a tree T G Seq,, such that

,

(40.3)

(x,, ..., X , ) E A - ( ~ Z E J ( ~ ) ( V ~ E N(n,(n), )

..., nr(n),z ( n ) ) E T

The formula (40.3) provides a more detailed analysis of C: sets in .P. Let HF = V, be the collection of all hereditarily finite sets. Let us call a subset X G HF arithmetical if X is a definable set in the model (HF, E). More generally, if Y,, . . ., & are subsets of HF, then X c H F is arithmetical in Y,, . . ., Yk if X is definable in the model (HF, E, Y,, ..., yk) where Y,, ..., yk are considered one-place predicates. We shall now define the lightface Z i hierarchy of relations over. Y . Let a be an element of .T,and let A C _ . V r .We say that the r-ary relation A over . Y- is Z: in a (A E C : ( a ) ) if there is a tree T E Seq,, that is arithmetical in a and such that (40.3) holds. If A is C: in 0,we say that A is C:. Clearly,

z;

u

=

q a )

a e .4

We call (40.3) the normalform for C: relations. For each n 2 1, we define the collections ZL(a), and follows (compare with (39.18)):

(40.4) A

E

if .Y‘ - A

n:(a)

E

if A = {(x,,

A E Zf+,(a)

Zi(a)

..., x,): (3E . V ) (x,, ..., x,,

for some ( r + 1-ary) B A

E

if

nf(a)

. tfr

-A

ni(a),and Ai(a) as

E

E

z) E B)

nf(a)

Zi(a)

Using the definition (40.4) and the normal form (40.3) for E: relations, we obtain the following characterization of Zf and IIfrelations: For instance, A G . V r is Zi in a if there is a tree T c Seq,, 2 , arithmetical in a, such that (x,, ..., x,)

E

A

-

32, Vz, 3n(~,(n),..., n,(n), z,(n), i2(n))$ T

(the variables z,, z 2 range over . V , n ranges over N). We shall now present a slightly different normal form. First, let us give a new definition of Zi(a) relations, and let me use this opportunity to be somewhat more general. Let A be an ( r + 1)-ary relation, A G -4.” x N’.We say that A is Xi in a E A’”if there is an (r + 1 + 2)-ary relation R c HF, arithmetical in a, such that

(40.5) (x,, ..., x,, k,, ..., k,)

E

A

-

32 Vn(X,(n), ..., X,(n), k,, ..., k,, F(n), n) E R

40. X,! AND

n,! SETS AND RELATIONS IN THE BAIRE SPACE

511

(z ranges over 1', n ranges over N). Either r or I can be 0; if r = 1 = 0, then the right-hand side of (40.5),i.e., 32 Vn(T(n), n) E R is either true or false; we call it a Z : ( a ) sentence. Accordingly, we say that A in (40.5) is a X:(a) property of (x,, ..., x,, k , , ..., k , ) or a Xt(a)formula. The new definition of X i ( a ) agrees with the old one: If T E Seq,, is such that (40.3) holds, then let R be the set of all (s,, ..., s,+,, n) such that (sl, . . ., s,+ ,) E T and n = length(s,); clearly R is arithmetical in T and (40.5) holds. Conversely, if R is an ( r 2)-ary relation such that (40.5) holds (with 1 = 0), then let T G Seq,, be such that

,

+

,

(s,,

..., s , + ~ )E T

V n Ilength(s,)(s,ln,,

iff

..., ~ , + ~ ln)n E, R

T is arithmetical in R and (40.3) holds. This leads to the following normal form for Zb(a) (and llb(a))relations over . Car x N': If n is odd, then A is Z,!(a) if

..., x r , k l , ..., k l ) € A 32, Vz, ... &,, Vrn(nl(m), ..., xr(m), k , , ..., k,, zl(m), ..., z,(m), m ) E R for some R arithmetical in a, and A is ll,!(a)if (40.6)

(x,,

(40.7)

(XI,

..., k l ) A* ~ h ( X , ( m ) , ..., X,(m), k l , ..., k l , zl(m), ..., ZJm), m ) E R

..., x,,

Vz, 32, ... Vz,,

kl,

for some R arithmetical in a. (The variables z,, ..., z, range over . Y , m ranges and 3m in over N . ) If n is even, then the quantifier Vm in (40.6)is replaced by h, (40.7) is replaced by Vm. [A student of recursion theory knows that we could obtain an even stronger normal form for Z,!(a)relations: The R in (40.6)can be required to be recursive in a, not just arithmetical.] We shall now use the normal form to prove some closure properties for Xb relations. Since a relation over .Y' x N' can be considered as an (r + I)-ary predicate, we can use the following notation AvB,

AAB,

lA,

A+B,

A - B

for Boolean operations (in fact, A v B is A u B etc), and

b A,

Vx A,

3m A,

Vm A :

..., x,, k , , ..., k,) E A

(x,,

..., x,, k , , ..., k , ) E k A

iff

3x

(x,,

. . ., x,, k , , . . ., k , ) E 3m A

iff

3mE N ( x l , .. ., x r , m, k l , .. ., k,) E A

E

.V(x,xl,

Lemma 40.1. Let n 2 1. (i) If A, B are X,!(a)relations, then so are %AA,

AAB,

AvB,

3mA,

VmA

7.

512

DESCRIPTIVE SET THEORY

(ii) If A, B are l l f ( a ) relations, then so are AAB,

VxA,

AvB,

Vm A,

h

A

(iii) I f A is Z,'(a),then 1 A is n,!(a); ifA is ll,'(a),then 1 A is Zf(a) (iv) I f A is l l f ( a ) and B is Zf(a), then A + B is Zf(a): if A is Zf(a) and B is llf(a), then A + B is nf(a). (v) If A and B are Aft(a),then so are 1A, A AB, A v B, A + B, A - B , Vm A, 3m A. [Compare this with Lemma 14.2.1 Proof. It suffices to prove the lemma for n = 1; the general case follows by induction. Moreover, the clauses (iik(v) follow fairly easily from (i). First, let A E Zi(a) and let us show that k A is Xi@). For simplicity, let r = 2 and 1 = 0:

(40.8)

(x, y ) E A

-

-

3 Vn(n(n), jj(n), Z(n), n) E R

where R is arithmetical in a. Thus y

E

3x A

h 3z Vn(n(n), jj(n), F(n), n) E R

We want to contract the two quantifiers 3.x 32 into one. Let us consider some nice homeomorphism between .V and . V 2 ,e.g., for u E .Y let u + and u - be u - ( n ) = u(2n

u + ( n )= u(2n),

+ 1)

(n E N) There exists a relation R' G Seq2 x N, arithmetical in R, such that for all u, y E . Y , Vn(D(n), ~ ( n ) n) , E R'

(40.9)

Namely, if n = 2k (or n = 2k length(t) = n and

iff

V k ( p ( k ) ,J(k),

u(k), k)

E

R

+ l), we let (s, t , n) E R' just in case length(s) =

((~(o),. .., s(2k - 2)),(t(O), . .., t ( k - l)),(s(l), ..., s(2k - l)), k ) E R Now (40.9) implies that y

E

3.x A

-

3u

V n (li(n),y(n), n) E R'

and hence k A is Xi(a). Next let A and B be Zi(a): xEA xEB

-

H

32 V n (K(n), Z(n), n) E R , 32 Vn (X(n), Z(n), n) E R2

where both R , and R , are arithmetical in a. Note that x

E

AAB

3,3z2 Vn[(n(n), .T,(n), n) E R 1 ~ ( x ( n ).T2(n), , n) E R2]

40.

AND

fli

SETS AND RELATIONS IN THE BAIRE SPACE

513

and hence, by contraction of 32, 3z2, there is R, arithmetical in R and R 2such that xEA

A

-

B

3 Vn(X(n), T(n), n ) E R

Thus A A B is E:(a). The following argument shows that the case A v B can be reduced to the case 3m C. Let us define R as follows (s, r E Seq, m, n E N):

-

(s, m, t, n) E R

either m = 1 and (s, t, n) E R ,

or m = 2 and (s, t, n) E R , R is arithmetical in R, and R 2 ,and x E AvB

++

32 V n (X(n),T(n), n) E R , v 32 V n (X(n),T(n),n) E R ,

H

3m 3z V n (X(n), m, T(n), n ) E R

4-b

XE3mC

where C is Xi (a). The contraction of quantifiers 3n 3z is easier than the contraction 3x 3z above. We employ the following homeomorphism between . Y- and N x . V : h(u) = (u(O), u’), where

u’(n)= u(n + 1)

If (x, m) E A

( n E N)

32 V n (X(n), m, T(n),n) E R

++

then we leave it to the reader to find a relation R’, arithmetical in R, such that for all u, x E . +p,

V n (X(n), a(n), n) E R‘

iff

V k ( X ( k ) , u(O), n’(k), k ) E R

Then x E 3m A

*--)

3u V n (X(n), ii(n), n ) E R‘

I t remains to show that if A is Z t ( a ) , then Vm A is Xi(a). Let (x, m ) E A

(40.10)

3z V n (X(n),m, T(n), n) E R

H

where R is arithmetical in a. Thus (40.11)

x

E

Vm A

H

V m 3z V n (X(n),m, T(n),n) E R

We want to replace the quantifiers Vm 32 by 3u V m and then contract the two quantifiers V m V n into one. Let us consider the one-to-one function r: N x N + N and the following homeomorphism between . C’ and w. C’: For each u E . t‘, let u,, m E N, be u,(n) = u ( r ( m , n))

(m,n

E

N)

514

7.

DESCRIPTIVE SET THEORY

Now we can replace Vrn 32 in (40.1 1) by 3u Vrn (note that in the implication we use the countable axiom of choice): (40.12) Vrn 32 Vn (X(n),rn, .T(n), n) E R

-

-,

3u Vrn Vn (X(n),rn, D,(n), n) E R

Let@:N - , Nand P: N-, N b e the inverses of the function r:if r ( m , n) = k , then rn = a ( k ) and n = P(k). From (40.11) and (40.12) we get x E Vm A

(40.13)

-

% Vk

(x(P(k)),a&), G ( P ( k ) ) , P ( k ) ) E R

Now it suffices to show that there exists a relation R' in R, such that for all u, x E . V ,

G Seq2 x

N, arithmetical

The relation R' is found in a way similar to the relation R' in (40.9), and we leave the details as an exercise. Hence Vrn A is Z : ( a ) because by (40.13) and (40.14), x

E

Vrn A

-

3u V k

(a&),ii(&), k ) E R'

H

One consequence of the closure properties proved above is that the Z: relations form a hierarchy as in Fig. 40.1. The inclusions in Fig. 40.1 are proper inclusions; the analog of Lemma 39.4 holds for the lightface X: too. In fact, for each n there is a X: set (in . Y ' + l )universal for Z; sets (in .Y').

The closure properties in Lemma 40.1 are very useful in practical computations of complexity of projective sets in . Y', and we shall use them extensively throughout this chapter. Before we start deeper investigations of the X!, hierarchy, let us make one more preliminary remark. If R E Seq x N is an arithmetical relation, then the relation A c . 1. x N (x, n) E A ++ (X(n),n) E R is both C: and ll:.More generally, if (40.15) (xl, . . ., xrr k l ,

. ..,k , ) E A-

..., X,(k,), . * . ) %(k,), X r ( k l ) , k l , ..., k,) E R where R is arithmetical is a A: relation in . Ir" x M. Let us verify this for the ( Z I ( ~ I )n1(k2), ,

case r

=

1, 1 = 2: (x, k l ,

k2) E

A

-

(n(ki), X(k2),

k1, k 2 ) E

R

40. zi

AND

flj

515

SETS AND RELATIONS IN T H E BAIRE SPACE

-

There is a relation R‘, arithmetical in R, such that for all x,

w,), k , , k 2 )

(~(k,), and hence

(x, k , , k , ) E A

--

E

r!

(X(r(k,, k2)), q k , , A,))

E

R’

(n(r(k,, k 2 ) ) , r(k,,k 2 ) ) E R‘ 3 Vn[n = T ( k , , k 2 ) + (X(n), n ) E R’] Vz 3n[n = T ( k , , k , ) ~ ( x ( n )n, ) E R’]

Thus A is both and fl:. Let us call a relation A c t.‘ x N‘ arithmetical if (40.16)

(x,,

-

..., xr, k , , ..., k , ) E A Vm, 3m2

... Vm,(Xi(m,), xi(kj),m,, k,)

E

R

where R is an arithmetical relation over HF. Arithmetical relations over 1.”‘ x N‘ have the following closure properties: If A and B are arithmetical, then so are 1A, A A B, A v B, A + B, 3m A and Vm A. Moreover, since A ! is closed under 3m and Vm and since (40.15) is arithmetical, it follows that every arithmetical relation is A ; . Also, if A is and Vz A is ll:. arithmetical, then 3 A is It goes without saying that the preceding discussion relativizes to A i ( a ) :A relation A is arithmetical in a if (40.16) holds for some R arithmetical in a. Every A arithmetical in a is A : in a.

The Sets WF and WO We shall now present two important examples of fl: sets in f “ . Let us associate with each x E 1’ the following binary relation E x over N: (40.17) m Ex n iff x(r(m, n ) ) = 0 (r is the canonical one-to-one mapping of N x N onto N); we say that x codes the relation E x . We define WF = {x E 1.” : x codes a well-founded relation} WO = {x E G’ : x codes a well-ordering of iV)

Lemma 40.2. The sets WF and WO are II:. Proof. We prove in some detail that WF is II;. By the principle of dependent choices (see (39.2e)), E x is well-founded if and only if there is no z: N + Nsuch that z(k + 1 ) E x z ( k ) for all k. Thus x E WF

* Vz 3k lz(k + 1 ) Ex ~ ( k )

In other words, WF = Vz A, where (x, z) E A

x(r(z(k

+ I), z(k))) # o

516

7.

DESCRIPTIVE SET THEORY

and it suffices to show that A is arithmetical. But (x, z)

E

A

++

3n, rn, i , j , k [i = (r(n))(k+ 1 ) rn = T(i, j )

A

j = (Z(n))(k)A

A

( ~ ( n ) ) ( r n#) 01

To show that WO is ll: it suffices to verify that the set LO = {x : E x is a linear ordering of is arithmetical (and we leave this easy verification to the reader). Then WO=WFALO~SII;. We shall show later in this section that neither W F nor WO is a Ci set; thus neither is a Borel set. For each x E WF, let llxll = the length of the well-founded relation E x

(40.18)

(see (2.1 1)). For each x, I(x(Iis a countable ordinal (and for each a < o1there is x E WF such that IJxJJ = a). If x E WO, then JIxJ( is the order type of the well ordering E x . Lemma 40.3. For each a c w l , the sets WF, = {x E WF : JJxJI I a},

WO,

= {x E

WO : JlxlJI a}

are Borel sets.

Proof. Note that the set {(x, n) : n E field(E,)} is arithmetical (and hence Borel). Let us prove the lemma first for WO,. For each a < o , , let

B, = {(x, n) : Ex restricted to {rn : rn E x n} is a well ordering of order type I a } We prove, by induction on a c wl, that each B, is a Borel set. It is easy to see that Bo is arithmetical. Thus let a c o1 and assume that all B,, /3 c a, are Borel. Then U, < ,B, is Borel and hence B, is also Borel because (x, n) E B,

++

Vrn rn Ex n + (x, rn) E

(

It follows that each WO, is Borel because x

E

WO,

++

Vn n

E

u B,)

,
field@,) -+ (x, n) E

u

B,)

8<,

To handle WF,, note that the rank function p E (see (2.12)) can be defined for any binary relation E ; namely: p E ( u )= a

iff

Vu(u Eu

+

p E ( u ) is defined) and

+

a = SUP{~E(U)1

uE

U}

40.

c,’ AND ni SETS AND RELATIONS IN THE BAIRE SPACE

517

For each a < wl, let C , = {(x, n) : pE,(n) is defined and I a )

p < a, are Borel, then

Again, C o is arithmetical, and if we assume that all C,, C , is also Borel: (x, n) E C ,

Vm m E x n + (x, m) E

(

++

UC , S
Hence each C, is Borel, and it follows that each WF, is Borel: x

E

-

WF,

V n n E field(E,) + (x, n) E

IJ C,) rn ,-=,

Corollary. The sets {x E W F : llxll = a } and {x E W F : JIxI(< a } are Boref (similarly for WO}.

Proof. {x E W F : IJxJI< a ) = U,.

WF,.

rn

Exercise 40.1. The set A = {(x, z ) : z $ WO v llxll 2 1 1 ~ 1 1 ) is Xi. Hence for each a, WO, is X ; ( z )for each z E WO such that llzll = a. [(x, z ) E A t-) z $ WO v (3: N + N)Vm V n ( m Ex n + h ( m ) E , h(n)).]

The Tree Representation of A C: set in

n: Sets in

+- is the projection of a closed set in

1”

l r 2 ,and every closed set in

I.” has the representation

[TI = {(al, a,)

E

1‘’ : V n

(a,(n), a,(n))E T }

where T is a tree, T G Seq, . The projection of [TI is the C; set (40.19)

A = {x E Y- : 32 E I.”V n (a,(n),n,(n)) E T }

Let us define, for a given T c Seq, and each x follows: (40.20)

t E

T(x)

E

. Y-, a tree T ( x )E Seq as

++

(X(n),t ) E T where n = length([)

H

3 Vn ~ ( nE) T ( x )

t ,

[ T ( x ) ]is nonempty

Thus (40.21)

xE A

Let us consider the partial ordering of Seq by inverse inclusion:

(40.22)

s1 IS,

iff

s1 2 s 2

Then we can reformulate (40.21) as follows: (40.23)

x

E

A

++

(T(x), I )is not well-founded

518

7.

Thus if C is a Ilt set, its complement A is and we have

x

(40.24)

E

C

++

DESCRIPTIVE SET THEORY

and has the representation (40.19),

T ( x )is well-founded

Moreover, if C is l l : ( a ) ,then there is T E Seq, , arithmetical in a, such that (40.24) holds. Conversely, if T G Seq, is a tree arithmetical in a, then the set C defined by (40.24) is l l i ( a ) ; this is because the complement of C satisfies the E i ( a ) definition (40.19). The formula (40.24) gives a useful representation of I l i sets. Before we proceed, let me show how this can be slightly improved. Let us extend the partial ordering s of Seq to a linear ordering < as follows: If s, t are incompatible finite sequences, let n be the least n such that s(n) # t(n);then let s < t just in case s(n) < t(n).Verify that 5 is a linear ordering and note that for every tree T E Seq, (40.25)

( T , I) is well-founded

iff

(T, I) is well-order

Thus if C is a

set, there is a tree T E Seq, such that

(40.26)

x EC

( T ( x ) ,I) is a well ordering

CI

Lemma 40.4. IfC is a ll: set, then there exists a continuousfunctionf . V + . 6’ such that C =f-,(WF), and there exists a continuous function g : . V + . V such that c = g- ,(WO). Proof. We shall give the proof for W F ; the proof for WO is similar. Let T c Seq, be such that x EC T ( x )is well-founded

-

Let { t o ,t . . . , t , , . . .} be an enumeration of the set Seq. For each x y = f ( x ) be the following element o f . V : y( r ( m , n ) ) =

0 11

E

. V , we let

if t,, t , E T(x),and t , < t , otherwise

I t is clear that E , is isomorphic to (T(x),<), and hence y E W F if and only if T ( x )is well-founded. Thus C =f- ,(WF) and it remains to show only thatfis continuous. But it should be obvious from the definitions of T ( x ) and of 1’= f ( x ) that for any finite sequence s = ( c 0 , . . ., c k - I), there is S E Seq such that if x 3 s‘ and y = f ( x ) , then y(&)= s. Hencefis continuous.

Corollary 1. WF is not Xi;W O is not Xi. Proof. Otherwise every ll: set would be the inverse image by a continuous function of an analytic set and hence analytic; however, there are ll: sets that are not analytic.

Corollary 2. (Boundedness Lemma). I f B G WO is Xi,then there is a < w 1 such that llxll < afor all x E B.

40. cf

AND

nf SETS A N D RELATIONS IN THE BAIRE SPACE

519

Proof: Otherwise we would have WO = ( x E . t- : 3z(z E B A llxll I

llzll))

Here (IxI(I Ilz(I for x , z E - 4 . means: either z 4 WO or llxll I llzll; this relation is E:; see Exercise 40.1. This would mean that W O is E:, a contradiction. Corollary 3. Every

nt set is the union of K, Borel sets.

Proof: If C is ll:, then C WF,, and hence WF =

u,<,,,l

= f-,(WF)

c=

for some continuous f: But

u f-I(WFu)

31<WI

Each f- ,(WF,) is the inverse image of a Borel set by a continuous function, hence Borel. Corollary 4. Assuming the axiom of choice, every fl: set is either at most coun'. table, or has cardinality KI,or cardinality 2O Proof: Follows from Corollary 3. Theorem 95 below improves on Corollary 3 by showing that every Zi set is the union of K , Borel sets. The following lemma is the first step toward that theorem.

Lemma 40.5. Every Xi set is the union of N, Borel sets. Proof: Let A be a Zi set. Let T E Seq, be a tree such that xEA

H

T ( x ) is not well-founded

If U G Seq is a tree and t E Seq, let U , denote the tree {s E Seq : tns E U } (hence U , = U). If (U, 3 ) is well-founded, let IIUII denote the length of the wellfounded relation ? on U . As in Lemma 40.3, we prove by induction on a that for each t E Seq and the set every a < ol,

I.

(40.27)

E L4

'

: II(T(x)),II I a1

is Borel. Namely, {x : IIT(x),1 50)= {x : (X(n), t ) q! T } and if a > 0, then 1 T ( x ) ,( 1 I a if and only if Vn( 3p < a ) 11 T(x),,1 I p. Let us define, for each a, the set B, as follows: (40.28)

x

E

B,

++

- w ( x ) l l < a ) ~ ~ ~ ( l l l ( W=) at )I l

u,<,,,,

Since the sets in (40.27) are Borel, it follows that each B, is Borel. We shall prove that A = B,. First let x E A. Thus T ( x )is not well-founded; hence IIT(x)ll # a for any a, and it suffices to show that there is a such that IIT(x),11 # a for all t . If there is no such a, then for every a there is t such that llT(x),1 ) = a, but there are N l a's and only KO t's; a contradiction. Next let x 4 A, and let us show that x q! B , , for all a. Let a < o 1be arbi-

520

7.

DESCRIPTIVE SET THEORY

trary. Since T ( x ) is well-founded, then either IIT(x)ll < a and x $! B , , or IIT(x)ll 2 a and there exists r E T ( x ) such that ll(T(x)),1) = a, and again x $! B , .

Theorem 95 (Sierpinski). Eoery Ci set is the union of K1 Bore1 sets. We postpone the proof until we have obtained a representation of Ei sets by trees over o x o I (see below). If we want to use the axiom of choice, then we set A is the projection of a ll: set can give a quick proof of Theorem 95: A i n . (. hence the continuous image of a fl set C in ,1., A = f [C].The set C is the union of K 1Borel sets B,, ct < ol, and hence A is the union of h', analytic sets A , = f [BJ. By Lemma 40.5, each A, is the union of h', Borel sets, and hence A is the union of Ec, Borel sets.

',,

Corollary. Assuming the axiom of choice, every or has cardinality H I , or cardinality 2"O. H Uniformization of a

set is either at most countable,

Relations

We have mentioned above (40.27) that if T E Seq, is a tree, then for every < wl, the set { x E . 6' : IIT(x)ll I a}

is Borel. This fact is also a consequence of the following general result:

Lemma 40.6. Let T be a tree, T G Seq,, arithmetical in a E . V . The following relations P, Q c 1.-' are E i ( a ) : (x, y ) E P

(40.29)

(x, y ) E Q

--

T ( y )is not well-founded or IIT(x)ll I IIT(y)ll T ( y )is not well-founded or 11 T(x)ll < 11 T(y)ll

Pro06 First we observe that if T and S are two well-founded trees G Seq and if IITII I IISII, then there is a mapping h : T -,S such that s c t E T ( x ) implies h(s) c h(r): This is proved by induction on as follows. For every ( n ) E T let s, be such that I(T<,)11 I IJS,, )I and let h, be an order-preserving mapping of T<,, into S,". Then let h ( 0 ) = 0 and h ( ( n ) " t ) = s,"h,(t). Conversely, if there is an order-preserving h : T + S, then either )I TI1 I IlSll or S is not well-founded. Thus we have

(40.30) ( x , y) E P H T ( y ) is not well founded or there is an order-preserving h: T(x)-+T ( y )

-

and by the same argument

(40.31) (x, y) E Q

T(y) is not well-founded or there is n and an orderpreserving h: T ( x )+ ( T ( y ) ) < " ) .

40. Z j

AND

nA SETS AND RELATIONS IN THE BAIRE SPACE

52 I

To verify that (40.30)and (40.31) are Zi(a), we first note that " T(y) is not well founded" is Zt(a). Secondly, we replace the quantifier "there is h: Seq -+ Seq"by the quantifier "there is z : N + N" by using an arithmetical enumeration { s n : n E lyl of the set Seq: For each z E 1 ., let h,: Seq + Seq be defined by Then we can express h,(s,) = "

there is an order-preserving h: T ( x )+ T(y)"

as 3z E 1 " R ( x , y, z ) where R is arithmetical in T; hence (40.30) is Z:(a). The same argument shows that (40.31) is Z : ( a ) . Remark. Note that (x, y) E P A ( y , x)

E P is equivalent to: Either IIT(x)ll = IIT(y)(l,or neither T ( x )nor T ( y )are well-founded.

Lemma 40.6 can be slightly generalized. Again, let T G Seq, be a tree arithmetical in a E tc, and let (2" : n E N } be an arithmetical enumeration of the set Seq. Then the following relations A, B G t > 2x N are Zi(u): (40.32)

( x , y, n ) E A

++

(T(y)),"is not well-founded or ll(~(X))l"Il

(x, y , n ) E

B

,*

I

5 (T(J4fnIl

(T(y)),"is not well-founded or Il(T(X))f"

II < I P ( Y N f " II

The proof of (40.32) is essentially the same as the proof of the lemma. The following theorem is a typical application of the representation (40.24) of Il: sets. Let P G . C x . C ' be a relation, and let F be a function such that F c 1.' x $., i.e., dom(F) c $", and ran(F) G V . . We say that F uniformizes P if dom(F) = dom(P). i.e., dom(F) = { x : 3y ( x , y ) E P } . Theorem % (Kondh's ll :-uniformization Theorem). Every ll:(a) relation P G t . x 1. can be uniformized by a l l i ( a )function f E 1. x V.

["'fis llt means that {(x, y) : y = f ( x ) ) is a l l i relation.] To prove the theorem, we have to construct, for each x E t', a subset Q,of the set P, = { ( J : ( x , y) E P ) such that if P, is nonempty, then Q, has exactly one element, and that Q = {(x, y) : y E Q,} is I l : ( a ) .This we shall d o uniformly for all x E , t.. The following lemma shows how the sets Q, G P , are obtained. "

Lemma 40.7. Let P c . 1. be a ll set. Then there is a ll i set Q 5 P such that Q has at most one element; and if P f 0, then Q f 0. '

Proof. Let T c_ Seq, be an arithmetical tree such that yEP

++

T ( y )is well-founded

Let {t,, : n E N } be an arithmetical enumeration of the set Seq and let us stipulate that to = 0.

522

7. DESCRIPTIVE SET THEORY

We shall construct, by induction on n, a sequence

P = Po 2 P , 2 * .2*P ,

?..*

and let Q = OX=, P , . It will be clear that Q has at most one element; we shall then Q # 0, and that Q is ll:. Let Po = P . For each then show that if P # 0, n, let (40.33) Y E

p,+1

(a) J E p , (b) y ( n ) I z(n)

ifl

for all z

E

(c) lI(TO9f"I 5 l I ( W ) ) l " II such that z ( n ) = y ( n )

P,

for all z

p,

E

,

I n other words, we let P,+ be a subset of P , such that all J E P , , have the 11. It is therefore clear that there is at most one same y(n) and the same JJ(T(y)),, 1. E t that belongs to all P , , n E N. Let P be nonempty. Then each P , is nonempty, and let y be the unique candidate for membership in Q ; i.e., for each n, y(n) is the unique k,such that x ( n ) = k, for all x E P , , 1. In order to show that Q is nonempty, we have to verify that y E P . In other words, we have to prove that T(y) is well-founded. Let t E T(J'),and let n E N be such that t = t,. For all x E P , + 1, the ordinal I/ ( T ( x ) ) 1~is" the same; let a ( t ) denote this unique ordinal. We shall show that if s c t E T(y), then a(s) > r ( t ) . This implies immediately that T ( y ) is well-founded. Let s, t E T ( y )be such that s c t , and let m and n be such that s = t , and t = 1,. Let k be such that k 2 m and k 2 n, and let x E Pk+l. Then x E P,+l and lience ~ ( s = ) II(T(x)),11; and similarly a ( t ) = ~ I ( T ( X (I.)But ) ~ because s c t , we have a(.$)= II(T(x))s II ll(T(X))f II = a ( [ ) . I t remains to show that Q is H i . The following equivalences verify that iQ is E:; A and B are the relations from (40.32): '

'

y$Q where YER

iff

-

J $ P v ~ ~ ( ~ E P , A J ~ $ Piff, , ~ )y $ P v y ~ R

M

Z

p

A

W A ll(T(Z))l, I1 = IlV(J))fk11) A M n ) < .V(n) " (44 = A n ) A II T(Z)f" I1 < I1

(Vk < n ) ( z ( k )= J

W

Now, if y E P , T(p) is well-founded;and it is easy to see that for all y iffy E S where J

E

s

++

3n 3Z[(vk < f l ) ( Z ( k ) = J'(k)A(J', A

2,

k)

E

A

A ( Z , J', f l ) E

E

t

"

I1))I

P, 1-E R

A)

(4.) < A n ) v ( z ( n )= y(n) A (z, y, n) E B ) ) ]

40.

z l A N D ni SETS A N D RELATIONS IN THE BAlRE SPACE

523

(Note that if L' satisfies the right-hand side, then T ( z )= (T(Z)),~ is well-founded and hence z E P.)Thus Y#

Q

iff

~#PvJES

Since T is arithmetical, the relations A and B are I;: and hence Q is

n:.

