Aspects of Constructibility

Lecture Notes in Mathematics Edited by A. Dold, Heidelberg and B. Eckmann, ZUrich

354 Keith J. Devlin University of Oslo, Oslo/Norway

Aspects of Co n stru cti b il ity

Springer-Verlag Berlin. Heidelberg • New York 1 973

A M S S u b j e c t C l a s s i f i c a t i o n s ( 1 9 7 0 ) : 02-01, 02-02, 02 K 99, 02 H 13

I S B N 3-540-06522-9 S p r i n g e r - V e r l a g B e r l i n • H e i d e l b e r g • N e w Y o r k I S B N 0-387-06522-9 S p r i n g e r - V e r l a g N e w Y o r k - H e i d e l b e r g • B e r l i n

This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically those of translation, reprinting, re-use of illustrations, broadcasting, reproduction by photocopying machine or similar means, and storage in data banks. Under § 54 of the German Copyright Law where copies are made for other than private use, a fee is payable to the publisher, the amount of the fee to be determined by agreement with the publisher. © by Springer-Verlag Berlin • Heidelberg 1973. Library of Congress Catalog Card Number 73-15453. Printed in Germany. Offsetdruck: Julius Beltz, Hemsbach/Bergstr.

PREFACE

This set of notes is essentially eonstructibility

theory.

a first draft of a proposed book on

As such it is probably more polished

notes" but lacking the finish of a true "textbook". should publish

this draft for two reasons.

for improvements

and enlargements

this way we publish something , whereas

It seemed desirable

Firstly, by exposing

general view, we hope to gain some reader-feedback suggestions

than a mere "set of that we

the material

in the form of corrections

(or otherwise).

Secondly,

the extra work required to produce a text of of the

for such an undertaking.

Basically, information

and/or

at least

reasonable quality might well take up several years due to the limitations time available

to

the book sets out to collect together most of the existing

on constructibility,

first draft, it is inevitable

much of which is fairly recent.

However, being a

that there will be several omissions.

(For example,

it

occurred to us after the manuscript was typed that we had not included the Levy absoluteness

theorem,

although the closely related Shoenfield

absoluteness

theorem

does in fact appear.) At the other end of the scale, we have included here some results which, because of their very detailed and technical nature, will most likely not appear in any future revision of these notes. forms of Kurepa's hypothesis,

(For example, we include here very general

involving pairs of cardinals,

forms would probably be more suited to a "graduate Nevertheless, pleased to hear),

Department

May 1973.

the book as it stands will be of some value.

of Manchester,

England.

the simplest

text".)

we hope that despite its many shortcomings

of Mathematics,

The University Manchester,

whereas

(of which we shall be

ACKNOWLEDGEMENTS

It is doubtful whether we would ever have come to write this book had it not been for the kind invitation

of Professor J°E. Fenstad to spend the Autumn of 1973

at the University of Oslo, Norway.

This enabled us to come into contact with

Professor Ronald Jensen, whose conversations with us led to our being able to give an account of much of his recent work. This book was written visiting

the University of Manchester,

K. Anderson, debt.

(rather quickly)

of Manchester University,

during the Spring of 1973 whilst we were

England.

The manuscript was typed by

for whose care and patience we owe a great

INTRODUCTION

Although still in a preliminary form only, this book is structured so as to provide one of those ever useful creatures, "a graduate text" on its subject. Accordingly, section i of the book is intended to provide just such a treatment of constructibility, as well as forming a starting point for the remainder of the book. Thus, in section I, we have tended to assume very little in the way of prerequisites, and have given most of the proofs in some detail.

Chapter 5 is somewhat of an

exception to this prerequisites policy, since we found it necessary to assume a fair knowledge of generalised recursion theory.

However, it is doubtful whether the

material in this chapter would be of interest to anyone not familiar with that theory, and in any case the assumed material is easily obtainable from another source (which we give at the outset of the chapter).

The sectioning of the book is really just a

grouping of the chapters into somewhat coherent units which are to some extent independent of the other sections. With the single exception mentioned above, therefore, Section i is intended to be readable for anyone with an elementary knowledge of mathematical logic, and forms a basis for the rest of the book. Section 2 requires no more prerequisites than those for section I, though the proofs tend to be fairly long and detailed.

Sections 3 and 4, though independent of each

other, depend heavily upon Section 2, and thus form an exception to our "sectional independence" rule above.

(Section 3 also really requires a knowledge of weakly

compact cardinals, though we do in fact state all the standard results we are assuming.)

In section 4, chapter 13 can be read independently of chapters 12 and 14,

since its inclusion in that section is primarily that chapter 14 provides both a motivation for, and a subsequent application of, the results of chapter 13.

Section

is fairly self-contained, though some knowledge of large cardinal theory would be an advantage.

Section 6 can be coupled with Section I, if desired, to form part of

the "graduate text", since it proceeds directly from the point where chapter 4 ends, and requires little extra in the way of prerequisites. this is literally true.

In the case of chapter 18,

For chapters 19 and 20, however, some knowledge of measurable

VIII

cardinals

and strongly compact cardinals,

chapter 19 requires

respectively,

is desirable.

some familiarity with indiscernability

It should be noted that the material presented part of the material of sections

In particular,

arguments.

in sections

1 and 6, is due almost entirely

2, 3, and 4, and to one person,

Ronald Jensen. Since the present version of the book is only a preliminary spent a great deal of time preparing a good list of references. glance will indicate other source.

that there are more references

This is purely because

one, we have not In fact, a cursory

to our own papers than to any

to give such references

is extremely easy; in

particu-lar the fact that six of our papers and only two papers of Jensen appear in the reference

table should on no account be taken to mean that our contribution

constructibility

to

theory is three times that of Jensenl

We should perhaps mention one rather glaring omission from the contents page. We have ignored admissibility entire personal knowledge secondly,

theory entirely.

This is because,

firstly, almost our

of this topic is already contained in our article [ 3 , and,

it would appear that several people are at present engaged upon producing

books on that very subject. Finally,

a word concerning notation.

We have tried to adopt the notation which

it seems to us is the most common in current usage amongst set theorists.

For the

benefit of those readers who like to skip all proofs at the first reading we use either QED or else the symbol | to denote the end of a proof of a lemma or theorem. When the symbol corollary

| occurs immediately

follows immediately

following,

say, a corollary,

from the theorem concerned.

a (proper~) initial segment of the set of positive fashion)

the set of all theorems,

integers

lemmas, and corollaries.

we mean that the

Within each chapter we use to index (in a one-one When we are outside of

chapter n, we refer to theorem m of chapter n as "theorem n.~', etc. We use the following abbreviations: 'iff'

for 'if and only if'.

's.t.'

for 'such that'.

'LHS'

for 'left hand side'

(of an equation,

equivalence,

etc.).

IX

'RHS'

for 'right hand side'.

'a.e.'

for 'almost everywhere' (in connection with ultrafilters).

'w.p.f.' for 'with parameters from'.

CONTENTS

SECTION I.

ELEMENTARY THEORY

Chapter O.

Zermelo-Fraenkel Set Theory

I.

The Constructible Universe

24

2.

The Axiom of Constructibility, The Condensation Lemma, and the GCH

29

3.

The Souslin Hypothesis

43

4.

The Kurepa Hypothesis

53

5.

Constructibility and the Continuum

59

SECTION 2.

i

JENSEN'S FINE STRUCTURE THEORY

6.

The Jensen Hierarchy of Constructible Sets

73

7.

The Fine Structure of the Jensen Hierarchy

86

8.

The Combinatorial Principles

9.

The Generalised Souslin Hypothesis for Successor Cardinals

SECTION 3.

104

~K

115

ON LARGE CARDINALS IN THE CONSTRUCTIBLE UNIVERSE

i0.

Ineffable Cardinals and the Generalised Kurepa Hypothesis

119

Ii.

Weakly Compact Cardinals and the Generalised Souslin Hypothesis

130

SECT%ON 4.

MODEL THEORY IN THE CONSTRUCTIBLE UNIVERSE; MORASSES

12.

The Gap-I Two Cardinal Theorem

138

13.

Gap-I Morasses

145

14.

The Gap-2 Two Cardinal Theorem

171

SECTION 5. 15.

LARGE CARDINALS AND CONSTRUCTIBILITY Small Large Cardinals. Ineffable)

(Inaccessible, Mahlo, Weakly Compact, 183

16.

Medium Large Cardinals (~ * (m)<m)

17.

Large Large Cardinals (K ÷ (ml)

<~

, Ramsey, measurable)

186 188

XII

SECTION 6.

RELATIVE CONSTRUCTIBILITY

18.

Relative Constructibility

206

19.

The Inner Model L[~]

214

20.

Strongly Compact Cardinals and Relative Constructibility

229

ZERMELO-FRAENKEL

SET THEORY

In this chapter, we give a brief sketch of the development results in Zermelo-Fraenkel throughout.

However,

set theory.

of the elementary

This is the system in which we shall work

since we shall have need to examine the system itself, we shall

carry out this development

in a fairly rigorous manner

which can easily be suitably rigorised). should consult the excellent

(or, at least,

in a manner

For a more complete account,

little book

([I~)

the reader

of Rotman and Kneebone.

Towards

the

end of the chapter, we cover some material not contained in [I~ , and in this case we give more details than elsewhere

in the chapter.

notions of formal systems in first-order systems, ([I~)

and interpretation

of Mendelson

these topics.

languages,

We assume familiarity with the logical deduction

of such systems in first-order

(up to chapter 3 ) still provides

structures.

(formal or otherwise).

denotes the satisfaction

relation

for

~

If ~

The book

one of the best accounts of

We use the symbols ~ and ~ (somewhat interchangeably)

logical consequence

in such

is a first-order

to denote structure,

~

in the language for ~[ having constants

for

each member of ~1. By the language of set theory, ~ predicates = (equality) V,

, we mean the first-order

and g (set membership),

language with

logical symbols, A, v , 7 ,

+, ~-~, 3 ,

and variables v O, v I, v 2, . . . . The primitive

formulas of ~

The formulas of ~ schemata. ~Vn~.

are strings of the form v m = v n and v m g v n .

are generated from the primitive

If ~, 4 are formulas,

so are ~ ^ ~, ~ v 4, - ~ ,

These have their usual meanings.

first-order

formulas.)

assumed known.

formulas by the following ~ ~ 4, ~ +-~ 4,

~Vn~,

(We generally use ~, ~, etc. to denote

The notions of free and (quantifier-)

bound variables

are

A statement is a formula with no free variables.

We write x ~ y for ~ x

E y and x # y for ~ x

etc. to denote arbitrary variables We shall assume ~

= y.

(We generally use x, y, z,

of ~ .)

also contains the bounded quantifiers

( I v m c v n) and

-2-

( ~v m

Vn), where m # n.

schemata. (We use -= to

identity. ) The bounded

the symbols

quantifiers

=-and

~'.

are assumed

These will

They are defined

and

= x ~-+ +(y)].

~'.x~(x)

which occurs

-

~x ~y[y

(We write

collection

"abbreviates"

y ~ z ~ (Vx

to

however,

~ y)(x s z)

+(x) to mean that x is a variable

~(y).

the formula

Clearly,

~'x(x c y ~+ ~(x)).

then,y = {xl#(x)}

means

Thus

that y is the

of all those x for which ¢(x) holds.

The formula On(x) ~ x)(~z

is defined

is defined

g x)(y ~ z

The Zermelo-Fraenkel

by Tram(x)

by On(x)

-- ( ~ y

s x)(~/z

~ y)(z e x).

~ Trans(x) A

~ z c y v y = z).

Axioms

The axiom system ZF consists

of the usual

calculus ~ ) together with

(I)

Extensionality.

(II)

Pairing.

(III)

Union.

(IV)

Infinity.

(V)

Power set.

(VI)

Foundation.

(VII)

Separation

V x ~y[ ~ z ( z

axioms

the following E x

for first-order

logic

(for the

axioms:

z ~ y) +-+ x = y] .

~ x ~y ~ z ~ w [ w e z +-+ w = x v w = y] . ~x~y~z[z

s y ~-+ ( ~ u

~x[On(x)

A (~y

~ x)(z ~ u)].

~ x)( 3 z e x)(y E z)].

~'x ~ y h/z[z e Y ~-+ z ~ x]. Vx[ ~ y ( y e x) ÷ (schema).

~y(y e x

~x ~y ~ z [ z

(VIII) Replacement

(schema).

V x ~y~(x,

is no finite

(countably)

y) ÷

~ ~(z)],

(h/u)(~v)(

~ an ~ - f o r m u l a

with

V x £ u)( ~y ~ v)#(x,

with free variables

infinitely

set of axioms which

e y)(z # x)].

z.

an ~ - f o r m u l a Note that there are

~ (~z

e y +-> z g x

free variable

that there

We also introduce

free in ¢.)

The formula Trans(x)

predicate

.

as abbreviations,

by the schemata

We use y = {xI~(x )} to abbreviate y e {xI~(x)}

to be part of ~

always be considered

and not as part of 4[ .

(Vy

by the following

~ v m Ivm g v n ~ 4] ' ( @ v m E v n )4 ~ ( ~/Vm) Ivm e v n ÷ 4] •

( ~ v m e Vn) ~ mean

These are introduced

many axioms

y),

x, y.

for ZF.

(We prove

imply all of the above axioms.)

later

Axiom I

-3-

tells us that ~ really is set membership. pair {x, y} of any sets x and y. sets.

Axiom II allows us to form the unordered

Axiom III allows us to form the union of any set of

Axiom IV will guarantee the existence of an infinite ordinal.

Axiom V allows

us to form the power set of any set, the set of all subsets of the set. tells us that ~ is a well-founded induction arguments on s.

relation,

and hence will enable us to carry out

Axiom VII allows us to form the set of all members of a

given set with the property 9, for all properties ~ expressible in ~ . will guarantee

Axiom Vl

Axiom VIII

(with axiom V~) that every "function" with a domain has a range.

By axioms IV and VII, there is a set, ~, such that

Vx(x # ~), the null set.

By axiom II, we can form the sets {x}, {x, y} for given x, y.

Hence, by a further

application of this procedure, we can form the ordered pair <x, y> = {{x}, {x, y}} of x, y, etc.

The sets 0 = @, I = {O5, 2 = {O, 15, 3 = {O, I, 25,... all exist, and

satisfy On(x). =

For each n so defined, we can form <Xl,..., x > = <Xl, <x2,..., x >> n n

..., the ordered n-tuple of xl,..., x n.

in axiom III is unique. are defined as usual.

We denote it by

By axiom I, the set y guaranteed to exist dx.

We write x u y for

We write x - y for {z c xlz ~ y}.

guaranteed to exist in axiom V is unique.

We denote it by

~{x, y}.

N and

By axiom I, the set y ~(x),

the power set of x.

The Axiom of Choice is the axiom: (AC)

'~x[(Vy ~ x)(y ~ ~ ) ~ ( ~ y , y '

c x)(y ~ y' ÷ y n y '

ZFC denotes the system ZF + AC. occasions

= 9) ÷ ~z(~y c x)(3~v ~ y)(v c z)]

ZFC is, to be precise, our metatheory.

On the

(all of them in section I) when it is important to note that AC is not

being used in an argument, we write, for example, ZF ~ 9, or ~ ZF P" For instance, it is easily seen that ZF ~ AC +-+ "Every set can be well-ordered". Elementary Theory of ZFC Objects which can be proved to exist in ZFC are called sets. the collection of all sets, denoted by V. It will be clear why when we do this.) of ZFC. variable,

The universe is

(We sometimes refer to V as the real world.

Since we are working in ZFC, V is a "model"

If ~ is some property of sets defined by an ~ - f o r m u l a with one free the collection of all x such that ~(x) is called a class.

We have already

-4-

fixed a n o t a t i o n for such a collection in the case where it is a set, n a m e l y {xI~(x)}.

We use the same n o t a t i o n even w h e n it is not a set, in this case calling it a

proper class.

Thus,

for example, V = (xlx = x}.

In the case of proper classes,

n o t a t i o n {xI~(x) } should be looked on purely as a notational y c {xI~(x)}

abbreviating ~(y).

(In real situations,

convenience, w i t h

this can actually be an

abbreviation~)

Hence, {xI~(x) } = {xI~(x)}

to p r o p e r t y ~.

We cannot write x = {yI~(y)} w h e n e v e r {yI~(y)}

simply means that p r o p e r t y ~ is equivalent

course, since the class {xlx ~ x} is clearly not a set. N, etc. to denote classes

the

(proper or otherwise).

is a p r o p e r class of

In general, we use X, Y, M,

We use x, y, z, etc. to denote sets

(as w e l l as variables "for sets", of course). A class M is transitive iff x c y c M ÷ x E M x E M ÷ x ~ M; w h i c h is to say

~M

~M

(which is the same as saying

if M is a set).

Hence, x is transitive iff

rrans (x). A n ordinal is a transitive set w h i c h is linearly ordered by c. ordinal iff On(x). ~(~)

We use ~, B, Y, etc.

instead of

~x[On(x)

~ ~(x)] and

to denote ordinals. ~(~)

is an ordinal, then by the a x i o m of foundation, If ~, B are ordinals,

instead of c well-orders

Moreover, we w r i t e ~x~On(x) ÷ ~(x)].

If

~ of course.

then it is easily seen that ~ = B or ~ ~ B or B ~ ~.

Hence the class On = {xI0n(x)} is w e l l - o r d e r e d by ~. N o t e that ~ < B + ~ ~ ~ B.

Thus x is an

We often write ~ < B for ~ ~ B.

It is easily seen that for any ~, ~ = {BI~ < ~}.

(This is

in k e e p i n g w i t h our e a r l i e r d e f i n i t i o n of O~ I, 2, 3,..., the natural numbers.) If e is an ordinal, so is ~ o {~}, and there is no ordinal y such that < ¥ < ~ ~ {~}.

We call ~ u {e} the successor of ~, and denote it by ~ + I.

ordinal of the form e + 1 is called a successor ordinal. is a successor ordinal. successor ordinal,

lim(~) +-+ (~B ~ ~ ) ( ~ y is a least it by ~.

We w r i t e succ(~) iff

Thus, succ(~) +-+ ( ~B c ~)(e = B ~ {B}).

it is called a limit ordinal, c ~)(B s y).

If ~ is not a

and we w r i t e lim(~).

Thus

By the axioms of infinity and foundation,

(in the sense of <) limit ordinal, w h i c h is unique of course. Clearly, ~ is just the set of all natural numbers.

by the formula On(x)

~ lim(x) a x ~ ~

~ (Vy

Any

there

We denote

Note that ~ is definable

~ x)(y ~ ~ ÷ succ(y)).

-5-

By means of the axiom of foundation, proof by e-induction

every instance

of the following

schema of

is a theorem of ZF:

V x [ ( ~ y E x)~(y) ÷~(x)] ÷ ~x¢(x) where ~(Vo) is a formula of ~

with free variable

v 0 and x and y do not occur in

¢(v0). A particular

case of the above is when

reduces to a transfinite

version of the "principle

(i.e. on the natural numbers). wish to establish separate

arguments)

the principle

~(~),

If X is a class

In applications

we are usually

~(0),

itself,

~(~) ÷ ~(~ + i), and

n e ~,

write X x Y for the class

Xn

denotes

for) sets.

In such cases,

[lim(~)

induction"

~ (~B

(often by quite

< ~)~(8)]

÷ ~(~)-

(By

~e~(~).)

the class

to denote

{<x I

,. .., Xn>IXl,...,

x n e X}.

We

Thus X 2 = X × X.

a finite string of variables

the length of the string

determined by the context.

of proof by mathematical

of the schema in such cases, where we

{<x, y>Ix e X a y e Y}.

We generally write v, x, etc.

in which case the schema

reduced to establishing

this clearly implies

,

~(x) + On(x),

or (names

(which may be arbitrary)

is

For example, we could write X n = {<~>I~ e X}.

An n-ar[ relation on a class X is a class R c_ X n. relation on X, we set dom(R)

= {<~>I ~yR(y,

accordance with our conventions, property

defining it.)

ran(R).

In any case, dom(R)

~)}

If R is an (n + l)-ary

and ran(R) = {Yl 3 ~ R ( y ,

we have here identified

~)}.

(In

the class R with the

If R is a set, then clearly so also will be dom(R) is the domain of R and ran(R)

the range.

and

R is total if

dom(R) = V. If R is an (n + l)-ary relation on a class X, and Z is a subclass denotes

the relation {
are R~Z and R"Z.

x> e RI~ e Z}, and R"Z denotes

If Z is a set

ran(R~Z).

of X, R~Z

If R is a set, so

(and even if X, R are proper classes)

then R~Z and

R"Z are sets. An n-dry function on X is an (n + l)-ary relation R on X such that -+

(~<x> cases.

e dom(R))(~'.y)R(y,

x).

We often write R(~) = y instead of R(y, ~) in such

-6-

The identity function

is defined by id = {<x, x>Ix = x}.

Thus id is a total,

unary function, with range V. A function whose domain is an ordinal domain,

is called a sequence;

if a is that

it is an a-sequence.

We often denote a function where A = dom(f). < ~, we write

In particular,

e A>,

such that f(~) = x~ for all

relation on V, there is a corresponding

by induction on that relation,

all instances

of which are provable

Using this fact, every instance

by induction on a w e l l - f o u n d e d (n + 2)-ary function, well-founded.

if f is an a-sequence

notation

f = <xgl ~ < ~>.

Given any w e l l - f o u n d e d

e-induction.

f by the pseudo-explicit

relation

of the following

is provable

h a total unary function

Then there is a unique,

total,

in ZF.

schema of proof

in ZF, as for

schema of definition

Let g be a total

such that the relation (n + l)-ary function

z e h(y) is

f such that

f(y, ~) = g(y, ~, ). A particular

case of the above is when h = id, when we obtain the schema of

definition by e-inductlon. The function TC is defined by induction x V by the e-induction

as follows.

Define

f(0, x) = x, f(n + I, x) = Uf(n, x).

f with domain

Define TC by TC(x)

=

Of"(m X {x}). It is easily seen that TC(x) x ~ Y. always

We call TC(x)

is the

the transitive

g-smallest

closure

of x.

transitive

class Y such that

Since TC is a function,

TC(x)

is

a set of course.

The function U{rank(y)

rank is defined by the e-induction:

+ lly e x}.

rank

(~) = O, rank

(x) =

Thus rank: V ÷ On in such a way that x e y ÷ rank(x)

(We use the usual notation

for functions,

f is a function and dom(f)

= X and ran(f) ~ Y.)

The cumulative hierarchy

so that,

for instance,

f: X + Y means that

of sets is defined by the E-induction

Vo

=

Va

= u{P(vs) IB_,

< a}.

< rank(y).

(on On)

-7-

By the axiom of foundation,

V =

~{V

le e On}.

It is easily seen that rank(x)

is just the least e such that x E V + I.

It is also easy to prove that:

(i) ~ < 8 + V~ c VB;

UB< e VB; and

(ii) lim(~) + Va =

The cumulative building

hierarchy

the universe

formalises

the manner

up from @ by collecting

(iii) V + 1 =

@(Va).

in w h i c h we intuitively

think of

all subsets of what we have so far, at

each stage. Ordinal Numbers The notion of an ordinal number has already played quite an important our development.

The arithmetic

set theory, however.

of ordinal numbers plays

We thus restrict

ourselves

role in

a fairly m i n o r role in our

to an extremely

cursory glance at

this. If ~, 8 are ordinals,

we define

+ (S + I ) + 6 =

the ordinal

sum ~ + B by the induction

(on B):

= (~ + ~) + I

U { ( ~ + B) IB < ~}, if lim(a).

The ordinal sum is not commutative.

For example,

2 + m = ~ but m + 2 > ~.

However, we clearly have ~ < e + B for all B. The ordinal product

~, B is defined by the induction

~-(B + i) ~.6 =

(~)

0{~.BIB

The ordinal product > m.

:

+ < ~}, if lim(6).

is not commutative.

Note that {m~I ~ E On} enumerates The ordinal power B

(on B):

For example,

2.m = m but m.2 = ~ +

all limit ordinals.

is defined by the induction

(on B):

~(~+i) : B . ~ 6

Ordinal

:

O{~SI 8 < 6 } ,

exponentiation

if

lim(a).

is not strictly increasing.

For instance

2 ~ = ~.

-8-

In the above, we have tacitly assumed that if A is a set of ordinals, is an ordinal.

This is obvious

from the definition.

ordinal ~ such that ~ E A + ~ < ~.

then

UA

We define sup(A) = the least

Thus if A has no m a x i m u m member,

then sup(A) =

If, on the other hand, ~ is the m a x i m u m member of A, then sup(A) = ~ + 1 and ~ = Note that rank(x)

UA.

= sup{rank(y) ly s x} for any x.

Let % be a limit ordinal, all ~ < %.

~A.

A is cofinal

return to these notions

A

~ %.

We say A is closed in % iff

in ~ (or unbounded

in %) iff

(~

< %)( ~ 6

D(A ~ )

s A for

e A)(6 > ~).

We

later.

Cardinal Numbers omT~

A cardinal 2,..., n,..., By AC, bijection Clearly,

is an ordinal ~ such that

~ are all cardinals.

-~(~B

< ~)(~f)(f:

We call < the cardinality

if ~ (= {~I~ < <}) is a cardinal,

If x, y are sets Xy denotes

particular,

(or power)

then

of x, denoted by

uncountable.

If K, X are cardinals

K+ denotes

If lim(B) and <~ le < 6> is a strictly

to be

All such cardinals

the least cardinal

the Cantor theorem above and the axiom of foundation, the form K + is " called a successor cardinal.

~

I~KI.

[15~), 2 < > ~ for all cardinals

greater than m.

If K is a cardinal,

Ixl.

I
the set {flf: x ÷ y}.

(see , for example,

there is a cardinal

O, i,

~ such that there is a

hence ~ = {~I~ < K} and % = {~I~ < ~}), we define the cardinal well known result of Cantor

Clearly,

We use K, ~, ~ etc. to denote cardinals.

for every set x there is a unique cardinal

f: < ~+ x.

B ÷ ~).

increasing

By a <.

In

are said to be

greater than K.

K + always exists.

Any other cardinal

(and

By

A cardinal

of

is a limit cardinal.

sequence of cardinals,

then

sup~
thus:

enumeration

<m I~ e On> of the infinite

cardinals

is defined by

~0 = ~ + ~+l

= ~

~6 = s u p ~ < ~

Clearly,

for each ordinal

form a proper class.

, if lim(6).

~ there is a cardinal

It is a closed class

(i.e.

~

(~ ~).

for any ordinal

Hence the cardinals ~,

O{
-9-

a cardinal } is a cardinal.

Note that the existence of limit cardinals

all such - but see later concerning ""inaccesszble " " cardinals) requires,

the axiom of replacement.

For example,

function f: ~ o+ On which sends n s w to ~ • n

m

follows

(in fact of from, and

exists because we can define the

And of course, ~

is a limit cardinal

iff lim(~). By definition of ~ e+l" 2w~ ~ ~ + i equality ever/always/sometimes GCH, or the generalised convention

holds here.

It is not decidable in ZFC whether

The assumption

continuum hypothesis.

for cardinals

in the above to mean

for all e.

(Note that we have used the same

as we did for ordinals, writing,

~K(<

(~<)(2 < = <+) is known as

for example,

~ K ( 2 K = <+)

a cardinal ÷ 2 K = K+).)

The sequence of power cardinals is defined by the induction

]0

= ~

]~+I

= 2]a

]6 = sup~<6]~'

if lim(6).

( ] is the Hebrew letter "beth".) Thus the GCH says

~(~

as is easily seen, 2K = I ~(K)I Cardinal Since, however,

= ~ ).

These are called "power cardinals"

because,

for all K (infinite or not),

arithmetic will be quite important to us, so we spend some time on it. it tends to be trivial, we shall not need to spend much of this time

on proofs. If <, % are finite, then cardinal arithmetic arithmetic.

In what follows,

therefore,

for <, ~ coincides with Peano

assume <, ~ are such that at least one of

them is infinite. The cardinal <
sum < + ~ is defined to be max(<, ~).

< B> is a sequence of cardinals,

g
the cardinal

sum [ e < B <

is defined to be l{flf: 8 ÷

(~

<sK ) & ( ~ e

if

is defined to be

Hence if lim(B) and <<~Is < 8> is strictly increasing, [ e < B <

The cardinal product K.% is defined to be max(<, %). Na<~<~

More generally,

= sup
More generally,

< 8)(f(~) e <e)}l,

the eardinality

-i0-

of the cartesian product of << I~ < B>.

In particular,

The following lemma is easily proved.

<% = ~ <%<, of course.

(Unless otherwise stated, all lemmas and

theorems are theorems in ZFC, formally provable from the axioms.) Lemma I Let K , ~ < B, be cardinals. (i)

Then:

~(%<2< = H <2< ~ = max(<0,
(iii) [ and ~ are commutative,

associative,

here means that E <71B<6< B = [feY6 ~ < y (iv)

<.[(%
(v)

for n g ~, Kn = <.

(By (i) & (iii).)

(The last one

and mutually distributive. K%f(~)') (Special cases of (iii).) I

Lemma 2 If < ~ %, then <% = 2 % • Proof:

It suffices to prove that <% ~ 2 %, since 2 % $ K% is trivial. suffices to prove that X% $ 2 %. one-one into

e(X).

For this, it

But there is a canonical way of mapping %%

Hence ~X = i%%1 ~ I ~(%)I

= 2%.

QED.

Lemma 3 If K~ < %~ for all ~ < B, then ~ < B K Proof: Let f: ( U
< ~
Since K (% < % (%, f(%~< (% cannot be onto % (%, so pick a f~ e % ~ - f ~ "< (%•

Define g by g((%) = a , ~ < B.

Then (identifying H
product which canonically defines it) g ~ H
Hence f is

QED.

Lemma 4 Let K , ~ < B, be cardinals with K = ~ < B K

and K + ~ IBI"

For any cardinal X,

K %K = E < B % Proof:

•

Since %K(% ~ %K for all ~ < B, we clearly have ~(%~B% Conversely,

~

"IBI ~

.K

=

%K ~ ~
-II-

of maps

f : <

÷

%.

We have seen what limit ~).

QED.

it means

If < is a cardinal,

for a subset A of an ordinal

we define

such that there is an e-sequence A cardinal see that cf(K)

is resular

whose

iff cf(<)

is a regular

the cofinality range

cardinal

of ~, cf(<),

is cofinal

= < and sinsular for any cardinal

~ to be cofinal

(for

to be the least

in <.

if cf(<)

< K.

It is easy to

<.

Lemma 5 Let < be an infinite there are cardinals Proof:

cardinal.

is the least ordinal

a such that

KS < <, ~ < ~, with < = [~< <E.

If < is so expressible, cf(<)

Then cf(<)

then clearly

<<~J$

< ~> is cofinal

in K, whence

= ~ by definition.

Conversely,

let cf(<) = ~.

Let be cofinal

K~ = 165J < < for all ~ < ~. < = [~<e<~.

But clearly,

in <.

by the cofinality

Then of the B's,

QED.

Le~na 6 < is regular

iff ~ < < + % K =

k)

~.

~
Proof:(-~learly,

f c U

x

÷ f e x<.

Conversely,

if f ~ x<,

then sup(f"X)

< < since

fZ
% < ef(<)

= <

so f ~

U

)~

C~
(+-) This

implication

is trivial

|

"

Lem~a 7 +

For any <, K Proof:

Suppose

.

Is regular.

cf(< +) $ <.

Thus < + ~ [ ~

Then we can write <+ = [ ~ < K

< = K.K = <, which

is absurd.

, where <

< K+, by len~a 5.

QED.

Lemma 8
Proof:

> <.

Let < = ~ < c f ( K ) K e ,

where < s

<

K.

By lemma 3, ~ < ~ < c f ( K ) < = Kcf(K)

Lemma 9 % < cf(K) & ( ~

< ~)(2 ~ .< <) ÷ K ~ = <.

QED.

-12-

Proof: % < cf(K) ÷ ~%

i% j = i ~ ~ < ~ %

jX

I~I.%

% so ~

~< <.< = K.

QED.

Lemma IO (i)

< regular & ( ~

(ii)

Assume GCH.

Then % < cf(<) + <

(iii) Assume GCH. Proof:

= K.

Then < regular + [~<<
By lemma 9. by <~.

< K)(2 ~ $ <) ÷ [~<<
(The cardinal

More generally,

[~<<<~ is called the w e a k power of < by K, denoted

we define

the weak power of < by %, <~, by

Lemma ii (i)

~< ~ <~ ~ 2 ~ ~ ~.

(ii)

~ = %+ ÷ <~ = 2 ~ = 2 %.

(iii) GCH +-~ ~ ( 2 ~ = ~). Proof:

(i) The last inequality = X +.

is the only non-trivial

Then K~~ = K % = J ~ < K

= 2 % = 2 ~ >. ~+ = <.

%~ J (by lemmas

(This proves

Now suppose ~ is a limit cardinal. >. ~ < ~ +

=

one.