Proof of Theorem 96. Although we stated the tree representation (40.24) explicitly only for fl: subsets of . I ', it is obvious that r-ary relations can be characterized in an analogous way: IfC c . I ',is ll:(a), then there is T c Seq,, arithmetical in a, such that (xl, . . ., x,) E C

-

where t E T(x,, ..., x,)

iff

T(x,, . . .. x,) is well-founded

( X , ( n ) , ..., x,(n), t ) E T

where n

=

length(t)

Thus let P G 1 x 1 be n:(a). Let T s Seq, be a tree arithmetical in a, such that iff T(x, y) is well-founded (x, y) E P '

'

For each x E I ., we let P" = (y : (x, y) E P ) . We define P: zP; 2 ... z P,"2 ... exactly as in (40.33) (with T(y) replaced by T(x, y)), and then Q" = P". If P" is nonempty, then Q" has a single element which we denotef(x). Thusf= {(x, y) : y E Qx}.

n;=o

It is obvious that the proof of Lemma 40.7 is uniform for all x ; and in particular, the proof of n:-ness provides a proof that the setfis in a. Exercise 402.

Every E:(a) relation R E

I' x

I'

can be uniformized by a Zi(a)

function. [Let P E I ' 3 be such that (.Y, J.) E R tt 3z(x, y. z ) E P. and that P is n:(a).There is a nt(a) relation Q E P such that for every x, if there is (j,,z ) such that (x. !', z ) E P, then there is a unique (j,,z ) such that (x. j'. z) E Q. L e t j = {(x, y) : 3z(x. y. z ) E Q);/uniformizes P.]

A collection % of subsets of I .' satisfies the reduction principle if for every pair A, B E %' there are disjoint A', B' E % such that A' G A, B' G B, and A' u B' = A u B. % satisfies the separation principle if for every pair of disjoint set A, B E % there is a set € such that both € and l€are in %, and that A E E and B G l€. Lemma 39.5 proves that the collection ofall analytic sets satisfies the separation principle. sets satisfies the reduction principle. The collection of [Let A and B be ni and let Tl and T2be such that x E A +-P T , ( x )is well-founded. and B - TJx)is well-founded. Let

Exercise 40.3.

x E

.YE A ' t * . x E A h

Exercise 40.4.

lIIT2(X)ll< ~ ~ T ~ ( .X ~E )B ~H X~ E, B A lIITi(x)ll5 11Tz(x)11.]

The collection of Zi sets satisfies the reduction principle.

524

7.

DESCRIPTIVE SET THEORY

[Let A and B be Xi, and let P and Q be l l i such that x E A - k (x, z) E P , x E B - 3z (x, z) E Q. Let j be a l l i function that uniformizes P u Q. Let x E A'3z((x, z) E P ~ f ( x )= z), x € B' H 3Z((X, Z ) € Q A (X, Z) $ P Af(X) = Z).]

The two exercises above hold also for Il:(a) and X:(a). Exercise 40.5. If a collection $9 satisfies the reduction principle then the collection W* = { A : 1A E %'} satisfies the separation principle. [If A, B E %'*are disjoint, then 1A u 1B = . Y' and so if A', B' E %' are disjoint such that A' G l A , B' G 1 B and A' u B' = 1 A u l B ,then B' = 1 A ' and both A'

and B' are in

%*.I

and for H i (and Z i ( u ) and

Hence the separation principle holds for

n:(a)). Exercise 40.6. There is no universal A,' set, for any n E N,i.e. no D E , 1 " such that D is A,' and that for every A,' set A G . .I' there is u E . C ' such that A = {x : (x, u ) E D}. [Assume there is such a D and let A = {x : (x, x) E D}.] Exercise 40.7. The collection of ni sets (or Zi sets) does not satisfy the separation principle. [The reason is that Ili satisfies the reduction principle (Ziis similar). Let h be a homeomorphism of , t ' x , 1 ' onto ,1', and let U c . + " be a universal l l i set. Let (x, h(u, u ) ) E A iff (x, u ) E U , (x. h(u, 11)) E B iff (x, E P, and let A', B' be disjoint ni sets such that A' c_ A, B' E B, and A' u B' = A u B. If there existed E E A; such that A' G E and B' G l E ,then E would be a universal A; set.] t l )

'Ihe Tree Representation of

Sets

Let us generalize the notion of a sequential tree. Let K be a set. We denote Seq(K) the set of all finite sequences in K. A tree ouer w x K is a set of pairs (s, h) E Seq x Seq(K) such that length(s) = length@) and such that for each

n I length(s), (qn),&)) is also in T. (For any sequence fof length Iw , f ( n )denotes (f(O), . . . . f ( n - I)).) Let T be a tree over w x K. For each s E Seq, we define T, G Seq(K) as follows: Let n = length(s), and let

(40.34) h E T,

c-*

length(h) I n and (s(k), h) E T where k

=

length@)

If h E T,, then h(k)E T, for all k Ilength(h). If s c t , then T, E T,; and in fact ?; is an end extension of T, because if length(h) = length(s). then h E T, iff h E T,. Thus for every x E . t * we let (40.35)

u

T ( x )= { K : s c x} T ( x )is a tree over K (i.e., T c S e q ( K ) ) .Notice that (40.35) is a generalization of (40.20) (where K = 0).

40. x,!A N D

n,! SETS AND RELATIONS IN THE BAIRE SPACE

525

If T is a tree over w x K ,we define

(40.36)

p[T] = { X = {X

.Y: (YEoK)Vn (X(n),f(n)) E . Y : 3fVnf(n) E T ( x ) }

E

E

T)}

= { x E . Y : T ( x )is not well-founded}

(“ T ( x )is well-founded” means the relation

=I over T ( x )is well-founded) and

[TI = { ( ~ , fE ). Y ‘ x “K : Vn (X(n),f(n)) E T } Trees over w‘ x K are defined analogously; if T is a tree over or x K,then p,[T] denotes the set

. . . , x , ) E t’‘ : 3 f V n (X1(n), .. ., X r ( n ) , f ( n ) )E T } The normal form for l2: sets states that a set A 5 . Ir is Z ! ( a ) if and only if {(xl,

there is a tree over w x w, arithmetical in a, such that A = p[T].We shall now show that sets admit a tree representation. (The representation generalizes easily from . Z. to . t .‘.)

Lemma 40.8. For every l2i set A c , t . there is a tree T over w x w1 such that = p[T].Moreover. ifA is Zi(a), then T is constructible from a.

A

Proof. Let A be a x:(a) subset of 1”. There is a tree U G Seq,, arithmetical in a, such that x EA

3y Vz 3n (X(n), J(n), F ( n ) ) # U

++

In other words, xE A

3y U ( x . y) is well-founded

c--)

A necessary and sufficient condition for a countable relation to be well-founded is that it admits an order-preserving mapping into wl. Thus x

E

A

H

3y (If : U ( x , y) + wl)[if u c c, thenf(u) > f(v)]

H

3y (3f : Seq + w , ) [ f I U ( x , y) is order-preserving]

Let {u, : n E N ) be an arithmetical enumeration of the set Seq such that for every n, length(u,) I n. Iffis a function on (a subset of) N,letf* be the function on (a subset of) Seq defined byf*(u,) =f(n). Thus

(40.37) x

EA

H

3v (3f : w -+ wl)[f* I U ( x ,y) is order-preserving]

Now we define a tree T’ over w 2 x w , as follows: If s, t E Seq and h E Seq(w are all of length n, we let

(40.38)

(s, t ,

h ) E T’

H

h* I Us., is order-preserving

Clearly, T’ is a tree over w 2 x wI.

526

7.

DESCRIPTIVE SET THEORY

Let x, y E .Y. I claim that if (X(n), y(n), h ) E T’,then h* I U ( x , y ) is orderpreserving. This is because if u, u E dom(h) n U ( x , y), then u = u i , u = u j for some i , j c n, hence length@), length(u) < n and hence u, u E U s , , where s = X(n), t = j(n). Thus (40.39)

f~ T ( x , y)

e-,

Vn[(f(n))*

I U ( x , y) is order-preserving]

But clearly a mapping f: w + w1 satisfies the right-hand side if and only if f * I U ( x , y) is order-preserving. Hence (40.37) and (40.38) give X E

A

e-,

3y

3f: w + w , f € T ( x , y ) 3f: w -,w l V n (X(n), y ( n ) , f ( n ) )E T

3y Now we transform T’ (over w 2 x wl) into a tree T” (over w x K where K = w x wl) such that we replace triples e-,

( ( ~ ( 0.). ~ . , s(n - I)), ) and we get x E A ++ (3g:w + K ) V n (X(n), g(n)) E T“ Since K = w x o1is in an obvious one-to-one correspondence with wl, it is clear that we can find a tree T over w x w1 such that (40.40)

x

E

A

c*

(39: w + w l ) V n (X(n), g ( n ) ) E T

that is A = p[T]. As for the “moreover” clause, it is obvious that the tree T so obtained is constructible from the tree U , which in turn is constructible from a. The tree representation of sets is very useful in applications. As a first application, we shall finish the proof of Theorem 95. Proof of Theorem 95. Let A be a set. By Lemma 40.8 there is a tree T over w x w 1 such that A = p[T]. For each y c w1 let Ty = {(s, h ) E T : h E Seq(y)). Since every$ w + w1 has the range included in some y c wlr it is clear that A =

u P[Pl

Y
-

For each y < w l , the set p[T] is analytic (because p [ T ] = p[T] for some T G Seq,) and is the union of N l Borel sets. I n fact, Lemma 40.5 gives a uniform decomposition into Borel sets for any p[U] where U is a tree over w x S with S countable. If we let x E B: 1(I1 w4l a ) A (vt Seqb)) ( -I II ( T ( x ) ) , II = a ) +

(cf. 40.28), then A

=

uz
41. PROJECTIVE SETS IN THE CONSTRUCTIBLE UNIVERSE

521

41. PROJECTIVE SETS IN THE CONSTRUCTIBLE UNIVERSE

We have seen in Chapters 2 and 3 that many problems in abstract set theory cannot be decided on the basis of the axioms of ZFC alone. The same is true for many problems in descriptive set theory. However, various questions about projective sets (e.g., which projective sets are Lebesgue measurable) can be settled in the constructible universe. In this section we shall investigate the relationship between constructibility and complexity of projective sets. All results hold interchangeably for projective sets (and relations) in the real line R, the Baire space . Y, or the Cantor space C.I shall usually consider the space for which the proof is technically easiest.

Theorem a7 (Godel). The set of all constructible reds is a Xi set. The relation < (the canonical well-ordering of constructible sets) is a Xi relation over R . The field of < is R n L. If all reals are constructible, then < is also ll: (because x < y iffy < x) and hence < is a A: relation over R. Theorem 97 easily generalizes to q a ] : If a E R (or a E w or a E . V ) ,then the set R n 4.1 is X i ( a ) ;also, the relation ‘‘x is constructible from y” is a Xi relation over R. We proved in Section 14 (see Theorem 35 and Exercises 14.16 and 14.17) that ‘‘x is constructible” and “ x < y ” are X 1 relations over the model HC of all hereditarily countable sets. Thus Theorem 97 follows from this lemma: Lemma 41.1. I f A G .+.is such that A is X I over (HC, !..

E),

then A is a Xi set in

Proof. If A is Z 1 over HC, there exists a Z, formula cp such that x

E

A

+-+

HC i= 3u cp(u, x)

t-)

(3u E HC) HC

,t

cp[u, x]

Since cp is Zo, it is absolute for transitive models and we have x

E

A

+-+

( 3 transitive set M

E

HC)(3u

E

M ) M b q [ u , x]

(e.g. M = TC({u, x))). By the principle of dependent choices every M countable and we have x

E

A

E

H

( 3 countable transitive set M) (3u E M ) M i= ~ [ ux],

H

( 3 well founded extensional relation E over w )

HC is

3n 3m [nE(m)= x and (w, E) i= cp[n, m]]

where nE is the transitive collapse of

(0, E)

onto (M,

E).

Recall thedefinition

7.

528

DESCRIPTIVE SET THEORY

(40.17) of E, for z E . V ; thus (41.1)

xEA

tt

(32 E . V ) [ ZE W F and ( w , E,) k Extensionality,

and 3n 3m (nE,(m)= x and (w, E , ) k cp[n. m])] We shall verify that (41.1) gives a Ci definition of A. Since W F is n:, it suffices to show that the relations “(0, E) k cp[nl, ..., nk]” and “ n E ( m )= x ” are arithmetical in E. I t is easy to see that (0, E) k cp is a property arithmetical in E. As for the transitive collapse, we notice first that if k E N, then nE(m)= k

t-)

3(r,,,

..., r k ) such that m

= rk

and (Vi < k ) ( w , E ) k Then for x

Gw

we have n E ( m ) =x

-

Vn[n E rn

and ( w , E) k ro = 0

= ri

-

u {ri})

nE(n) E x ]

and a similar formula, arithmetical in E, defines n&) Hence A is Ci. W

=x

for x E .1‘.

The set A = -N n L is a C: set in P.. In case that w t = w , and 2N0> K,,A is an example of a C: set of size K 1 although 2N0> K 1. We can obtain an even better example. Since A is Ci,A is the projection of a ll; set B in . V z .By the uniformization theorem, B has a ni subsetfthat is a function and has the same projection A. Since I A I = K1 (under our assumption that 04 = w l ) , the setfhas also size K,.Thus it is consistent that 2”’ > K, and there is a ll; set of size K1. Theorem 97 provides examples concerning Lebesgue measure, the property of Baire, and perfect sets. It is particularly useful that the order-type of < on L,, is ol, and hence for every constructible x E HC, the set { y : y < x } is at most countable. ~

Corollary 1. If V = L then there exists a A: set that is not Lebesgue measurable and does not have the Baire property.

Pro05 Let A = { ( x ,y ) : x < y } . For every y , the set {x : (x, y ) E A } is countable, hence null and meager, and by Lemmas 39.6 and 39.8, if A is measurable, then it is null; and if it has the Baire property, then it is meager. Let B be the complement of A in R2,B = {(x, y ) : y I‘ x}. Again, for every x, the set { y : (x, y ) E B} is countable, and hence null if measurable, and meager if it has the Baire property. It clearly follows that A neither is Lebesgue measurable nor has the property of Baire. W The following lemma tells even more than that < is a C\relation. For any z E .N, let zm, m E N,be defined by zm(n)= z ( r ( m , n ) ) (the canonical homeomorphism between ./tr and ON).

41.

529

PROJECTIVE SETS IN THE CONSTRUCTIBLE UNIVERSE

Lemma 41.2. The following relation R over . .I. is

( z , x ) ~ R iff

{ z , , : n ~ N ) = { y :xyL x ]

Proof. Since the relation { z , : n E N ) G { y : J' show that Vy <

(41.2)

Zi:


x } is clearly Xi, it suffices to

x 3n(y = z,)

is Xi. There is a sentence O such that if M is a transitive model of 0, then < is absolute for M; and if x E M is constructible, then every y < x is in M . Thus (41.2) is equivalent to 3 countable transitive model M that contains x, z, and all z,, and M 1 [0and V y < L x 3n(y = z,)].

This last property is Zi by a proof similar to Lemma 41.1. Corollary 2.

W

If V = L then there exists an uncountable ll: set without a perfect

subset.

Proof. First we construct an uncountable Zi set without a perfect subset. Let x € A

x E W O A V J '< L x ( l I I y I I = I I ~ I I )

c-)

The set is uncountable: A is a subset of W O and for every a < o1there is exactly one x in A such that (IxI(= a. Let me show that A is Xi:Let R be the Ci relation from Lemma 41.2; thus x

E

A

++

x

E

W O A 3z[R(z, x ) ~ V n1( )Iz,I( = 11x11)]

and since lllz,ll = llxll is ll:, A is Ci. The set A does not have a perfect subset; in fact, it does not have an uncountable analytic subset. This follows from the boundedness lemma: For every analytic set X G A, the set { J ( x (: (x E X } is bounded, and hence countable (because of the definition of A). Now A is Ci and hence is the projection of some ll: set B G . CT2. By the uniformization theorem, B has a ll: subsetfthat is a function and has the same projection A. The setfis uncountable, and I claim thatfdoes not have a perfect subset. Assume that P ~f is perfect. The projection of P is an analytic subset of A. Since P G J P is itself a function and because P is uncountable, the projection A = dom(P) is also uncountable. This is a contradiction since we proved that every analytic subset of A is countable. Corollary 3. Assume that there is a E o such that Nit"]= N,. Then there is an

uncountable l l i set without a perfect subset. Proof. This is just a generalization of Corollary 2. Let xEA

4+

X €

4.1 A X E wo A v y


( 1llyll

= 11x11)

530

7.

DESCRIPTIVE SET THEORY

The same argument as before shows that A is a Zi(a) subset of W O and that for all u < wI, A has exactly one element x such that llxll = u. The rest of the proof proceeds as before, and we obtain a ll:(a) set of cardinality K, without a perfect subset.

+

Exercise 41.1. Let u be a theorem of ZF DC such that u implies Extensionality, and let P ( x ) be a property (of reals) upward absolute for transitive models of u. Then P is a xj property. [Use Exercise 14.18 and Lemma 41.1, or prove directly, as Lemma 41.1.1 Exercise 41.2. If A is a Xi set in . t‘, then x E A is Z 1 over HC. [Let T G Seq3 be arithmetical such that x E A iff (3y E t’) T(x, y) is well-founded. Show that { ( T ,x, y ) : T(x. y ) is well-founded} is At over HC (cf. Lemma 14.3)and use the fact that T is arithmetical.]

Absoluteness of Zi Relations Let A c . x N’ be a Z i (or H i ) relation and let us assume that A is presented in a normal form (see (40.6), (40.7) or (40.3), and (40.4)).Let 911 be a transitive model and assume that 911 satisfies enough axioms of ZF + DC. We denote by A” the relation (in 911) which is defined in 91 by the same normal form formula as A is in the universe. (Thus A” depends on the normal form for A, and we shall always implicitly consider such a normal form.) For instance, if A = {x : 3y Vn (X(n), y(n), n) E R} then A“

= {x E

91 : (3y E 91)Vn ( X ( n ) ,y(n), n ) E R }

Note that we also assume implicitly that if R E HF is arithmetical, then R E 911. But that is subsumed in the assumption that 91 satisfies enough axioms of ZF. More generally, let A be Z i ( a ) (or lli(a)) and let 911 be such that a E 91. Then A“ denotes the corresponding Z:f(a)(or llj(a)) relation in 911. We say that a relation A is absolute for 911 if A“ = A n 911; i.e., if for every x,,

..., x,

E . 1”

(XI,

n 911,

..., x,, k , ,

f . . , k,) E

A

++

911 b

(XI,

..., x,,

kl,

...) k , ) E A

where the formula to the right of b is the defining normal form for A. For instance, if xEA V y 32 Vn ( X ( n ) , J(n), T ( n ) ) E T

-

then “A is absolute” means that for every x xE A

++

E

. + ” n 911,

(Vy E . 1” n 911) (32 E . 1’’ n 911) Vn (T(n), j ( n ) , 7 ( n ) )E T

Theorem 98. (Shoenjeld’s Absoluteness Theorem). Every Z :(a) relation and every lli(a) relation is absolutefor all transitive models 91 ofZF + DC containing all ordinals and such that a E 911. I n particular, Zi and relations are absolute for L.

41.

53 1

PROJECTIVE SETS IN THE CONSTRUCTIBLE UNIVERSE

Proof: It is enough to prove absoluteness of 72; relations since a relation is absolute if and only if its negation is. Let us restrict ourselves to one dimension: Let a E Y and let A be a X:(a)subset of C’. Let 911 be a transitive model of ZF + DC containing all ordinals such that a E 91. We shall prove that A ~ A. = Let U G Seq, be a tree, arithmetical in a, such that for all x

x

E

-

A

E

Y,

3y U ( x , y ) is well-founded

Thus for all x E 1.’ n !Ill, x

E

A”

( 3 y E 911) 911 k U ( x , y) is well-founded

H

However, for all x , y E 9l,U ( x , y ) is obviously the same tree in 911 as in the universe; and since well-roundedness is absolute, we have x

Thus, if x X E Q

E

E

A”

++

A”, then x

E

(3y

E

911) U ( x , y) is well-founded

A, and it suffices to prove that if x

E

A n 91l, then

A“.

We shall now use the tree representation of 72; sets. Let T be the tree over x w 1 constructed in the proof of Lemma 40.8. Hence T E 4 1 . and for every

X E

1’,

(41.3)

x

E

A

H

T ( x ) is not well-founded

Now if x E 911 is such that x E A, then T ( x ) is not well-founded, and by absoluteness of well-foundedness, 911 k T ( x ) is not well-founded

In other words, there exists a function g E 91 from 1- into the ordinals such that V n (n(n), g ( n ) ) E T. Now following the proof of Lemma 40.8 backward, from (40.40)to the beginning, and working inside 911, one finds y E 911 such that 911 k U ( x , y) is well-founded

Hence if x

E

Exercise 41.3.

ZF

A n 911, then x E A” and we are done. Show that

relations are absolute for all transitive models 91 of

+ DC containing all countable ordinals and such that a E 91.

[Go through the proof of Theorem 98 and notice that we needed only that T a class in 91.1

91 is

Exercise 41.4. Prove Theorem 98 using the Levy-Shoenfield absoluteness lemma (Theorem 36). [Use Exercises 41.2 and 15.15.1 Exercise 41.5. Prove Theorem 36 in ZF theorem. [Use Lemma 41.1.1

+ DC,

using Shoenfield’s absoluteness

532

7.

DESCRIPTIVE SET THEORY

+

Exercise 41.6. There is a theorem u of ZF DC such that every X i ( a ) and every nt(a) relation is absolute for all transitive models 91 of u such that a E 91I. [If A is ni(a), then x E A t . T ( x ) is well-founded; use absoluteness of well-foundedness.]

Corollary. Every Xi subset of w is constructible. Proof: If A E w is Xi, then A’- = A and hence A

E

L. W

In contrast to the corollary, it is consistent that there is a A \ subset of w that is not constructible; an example will be presented in Section 44.Also, we have Lemma 41.3. If 0’ exists, then 0’ is a A! subset of w. Proof: We of course identify the set 0’ of sentences with the set of the Godel numbers of the sentences, and so 0’ G w. Let us also identify 0’ with its characteristic function, i.e., z E . 6‘ such that z ( n ) = 1 if n E 0” and z ( n ) = 0 if n 4 0‘. By Exercise 30.1, the property “C is a well-founded remarkable E.M. set is Il, over HC. Hence by Lemma 41.1, the set {O’} is a IIi subset o f . V. Thus n E 0’ ++ 3z(z E (0‘) and z(n) = 1) ”

CI

Vz(z

E

(0”)+ z(n) = 1)

shows that 0’ is a A: subset of o. The following lemma is an interesting application of Theorem 98: Lemma 41.4. Lat S be a set of countable ordinals such that the set A = { x E WO : llxll E S } is Xi. Then S is constructible. (And more generally, i f A is Zi(a), then S E q a ] . ) Proof: Let A ( x ) be the Z \ property such that A = { x : A ( x ) } .For each countable ordinal a, let P, be the notion of forcing that collapses a ; i.e., the elements of P, are finite sequences of ordinals less than a. Each P , is constructible; let us consider, in L the forcing languages associated with the P,, and the corresponding Boolean valued models Lpl. We shall show that for every a < wl,a belongs to S if and only if (41.4)

L l=every p

E

P , forces 3 x ( A ( x )llxll ~

= a)

This will certainly show that S is constructible. In order to prove that a E S is equivalent to (41.4), let us consider a generic extension 91 of the universe Y such that in 9,w‘; is countable. Let us argue in 9. The notion of forcing P, has only countably many constructible dense subsets, and hence for every p E P , there exists G G P, such that G is Lgeneric and p E G. It follows that for every a, every cp and every p E P a , (41.5)

L C (p

Ik

cp)

iff

for every Lgeneric G

3

p, 4G] t= cp

41.

533

PROJECTIVESETS IN THE CONSTRUCTIBLE UNIVERSE

Let a < uy,and let I E V be such that ilzll only if V satisfies 3 X ( A ( X ) All-~ll=

(4 1.6)

= a.Clearly, ct

belongs to S ifand

II=ll)

The property (41.6) is C:; and by absoluteness, (41.6) holds in V if and only if it holds in 91, and therefore 91 satisfies (41.6). Let G be an arbitrary Lgeneric filter on P I , and let u E YG] be such that llull = a. Since 91 satisfies (41.6) if and only if it satisfies (41.7)

3 x ( A ( x ) Ailxll =

1.1)

and (41.7) is Zi,it follows that a E S ifand only if LEG] satisfies (41.7). Since an Lgeneric filter on PI exists (in 91), we conclude (still in 91), that ct E S is equivalent to (41.8)

for every Lgeneric G

G

P,,

qG] k

3 x ( A ( x )A llxll = a )

But in view of (41.5) this last statement is equivalent to (41.4).

Perfect Subsets of

Sets

Theorem 99 (Mansfield, Solovay). Lat A he a X:(a) set in . 1 '. If A contains an element that i s not in q a ] , then A has a perfect subset. The proof uses the tree representation of Z: sets (Lemma 40.8). The theorem clearly follows from this more general lemma:

Lemma 41.5. L.et T h e a tree over w x K and let A = p[T].Either A c LET].or A contains a perfect subset; moreooer, in the latter case there is a perfect tree U E Y T ] ouer w such that [ U ] G A. Proof: The proof follows very closely the Cantor-Bendixson argument; compare also with the proof of Theorem 94c. For each tree T over w x K , we define

(41.9) T' = {(s, h ) E T : there exist (so, ho), (sl, h , ) E T such that so 2 s, s, 3 s, ho 3 h, h , 2 h, and that so and s1 are incompatible) Let T be a tree over

(0

x K . Let

7"' = T , 7'")= P

p + l )

TcS)

=

(T(1))1

if c1 is a limit

a

It is obvious that the definition (41.9) is absolute for all models that contain T, and hence 7")E Y T ] for all a. Let a be the least ordinal such that 7'" = T'". Let us assume first that T'")= 0; we shall show that A G q T ] . Let x E A be arbitrary. There existsfe "K such that ( x , f ) E [TI. It is obvious that there

534

7. DESCRIPTIVE SET THEORY

exists y < a such that (x, f )E [P')] but (x, f )@ [ F Y + ' ) ]Thus . there is (s, h ) E F y )such that s c x, h C J and (s, h ) @ FY+'); this means that for any (s', h') E T"), if s' =I s and h' I> h, then s' c x. Now it follows that x E Y T ] ;in L [ T ] ,x is the unique x = {s' 2 s : (s', h') E T(y)for some h 3 h'i. Now let us assume that T'")# The tree TO')has the property that for every ( s . h ) E T'")there exist two extensions (so, h,) and (sir h , ) of (s, h ) that are incompatible in the first coordinate. Let us work in q T ] . Let (so, h,) and (sl, h , ) be some elements of F') such that so and s, are incompatible. Then let ( s o 0 , boo), (sol,h,,), (slo,h,,), and ( s l l rh l l ) be elements of F') such that sij 3 s i , hij 3 hi and that the sij are incompatible. In this fashion we construct (sI , h,) E T'") for each finite 0,l-sequence r. The sI generate a tree U = {s : s E s, for some t } . It is clear that U is a perfect tree, that U E q T ] , and that [U] c p[T] = A . I Corollary. The following are equivalent:

u

a.

(i) For every a C_ o,Kf;'"]is countable. (ii) Eoery uncountable Hi set contains a perfect subset. (iii) Every uncountable Zi set contains u perfect subset. Proof. Obviously, (iii) implies (ii). By Corollary 3 ofTheorem 97 (ii) implies (1). I n order to show that (i) implies (iii), let us assume (i) and let A be an uncountable C: set. Let a E -2.' be such that A E Z ;(a). Since N4'"I is countable, there are only countably many reals in Qa], and hence A has an element that is not in 4.1. Thus A contains a perfect subset. I Projective Well-Ordering of Sets of Reals

By Theorem 97, if all real numbers are constructible, then there exists a A: well-ordering of the set of all reals. We shall now show that a converse is true: If a Zi relation well-orders the set of all reals, then all reals are constructible. We shall also show that there is no I:: well-ordering of R, and that every well founded C: relation over R has length less than w 2 . As usual, it is more convenient to prove these statements for the Baire space . 4.. Theorem 100. (a) 1fE c_ .1" x . b' is a Zi well-founded relation, then length ( E ) < wl. (b) lf E E . V' x . Y is a Ei well-founded relation, then length(E) w 2 . (c) l f < is a Xi well-ordering of Ir, then every x E .V is constructible. (More generally, if < is Xi(a), then. I/" E qu].)

-=

I

Note that (a) implies that there is no C: well-ordering o f . Y. Proof of (a). This is an application of the boundedness lemma, or rather of the fact that the set WO is not 72;. Assuming that a C: well-founded relation over . Y has length 2 wl, we shall describe the set WO in a 2;way, a contradiction. Let me first do the easier case when E is a well-ordering. Thus let E be a Ei

535

41. PROJECTIVESETS IN THE CONSTRUCTIBLEUNIVERSE

well-ordering and let us assume that its order-type is 2 wl. Then for every < w 1 there is an order-preserving mapping of (a, < ) into ( 1-,E ) . Conversely, if a countable linearly ordered set (Q, < ) can be embedded in ( 1’, E), then (Q, < ) is a well-ordering. Hence let E x be, for each x E t p , the relation coded by x (see 40.17), and let LO be the arithmetical set of all x that code a linear ordering of N. Then a

(41.10)

x E WO

++

-, 1”) V n Vm(n E x m

x

E

LO A (3f: 1

x

E

LO A 32 E t ” V n Vm(n E x m --* (z,,, z,)

‘

--*

(f(m),f(m)) E E ) E

E)

where for each z E 1.” and each n, z, is the element of t - defined by z,(k) = z(r(n, k ) ) for all k E N. Now (41.10) gives a Ei description of WO, a contradiction. In the general case when E is a 72: well-founded relation we observe that if a is a countable ordinal such that a < length(E), then there exists a countable set S G V’ and a functionfof S onto a such that for every u E S and every fi
x E WO

c--)

x E LO A ( 3 countable S = { z , : n E w)(3$S Y N ) V n Vk [if (k,f(z,)) E E x , then 3m such that (z,, z,) E E and either k =f(z,) or (k, f ( Z m ) ) E E x 1

Again, (41.1 1) can be written in a 72; manner, and we get a contradiction. Proof of (b). Let E G 4 ” x Y be a Ei relation. Let T be a tree over w 2 x w 1 such that for all x, y E V , (41.12)

(x, y) E E

-

(3f: w

--$

w l ) V n (X(n),y ( n ) , f ( n ) )E T

As usual, for each z E 4f and each n E 1,let z, E .V be such that z,(k) = Z(r(n,k ) ) for all k ; similarly, for each$ w -,w 1 and each n, let f,:w -,w 1 be k ) ) for all k . such thatf,(k) =f(r(n, Each of the following formulas is equivalent to the statement that the relation E is not well-founded: 3 Vm ( x m +

IX Vm 3fvn

17 xm)

* Y v m Vn

E

E

(a+ l(n), xm(n),T(n)) E T (Xm+

1(n)y

Xrn(n),fm(n)) E

T

536

7.