Suppose first that

7 and 6) = ~%.<~<~J%~J = K.% %

(ii) also.) Thus K = ~%<<%.

Hence,

2~ = Z%
K.

(iii) If GCH holds,

then 2 ~ = ~%~
( V < ) ( 2 ~ = K)

+

follows

from (i).

The converse

is trivial,

since 2 K = 2 ~

= ~+.

|

Lemma 12 K~ Proof:

~ iff < is regular and 2 ~

Assume

~

K

= K.

If cf(K)

K

< ~, then K~ > ~ by lemma 8.

lemma II, we know that 2 ~ ~ ~. to the assumption. K~ ~ K.

that ~

=

K.

Now assume

If K were regular,

> <, whence < < ~

is singular.

QED.

%

K is regular.

By

If 2 ~ > K, then by lemma II, ~ K > K, contrary

We show that ~ must be singular.

such that <% > ~.

%

Hence 2 6

Hence

< (~.%)

2~

= K.

Now, as ~

Suppose,

in fact

> ~, we can find % <

then we could find a cardinal ~ < K such

(~.%)

= 2

(~.~)

~ 2~, a contradiction.

Hence

-13-

C o r o l l a r y 13 K

Assume GCH.

T h e n Kw = K iff K is regular.

(Immediate, by lemmas ii, 12.)I

By lermna 7, all s i n g u l a r cardinals are limit cardinals. limit cardinal.

However,

not provable in ZFC.

And m is a regular

the existence of an u n c o u n t a b l e regular limit cardinal is

(We shall indicate w h y in a short while.)

said to be w e a k l y inaccessible.

Such a cardinal is

We give this concept a name b e c a u s e it is b e l i e v e d (if assumed as an a d d i t i o n " a x i o m of i n f l"n l"t y " ,

that the existence of such a cardinal

say) does not lead to a c o n t r a d i c t i o n w i t h ZFC.

A cardinal K > ~ such that K is

regular and % < K ÷ 2 % < K is said to be inaccessible. cardinal is w e a k l y inaccessible.

Clearly,

an inaccessible

The converse holds if we assume GCH, of course.

Closed U n b o u n d e d Sets In the later parts of this section, and in the later sections of this book, the concept of a closed u n b o u n d e d set will play an important role.

Let

u n c o u n t a b l e cardinal.

~ < to be closed and

u n b o u n d e d in K. so is A n B. emptiness,

We have already seen what it means for A

K be a regular,

It is e a s i l y seen that if A, B are closed and u n b o u n d e d in K, then

(Closure is obvious.

For unboundedness,

and in p a r t i c u l a r for non-

given any ~ < ~ one picks an increasing sequence

<8

In < ~> from K such

n

that

80 > %

B~n ~ A,

82n+i ~ B, w h e n c e

B = SUPn<wBn ~ A

idea is, incidently, used time and a g a i n in our arguments.

~B

and 8 > ~.

This simple

In fact, w h e n e v e r we say

" C l e a r l y C is closed and unbounded in <" for some C, this type of argument is the one w h i c h the reader must t h e n supply.) filter-base - w h i c h is to say filter on

Hence the closed u n b o u n d e d subsets of K f o r m a

5 = {X ~ KIX ~ C for some closed u n b o u n d e d C ~ ~} is a

~(K).

A function f: K ÷ < is continuous if lim(6) ÷ f(~) = sup~<6f(~). function on ~ is a s t r i c t l y increasing

A normal

(i.e. ~ < B ÷ f(~) < f(~)), continuous

function f: < ~+ <. It is easily seen that if f is a normal f u n c t i o n on <, t h e n f(~) ~ ~ for all

Clearly,

if f is a normal function on K, then f"< is a closed unbounded subset

-14-

of K.

Conversely,

if C ~ K is closed and unbounded

in <, then the monotone

enumeration of C is a normal function on K. Clearly, unbounded

if f is a normal function on ~, then {~ c KIf(~) = ~} is a closed

subset of <.

Metamathematics

of ZFC.

An g-structure E ~ X 2 interprets occasions.

(Recall our "clearly"

~.

convention above for such situations~)

Inner Models

consists of a pair <X, E> where X is a non-empty In case E = ~ n X 2, we write

set and

'X' for '<X, E>' on many

(Note the dual use of ~ here.)

If X is a set (or a class, modulo some extra care in what follows), the first-order each x c X. ~

expansion of ~

The interpretation

denotes the ~

satisfaction

~X

denotes

obtained by introducing

an individual

of ~ X

<X, e> is the obvious one.

in the structure

constant ~ for

relation for the structure <X, e> under this

interpretation. For the following discussion, completeness

theorem,

the Godel incompleteness

statement "T is consistent" consistent". model.

Roughly speaking,

Con(T).

the completeness

Hence, we know that Z F ~

(see later).

We write Con(T)

Con(ZF).

However, we can prove,

of ZF, and in particular

that ZFC ~ "(ZF--Replacement)

all finitely axiomatisable axiom, so this

is consistent".)

But we know for certain, by the above, If K is an infinite cardinal, we set

H K = {xJJTC(x) I < ~}, the set of all sets hereditarily is transitive

and V = ~

K

H .

of cardinality

If ~ is uncountable

less than ~.

and regular,

then H

KK

is a model of the important axiom from ZF.

in ZF,

(We are tacitly assuming Con(ZFC) here, of course, since

otherwise we could prove 0 = i if we wished.

H

to mean "T is

theorem says that Con(T) iff T has a

For example, Vm~ is a model of ZF --Replacement

theory is consistent.

Clearly,

and the meaning of the

theorem says that if T contains Peano arithmetic,

the consistency of most subtheories ones

theorems,

for T a formal theory.

The second incompleteness

then T ~

we assume the reader is familiar with the Godel

theory ZF-.

(The theory ZF

ZFC-is defined analogously

K

This is obtained omitting the power set

plays an important role in constructibility

starting from ZFC instead of ZF.

is easily seen that V K = H K and that V K is a model of ZFC.

theory.)

If ~ is inaccessible, Hence, by the second

it

-15-

incompleteness ZFC.

theorem,

the existence of an inaccessible

Since GCH implies that weak inaccessibility

cardinal is not provable in

and inaccessibility

coincide,

it

will follow from results in chapter 2 that the existence of a weakly inaccessible cardinal is not provable

in ZFC either.

Such cardinals

later sections of this book, we consider

further

Note that if K is weakly inaccessible, ment cannot result in the contradiction

In

"large cardinals".

then ~

~ g V

are said to be large.

= K.

Thus the axiom of replace-

when V K is a model of ZFC, in the K

obvious way. Suppose M is a class.

For each ~ - f o r m u l a

M, is defined by induction on ¢. finitely many applications induction procedures ¢.)

on this definition.

¢ primitive.

~ v. ~ M.

If ¢ ~ v. = v., then cM ~ v. = v.

^v. E M.

Case 3.

j

Otherwise.

1

j

¢ z ~I ~ ~2" ¢ ~ ~1 v ~2' ¢ ~ - ~ ' Then cM ~ @I

We call such an induction an induction o__n_n

If ¢ ~ v. e v., then ~M z v. ~ v. ~ v. E M 1

M

1

j

j

•

A v. ~ M 1

~

~2' etc.

If ¢

3Vm~ , then cM ~ ( 3 v m c M)~.~

If ¢

VVm~ , then cM _ ( V v m ~ M)~I

If ¢

( 3 v m ~ Vn)~, then cM _ Vn g M

~ ( ~ v m s M)(v m ~ V n A ~ M ) .

If ¢

( ~ v m s Vn)~ , then cM _ Vn e M

A ( ~ v m ~ M)(v m ~ v n + ~M).

If M is a transitive

Otherwise, we must

of course.

set, then clearly, M will be a model of ¢ iff cM holds

In particular,

(in

M will be a model of ZF iff cM holds for each axiom

Suppose now that M is a proper class.

(i) M is transitive;

j

¢ ~ ~I ÷ ~2' ¢ ~ ~I +-~ ~ 2 "

identify M with its defining formula in the above definition,

¢ of ZF.

J

M

If M is a set, then we can regard cM as a formula of ~{M}"

the real world).

formulas by

is as follows.

1

Case 2.

(Since ¢ is built up from primitive

cM, of ¢ to

of the formula generating schemas, we can carry out

The inductive definition

Case I.

¢, the relativisation,

We say M is an inner model of ZF iff:

(ii) On ~ M; and (iii) for every axiom ¢ of ZF, cM is true.

concept of an inner model of ZF is of particular

importance

this book is devoted to the study of a particularly

in set theory.

The

Indeed,

good inner model, the class of

-16-

construetible sets.

One important aspect of inner models of ZF follows from the fact

that we can obtain relative consistency results in this way.

More precisely, we have:

Lemma 14 (Metatheorem) Suppose M is an inner model of ZF.

More precisely, M is a transitive class,

definable in ZF by a formula with one free variable, containing all the ordinals, such that for each axiom ~ of ZF, ZF ~ ~M.

Let ~ be some particular statement in

(e.g. AC), and suppose that ZF ~ ~M.

Then ZF ~ Con(ZF) + Con(ZF + ~).

Proof: Argue in ZF.

Then there is a model <X, E> of ZF.

Assume Con(ZF).

Since

<X, E> is a model of ZF, there is a class K of <X, E> which is definable in <X, E> by the formula which defines M in the real world.

Clearly, K will be

an inner model of ZF + ~ in the sense of <X, E>, since our proof of this fact for M took place in ZF, of which <X, E> is a model. Consider the structure .

Let Y = {x ¢ X]x E K}.

By induction on ~, we can easily prove

that for any ~y-Statement ~, if in <X, E> it holds that K the real world is a model of ~. transitive in the sense of <X, E>.) ZF + ~. Remarks. I°

is true, then in

(Essentially this is because K is

It follows that

is a model of

Hence Con(ZF + ~). ]

We emphasised in the above that we were stating a metatheorem.

This is

to distinguish it from our previous results, which were all supposed to be theorems of ZFC, formally provable from the axioms of ZFC°

E0f course, our metatheorem was proved

in our metatheory (which is also ZFC).~2. in the proof of the above, we mentioned induction on # for ~ an ~y-Statement. of

It should be clear how the primitive formulas

~ y are defined, and how the ~ y-formulas are generated from the primitive formulas,

by analogy with the

~

situation.

Absoluteness. The Levy Hierarchy The concept of absoluteness plays an important role in the study of inner models Suppose M is a transitive class, and ~(x) is an ~M-formula. absolute, or D-absolute, iff ( ~ absolute, or U-absolute,

iff ( ~

¢ M)(~(x)

.....~M(~)).

M÷ s M)(~ (x) ÷ ~(~)).

We say ~(~) is downward

We say ~(~) is upward We say ~(x) is absolute iff

-17-

it is both D-absolute and U-absolute. U-absolute for M, etc.

In the case where the M is not clear, we say

Where # defines a class, we sometimes talk of the class being

D-absolute, etc., without referring explicitly to $. It is clear that primitive ~ M - f o r m u l a s will be absolute for M for any transitive class M.

For other formulas, absoluteness depends upon the logical complexity of the

formula to a large extent.

L~vy has introduced a useful notation for the classifica-

tion of logical complexity of formulas.

This is where it is convenient to have

bounded quantifiers as part of the definition of ~ , Let ~ be an ~-formula. quantifiers.

as we stated earlier.

We say $ is ~0 (or N 0) if it contains no unbounded

For n ~ I, we say (inductively) ~ is In if it is of the form

where ~'(~) is ~n-l' and ~ is Hn if it is of the form ~ ' ( ~ ) There are analogous definitions for formulas of ~ X '

3~@'(~)

where #'(~) is ~n-l"

any class X.

In particular,

if Y £ X and ~ is a In (or Nn ) ~ x - f o r m u l a with constants from the set {~IY E Y} only, we say ~ is ~n(Y) (resp. Hn(Y)). A formula ~ of Similarly H ZF. n

~ is ~ F

iff there is a In ~-formula ~' such that ZF ~ ~ +-+ ~'.

For n ~ I, a formula ~ of ~

is AZF iff it is both ~ F n

and

~ZF. n

These notations are also used to refer to classes so defined, and also when ~ is an ~ x - f O r m u l a for some X. Lemma 15 (Metatheorem) The following formulas are ~ F . x = y; x ~ y; x ~ y; y = {x, z}; y = <x, z>; y = (x)0; y = (X)l; y = x o z; y = x q z; y = ~ x ;

y = ~x; y = x - z; x is an ordered pair; x is a relation; x is

a function; y = dora(x); y = ran(x); y = x"z; y = x~z; y = x x

z; y = x

-i

; y = x ~ {x};

On(x); lira(x); suee(x); x = ~; x e w; m s x. Proof: The proofs are all trivial.

Note that (-)0'

(-)i are the functions defined on

ordered pairs so that y = <x, z> implies (Y)0 = x and (Y)I = z. [~F definitions: I (w) 0 =

D f~w, if w is an ordered pair undefined, otherwise.

Theyhave the

-18-

U•W,

if w is an ordered pair and Dw = ~ w

(w) 1 = j ~(~Jw - f~w), if w is an ordered pair and kJw ~

I

L undefined, otherwise. And x -I is defined as {<(Z)l,

(z)0>Iz is an ordered pair & z e x}. I

Lena

16 (Metatheorem)

(i)

ZF If ~, ~ are ZO , so are ~ ^ ~, ~ v ~, --~, $ ÷ ~,

(ii)

ZF ZF If ~ is IzF, -7 ~ is ~ZF. n ' if ~ is Nn '--~$ is In .

ZF (iii) ~ is AZF iff ~ and-1~ are _~n " n

(iv)

ZF If ~, ~ are Ii , so are ~ ~ *, ~ ~ ~,

~x$,

(~x

~ z)~, ( ~ x

(v)

ZF If ~, ~ are N1 , so are ~ ~ ~, ~ v ~, ~x$,

(~x

c z)~, ( 3 x

Proof:

(i) - (iii) are trivial, and (iv) implies

(v).

s

We prove (iv).

The only non-+

trivial part is ( ~ x

e z)~.

->

Let ~ be a 10 formula such that ZF ~ # +-+ 3y@(y).

By means of the I zF functions <-, ->, (-)0, (-)i, we can assume in fact that~ is chosen so that ZF ~- ~ ~-+ 3y~(y). ZF 10 "

By the replacement axiom, ZF ~- [ ( V x

But ( ~ x Remark.

~ z)(~y

~ u)~(y) is 10"

~ z) ~y~(y) +-+ ~ u ( V x

is

e z)(~y ~ u)~(y)].

QED.

In the above proof we introduced the concept of "contraction of quantifiers"

to simplify matters. mention.

We must show that ( ~ x s z)~y~(y)

This is a frequently used device, and is usually invoked without

ZF If ~ is ~n " we will just say, "pick ~ Nn-I so that ZF ~ ~ ++ ~y~(y)",

Corollary 17 (Metatheorem) ZF If ~, ~ are A 1 , so are $ ~ ,

~ v ~, --~, ~

->

~, $

~vx

z)~

( ~ x £ z)~. | Lermma 18 (Metatheorem) -+ ÷ ZF If +(y, ~) is ~I F and ZF ~- ~ x ~'y~(y, x), then ~ is AI " Proof: ZF ~- [~(y, x) ++ ~z(+(z, x) ÷ z = y)].

QED.

Len~a 19 (Metatheorem) If f is a ~ F

ZF function and dora(f) is H~F, then f and dora(f) are ~I "

etc.

-19-

Proof:

(a) y = f(x) +-~ ~ e dora(f) ~ ~ z ( z = f(x) ÷ z = y) (b) <~> e dora(f) +-+ B y ( y = f(~)).

QED.

Lemma 20 (Metatheorem) Let M be a transitive model of ZF or else an inner model of ZF. -formula

(or possibly an

Let # be an

M formula).

(i)

If ~ is ZZF, then ~ is absolute

(for M).

(ii)

If ~ is zZF, then ~ is U-absolute.

(iii) If ~ is HZ F, then ~ is D-absolute. ZF If ~ is A I , then ¢ is absolute.

(iv) Proof:

(i) By induction on ~.

(As M is transitive,

the bounded quanifier steps cause

no problem. ) (ii) - (iv) follow easily from (i). Remark.

QED.

On occasions we shall use sharper forms of lemma 20.

For example,

if M is

not a full ZF model, but is a model of some fragment T of ZF, then the above results hold for any ~ which is [0ZF ' 11ZF , ~IZF ' AIZF (resp.) and such that only axioms in T are used in proving the corresponding

logical equivalences.

For example, all of the

formulas mentioned in lemma 15 (and in fact most formulas we have occasion to use) are [ZF by virtue of ZF- alone, and le~na 16 holds with ZF- everywhere replacing ZF. In connection with lemma 20, let us state, by way of an example, the following result. Lemma 21 (Metatheorem) (a) The following are [ZF 1 : ,,~ is not a cardinal";

cf(~) ,< 8; "~ is singular";

Jxl .< lyl. (b) The following are NIZF : "e is a cardinal"; inaccessible";

"~ is regular"; "~ is weakly

y = 6D(x).

Proof: Trivial. The Montague-L~vy Reflection Principle Closely connected with our above discussions

is the reflection principle of

-20-

Montague and Levy. down to some V

Loosely speaking, this says that any property valid in V reflects

(i.e. is already valid in some V ).

We prove this first of all in a

very general way. Theorem 22 (Metatheorem) Let <W~I~ ~ On> be any hierarchy of transitive sets such that (i)

~ < E + W

=- WE;

(ii) lim(6) ÷ W6 = 0 <6W ~. ÷

Set W = ~ eonW • ZF~

(~a)(~8

Let $($) be any ~ -formula with free variables v.

> ~)[lim(E) m ( g x

s WE)(¢W(x) +-+ $

Then

(x))].

Proof: For the duration of this proof, we shall assume that the bounded quantifiers are not a part of ~ , hut are in fact defined. define ~ , v ,

÷, ++ in terms o f ~ , ~

Also, since it is easy to

,-7, we shall assume that

3, ^, -~ are the

only logical symbols of ~ . ÷

Let us call an ~-formula @(v),÷ with free variables v, hereditarily absolute -~

for 8, and write Habs(@(v), $(v) itself), Claim.

÷

÷

B), iff for all subformulas $(v, w) of @(v) (including

(gx, ~ s WE)(@ (x, y) ++

(x, y)).

If <En In < ~> is a strictly increasing sequence of ordinals such that Habs($(~), E n) for all n, and if B = SUPn<mB n, then Habs($(v), -~

We prove the claim by induction on ~.

÷

÷

Let $ (v, w) be a subformula of ¢ (v) such ÷

-~

÷

that the claim is proved for all subformulas of $(v, w). result is trivially true.

-~

÷

÷

÷ ÷y) +-+ ~W E (x, ÷ -~ We must show that ~ W (x, y).

e W~ )~'

WE

÷ ÷ n(u, x, y).

-~

Suppose @(v, w) --- ~u@' (u, v, w).

the

÷

Let x,

Pick n < ~ so that ÷x, +y e WEn.

W + ÷ ~WEn( By the assumption of the claim for En, @ (x, y) +-+ ~, y). (~u

-~

If $(v, w) is primitive,

If @ = @l ~ $2 or if $ = -i$', then the claim for @ follows

trivially from the induction hypothesis. ÷y e W B.

E).

W ÷ + Hence $ (x, y) ~-+

But ~' is also a subformula of ~, so again, for

~n

÷

u ~ W~ , ~'WBn(u, I, ~) +-+I/W(u, x, ~).

+

Hence ~W(x, y) ++ ( ~u e WEn)@'W(u , x, y).

~n

Thus, by induction hypothesis,

÷ ÷y) +-+ ( ~ u ~ W (x,

WE ÷ ÷ @W(~, ~) ++ ~ (x, y), as required.

g WEn)~,WE(u ' +x, ÷y).

It follows that

The claim is proved.

->

To prove the theorem, let $(v) be given as stated.

We prove by induction on

-21-

that for any ~ there is 8 > ~ such that lim(B) and Habs(#(v), B). primitive, the result is trivial, since ~ = ~ + ~ suffices. = ~,

the induction step is trivial.

~-sequence <8nln < ~> as follows.

replacement axiom.

If ~ = ~i A ~2 or

Suppose ~(v) = ~u~(u, v).

Define an

Let ~0 > ~ be the least ordinal such that

Habs(~(u, ~), 80), by induction hypothesis. be the least y > B~n such that ( ~

If ~ is

Suppose now B2n is defined.

+ a WBln)[¢W(~) + ( ~y e Wy)~ (y, x)], by the

Let B~n+2 be the least y > B2n+l such that Habs(~(u, v), ~).

This defines <8nln < ~>.

Let B = SUPn< Bn.

Then B > ~ and lim(B).

We show that

Habs(~(~), B). Since B = SUPn< B2n, then by virtue of the claim, we have + (~W(y, ÷ W~ + Habs(~(u, v), B) and in particular, (~y, ~ s W B) x) +~ ~ (y, x)).

(~÷x

c

w~)(+wB (~)

with ~ e W B (~y c

.

Let Bin+l

W

÷

cw (~)).

Conversely, suppose

÷x

~ W B and that ¢W(~).

Hence Pick n <

Then, by definition, ( ~ y E W~2n+l)~W(y, x), whence

~ W~)~ ~(y,

÷

w~

x), i.e. ¢

÷

÷

w ÷

w

(x). Thus ( ~ x s W~)(¢ (x) + ¢ g(x)), and the proof is

complete. I The above metatheorem is known as the Generalised Reflection Principle, GRP. As an immediate corollary, we have the Reflection Principle: Corollary 23 (Metatheorem) Let ~(~) be any ~ - f o r m u l a with free variables ~. ZF ~ ( ~ ) ( ~ B

> e)(lim(8) A ( ~

c Vg)(¢(~) ~+ cVB(~))).

Then I

As an immediate consequence of this, we have: Corollary 24 (Metatheorem) (i)

If T is a finite subtheory of ZF, then ZF ~Con(T).

(ii) ZF is not finitely axiomatisable. I Structural Definability Suppose we have a first-order structure of the form M = <M, ~, A>, where M is a non-empty set and A ~ M. is the language in ~.

~M

(We usually write M = <M, A> in this case.)

The M - I ~ _ ~

with the extra unary predicate symbol A, so that A interprets

The notions of primitive formula, formula, ~M' and In-fOrmula (ete) of the

M-language are defined in the obvious way, by analogy with the case where A ~ #,

-22-

considered earlier. formulas. )

(In particular, A(Vn) , n c ~, and A(~), x c M, are primitive

In this context, those formulas of the M-language in which the predicate

does not appear are called E-formulas.

We denote by ~

the restriction of ~M to

all In-formulas of the M-language. Let N =_ M.

M

A set R =-Mm is In(N ) (resp N (N)) if there is a In(N) (resp Hn(N))

formula $($) of the M-language (in particular, therefore, all constants in ~ lie in {xlx e N}) such that ->

M)(<x> ~ R ~ + ~M + (~)). We also use the symbol ~M(N) (resp. ~M(N)) to denote the set of all such R (for varying m s ~).

We set AM(N) = ~M(N) ~HM(N).

We write ~M for ~M(~) and ~n(M) for ~M(M).

Similarly for H and A.

÷

If $(v) is a formula of the M-language, II$II~M denotes the relation {I~ E M & l=M$(~)} Let

~ be a class of structures of the form M = <M, A>.

uniformly ~M for M E ~

A relation R -~Vm is

iff there is a ~n(~) formula ~ such that M ~ ~ -> R n M m = II~I~.

The interpretations of the constants, ~, occurring in a formula of an M-language are usually called parameters. from".

We use the abbreviation wpf to mean "w1~h parameters

Hence if we say $ is a formula ~

X, we mean that every constant in ~ is of

the form ~ for some x c X. We use Fml to denote the set of formulas of a given M-language. those ~n members of Fml.

Fml ~n denotes

If N -~ M, Fml N and Fml~ n mean that the members of these sets

of formulas have parameters only from N. Let M = <M, A> be given.

We write X -
and for all In(X)

statements $ of the M-language, ~<X, A ~ X > $ iff ~M $" If M, X are transitive, then X =-M implies X -<~0 M by lemma 20 (i) (or more precisely by an obvious analogue of lemma 20 (i).).

-23-

If M is transitive M R s ~(M) ~ ~ ( X ) ,

and n > O, then X - < ~

R ~ ~ ÷ R ~X

+ ~.

~ iff X ~ M

and for all

(This is easily seen.)

The Collapsing Lemma We have kept until last one of the most important will be clear from all of the previous discussions particularly Mostowski

good ones to deal with.

and Shepherdson,

transitive

structures

lemmas for our purposes.

that transitive

The collapsin$

structures

are

lemma, due independently

tells us that in a great many situations,

It

to

there are

at our disposal.

Theorem 25 Let <X, E> be a model of the axiom of extensionality Then there is a unique transitive And if Y ~ X is E-transitive

such that E is well-founded.

set M and a unique isomorphism ~: <X, E> ~ <M, ~>.

and E ~ Y 2

= E ~ y2, then ~ Y

= idlY.

Proof: Define a function ~ on X by the E-induction ~(x) = {~(Y)IY E X & yEx}. Let M = ran(~). Clearly,

Since <X, E> satisfies

~: <X, E> m < M ,

E>.

the axiom of extensionality,

By definition of v, M, M is transitive.

~': <X, E> ~ <M', e> where M' is transitive, prove that x E X + ~(x) = ~'(x). Y ~ X is E-transitive

We call <M, E> the transitive

then by E-induction

Hence ~ and M are unique.

and E n y2 = s N Y 2 ,

prove that ~(y) = y, so ~ Y

= idlY.

such that ~: <X, ~> ~ <M, e>. rank(x)

for all x e X, as well.

If

it is easy to

Finally,

then by E-induction

if

on y s Y we can

QED.

collapse,

or transitivisation

A particular case of the above is when E = e ~ X 2. of the axiom of extensionality,

~: X +-~M.

Thus,

of <X, E>.

for any model <X, E>

there is a unique ~ and a unique transitive set M In this case, it is easily seen that rank(~(x))

Ch__~_pter I

THE CONSTRUCTIBLE

Throughout

UNIVERSE

the 19th century, mathematicians

were, without apparently being aware

of the fact, liable to assume that given any infinite collection of nonempty sets, one could simultaneously

pick a member from each set.

What do we mean by "being aware of

the fact", since it is plain that they must surely, in a sense, have been aware of what they were doing?

Well, we mean that they never for a moment doubted

validity of this assumption. ordinary differential this procedure.

set theory) did not exist. whether there exist, assumes a prominent

It was Peano who, in 1890, in a paper on (of all things)

equations,

Of course,

first appears to have expressed a doubt concerning

in those days, set theory as we understand

it (i.e. axiomatic

But, even when we assume the axioms ZF, the question as to

in the ZF-universe, role.

(in public) the

"choice sets" for infinite collections,

We accept ZF because

of what the universe of sets should be like.

it corresponds

to our intuitive

(See the Scott article,

is not true (at least to the same extent) of AC.

Nevertheless

useful assumption to have, for many branches of mathematics.

[16].)

still ideas

But this

it is, indeed, a very So, if we cannot prove

AC in ZF, at least we should hope that AC is consistent with ZF, so that we can work with ZF + AC and not pause to ponder the legitimacy of such an action. did, in fact, prove that if ZF is consistent, that we knew that Con(ZF) ÷ Con(ZF + ~ A C ) . is a little more complicated,

so is ZF + AC.

In 1936, GSdel

It was not until 1963

(Actually the situation in this direction

and is anyway described in detail in [ 9].

The 1963

result we mean is, of course, Cohen's result that AC could fail in a "real" ZF-universe.) Meanwhile,

back to the Godel proof.

pairwise disjoint sets.

Suppose

A choice set for

~ is an infinite collection of non-empty,

~ will be an element of

we have the axioms of union and power set available,

~(0~).

Thus, as

the axiom of separation will yield

the existence of a choice set providin$ we can find a formula ¢ which defines such a set.

The idea behind Godel's proof is to take an inner model of the universe in which

every set is definable by a formula in a certain ramified language, so that, in this inner model a formula ~ as above can always be found.

Another way of looking at it, is

to say that we take for our inner model only those sets which we can specifically

assert

-25-

to exist.

We formalise

definability

procedure,

carried out in ZF. definition

this last, rather vague notion by means of a ramified and in this way obtain our model.

In particular,

Theorems

From now on, everything

2 and 3 below are ZF theorems.

is

The precise

of the inner model, L, is given below.

Let X by any set.

By Def(X) we mean the set of all subsets

some formula ~(v0) of ~ X

with one free variable v 0 ,

¢(~)}.

a = {x E E l k x

In words, Def(X)

a of X such that for

is the set of all first-order

definable

(in ~ X) subsets of X.

By induction on a e On, we define: L0 = L+ I = Def(L) L6 =

U

<6La,

if lim(6).

is the constructible constructible

universe.

The class L =

A set x is constructible

the formula "v 0 is constructible" class.

hierarchy.

is expressible

iff x s L. in ~ .

U EonL ~ is the

Since x E L +-+ ~ a ( x s L ),

Hence L is a ZF-definable

We shall prove that L is an inner model of ZF + AC.

Lemma I

(i)

a < 6 ~ L

u { L a} c L .

(ii)

For all a E On, L

is transitive.

(iii) For all a s On, a ~ L Proof:

(i) By induction reduces

= a n L

on B > a.

of L

Limit

L

+ 1 •

Hence O n ~ L .

stages are trivial.

NOW,

of the hierarchy,

as y c L T + I ,

Hence z s y ÷ z c L . Y

(iii) By a trivial

This follows

Since y s L , there is a least ordinal

By definition

x ~ y ~ Ly+ 1.

The successor

step

easily from

~+l"

(ii) Let x s y e L .

~(v0).

a and a c

=

to proving that L a U { L a} ~ L a + 1 for all a.

the definition

y s L 8.

Hence L is transitive.

induction

B ~ a such that

8 = y + I for some y.

Hence

*(Z)} for some ~ L y y = {z ~ Lyl~Ly .

In particular, on a. I

x c L . Y

So, by

formula

(i), x s L .

-26-

T h e o r e m 2 (Godel) L is an inner model of ZF. Proof:

Working

In fact,

for each axiom ~ of ZF, ZF ~ ~L.

in ZF, we show that for each axiom ~ of ZF, ~L holds.

In view of lemma

i, L will thus be an inner model of ZF. Extensionality.

Let x, y ~ L.

We must show that x = y *-+ ( ~ z

But by lemma i, L is transitive. itself

e L)(z E x *-+ z E y).

Hence this holds by the axiom of extensionality

(in the real world).

Pairin$.

Let x, y s L.

e Def(L

) = L + 1 ~ L.

Union.

Let x s L.

formula ~(v0) y s Def(L

~

Pick ~ with x, y E L . Hence the pairing

axiom holds

Pick ~ so that x s L .

Since L

itself for x to verify

Clearly, [( ~ y ) ( y =

By definition,

(y =

U x ) L.

I~L ~ = ~ v ~ = ~}

in L. is transitive,

~ V l ( V 0 ~ v I ^ v I ~ ~) defines y as a subset of L~

) = L + 1 ~ L.

Infinity.

Then {x, y} = {z e L

y =

dx

~ L .

The

(in L ), so we have

Thus, we have used the axiom of union

D x ) ] n.

~ s L + 1 ~ L.

And clearly,

(w satisfies

the axiom of

infinity) L. Power Set.

Let x c L.

Let y = {z E ~(x) Iz E L}.

of power set and separation.) z E L .

(The set y thus exists by the axioms

For each z a y, let f(z) be the least ~ such that

By the axiom of replacement,

let ~ exceed all f(z), z c y.

Thus y ~ L

The formula ~(v0) ~ ( ~ V l ) ~ V 1 a v 0 +~+ v I ~ xJ defines y as a subset of L y s L + I.

Clearly,

(y =

Regularity.

Immediate,

"E-relation"

of L.

Separation.

Let x E L.

must show that hierarchy,

[(~y)[y

~(Vo) ~ v 0 s ~

in L , so

~ ( x ) ) L. since c is w e l l - f o u n d e d

Pick ~ so that x s L . = {z c xI~(z)}]] L.

we can find B > ~ such that

y = {z E xJ#(z)}.

in the real world,

and ~ ~ L 2 is the

Let ~(v 0) be a given

By applying

( ~ z c L~)(~L~(z)

~-formula.

+-+ ~L(z)).

Let

Then x g LB and y ~ x ~ LB, so by choice of B, the formula

~ ~(Vo)

Suppose

defines y as a subset of LS, in LB, so y ~ LB+ I, and we are

~ is an ~ - f o r m u l a

such that

We

the GRP to the constructible

done. Replacement.

.

[(~x)(~y)~(y,

x)] L, and let

-27-

a e L be given. a s L .

We seek b s L such that [ ( ~ x e a ) ( ~ y

e b)~(y, x)] L.

Pick ~ with

For each x ~ a, let f(x) be the least ordinal B >. ~ such that (-~y e LB)~L(Y, x).