DESCRIPTIVE SET THEORY

It is easy to construct a tree U over o x o1such that for all x E . t - and all

f: O + W ] ,

(41.13)V m V n (Z,+ l ( n ) , X,(n),f,(n))

E

T

iff

V k (T(k),f(k))

E

U

(see (40.9) for a similar construction). I t follows from (41.13)that

(41.14)

iff

E is well-founded

U is well-founded

Now let E E . t' x . V' be a well-founded relation; we want to show that length(E) < w 2 . Let T be a tree over o2x w 1 such that (41.12)holds for all x, y E . and let U be the tree over o x w 1 constructed from T a s above; since E is well-founded, U is well-founded. Let us consider a generic extension W [ G ] of the universe (which we henceforth call 91) in which w'f is countable and o'f = wFG1. Let us argue in

+"

U[G].

Let E* be the relation over . f a defined by (41.12).First we observe that E E E*: If x, y E 9J1, then

(x,y) E E

iff

91l k T(x,y) is well-founded

iff

!Ul[G]k T(x,y) is well-founded

iff

(x,y) E E*

(because well-roundedness is absolute). We notice further that E* is wellfounded: This is because by the construction of U (which is absolute) and the definition of E*, W [ G ]satisfies (41.14),i.e., E* is well-founded

iff

U is well-founded

Hence E* is well-founded, and length(E) s length(E*). The tree T is a tree over o x coy and oy is a countable ordinal. Since E* = p[T],it follows that E* is Z{ relation. By part (a) of the theorem, the length of E* is countable; i.e., length(E*) < wFG1. I t follows that length(E) < wFG1= o'f, Now we can step back into the ground model and look at the result of the above argument: length(E) < w 2 . Proof of (c). Let < be a Xi well-ordering of. 1- and let us assume that there is a = "(0, 1) be the nonconstructible real. Let To = Seq((0, If), and let C = [To] Cantor space. Let us consider trees T E Toand functionsf: T + Tosuch that (i) s G t implies f (s) G f ( t ) and (ii) for every x E [TI, f ( a ( n ) )E C. Every such function induces a continuous function from [TIinto C, which we denote

u;=,

f *. Lemma 41.6. If T E To i s a constructible perfect tree and iJJ T + To i s a constructible function such that f * is one-to-one, then there exists a constructible pegect tree U G T and a constructible g : U + Tosuch that g* is one-to-one, and g*(x)
42.

537

A MODEL WHERE ALL SETS ARE LEBESGUEMEASURABLE

It suffices to prove this lemma because then we can construct a sequence of trees To 2 TI z ... 2 T,, 2 ... and functions fo,fl, ..., f,,, ... such that f .*+ l(x) < f , * ( x )for all x E [T,,, ,I. Since all [TJ are compact sets, their intersection is nonempty and therefore there exists x such thatfz(x) >f:(x) > ... > .f.*(x)> ... contrary to the assumption that < is a well-ordering. Proofof lemma. Let T c To be a constructible tree and let f: T + To be constructible, such that f* is one-to-one. Since T is perfect, there exists a constructible function h : T + To such that h*: [TI + C is one-to-one and onto. For each s E To, let S be the "mirror image" of s, namely ifs = (s(O), . . ., s ( k ) ) ,let S = ( I - s(O), ..., 1 - s ( k ) ) ; for x E C, x is defined similarly. We claim that at least one of the sets ~

A = {X

E

[TI :f *(x)> h * ( x ) ) ,

B

={X E

[TI :f * ( x )> h*(.x))

contains a nonconstructible element. Let z be the least nonconstructible element of C, and let x , y E [TI be such that h*(x)= z and h*(y) = 2. Then both x and J* are nonconstructible and hence f * ( x ) 2 z and f*(y) 2 z. Thus either f * ( x )> z or f*(y) > z and so either A or B contains a nonconstructible element. For instance, assume that A does. (The other case is similar.) Since < is E 2 , and T , f ; and h are constructible subsets of HF, the set A is El(a) for some a E L. By Lemma 41.5, there exists a constructible perfect tree 15' such that [ L i ] c A. If we let g = h I U , then U and g satisfy the lemma. 42. A M O D E L WHERE ALL SETS ARE LEBESGUE MEASURABLE

This section is devoted entirely to the proof of the following theorem:

Theorem 101 (Solovay). As.sutne that there exists an inaccessible cardinal. (a) There i s a moilel of ZF + DC in which all sets of real nurnbers are Lvbesgire tnea.suruhle ant/ haw the propert!* of Baire, and ererj' uncountable set of rcw1.s has a perfect subset. (b) There is a model of ZFC in which ererj. projectire set of reds i s Lrhesgire measurable, has the Baire propert!', and lf uncountable, then i t contains a perfect subset. The proof makes substantial use of absoluteness of measure and category, and we shall deal with that first. We begin with a finer analysis of the hierarchy Zy of Borel sets. Codes of Borel Sets

steps, from the basic Every Borel set of reals is obtained, in fewer than open sets by taking complements and countable unions. We shall show how this procedure can be coded by a function c E '"(I).We shall define the set BC of

538

DESCRIPTIVE SET THEORY

7.

Borel codes and assign to each c E BC a unique Borel set A,. The code c not only describes the Borel set A, but also describes the procedure by which the set A, is constructed from the basic open sets. Let I Iz , .. ., Ik , . . . be an arithmetical enumeration of open intervals with rational endpoints (i.e., the sequence of the pairs of endpoints is arithmetical). For each c E . 2 ' , let (42.1)

and

U(C)

(i E N )

ui(c)

+

be elements of 1 defined as follows: If d = u(c), then d(n) = c(n 1) for all n; if d = ui(c), then d ( n ) = c(T(i, n) + 1) for all n (where r is the canonical one-toone correspondence between N x N and N). For 0 < a < wI,we define sets Z, and f7, G .1- as follows: "

(42.2)

C E

C1

c E f7,

if c(0) > 1 if either c E C, u f7, for some f l < a or c(0) = 0 and

(a > 1)

c E Z,

U(C) E

C,

if either c E 1,u I7, for some f l < a

or c(0) = 1 and ui(c) E

U (C,

u f7,) for all i

P<,

I f c E Cl(if c E n,), we call c a X:-code (a lit-code). Let BC, the set of all Borel codes, be BC=

u C,= u I7,

,
Z
For every c E BC, we define a Borel set A, as follows (we say that c codes

4): (42.3)

then A, = U { I , : c(n) = 1)

if c E C1

if c E f7, and c(0)

=0

then A, = R

- A,,(,,

u m

if c E 1,and c(0) = 1

then A, =

A,i(,)

i=O

I t is clear that for every a > 0, if c E C, (if c E H,), then A, E Z : (A, E rl:). Conversely, if B is a E : set (a rl," set), then there exists c E L,(c E f7,) such that B = A,. This is proved by induction on a using facts like: If ci, i E w are elements of U,. n,, then there is c E C, such that ci = ui(c) for all i E w. Thus { A , : c E BC} is the collection of all Borel sets. Lemma 42.1. The set BC of all Borel codes is

ll:.

Proof: Let us consider the following relation E over . i .:

(42.4) x E y

iff

either y ( 0 ) = 0 and x = u ( y ) or

y(0) = 1 and x = vi(y) for some i

E

w

42.

539

A MODEL WHERE ALL SETS ARE LEBESGUEMEASURABLE

The relation E is arithmetical. If y E Z1, then y is E-minimal (i.e. ext,(y) = 0) and vice versa; if y E n, and x E y , then x E Z,, and if y E Z. (a > 1) and x E 4: then x E Ua<.(Z, u no). We claim that

(42.5) y

E

BC

--

E is well-founded below y there is no (zo, z,,

. . ., z,,

and that Vn(z,+ E

2,)

...) such that zo = y

By the remark following (42.4), if y E BC, then there can be no infinite descending sequence zo = y , z1 E zo, z 2 E z , , etc. Conversely, if E is well-founded below y, let p denote the rank function for E over ext,(y). By induction on p ( x ) , one can see that every x E ext,(y) is a Borel code, and finally that y is itself a Borel code. Now (42.5) gives a Il: definition of the set BC and the lemma follows. W Lemma 42.2. The properties A , c A , , A , = A,, and A , = 0are ll properties of Borel codes. Proof: We shall show that there are properties P, Q E R x +'such that P is Il! and Q is Ct and such that for every c E BC,

(42.6)

a

E

A,

c-)

(a, c ) E P

c*

(a, C) E Q

(I hope that the reader does not mind that I talk about Il: relations over R x 1 '). Then c, d E BC A va((a, C) E Q + (a, C) E P) A, G A ,

-

d

BC A A c G

A, = A ,

H

C,

A,

++

c E BC A V U (u, C) $ Q

= (21

E

E A,

To find Pand Q, let x E , 1 ' be fixed. Let T be the smallest set T x E T, and if y E T and z E y, then z

(42.7)

E

E ,1'such

that

T.

The set T is countable. Let h : T + (0, 1) be a function with the following property: for all y E T,

(42.8) if y ( 0 ) > 1, then h(y) = 1 iff for some n, y(n) = 1 and a E I , : if y(0) = 0, then h(y) = 1 iff h ( u ( y ) )= 0;

if y(0) = 1, then h ( y ) = 1 iff for some i, h(u,(y))= 1. Note that if x is a Borel code then there is a unique countable set T C 1. with the property (42.7), and a unique function h with the property (42.8); moreover, for every y E T we have h ( y ) = 1 just in case a E A , . Thus we let

(42.9) (a, x ) E P

H

(V countable T

c I'-)(Vh: T - ,

(0, 1))

[if (42.7) and (42.8) then h ( x ) = I]

7. DESCRIPTIVE SET THEORY

540

and

(42.10) (a, x) E Q

(3 countable T E . Y)( 3h: T + {0, I})

c--)

f(42.7)A (42.8)A h ( x ) = 11 and it is clear that if c E BC, then a E A , iff (a, c ) E P iff (a, c) E Q. It is a routine matter to verify (with some effort) that (42.9)can be written in a fli way and (42.10) in a Zi way. (The quantifiers VT, Vh, and 3T 3h are the only ones for which one needs quantifiers over .4”; note that for instance, Vz(z E y + z E T) in (42.7) can be written as (y(0) = 0 + u ( y ) E T ) A (y(0) = 1 -+ Vi(u,(y) E T ) ) ) .

Absoluteness of Measure and Category We shall now show that certain properties of Borel codes are absolute for transitive models of ZF + DC. If 911 is a transitive model of ZF + DC and c E “‘ois in 91, then because the set BC is l l i , c is a Borel code if and only if 91 k c is a Borel code. By Lemma 42.2, the properties of the codes A , E A d , A , = A,,, and A , = 0 are fl: and therefore absolute: A, = Ad holds if and only if A T = A?, etc., where A: denotes the Borel set in 911 coded by c. Moreover, since a E A , is ll:, it follows that A T = A , n 9l,for every Borel code c E 911.

Lemma 42.3. The following properties (of codes) are absolute for all transitive models 91 of ZF + DC: A, = A ,

U

A , = A, n Ad,

Ad,

5

A, = R

-

A,,

A , = A , A Ad,

A, =

UA,

n=O

(we assume that the codes c, d , e are in 91k as is the sequence (c,: n = 0, 1, . ..)). We say that the operations u, n, -, A, absolute for 911.

u:=,on Borel sets with codes in

911 are

Proof. If c o , cl, . .., cn, . . . is a sequence of Bore1 codes in 911, let c E . i’.be such that c(0) = 1 and that ui(c)= ci for all i E o.Clearly, c is a Borel code, c E 91\ and c codes (both in the universe and in 91)the Borel set Acm.Hence for any Borel code e E 9l,we have

u;=o

u AT 5

A?=

n=O

u A, 31

iff

A?= A?

iff

A , = A,

u;=o

iff

A,=

n=O

because A, = A, is absolute for 911. Thus A , = A,” is absolute. An analogous argument shows that R - A , is absolute, and the rest of the lemma follows easily because the operations n and A can be defined from u and -.

42.

A MODEL WHERE ALL SETS ARE LEBESGUE MEASURABLE

541

Lemma 42.4. “ A , is null” and “ A , i s meager” are properties absolutefor all

transitive models 911 of ZF

+ DC.

+

Pro05 Let 911 be a transitive model of ZF DC. Let p denote the Lebesgue measure. First we claim that if c E 911 is a X:-code, then $”(A:) = p ( A , ) . Let k,, k , , .... k,, ... be all the k E N such that c ( k ) = 1 ; thus A , is the union f k m of open intervals with rational endpoints. For each n, let X, = I,” ( l k ou ... u l k n . ,); hence A , = u i y 0 X, and p(A,) = p ( X , ) . However, each X , is a finite union of disjoint rational intervals, and clearly p ( X , ) is absolute. Hence p ” ( A Y ) = p(A,). A similar argument shows that if c E 911 is a ll:-code, then pc”(AY)= p(A,). Next we claim that if c E 911 is a ny-code, then A , is rare if and only if 911 1 A , is rare. This is because d = u(c) E Z, and it is easily verified (using open rational intervals) that “Ad is dense is absolute. Now we are ready to prove the lemma. Let us consider first the property “ A , is null.” We use the following properties of Lebesgue measure: ( 1 ) X is null if and only if for every n E N, there is an open set G 3 X of measure I l/n, and (2) p ( X ) > 0 if and only if there is a closed set F G X of positive measure. If 9J1 1 A , is null, then 911 satisfies

on‘=,

”

(42.1 1 )

Vn L ( e E Cl and A , 2 A , and ,u(A,) 5 l/n)

Since the part (.-.) of (42.1 1 ) is absolute, it is clear that (42.11 ) holds in V , and hence A , is null. I f 911 k A , is not null, then 911 satisfies (42.12)

L(rE

I7, and A ,

G

A , and p ( A , ) > 0)

Again, (...) is absolute, thus (42.12) holds in V and hence Ai is not null. Finally, we consider the property “ A , is meager.” If 911 1 ( A , is meager), then 9Ji satisfies: (42.13) There exist c, E n,, n A, c ACn.

u;=,

= 0,

1, . . ., such that each A,” is rare, and

Then (42.13) holds in V and so A , is meager. A Bore1 set B is not meager if and only if there is a nonempty open set G such that B A G is meager. Thus if 911 k A , is not meager, then 911 satisfies

(42.14) 3tI 3r(d E L‘, and Ad # 0and A ,

=

A, A A, and A, is meager)

Then (42.14) holds in V and hence A , is not meager. H Exerciscp 42.1. “ A , is null” and ” A , is meager” are absolute for all transitive models of

ZF-

+ DC.

[Because

ni properties are.]

7.

542

DESCRlPTlVE SET THEORY

Random and Generic Reals Let us consider generic extensions of a ground model %I, using either the algebra of Borel sets modulo the ideal of null sets or the algebra of Borel sets modulo the ideal of meager sets. Let 9 be the o-algebra of Borel sets of reals, let I , and I , (m for measure, c for category) be the a-ideals I,

=

I , = { B E J : B is meager}

{ B E J : p ( B ) = 0},

and let (42.15) .?dm= &/I,,, = { [ B ] , : B

E

.d}, A?, = J / l , = { [ B ] ,: B E A?}

where [ B ] , and [BIc denote equivalence classes mod I , and mod I , , respecare complete Boolean algebras (by Exercise 17.23) and if tively. .a, and -3, E n , n E o,are Borel sets then (in either 33, or a,), n=O

Exercise 42.2. The algebra . ~ is9 the ~ unique atomless complete Boolean algebra that has a countable dense subset. [If B is a nonmeager Borel set, then there is a nonempty open set U such that U A B is meager; hence there is a rational interval f such that [fIc 2 [BIc.]

Thus to form a generic extension using a, is the same as to adjoin a Cohen generic real. This observation justifies the definition below of a real number generic over 911. Let 911 be a transitive model of ZFC. Let us consider the Borel sets in 911; let .d denote the collection of all Borel sets in 91\ and let .M, and . d , denote the complete Boolean algebras (42.15) in 91. Let B be a Borel set in 911. B has a Borel code c E 911, B = A,. Let us denote B* the Borel set in the universe coded by c. This definition does not depend on the choice of c E BC" because by Lemma 42.2, if A , = A d . then A: = A:. We recall that B = B* n 911, for every B E A?. Lemma 42.5. (a) I f G is an 911-generic ultraflter on d , , then there is a unique real number x c such that f o r all B E -1, (42.16)

x G e B*

-

[B],,, E G

The formula (42.16) determines G and hence 9ll[G] = 911[xG] (b) If G is ail 911-generic ultraflter on .d,, then there is a unique real number x G such that f o r all B E .d,

(42.17)

X

~

B* E

*--)

[ B ] , EG

Theformula (42.17) determines G and hence 911[G] = 911[xG].

42.

543

A MODEL WHERE ALL SETS ARE LEBESGUEMEASURABLE

If x is a real number and if x = x c for some G G W, generic over Yll, then x is random over 911. If x = x G for some G c Bc generic over 911, then x is generic over 911. Proof. The same proof works for both (a) and (b); let [B] denote [B],,,in case (a) and [B], in case (b). To start, we claim that there is at most one real number x that satisfies (42.18)

x

E

B*

(for all B E 99)

[B] E G

t*

If x satisfies (42.18),then x belongs to all B* such that [B] E G. If x < yare two real numbers, let r be a rational number such that x < r < y, and let A be the interval (r, co) = {z E R : z > r}, Either [A] or [R - A] belongs to G but x # A* and y # (R - A)*. In order to show that there exists a real number x that satisfies (42.18),let (42.19)

x = sup{r : r is a rational number and [(r, a ) ]E G}

By the genericity of G, there exists r such that [(r, a ) ]# G, and hence the supremum (42.19) exists. Note also that x # 911 (by the genericity of G). We shall show that x satisfies (42.18).We shall show, by induction on Bore1 codes in %I, that for every c E BCm, (42.20)

A:

X E

H

[A,] E G

First we consider Xy-codes (in W), and let us start with those c E ,TI n 911 that code a rational interval, i.e., such that ~ ( n=) 1 for exactly one n; then c codes the interval I,. Let In= (p, 4). We have x

E

A:


< sup{r : [(r, a)]E G} < q

iff

p <x

iff

[(P, a ) ]E G and [(q, co)I

iff

[(p, q)]

Now if c E Z,,then A, = { k : c ( k ) = I ) , and we have

u

iff

G

u.'=o

p

iff

/kn,

#G

[A,] E G

where {kn : n = 0, 1, . . .} is the set

rx

x

E

iff

A,*

x

E

1;"

iff

3n(x E 1:")

iff

3n([1,J E G )

n=O

[IkA E G

iff n=O

iff

[

5

Ikn]E G

Iff

[A,] E G

n=O

Next let a < oy'and let c E fl, A 911, and let us assume that (42.20)holds for all c E Cz n 91i. We may assume that c(0) = 0; then U(C) E C, n 91 and AUlc,= R - A,, and we have iff x 6 A&, iff [AUlE,] 6G iff [A,] E G Finally, the induction step for Cz is handled in a way similar to the case for c E ,Yl. Thus (42.20) holds for every c E BC", and thus x is the unique real number that satisfies (42.16) (in case of .&,,) or (42.17) (in case of .&,). x

E

Af

7.

544

DESCRIPTIVE SET THEORY

The following lemma provides a characterization of random and generic reals.

Lemma 42.6. A real number is random over 911 ifand only i f i t does not belong to an)' null Borel set with code in 91t and is generic over 911 ifand only if it does not belong to any meager Borel set with code in 911. Hence if Ra(911) and Ge(Yl1) denote the sets of all random and of all generic reals over 9t we have (42.21)

Ra(Yl1) = R*

-

Ge(911) = R*

-

u u

{A: : c E BC" and Aris null) {A:: c E BCm and A f is meager)

Note that by Lemma 42.4, A, is null (in 911) if and only if A: is null (in V ) . Pro05 On the one hand, if x is random over 9t let G be an 91-generic ultrafilter on .dmsuch that x = x c . Then if A , is null then [ A d 4 G, and by (42.16),x 4 A : . Similarly for x that is generic over 911. On the other hand, let x be such that x # A: whenever A, is null (and c E 911). First we observe that if [A,] = [Ad] then A, A Ad is null, hence A: A A: is null and it follows that x belongs to A: if and only if x belongs to A:. Let (42.22)

G = {[A,] : x E A:)

It is easy to see that G is a filter on .dm:If [A,] E G and [Ad] E G, then x E A: n A: and hence [ A , n Ad] E G ; similarly, if [A,] 5 [Ad] and [Ac] E G, then [Ad] E G. We shall show that G is 911-generic. Since .dmsatisfies the c.c.c., it suffices to show that if { A E n: n E w } E 91 is such that [A,,] E G, then some [ A J is in G. But this is true because

i

n=O

a

[A,"] =

[ U4 1

and

n=O

( ii n=O

A,")' =

ii

n=O

A:

Finally, we claim that x = xc . But this follows from (42.22), by the genericity of G. Thus a real number x is random over 91 if and only if x 4 A: for any null Borel set A, E 911. The proof is entirely similar for generic reals. H

Solovay Sets of Reals Let 911 be a transitive model of ZFC. Let S be a set of reals. We say that the set S is Solovay over 911 if there is a formula cp(x). with parameters in 91t such that for all reals x , (42.23)

x ES

iff

911[x] k cp(x)

The following lemma is the key step in the proof of Theorem 101:

42.

545

A MODEL WHERE ALL SETS A R E LEBESGUEMEASURABLE

Lemma 42.7. Let S be a Solovay set ofreals over 911. There exist Borel sets A and B such that S

n

Ra(911) = A n Ra(911)

and

S n Ge(911) = B n Ge(911)

Corollary. (a) I f the set of all reals that are not rarrtiom over 911 is null, then S i s Lebesgue measurable. (b) If the set of all reals that are not generic oaer 911 is meager, then S has the property of Baire. [Proof. Under the assumptions of the corollary, S A A is null and S A B is meager .]

Proof: Let me prove the lemma for random reals. Let us consider the forcing language in 911 associated with .d,. Let G be the canonical name for a generic ultrafilter on .d,,and let a be the canonical name for a random real; i.e., let a be the .J,-valued name defined in 9W . from G, by (42.16): 110 - xcll = 1. Let q ( x ) be a formula with parameters in 911 such that (42.23) holds for all x . Let A, E .dbe such that and let A

=

A:. The set A is a Borel set (in the universe); I claim that for all

x E Ra(911), x belongs to S if and only if x belongs to A. But ifx is random over 91t let G be 911-generic on .d, such that x = x,; then a is a name for x and we

have x E S

iff

911[x] b q(x)

iff

li(p(a)ll E

G

iff iff

911[G] k q ( x )

[A,]

E

G

iff

x E A:

H

The Models 911[G] and 91 Let 911 be a transitive model of ZFC and let ti be an inaccessible cardinal in 911. Let B be the Levy algebra for K , i.e., B = r.0. P where P is the notion of forcing (20.7) that collapses each cx. < ti onto No:The conditions are functions p on subsets of K x cu such that each dom(p) is finite, and p(a, n ) < a whenever (a, n ) E dom(p). Let G be an 911-generic ultrafilter on B. We shall show that in 911[G] every projective set of reals is Lebesgue measurable, has the property of Baire, and if uncountable, then it contains a perfect subset. In 911[G], let S be the class of all infinite sequences of ordinal numbers, S = “Ortl, and let ‘91 = HOD(S) be the class of all sets hereditarily ordinal definable over S. The class 91 is a model of ZF (see Exercise 15.21). In fact, 91 is a model of ZF DC (see Lemma 42.8), and we shall show that in 91 every set

+

7.

546

DESCRIPTIVE SET THEORY

of reals is Lebesgue measurable, has the Baire property, and if uncountable, then it contains a perfect subset. Let s be an infinite sequence of ordinals (in %1[G]),let cp be a formula, and let X E G be a set such that

X

(42.24)

= {X : 91[G] k V(X, s)}

The method of Section 15 shows that X is (in 911[G]) ordinal definable over S = “Ord. Conversely, if X E OD(S), then for some formula I(/ and a finite sequence (sir ..., sk) of elements of S, X = {x : %1[G] 1 $(x, (sl, ..., sk)). Then clearly there is cp and s E S such that (42.24)holds. Hence the class OD(S) consists of all sets X of the form (42.24)-sets definable in %1[G]froma sequence of ordinals. Note that every projective set of reals is definable from a sequence of ordinals: If A is Z i ( a ) for some a E “0, then A is definable in HC from a, and therefore A E OD(S). Lemma 42.8.

(a) Iff€ 911[G] is afunction on w with values in 91, thenfE 91. (b) The model 91 satisfies the principle of dependent choices.

u

Proof. (a) We show that iff is a function from w into OD(S) thenfE OD(S). by (42.24), OD@) = {OD(s): s E S}; by Lemma 15.8 there is a definable function F on Ord x S such that for each s E S, the function F,(a) = F(a, s) maps Ord onto OD(s). Let$ w -,OD(S). For each n, we choose a,, and s,such that f ( n ) = F(a,, s,). Clearly,f is definable from (a,, : n E w ) and (s,, : n E w ) . I t is easy to find a single sequence u of ordinals such that both ( a , : n E w ) and (s, : n E w ) are definable from u. Hencefis definable from u, and sof E OD(S). (b) In 91, let p be a relation over a nonempty set A such that for every x E A there is 4’ such that y p x. Since 911[G]satisfies the axiom of choice, there exists in 911[G]a sequence a. , a,, . . .,a,, . . . such that a,, p a,,for all n. However, by part (a) of this lemma, the sequence ( a , : n E w ) is in 91. Measure and Category in 91[C] and 91 We shall now prove the part of Theorem 101 dealing with Lebesgue measure and the Baire property, using Lemma 42.7. Lemma 42.9. Ler s E 931[G]be an infinite sequence ofordinals. The set ofall reals (in 911[G]) that are random ouer 911[s] is null; the set of all r e d s that are nor generic ouer 9lJ[s] is meager. Proof. Since the algebra B is K-saturated, there exists a subalgebra D G B such that ID < K and 91[s]= 91[D n GI. It follows that K is inaccessible in 91[s]; and since K = h’FCJ, 9l[.s] has only countably many subsets of w. Thus there are only countably many Bore1 codes in 9ll[s]; and by (42.21),the complement

I

42.

A MODEL WHERE ALL SETS ARE LEBESGUE MEAPIJRABLE

547

of the set Ra(911[s]) is the union of countably many null sets and hence null. Similarly, the complement of Ge(91[s]) is meager. W

Lemma 42.10. Let X E 911[G] be a set ofrrals that is deJnable in M[G]from a sequence s ofordinals. Then X is (in W1[G]) Solovay ouer 91[s]. Proof. The proof uses the properties of the Levy algebra discussed in Section 25, in particular the factor lemma (Exercise 25.11). We shall first prove the following: Given a formula cp, there is a formula @ such that for every sequence of ordinals x E W[Gl, (42.25)

911[G] 1 ~ ( x ) iff

911[x] k @(x)

We recall that P is the notion of forcing (20.7) that corresponds to the Levy algebra for K ; the forcing conditions are finite and so the definition of P is absolute for all models. If M is a model, we denote M P the Boolean valued model constructed in M using P and if II/ is a formula and z E M P , we denote ~ ~ I I / ( z ) the [ ~ ~ Boolean value (computed in M using P) of II/(z). I f a E M , then Z EM P is the canonical name for a. Let @(x)be the following formula

~ ~ c p ( z )=~ p1 '

(42.26)

Let x be a countable sequence of ordinals in 911[G]; we shall show that 91JqGI 1 cp(x) if and only if %I[x] b q(x). By the factor lemma (Exercise 25-11), there exists an 911[x]-generic filter H on P such that 911[G] = 911[x][H]. Arguing in 911[x], we invoke the homogeneity of the Levy algebra (see the corollary of Theorem 63): the Boolean value b = Il~p(2)ll'O"~' is either 0 or 1. Since H is generic on P over 911[x], cp(x) is true in !W[x][H] if b = 1 and false if b = 0. Hence q(x) is true in 911[G] if and only if @(x)is true in 911[x]. Clearly, the above argument works also for a formula cp with two variables: There is S, such that for all x, y E 911[G] n "Ord, 911[G] k ~ ( xy),

iff

911[x, y] 1 @(x,y)

Now let X E 911[G] be a set of reals that is definable in 911[G] from a sequence of ordinals s . For some formula cp x

E

X

++

9ll[G] k q(x, S )

for all reals x E 911[G]. Since every real can be considered a countable sequence of ordinals, we have, for all x E Itmc1, x

E

x

c-)

9qs, X I I=@(% x)

H

911[S][X] 1 @(s, x)

which shows that X is Solovay over 9ll[s]. W

548

7.

DESCRIPTIVE SET THEORY

Corollary 1. In 911[G]every set of reals definable from a sequence of ordinals (and in particular, every projective set of reals) is Lebesque measurable and has the property of Baire. Proof: This follows from Lemmas 42.7, 42.9, and 42.10.