Let y exceed all f(x), x e a. that [ ( ~ x

~ a)(~y

Thus ( ~ x

s a)(~y

~ L ) L(y, x). It follows at once Y is (by lemma i) as sought.

~ Ly)~(y, x)] L, so b = L Y

That completes the proof of Theorem 2. We have shown above that ZF ~ ~L for every axiom ~ of ZF. ZF ~ AC L.

We show next that

In fact, we do this in a very strong way, by exhibiting an ~ - f o r m u l a

which defines

(in L) a well-ordering of the entire class L.

(We can then obtain choice

sets by picking the least members of each set under this well-ordering, This formula will define a well-ordering

of course.)

of L by fixing an "order of construction"

for the members of L. Theorem 3 (G~del) There is a ~ - f o r m u l a

~(v0, vl) with free variables v0, vl, which is absolute

for L, such that ZF ~ [{<x, y~Ix, y e L & ~L(x, y)} well-orders L]. Proof: We first of all define, in a canonical way, by induction on ~, a well-ordering <

o f La.

I f a = O, t h e r e

is nothing to define.

For the remaining cases,

first of all fix a recursive enumeration <~nln < ~> of all ~-formulas least the free v a r i a b l e Suppose, then, <

~

(and p o s s i b l y

is defined.

we

with at

others).

Then we can define a well-ordering <* of all the

finite sequences of elements of L a by ~ <~T~ +-+ [length(~) < length(T~ v [length(~) = length(z) & (if n is the least integer such that ~(n) ~ ~(n), then ~(n) <

~(n))].

We can then define a well-ordering

<' of the formulas of

~ L ~ with free variable v 0 by e(v0) <'~ ~(v0) +-~ [the least n such that O(Vo) 5 ~n(VO, ~) f o r some x e L a i s s m a l l e r ~(v 0) ~ ~m(V0, ~) for some y e L _

~

E L

A y ~ e

but the

~) <~-precedes the <~-least ÷ such that

Finally, we can then define <~+i by x
^ x <

y] v ~

e L~ ^ y ~ La] v ~

least ~ L - f o r m u l a w i t h f r e e v a r i a b l e <'-least

m such that

V [the above n and m are equal

<~-least <~> such that 0(v0) 5 ~n(V 0 ~(v0) ~ ~n(V0, ~)].

than the least

~ e

v 0 which defines

^ y ~ L

^ the <'-

x i n La s < ' - p r e c e d e s t h e

~L-formula with free variable v 0 which defines y in L .]

-28-

Since L + I = Def(L ), this clearly defines a well-ordering of L + I. Finally, if lim(~) and
< ~)(x e L B ^ y ~ LB) ] V [ ( V B < ~)(x e e B ~-+ y ~ LB)

by x <

y +-+

^ (if B < ~ is least such

that x, y s LB, then x
<~'

from

<6'

B < ~, for lim(e).

Clearly,
~L

Let ~(v0, v I) be the obvious (though

~-formula which describes the above definition of
It is easily seen that the canonical choice of

will be absolute for L, so that we are justified in writing ~L here.

(To "easily

see this", it is not, of course, necessary to try to write out ~ explicitly; a careful look at the construction of
(that

Thus, we can paraphrase~x
It should now be clear why the axiom of choice holds

(2.) The proof of theorem 3 was somewhat informal, of course;

later on we shall give a more rigorous proof. Theorem 4 (Godel) Con(ZF) ÷ZF Con(ZFC). Proof: By theorems 2 and 3. ~

Chapter

2

THE AXIOM OF CONSTRUCTIBILITY~ THE CONSISTENCY

OF THE GENERALISED

The axiom of constructibili~y constructible.

We generally

V = L +-+ ( ~x)( ~ ) ( x

CONTINUUM

is the assertion

V L = L, it is clear that this boils down to showing

that

(L)L

down to showing

= L

to an examination

of the constructible discussion and Karp Primitive

of a certain

For this,

(L) L = L.

~-formula complexity

it is convenient

class of set functions,

Recursive

We shall Since

This,

in turn,

And, by our discussions "v 0 = Lvl"

is AIZF"

of the defining

formulas

to commence with

first examined

Thus,

a

in detail by Jensen

Set Functions

f: V n ÷ V is primitive

recursive

(p.r.)

iff it is generated

by the

schemata:

(i)

f(x I .... , Xn) = xi, 1 $ i ~ n.

(ii)

f(~) = {xi, xj},

(iii) f(~) = x i - xj, (iv)

f~)

(v)

f(y, ~) =

(vi)

f(~) =

= h(g~)

1 ~ i, j ~ n. 1 $ i, j $ n.

..... ~ ( ~ ) )

D z e y g(z, ~)

(vii) f(y, ~) = g(y, x,
~ h(y) > ), providing

(vii) is known as the recursion

(i) - (v) alone, schema

that

(V = L) L.

Since

([12~).

A function following

hierarchy.

i.e. a l l sets are

an ~ - f o r m u l a .

for each ~, of course.

of the logical

AND

for this axiom.

(in ZF) that

in Chapter O we know that this will be so if the we are reduced

( ~x)(x e L),

e L ), we see that V = L abbreviates + V = L) by showing

LEMMA,

HYPOTHESIS

write V = L as an abbreviation

prove that Con(ZF) + Con(ZFC

boils

THE CONDENSATION

schema.

it is said to be rudimentary

(vii) is when h = id.

Hence,

(rud).

in particular,

z g h(y) ÷ rank(z)

If f: V n ÷ V is generated Note that a particular rank is itself

If p is a set, we say f: V n ÷ V is p.r. in parameter (i) - (vii) and the schema

< rank(y).

a p.r.

by

case of

function.

p iff f is generated

by

-30-

(viii) f(~) = p. Similarly for f rud in parameter p. If A is a class, we say f: V n ÷ V is p.r. in A iff f is generated by (i) (vii) and the schema (ix)

f(~) = x. ~ A, I $ i ~ n. i

Similarly for f rud in A. We say R e V n is p.r.

(resp. rud) iff there is a p.r.

(resp. rud) function

f: V n ÷ V such that R = {<~>If(~) + ~ }. For example, the following functions are clearly rud (and hence p.r.): ÷

÷

f(x) =

l]xi; f(x) = x.1 u x.j =

<]{xi, xj}; f(~) = {x}; f(~) = <x> =

{{El}, {x I, <x2,..., Xn>}}; f(~) =~ y ~ x +-+ {y} - x +~ .

= x i - x i.

Hence, the relation ¢ is rud, since

And if f(y, ~) is rug (resp. p.r.), then so is g(y, ~) =

, since g(y, ~) =

0zey{}, of course.

Len~na I (I)

If f, R are rud (resp. p.r.), so is g(x) =

{f(;), if R(~) ÷ , if ~ R ( x ) .

Proof: g(x) ÷ = (2)

U

÷ f(x), ÷ where R(x) ÷ ~-+ r(x) ÷ + ~.] yEr (x)

Let~, R be the characteristic

function of R.

Then R is rud (resp. p.r.) i f f % R

is. [Proof: By (i).] (3)

R is rud (resp. p.r.) iff -,R is.

[Proof: By (2), since ~ R(~) = i - %R(~).] (4)

Let f.: V n ÷ V be rud (resp. p.r.) for i = i ..... m.

Let R. _= V n be rud (resp.

i

i

p.r.), i = i ..... m, such that i + j ÷ R i ~ Rj = ~ and f: V n ÷ V by~f(~) = fi(~)]+÷ Ri(x).÷

Then f is rug (resp. p.r.).

{ fi(~), if Ri(~) , [Proof:

~l<.i.<mRi = Vn"

i = i ..... m.

Set fi (~) = , if -~Ri(~)

By (i), each ~.l is rud (resp. p.r.).

But f(~) = ~l
Define

-31-+

(5)

If R(y, x) is rud (resp. p.r.), then so is f(y, ~) = y n {zlR(z, x)}. x)

[Proof: Set h(y,

= I {Y}' if R(y, ~)

[

, otherwise.

By (I), h is rud (resp. p.r.). (6)

But f(y, ~) = UzEyh(Z, ~).]

Suppose R(y, ~) is rud (resp. p.r.) and that (~x)( ~'y)R(y, x).

Set

~) = I that z ~ y such that R(z, x), if ( 3z ~ y)R(z, ~). f(y,

[ ~,

otherwise.

Then f is rud (resp. p.r.). [Proof: f(y, x) = ~ [y ~{zIR(z , x)}], so (5) gives the result.] (7)

If R(y, x) is rud (resp. p.r.), so is ( 3z 6 y)R(z, x).

[Proof: Let r be rud (resp. p.r.) such that R(y, x) *-+ r(y, x) ~ @.

Then, clearly,

( ~ z ~ y)R(z, ~) +-+ UzEyr(Z , ~) # ~.] (8)

Suppose Ri(x), i = I,..., m are rud (resp. p.r.).

Then so are ~l.
(~I.
(-)0' (-)I are rud (and hence p.r.) . More. generally, . . so are (-)O,"

(-)n-l'n

that z e • x such that (~u I, u 2 e Ux)(x = ~ ul=W) [Proof: For example, (x) 0 = ~, if no such z exists.] that z E LJ(Jx such that ~ x (I0) Set x(y)

[

~, if no such z exists.

Then f(x, y) = x(y) is rud. [Proof: By definition.] (ii) dom and ran are rud. [Proof: dom(x) = {z c ~)lJxI(~w ~ ~)lJx)(<w, z> c x)} ran(x) = {z e 0[JxI(~w e l)Ox)( ~ x)}.] (12) x x y is rud. [Proof: x × y = ~uex

~vey {}.]

-32-

(13) x~y is rud. [Proof: x~y = x n(ran(x) × y).] (14) x"y is rud. [Proof: x"y = ran(x~y).] (15) x -I is rud. [Proof: Set h(z) = <(Z)l, (z)0>.

Then x -I = h"(x ~(ran(x) x dom(x))).].

I

ZF As we shall show,the rud predicates are just those predicates which are 10 . this, it is convenient to make the following definition.

For

Call a function f: V n ÷ V

simple iff, whenever ~(z, ~) is a ~0 ~-formula, there is a Z0 ~-formula ~ such that ZF ~ [~(f(~), ~) ~+ ~(~, y)].

Note that if f, g are simple, so is fog.

Lemma 2 f: V n ÷ V is simple iff: (i)

the predicate x ~ f(y) is z~F; and

(ii) whenever A(z) is ~ F

then so is ( ~x c f(~))A(x).

Proof: (÷) Trivial. (÷) By (i), (ii), it is easy to prove by induction on the length of ~ that for any Z0 ~-formula ~(z, ~), ~(f(x), y) is ZF equivalent to a ~0 ~-formula. I Lemma 3 If f is rud, then f is simple. Proof: Let go"'''

gk be a definition of f from the schemata for rud functions.

By

induction on n $ k, using lemma 2 and the closure of simple functions under composition, we see that each gn is simple, and hence that f is. I We refer to an induction of the above form as an induction on a rud definition of f. Lermna 4 R ~ V n is ~0ZF iff it is rud. Proof: (÷).

By lemma 1 (3), (7) & (8), and an easy induction on the ~ F

definition

-33-

o~ R. (÷) By ind. on a rud definition of ~R' using lerm~as 3 and 2, we see that ~ [ZF, whence so is R.

is

I

Corollary 5 (i)

ZF then R is p.r. If R ~ V n is [0

(ii) If R _c V n is p.r., then R is [ZF. Proof:

(i) follows from lemma 4. definition of ~R"

(ii) is proved by an easy induction on a p.r.

~

ZF Since the predicate On(x) is [0 , and hence (by lemma 4), rud, lemma I (I) leads us to call a function f: On n + V rud (resp. p.r.) iff it is the restriction to On n of a rud (resp. p.r.) function f: V n + V.

And in view of lemma I (i), we shall often

"define a rud (resp. p.r.) function f: o n n ÷

V" by simply specifying the behaviour of

f on arguments in On n. Lemma 6 The functions ~ + i, a + B, ~.~, ~ B ,... are p.r. Proof: By the recursion schema. Lemma 7 Let f(y, ÷x) be p.r.

Define g(~, y, ÷x) = f~ (y, ÷x), where f0 (y, ÷x) = y,

fv+l (y, ~ -~ ÷ +x), fl(y, ÷x) = x) = f(f~(y, x),

O ~ < 1 f~ (y, ÷x) if lira(1). Then g is p.r.

Proof: Since lim(1) is [ZF 0 ' and hence p.r., this is proved by the recursion schema.~ Corollary 8 TC is p.r. Proof: Trivial. | Using the apparatus of p.r. functions, we shall investigate the notion of firstorder satisfaction, and thence the definition of L.

In the meantime, there are a few

points concerning p.r. functions which we may as well dispose of whilst we're at it. A class U is said to be p.r. closed (resp. rud closed) iff f"U n ~- U for every

-34-

(n-ary) p.r. (resp. rud) function f. Lemma 9 Suppose U is a p.r. closed (resp. rud closed) and transitive class.

Let

f: V n + V be p.r. (resp. rud) and defined by the p.r. (resp. rud) schemata go"'''

gp"

Then the same sequence relativised to U defines f~U n,

More generally,

this holds for any transitive class U which is closed under go .... ' gp" Proof: By induction on i $ P.I Lemma IO Suppose U is p.r. closed (resp. rud closed), and that ~: U ~ W, where W is transitive. x e U°

Then for any p.r. (resp. rud) function f, ~(f(x)) = f(~(~)) for all

More generally, this holds for any class U which is closed under a p.r.

(resp. rud) schema defining f. Proof: By induction on the requisite definition of f. I Lemma iO is sometimes referred to as the condensation lemma.

We shall, however,

reserve this title for a somewhat more important consequence of lemma iO, to be proved later. Lemma Ii Let X be any class, and let U be the p.r. closure of X (i.e. the smallest p.r. closed class U containing X).

Let n: U ~ W, where W is transitive.

If ~ X

= idlE,

then ~ = id~U. Proof: Since p.r. functions are closed under composition, U = {f(x) l~ g X and f is +

p.r.}.

By lemma I0, for x E X and f p.r., n(f(x)) = f(~).

The lemma follows

at once. I That completes our discussion of p.r. (and rud) set functions. to the study of the logical complexity of the satisfaction predicate. Arithmetisation of ~ v, Semantics o f ~ . This will amount to a formal definition of the "language" ~ .

We now turn

-35-

Variables:

v

n

= <0, n>, n e ~.

The predicate Vbl (v 0) -- "v 0 is a variable" Constants:

{ = , x ~ V.

The predicate Primitive

is clearly ~ F .

ZF is clearly ~0 "

Const(v 0) - "v 0 is a constant"

formulas:

"x ~ y" = <2, <x, y>>, x, y E Vbl o C o n s t . "x = y" = <3, <x, y>>, x, y s Vbl u Const.

The predicate PFml(v 0) =- "v 0 is a primitive Formulas:

formula"

is clearly ~ F .

~ ~ ~ = <4, <¢, ~>> v ¢ = <5, <¢, ¢>> ÷ ~ = <6, <¢, ~>> ~ + ~ = <7, <¢, 4>>

~¢

=

<8,

¢>

~ u ¢ = <9, >

,

~u¢

, uE

=
¢>>

u

~

Vbl Vbl

u, v c Vbl, u # v . (Vuev)¢

= >, u e Vb[, v ~ Const. u, v E vbl, u # v ,

( 3ueV)¢

= <12, >, u g Vb[ , v E Const.

The class of formulas

of

~/

is the closure of PFml under the above operations.

We shall show below that the predicate Fml(v 0) = "v 0 is a formula"

is p.r.

Note that the functions Vbl u = {vili E ~}, Const u = {~Ix ~ u} are p.r.

functions

of u. Let PFmlu = {tl e t21t I, t 2 c Vblu u Const u} {t I = t21tl,

t 2 ~ Vbl u ~ Const u}

Thus the function PFml u is a p.r. Note that Const = PFml =

~uPFmlu

~uConstu

and PFml(x)

Define F(v, u) = {x

function of u.

and that Const(x)

+-+ x ~ ConStTc(x);

and likewise

+-+ x ~ PFmlTc(x). ~ylx,

y E v} o {x v ylx, y ~ v } ~ { ~ x l x

E v} ~

-36-

U{x÷

ylx, y E v } U { x ~-+ ylx, y c

U{(Vx)ylx

v}

~ Vblu, y e v } U { ( ~ x ) y l x

e Vbl u, y e v}

~J{(~x e z)ylx E Vblu, y ~ v, z e Vbl

O Const U

- {x}} U

0{ ( ~ x ~ z)ylx e Vblu, y e v, z e Vbl u ~ Const u - {x}}. Clearly, F is p.r. In particular, Fml

Hence, so is Fml~u = F~(PFMIu' u) (as a function of ~, u).