Corollary 2. In 93 every set ofreals is Labesgue measurable and has the property of Bairr. Proqf Clearly, the model 91 has the same reals and the same elements of "(u as the model 911[G].In particular, 91 and %1[G]have the same Borel codes, and since A:' = Alm[cl n 91 = A?'' for every c E BCN"1, the two models have the same Borel sets. If X E 91 is a set of reals, then X is definable in 911[G] from a sequence of ordinals and hence 911[G] k ( X is Lebesgue measurable and has the Baire property). Thus there are (in 911[G]) Borel sets A , B, N, M such that X A A E N , X A B G M , and N is null and M is meager (in W [ G ] ) .By Lemma 42.4, 91 satisfies that N is null and M is meager, and hence 91 satisfies that X is Lebesgue measurable and has the Baire property. A Remark on C: Sets of Reals

The method of the proof of Theorem 101 can be used to prove the following result on sets:

Lemma 42.11. Assume that for every a G w, K;'"' is countable. Then every Xi set of reals is Lebesgue measurable and has the Baire property. [Compare this with the Corollary of Theorem 99.1 Proof: We use the tree representation of C: sets. Let A be a C i set. (We may assume that A is a Xi subset of the Baire space which is homeomorphic to the irrational members.) Let a E Ow be such that A is X\(a), and let T be a tree over w x w 1 such that T E 4 1 . and xEA

iff

T ( x ) is not well-founded

for all x E Ow. Since well roundedness is absolute, we have

x

E

A

iff

q a ] [ x ]b T ( x ) is not well-founded

and hence A is Solovay over q a ] . Now each q a ] has only countably many subsets ofw, hence only countably many Borel codes, and by (42.21) the complement of R a ( q a ] )is null and the complement of G e ( q a ] ) is meager. Hence by Lemma 42.7, A is Lebesgue measurable and has the Baire property. Exercise 42.3.* Use the above method to show that all and have the Baire property.

sets are Lebesgue measurable

42.

A MODEL WHERE ALL SETS A R E LEBESGUE MEASURABLE

549

[Given a Z; set A , find a countable transitive model M of ZF- + DC such that A is Solovay over M . Use absoluteness o f Z f for countable models, Exercise 42.1 and the fact that M has only countably many Bore1 codes.]

Perfect Subsets in 911[G] and 91

We shall now finish the proof of Theorem 101 by showing that in 911[G] every uncountable set of reals definable from a countable sequence of ordinals contains a perfect subset. Then it follows that in 91, every uncountable set A of reals has a perfect subset: If A is uncountable in 91, then A is uncountable in 911[G] (by Lemma 42.8); and since A is definable from a sequence of ordinals, A has a perfect subset F (in 911[G]);but then 91 1 F is a perfect set. By Lemma 42.10, every set of reals definable in 911[G] from s is Solovay over 911[s]; thus it suffices to prove that in 911[G] every uncountable set of reals, Solovay over some 9ll[s]. contains a perfect subset. Furthermore, it suffices to give the proof only for sets of reals Solovay over 911 since the general case (Solovay over 911[s]) follows from the special case by the factor lemma: 911[G] = 9ll[s][H] is a generic extension of 911[s] via Levy forcing. And finally, we can consider subsets of the Cantor space "(0, 1 ) instead of sets of reals. Thus let A be, in 911[G], an uncountable subset of "{O, l ) , and let cp be a formula (with parameters in 911) such that for all x E "(0, 1) in 911[G], x

E

A

iff

911[x] 1 cp(x)

Since A is uncountable. there exists x E A such that x # 911. There exists (in 911) a complete subalgebra C c B such that 1 C I < K and that x E 911[G n C]. Let us consider the Boolean-valued model 911' and the corresponding forcing relation IF. There exists a name x E 911' and a condition p E C n G such that (42.27)

p

ik x E "'(0,1)

and x

4 91 and

(911[x] 1 cp(x))

Since P"(C) is countable in %1[G], let D o , D . .,D,,. . . be an enumeration (in 9li[G]) of all open dense subsets of C in 911. We shall construct conditions ps E C, for all finite 0, 1-sequences s, as follows: Let p D I p be such that p D E D o . Given p s , there exists n, E o such that p , does not decide x(&) (because p IF x r$ 919 and we let pso and psl be such that pso 11x(Q = 6, and p,, 11.(is) = i; moreover, we choose pso and psl so that both are in the open dense set D,, where k is the length of s. For every z E "{O, l}, let G , = { p E C : p 2 p s for some s c z}. Clearly, G , n D, # 0 for every n, and hence G , is an 91-generic ultrafilter on C. Let f(z) = iG,(x)be the interpretation of x by G , . Since G , is generic, and by (42.27). we havef(z) E A. Thusfis a function from "(0, 1) into A. It follows from the construction off that f is one-to-one and continuous. Thusf["(O, l}], the one-to-one continuous image of a perfect compact set, is a perfect subset of A.

7.

550

DESCRIPTIVE SET THEORY

43. THE AXIOM OF DETERMINACY

we associate the following game G A ,played by With each subset A of two players I and 11. First I chooses a natural number a o , then I1 chooses a natural number bo, then I chooses a,, then I1 chooses b,, and so on. The game ends after o steps; if the resulting sequence ( a o , b o , a b . . .) is in A, then I wins, otherwise I1 wins. A strategy (for I or 11) is a rule that tells the player what move to make, depending on the previous moves of both players. A strategy is a winning strategy if the player who follows it always wins. The game G , is determined if one of the players has a winning strategy. The axiom of determinacy (AD)states that for every A C _ “‘0, the game G, is determined. The axiom of determinacy would be an attractive hypothesis if it were not for the fact that it contradicts the axiom of choice (see Lemma 43.1). In a sense, some consequences of the axiom of determinacy for sets of real numbers are more desirable than the consequences of the axiom of choice; for instance, AD implies that all sets of reals are Lebesgue measurable. However, the consequences of AD for cardinal arithmetic are quite undesirable: For instance, AD implies that all K,, 3 I n < w, are singular cardinals. More interesting is the relation between determinacy and the theory of large cardinals. It has been established that AD implies existence of models with many measurable cardinals, and thus in order to prove the consistency of AD with ZF, one has to make some large cardinal assumption (consistency of AD is an open problem). A more plausible hypothesis than AD, widely used by descriptive set theorists, is projecriuedeterminacy (PD)which states that G, is determined for every projective set A c Oo.Although consistency of PD is also an open problem, PD does not seem to contradict the axiom of choice and has pleasing consequences in descriptive set theory.

,, ,,

Some Consequences of the Axiom of Determinacy

First some definitions: Let A C_ Oo be given and let G , denote the corresponding game. A play is a sequence (a, boa , b , ...) E Ow; for each n, a,is the nth moue of player I and b, is the nth move of player 11. A strategy for I is a function a with values in w whose domain consists of finite sequences s E Seq of even length. The player I plays ( a o b o a ,b , ...) b y the strategy a if a, = a(@), a , = o((aobo)), a2 = a((ao b, a, b , ) ) ,and so on; it is clear that if I plays by a, then the play is determined by a and the sequence (b, : n E 0).We denote the play by a * b. A strategy a is a winning strategy for I if (43.1)

{a * b : b

E

. Y} !z A

in other words, if all plays that I plays by a are in A. Similarly, a strategy for I1 is a function T with values in o,defined on finite sequences s E Seq of

43.

55 1

THE AXIOM OF DETERMINACY

odd length. If a E . 1 and if T is a strategy for 11, then a * z denotes the play in which I plays a and I1 plays by T . A strategy T for I1 is a winning strategy if '

(43.2)

{ a * ~ : 1a' ) g ~ .1'- A

Exercise 43.1.

(a) The f'unctionJ(h) = Q * b is a one-to-one continuous function. (b) The set {o * b : h E l } contains a perfect subset.

[We shall sometimes consider games G, whose moves are not natural numbers but elements of an arbitrary set S . A play is then a sequence p E "S, and the result of the game depends on whether p E A or p $ A (here A is a subset of "S). The other relevant notions are defined accordingly.] Since the number of strategies is 2'O, an easy diagonal argument shows that the axiom of choice is incompatible with the axiom of determinacy: Lemma 43.1. Assuming the axiom of choice, there exists A E "w such that the game G Ai s not determined.

Proqf Let {a, : a < 2'O} and { T , : a < 2'O) enumerate all strategies for I and all strategies for 11. We construct sets X = { x , : a < 2'0) and Y = {y, : a < 2'O}, subsets of 1 as follows: Given { x E:, 5 < a } and { y E:, 5 < a}, let us choose some y , such that y, = a, * b for some b and y , $ { x r : 5 < a } (such y , exist because the set {a, * b : b E . t - ! has size 2'O); similarly, let us choose x , such that x, = a * T , for some a and x , # { y c : I a}. It is clear that the sets X and Y are disjoint, that for each a there is b such that a, * h $ X, and there is a such that a * T , E X . Thus neither I nor I1 has a winning strategy in the game G,, and hence G, is not determined.

'.

<

In contrast with this lemma, the axiom of determinacy implies a weak form of the axiom of choice: Lemma 43.2. The axiom of determinacy implies that every countable family of nonempry sets of real numbers has a choice function.

Proof: We prove that if Y = {X, : n E w } is a family of nonempty subsets of. .I -, then there exists f on 2" such that f ( X , ) E X, for all n. Let us consider the following game: If I plays ( a o a l a2 ...) and I1 plays ( b q b i b z ...), then I1 wins just in case b E X a o . It is clear that I does not have a winning strategy: Once I plays a , , the player I1 finds some b E X,,, plays b = (b,b,b, ...) and wins. Hence I1 has a winning strategy T , and we can define f on S as follows: f(X,) = T * ( n o 0 0 ...). H [Regardless of the above lemma, we assume the principle of dependent choices throughout this section.] The first theorem of this section shows that the axiom of determinacy

552

7.

DESCRIPTIVE SET THEORY

eliminates some of the “undesirable” counterexamples that can be obtained if one uses the axiom of choice. Theorem 102. Assume the axiom of determinacy. Then

(a) E w r y set of r e d s is Lebesgue measurable. (b) Eaery set of r e d s has the property of Baire. ( c ) Every uncountable set of reals contains a perfect subset. Let us deal first with the Lebesgue measure. We shall prove the following lemma: Lemma 43.3. Assuming AD, let S be a set of r e d s such that e i w y measurable Z c S is null. Then S is null.

Recalling Lemma 39.7, it is easy to see that Lemma 43.3 implies that every set X is Lebesgue measurable: Let A 3 X be a measurable set with the property that every measurable Z c A - X is null. Then A - X is null and hence X is measurable. Thus let S be a set of reals with the property (43.3)

if Z

G

S is Lebesgue measurable, then Z is null;

we shall use AD to show that S is null. It is clear that we can restrict ourselves to subsets of the unit interval; thus assume that S c [0, 13. In order to show that S is null, it suffices to show that the outer measure p*(S) is less than any E > 0. Thus let E be a fixed positive real number. The Covering Game

Given S and E , let us set up a game as follows: If (aoa a 2 . . .) is a sequence of 0’s and l’s, let a be the real number (43.4)

For each n E w, let GI, k = 0, 1, 2, ..., be an enumeration of the set K n of all sets G such that (43.5) (i) G is a union of finitely many intervals with rational endpoints; (ii) p ( G ) I~ / 2 ~ (’).” + The rules of the game are that player I tries to play a real number a E S , and player I1 tries to cover the real a by the union uIp=o H nsuch that H , E K , for all n. More precisely, a play

>

( a o b o a , b , ...

43.

553

THE AXIOM OF DETERMINACY

is won by player I if (i) a, = 0 or 1, for all n ; (ii) a E S; and (iii) a # IJ;=, Gin.

I claim that player I does not have a winning strategy in the game. To show this, notice that if 0 is a winning strategy for I, then the functionfthat to each b = ( b o b l b2 ...) E . 1 ' assigns the real number a = f ( h ) such that (ao 6, a , b , ...) = (i * b is continuous and hence the set Z = f [ . i '1 is analytic and hence measurable. Moreover, Z G S, and therefore Z is null. Now a null set can be covered by a countable union H , such that H,E K, for all n (see Exercise 39.8), and therefore there is b E . i . such that if 11 plays ( b , b,b2 ...) and I plays by 0, then I1 wins. Thus 0 cannot be a winning strategy for I . Assuming AD, the covering game is determined, and therefore player I1 has a winning strategy. Let T be such strategy. For each finite sequence s = ( a o , .. ., a,) of 0 s and l's, let G , E K , be the set G:., where ( b o , ..., 6,) are the moves that I1 plays by T in response to a o , . . ., a,. Since t is a winning {G, : s c a} and hence strategy, every a E S is in the set

u;=,

(43.6)

sG

u

u {G,:

u

co

s E Seq{O, 1)) =

lJ

G,

n = l ssnl0. 1)

Now for every n 2 1, if s E "(0, l}, then p(G,) I ~ / 2 and ~ " hence

u

P( S E " ( 0 .

E

&

G , ) I#"= 2"

I)

It follows that p(u,"= UsEnlO, G,) I 42" = E and thus p * ( S ) I E . Since > 0 was arbitrary, S is null. This completes the proof of Lemma 43.3 and also of part (a) of Theorem 102. Next we consider the property of Baire:

E

The Game G!* Let X be a subset of the Baire space . ip, and let us consider the following game G!*: Player I plays a finite sequence so E Seq; then I1 plays a proper extension to I> so; then I plays s 1 2 t o , etc.: (43.7)

so c to c s1 c f l c

.'.

The sequence (43.7) converges to some x E . C'. If x E A, then I wins, and otherwise I1 wins. First we verify that this game can be reformulated as a game G Aof the kind introduced at the beginning of the section (i.e., when the moves are natural numbers). Let u k , k E N,be an enumeration of theset Seq. If ( a o b o a l b l...) is a sequence of numbers, then consider the sequence (43.8)

uao,

ubo?

ua13

ubl,

554

7.

DESCRIPTIVE SET THEORY

and let A be the set of all ( a o b o a l b , ...) E . I'such that: Either there is n such that uao c ub,,

"'

u,. $ ubm

or the sequence (43.8) is increasing and converges to some x E X. It is clear I wins the game GZ* if and only if I wins the game G A. Thus if AD holds, the game G;* is determined, for every X G . V . We will use the **-game to show that every X c . V has the Baire property. Lemma 43.4. Player I1 has a winning strategy in Gf* ifand only i f X is meager. Proof. Let Y be the complement of X in ..( '. For each s E Seq, U s denotes the basic open set {x E . I ' : s c x}. (a) If X is a meager set, then there exist open dense sets G,, n E N, such that Y2 G,. It is easy to find a winning strategy 7 for 11: If I plays so, let to = SO)) be some to 3 so such that U f 0c Go; such to exists because Go is dense. Then if I plays s1 3 t o , let t l = 7 ( ( s 0 to sl)) be some t l 3 s1 such that U , , c GI, and so on. It is clear that every such play so c to c s1 c ... converges to x E G,, and hence 7 is a winning strategy for 11. (b) Conversely, assume that I1 has a winning strategy T. A correct position is a finite sequence (so, t o , . . . , s,, t,) such that so c t o c ... c t , and t o = ~((s,)), t , = r((so to s,)), etc. We shall first prove the following claim: Let x E .1" and assume that for every correct position p = (so ... r , ) with t , c x there exists s 3 r, such that r(p"s) c x. Then x E Y. To prove the claim, let x satisfy the condition. To begin, there exists so such that 7(.s0) c x; let to = SO)). Then there exists s 1 3 t o such that f l =t((sotO.sl))cx;thenthereiss2x t , s u c h t h a t r ( ( s o t o s l t l s 2 ) ) c x ; a n d so on. The sequence so c to c sl c t l c ... converges t o x and is a play in which I1 plays by 7. Hence x E Y. For every correct position p = (so . . . t,), let

n;l"=o n:=o

F,

= {x E . 1.'

: t , c x and (Vs

3

t , ) r(p"s) @ x)

By the claim, for every x E X there is a correct position p such that x E F,; in other words,

X

G

u {Fp

: p is a correct position}

It is easy to see that for each p, U,, - F , is an open dense set in U,";hence F, is a closed rare set. The number of correct positions is countable and hence X is meager. H Corollary. Player I has a winning strategy in GZ* if and only iffor some s E Seq,

U s- X is meager.

Proof. Note that I has a winning strategy in G*,*if and only if there exists s E Seq (the first move of 1 in G*,*)such that player I1 has a winning strategy in

43.

555

THE AXIOM OF DETERMINACY

,

the following game Gt:?,: I plays t o 3 s, I1 plays so 3 t o , I plays t 3 so, etc.; and I wins if t o c so c t c ... converges to x E U s- X.By the lemma, I1 has a winning strategy in this game just in case U s- X is meager. H

,

Now part (b) of Theorem 102 follows. I f X G . V, then since G*,* is determined, either X is meager or for some s E Seq, U , - X is meager. Thus let X G . t " be arbitrary. If X is meager, then X has the Baire property. If X is nonmeager, then let G = {Us: U s - X is meager}. Clearly, G - X is meager, and X - G must be meager too because otherwise there would be s such that U s - (X - G ) is meager, which contradicts the definition of G. It follows that X has the Baire property.

u

Now we turn our attention to part (c) of Theorem 102. We will use AD to prove that every uncountable set in the Cantor space C = "(0, I} has a perfect subset. We consider the following game.

The Game G i Let X be a subset of "(0, 1). The game G*,is defined as follows: Player I plays a sequence so E Seq{O, l} of 0's and 1's (possibly the empty sequence), then player I1 plays no E (0, l}, then I plays s , E Seq(0, l}, and so on. Let x = so "no "s, "n " . . . . Player I wins if x E X,and I1 wins if x 4 X.The game G i can be reformulated as a game G , , for some A E "w.

,

Lemma 43.5. If11 has a winning srrategy in Gg, then X is countable. Proof. The proof follows closely the proof of Lemma 43.4b. Let T be a winning strategy for 11. A correct position is a finite sequence (so, no, . . ., sk, nk) such that no = ?((so)). n , = ?((so, n o , s,)), etc. By exactly the same argument as in Lemma 43.4, we get the following claim: Let x E "(0, 1) and assume that for every correct position p = (so ... nk) if sonnon...nnkc x, then there exists s E Seq(0, 1) such that ~ ~ " n ~ " ~ ~ ~ " n , " s " r ( p "cs x. ) Then x 4 X. I t follows that X E {F, : p is a correct position}, where

u

F , = {X E c :Son'""nk

C X

and ~S(SOn'""nknSnT(pnS)$ X)}

The lemma will follow if we show that each F , has exactly one element x E C. This element x is uniquely determined as follows (because each x ( m ) is either 0 or l):firstly,forsomeI~N , ( x ( O ) ,..., x(1- 1)) = so"no"~~~"n,;thenx(l) = 1r(pnO),x(l+ 1) = 1 - t ( p " ( x ( l ) ) ) , x(I + 2) = 1 - s(pn(x(l),x(I+ l))), and so on. H Now part (c) of Theorem 102 follows. If X G C i s uncountable, then I1 does not have a winning strategy in G i ; and since the game is determined, I has a winning strategy 6. For each x = ( n o , n,, . . .) E C,let F(x) E C denote the 0, 1-sequence n

,?

n

so no s, n l " . . .

7.

556

DESCRIPTIVE SET THEORY

where so = a(0), s1 = a((sonno)), sz = a ( ( s o n n l n s l n n l ) )etc. , The functionf is continuous and one-to-one, and hencef[q is a perfect set. But X 2 j [ q and hence X has a perfect subset. Exercise 43.2. I has a winning strategy in G+,if and only if X has a perfect subset. I 1 has a winning strategy if and only if X is countable.

We proved in Section 41 that if K l = K4'"' for some a c w, then there is an uncountable set without a perfect subset. Thus if every uncountable set of reals contains a perfect subset, then every Kf'"]is countable, and it follows that for every a c w, K1is an inaccessible cardinal in 4.1. (Because if K 1= K;!"',, then there exists x s w such that Kktal is countable in q a , x], and hence for some b c_ w, K 1= Kftbl.) Thus we have: Corollary to Theorem 102. IfAD holds, then K l is inaccessible in L[a],for every a c w. AD and Large Cardinals To illustrate the relationship between the axiom of determinacy and the theory of large cardinals, we shall show that AD implies that K 1and K 2are measurable cardinals. We recall that measurability of successor cardinals (in ZF without choice) is a plausible property (see Theorem 90). Theorem 103 (Solovay). The axiom of determinacy implies lhaf (a) K1 is a measurable cardinal, and moreover, the closed unbounded jilter over K1 is an ultrajilter. (b) K2 is a measurable cardinal. We shall first show that AD implies that w 1 is measurable. We already know that w 1 is inaccessible in every q a ] , a G w. Let us consider the following partial ordering of the Baire space: x
(43.9)

iff

x ~ q y ]

and the corresponding equivalence relation (43.10)

x=y

iff

x
and y < x

We say that A G .V is =-closed if (x E A and y = x) implies y E A. Note that the collection of all =-closed sets in .V is a complete Boolean algebra. If xo E .V, then we let (43.1 1)

cone(x,) = {x E . V :xo a x } = { x :xo E qx]}

and call cone(x,) a cone. Clearly, every cone is =-closed. Let

9, ={AE

: A contains a cone}

43. THE AXIOM OF DETERMINACY

557

We claim that Fcis a a-complete filter on B. Let A o , A I, . . ., A,, . . . be elements of 3,. For each n, we choose x, E 1 such that A, 2 cone(x,). Let x E t . be defined as follows: x ( r ( n , m))= x,(m), for all n, m E N (where r is the canonical r: u x w -,w). It is clear that for each n, x, E L[x] and hence cone(x) E cone(x,) G A,. Thus A, is in T,. '

n;=,,

Lemma 43.6. AD implies thatfor every =-closed A c 1 ., either A or its complement contains a cone. Hence S, i s a a-complete ultrajilter on .&.

Proof: We show that if I has a winning strategy in the game G,, then A contains a cone (and similarly, if I1 has a winning strategy, then - A E .FJ. Let a be a winning strategy for I. It suffices to show that A contains the cone {x E .I. : d E fqx]}. Let x E 1 " be such that a E 4x1. Then a = a * x is in A because a is a winning strategy. Clearly, x E aa], and because E a x ] , we also have a E 4 x 1 and hence x = a. Since A is r-closed, we have x E A.

Now we can use AD to find a nonprincipal a-complete ultrafilter U over w l . For each x E i',letf(x) = N';'"l;f(x) is a countable ordinal. Note that if x = y, then f ( x ) = f ( y ) , and hence for every X c ul,the set f - I(X) c i is =-closed. Let

u = {X c w1 : f - I(x)E ,FC} Since ,Scis a c-complete ultrafilter on 1,U is a a-complete ultrafilter over w It remains to show that Lr is nonprincipal. But for every a -= wlr if x E 1" is such that N:'"] = a, then there is y * x such that h';'' > a ; and hence

f - 1 ( { a } )#

,gc.

Thus AD implies that u1is a measurable cardinal.

Lemma 43.7. AD implies that for every S

n:.

Corollary. AD implies that for every S

G ul, the

G w1 there

set {x E WO : (Ix((E S} is

is a

Gw

such that S

E

43 ..

[Proof of Corollary: Lemma 41.4.1 Proof of Lemma 43.7. If x E 6', then for each n E N we let x, that x,(m) = x(r(n, m ) ) for all m E N . ~

E

I

1.' be such

The Solovay Game Let S E wl. Player I plays a = (a(O), a(l), ...), and I1 plays b = (b(O), b( l), . . .). If a 4 WO, then I1 wins; if a E WO, then I1 wins just in case {a E S : a IIlal[)E (((6,( 1 : n

E

wf E S

We claim that I does not have a winning strategy in the Solovay game. Let be a winning strategy for I ; for each b E .Af, letf(b) be the a E .A" such that

558

7.

DESCRIPTIVE SET THEORY

* b. The setf[. 1'3 is a 72; subset of WO, and by the boundedness lemma, there is a < o 1such that Ilf(b)ll < a for all b E . t .. Hence let b E i p be such that { llbn 1 : n E o}= S n a. Then a * b is a play won by player 11, and hence a cannot be a winning strategy for 1. Now the lemma follows: Let S E 0 , . By AD, player 11 has a winning strategy T in the Solovay game. For each a, let g(a) be the b E . 1 " such that (a(O), b(O), . ..) = a * 7 . It follows that for each a E WO,

(a(O),b(O), a( l), b( l), .. .) = a

llall E

s

ifl

* ( a = (s(a))J

and consequently the set {x E WO : llxll E SJ is

n:.

H

We can now complete the proofofTheorem 103a.If X G w , , then X E y a ] for some a G o.Since K, is a measurable cardinal, a # exists, and it follows that either X or w , - X contains a closed unbounded subset (see Exercise 30.3). Thus the closed unbounded filter over o1is an ultrafilter. By the countable axiom of choice, the closed unbounded filter is acomplete, and we therefore conclude that AD implies that the closed unbounded filter over w1 is the unique a-complete normal ultrafilter over 0,.

Proof of Theorem 103b. We shall now show that, assuming AD, N 2 is a measurable cardinal. For each x E 1', let f ( x ) denote the successor cardinal of the (real) cardinal K , in 4x1: f ( x ) = ((N,)+)L["l

If x G w, then because x* exists,f(x) is an ordinal less than K 2 .Moreover, if x = y, thenf(x) =f(y), and hencef- ,(X) is =-closed for each X G w 2 .Thus let us define an ultrafilter U over w 2 as follows:

u = { X E w2 :f-l ( X ) E .Fc} We wish to show that U is an X,-complete nonprincipal ultrafilter over w 2 . Since .Fc is a-complete, U is a-complete. It is also easy to see that U is nonprincipal: If a < w 2 and f(x) = a, then there exists S E w l such that a is not a cardinal in L [ S ] ;then by the corollary of Lemma 43.7 there is y E . I such that x E t[r,] andf(y) > a. Hencef- * ( { a ) )does not contain a cone. I t remains to show that U is &complete. Since U is a-complete, it suffices to show that if '

(43.12)

X, 2

x,2 . . .2 X I z...( a < w l )

is a descending sequence of subsets of w 2 such that each f- , ( X , ) contains a cone thenf- ,((I,
43.

THE AXIOM OF DETERMINACY

559

We claim that I does not have a winning strategy in this game: If o is a winning strategy for I, then the set of all a E I . that I plays by n against all possible b E f *, is a Xi subset of WO and hence there is a such that llall < a for all these a's. Now I1 can beat I simply by playing some b E 1 . such that cone(b) z f - ,(XI). Thus I1 has a winning strategy T, and we intend to show that f - I(X) contains the cone {x E I ' : T E ax]}. Let a < w1 and let x E I . be such that T E a x ] ; we want to show that f(x) E X,. Let P I be the standard notion of forcing that collapses a onto w : The conditions are finite sequences of ordinals less than a. Since S, is inaccessible in 4~1% 4 x 1 has only countably many subsets of P I , and therefore there exists an 4x1-generic filter G on P , . Let a E WO be such that llall = a and that 41.1= q C ] and let y E I . be such that ~ I J=' ]4x][G] = Yx][a]. Since G is generic on P I over 4x1, all cardinals in a x ] greater than a are preserved in qx][G]. In particular, (S:)L[xl is preserved and hencef(y) =f(x). Now if I plays a = (a(O), a( I), ...) and if I1 plays against a by his winning strategy T, I1 produces b = (b(O), b(l), ...) such that cone(b) s f - l ( A I ) .But since b E q ~a ], and T E a x ] , we have b E a x , a] = qy] and therefore y E cone(b). I t follows that f ( y ) E X I ;and because f(x) =f(y), we have f(x) E X,,as we wanted to prove. This completes the proof of Theorem 103b. Already Theorem 103 indicates that AD is a very strong large cardinal axiom: The fact that the closed unbounded filter over w I is an ultrafilter has alone strong consequences,namely existence of models with many measurable cardinals. The AD also implies existence of a model of ZFC having a measurable cardinal that is a limit of measurable cardinals. And S, and S, are not the only measurable cardinals if AD holds: There are many more measurable cardinals (but not S , ; all S,, 3 5 n < to are singular and have cofinality K,; the next two measurable cardinals are K,,,,and S,,, ,).

Projective Determinacy Since the assumption that the game G, is determined for every A c I is inconsistent with the axiom of choice. and perhaps too strong to be plausible, it is natural to ask whether one can assume that G, is determined whenever A is a simple set; i.e., if A is definable or has a simple structure. One immediate observation is that all open games are determined. '

Lemma 43.8. !f A G I is '

mi

open sef, then G,4is rleterniined.

Proqf Player 1 plays ( a o a , ...), player I1 plays ( h o b , ...), and 1 wins just in case ( a o h o a ,h , ...) E A. Let us assume that player I does not have a winning strategy, and let us show that I1 has a winning strategy. The strategy for I1 is as follows: When I plays a,, then because I does not have a winning strategy,

7.

560

DESCRIPTIVE SET THEORY

there exists b, such that the position (aobo)is not yet lost for 11. That is, 1 does not have a winning strategy in the game Gyd0)that starts at (a, b,), in which I plays (a1a2 ...), I1 plays ( b , b 2 ...), and in which I wins just in case (aoboal bl ...) E A. Let I1 play such bo . When I plays a then again, because I1 is not yet lost at (a, b,), there exists b l such that I1 is not yet lost at (ao boal b , ) . Let I1 play such b,. And so on. We claim that this strategy for I1 is a winning strategy. Let x = (aob,al b , ...) be a play which I1 plays by the above strategy; we want to show that x # A. If x E A, then because A is open, there is s = (a,b, ... a, b,) c x such that U s c A. But then it is clear that I1 is lost at s ; a contradiction. W The same argument (interchanging the players) would show that every closed game is determined. Or, we can show that every closed game is determined as follows: I f A is closed, then I has a winning strategy in G, ifand only if there is a, E N such that I1 does not have a winning strategy in the open game GY in which I1 makes a first move 6 0 , then I plays a 1, etc., and 11 wins iff (ao b o a , ...) is in the open set. 1. - A. Since G?O is determined for all a, E N, G Ais determined. A similar argument shows Exercise 43.3. Let n > 0. I f G Ais determined for every Xf set A, then G, is determined

for every Ili set A, and vice versa.

A useful conjecture for descriptive set theorists is the axiom of projectiw determinacy (PD) which states that G, is determined for all projective sets A E f ’. Unlike AD, the axiom PD does not seem to contradict the axiom of choice. The reason projective determinacy is so useful in descriptive set theory is that it enables one to prove various properties of projective sets and decides various questions on projective sets which are undecidable in ZF + DC. For instance, the proof of Theorem 102 shows that PD implies that all projective sets are Lebesgue measurable, have the Baire property, and are either countable or contain a perfect subset. Another application of PD is a generalization of the uniformization theorem (Theorem 97): if PD holds then every fl:n+l relation can be uniformized by a rIi,,+ function. A similar generalization applies to the reduction principle (cf. Exercise 40.3). Let us say that all X: (rl!,, A!,) games are determined if G, is determined for every E,!(ll,!,A:) set A G . f ’. The following two theorems solve the problem of determinacy of Borel and analytic games. Theorem 104 (Martin). All Borel games are determined. Theorem 105. All analytic games are determined xG

0.

if and only (f x ’ exisrs .for all

43.