= Fml m is p.r. U

U

Then Fml -- ~ u Fmlu and Fml(x) ~-+ x~

FmlTc(x), whence Fml is a p.r. predicate.

The set of variables occuring free in ~, if Fml(~) Define Fr(~) @, otherwise. Fr is definable by a recursion of the form Fr(~) = G(~, ), where G is p.r., and hence Fr is p.r. If ~ is a formula, x is a variable, and t is a constant, ~(x/t) denotes the result of replacing each free occurrence of x in ~ by t.

Define Sub(~, x, t) = ~ ~(x/t), if Fml(~) & Vbl(x) & Const(t)

[ ~,

otherwise.

Sub is definable by a recursion of the form Sub(~, z) = G(~, z, <Sub(y, z) 1 y ~ TC(~) >), where G is p.r., and hence Sub is p.r. Formalising our earlier notation, let ~u ~ mean that ~ is a statement of ~ u which is true in under the obvious interpretation of the members of Const u . Noting that the map which sends x ~ Const u to x c u is p.r., we have: Lemma 12 The function f(u) = {~l~u¢} is p.r. Proof: Define ~ by ~(~, u) = {~ e Fml~I~u~}.

Since f(u) = ~(~, u), it suffices to

show that ~ is p.r. Define g by g(u) = {~ c PFmlulFr(~) = @ & ~u~}.

Clearly, g is p.r. (by our

remark just prior to the lemma). Define h by h(v, w) = v ~ { #

...O{Vx¢l(~y

^ ~]~, ~ e v} O ...

~ w)[¢(x/~) ~ v J } 0 f ( V x

~ ~)+l(Wy

~w

n z)[¢(x/9) E v]} ~ ....

-37-

Clearly, h is p.r. But ~(v, u) = h v(g(u), u).

Hence ~ is p.r.

QED.

Lemma 13 The function Def is p.r. Proof: Define h by h(u, ~) = {x e Ul~u~(V0/X)}.

Taking f as in lemma 12, But Def(u) =

h(u, ~) = u n {x[Sub(#, v0, ~) e f(u)}, so h is p.r. h"({u} x Fmlu).

QED. We may now study the

That completes our analysis of 2 v and its semantics. constructible hierarchy in some detail. The Cons tructible H i e r a r c h ~ T h e Consensation Lemma Lemma 14 (Jensen, Karp)
On> is p.r.

Proof: Since L

= DefV(~), the result follows from lemma 13. [

Corollary 15 The predicate "y = L x

,,

ZF is A 1 .

Corollary !6 The predicate "x is constructible" Proof: x e L +-+ (~e)[On(~)

ZF (i.e. "x ~ L") is El •

~ x e L ]. I

Corollary 17 Let M be a transitive model of ZF, a ~ M ~On.

Proof:

By c o r o l l a r y

15, t h e p r e d i c a t e

a e M-+ ( L ) M

e M.

" y = Lx

vl

is absolute

I

Corollary 18 For all ~ e On,

(L)L

=L

.

Hence (L) L = L. [

C_orollary 19 (V = L) L.

Then L

Hence Con(ZF) + Con(ZFC + V = L). I

~ M and L

f o r M.

= ( L ) M.

But as M ~ ZF,

-38-

Corollary 20 La For all infinite ~ a On, is

@

Proof: L~ is evidently closed under the function . is just the restriction

to a of a p.r. function.

that the canonical p.r. definition

And, by lemma 14, this

Moreover,

it is easily seen

of this function does not lead out of L .

We can thus repeat the (trivial) proof of corollary 5(ii) for L to show that ~ e ) e~ is u. Since a(= On is a ~0 predicate, the result follows at once. I The following lemma is due, in the form stated, Corollary

21 (Condensation

Suppose lim(~). Proof: Let X ~ Z I

Lemma) L , then there is B ~ ~ such that <X, ~> ~ .

If X ~ [ I

L , lim(a).

Then <X, ~> is extensional,

then (as L~ is transitive) ~x

to Jensen and Karp.

~z(z c x *-+ z ~ y).

~e

for if x, y g X and x + y,

~z(z c x *-+ z ~ y), so as X ~ El L ,

So, by the collapsing

lemma, let ~: <X, E> ~ <W, ~>,

where W is transitive. Now, is p.r., and as we remarked during the proof of corollary L~ is closed under its obvious p.r. definition. 5(ii) tells us that u e X ÷ L definition of . Set B = ~"(X course.)

~ On).

(Since ~ is a collapsing map,

c X & ~(L~) = L (v).

B must be an ordinal, of

Note that lim(B), since X ~ I

Then x s La, so ~L ( ~v)(x s L ).

~X ( ~u)(x e L ).

, corollary

s X and that X is closed under the p.r.

So, by lemma I0, u E X ~ L

We show that W = L B .

Let x s X.

So, as X ~ Z I L

20,

Pick u E X with x s L .

La and lim(~).

So, as X ~ I L

,

Then ~(x) E ~(L ) = L (u) ! L B •

Hence W = ~"X ~ L B . Conversely,

let y ~ L B .

Then for some ~ ~ X ri0n, y ~ L (u).

the definition of ~, L (~) is the transitive y = ~(x) for some x s L

~ X.

collapse of L

Thus y ~ ran(~) = W.

The proof is complete. I We now give a rigorous proof that AC holds in L.

But look, by

~ X.

Hence

In other words, L~ ~ W.

-39-

Lemma 22 Let a ~ ~.

There is an L -definable

function

<-, ->e on L

such that:

e

(i)

L

is closed under <-, ->~;

(ii) for all x, y, x', y' e L , <x, y>e = <x', y'>~ ~-+ x = x' ~ y = y'. Proof:

If lim(~),

then <x, y>~ = <x, y> = {{x},

we construct

<-, ->e by induction

it is L -definable, define

it is an element

x, y 5 L , this is sound.)

(-)O'

u (y ×~+I{i}).

We call <-, ->

the e-pairing

Thus,

Le-definable

of Le+ I.

(w)~ = x,

(w)0, ~

(w)~ = y.

<-, ->e is defined.

e x ^ v e y}.

function.

For successor

e,

Thus,

So, for x, y e L +i, we can

The lemma follows

if w = <x, y>~ , then

from w) and

Suppose

We can then define,

= (x ×~+i{0})

(-)I •

on ~.

in L + 1 , x x e+l y = {~lu

,

{x, y}} suffices.

(Since x, y e L + 1 +

in L +i, <x, y>

~+I

at once, by induction.|

Its inverse

functions

(w) I are defined

are denoted by

(in fact they are

(If w is not an a-pair,

then

(w)~,

(w)~ are not defined.) Len~na 23 Let e >~ ~. ~n ¢(Vo/~, Proof:

If x c L + I then for some ¢ ~ Fml@

and some p e L , x = {z ~ Lel

vl/P)}.

Let x e L + I.

Then

for some ¢0 e FmlL

, x = {z ~ L

I~L ~ ¢0(v0/~)}.

Let

0

Pl'''''

Pn b e t h e m e m b e r s o f C o n s t L w h i c h o c c u r

Then p =
~ pn >

Pn a r e L c ~ - d e f i n a b l e

~La(~v0)[¢l(Vl/~) Lermna 24 ( J e n s e n ,

Karp)

Let e >. m.

There


(= < P l '

~-+ ¢0].

from p, Thus,

there

is

x = {z e L

is a well-ordering

<

of L

i n ¢0"

> ' etc.)

Thus Pl'" ~ Lc~•

" ' Pn e L

Since

a ¢1 e Fml~ s u c h t h a t IJ=L ¢i(v0/~,

Vm/~)}.

QED.

such that:

(i) (ii) (iii)

(iv)

is

<

x <

y e L

x c y ~ L

of <~;

c~

-+ x <

We define

on L ) ;

÷ x e L ;

e

e

Proof:

predicate

< 8 ÷ <8 is an end-extension C~

(iv)

L Aj. cL ( a s a t w o p l a c e

y . ~

<

by induction

on e.

(The induction will

satisfy

the recursion

as

-40-

schema, whence

(i) will be immediately true.)

Let <4nln < ~> be a recursive enumeration of Fml#. (inductively) speak.

a canonical well-order of L~, "in order of appearance",

L~ Let this ordering, which is clearly A I , be <~.

< is defined, set <~ = defined.

~<~
so as to

If ~ > m, lim(e), and

Finally, suppose ~ ~ ~ and <~ is

For x s L +i, set i(x) = the least n a ~ s.t. ( ~p a L )Ix=

= {z ~ e I~e ~ ~n(V0/~

vl/P)~,

j(x) = the < -least p ~ e

~ vl/P)}. x = {z ¢ ee[~e ~ ~i(x)(V0/Z, v Ix ~ e

This enumeration defines

~ y ~ e

Set x <~+I y ~

such that

(x <

^ [(i(x) < i(y)) v (i(x) = i(y) A j(x) <

y)

v (x ¢ L

j(y))]].

~y ~ L )

The lemma

is easily checked, l Lemma 25 (Jensen, Karp) Assume V = L. universe.

Then there is a A ZF formula ~ which well-orders 1

Hence V = L ÷ZF AC.

Proof: Set x
y).

Clearly,
the universe

(L) in a

QED.

By the above lemma 24, we can add to the statement of corollary 21 (the

Condensation Lemma) the extra clause: ~(x) E L x for all x e X. <e is

the entire

~[definable,

Note also that since

it is always preserved in applications of the condensation lemma.

For later use, let us state the following extremely important result: Lemma 26 Suppose ~ >. ~ and L

is p.r. closed. Then ~L I (the restriction of ~L e to ZI £~ ct statements of ~ L ) is EL~ (and hence U-absolute). Proof: It suffices to prove that it is p.r.

is El .

And for this, it suffices to prove that

But for any ~0 statement , of ~ L

so the result follows immediately.

' ~0,

+-+ [~TC(,), ~

Lc~

~-+ ~TC(,)*'

(for the last equivalence, we have used

ZF corollary 5 to see that p.r. functions are E1 , lemma 0.19 to see that the ones concerned are in fact &lZF (having HIZF domains), and lemma 0.20 to conclude that these functions are thus absolute for

La.) I

Remarks. (i.)By a close examination of the proof of the above, it is possible to

-41-

conclude

that the assumption

that L

(This boils down to showing that L is not particularly Any p.r. shown,

difficult.)

closed transitive

therefore,

be p.r.

closed in the above can be omitted.

is closed under the relevant p.r.

12.)There

is nothing special

We conclude w i t h the following

is U-absolute

and

about La in the above.

set M will have the same property.

is that [1-satisfaction

functions,

What we have really

under very general

conditions.

important

result.

Let X ~ L

, and let M be the set of all elements

Lemma 27 Let e ~ ~ be a limit ordinal. of

L

Proof:

which

are

L -definable

By lemma 24, L

Corollary

from

parameters

has a complete

in

X.

Then M ~ L .

set of definable,

canonical

skolem functions.

28

Let ~ ~ ~ be a limit ordinal. sense of ~ )

M ~L

The Generalised

Then for each X ~ L

such that X ~ M;

there is a smallest

(moreover~IM I = I X I + ~ ) .

Continuum Hypothesis.

We have seen that L is an inner model of ZF, definable AC + V = L are (ZF-provably)

(in the

(ZF-provably) consistent

true.

relative

Hence,

any consequence

to Con(ZF).

in ZF, in which

of ZFC + V = L will be

One such consequence

is GCH,

as we now

show. T h e o r e m 29 (G~del) Assume V = L. Proof:

Then GCH holds.

By induction,

it is easy to prove that for all infinite

Hence it suffices So, let a ~ K.

such that K u {a}

lemma,

let ~: M ~ L .

transitive,

that for any such K,

Pick a cardinal

M ~ L

Now,

to prove

m M.

IL I = K, so we must have ~ < ~+. ~<

= ida<.

Hence ~(a) = {~(~)I~

This means that a c ran(q)

= Le E LK+.

~(K) ~ L +. K

% such that a e L%.

Thus, by corollary

Hence

28,

cardinals

ILKI

= K.

--I~)

Let M be the smallest

IMI = <.

By the condensation

Hence L

_= LK+.

~ a nM}

= {~(~)I~

~(<) _c LK+.

<,

Since < ~ M

QED.

is

s a} = {¢I~ s a} = a.

-42-

We conclude this chapter by remarking, L

= H

K

for all infinite cardinals K.

in the sense of L, then L world).

K

Hence,

is a ZF--model

for later use, that if V = L, then if K is an uncountable regular cardinal

(both in the sense of L and in the real

This implies that (regardless of whether or not V = L) L

whenever K iS an uncountable regular cardinal of L. L(~)

is a countable ZFC--model.

K

is a ZFC--model

In particular, if ~

< ~I' then

Chapter 3

THE SOUSLIN HYPOTHESIS

By a linear ordering we mean a structure <X, <> , where transitive, antisymmetric binary relation with field X. (~x,

y s X)(x < y + ( ~ z

ordering

X

s X)(x < z < y)).

such that wherever Y ~

<

is a reflexive,

It is dense if

An ordered continuum is a dense linear

X is such that there are x,-y ~ X

with

x ~ z ~ y for all z s Y (in which case we say Y is bounded in X), then there is a maximum such x and a minimum such y, in the sense of < (in which case we call x the infimum of Y and y the supremum of Y). end-pointSo (Vx,

An ordered continuum is open if it has no

A subset of Y of a linear ordering X is dense in X if

y g X)(x < y ~ ( 3 z

~ Y)(x < z < y)).

It is a well-known theorem of elementary

real analysis that if X is an open ordered continuum having a countable dense subset, then X is isomorphic to the reals,

5.

Souslin asked the following question. open ordered continuum,

Can we weaken the above condition on an

and still characterise the reals uniquely?

More precisely,

let us call a subset Y of a linear ordering an interval if it is of the form {z ~ X I x < z < y} for some x, y E X, x+y. By the Souslin property on a linear ordering X we mean that every collection of pairwise disjoint intervals of X is countable.

Clearly, if X is an open ordered continuum with a countable dense subset,

then X has the Souslin property, the countable dense subset°

since any interval of X must contain a member of

The Souslin hypothesis is the assertion that any open

ordered continuum with the Souslin property is isomorphic to the reals° It is known from work of Tennenbaum and Solovay that the Souslin hypothesis undecidable in ZFCo

In fact, by later work of Jech and 3ensen, the Souslin hypothesis

is known to be still undecidable if we assume the GCH as well. prove below,

is

the Souslin hypothesis is false if we assume V = L.

the work of Tennenbaum, etc., the reader should consult

[10].)

However,

as we shall

(For references to Before we commence,

let us observe that the Souslin hypothesis as stated above is clearly equivalent to the following assertion, which we denote by SH:

Every dense linear ordering with

-44-

the Souslin property has a countable dense subset. The idea behind our proof that the Souslin hypothesis is false in L is the construction of a counterexample to SH, i.e. the construction of a dense linear ordering having the Souslin property but no countable dense subset.

To do this, it

is helpful to have some sort of well-ordering structure on the proposed counterexample, so that the construction may be done inductively,

For, the strength of the

axiom V = L is that, by virtue of the condensation lenmm, transfinite induction arguments can be carried out so that one has considerable control over the eventual result.

This should become clear when we come to conStruct the counterexample.

First we consider the "well-ordering structure" we mentioned above.

We commence

with some definitions. A tree is a partially ordered set T = such that for every x E T, the set ~ = {y s T I Y
is well-ordered by < .

is called the heisht of x in T, htT(x). the set Te = {x e T I htT(x) = ~}. set ~8< e

TBo

If e is an ordinal, the ~'th level of T is

We write T~e for the restriction of T to the

A branch of T is a linearly ordered initial segment of ~.

s-branch if ~ is its order-type under < o @~.

The order-type of ~ under
It is cofinal in T if T

~ ~+

It is an T

~ b

(Note that if b is a branch of T, then Ib N T I ~ 1 for all ~. Also, ~ is a

branch of T for every x e T.)

An antichain of T is a pairwise incomparable subset

of T. Let K be any ordinal, ~ a cardinal.

A (<, l)-tree is a tree such

that:

(i)

(V~

< <)(T

& (T< = 9 ) ;

#~)

(ii) L e t T be a (<, l ) - t r e e .

(i)

T is normal iff:

IT0i = 1 ;

(ii)

(~+I

(iii)

(V~

< B < m)(VX

(iv)

(Ve

< <)(lim(~) ÷ ( M x ,

< K)(Vx

g T)(~yl, e T)(

Y2 e T +l)(y I ~ Y2 & x
~y s T~)(x <

y);

y E T )(~ = ~-> x = y).

-45-

The well known K~nig Infinity Lemma says that every (~, ~)-tree has a cofinal (i.e. infinite) branch.

However, Aronszajn has shown that, in ZFC, there is a

normal (ml' ~ )-tree with no cofinal branch. 1 the reader to []O]-).

(For a proof of this result, we refer

Such a tree is called an A~onszajn tree.

By a Souslin tree we mean a normal (~i" ~I )-tree with no uncountable antichain. Lemma I If T is a Souslin tree, then T is an Aronszajn tree. Proof: Suppose b is a cofinal branch of T. A = {y s T + 1 I a < ~I & ( ~ x

E T

But A is clearly an antichain of T.

By normality, the set N b)(x < y) & y ~ b} has cardinality w • T I Hence, if T has a cofinal branch, it

cannot be Souslin. QED. Lenmm 2 Suppose there is an (~I' ~I )-tree" T, with no cofinal branch and no uncountable antichain.

Then there is a Souslin tree (embedded in T).

Proof: Let T be as stated.

Let T' = {x e T I ( ~ @

> h t ( x ) ) ( ~ y e T~)(x
It is easily seen that T' is an (~ , ~ )-tree with no eofinal branch and no 1 1 uncountable antichain, such that T' ~ T

for all ~, and such that T'

satisfies condition (iii) of the normality requirements.

Let x 0 be an

arbitrary member of T O' , and set T" = {x s T' I x0 ~m x} .

Then ~ T" has the

same properties as those of T' just mentioned, and in addition T" satisfies condition (i) of the normality requirements.

By induction on ~ < ~I' define

a strictly increasing function f = ~I ÷ ~I so that f(O) = 0, f(~) = supB f(a) such that (~x

~ T'~(~))( ~Yl ' Y2 ~ T")" B tYl 4 Y2 & x
T* = ~a<~ I

T"f(a+l).

Set

Then T*~ Is an (~I , ~l)-tree with no cofinal branch and

no uncountable antichain, such that T* satisfies (i) - (iii) in the normality definition.

(In particular, since T" had no cofinal branch and yet

satisfied condition (iii), T* is well-defined.) the levels of T*

Finally, by induction on

it is easy to obtain a subtree of T* which satisfies

-46-

condition (iv) of normality as well (This tree will be a Souslin tree, of course.).

One simply picks (by induction on the levels) one limit extension

of each limit branch which has an extension.

The lemma is proved.

Remark By examining the above proof, one can easily amend the definition of the function f so that in the resulting tree, every point has infinitely many successors on the next level.

Such a tree is said to be prolific.

The next lemma justifies our present interest in trees, and the terminology we employed. Lemma 3 $H iff there is no Souslin tree. Proof: Suppose T is a Souslin tree. is prolific.

By the above lemmas and remark, we may assume

For each ~ < ~I' T

is countable, and for ~ > O, infinite.

So, for each non-zero e < ml, let < that of the rationals.

be a linear ordering of T

Let X be the set of maximal branches of T.

linear ordering on X by b < d +-+ ( ~ ) [ ( ~ x (b(e) <

of order-type Define a

c TI~)(x c h +-+ x e d) &

d(~))], where b(e) denotes the unique element of r

N

b, and

similarly for d(e)o

It is easily seen that <X, <> is a dense linear ordering

without end-points.

Since T is a Souslin tree, <X, <> has the Souslin

property.

For suppose I is a non-trivial interval of Xo

I = {c ~ X I b < c < d} b(e) ~ d(~).

Let

xI s r

for some b, d E X.

Let ~ be least such that

be such that b(~) <

maximal branch of T containing x I.

Then

xI <

Thus b(1) e I.

d(~).

Let b(1) be a

If, now, J is an

interval of X disjoint from I, then clearly x I and xj are incomparable in T.

Hence an uncountable collection of pairwise disjoint intervals of X would

give rise to an uncountable ~ t i e h a i n

of T, which is impossible.

Finally, we show that X has no countable dense subset, which will establish "~SH. subset of X.

Suppose, on the contrary, that D is a countable dense

For each b, d ~ D, let e(b, d) be the least ~ such that

-47-

b(~) ~ d(~).

Let ~ = sup{~(b, d) I b, d e D}.

As D is countable, 6 < ~ . I

Let x ~ T~, and pick YI' Y2 ~ T6+I such that x
Suppose, for the sake of argument, that Yl <6+I Y2"

Then

But since b I and b 2 coincide on T~6, no member of D can lie

between b I and b2, by choice of 6.

This contradicts the supposed density of

D in X, and completes this part of the proof. For the converse, assume ~ S H ,

and let <X, <> be a counterexample to SH,

i.e. suppose X is a dense linear order without end-points, satisfying the Souslin property, but having no countable dense subset. of non-trivial intervals of X. defined, let A

Io is arbitrary.

If is

be the set of end-points of all IB, B < ~ (i.e. the points which

determine each interval)° I

By induction, we define a sequence

Since

A

is countable, it cannot be dense in X, so pick

so that it and its end-points are disjoint from A .

partially order T by I
Let T = {I

I ~ < ml}, and

Since, by construction, Is ~

The existence of an ~l-branch < I

I B implies

I ~ < ~i >, of T, would imply

the failure of the Souslin property for X, since, if we let x~ be the lower end-point of I

, each E, then the pairs <x~, x~+l> determine an uncountable collection of

pairwise disjoint intervals of Xo

The existence of an uncountable antichain of T

would likewise contradict the Souslin property for X, since I and J will be Tincomparable iff they are disjoint intervals of X, of course, Thus T must be an (~I' ml )-tree with no cofinal branch and no uncountable antichain. a Souslin tree now follows by lemma 2.

The existence of

The lemma is proved.

By virtue of lemma 2, by SH we shall, from now on, mean the assertion that there is no Souslin tree° of a Souslin tree in L.

The task of this chapter thus reduces to the construction This can be done directly by means of an inductive procedure

on the levels of the proposed tree, using the condensation lemma (in a very ingeneous manner) to ensure that the resulting tree has no uncountable antichain.

However, it

is perhaps more instructive to isolate the fundamental combinatorial property of L which makes the whole thing work.

-48-

Recall (~ q

that a subset

C of a limit ordinal

C) ~ C for all ~ < %.

every closed cofinal cardinal,

(in %) iff

A subset E of % is stationary

subset of %.

In particular,

then every closed cofinal

of a stationary

% is closed

in % if it intersects

if ~ is a regular uncountable

subset of % is stationary

in %.

subset of such a % would be the set E = {e ~ % I cf(~)

note that, w i t h % still regular and uncountable,

if E is stationary

E

subset C of %.

n C is stationary

in % for any closed

for any limit ordinal Classically, "regressive"

%, a stationary

stationary

functions.

f: A ÷ % is regressive

cofinal

discovered

Also,

in %, then

Finally,

note that in %.

sets were studied in connection with the so-called

(If % is a limit ordinal if ( ~

= ~}.

subset of % must certainly be cofinal

c A)(f(~)

< ~).).

the following one, later adapted by Rowbottom cardinals.

Another example

and 0 ~ A

~

%, a mapping

The main result in this area,

is

to figure in the theory of large

The proof given is not Fodor's but it is not certain who first it.

Theorem 4 (Fodor) Let K be a regular uncountable in K.

If f: E ÷ K is regressive,

stationary Proof:

cardinal,

and let 0 ~ E

~

then for some 8 e K, {~ ~ E

< be stationary

I f(~) = B} is

in K.

Suppose,

on the contrary,

stationary CB ~

in K.

Then,

that for each B E ~, {~ s E

for each 8 s < we can find a closed cofinal set

< such that ~ c C B q

Clearly,

I f(e) = B} is not

E ÷ f(~) ~ B.

C is closed and cofinal

B < ~, ~ e C B, so f(~) # B.

in K.

Hence

Let C = {~ c ~ I ~ E ~ B < ~ CB}.

Let ~ s C

f(e) ~ ~.

q E.

Then,

for each

But this contradicts

the

regressive nature of f. QED For the present purpose, following By S

~

combinatorial

stationary

principle

0 we mean the assertion

of Jensen. that there is a sequence

~ for each e, and, whenever X

is stationary

in ~io

sets are important by virtue of the

~ w I is given,

<S

I ~ < ~i > such that

the set {~ s ~ I X

q ~ = S }

-49-

Note that

0

immediately

implies

2m

, since every subset of ~ must figure = ~I

(~i many times,

in fact)

We shall prove We require

in the 0 - s e q u e n c e •

that

0 holds

two simple

in L.

lemmas on the constructible

hierarchy.

Lemma 5 Suppose V = L. suffices

If X ~ L i' then X = L~ for some ~ E ~I"

that X ~

L i , as the proof will

(In fact,

it •

indicate•)

I Proof:

By the condensation

lemma,

Let x E X be arbitrary. f: m +-~ x.

Since x ~ L

to prove

L 1-definable

, f ~ X.

But ~

other words x = f" ~

~

, f is ~ X, o f

X.

that X is transitive.

, x is countable.

Let f be the
since
it clearly suffices

L01-definable course,

By 2.29(*) 9 f E L i from x.

so f(n)

Thus X is transitive•

Hence there is

E X for

So,

And,

as

e a c h n ~ w.

In

QED.

Lem~a 6 Suppose V = L.

If X ~ L

, then X

~ L

~2 Proof:

L 1 ~s a definable

point of L 2

the same formula that defines follows easily that L i A the s-extension L

.).

~

for some ~ Z ~ . I

Hence L ] E X and L i is X-deflnable by

L

as a point in L

~I X ~ Lwl

of the point L

The result

= L ~I

Since X ~ L , it ~2 ~2 (This is because L i ~ X is of course,

in X and L

is the s-extension

of L

in

follows by lemma 5. I

~2 Note that, in particular, use in our proof of to ensure that ~I' ~ results

~ 2

in L.

X

This is the fact we shall

We remark that the assumption

are absolute

in ZFC with ~I' L ~

n ~i = ~ in the above•

in the above

two lemmaso

V = L was only necessary There are corresponding

in place of ~i' ~2' of course•

T h e o r e m 7 (Jensen) Assume V = L. Proof:

Then

~

holds.

We define, by induction Set S O = C O = ~ and S

~+I

on ~ < ~i' a sequence = C

~+I

<<S , Ca>

= ~ + i, each ~ < el"

I e < ml > as follows• If lim(~),

let

-50-

<Sa, Ca> the the
~ Ca)(S a ~ y # Sy); if no such pair exists, set

S 5 = C 5 = a. We claim that <S

a

I a < ~ > satisfies 1

O"

Note that the sequence

<<S a, Ca> I a < ~i > is (by the above definition) <$5 I a < ~i> does not, in fact, satisfy ~ .

L 2-definable.

Suppose

Then we can find a pair <S, C> of

subsets of ~I such that C is closed and cofinal in ~I and ( ~y e C)(S

O Y # Sy).

Let <S, C > be, in fact, the .

Let X ~ L

be countable.

Since <Sa I a < el > is L~ -definable, 2 By the condensation lemma, let ~: X m LI.

2 By lemma 6, let a = ml ~ X.

Clearly, ~(el ) = a, ~(<S, C>) = <S

~(<<S B, C~> I ~ < ~I>) = <<S 8, CB> I ~ < a>.

~ a, C

~ a>, and

(All of this is extremely easy to see,

and simply uses the fact that X ~ L

, and the definition of the collapsing ~2 isomorphism - which is, of course, the identity on La.) Since - I : L%-< L 2 , ~L% "<S

~ a, C

~ 5> is the
closed and cofinal in a and ( ~ y considerations,

E C

~ ~)(S

~ a n y ~ Sy)."

By absoluteness

we see that since this sentence holds in L%, it must in fact hold in

the real world (i.e. L).

Thus, by definition~<S

particular,

But look, since - I

S

n a is

n a = Sa •

:

n a, C n 5> = <Sa, Ca>.

L% ~ L~ 2 ' I=L% "C n a is closed and

unbounded in a", so again this sentence must really be true. in ~i" a = sup(C

~ ~) e C.

In

By definition of C, therefore,

Hence, as C is closed S ~ a ~ Sa.

This

contradicts our earlier result that S N 5 = S , and the proof is complete. I Finally, we show how "condensation principle",

O, which, as the above proof shows, is a kind of implies the falsity of the Souslin hypothesis.

Theorem 8 (Jensen) Assume <>o

Then there is a Souslin tree.

Proof: We shall construct, by induction on the levels, a Souslin tree T = <~i' " The induction will be carried out so that ~
If ~ ( a

+ i) is defined, obtain Ta+ 1 as follows.

The

-51-

elements of T ordering.

are countable ordinals,

Proceeding

along T

and hence have a canonical well-

under this ordering,

appoint,

to each x s T

in turn, the next two ordinals not already used, as immediate x on T + I.

This defines T~(e + 2).

If T~(e + I) was a normal

tree, then T~(~ + 2) will clearly be a normal remains the case where T ~ < ~i a limit ordinal.

successors

((e + l),~l)-

((~ + 2), ~l)-tree.

is defined, and is a normal

to extend

(By normality and the fact that cf(~) = ~ , for

there will of course b e a cofinal branch of T ~

The problem is to decide which branches of T ~

to extend.

this decision, we use our ~-sequence, not a maximal antiehain of T ~

then we let b

x

( = ~i) induces a well-ordering

is a maximal antichain of TI~.

be an arbitrary

For each x e T~e, let b

x

If S

is

cofinal branch

consist of one-point of the points

of the b 'x s , we can do this in a interval of ordinals.

Then, for each x e T ~

such that x and x' are T-comparable.

To make

Look at S .

Since the canonical well-ordering

canonical way, using the next available

Let T

I ~ < ~ >.

containing x, for each x E Tra, and let T

extensions of each b . x x e T~

<S

containing

Since we muSt

have ~ Tel = ~, we cannot extend them all, for there are 2 m such.

of T ~

for

Since we shall want T l'(e + I) to be a normal

each cofinal branch of T ~ .

x.).

There

(~, ~l)-tree

((~ + i), ~l)-tree, we know that we can only add one point in T

every point x s T ~

of

Suppose

S

we can find x' e S

Let x" be the T-maximal

of x, x'.

be an arbitrary cofinal branch of Tre containing x".

consist of (canonically

appointed)

one-point extensions

of each b . x

Note that, in either of the above cases~ T~(e + I) is clearly a normal ((~ + 1), ~l)-tree.

Furthermore,

T~e, then every point in T That completes

if S

happens

to be a maximal antichain of

is comparable with some point of S

the definition.

Set T =

T~o Clearly, T is a normal I We show that T is Souslin. It suffices to

(~I' ~I )-tree with domain ~I"

~ <~

show that every maximal antichain of T is countable. let A be an uncountable maximal

antichain of T.

C = {e e ~I

~ a & A ~ ~

I lim(c0 & T~e = T

is closed and cofinal in ml"

below it.

Well, suppose not, and

Clearly,

the set

is a maximal antichain of T ~

(This is easily proved, using only the facts

}

-52-

that each T ~

is countable

s C such that A is a maximal

and cf(~ I) > ~.).

n ~ = S .

antichain of T ~ .

lies above Some element of S . such that htT(x)

> ~.

Since ~ e C, lim(~),

~ , we can find

T~

= T

Thus, b y construction, Since A is uncountable,

Then x lies above some y s T

above remark, we know that there is z e S z c A

So, by

The proof is complete. I

every element of T we can pick x E A

of course.

lying below y.

q ~ and x ~ A lies above z, so this contradicts

antichain of To

N ~, and S

And, by our

But look,

the fact that A is an

Chapter 4

THE KUREPA HYPOTHESIS

Suppose T is a normal (Wl, ml)-tree. ~l-branches does T have?

One can ask the question: How many

By the result of Aronszajn mentioned in the previous

chapter, the answer can be "none".

By making simple-minded alterations to an

Aronszajn tree, it is easy to obtain normal (~I' ~I )-trees with m ~l-branches or with ~I ~I -branches"

Now, since a branch of an (~i' ~I )-tree is certainly a

subset of the tree, and since such a tree has cardinality ~I' 2~I bound to the number of branches T can have.

• is an upper

So, assuming GCH, is there a normal

(~I' ml )-tree ~ in which this upper bound of ~2 Wl-branches is attained.

Such a

tree is called a Kurepa tree; and the Kurepa hypothesis is the assertion that there is a Kurepa tree.

It is known from work of Silver that Con(ZFC + "there is

an inaccessible cardinal") ÷ Con(ZFC + GCH + "there is no Kurepa tree"). prove below that if V = L, then there is a Kurepa tree.

We shall

This result is due to

Solovay, although we shall obtain it by a route similar to that employed in Chapter 3, by first abstracting from L the salient fact we require. due to Jensen.