561

T H E AXIOM OF D E T E R M I N A C Y

Corollary.

If there

exists a measurable cardinal, then all analytic games are

determined.

[The “ i f ” part of Theorem 105 is due to Martin, the “only i f ” part is due to Harrington.] Theorem 105 shows that PD is essentially a large cardinal axiom. In fact, it is known that determinacy of A; games implies existence of a model with many measurable cardinals (a measurable cardinal that is a limit of measurable cardinals). Consistency of PD is an open problem; the best result to date is this theorem of Martin: A; determinacy follows from the existence of an elementary embeddingj : V -+ M with the property that ‘M c M for every A < sup{ti,j ( K ) , j ( j ( K ) ) , ...), where K is the least ordinal moved. (Compare this with Exercise 28.9 and the concluding remark in Section 33.) Theorems 104 and 105 are stated here without proof. I shall prove only the corollary of Theorem 105; the proof can be converted without much effort into a proof of the “ i f ” part of Theorem 105. The proof features an important technique of determinacy, the use of an auxiliary game. Let K be a measurable cardinal and let A c 1 be an analytic set. We want to show that the game G , is determined. Let us use the tree representation of analytic sets. There is a tree T c Seq, such that for all x E 1 ., x E A H T ( x ) is not well-founded ’

(see (40.23)). Let 4 be the linear ordering of the set Seq that extends the partial ordering 2 : If s, t E Seq, then s 4 t either if s 2 t , or if s and t are incompatible, and s(n) < t(n) where n is the least n such that s(n) # t(n). Thus (see 40.26)

<

T ( x )is not well-ordered by We also recall that T ( x )= {t : (s, t ) E T for some s E Seq} and that the first n levels of T ( x )depend only on %(n). For s E Seq, we let x

E

A

H

T(s)= { t : (u, t ) E T for some u G s}

then T ( x ( n ) )is exactly the first n levels of the tree T(x).We need some further notation. Let t o , t , , . . ., t , , . . . be an enumeration of the set Seq. If s E Seq is a sequence of length 2n, let K, be the finite set { t o ,. . ., t n - ,}n T ( s )and let k, be the size of K,. We shall now define an auxiliary game G*: Player I plays natural numbers a,, a,, a , , .. ., and player I1 plays pairs ( b , , h,), ( b , , h l ) , (b2, h 2 ) . ... where b,, b,, b2 , . . . are natural numbers, and for each n, h, is an order-preserving mapping from ( K s , <) into K where s = ( a , , b,, . .., an,b”), such that ho c h , E h2 G ... c h, E .... If player I1 is able to follow these rules throughout the game, then he (or she) wins. Otherwise, I wins. I t is clear that the game G* is determined: If I does not have a winning strategy, then he cannot prevent I1 from following the rules and thus I1 has a

7.

562

DESCRIPTIVE SET THEORY

winning strategy, namely his each move is to reach a position in which I does not have a winning strategy. (The argument is the same as in the proof of determinacy of open games; in fact, G * is an open game in a suitable topology.) If I1 wins a play in the game G*, then he has constructed an orderpreserving mapping h = U;=, h, of (T(x),<) into K, where x = (a,, b,, a,, b , , , , .); hence well-orders T ( x )and so x # A. Thus we can view the game G* as a variant of GA , but more difficult for player 11: I1 tries to make sure that x $ A, and in addition, he tries to construct an embedding of (T(x),<) in K. Hence it is fairly obvious that if I1 has a winning strategy in the game G*, then I1 has a winning strategy in G,: If t* is a winning strategy for I1 in G*, let t be ) 60 where (b,, h,) = r*((ao)); then as follows. When I plays a,, let ~ ( ( a , ) = when I plays a,. let s ( ( a , b , a , ) ) = b , where (b,, h , ) = r*((ao, (b,, ho), a , ) ) ; etc. Since G* is determined, it suffices to prove the following lemma in order to show that GA is determined:

<

Lemma 43.9. I has a winning strategy in G*, then I has a winning strategy in GA. Proof. Let a* be a winning strategy for I in G*. After 2n + 2 moves, the players have produced a sequence s = ( a , , b,, . . ., a,, b,), and I1 has constructed order-preserving functions ho E ... E h,; the strategy a* then tells player I what to play next. Let E be the range of h,; E is a finite subset of K, and in fact its size is k,. We observe that there is one and only one way I1 could have constructed h , , .. ., h, so that ran@,) = E; the reason is that h, is the unique order-preserving one-to-one function between (Ks,<)and E. Thus a* depends (as long as I1 plays correctly) only on s E Seq and the finite set E c K. For each s E Seq of even length, let F, be the following function from [ K ] into o: (43.13)

F,(E) = a*(& E )

Each F, is a partition of [ K ] ~ ’into o pieces; and because K is measurable, there ) exists a set H c K of size K homogeneous for each F,. Let us denote by ~ ( s the unique value of F , ( E ) for E E [HIks. We shall complete the proof by showing that a is a winning strategy for I in the game GA. Let x = (aob, a , b , ...) be a play in which I plays by cr. We shall show that x E A. Assume that on the contrary, x 4 A. Then (T(x),<)is well-ordered and has order-type c wl. Since H is uncountable, there exists an embedding h of (T(x), <)into H.Let us consider the following play of the game G*: I plays a,. Then I1 plays (bo, h,) where ho is the restriction of h to K C a o b oThen ). I plays a , and I1 plays (b,, h , ) where h , is restriction of h t o K<4d04,bl).And so on. We show that in this play, player I plays by the strategy cr*. Clearly, a, = a(@) = a*(@, fa). Then a , = cr((a,, b,)), and by the definition of cr it is

~ ~

44. SOME APPLICATIONS OF FORCING IN DESCRIPTIVE SET THEORY

563

clear that a ( ( a o b o ) )= a * ( ( a o b o ) , h[K,,,,,]) and therefore a , is a move according to a*. And so on: All the moves a o , a,, ..., a,, ... are by a*. This is a contradiction because a* is a winning strategy for I in G*, but the play we described is won by player 11. It follows that x E A and hence a is a winning strategy for I in the game G,. H This completes the proof of Z: determinacy assuming a measurable cardinal. This assumption can be weakened to the assumption that x # exists for all x G o.The above proof is then modified as follows: We play the auxiliary game as before; K is an uncountable cardinal. The definition of the auxiliary game is absolute for the model L [ T ] ,and it follows that either I or I1 has a winning strategy for G*, which is in q T ] . In particular, in Lemma 43.9, we may take a* E Y T ] .Then the collection { F , : s E Seqf, where F , is defined by (43.13), is in q T ] , and an indiscernibility argument shows that there is an uncountable set H G K of indiscernibles for q T ] such that each F , has the same value for all E E [HIks.The rest of the proof is the same. 44. SOME APPLICATIONS OF FORCING IN DESCRIPTIVE SET THEORY

This last section contains several consistency results in descriptive set theory. Since we use Martin's axiom and generic models, we work again in set theory with the axiom of choice.

Some Consequences of Martin's Axiom Theorem 106 (Martin-Solovay). If Martin's Axiom holds, then the union of less than 2K0null sets is null, and the union of less than 2"O meager sets is meager. Corollary 1. If MA holds, then both the algebra of Labesgue measurable sets and the algebra of sets with the Baire property are 2No-complete,and moreover, Lebesgue measure is 2"O-addirive, i.e., if K < 2"O and A , , a < K , are pairwise disjoint, then (44.1)

[See (27.10) for the definition of the sum on the right-hand side of (44.1).]

Corollary 2. If MA holds and if2H0> H I , then every able and has the property of Baire.

set is Lebesgue measur-

To deduce Corollary 1 from the theorem, we prove by induction on K < 2 " O that if A , , a < K , are Lebesgue measurable, then A = Ub<, A, is Lebesgue measurable. Given A , , a < K , let B, = A, A , , for each a < K. The sets B, are Lebesgue measurable (by the induction hypothesis), and being pairwise

u,<,

564

7.

DESCRIPTIVE SET THEORY

disjoint, all but countably many are null. It follows from the theorem that A= B, is Lebesgue measurable. The same argument proves (44.1) (see also Lemma 27.4), and the property of Baire is analogous. As for Corollary 2, we use Lemma 42.7. If A is Z\(a), then A is Solovay over q a ] (see Lemma 42.1 1 ) . Since (2No)L"4 = K4["1 I K1,it follows that the set of all reals that are not random over 401is the union of at most K 1null sets, hence null (by the theorem). By the corollary of Lemma 42.7, A is Lebesgue measurable. The Baire property is analogous.

ua
Proof of Theorem 106. First we prove that the union of less than 2"O null sets is A,. In null. Let K < 2"O and let A,, a < K, be null sets of reals. Let A = order to prove that A is null, it suffices to find, for each E > 0, an open set U 2 A such that p( U) 5 E. Let E > 0. We apply Martin's axiom as follows. Let P be the set of all open sets of measure < E , and let p E P be stronger than q E P if p 2 q. We claim that the notion of forcing ( P , 2 )satisfies the countable chain condition. It suffices to show that if W is an uncountable subset of P, then there are p, q E W ,p # q, such that p ( p u q) < E. Let S be the countable set ofall unions of finitely many open intervals with rational endpoints. If W c P is uncountable, then there is n E Nand an uncountable Z G W such that p ( p ) < E - I/n for all p E Z . For each p E Z, let p* E S be such that p* c p and p ( p - p * ) < l / n . Since S is countable, there exist p , q E Z , p # q, such that p* = q*. Then P ( P u 4 ) < E* For each a < ti, let D, = {p E P : A, G p}. Each D, is a dense subset of P: If p E P, then since A, is null, there exists an open set q 2 A, such that p ( p ) + p(q) < E , and hence p u q E D, and p v q 2 p. By Martin's axiom, there exists a filter G G P such that G n D, # 0 for all a < K. Let U = U { p : p E C}. It is clear that A G U and it remains to show that p ( U ) I E. We use the well-known fact (easy to verify) that if I! = { p : p E G}, then there is a countable H E C such that U = { p : p E H}.Thus if p( V )> E , there exist p l , . . ., p n E H such that p(pl u ... u p , ) > E. But this is impossible: Since G is a filter on P, we have p u q E G whenever p E G and q E G ; thus p1 u ... u pn E G and hence p(pl u ... u pn) < E. This completes the proof that the union of c 2"O null sets is null if MA holds. In order to show that the union of less than 2'O meager sets is meager, it suffices to show that the union of less than 2"O closed rare sets is meager. The following lemma will complete the proof:

ua
u

u

Lemma 44.1. Assume Martin's axiom. L4t ti < 2"O and let A,, a < ti, be closed rare sets of reals. Let A = A,. Then there exists a countablefamily of dense Hi. open sets H i , i = 0, 1, 2, ..., such that A is disjoint from

ul
ni.=o

Proof. We apply Martin's axiom as follows. Let P be the set of all finite

44. SOME APPLICATIONS

OF FORCING IN DESCRIPTIVE SET THEORY

565

such that (44.2) (i) each Ui is the union of finitely many open intervals with rational endpoints ; (ii) each Ei is a finite subset of ti; and (iii) for each i, U i is disjoint from E, A,.

u,

A condition p' = ((U',,, I?,), ..., ( U L , I?,,,)) is stronger than a condition p = ( ( U , , Eo). . . . ( U , , E n ) ) if

.

(44.3)

(i) rn 2 n ; and (ii) for each i I n, Ui

2

Ui and EI 2 E i .

I t is clear that this notion of forcing satisfies the countable chain condition: I f two conditions have the same U , , . . . , U , , then they are compatible, and there are only countably many sequences ( U o , . . ., U , ) . Let I,, , k = 0, 1, 2, . . ., be an enumeration of all open intervals with rational endpoints. We let, for each a < ti and all i, k = 0, I, 2, . . . (44.4)

D, = { p : p

=

( ( U o , E,), ..., ( U , , &))and a

E

Ei for some i I n ]

E i k = { P : p = ( ( U o , E o ) , ...t(U,,En))and Ui n l k # O )

Since each A , , a < ti, is rare, it is clear that for all i and k, every condition can be extended to a condition p E E , , and hence each Ei, is dense in P. Also. each D, is dense in P. By Martin's axiom, there exists a filter G G P such that G n D, # 0 for all a < ti, and G n Eik # 0 for all i, k E w. For each i = 0, 1, 2, . .., we let

Since Elk is a dense set of conditions, for all k. H , is a dense open set of reals. Now if ct < K , then because D, is dense, there exists i E (0 such that H i is disjoint from A , , and hence A , is disjoint from H , . Therefore A is disjoint from H,.

n:=,

n,71=,,

We showed in Section 27 (Corollary 2 of Lemma 27.9) that if Martin's axiom holds, then 2"" is not real-valued measurable. Using Theorem 106, one can get a better result: Corollary 3.

If MA holtl.~,then there is

110

real-valued measiiruhle cardinal

I 2"o. Proof If there is a real-valued measurable cardinal I 2"9 then there exists a a-additive measure v on R that extends the Lebesgue measure p (see the remark following Lemma 27.5). Theorem 106 implies that every set B G R such that v ( B ) > 0 has size 2"". Using this fact and u-additivity of v, the proof of Lemma 27.9 shows that there is no 2"O-scale; but MA implies that a 2%cale exists.

566

7.

DESCRIPTIVE SET THEORY

A set of reals A has strong measure 0 if for every sequence a. 2 a I 2 * . . 2 ... of positive reals, there exists a sequence of open intervals I,, n = 0, 1,. . .,such that length (I,) I a, and that A c U;=o I, .It is clear that every set of strong measure 0 is null, but not every null set has necessarily strong measure 0 (compare this with Exercise 39.8, where intervals are replaced by finite unions of intervals). For instance the Cantor set C (cf. Fig. 4.3) is null but does not have strong measure 0; if a. = a = & a2 = &, etc., then Ccannot be covered with open intervals of lengths a o , al, a 2 , ... . a, 2

4,

Exercise 44.1. If A E R contains a perfect subset, then it does not have strong measure

0. [Either prove directly, or use the fact that A contains a subset homeomorphic to the Cantor set and that a uniformly continuous image of a set of strong measure 0 has strong measure 0.1

Clearly, every countable set has strong measure 0. Martin's axiom generalizes this. Exercise 44.2. Martin's axiom implies that every set A E R of size < 2'O

has strong measure 0. [May assume 2N0> XI.Consider the forcing notion (P,< ) = (Seq, 2 ) which adjoins a Cohen generic x E Ow. Since I A I < 2N0, MA implies that there is x E P-generic over every Ya], a E A. Let r o , r l , . .. , r,, . .. be an enumeration of all rational numbers. Given positive reals a. 2 a l 2 .. . 2 a, 2 .. ., let for each n, I, be the interval I., for with center rx,,)and diameter a,. Use the genericity of x to show that a E each a E A.]

u."=,,

While every countable set has strong measure 0, the problem of whether only countable sets have strong measure 0 (Borefsconjecture) is independent of ZFC. The following two exercises show that the continuum hypothesis implies that there is an uncountable set of strong measure 0; and there is a generic model in which no uncountable set of reals has strong measure 0. Exercise 44.3. If 2'" = K l , then there exists an uncountable set E E R such that for every rare set F, E n F is at most countable. (E is called a Luzin ser.) More generally, MA implies that there is a set E of size 2'O whose intersection with every rare set has size < 2%. [Let F o , F , , ..., F , , ..., a < 2*, be an enumeration of all closed rare sets. Let E = (e, : c( < 2"O), where for each a, e, $ U p < ,F , . Each e , exists; the MA case uses Theorem 106.1 Exercise 44.4. Martin's axiom (and in particular the continuum hypothesis) implies that there is an uncountable set of reals of strong measure 0. [Let E be the set from Exercise 44.3. Let a. 2 a l 2 ... 2 a,, 2 ... be given positive reals. For each n, let Izn be the interval of length a2, around the nth rational. The set U= 12" is open dense and hence E - U is countable, E - U = {xo, xl. ..., x,, . . .). For each n, let 12,,+ be the interval of length a 2 , + , around x, .]

u:=l

44. SOME APPLICATIONS OF FORCING IN

567

DESCRIPTIVE SET THEORY

I shall mention one more consequence of Martin's axiom. Let g,,, be the complete Boolean algebra of Borel subsets of the unit interval modulo null sets. The algebra B, does not have a countable dense subset. (To see this, let [AJ, n = 0, 1,2, . . ., be nonzero elements of W,.For each n, let B , G A, be such that 0 < ,u(B,) I1/2"+', and let B = u2'oB,. The complement C of B has positive measure, and clearly [An] $ [C], for all n.) Exercise 44.5. Martin's axiom implies that every dense subset of .am has size 2u0. [Let K = 2 N ~Let . x, ,a < K , be an enumeration ofall reals. MA implies that for every a, { x a : /I2 a } has positive measure; let K , be a compact subset of {xa : /3 2 a } such that p ( K , ) > 0. If .d, has a dense subset of size < K, then since K is regular, there exists

W c K of size K and a set X of positive measure such that X - K, is null for all a E W. Hence every finite subset of ( K , : a E W } has nonempty intersection and so K, is nonempty; a contradiction.]

0,.

In contrast to Exercise 44.5, there is a model of set theory in which A?, has a dense subset of size < 2'O. Recall that a complete Boolean algebra is weakly (o, o)-distributive if it satisfies

(see (25.8)). The algebra A?, of Borel sets modulo meager sets is not weakly (0, o)-distributive (cf. Exercise 25.4). Exercise 44.6.* The Boolean algebra ,dmis weakly (o, w)-distributive. [Let un, (n,m = 0, 1, 2, . . .) be given. Clearly, the right-hand side is 5 the left-hand side. Conversely, let 0 # u < LHS, and let r = p(u). Let g: o + o be defined as

follows: g(n) is such that p(u-

Then u .

(n;=o

1

rn < e(n)u n - )

r <2n+2

urn,,,)# 0. Hence LHS I RHS.]

(Compare with Lemma 27.9.)

Two Examples on Measure and Category If Martin's axiom holds and 2'O > K 1, then every set of reals of size K is null and meager; in particular, the set R n L of all constructible reals is null ' is large and and meager. We shall now give an example of a model where 2O the set R n L is not Lebesgue measurable, and another example where 2O ' is large and R n L does not have the Baire property.

Lemma 44.2. If there exists a nonconstructible real, then: (a) R n L is either null or not Lebesgue measurable. (b) R n L is either meager or does not have the Baire property.

568

7.

DESCRIPTIVE SET THEORY

Proof. We use the fact that both measure and category are preserved under translation. (a) Let S be the set of all constructible reals in the unit interval [0, 11. Let a be a nonconstructible real. For each n > 0, let S, be the translation of S by a/n, S, = {x + (a/n): x E S}. The sets S, are pairwise disjoint, p ( S , ) = p ( S ) for all n, and S, is a bounded set. It is therefore clear that if S is measurable, then p(S) > 0 is impossible. (b) Let S be the set of all constructible reals. First we prove that R - S is not meager, Let a be a nonconstructible real and let S, = {x + a : x E S}; clearly, S n S , = 0. Thus R = (R - S ) u ( R - S,) u ( S n S,) = ( R - S ) u (R - S,), and if R - S were meager, then R - S, would also be meager, a contradiction. It follows that for any nonempty interval I, I - S is not meager: For each rational r, let A, = {x + r : x E I - S ) ; if I - S is meager, then each A, is meager, and R - S = {A, : r rational}. If S has the Baire property, then because U - S is not meager for any nonempty open set U , S is meager. H

u:=-,

u

Example I. A model where 2N0> h', and the set of all constructible reals is not Lebesgue measurable.

Let 1be a regular uncountable cardinal and let B be the measure algebra m) be the product measure space, where S is the defined in (34.3); i.e., let (S, P, product of 1x w copies of {O, 1). 9 is the least a-complete field of subsets of S containing all { r E S : t(a, n) = 0}, and m is the product measure, and let B be the measure algebra, B = 9/sets of measure 0. Let us consider the generic extension of the constructible universe by the measure algebra B. As we proved in Section 34, the generic extension qG] satisfies 2O ' = 1.We shall show that in U G ] the set of all constructible reals is not Lebesgue measurable. In view of Lemma 44.2, it suffices to show that the set of all constructible reak is not null. Thus assume that it is null and let I , , k = 0, 1, 2, . . ., be an enumeration (in L) of all intervals with rational endpoints. Let p denote Lebesgue measure. Assuming that 4G] C p ( R n L) = 0, there is a B-valued name X for a subset of w, and a rational E > 0 such that the Boolean value (44.5) I( { I , : k E X}contains all constructible reals, and has Lebesgue measure I E ~ (

u

is in G. We may assume, without loss of generality, that the Boolean value (44.5) is 1. For each k E A', let A, E 9 be such that

1

1'

=

44.

SOME APPLICATIONS OF FORCING IN DESCRIPTIVE SET THEORY

569

Let us consider (in L ) the product measure space (R, p ) x (S, 3,m ) with the product measure v = p x m. Let E E R x S be the set

u

n.

=

(lk

Ak)

k=O

We claim that v ( E ) I E . It suffices to show that for every k o , v ( u k < k o (1, x Ak)) IE . Let ko E N. By (44.5), for every condition a = [ A ] there exists a stronger condition c = [C] and a set Y c ko such that

x f Clearly, [c]I [Ak] if c 11 n Co =

and that p ( u k s 1,) I k E ko - Y and hence

E.

k

E

Y , and [ C ] .[ A t ] = 0 if

Thus the set of all [ C ] for which (44.6)holds is dense in the algebra Band hence V

UlkXAk I& (k
Since v ( E ) IE , the complement of E has positive measure and hence there exists, by Fubini’s theorem, a number x E R such that

m({f E S : (x, t ) $ E } ) > 0

It follows that there exists A

E 9 of

(44.7) (x, t ) 4 E, We shall show that

(44.8)

[ A ] I=X 4

positive measure such that for all t

u {Ik k :

E

E

A

X}

thus completing the proof. If (44.8) were not true, there would exist some k E Nand some C c_ A of positive measure such that x E 1, and C I1 CE X. But then [C] I [ A d and hence there is t E C such that (x, t ) E E, contrary to (44.7).

Example If. A model where 2N0> HIand the set of all constructible reals does not have the property of Baire. Let 1 be a regular uncountable ordinal and let P be the notion of forcing that adjoins I Cohen reals: A condition is a finite 0,l-function whose domain is a subset of 1. Let us consider the generic extension of the constructible universe by P. In YG],2N0= 1.We shall show that in LEG] the set of all constructible reals does not have the Baire property. In view of Lemma 44.2, it suffices to show that the set of all constructible

570

7.

reals is not meager. For every S let Gs = G n Ps.

GI

DESCRIPTIVE SET THEORY

(in L), let Ps = {p E P : dom(p) E S},and

Lemma 443. If 461 k R n L is meager, then there exists a countable S in L ) such that L[G,] k R n L is meager.

C

I

Pro05 Let It, k E N, be an enumeration of all open intervals with rational endpoints. If Y G ] k R n L is meager, then there exists a sequence (U, : n E N> E Y G ] such that for every n E N, LEG] k U , is dense open, and that R n L E (R - U,). Let A = {(n, k ) : I k E U,}, and let A be a name for A. Since P satisfies the countable chain condition, there is a countable S c 1 such that A is r.o.(Ps)-valued. It is easy to verify that if Uk = U , n LEG,], then (R - U',,). for each n E N, QG,] k Uk is dense open, and that R n L G Thus q G s ] k R n L is meager.

u;=,

u:=,

Since P , is countable, it suffices to prove the following lemma:

Lemma 44.4. If P is a countable notion of forcing in L and fi G is an Lgeneric filter on P, then 461 k R n L is not meager. Proof. I t suffices to show that if 4! = (U,: n E N> is (in q G ] ) a sequence of dense open sets of reals, then there is a constructible real a such that aE U , . Let 4V be a name for 42 and let us assume, without loss of generality, that every condition forces that each 42, is dense open. It is enough to find (in L) a real number x such that for each n and each p E P, there is q 5 p such that q ~f3 E en. Let f k , k = 0, 1, .. ., be an enumeration of all pairs t = (n, p ) where n E N ... of closed and p E P. Let us construct a sequence I, 2 I 2 ... 2 bounded intervals as follows: Let t k = (n, p ) . Since p If "71, is open dense, there is an open interval J G I k - with rational endpoints such that some q I p forces J E @,, . Let 1, G J . The intersection I , is nonempty; and if x is in it, then for each n and each p E P there is 4 I p such that q IF 1 E %,.

n;=,

n&,

A Nonconstructible A:-Set of Integers By Shoenfield's absoluteness theorem, every IIi or Ci subset of N is constructible. The model presented below (due to Jensen) shows that it is consistent with ZFC that there exists a nonconstructible subset of N that is A;. (Recall that another example of a nonconstructible A: subset of N is O#.) Example 111. A generic extension of L which has a nonconstructible A: subset of N.

Let us assume the axiom of constructibility, i.e., the ground model is the constructible universe L. We shall construct a notion of forcing P such that if G is an Lgeneric filter on P, then q G ] k there exists a nonconstructible hi-set A c N. The construction is a combination of perfect set forcing (Section 26) and

44. SOME APPLICATIONS

OF FORCING IN DESCRIPTIVE SET THEORY

571

arguments using the 0-principle (Section 22). Let us consider perfect binary trees: Seq(0, 1) denotes the set of all finite 0,l-sequences; a nonempty set p E Seq(0, 1 ) is a perfect tree if: (a) if s is an initial segment oft and r E p, then s E p ; and (b) for every s E p, there exists t 3 s such that both t ^O and t "1 are in p. We denote by [ p ] the set of all branches in p :

(44.9)

a

E

[p]

H

a

E

"{0, 1) and Vn(ir(n) E p )

The set [p] is a perfect subset of the Cantor space. The stern of a perfect tree is the maximal s E Seq{O, 1) such that for every t E p, either t E s or s E t. If p is a perfect tree and if s E p, we denote by p(') the perfect tree

(44.10)

p(') = ( t E p : t G s or t 3 s}

Assume that P is a set of perfect trees, partially ordered by E,such that if P and s E p, then p(') E P,and let G be an Lgeneric filter on P. Then there is a unique a E "(0, 1) which is a branch in every p E G; and conversely, G = { p E P : a is a branch in p}. Therefore L[G] = L[a],and we call a P-generic over L A set X G P is a couer (see Exercise 16.4) if for every p E P, there is a stronger 4 E P such that q I x for some x E X. (Clearly, every dense set is a cover.) Note that a E "{0, l} is P-generic over Ljust in case for every constructible cover X G P,a is a branch in some x E X. Let P x P denote the product of P with itself; i.e., (p', q') 5 (p, q ) means p' E p and q' E q. Similarly as for P , a generic filter G on P x P corresponds to a unique pair (a, b) such that for each (p, q ) E G, a is a branch in p and b is a branch in q. A pair (a, 6) is (P x P)-generic over L just in case for every constructible cover X c P x P, there is (x, y ) E X such that a is a branch in x and b is a branch in y. When we used perfect sets as forcing conditions in Section 26, we used Lemma 26.1 (the fusion lemma) which enabled us to apply the operation of fusion to perfect sets. Let T = { T(s): s E Seq{O, I}} be a collection of perfect trees such that for every s, p

E

(44.11) (a) T(s)is a perfect tree whose stem has length 2 length(s). (b) T(s^O)G T ( s )and T(s"1) G T(s). (c) T(s^O)and T(s"1)have incompatible stems.

If T satisfies (44.1 l), we say that T is fusionable and we let (44.12)

For each fusionable T, p = . f ( T )(thefusion of T) is a perfect tree; and for each r is the stem of ps = T(s),then p(') is stronger than both p and p s . We shall not use the set of all perfect trees as the notion of forcing; rather

s, if

512

7.

DESCRIPTIVE SET THEORY

we shall construct a set P of perfect trees with the property that if p E P and s E p , then pcs)E P. We shall construct P such that if a is P-generic over L then a is the only P-generic set in q a ] , and such that {n E N : a(n) = 1) is (in 4 .1) a A; subset of N. We shall construct P as the union of countable sets

Po c P1 c ... c PI E ...

(a < al)

of perfect trees. The construction uses the 0-principle (cf. 22.15):

( 0 ) There exists an a,-sequence (S,: a -= w l ) such that for every X c w , , the set { a : X n a = S,} is stationary. By Exercise 22.17, there is a sequence (S,: a c a,)that satisfies 0 such

that the function X W S , is A 1over L,,; let us fix such a sequence. Also, let us fix a A , over L,, function 5 that is a one-to-one mapping of L,, onto al. We shall now construct the sequence Po C PI G * * . E PI E '":

(44.13) Po = the set of all p g ) where po is the full binary tree, p o = Seq{O, l}, and s E p o ; PI = P, if a is a limit ordinal.

u

P,,

u Q,+ where Q,+ is a set of perfect trees defined as follows:

= PI

Let PI = { p i : n E a}; and let us consider the <,-least such enumeration. Let .TI and Y, be the following countable collections of subsets of P I and PI x P, respectively: (44.14)

X z contains: (a) all Q p .B 5 a, (b) all X E PI such that r [ X ] = S, for some /? I a.

!Y, contains: (a) Q, x Q, for all B 5 a ; (b) all Y E PI x PI such that r[ Y] = S , for some /? Ia. There exists a family { T,; n E a}of fusionable collections of elements of PI such that: (44.15) (a) T,(a) = p i for all n ; (b) for every X E XI,if X is a cover in PI, then for every n E Nand every h E N, there is k 2 h such that for each s E '{O, l}, there exists x E X such that T , ( s ) I x; (c) for every Y E Y,,if Y is a cover in PI x PI, then for every n, every m,and every h there is k 2 h such that for each s E "0, 1) and each I E '{O, I}; ifeither n # m or s # 1, then there exists (x, y) E Y such that (T,(s), T,(f)) I(x, y).

44. SOME APPLICATIONS OF FORCING A family { T', : n

E

IN DESCRIPTIVE SET THEORY

573

o}with properties (44.15a-c) is easily constructed because : n E w } the < ,-least such family, and

XIand !?II are countable. We denote

{c

let (44.16)

a

QI

c)for some n. and s

= {p'" : p = .F(

E

p}

uI<w,

E

We let P = P I . The following sequence of lemmas will show that if "(0, I ) is P-generic over L then in y a ] the set { n : a(n) = 1) is A:.

Lemma 44.5. For each a, QIt is dense in P I , I , and Q1+ x Q1,I is dense in x PIt1.

PIt1

Proof. I t suffices to show that below each p E P I there is some 4 E Q , , p E PI, then p = p i for some n, and .F(T,)E T,,(a) = p. H

If

Lemma 44.6. For each u, i f X E .TIis a corer in P I , then X is a cover in PI, ; if Y E Y l is a c o w in P I x PI, then Y is a corer in P I +I x PI+ Consequentlj~,if X E .TI is a corer in P I (if Y E Y 1is a cover in P I x PJ, then X is a corer in P (then Y is a cowr in P x P). Proof: Let X E .TIbe a cover in PI; we have to show that for each p(") E Q1, there is a stronger 4 E Q1+I such that 4 I .Y for some .x E X. Let p = .F(T',)and let ti E p. Let h = length(u). There is k 2 h such that n and k satisfy (44.15b). There is s E '{O, 1) such that u E T , ( s ) ; let r be the stem of T,(s). Then p(") I T , ( s ) and T,(s) 5 ?I for some x E X. A similar argument, using (44.15~).shows that if Y E !!I is,a cover in P I x P I , then Y is a cover in P I , I x P I , Since the sequences X I , 2 < wl, and !YI,o! < w,, are increasing, it follows by induction that X is a cover in every P,, < ( u I , and hence in P.Similarly for Y.

Lemma 44.7. P x P sarisjies rhe countable chain contlirion (and hence P also satisfirs the cotinruble chain condition). Proof: Here we use 0. Let us assume that Y c P x P is a maximal incompatible set of conditions in P x P and that Y is uncountable. Since each P I is countable, it is easy to see that the set of all a < (ul such that T[ Y n ( P I x P I ) ] = T[ Y ] n w I , is closed unbounded (T is the one-to-one function of L,,,, onto ( u I ) . Then it is not much more difficult to see that the set ofall Y < (uI such that Y n ( P I x P I ) is maximal incompatible in P, x P I , is closed unbounded (compare this argument with the ()-construction ofa Suslin tree in L). By 0. there exists a such that Y' = Y n ( P I x P I )is a cover in P, x P,and that T[ Y ] = S , . Therefore Y E 9 , and by Lemma 44.6, Y is a cover in P x P. I t follows that Y' = Y . Thus Y is countable. H

7.

574

DESCRIPTIVE SET THEORY

Lemma 44.8. (a) l'fa E "(0, l}, then a is P-generic ooer L ifand only iffor eoery a < w there is n E N such that a is a branch in 9( (b) l'fa # b E "(0, l}, then (a, b ) is (P x P)-generic over L ifand only iffor eoery a < w1 there exist n, m E N such that a is a branch in a(Pn) and b is a branch Tm). in 9(

c).

Proof. (a) Let a be P-generic and let a < wl. Since Q a + is dense (hence a cover) in P,+ and because Q a + E Z Z + Q a + is a cover in P. By the genericity of a, there exists q E Qm+ such that a is a branch in q. But q = pcs) where p = .S( for some n and s E p , and clearly a is a branch in p. Conversely, let us assume that the condition is satisfied. Let X c P be maximal incompatible; we wish to show that a is a branch in some x E X. By Lemma 44.7, X is countable, and (by the proof of Lemma 44.7) there is tl such that X E Zz.Let n E N be such that a is a branch in By (44.15b),there is k E N such that each T',(s), s E "0, l}, is stronger than some x E X. Since a is a branch in 9(T,,), it is clear that there is a unique s E k{O, 1) such that a is a branch in T,(s).But if x E X is such that 7'4s) E x, then a is also a branch in x. (b) The proof that the condition is necessary is analogous to (a). Thus let us assume that the condition is satisfied and let Y C P x P be maximal incompatible; we want to find ( x ,y ) E Y such that a is a branch in x and b is a branch in y. Again, there is a such that Y E Yz. Let n and m E N be such that a is a branch in and that b is a branch in .F(Tm). Let h E N be such ii(h) # 6 ( h ) ;by (44.15c),there is k E N such that for each s, t E k{O, I}, if either n # m or s # t , then ( K ( s ) , T,(t)) I ( x , y ) for some (x, y) E Y. There is a unique pair s, t such that u is a branch in T'(s) and h is a branch in Tm(t): It is clear that either n # m or s # t. Then a is a branch in x and b is a branch in y where ( x ,y) is some element of Y such that (T,,(s)* Tm(t)) I (-x, y). Corollary 1. l'f a and b are P-generic over L and a # b, then (a, h ) is ( P x P)-generic over L.

c)

.F(c).

J(c)

Corollary 2. Ifa is P-generic over L rhen y u ] k a is the only P-generic x over L.

Proof. If u # b and if both a and h are P-generic over L then by the product lemma 20.1, b is a P-generic over L[a] and hence b $ L[a].

Lemma 44.9. The set H = { a : a is P-generic oiler L) is I l l oiler HC. Proof. It follows from the construction of P that the function a H (71, : n E o) is A, over Lu;. Since Lo: is a Z, set over HC, the function is A, over HC. By Lemma 44.8, a E H c* (Va < 04) ( 3E w ) [a is a branch in .F( and hence H is I l l over HC.

c)]

44. SOME APPLICATIONS OF FORCING IN

DESCRIPTIVE SET THEORY

575

Corollary. I f a is P-generic ouer L a n d i f A = { n E N : a(n) = l}, then q a ] k A is a A; subset of N.

-

Proof: We have in a a ] :

(44.17)

nE A

++

( % a . I . ' ' ) [ a ~ H ~ a ( n11) = (Va E . +*) [a E H

+ a(.)

=

13

By Lemma 41.1,since H is ll, over HC, it is a ll: subset of. V . Hence (44.17) shows that A is A;. W A Long Il: Well-Ordering By Theorem 100, every 72: well-ordering of a set of reals has order-type less than w 2 .The model given below (due to Harrington) shows that it is consistent with ZFC that there exists a II: well-ordering of a set of reals whose order-type is as large as prescribed. Example IY. A generic extension ofL having a Il: well-ordering of a set of reals of order type 2 w 2 . A long ll: well-ordering is obtained as follows: Let 8 be an arbitrary ordinal. First we extend the constructible model generically by adjoining at least 8 Cohen reals: in L[G], 2O ' 2 8. In %I = L[G] (which has the same cardinals as L), we consider a well-ordering A s R x R whose order-type is 8. The following lemma then shows that there is a generic extension 9lI[H] of 91, with the same cardinals, in which A is a II: relation. Thus in 911[H], there is a II: well-ordering of a set of reals of order type 8. Lemma 44.10. Let 911 be a model ofZFC such that W? = w:. Let A be (in 911) a subset of There is a generic extension of 911 that preserves all cardinals and in which A is a H i set. The generic extension will be obtained by a two-step iteration of forcing, and both steps will use a variant of the almost disjoint forcing introduced in Sections 23 (Lemma 23.10) and 26 ((26.1)-(26.3)). Let {s, : n E w ) be an arithmetical enumeration of the set Seq. If x E . 1 ., let

(44.18)

S(X)

= {n E

0

: s, c x)

Each S(x) is an infinite set; and if x # y, then S(x) n S(y) is finite. Thus {S(x) : x E I '1 is an almost disjoint family of subsets of w . With each X c ww we associate the following notion of forcing P,: A condition is a pair p = (u, E ) where u is a finite subset of w, and E is a finite subset of X. A condition ( u , , E 2 ) is stronger than (ul, E l ) if

(44.19)

El

c E,,

u1 G

u 2 , and for every x

E

E , , S(x) n u2 G u1.

Note that if (u, E) and (u, F) are compatible, then ( u u u, E u F) is a condition. F) are incompatible, then u # 0 ;and since there are only countably many u's, P , satisfies the C.C.C. Also, if (u, E) and

(11,

7.

576

DESCRIPTIVE SET THEORY

Let G be a generic filter on P , , and let (44.20)

a = { n E w : n E u for some (u,E) E G}

It is easy to see that G = {(u, E) E P,: for every x E E, S(x) n a G u } and hence 911[G] = 911[a]. If x E X,then the set of all (u,E) such that x E E is dense in P, and hence there is (u, E) E G such that x E E. Since S(x) n u G u for every stronger condition (0, F) I(u, E), it follows that S(x) n a G u and hence S(x) n a is finite. If x E "w - X (in the ground model), then for every condition (u, E) and every k E u there is n 2 k such that n E S(x) and that (u u { n } , E) is stronger than (u,E); namely, since S(x) n S(y) is finite for every y E E, we take n L k such that n E S(x), and n # S(y) for ally E E. It follows that S(x) n a is infinite. Thus we have shown that for every x E Ow in the ground model, (44.21)

xE

X

iff

S(x) n a is finite

Let 911 be a model of ZFC such that cu'f = uk, and let A E 911 be a subset of '"w. We consider first the notion of forcing P that is the product of N l copies of P A : A forcing condition is a sequence P = (p, : a < wl)such that: for each a, p ,

E

P A , and { a : p , #

(a,0)) is finite.

A condition (p, : a < w l ) is stronger than (4, : a < w l ) if for each a, p , is stronger than q,. It is easy to see that the product of a finite number of copies of P Asatisfies the countable chain condition; hence by Lemma 23.1 1, P satisfies the C.C.C.Let G be an 911-generic filter on P and let for each a < wl, G, = { p , : p E G}, and (44.22)

a, = { n E w : n E u for some (u, E) E G,}

Clearly, 911[G] = Yl1[(a, : a < wl)]

Lemma 44.11. 9ll[G] sarisjes thatfor every x E Ow, (44.23)

xEA

iff

(Va < ol) S(x) n a, is finite

Proof. If x E A, then for every a, G, is an Yli-generic filter on P A ,and hence by (44.21), S(x) n a, is finite. If x E Ow - A, then, because P satisfies the c.c.c., there exists a < w 1 such that x E Yll[(ap:p < a)]. Then (by the product lemma 20.1) G, is an 91[(ap: p < a)]-generic filter on P A , and since x 4 A, (44.21) implies that S(x) n a, is infinite. H

Let (d," : a < u1and n E 0)be a constructible one-to-one sequence of constructible reals. Moreover, let us choose the sequence so that the function

44. SOME APPLICATIONS

OF FORCING IN DESCRIPTIVE SET THEORY

577

I,<,,,.In W[G] let

(a, n ) ~ d , is , A 1 over

(44.24)

B

=

{d,, : n

E

a,)

and let us consider, in 91[G], the notion of forcing P , . Let H be an 91[G]-generic filter on P , and let a = {n E w : n

(44.25)

For every x

E

E

u for some (u, E ) E

”w in 91[G], we have

xEB

iff

S(x) n a is finite

and therefore, for every a < w , and every n

(44.26)

H}

n

E

a,

iff

It follows from (44.26) that 911[G][H] = 91[~].

E

w, we have

S(d,,) n a is finite (a, : a < 0,) E 911[a]

Lemma 44.12. 91[a] sarisjies thatfor eoery x

E

and

therefore

“w,

(44.27) xEA iff (Va < w l ) S(x) n a, is finite Proof. If x E A, then S(x) n a, is finite for every a, by (44.23).Thus it suffices to show that if x E 911[a] is such that x E “‘w - A, then S(x) n a, is infinite for some a. The model 911[u] = 911[G][H] is obtained from 911 by a two-step forcing, first using P and then P,. We note that the model 911[a] can be obtained as a generic extension via the following notion of forcing: For each a < wl, let u, be the canonical r.0. P-valued name for a,. Let Q be the set of all pairs (p, 4) such that: (44.28) (i) p E P ; (ii) q = (u, E) where u is a finite subset ofw and E is a finite subset of w I x (I)such that for every (a, n) E E, p If ti E u,. A condition (p’, (u’, E’)) is stronger than (p, (u, E)) if p‘ Ip and for every (a, n) E E, u’ n S(d,,) G u. Since B = {d,, : n E u,}, it is clear that the generic extension by Q corresponds to the two-step extension by P and P , . The following is a trivial but important observation: If (p, 4 ) and (P, 4) are conditions in Q and if p = ( ( u , , 15,):a < o,),P = ((U,,E , ) : a < w,>, q = (u, E) and ij = (a, E ) ; and if ( p , q ) and (p, q) are compatible, then (((u~ u u,, E, u E , ) : a < w , ) , (u u ii, E u E ) ) is a condition in Q; let us denote this condition ( p v p, q v q). Let x be an r.0. Q-valued name for an element of “w. For each n E w, let W, be a maximal incompatible set of conditions in Q that force ti E S(x). Since Q satisfies the c.c.c., each W, is countable and therefore there exists ct < o 1such 0). that for every n E Q and every ( p , q ) E W,, p , = (0,

578

7.

DESCRIPTIVE SET THEORY

We shall show that for every ( p , q ) E Q that forces x @ A and for every k E a,there exists a stronger condition (p‘, q’) and n 2 k such that (p’, q’) IF rf E S(x) n (I,. Let (p, q ) E Q be such that (p, q ) IF x # A, and let k E w. Let p , = (u,, E,) be the ath coordinate of p. If y E E , , then ( p , q ) IF x # !, and hence ( p , q ) IF S(x) n S(j7) is finite. Hence there is n 2 k such that n $ S ( y ) for all y E E , and such that (p, q ) does not force ti $ S(x). Thus there exists (p,q) E W, that is compatible with ( p , q). By the important remark that we made earlier in the proof, (p v p, q v 4) is a condition, and forces n’ E S(x). Since (p,4) E W,, the ath coordinate of p is (0, a),and hence the ath coordinate of (P v v 4) is p a = (u, Em)* Let (p’, q’) be the condition obtained from (p v p, q v q ) by changing the ath , to p i = (u, u {n}, E,). Clearly, p i is a stronger coordinate from p , = ( u ~E,) condition than p , (in P A ) and hence (p’, q’) I ( p v p, q v q). Moreover, (p’, q’) IF ti E (I,, and therefore (p’, q ’ ) IF n’ E S(x) n (I,. Thus we have shown that if x E Ow - A, then S(x) n a, is infinite. W 9

~9

Lemma 44.13. In 9J[a] the set A is ll:(a). Proof: First we recall that the function (a, n)t+dnn is A 1 over HC. Then we use (44.26) to show that the function a w a , is A 1 in the parameter a over HC: y = a,

-

(Vn E w ) [n E y

-

-

S(d,,) n a is finite]

Hence by (44.27), A is ll, in a over HC: x

E

A

Va(S(x) n a, is finite)

By Lemma 41.1, A is a n : ( a ) subset of Ow. W This completes the proof of Lemma 44.10.

HISTORICAL NOTES AND GUIDE TO THE BIBLIOGRAPHY

CHAPTER I

Set theory was invented by Georg Cantor. The first attempt to consider infinite sets is attributed to Bolzano (who also introduced the term Mengp). I t was however Cantor who realized the significance of one-to-one functions between sets and introduced the notion of cardinality of a set. Cantor originated the theory of cardinal and ordinal numbers as well as the investigations of the topology of the real line. Much of the development in the first four sections follows Cantor's work. The main reference to Cantor's work is his collected works, Cantor [ 19321. Another source of references to the early research in set theory is HausdorR's book [1914]. Cantor started his investigations in [ 18741. where he proved that the set of all real numbers is uncountable, while the set of all algebraic reals is countable. In [ 18781 he gave the first formulation of the celebrated continuum hypothesis. Section I The axioms for set theory (except Replacement and Regularity) are due to Zermelo [ 19081. Prior to Zermelo. Frege attempted to axiomatize set theory, using the false comprehension schema; Zermelo formulated his axioms following the discovery of several paradoxes by Burali-Forti. Cantor. Russell, and Zermelo. The Replacement schema is due to Fraenkel [1922a] and Skolem (see [1970. pp. 137-1521). Exercises 1.12 and 1.13: Tarski [1925a]. Section 2 The theory of well-ordered sets was developed by Cantor, who also introduced transfinite induction. The idea of identifying an ordinal number with the set of smaller ordinals is due to Zermelo and von Neumann. Section 3 Cardinal numbers and alephs were introduced by Cantor. The proof of the Cantor-Bernstein theorem is Bernstein's; see Bore1 [ 1898. p. 1031. The first proof of 6,. XX= 6,appeared in Hessenberg [1906, p. 5931. Regularity of cardinals was investigated by HausdorN, who also raised the question of existence of regular limit cardinals. D-finiteness was formulated by Dedekind. 519

580

HISTORICAL NOTES AND GUIDE TO THE BIBLIOGRAPHY

Section 4 Theorems 10 and I I are due to Cantor. The construction of real numbers by completion of the rationals is due to Dedekind [ 18721. Theorem 13: Baire [1899]. Theorem 14: Cantor. Bendixson [ 18831. Exercises 4.3, 4.4. 4.5a: Cantor. Suslin’s problem: Suslin [ 19201.

Section 5 The axiom of choice was formulated by Zermelo, who used it to prove the well-ordering theorem in [ 19041. The present form of the Kuratowski-Zorn lemma is as in Zorn [ 19351; for a related principle, see Kuratowski [1922]. (Hausdorff in [1914, pp. 140-141] proved that every partially ordered set has a maximal linearly ordered subset.) The principle of dependent choices was formulated by Bernays in [1937-1954, Ill]. Exercise 5.1 : Felix Bernstein.

Section 6 Konig’s theorem appeared in J. Konig [1905]. Theorem 18 was found independently by Bukovsky [I9651 and Hechler. The discovery that cardinal exponentiation is determined by the gimel function (Theorem 20) was made by Bukovsky; cf. [ 19651. The inductive computation of X> is as in Jech [1973b]; that paper also investigates properties of the gimel function. Exercise 6.6: L. Patai. HausdorB’s formula (6.18): Hausdorff [ 19041. Exercise 6.8: Tarski [1925b]. Exercises 6.15 and 6.19: Jech [1973b]. Inaccessible cardinals were introduced in the paper by Sierpinski and Tarski [ 19301; see Tarski [ 19381 for more details. Section 7

The notion of filter is, according to Kuratowski’s book [ 19661, due to H. Cartan. The ultrafilter theorem was first proved by Tarski in [ 19301. Fodor’s theorem appeared in Fodor [ 19561. Reduced products were first investigated by to S in [1955]; see Section 28. Lemma 7.6 is a special case of Theorem 85 of Solovay proved in Section 35. Exercise 7.12: Friedman [1974]. (For related problems, see Prikry and Solovay [1975]). Mahlo cardinals are named after P. Mahlo, who in 191I- 1913 investigated what we call weakly Mahlo cardinals.

Section 8 Theorem 23 is due to Silver. Silver’s proof [I9741 combines forcing with the ultrapower method; see Section 35. An elementary proof was subsequently found by Baumgartner and Prikry [ 1976, 19771 and Jensen. The combinatorial argument of Baumgartner and Prikry was extended by Galvin and Hajnal in [ 19751 to obtain Theorem 24. The proof of Lemma 8.7 is a modification of Galvin-Hajnal’s method. Lemma 8.5: Erdos, Hajnal, and Milner [ 19681. Models where the singular cardinals hypothesis fails have been constructed by Magidor, cf. [1977b] and [ 1977~1.

Section 9 The axiom of regularity was introduced by von Neumann in [1925], although a similar principle had been considered previously by Skolem (see [ 1970, pp. 137-1521). The transitive collapse is

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defined in Mostowski [ 1949); see Theorem 28 in Section 11. Induction o n well-founded relations (Theorem 25') was formulated by Montague in [ 19551. Appendix The axiomatic system BG was introduced by Bernays in [ 1937- 1954. I] and used by Godel in his consistency proof [ 19401. Shoenfield's result was published in [ 19541. For more references on the history of axioms of set theory consult Fraenkel et a/. [ 19731. CHAPTER 2

The main results. namely consistency of the axiom of choice and of the generalized continuum hypothesis. are due to Kurt Godel. as is the concept of constructible sets. The results were announced in [ 19381, and an outline of proof appeared in [ 19391.Godel's monograph [ 19401contains a detailed construction of L and the proof that L satisfies AC and GCH. Section 10

For concepts of model theory. the history of the subject and for model-theoretical terminology, I refer the reader to Chang and Keisler's book [ 19731. The impossibility of a consistency proof of existence of inaccessible cardinals follows from Godel's theorem [ 19311. An argument that more or less establishes the consistency of the axiom of regularity appeared in Skolem's work in 1923 (see [1970. pp. 137-1521), Exercise 10.6: Montague and Vaught [ 19591. Section 11 The investigation of transitive models of set theory was of course motivated by Giidel's construction of the model L. The first systematic study of transitive models was done by Shepherdson in [ 1951- 19531. Bernays in [ 1937- 1954, I], employed a finite number of operations on classes to give a finite axiomatization of BG (see Exercise 11.9).Theorem 31 is explicitly stated by Hajnal in [1956]. The collapsing theorem 28 appeared in Mostowski [1949]. The reflection principle was introduced by Montague; see [I9611 and k v y [1960a]. Exercise 11.2: Calvin. Section 12 In [ 19391 Godel defined constructible sets using L,, = {all subsets of L, definable over L,}; in [ 19401 he used finitely many operations (and worked in the system BG). Exercise 12.5: The J , hierarchy was used by Jensen in [ 19721. Section 13 Both Theorems 33 and 34 are due to Giidel. Section 14 The hierarchy was introduced by LRvy in [1965b]. LRvy proved in [1965b] Theorem 36 (the absoluteness lemma), using the principle of dependent choices and Shoenfield's absoluteness theorem (Theorem 98 in Section 41)-see Exercise 41.5; the present proof is due to Barwise and Fisher [ 19701. Another result of Levy [I9651 is that ihe truth predicate C n + l is Z,, (Exercise 14.20); for Tarski's theorem, see, e.g., Tarski [ 39391.

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Karp's paper [I9671 investigates El relations and gives a detailed computation verifying that constructibility is E l . Primitive recursive functions appear in Jensen and Karp [ 19711. The characterization of the sets L, as transitive models of a single sentence u is a result of Boolos [1970]. Exercises 14.23 and 14.24: Jensen [1972]. Section 15 Relative constructibility was investigated by Hajnal [ 19561. Shoenfield [ 19591 and most generally by Levy [ 19571 and [ 1960bl. Lemma 15.1: Voptnka and Balcar [1967]. Lemma 15.2 and Exercise 15.10: Levy [I9571 and [1960b]. Exercise 15.16: see Theorem 98. The concept of ordinal definability was suggested by Giidel in his talk [1946]; the theory was developed independently by Myhill and Scott in [I9711 and by VoMnka, Balcar, and Hajek in [1968]. The former authors use the definition from Exercise 15.18, the latter use (15.16). CHAPTER 3

The method of forcing was invented by Paul Cohen who used it to prove the independence of the continuum hypothesis and of the axiom of choice (see [1963-19641 and the book [1966]). The present formulation of Cohen's method using Boolean-valued models has been developed by Scott, Solovay, and Voptnka. Section 16 The two fundamental theorems of the method of forcing, the forcing theorem and the generic model theorem, are due to Cohen. (To be accurate, Cohen's original method was formulated for particular examples of a notion of forcing, and under the assumption that 91 is a countable transitive model of ZFC. The definition (16.3)ofan 91-generic filter on a partially ordered set is due to Solovay.) The forcing conditions defined in (16.4) are the ones used by Cohen in his proof of independence of the continuum hypothesis; the "typical example," a collapsing mapping ofw onto to, is also due to Cohen. Section 17 Most of the facts about Boolean algebras can be found in Sikorski's book [1964], which also contains an extensive bibliography. The representation theorem for Boolean algebras (Theorem 4 1) is due to Stone [ 19361. The important lemma 17.6 on the chain condition was proved by Erdos and Tarski in [1943]. For a more detailed discussion of the role of the prime ideal theorem, see my book on the axiom of choice [ 1973~1;for the model-theoretical concepts involved, consult Chang and Keisler [ 19731. The independence of the prime ideal theorem from AC is due to Feferman [ 19651 and is proved in Section 21 (Example V); the independence of the axiom of choice from the prime ideal theorem is a result of Halpern and Levy [1971]. Lemma 17.5: Rasiowa and Sikorski [1950]. Exercise 17.14: Makinson [1969]. Exercise 17.2I : Pierce [ 19581. Section 18 The Boolean-valued version of Cohen's method has been formulated by Scott, Solovay. and Voptnka. Following an observation of Solovay that the forcing relation can be viewed as assigning Boolean values to formulas, Scott formulated his version of Boolean-valued models in [ 19671. Voptnka developed a theory of Cohen's method of forcing, using open sets in a topological space as

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forcing conditions (in [ 19641, [ 1965-19661, [ 1967al and Vophka-Hajek [ 19671). eventually arriving at the Boolean-valued version of forcing more or less identical to Scott-Solovay's version (VopCnka [ 3967bl). VoNnka's paper [ 19641 contains the observation that Cohen's construction can be modified so that one can construct a (nontransitive two-valued) model directly, without assuming existence of a countable model or a generic filter (see Lemma 18.1). Shoenfield's article [I9711 uses a slightly different approach to forcing. proving the main theorems without using Boolean-valued models.

Section 19 The two independence results, independence of the continuum hypothesis (Theorem 44)and of the axiom of choice (Theorem 45) are due to Cohen; cf. [1963-19641 and [1966]. Our method of proof of Theorem 45 (using 91 = HOD(A)) is basically that of Cohen (who used a model !U1(A)adjoining an infinite set of mutually generic reals to a countable transitive model of ZFC). Shortly after Cohen's discoveries, Solovay (in [ 19631) noticed that Cohen's construction of a model for O 2' = S, can be generalized so that for a regular K one obtains a model with 2" = I (Lemma 19.7); see also Vopenka [1965-1966]. (For the general result. see Easton's theorem in Section 20.) The notion of forcing that collapses a given cardinal (lemma 19.9) is due to Cohen (see [ 19661). The relation between the chain condition and preservation of cardinals (Lemma 19.2) is basically due to Cohen; the observation that a I-closed notion of forcing does not produce new subsets of I (Lemmas 19.5 and 19.6) is due to Solovay. The combinatorial lemmas used in computing the chain condition (Lemmas 19.3 and 19.8) are borrowed from point set topology: Marczewski [ 19411 and Engelking and Karlowicz [1965]. Exercises 19.3-19.7 contain facts noticed by many people including this author.

Section 20 Easton's theorem (Theorem 46) was published in [ 19701. The generalization of Cohen's method allowing a class of forcing conditions is due to Easton. The model in Lemma 20.4 (the Levy collapse) was constructed by Levy; cf [ 19701. The product lemma (Lemma 20.1) is due to Solovay [1970]. Lemma 20.3 follows from the aforementioned combinatorial result of Engelking and Karlowicz. The chain condition of a product of two C.C.C. partial orderings: Using a construction of Galvin. Fleissner showed that it is consistent that 2'" is large and that there are C.C.C.P and Q such that sat(/' x Q) = Exercise 20.8: Jensen [ 19651. The concluding remark on the axiom of choice in BG uses an observation made by various people, e.g., Felgner [ 19711. ('*I)'.

Section 21 The idea of using symmetry arguments to construct models in which the axiom of choice fails goes back to Fraenkel [ 1922hI; the two examples of models of ZFA (an infinite set of atoms without a countable subset. and a countable set of pairs that has no choice function) are basically due to him. Further examples of permutation models were given by Mostowski who (in [ 19391)developed a theory of such models. The present definition using normal filters (21.5) was given by Specker [ 19571. Cohen incorporated the symmetry arguments into his method and constructed the models that are presented in Examples I and I I . The formulation of Cohen's method in terms of symmetric submodels of Boolean-valued models is due to Scott (unpublished) and Jech [1971a]; the latter's version was a reformulation of a topological version of Vophka and Hajek [ 19651. The embedding theorem (Theorem 47) is due to Jech and Sochor [1966a, b]. Numerous

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applications of the theorem are given in the second paper [ 1966bl. The method has been generalized by Pincus in [ 19711 and in [ 19721, extending further the analogy between permutation models of ZFA and symmetric models of ZF. Levy showed that in Cohen’s model in Example I every set can be linearly ordered; consequently, Halpern and Levy [ 19711 proved that the prime ideal theorem holds in the model. Example 111 (a prescribed partially ordered set of cardinals) is due to Jech [1966]; see also Takahashi [1968]. Example IV (singularity of K,)is due to Feferman and Levy [1963]. Example V (independence of the prime ideal theorem) is due to Feferman [ 19651; the same model establishes the independence of the principle of dependent choices from the axiom of choice; see Feferman [1964]. Jech’s paper [1971a] uses a similar construction to obtain two models of Z F that have the same sets of ordinal numbers. A. Blass constructed in [I9771 a model. similar to Feferman’s model, in which every ultrafilter is principal. CHAPTER 4

Section 22 Suslin’s problem was formulated by Suslin in [ 19201. Tennenbaum [ 19681 and Jech [ 19671 discovered models of set theory in which a Suslin line exists; Solovay and Tennenbaum [ 19711 proved that existence of a Suslin line is not provable in ZFC. Subsequently, Jensen proved that a Suslin line exists in the constructible universe (cf. [1968, 19721). The present proof of Theorem 48 is as in Jech [ 19671 (countable forcing conditions); Tennenbaum’s proof (finite conditions) is presented in Exercises 22.9 and 22.10. Suslin’s problem is independent of the continuum hypothesis: Tennenbaum’s construction yields a Suslin line, with or without the continuum hypothesis (see the remark following Exercise 22.10); the continuum hypothesis is false in Solovay-Tennenbaum’s model (cf. Section 23). but an alternative consistency proof of Jensen gives a model where there are no Suslin lines but the continuum hypothesis holds. For more on Suslin’s problem and trees in general, consult Jech [1971b] and Devlin and Johnsbriten [1974]. The first systematic study of trees appears in Kurepa [1935]; Kurepa’s paper states that Suslin’s problem is equivalent to the problem of existence of a Suslin tree. This equivalence (Lemma 22.1) was later rediscovered by Miller [I9431 and Sierpinski [1948]. The first construction of an Aronszajn tree was given by Aronszajn and appears in Kurepa [ 19351; see also Specker [1951]. The Boolean algebraic reformulation of Suslin’s problem (Lemma 22.2 and Exercise 22.3) is due to Maharam [ 19481. The construction of rigid and homogeneous Suslin trees in L (Exercises 22.15 and 22.16)isdue to Jensen [ 19691. Automorphisms of Suslin trees in general are investigated in Jech [ 1972al. Exercise 22.8: Fukson [1971]. The A-lemma was formulated and proved by Sanin in [ 19461. Exercise 22.1 : D. Konig [ 19271. Exercise 22.4: Kurepa [ 19501. Section 23 Following Solovay and Tennenbaum’s construction (by iterated forcing) of a model in which there are no Suslin trees (Theorem 50), Martin formulated an axiom (MA,,) which implies that there are no Suslin trees (Lemma 23.1) and whose consistency was obtained by an easy generalization of Solovay-Tennenbaum’s construction. The consistency proof of MA + 2& > K, (Theorem 51, and also Exercise 23.9) appears in Solovay and Tennenbaum [1971]. Martin’s axiom is investigated in detail in the paper [1970] of Martin and Solovay. The paper contains various equivalent formulations of Martin’s axiom and numerous applications (including

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Theorem 52); for applications of MA in descriptive set theory, see Section 44. In recent years, MA has been very popular with point set topologists; cf. Juhasz's book [ 19711. Iterated forcing, using direct limits of complete Boolean algebras, is developed in Solovay and Tennenbaum's paper [1971]; the paper contains, e.g., Lemmas 23.6 and 23.7 on the C.C.C. The proof of Lemma 23.10 (using the notion of forcing (23.32)) is a variation of Solovay's method of "almost disjoint forcing.'' This method was introduced in Martin and Solovay [I9701 and Jensen and Solovay [ 19701; see also Sections 26 and 44. Marczewski proved in [ 19471 that the strong C.C.C. is preserved under (topological)products (of any number of copies). Lemma 23.13 (MA,, implies that every C.C.C.space satisfies the strong c.c.c.) is due to Kunen, Rowbottom, and Solovay; Theorem 53 is a consequence of this and of Marczewski's result. Exercise 23.7: Krivine [ 19691.

Section 24 Baumgartner, Malitz, and Reinhardt [ 19701 proved that MA,, implies that every Aronszajn tree can be embedded in the rationals (Theorem 54). Special Aronszajn trees have applications in model theory (this fact is due to Rowbottom and Silver) and are investigated in Mitchell's paper [1972]. The equivalence in Lemma 24.1 is due to Calvin. Exercises 24.2 and 24.3: Baumgartner [ 19701. Exercise 24.4: Jech [1972b]. Construction of a rigid Aronszajn tree is an unpublished result of Baumgartner. Existence of a Kurepa tree was a problem formulated (in terms of a family of sets, see Lemma 24.6) by Kurepa in [ 19351. A model in which there is a Kurepa tree (using the notion of forcing (24.5)) was constructed by D. H. Stewart (unpublished). The model (24.7) in which there is no Kurepa tree is due to Silver [1971b] (for a modification of Silver's model such that 2," > K,, see Devlin [ 19751).The construction of a Kurepa tree in L is due to Solovay; as Jensen's construction of a Suslin tree in L it is an example of application of the fine structure of L, a method extensively developed by Jensen in [ 19721. The observation that Kurepa trees exist unless K, is inaccessible in L is due to Silver. For more on Kurepa trees, consult Jech [1971b] and Devlin's book [1973]. Lemma 24.4: Erdos and Rado [ 19601 and Michael [ 19621. The notion ofalmost disjoint sets is due to Sierpinski [1928a] (Lemma 24.7). Exercise 24.6: Tarski [1928]. Exercise 24.7: Erdos and Hajnal [1971]. The model where 2'" = 2'' = K, and no tree of size K, has K, branches of length w 1 is due to Mitchell [1972]; so is the observation that in the model obtained by adjoining Nu,Cohen reals there is no dense linearly ordered set of size 2'' with a dense subset of size h', (the equivalence of the dense linear ordering problem with the problem of existence of a tree with a large number of branches is due to Baumgartner). Jensen's paper [ 19721introduces the fine structure of 15;the combinatorial principles 0, 0 and (24.17) are proved (in L ) in that paper. Silver's principle (W) replaces (in several proofs) another, quite complicated, combinatorial principle of Jensen (existence of a morass-see Devlin [ 19731) true in L Exercises 24.10 and 24.1 1: Baumgartner, Hajnal, and Mate [1973]. The implication 0= (24.17) is due to Magidor. By Jensen [1972], holds unless K, is a Mahlo cardinal in L; Baumgartner in [ 1976al uses a weakly compact cardinal to construct a model in which (24.17) (and hence 0 too) fails. Theorem 56 is due to PospiSil[1937]; the present proof uses independent sets (Lemma 24.8); cf. Fichtenholz and Kantorovich [I9341 (K = o)and Hausdorm [1936]. W. Rudin [I9561 proved that p-points exist if 2,o = K, (Theorem 57a); a recent result of Shelah shows that existence of p-points is unprovable in ZFC. Galvin showed that 2% = Kl implies existence of Ramsey ultrafilters; construction of p-points and Ramsey ultrafilters (and

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p-points that are not Ramsey) under the assumption of Martin's axiom is due to Booth [1970]. Our proof of Theorem 57 follows Ketonen [1976a]. Exercise 24.15: Booth [1970]. Scales were investigated extensively by Hechler [ 19741. Hechler introduced the notion of forcing used in the proof of Lemma 24.17 (the lemma itself was explicitly formulated by Kunen). 2' > K Hechler, among others, showed that if cf K > w, then there is a generic extension in which O and a K-scale exists. Section 25

The relationship between automorphism properties of complete Boolean algebras and definability in generic extensions had been studied by several authors. Lkvy in [ 1965al showed consistency of L = HOD # V virtually as in Exercises 25.8 and 25.9. McAloon in [1971a] constructed a model in which L # HOD = V and remarked that according to Solovay the algebra used in the construction must be rigid. (A direct construction of a rigid complete Boolean algebra is more recent and is due to Shelah; cf [1974].) Theorem 59 which gives the exact description of definable sets in generic extensions is due to Vop4nka and appears in the book [1972] by VopL'nka and Hajek. The relationship between distributivity laws in Band the size of"new" sets in 9t[G] (Theorem 58b) had been known already to Scott, Solovay, and Vop6nka; for instance, cf. Vop6nka [1965-19661. Exercises 25.1 and 25.2 (reduced to a triviality by the forcing argument) had been known before forcing; see Sikorski [1964]. McAloon [ 1971a] constructed models where either L # HOD = V or L # HOD # V and noted that it is also consistent that HODHoD# HOD. McAloon's paper [ 19741has an example of a strictly descending sequence of models 91, 3 9l13 ... 2 91, 3 ..., n < w. such that 9tn+, = HOD"; Jech's paper [1974b] has an example of a transfinite sequence 9to3 ... 2 9tz3 ..., of arbitrary length such that % l a + , = HOD", with intersections at limit steps. Thesequence HOD' = V, HOD' = HOD, HOD' = (HOD)"OD, . . ., HOD", .. . is not necessarily expressible in ZF (by an unpublished result of Harrington); but if V = 4x1. then the sequence HOD' is expressible (GrigorieN [1975]). S. Roguski announced recently that for any transitive model 91 of ZFC there exists a generic extension 91 (via a proper class of conditions) such that 91 = HOD". The construction of a rigid complete Boolean algebra in L is a modification of Jensen's construction of a rigid Suslin tree in L [ 19691. Simple complete Boolean algebras were investigated by McAloon [1971b] and Jech [1974a]. Lemmas 25.8 and 25.9 are due to McAloon [1971b]. Existence of simple complete Boolean algebras (in ZFC) is an open problem. Jensen's Example 111 in Section 44 (a minimal A: real) (cf. [ 19701)gives, by McAloon, a simple complete Boolean algebra in L.The construction ofa simple complete Boolean algebra using 0 mentioned in the text appears in Jech [1974a]; it has subsequently been used in Harrington and Jech [ 19761to construct a generic extension of L that admits a Z I well-ordering. Theorem 60 is an unpublished result of Solovay; prior to Solovay's proof, this author had an alternative proof using the continuum hypothesis. Our paper [ 1973a] shows that it is consistent that a Suslin algebra exists of arbitrarily prescribed size K. Subsequently, Laver showed (using 0 and W)that in L there is a Suslin algebra of size K, . Theorem 61 is due to Gaifman [ 1964a] and Hales [ 19641;the present proof using Lemma 25.10 is Solovay's; see [1966]. Theorem 62 is due to Kripke [1967]. I n [1970] Solovay proved the factor lemma for Levy collapse (Exercise 25.1 I ) as the main technical lemma for his consistency proofon Lebesgue measure (see Section 42). Jensen used the factor lemma to prove Theorem 63. Subsequently, Krivine proved the more general factor lemma (Exercise 25.10) in [ 19691. Lemmas 25.13 and 25.14: Jech [1972c] and [1971c]. Section 26

The method of almost disjoint forcing appears in Martin and Solovay [ 19701 and Jensen and Solovay [ 19701. Theorem 64 is due to Sacks [ 19711. Example II (changing the cofinality of w2 tow)

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is due to Namba [1970]; see also Buk'ovsky [1976]. Theorem 65 is due to Vop6nka and appears in the book [1972] of Voptkka and Hajek. A recent result of Z. Adamowicz provides various examples of well-founded initial segments of degrees of constructibility; Jech's paper [1974b] contains an example of an initial segment of Ldegrees of order type to*. Hajek in [I9741 constructs two Cohen generic reals x. y without greatest lower bound: q.x] n y y ] is not a model of ZFC. Lemma 26.6: Jech [ 1974bl. Exercise 26.5: Hajek [ 19741. Exercise 26.6: Grigorieff [ 19751. CHAPTER 5

The study of measurable cardinals originated around 1930 with the work of Banach, Kuratowski, Tarski. and Ulam. Ulam showed in [I9301 that measurable cardinals are large, that the least measurable cardinal is at least as large as the least inaccessible cardinal. Around 1960 Keisler and Tarski introduced the method of ultraproducts in the study of measurable cardinals, and it was established that the least measurable cardinal is greater than the least inaccessible cardinal. Scott [I9611 used the method of ultrapowers to prove that existence of measurable cardinals contradicts the axiom of constructibility. In the 1960s two methods have been successfully applied to measurable cardinals: Rowbottom and Silver initiated applications of infinitary combinatorics (developed by Erdos and his collaborators), and Gaifman and Kunen introduced the method of iterated ultrapowers.

Section 27 The main result on measurable and real-valued measurable cardinals (Theorem 66) is due to Ulam [ 19301. The fact that a measurable cardinal is inaccessible (Lemma 27.2) was discovered by Ulam and Tarski (cf. Ulam [1930]). Prior to Ulam, Banach and Kuratowski proved in [I9291 that if 2& = N,,then there is no measure on the continuum; their proofis as in Lemma 27.9. Real-valued measurable cardinals were introduced by Banach in [ 19301. The notion of a x-saturated ideal was introduced and studied by Tarski in [1939-1945, 111. The example of a Lebesgue nonmeasurable set in Exercise 27.1 was given by Vitali in [ 19051. For further information on measure theory, we refer the reader to Halmos' book [1950]. Section 28

The fundamental theorem on ultraproducts (Lemma 28.1) is due to ZoS [1955]. Keisler and Tarski [I9641 introduced ultrapowers in the study of large cardinals. Theorem 67 is due to Scott [ 19611. and Theorem 68 is due to Kunen [ 1971aI. In [ 19644 Hanf studied compactness of infinitary languages (see Sections 32 and 33); his work led to the systematic study of Keisler and Tarski. Hanf proved that the least inaccessible cardinal is not measurable (in fact not weakly compact); Erdos and Hajnal then pointed out (cf. [1962]) that the same result can be proved using infinitary combinatorics. Keisler and Tarski introduced the Mahlo operation (28.23)and showed that the least measurable cardinal is much greater than, e.g.. the least Mahlo cardinal. etc. (Lemma 28.12). In [ 19621 Voptnka gave another proof of Scott's theorem: Observing that I(2')"'' I = Ij(.) 1. it follows that the cardinals in the model Ulr are not absolute and hence V # L Lemma 28.8: Erdos and Hajnal [ 19661. Exercise 28.12: For a related result see Ketonen [1973]. Exercise 28.14: Solovay. Lemma 28.13 follows from Lemma 28.15. which appears in the abstract [I9611 of Hanf and Scott. The result in Exercise 28.18 was announced by Vop6nka in [1970]; more applications of a measurable cardinal to the singular cardinal problem (including Exercise 28.20) are in Jech [ I973bl.

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Lemma 29.1 (Ramsey’s theorem) is due to Ramsey [ 19301. The theory of partition relations has been developed by Erdos, who has written a number of papers on the subject, some coauthored by Rado, Hajnal, and others. The arrow notation is introduced in Erdos and Rado [1956]. Other major comprehensive articles on partition relations are Erdos, Hajnal, and Rado [I9651 and Erdos and Hajnal [ 197I]. Theorem 69 appears in Erdos and Rado [1956]. The present proof incorporates ideas from a model-theoretical proof of the Erdos-Rado theorem due to Simpson [1970]. as published in Chang and Keisler’s book [ 19731. Weakly compact cardinals are investigated in the paper [I9611 of Erdos and Tarski. The investigations started in that paper were subsequently completed by others; see the notes for Section 32. For applications of partition relations in model theory, we refer the reader to Chang and Keisler’s book [ 19731. Lemma 29.3: Sierpinski [1933]. Exercise 29.2: Kurepa [1962]. Exercises 29.3 and 29.4: Erdos and Rado [ 19521. The partition relation N, + (a)’ for a < w , was proved by Baumgartner and Hajnal [1973]; the consistency N, (wI + w)* by forcing is due to Hajnal. and in L due to Rebholz [ 19741. The partition relation w + ( o r i s consistent with ZF (without AC), by Mathias [I9681 and [I9771 (using an inaccessible cardinal). For further results on the partition relation w + (w)“ for Bore1 and analytic partitions, see Galvin and Prikry [1973], Silver [ 1970b1, and Ellentuck [ 19741. That every measurable cardinal is a Ramsey cardinal had been proved by Erdos and Hajnal (see their paper [1962]). The stronger version of that result (Theorem 70) is due to Rowbottom [1971], as are Theorem 71 and Lemma 29.10. No N, is a Jonsson cardinal: Erdos and Hajnal [ 19661. The least Jonsson cardinal is weakly inaccessible or has cofinality w : Rowbottom. If V = L then there is no Jonsson cardinal: Keisler and Rowbottom [1965]. A model where every Jonsson cardinal is a Ramsey cardinal: Kunen [1970]. A model in which there is a Rowbottom cardinal of cofinality w : Prikry [1970a]. Consistency of Chang’s conjecture: Silver. Exercise 29.14: Silver; cf. Benda and Ketonen [ 19741. The generalization of Aronszajn’s construction for successors of regular cardinals (assuming the GCH) is due to Specker [1951]. A model in which there exists no special Aronszajn tree on w , was constructed by Mitchell [1972], who also showed that Mahlo cardinals are necessary for the construction. An analogous result for Aronszajn trees and weakly compact cardinals is due to Silver, see Mitchell [1972]. The generalization of forcing construction producing Suslin trees of larger heights is due to HrbaEek [1967]. Jensen’s construction of a Suslin tree in L generates to larger cardinals. using 0 and 0 ;see Jensen [1972]. The construction of a Suslin o,-tree using 0 and the GCH is due to Gregory [1976]. The model without Suslin w,-trees is due to Laver.

+

Section 30

Theorem 72 was discovered by Gaifman. Gaifman’s results were announced in [ 1964b] and the proof was published in [1974]. Gaifman’s proof used iterated ultrapowers (see also Gaifman [1967]). Silver in his thesis (1966, published in [1971a]) developed the present method of proof, using infinitary combinatorics, and proved the theorem under the weaker assumption of existence of K with the property K + (K,)‘”.Gaifman proved that if there is a measurable cardinal, then there exists A E w such that the conclusion of Theorem 72 holds in LEA]. Solovay formulated O x and proved that it is a A: set of integers; Silver deduced the existence of O * under weaker assumptions. Construction of models with indiscernibles was introduced by Ehrenfeucht and Mostowski in [1956]; for further details and applications in model theory; see Chang and Keisler [1973].

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589

The equivalence of existence of O * with existence of a nontrivial elementary embedding of L (Theorem 73) is due to Kunen; the present proof is due to Silver. Theorem 74 (and its corollaries) is due to Jensen. A proof of the theorem appeared in Devlin and Jensen [ 19751. Exercise 30.5: Solovay. Exercise 30.7: Magidor. For further work on Silver indiscernibles, see Paris [ 19741. Section 31

Most of the results in this section are due to Kunen, who in [I9701 developed the method of iterated ultraproducts invented by Gaifman (cf. [ 1964b] and [ 19741). Kunen found the representation of iterated ultraproducts (Lemma 31.1 1) and generalized the construction for M-ultrafilters. Kunen applied the method to obtain the main results on the model L [ D ] (Theorem 76). Theorem 75 (the proof of the GCH in YO])is due to Silver [ 1971~1;using a similar technique. Silver proved in [1971d] that in q D ] there is a A; well-ordering of the reals. The description of h--complete ultrafilters over ti in Y O ](Lemma 31.18) is due to both Kunen [ 19701 and Paris [ 19691. Lemma 31.3 was first proved by Solovay. Lemma 31.5 is due to Gaifman; cf. [1974]. Lemmas 31.17 and 31.20 are results of Kunen [1970]. Kunen generalized the basic results on YO] to the model L(D, : a i0) constructed from a sequence of measures (with 0 < the least measurable cardinal in the sequence). Mitchell in [ 19741 generalized the construction further; one of the results in Mitchell's paper is that the model from Exercise 31.13 has a measurable cardinal with exactly two normal measures. Jensen extended some of his results (e.g., 0 )from [ 19721from L to YO]. 0' was formulated by Solovay, in analogy with 0'. The model y j ) was considered by Jech in [1971d] and also by HrbaEek. The result that in 401 the unique normal measure is the only measure that extends the closed unbounded filter (Exercise 31.12) appeared in Jech [1973a]. For further work on iterated ultrapowers of YO]. see Bukovsky [ 19731and Dehornoy [ 19751. Section 32

The equivalence of various formulations of weak compactness is a result of several papers. I n [ 1964al Hanf initiated investigations of compactness of infinary languages. Erdos and Tarski listed in [ 19611 several properties that were subsequently shown mutually equivalent (for inaccessible cardinals) and proved several implications. These properties included the partition property ti + (ti):, the tree property from Section 29. and the properties from Exercises 32.3, 32.4, and 32.5. Hanf and Scott [I9611 introduced Wmindescribability and announced Theorem 77. Further contributions were made in the papers Hanf [ 1964b], Hajnal [ 19641. Keisler [1962], Monk and Scott [ 19641, Tarski [ 19621, and Keisler and Tarski [ 19641. A complete list of equivalent formulations with the proofs appeared in Silver [ 1971al. Without inaccessibility, some of the formulations of weak compactness are no longer equivalent; this is discussed by Boos in [1976]. Silver showed that every weakly compact cardinal is weakly compact in L. that every indiscernible is weakly compact in L (both in [1971a]), and that ifh', has the tree property, then N, is weakly compact in L (cf. Mitchell [1972]). Mitchell pointed out that a weakly compact cardinal is not necessarily weakly compact in a submodel. Nonexistence of simple complete Boolean algebras of weakly compact size is proved in Jech [1974a]. Lemma 32.2: Hanf and Scott [1961]. Exercise 32.6: Kunen. Ineffable cardinals were defined by Jensen (using the property from Lemma 32.7); Kunen proved the equivalence in Lemma 32.7. Kunen also showed that ineNable cardinals are n: indescribable, and Jensen and Kunen proved that every inenable cardinal is ineNable in L

590

HISTORICAL NOTES AND GUIDE TO THE BIBLIOGRAPHY

The main results on partition cardinals are due to Rowbottom. Reinhardt, and Silver. Rowbottom proved that if q,, exists. then there are only countably many constructible reals (see [1971]); Silver then used the same assumption to prove Theorem 72 (i.e., that O* exists; see [1971a]). Theorem 78e was proved by Silver in [1970a]; Theorem 78c and Exercise 32.13 were proved by Reinhardt and Silver (see [ 19651). Lemma 32.8 is due to Rowbottom [1971], and Lemma 32.9 is due to Silver. Lemma 32.10 is in effect proved in Jech and Powell [1971]. Kunen showed that both the existence of Jonsson cardinals and Chang's conjecture imply that O x exists. Kunen also showed that in q D ] every Jonsson cardinal is a Ramsey cardinal; see [ 19701. Kleinberg in [I9721 and [I9731 showed that existence of Jonsson and of Rowbottom cardinals are equiconsistent. Silver used a Ramsey cardinal to construct a model in which Chang's conjecture holds and showed that if h', is a Jonsson cardinal and a strong limit. then there exists a transitive model with a measurable cardinal. CHAPTER 6

!kction 33 Compact cardinals were introduced by Keisler and Tarski in [ 19641; supercompact cardinals were defined by Reinhardt and Solovay. Theorem 79 is due to Vophka and Hrbaeek [1966]; Theorem 80 is due to Kunen [1971b]. Solovay discovered that the singular cardinal hypothesis holds above a compact cardinal (Theorem 81); see [ 19741. Menas and Magidor obtained several results on the relative strength of compact and supercompact cardinals. Menas in [I9751 showed that it is consistent (relative to existence of compact cardinals) that there is a compact cardinal that is not supercompact. Magidor in [I9761 improved Menas' result by showing that it is possible that the least measurable cardinal is compact (while by Lemma33.10 it is not supercompact) and also showed that it is consistent (relative to supercompact cardinals) that there exists just one compact cardinal and is supercompact. Kunen's proof of Theorem 80 uses a lemma of Ketonen (Lemma 33.3); Lemmas 33.5 and 33.6, which Solovay used in his proof of Theorem 81, are also due to Ketonen; see [1972]. Most results on supercompact cardinals (e.g., Lemmas 33.10 and 33.1 I ) are due to Solovay; see Solovay, Reinhardt, and Kanamori [1978]; Magidor's paper [1971a] gives a generalization of the corollary of Lemma 33.11: the number of normal measures on P$). The example of a compact nonsupercompact cardinal (Lemma 33.12) is due to Menas [ 19751. Exercises 33.7 and 33.8: Solovay [1974]; for the existence of an E-H function, see Erdos and Hajnal [ 19661. The closed unbounded filter over P m ( l )(Exercise 33.4) was introduced in Jech [ 1973al.So was the partition property for P#). The equivalence of (iii) and (iv) in Exercise 33.9 as well as the results in Exercises 33.10 and 33.11 are due to Solovay. Menas in [1976a] showed that if K is supercompact, then a strongly normal measureexists on every P =(A). Further results on the partition property and on stationary sets in P . ( l ) are in Magidor [1974]. Extendible cardinals were introduced by Reinhardt ; he proved that extendible cardinals are supercompact; see [1974]. The present proof of Lemma 33.14, as well as Lemmas 33.13 and 33.15, are due to Magidor [1971b]. Lemma 33.16: Powell [1972]. Exercise 33.12: Solovay, Reinhardt, and Kanamori [19781. Exercise 33.13: Morgenstern [ 19771. seetion 34

Theorem 82 is due to Solovay [1971]. Theorem 83 is due to Prikry [1975], and so is the model in which K carries a a-saturated ideal; see [1970a].

HISTORICAL NOTES AND GUIDE TO THE BIBLIOGRAPHY

591

Exercise 34.4: Silver. Exercises 34.5, 34.6. and 34.7: Prikry [ 1970al. Section 35 Saturation of ideals and the problem of almost disjoint sets have been investigated by various authors. e.g., Solovay [1971], Kunen [1978], Baumgartner [1976b], and Jech and Prikry [ a ] . The Jech-Prikry paper gives applications of generic ultrapowers, as well as of almost disjoint functions. Theorem 84a on almost disjoint sets is due to Baumgartner; cf. [1976b]. The present proof For Theorem 84b, see Jech and Prikry [I9761 and [E]. Theorem 85 is follows Jech and Prikry [a]. due to Solovay; cf. [1971]. The present proof of Theorem 85 uses ideas of Mate and Simpson; Solovay’s proof uses Lemmas 35.3 and 35. I 1. (The present proof of Lemma 35.1 1 is as in Baumgartner, Hajnal, and Mate [ 19731.) Examples I and I I are due to Baumgartner (Taylor showed that the thin ideal in Example I I is in fact h’,-saturated.) The result that if K carries a K-saturated ideal, then K is a weakly Mahlo cardinal is due to Solovay [1971]. If X I carries an X2-saturated ideal then there are models with many measurable cardinals: Kunen [I9701 and [1971b]. There are several consistency results on saturation of ideals using measurable cardinals: A model where w = 2% and sat(/) = w (Kunen and Paris [1971]), a model where K is inaccessible and sat(/) = K (Kunen [1978]), models where K is inaccessible and sat(/) = K’ (Kunen and Paris [I9711 and Boos [1974]). In [I9781 Kunen constructs a model (using a huge cardinal) in which N l carries an &saturated ideal. Lemma 35.5 is in Baumgartner, Taylor, and Wagon [1977]; Corollary 2 of the lemma appeared in Namba [1974]. The problem of existence ofw-closed uncountable subsets ofsets A & w 2 isdiscussed in Prikry and Solovay [ 19751. Exercise 35.1: Smith and Tarski [1957]. Exercise 35.2: Solovay [ 19711. Exercise 35.4: Kanamori [1976]. Exercise 35.5: Levy, Silver. Exercise 35.6: Prikry [ 19731. Exercise 35.7: Baumgartner, Hajnal, and Mate [ 19731. Exercise 35.8: Baumgartner, Taylor, and Wagon [ 19771. Generic ultrapowers are studied in detail in Jech and Prikry [ a ] . Solovay [I9711 used the Boolean-valued ultrapower in the case of saturated ideal. Precipitous ideals were introduced in Jech and Prikry [ 19761. Theorem 86a is due to Jech and Prikry; the proof generalizes an earlier proof of Kunen, who in [ 19701 obtained the result for K+-saturatedideals. Theorem 86b was proved by Mitchell. Magidor subsequently constructed a model where the thin ideal is precipitous. Cf. Jech rr al. [co]. The proof of Silver’s theorem (application 1) is a simplified version of Silver’s original proof in [ 19741. Applications 2, 3, and 4 are from Jech and Prikry [ 19761 and [ a ] ; for a related result on almost disjoint stationary sets, see Ketonen [ 19741. The result (35.15) on w-saturated ideals is due to Solovay [1971]. The method of changing cofinalities of cardinals is due to this author, and Namba [ 19741. Lemmas 35.8, 35.9, 35.10: Jech and Prikry [ 19761. Exercises 35.10, 35.1 I : Solovay [ 197I]. Exercises 35.12, 35.13: Jech and Prikry [m]. The observation that the conclusion of Exercise 35.13 contradicts Chang’s conjecture is due to Benda; previously, Magidor had shown by a different argument that Chang’s conjecture implies 2nwl
Section 36 Theorem 87 is due to Kunen [1971b]. Theorem 88 is an unpublished result of Silver. Lemma 36.2 (with the proof as in Exercise 36.1) is due to U v y and Solovay [ 19671. Consistency

592

HISTORICAL NOTES AND GUIDE TO THE BIBLIOGRAPHY

of measurable cardinals with the GCH by forcing (using the construction from Exercise 20.8) is due to Jensen; cf. [1974]. Lemma 36.3: Kunen and Paris [1971]. The method of inverse limits in interated forcing was developed by Silver and worked out in detail and applied by Menas in [ 1976bI. Section 37 Theorem 89 is due to Prikry [1970a]. Theorem 90 is due to Jech [I9681 (see also Takeuti

[ 19701). Magidor in [ a ] generalized Prikry's construction, to change a supercompact cardinal into a singular cardinal of cofinality (uI. Another construction of Magidor (using a supercompact cardinal) gives a model (of ZF) in which cf wI = cf (u2 = (0. The proof of Theorem 86b is due to Mitchell. The construction of a model with many normal measures (the example) is due to Kunen and Paris; cf. [ 19711. The core model K has been constructed by Jensen; see Dodd and Jensen [ w]. Mitchell in [m] shows that every Ramsey cardinal is Ramsey in K. I n [1976b] Menas used Silver's technique of inverse limits to construct models with a supercompact cardinal K. in which either the GCH holds, or 2" > K + . Silver's method also proves the consistency of a weakly compact (or Ramsey) cardinal K with 2" > K + . In [1977b] and [1977c]. Magidor uses a supercompact (respectively. huge) cardinal to construct a model where 2N-< X,, (respectively. 2'(- = X,, ,) for all n. and 2". > X,,,+I . Martin and Mitchell prove in [ a ] that if the closed unbounded filter is an ultrafilter, then there exist models with many measurable cardinals. The measurable case of Lemma 37.1 is due to Levy and Solovay [ 19671; Lemma 37.2a: Kakuda [ 19721, 37.2b: Kunen [ 19781. Exercise 37.1 : Jensen. Exercise 37.3: Prikry [ 1970al. Exercises 37.5. 37.6: Ketonen [ 19731. Section 38 Ramsey ultrafilters are investigated in Booth [ 19701. Theorem 91 is due to Kunen [ 19761. Exercises 38.1 and 38.2: Frolik [1968]; M. E. Rudin [1971]. The notion of a regular ultrafilter was introduced by Keisler who initiated its applications in model theory. Lemma 38.6: Frayne, Morel, and Scott [ 19621; Keisler [ 19641. Exercise 38.6: Frayne, Morel, and Scott [ 19621. Exercise 38.7: Keisler [ 19641. Exercise 38.8: Keisler [ 19631. Exercise 38.9: Jorgensen [ 19751. In [1970b] Prikry showed that a nonregular ultrafiller over (ul contradicts the axiom of constructibility. Benda and Ketonen [I9741 improved on his result, proving Lemma 38.9. Suhsequent work of Kanamori (Lemma 38.8). cf. [I9741 and [1976]. and Ketonen (Lemma 38.1 1) led to Ketonen's Theorem 92; cf. [l976b]. In [ x z ] . Magidor constructed a model that has a nonregular ultrafilter over U J ~ . Corollary (a) of Lemma 38.9: Benda and Ketonen [ 19741. Corollary (b) of Lemma 38.9 and Corollary of Lemma 38.10: Jech and Prikry [cc]. CHAPTER 7

Descriptive set theory was developed by Luzin and others following the work of Bore1 and Lebesgue on real-valued functions. After Kleene's investigations of hierarchies of number theoretic predicates it became clear that these hierarchies are closely related to Luzin's hierarchy of projective sets. The introduction of recursion theoretical methods brought about new developments in descriptive set theory.

HISTORICAL NOTES AND GUIDE TO THE BIBLIOGRAPHY

593

"Classical" descriptive set theory is treated in the books of Luzin [I9301 and Kuratowski

[ 19661; both books have an extensive bibliography. For "modern" descriptive set theory. we refer the reader to Martin's expository article [1977]. Section 39

Borel sets were introduced by Borel in [ 19051. Lebesgue in [ 19051 proved in emect Lemma 39.1. Suslin's discovery of an error in a proof in Lebesgue article led to a construction of an analytic non-Bore1 set and introduction of the operation 4.The basic results on analytic sets as well as Theorem 93 appeared in Suslin's article [1917]. Projective sets were introduced by Luzin [ 19251and [ 1927b],and Sierpinski [I9251 and [ 19271. The present notation (E and fl) appeared first in the paper [ 1959b] of Addison who noticed the analogy between Luzin's hierarchy of projective sets and Kleene's hierarchy of analytical predicates [ 19551. Lemma 39.4: Luzin [ 19301. Lemma 39.5: Luzin [1927a]. Exercise 39.7: Suslin. Lebesgue. For detailed treatment of Lebesgue measure, we refer the reader to Halmos' book [1950]; Lebesgue introduced his measure and integral in his thesis [ 19021. Sets of first and second category were introduced by Baire in [ 18991. Lemmas 39.7 and 39.9: Marczewski [1930a]. Lemma 39.8: Kuratowski and Ulam [1932]. Exercise 39.1: Marczewski [ 19341. Theorem 94a (measurability of analytic sets) is due to Luzin [1917]. Theorem 94b (Baire property) is due to Luzin and Sierpinski [ 19231 and Theorem 94c (perfect subset) is due to Suslin; cf. Luzin [1930]. The present proof of (a) and (b) follows Marczewski [1930a]. Prior to Suslin (and following the Cantor-Bendixson theorem for closed sets) Young proved in [I9061 the perfect subset result for G, and F , sets; and Hausdorm [I9161 and Aleksandrov [I9161 proved the same for Borel sets. Section 40

The subject of this section is basically classical descriptive set theory, as in the books of Luzin

[ 19301 and Kuratowski [ 19661; the terminology is that of modern descriptive set theory based on the analogy with Kleene's hierarchies (see [ 19551). Theorem 95 appeared in Sierpinski [ 19251. Theorem 96 (as well as Exercise 40.2) is due to Kondb [ 19381. The basic facts on Z;: and sets (Lemmas 40.2-40.5) are all in Luzin's book [ 19301and some are of earlier origin: Lemma 40.3 was in emect proved by Lebesgue in [1905], and Corollary I of Lemma 40.4 and Lemma 40.5 were proved by Luzin and Sierpinski in [ 19231. Reduction and separation principles were introduced by Kuratowski; they are discussed in detail in Kuratowski's book [I9661 and in Addison [ 1959bl. Exercise 40.3: Kuratowski [ 19361. The tree representation of Z: sets is implicit in Shoenfield's proof in [ 19611. Section 41

Theorem 97 and corollaries: In his announcement [I9381 Godel stated that the axiom of constructibility implies that there exists a nonmeasurable A: set and an uncountable n: set without a perfect subset. Godel did not publish the proof but gave an outline in the second printing (in 1951) of his monograph [1940]. Novikov in [I9511 gave a proof of the corollaries (Kuratowski's paper [ 19481 contains somewhat weaker results) and Addison in [ 1959al worked out the details of Godel's outline of the proof of the theorem. Theorem 98 is due to Shoenfield [ 19611. Previously. Mostowski had established absoluteness of Xi and fli predicates (Exercise 41.6). In [1965b] Levy used Shoenffield's theorem to prove Theorem 36.

594

HISTORICAL NOTES AND GUIDE TO THE BIBLIOGRAPHY

Lemma 41.3: Solovay [1967]. Lemma 41.4 is due to Kechris and Moschovakis [ 19721 (earlier. Solovay proved the lemma under the assumption of the axiom ofdeterminacy); the present proof is due to Blass [1975]. Theorem 99 and its corollary are due to Mansfield [ 19701and Solovay [1969]. Lemma 41.5 was formulated and first proved by Mansfield: the present proof is Solovay's. Part (b) of Theorem 100 is due to Martin; both (a) and (b) appear in Martin's paper [m], where (a) is attributed to the classical descriptive set theorists. Theorem lOOc is due to Mansfield [1975]; the present proof is as in Kechris [XI. Section 42 The results in this Section are due to Solovay; cf. [ 19701. &tion 43 Infinite games were first considered in the 1930s when Banach formulated the game G** and pointed out its relation to the Baire property. (Mazur then proved part (b) of Theorem 102; the first published proof of this result appeared in Oxtoby [1957].) I n [ 19531 Gale and Stewart investigated infinite games in general and proved that the axiom of choice implies that there exist undetermined games (Lemma 43.1) and that open games are determined (Lemma 43.8). I n [I9621 Mycielski and Steinhaus proposed an axiom and called it axiom of determinateness (AD). In [1964- 19661 Mycielski gave a comprehensive account of consequences of AD and related open problems. Following Solovay's discovery that AD implies that H I is a measurable cardinal, attention has been turned to the relation between determinacy and large cardinals. There have been numerous results in this direction, and I shall return to subject below. Concerning determinateness of Borel games: Wolfe in [1955] proved that E! games are determined. Then Davis in [I9641 proved determinacy for E:, and in [1972], Paris proved E2-determinacy. Finally, in [ 19751 Martin showed that all Borel games are determined (Theorem 104). Theorem 102a is due to Mycielski and Swierczkowski [1964]; the present proof (and the covering game) is due to Harrington. Theorem 102c and its corollary (and Exercise 43.2) are due to Davis [ 19641. Theorem 103 is due to Solovay; the present proof of measurability of K, (Lemma 43.6) is due to Martin [1968]. Theorem 105: In [ 19701 Martin proved that analytic games are determined if x ' exists for all x & w ; the converse was proved by Harrington [m]. Various consequences of AD and PD are investigated in the forthcoming paper [w] of Martin. Martin proved, among others, that AD implies that each K,, 3 5 n < w, is singular and that Hut is measurable. Other consequences of AD: There exist exactly two normal measures on K, (Martin and Paris); Not, is measurable (Kunen); various other cardinals are measurable (Martin, Kunen); the partition property w , + ( o , )(Martin; ~~ see Kleinberg [I9701 for further details on the partition property); there exist transitive models of ZFC with many measurable cardinals (Solovay); there exists a transitive model in which K, is a measurable limit of measurable cardinals (Kunen). Further results relating determinacy with large cardinals are in Moschovakis [1970] and Friedman [1971]. Martin and Mitchell [a)] showed that if the filter over w , is an ultrafilter. then there exist models with many measurable cardinals. Among applications of projective determinacy in descriptive set theory are the results of Addison and Moschovakis [ 19681 and Martin [ 19681 (reduction principles), and Moschovakis [ 19711 (uniformization).

,

HISTORICAL NOTES A N D GUIDE TO THE BIBLIOGRAPHY

595

Section 44

Theorem 106 is due to Martin and Solovay [1970]; Corollary 3 is due to Kunen. Example I : Solovay. Example I I : VopCnka and Hrbafek [1967], and Solovay. Example 111: Jensen [1970]; for another construction of a A: nonconstructible real see Jensen and Solovay [ 19701. Jensen and Johnsbriten [ 19741. Example IV: Harrington [1977]. Strong measure zero and Borel conjecture: Borel [ 19191. Exercise 44. I : Marczewski [ 1930bl. Exercise 44.2: Kunen. The construction ofa Luzin set using the continuium hypothesis (Exercise 44.3) is due to Luzin [1914]; its application to get an uncountable strong measure zero set if the CH is true (Exercise 44.4) is due to Sierpinski [1928b]. For more on the Borel conjecture, see Sierpinski [1937]. Laver in [ 19761 constructed a model in which the Borel conjecture is true. The argument in Exercise 44.5 is due to Erdos. The model in which .d, has a dense subset of size < 2% is due to Kunen and Solovay. Exercise 44.6: Banach and Kuratowski [ 19291.

John W. Addison [ 1959al Some consequences of the axiom of constructibility, Fund. Math. 36 (l959), 337-357. [ 3959b] Separation principles in the hierarchies of classical and effective descriptive set theory, Fund. Math. 46 (1959), 123-135. John W. Addison and Yiannis N. Moschovakis [ 19681 Some consequences of the axiom of definable determinateness, Proc. Nat. Acad. U.S.A. 59 (1968). 708-712. Pave1 S. Aleksandrov [I9161 Sur la puissance des ensembles mesurables B, C.R. Acad. Sci. Paris 162 (l916), 323-325. Rene Bake [I8991 Sur les fonctions de variables reelles, Annali di Math. (3) 3 (1899), 1-123. Stefan Banach [I9301 ober additive Massfunktionen in alstrakten Mengen, Fund. Math. 15 (1930). 97-101. Stefan Banach and Kazimierz Kuratowski [I9291 Sur une generalisation du probleme de la mesure, Fund. Math. 14 (1929). 127-131. Jon Barwise and Edward Fisher [ 19701 The Shoenfield absoluteness lemma, Israel J. Math. 8 (l970), 329-339. James E. Baumgartner [I9701 Decompositions and embeddings of trees (Abstract), Norices h e r . Math. SOC. 17 (l970), 967. [ 1976al A new class of order types, Ann. Math. Logic 9 (l976), 187-222. [1976b] Almost disjoint sets, the dense set problem and the partition calculus, Ann. Math. Logic 10 (1976). 40-439. James E. Baumgartner and Andras Hajnal [I9731 A proof (involving Martin's Axiom) of a partition relation, Fund. Math. 78 (1973). 193-203. James E. Baumgartner, Andras Hajnal, and Attila Mate [I9731 Weak saturation properties of ideals, I n "Infinite and Finite Sets." Coll. Marh. SOC.J . Bolwi, 10 (1973). 137- 158, Keszthely (Hungary). James E. Baumgartner, Jerome 1. Malitz, and William Reinhardt [I9701 Embedding trees in the rationals, Proc. Nar. Acatl. Sci. U.S.A. 67 (1970). 1748-1753. James E. Baumgartner and Karel L Prikry [I9761 On the theorem of Silver, Discrete Math. 14 (1976). 17-22. [ 19771 Singular cardinals and the generalized continuum hypothesis. Aitiur. Math. Monrh/I 84 (1977). 108- 113. James E. Baumgartner, Alan D. Taylor, and Stanley Wagon [I9771 On splitting stationary subsets of large cardinals. J . Symbolic Logic 42 (1977). 203-214. 5%

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Miroslav Benda and Jussi Ketonen [I9741 On regularity of ultrafilters, Israel J . Math. 17 (1974). 231-240. lvar Bendixson

[I8831 Quelques theoremes de la thebrie des ensembles de points, Acta M a t h . 2 (1883). 415-429. Paul Bernays [1937-19541 A system of axiomatic set theory, J . Symbolic Logic, 1. 2 (1937). 65-67; II. 6 (1941). 1-17; lll.7(1942),65-89;lV.7(1942), 133-145; V.8(1943).89-106; VI. 13(1948),65-79;VII. 19 (1954). 8 1-96. Andreas Blass

[ 19751 A forcing proof of the Kechris-Moschovakis constructibility theorem. Proc. Amer. Math. SOC. 47 (1975). 195-197. [I9771 A model without ultrafilters, Bull. Acad. Polon. Sci. 25 (1977). 329-331. George Boolos

[I9701 On the semantics of the constructible levels, Z. M a t h . Logik 16 (1970). 139-148. William Boos

[ 19741 Boolean extensions which emace the Mahlo property. J . Symbolic Logic 39 (1974). 254-268. [ 19761 lnfinitary compactness without strong inaccessibility. J . Symholic Logic 41 (1976). 33-38. David Booth

[I9701 Ultrafilters over a countable set. Ann. Math. Logic 2 (1970). 1-24. BrniIe Bore1

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Lecons sur la theorie des fonctions." Gauthier-Villars. Paris, 1898. Lecons sur les fonctions de variables reelles." Gauthier-Villars, Paris, 1905. [I9191 Sur la classification des ensembles de mesure nulle. Bull. Soc. Marh. France 47 (1919). 97- 125. I'

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[I9651 The continuum problem and the powers of alephs, Comment. Math. Unir. Carolinae 6 (1965). 181-197. [ 19731 Changing cofinality of a measurable cardinal. Cornmenr. Marh. Unit,. Carolinae 14 (1973). 689-697. [I9761 Changing cofinality of Nz.I n "Set Theory and Hierarchy Theory," Lecture Notes in Math. 537, pp. 37-49. Springer-Verlag, Berlin and New York. 1976. Georg Cantor

[ 18741 uber eine Eigenschaft des Inbegriffs alter reellen algebraischen Zahlen. J.1: Math. 77 (1874). 258-262. [I8781 Ein Bertrag zur Mannigfaltigkeitslehre, J . j : Math. 84 (1878). 242-258. [ 19321 "Gesammelte Abhandlungen *' (E. Zermelo. ed.). Springer-Verlag. Berlin and New York. 1932. Chen-Chung Chang and H. Jerome Keisler [ 19731 "Model Theory." North-Holland Publ. Co. Amsterdam. 1973. Paul J. Cohen [ 1963- 19641 The independence of the continuum hypothesis, Proc. Nut. Acatl. Sci. L'.S.A. 50 (1963). 1143 1148; 51 (1964). 105--110. [ 19661 "Set Theory and the Continuum Hypothesis." Benjamin, New York. 1966. Morton Davis

[I9641 Infinite games of perfect information. I n "Advances in Game Theory." Ann. Math. Stud. 52 (1964). 85-101. Richard Dedekind

[ 1x721 "Stetigkeit und irrationale Zahlen." Braunschweig, 1872. Patrick Dehorno)

[I9751 Solution d'une conjecture de Bukovsky. C.R. Acarl. Sci. Paris SGr. A - B 281 (1975). 821 824.

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[I9661 On strongly measurable cardinals. Bull. Acad. Polon. Sci. 14 (1966). 587-591. [I9671 The consistency of some theorems on real numbers, Bull. Acad. Polon. Sci. 15 (1967). 107- I 1 I. Philip Wolfe [I9551 The strict determinateness of certain infinite games, Pacific J . Math. 5 (1955). 891-897.

William H. Young [I9061 “The Theory of Sets of Points.” Cambridge Univ. Press, London and New York. 1906. Ernst Zermelo [I9041 Beweis. dass jede Menge wohlgeordnet werdern kann. Marh. Ann. 59 (1904). 514-516. [I9081 Untersuchungen iiber die Grundlagen der Mengenlehre, M a r k Ann. 65 (1908). 261-281. M a x Zorn [ 19351 A remark on method in transfinite algebra, Bull. A w r . Marh. Soc. 41 (1935). 667-670.

NOTATION

Page

Symbol

Page

3 4 4 4 6 6 7 7 8 8 8

25

8 8 9 9 9 9 9 9 12 12

35 35 36 37 38 41 43 43

20 20 21 21 22 25

26 29 31 31 31

43 44

14 14

46 47

16 16 16 16 18 18 19

48 57

67 71

71 72 611

612

NOTATION

Page

Page

72 74 74 78 81 86 86 93 96 100 100 109 112 114 114 114 133 I34 142 I48 148 150 I50 153 157 I58 159 161 I65 166 167 167 169 170 187 187 189 198 199 200 202 217 217 226 230

230 233 235 237 237 238 254 255 251 259 269 276 299 300 305 305 307 308 308 308 309 311 3 14 317 3 18 318 32 1 322 324 325 327 327 327 337 339 342 342 356 366 366 366 369 369 369 37 1

NOTATION

S.rtnhol

613 Page

-

Synbol

Page -

380 381 38 I 392

A;(a) WF WO LO

5 10 5 15 515 5 16 5 16 5 16 5 16 525 530 538 538 538 538 542 542 544 544 5 50 5 50 5 50 55 1 553 555 556 556 560 57 1 575

398 408 409 409 417 417 428 438 438 439 456 457 495 495 499 499 500

500 500 503 504 505 509 5 10 5 10

Il.yll

WF, WO, P[Tl A""

r, n,

BC

4 .&In

.Hc

Ra(9llJI) Ge(9lt)

AD G, a*h

a*r G,** G: cone(.u) .F,

PD . F (T ) S(Y)

This Page Intentionally Left Blank

INDEX A

B

Absolute, 530 Absoluteness, 101 Aleph, 25 Almost all. 306 Almost disjoint family, 64 Almost disjoint forcing, 283 Almost disjoint functions, 64,252,426 Almost disjoint sets, 241,252,430 Analytic set, 497 Antichain, 217 maximal, 223 AronszajnK - tree, 326 Aronszajn tree, 220 regular, 244 special, 244 Arrow, 322 Atom. 146.296.416 Atoms. 198 Automorphism, I46 Axiom of choice, 38 ofconstructibility, 101 countable axiom of choice, 41 of determinacy, 550 of extensionality, 4 of infinity, 10 Martin's axiom, 229, 563 of pairing. 4 of power set, 7 of projective determinacy, 560 ofregularity, 70 of union. 6 Axiom schema of replxement. 1 t of separation. 2. 5

Baire category theorem, 33 Baire property, 505 Baire space, 35 Bernays-Godel axioms, 76 Boolean algebra, 144 atomic, 147 atomless, 147 complete, I50 K-complete. 150 K -distributive, IS8 K -saturated, 157 u -complete, I50 Boolean universe, 159 Boolean value, 159 Borel code, 538 Borel's conjecture, 566 Borel set, 495 Boundedness lemma, 518 Box principle, 255

C Canonical embedding, 154 Canonical generic ultrafilter, 170 Canonical names, 169 Canonical projection, 238,456 Canonical Skolem functions, 341 Canonical well-ordering o f a x a ,20 Canonical well-ordering of L , 109 Cantor-Bendixson theorem. 34 Cantor-Bemstein theorem, 23 Cantor set. 3 I 615

616 Cantor's theorem, 22 Cardinal bound singular, 51 extendible, 413 huge.415 inaccessible, 52 ineffable, 390 Jonsson, 333 A - supercompact, 413 limit, 25 Mahlo. 61 measurable. 297 partition cardinal, 392 n':,2- indescribable, 385 Ramsey, 328 real-valued measurable, 300 regular, 27 Rowbottom, 333 singular, 27 strong limit, 50 strongly compact, 398 successor, 25 supercompact, 408 weakly compact, 325 weakly inaccessible, 28 weakly Mahlo, 61 Cardinality, 22 Cardinal number, 22.24 Cartesian product, 7,44 Cauchy sequence. 494 Chang's conjecture, 334 Choice function, 38 Class almost universal, 97 closed, 93 definable, 3 6 -closed, 353 extensional, 87 proper, 4 z,, - class, 114 universal, 3 Closure, 96 Cofinal, 26 Cofinality, 26 Collapsing cardinals, 182 Collection principle, 72 compact cardinal, 398 Compactness theorem, 81.149.382 Compatible, 138,147 Complete Boolean algebra, 150 K - generated, 262

INDEX

- distributive, 262 locally rigid, 262 rigid, 262 simple. 273 weakly homogeneous, 270 weakly (K,A)- distributive, 264 Completeness theorem, 81, 149 Completion of a Boolean algebra. 152 Cone, 556 Consistent, 79 Constant function, 308 Constructible hierarchy , 100 Constructible universe, 99 Continued fraction. 36 Continuum hypothesis, 3 I Core model, 469 Correspondence, 9 Countable chain condition (c.c.c.). 32,157 strong, 243 Cover, 139 Cumulative hierarchy. 92 ( K , A)

D Degree of constructibility, 291

A -lemma, 225

de Morgan laws, 144 Dense, 30,139.152 Derivative, 34 Diagonal function, 3 I 1 Diagonal intersection, 57 Diamond, 226 Direct limit, 237,456 Direct product, 187 Direct sum, 187 Disjoint, 6,144 family, 9 Domain, 8 Downward absolute, I17

E Easton product, 193 Easton's theorem, 192 Elementary embedding, 82,308 Elementary equivalence, 307 Embedding, 81,146 complete, 15 I elementary, 82,308 natural. 308

INDEX

617

E.M. set, 342 induction, 73 Equivalence class, 9 Equivalence relation, 9 Erdos-Rado partition theorem, 323 Extension of a function, 9 of a tree, 222 E

-

regressive. 58

I. - function, 114 Functional, 438 Fundamental theorem of ultraproducts, 306, 372 Fusion, 285,571 Fusion lemma, 285

G

F Factor lemma, 280,366,461 Field of a relation, 8 Field of sets, 144 K -complete, I51 Filter, 52, 139, 147 closed unbounded, 57,408 countably complete, 55 P - generic, 139 Frechet, 53 K -complete, 56, 151 normal, 60 normal on a group, 199,201 principal, 53 C T - complete, 55 Finite intersection property, 54, 147 Fixed point, 18 Fodor's theorem, 59 Forcing, 139, 165 iterated, 232,458 Forcing condition, 139 Forcing language, 142, 166 Forcing relation, 142 Forcing theorem, 142, 167, 196 Formula, 2.80 atomic, 2 nth order, 384 restricted, 82 I, - formula, I14 2. - formula, 114 n,, - formula, I14 Fubini theorem, 504 Function, 8 increasing, 13 inverse, 9 monotone, 318 n-ary,8 normal, 57 one - to - one, 9 order-preserving, 13 primitive recursive, 119

Game, 550 covering, 552 determined, 550 Gx*, 555 Gx**, 553

Solovay, 557 Generalized continuum hypothesis (GCH), 46 Generic class, 195 Generic extension, 138, 140 Generic model theorem, 140 Generic real, 543 Generic ultrafilter, 154 Generic ultrapower, 438 Gimel, 48 Godel number, 125 Godel operations, 92 Godel's completeness theorem, 81 Godel's second incompleteness theorem, 79 Ground model, 138

H Hahn-Banach theorem, 41 Hartogs number, 25 H a u s d o e s formula, 48 Homogeneous set, 321,327 Homomorphism, 146 complete, 151

I Ideal, 53, 147 atomless, 416 countably complete, 56 K -complete, 56, 151 K - saturated, 298 A - saturated, 428 normal, 60 precipitous, 439

618

INDEX

prime, 55, 148 u - ideal, 56 u - saturated, 298 thin, 58 Image, 9 inverse, 9 Inclusion, 4 Infimum, 12 Indiscernible, 329 Silver, 337 Initial segment, 13 Interpretation, 80, 167 Inverse limit, 456,457 I - partition, 438 Isomorphism, 13, 81, 146 Isomorphism theorem, 74 Iterated forcing, 458 Iterated ultrapower, 362

J Jech-Sochor embedding theorem, 208 Jensen's covering theorem, 356 Jensen's principle, 226 Jonsson cardinal, 333

K -chain condition, 157 - collapsing algebra, 276 K - saturated partially ordered set, 157 Kernel, 148, 199 Kond8'sIIt - uniformization theorem, 521 Konig's theorem, 45 Kuratowski-Zorn lemma, 40 Kurepa tree, 247 K

( K , A)

Lowenheim-Skolem theorem, 81 Lower bound. 12

M Mahlo operation, 314 Mapping, 9 Martin's axiom (MA), 229, 563 Maximal, 12 Measurable cardinal, 297 Measure, 295 atomless, 296,480 fine, 398 finitely additive, 480 K - additive, 299 Lebesgue, 295 normal, 316,416 normal fine, 408 product, 422 u - additive, 295 strongly normal, 412 translation invariant, 295 two-valued, 296 Measure algebra, 421 Measure space, 421 Minimal, 12 Model, 78,80 Boolean-valued, 159 extensional Boolean-valued, 161 full Boolean-valued, 160 ground, 138 permutation, 199 ( 2 ,a)- model, 343 transitive, 82 Mostowski's collapsing theorem, 88 Move, 550

N L Language, 2.80 Lebesgue measure, 295,503 Length of a well-founded relation, 21 Levy algebra, 280 Levy collapse, 191 Levy-Shoenfield absoluteness lemma, 120 Lexicographic ordering, 18 Limit, 18 Limit ordinal, 15 Lindenbaum algebra, 145

Name, 166 hereditarily symmetric, 202 symmetric, 202 Natural embedding, 308 Natural number, 16 Norm, 67 Normal form, 5 10 Normal form theorem, 92 Notion offorcing, 139 A -closed, 180 A -distributive, 180 stronglyA -closed, 459

INDEX

619

Null set, 298 Number algebraic, 32 transcendental, 32

Q Quantifier, 3 restricted, 83 Quotient, 10, 148

0 Operation, 9 Ordering linear, 12 partial, I2 Order-type, 14, 18 Ordinal, 14 Ordinal number, 14

P Pair, 4 ordered. 4 Partition, 9, 139, 156 property, 321 Perfect set forcing, 284 Permutation, 199 Play, 550 Point condensation, 35 isolated, 33 limit, 33 Polish space, 494 Positive measure, 416 Power set, 7 p-point, 257 Precipitous ideal, 439 Preservation of cardinals, 177 Prime ideal theorem, 148 Principle of dependent choices, 41 Product, 19. 156 Cartesian, 7 K - product, 189 of measures, 42 I of partially ordered sets, 187, 189 reduced, 54 Product lemma, 188 Projection, 317,497 Projective determinacy. 550.559 Projective set. 500 Property of Baire, 505 Property,A,,, I14

R Ramsey cardinal. 328 Ramsey's theorem, 321 Ramsey ultrafilter, 259,478 Random real, 543 Range, 8 Rank, 21,72 Real-valued measurable cardinal, 300 Reduction lemma, 97 Reduction principle, 523 Refinement, 158,438 Reflection principle, 89 Relation, 8 arithmetical, 515 extensional. 87 n - ary. 8 reflexive, 9 symmetric, 9 transitive, 9 well-founded, 21, 75 Relative constructibility, 127 Representation lemma, 37 1 Restricted separation, 93 Restriction of a function, 9 Rowbottom cardinal, 333 Rudin-Keisler order, 318 Russell's paradox, 2

S Satisfaction. 78 Scott's theorem, 31 I Sentence, 3 Separation principle, 523 Separative partially ordered set. 152 Sequence, 16 finite, 16 nondecreasing, 18 normal, 18 transfinite, 16 Set adequate. 110

INDEX

analytic, 497 arithmetical, 5 10 at most countable, 25 Borel, 495 closed, 32,60 closed unbounded. 56,408 Cohen generic, 141 compact, 33 complete linearly ordered, 30 constructible, 100 countable, 25 definable, 82 dense linearly ordered, 30 D-finite. 28 empty. 6 finite, I I of first category, 505 generic, 139 hereditarily countable, 86 hereditarily ordinal-definable, 134 hereditarily symmetric, 199 homogeneous, 321,327 of indiscernibles, 329,342 Luzin, 566 '%-definable, 136 meager, 147,505 minimal overW, 284 nowhere dense, 147,505 null. 298, 503 open, 32, 139 open dense, 158 ordinal-definable, 132 perfect, 33 projective. 500 rare, 505 regular open, 145 remarkable, 346 of second category, 505 separable linearly ordered, 32 stationary, 58,408 of strong measure 0, 566 symmetric, 199 7'- finite, 10 thin, 58 transitive, 10 uncountable, 25 Set of indiscernibles. 329,342 Shoenfield's absoluteness theorem, 530 u - algebra, 494 Silver indiscernible, 337 Silver's principle W ,254

Singleton, 4 Singular cardinals hypothesis, 61 Skolem constant, 382 Skolem function, 82, 331 Skolem hull, 331. 342 Skolem sentence, 382 Skolem's normal form theorem, 121 Skolem term, 331 Solovay set of reals, 544 Space complete, 494 separable, 494 Stone's representation theorem, 149 Strategy, 550 winning, 550 Strong countable chain condition, 243 Subalgebra, 146 complete, 151 completegererated by X, 152 generated b y X , 146 Submodel, 81 elementary, 81 2 , - elementary, 126 Subset, 7 proper, 7 Successor, 25 Successor ordinal, 15 Sum, 18 Supercompact cardinal. 408 Support, 371 Supremum, 12 Suslin algebra, 220.274 SuslinK -tree, 335 Suslin line, 216 Suslin operation, 499 Suslin's problem, 32,216 Suslin tree, 217 Symmetric submodel, 202 Symmetry, 214

T Tarski's ultrafilter theorem, 55 Term, 80 Transfinite induction, 16 Transfinite recursion. 17 Transitive closure, 71 Transitive collapse, 76.88 Transitive set, 14 Tree, 37,216,289,524 Aronszajn, 220

INDEX

621

automorphism of, 224 binary, 334 branch of, 2 I7 height of, 217 homogeneous, 224 isomorphism of, 224 Kurepa, 247 level of, 217 normal, 218 perfect, 37,289,571 regular Aronszajn. 244 rigid, 224 special Aronszajn, 244 Suslin, 217 Tree property, 326 Truth definition, 125 Tychonoff theorem, 41

U Ulam matrix, 302 Ultrafilter, 55, 148 generic, 154 iterable M - ultrafilter, 372 ( K , A) - regular, 406 - complete, 154 M - ultrafilter, 352 minimal. 480 p-point, 257 Ramsey, 259

w

regular, 482 uniform. 256,405 .T -complete, 156 Ultrafilter theorem, 55 Ultraproduct, 306 Ultrapower, 305,307 iterated, 362 Union, 6 Upper bound, 12 Upward absolute, 117

V Vope'nka's principle, 414

W Weak compactness theorem, 382 Weakly compact cardinal, 325 Well-ordering, 13 Well-ordering theorem, 39

Z Zermelo-Fraenkel axioms, I Zermelo's well-ordering theorem, 39 Zero-dagger, 380 Zero-sharp, 339,347 Zorn's lemma, 40

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