This proof is

Silver's result, and the proof of Solovay, may be found in [I0].

It will be shown in Chapter 18 that a slight modification of the proof works for the universe L[A], where A ~ ml L[A].

From this result (noticed first by Solovay

himself), it follows that if ~2 is not inaccessible in L, then there (really) is a Kurepa tree.

Hence Silver's assumption of an inaccessible

necessary one.

in the above

is a

All of the details are in chapter 18.

Before we commence our proof, we reformulate the Kurepa hypothesis as a simple set-theoretical assertion. family

By KH, we mean the assertion that there is a

~ ~ @(~i ) such that I~ I = ~2 and for all ~ < ~I' {x n ~Ix ~ ~ } is

countable. collection

(Thus, if an observer of countable height were to look at the ~ , he would conclude that

~ was a small (i.e. countable) collection,

whereas an observer of height ~I would see a very large (i.e. of cardinality ~2) collection.)

-54-

Le_____~a I KH iff there is a Kurepa tree. Proof: Let T be a Kurepa tree. that ~ < T ~ ÷ ~ < ~"

Since I~I = e I, we may assume ! has domain e I and Let ~ be the set of all wl-branches of ~.

Clearly,

satisfies KH. Conversely, let ~ by fx(~) = x o ~.

~(w I) satisfy KH.

For each x ~ ~, define a map fx : el ÷ ~(el)

Let T = {g I (~x e ~)

say gl < T g2 iff gl = g2"

(~

Clearly, T = < T ,

that it is, in fact, an (el, el)-tree.

< el)(g = f x ~ ) } . ~ T >is

a tree.

For gl' g2 ~ T, It is easily seen

(In particular, since {x q six e ~} is

countable for each ~ < el, each T~ is countable).

Each set {fx~eI~ < e I}

ml-branch of T, unique for that x ~ ~, so T has w 2 el-branches. in lemma 3.2, it is easy to obtain a Kurepa tree from T. requirements easier.) Remark.

is an

By an argument as

(Note that the normality

(i) and (iv) hold already for T, so this time the argument is even

The lemma is proved. I KH is, in fact, the formulation of Kurepa hypothesis which Kurepa

considered. In view of the previous chapter, it may be somewhat surprising to find out that it is easier to establish KH in L rather than to construct a Kurepa tree in L. Indeed, one can construct a Kurepa tree outright in L.

It just happens that, in

this case, and unlike the Souslin hypothesis, the tree structure is not particularly important.

(This holds both for Solovay's proof and the proof which we shall give.)

The following assertion, due to Jensen, is known as <S

I ~ < ely such that S

0÷: There

is a sequence

_ ~(~) and IS I ~ w for each ~ < e I, and, whenever

X ~ e I is given, there is an unbounded set B ~ e I such that ~ = sup(B q ~) implies X ~ ~, B ~ ~ ~ S . Firstly, let us see why this principle is called ~+. <S

I~ < e l > i s

closed

cofinal

Clearly, if

as above, then for any X =- el' {~ ~ ~I I X n ~ ~ Se} contains a subset of e I, namely the set of all limit points of the

corresponding set B.

Hence, if by ~' we mean the assertion t h a t ~ t h e r e

is a

-55-

sequence < S whenever

J ~ < ~I> such that S

X ~ ml

consequence

then {~ s ~i

of ~+.

~ ~)

I X ~

The next

Is~l ~ ~

and

for each ~ < ~I' and,

E S } is stationary],

lemma thus explains

then 0' is a

our notation.

Lemma 2 (Jensen) 0 and 0' are equivalent. Proof: Let < S Conversely,

I ~ < ~°i> satisfy 0. assume ~'.

T~¢- ~(~ × ~) and {~ e ml

I A • 2

for infinite Let
(n)

~,

Then we can find a sequence

g T } is stationary. I~ × ~I = I~l.)

Claim i.

For each X c 2

T (n(X))}

is stationary.

ml"

Define

Claim 2.

There

not.

there is n(X)

(trivially),

and {~ ~ ml

S

by means

< ~ such that {~ E ~I

of two claims.

I X n~2

=

& X n ~2 e T }, a stationary

Theorem

Then for each n < m there

contrary

= T (n)"{n}.

claim 2) stationary. 0' ÷ 0"

is the fact that

3.4 gives

QED.

subset of Since

us an integer n < ~ and a

Then n is as claimed.

I Y ~ ~2 = T (n)} is stationary.

I Y ~2

= T~(n)}

is X n c_ coI such that whenever is not stationary.]

Then for all n < m, {~ ~ ~i

2 Y c_ ~i'

Define Y c ~0~ by

J Y n ~2 = T (n)} is not

to claim i.

The lemma is now easily proved. set

× Wl'

2 is n < m such that for all X G m I there is Y c ml such that

= Xn, each n < ~0.

stationary,

~'.

T , each ~ < ml"

set H =_ E such that ~ s H -~ f(e) = n.

X n = Y"{n} ÷ [{~ s ml Y"{n}

A ~i

f : E ÷ ~ by f(~) = the least n such that X ~ ~2 = T (n).

X = Y"{n} Suppose

The proof now proceeds

0', so ~ +

I ~ < ~i ~ such that

(All we are using here

Set E = {~ g WlJ ~ > ~

f : E -~ ~I is regressive stationary


IT J .< w for each ~ < ~i' and, whenever

I n < ~enumerate

Let X c _ ~ .

Then <{S } I ~ < ml > satisfies

Then

S

Hence < S

c ~

Pick and, if

n

as in claim 2.

X -ca I, {~ E ~i

I ~ < ~i > satisfies

0.

For each ~ < ml '

I X n ~ = S } is (by

This proves

that

-56-

Thus, ~ + is a strong version

of 0.

In particular, O + ÷ CH & ~SH.

Theorem 3 (Jensen) Assume V = L.

Then O + holds.

Proof: For each a < Wl, 3.5, M s = LB(s)

let M e be the smallest M ~ LWl such that a c M. for some B(s) < w 1 .

w h i c h is Lwl-definable Thus Ss q; ~(s) and that the sequence

Note

that as there is a map f :

We show that < S

<S s I a < w I > is (by the above definition)

L 2-definable.

subset of Wl such that for all unbounded sets B ~ el' there is a limit

Lw -definable. Using Lermna 3.6, define a chain < N 2 models of L and a continuous, cofinal sequence < s ~2 ordinals, by induction, as follows. Let N O be the smallest N < L 2.

Let N~ = ~ <6N

,

s~

= sup~<~s

n :

Then v (Wl) = a

B = {$(~)

I ~ < ~i }, an unbounded

We show that B satisfies

let a be an arbitrary

Claim I.

<

Clearly, transitive

I ~ < Wl> of countable

Set s + I = ~i ~ N~÷I "

and s

= w I ~ N~ for all ~ < w I .

and ~y(X)

subset of Wl"

Let

= X ~ a , each ~ < Wl"

(Each Nv is clearly

of O + for X, contrary

Set

countable).

to the choice of X.

limit point of B.

~(~) < s +i, all v < ~i" s

< 6(~)

collapse

s N + I.

(~ < w~

the requirements

To this end, s

sub-

, if lim(6).

... ~ Lw2

each ~.

that X is

Set s 0 = w I ~ N O .

Thus N0~< N 1 -< .. . - < N ~ ~ ~ L6(v),

Note

I v < W l > of elementary

Let N + 1 be the smallest N ~ Lw2 such that N v o {N~} ~ N.

L6(~)

Note

Let X be the

point a of F such that it is not the case that X q s, B q s e S . s

N

L~(a)"

; s < wl> satisfies 0 +.

Suppose that < S s I s < w I > does not, in fact, satisfy 0 +.
s

Let S s = ~(s) q

from s, ~LB(s)'~s is countable".

Issl ~ ~.

By Lemma

of N

for each ~ < w I. is an element

But 6(v) < ~i"

Also,

of N

~+i

.

since N~ e N~+I ~ Since N

Lw2 , the

c N , this means ~ - ~+i

Hence ~(v) e w I q N + 1 = e + I.

that

This proves the

claim. By the claim,

it follows

that

s = s

for some limit ordinal n < ml"

The

-57-

next claim is the guts of the proof, and emphasises the "condensation lermna" nature + of the ~ -sequence we have defined. Claim 2.

B n ~ is ZF--definable from L6(~).

Define,by induction on ~ < ~, a chain of elementary submodels of L6(n) as follows. Let Ng be the smallest N ~ L@(D). Let N'~+I be the smallest N ~ L ~ ( q )

such that N'~ o {N~} c_ N.

Let N~ = U < 6 N'~ if lim(@). Now, for

~ < ~, N

-< L , so we can equally well define the sequence D 6a2
-~ N

(Limit stages in the induction are trivial.

For successor

stages we are simply taking skolem hulls, inside isomorphic structures, of sets which are isomorphed by that isomorphism, so the result is easily seen. initial stage is a degenerate case of the successor step.) the same transitive collapse, namely L~(~), for each ~ < n.
i

I ~ < ~>is

<e6(~)

Hence N' and N

and, in particular, so is <6(~)

<$~(~) I ~ < ~ > i s

cofinal in ~ = ~ .

I ~ < n~.

have

But the sequence

ZF--definable from L6(q) (by the above definition).

I ~ < ~,

The

Hence so is

But by claim I,

Hence {6(~) I ~ < ~} = B ~ 6 ,

and the claim

is proved. Claim 3.

~(n) < B(~).

We have observed that ~LB

"6 is countable".

But ~ = 6

(6)

e~ (n)

"6 is uncountable " .

= ~ (6oi), so n

n

Hence 6(n) < B(~), as claimed.

By claims 2 and 3, we see that, as LB(6) is a ZF--model (by its definition), B n6

g LB(6).

Hence X ~ ,

Also, X O 6 = ~q(X) g L6(~) ~-LB(6)

B n~

E S .

c LB(6).

But this contradicts the choice of X, and the theorem is

thus proved.| Theorem 4 (Jensen) + Assume ~ .

(by claim 3), so X n 6

Then KH holds.

-58-

Proof: Let < S e I e < m l > s a t i s f y

0 +.

u(U~seS6) ~

M -< H001 such that

(e +I )

Set

(~e < Wl)(X

5=

{xC_~0 1 I

For each ~ < ~i' let M

qe

M.

{x q e I x e ~ } ~ M , w h i c h is countable,

only necessary

to prove

repetitions)

all unbounded members

closed unbounded is unbounded

Let<x

set X c w I such that

in e)].

X o e, B q e e S e •

we suppose

I v < ~ l>

[~ e X -~ lim(e)~

e~ ~< ~

and eofinal in ~I"

< ~V+l "

Set x = { B

enumerate

(~

< ~l)(X =~x ).

this is trivial, limit point

e

e

~ Z~-.

definable

by showing .

Suppose x n e is infinite.

Since

Thus Since

that x e ~. If x q a is finite,

Let B be its greatest

Then x q ~ = (x q 6) O (x ~ e - B).

and hence

B q B ~ SB c Ms .

an element of M s . Now,

But

So, as

B is a limit point of x ~ 6,

But look, x n 6 is clearly gF--

from X q 8, B q ~ in the same way that x was defined

x q 6 ~ M s, and the proof is complete. I

qe)}.

c~ B~ + ~ < e +i, each v < ~I' w e see

to prove that x q B e M . e

and x ~ B, so by 0+, X q B ,

q e

I 9 < ~°I> is

subset of ~i"

We must show that x q e e M

X n ~ - 6 is finite,

~ ZF-, it suffices

e X)(~v < e ) ( x

in ~01 , < a

We obtain our contradiction

(which may be ~, of course).

by choice of $, M

since M

& [(~

= min(B - ~ ), each 9 < ~I"

but x N ~ v + l

Let e e ~i be arbitrary.

(possibly with

{~ ~ X I ~ = sup(B

I v < ~i } , an unbounded

xv q e + I is unbounded in ~ + I that

Set B

I~I .< el,

and such that ~ = sup(B q ~) implies

X is closed and unbounded in ~i and B is unbounded continuous

that, on the contrary,

Since cf(~l) = ~i' it is easy to find a

Let B c el be unbounded Let < a

KH.

for each e < ~I' so it is

I ~ < ~ I > enumerate

of ~.

be the smallest

~01

We show that 7 s a t i s f i e s

By definition,

and w o r k for a contradiction.

-< H

Note that each M s is a ZF--model.

£ M)}.

that I~I = w 2.

e

from X, B.

Hence

Chapter 5

CONSTRUCTIBILITY

AND THE CONTINUUM

In this chapter we shall consider some of the effects of V = L upon the descriptive

set theory of the continuum.

the linear ordering f < g iff: f(n) < g(n)].

(By the 'continuum'

we mean the set w

[f # g & if n is least such that f(n) # g(n) then

This is the most convenient

choice of 'continuum'

since it allows us to make use of standard results in recursion although the results we shall prove are not primarily it will be necessary theory.

we require already available

clear and concise description

in the literature,

account is so good (in our opinion)

denote numbers

generalised

~

of [17].

For this reason, we shall here as there.

By 'number' we mean 'natural number'. By 'function'

Now,

recursion

of exactly the material

namely chapter

chapter only) use the same notation and conventions

subset of some n .

theory.)

This

that we shall not even try to surmnarise it here,

but will assume it as a prerequisite.

the following.

for our purposes,

recursion theoretic results,

to assume some familiarity with elementary

There is an extremely

with

(or sometimes number variables,

In particular,

By 'relation'

we mean an element of mm.

(i.e. for this we use

we mean a

Small Latin letters

according to context)

and small Greek

letters denote functions (or variables for such). P, Q, R, S, T denote relations. m n We set N = ( K ~ ) × ( ~ ~). A map from some N into ~ is called an (m, n)m,n i=l i=l m,n dry functional.

We use F, G, H to denote functionals.

called an (m, n)-ary relational. also to denote relationals, N m,n is described assuming here.

in chapter

context.

(Thus

the (recursive)

to context.

] of [17]

al"'''

function

,~

an' whose

functions

, etc. to denote arbitrary

(m, n)-length

is determined by the

For any n, ~ x I,... , x n ~ denotes

(defined in []7]) which associates with each

n-tuple of numbers,~, a unique number,

contraction

The recursion theory of the spaces

<~i> E Nm,n in the above example.)

are all reeursive partial

is

and it is this theory which we are primarily

We use capital gothic letters, ~

contraction

m,n

We double-up on notation by using P, Q, R, S, T

according

strings of the form e l " ' ' ' ~m'

A subset of some N

functions•

(for varying n).

such that the inverse functions Seq is the (recursive) If e s

(X)o,...,

range of all the

~, we define ~ s ~

by

(X)n_ 1

-60-

~(n)

=~(O)

..... ~(n-l)~

.

string ~l(X) ..... L ( x ) , two notions

.)

m,n

for each N

the resulting

notions will

a functional

F: N

m,n

topology.

in B.

topology

initial

It follows

that F is continuous.

Then,

~r

Baire

result holds Theorem

space.)

last

for spaces

a topological

space

We obtain a topology factors.

All

It is not too hard to show that

m,n

for any ~ E w , F(~)

is determined

by some

of the Baire

suppose F is continuous. ~ B s)

We

= ~w) as an illustration.

from the definition

Define

topology

B c mw by

(F(y) =r)

otherwise,

s = ~s0,... , Sp~ ÷ B s = {f s ~w I f extends

t = ~x[B(~(x))

of recursiveness

over the various

+ I, if s ~ Seq & ( V y

B(s) = J~ O ,

the

these

in some B just in case it is continuous.

easily

Conversely,

[17]that

(This gives

for the case m = I, n = O (whence N

of ~.

from

~(x)

is defined by giving m the discrete

refer to these topologies.

Suppose F is recursive

where

on ~

It is called

÷ ~ is recursive

sketch the argument

segment

(It will be recalled

product

by taking the product

m,n

al ..... an, we denote by

role in even the definition

to the irrationals.

topological

= ~I ..... ~ m '

The Baire topology

and taking

homeomorphic

~

a I ..... an .

play an important

of the form N topology

If

> 0], we have F(~) = B(~(t)) if F is a mapping

from some N

<So, .... s>} P

.

Then if

~ i, so F is recursive

m,n

into w .

in ~.

(Such mappings

A similar

occur in

7 below.)

Analytical

and Projective

The analytical by quantification propositional

Relationals

relationals

over function

connectives.

in terms of B-recursiveness.

are those obtained variables

Similarly

and number variables

we define

A projective

from recursive

relationals

relational

relationals

and by the use of

analytical

is any relational

in B, B ~

w

~,

analytical

some ~. The above definitions

include,

(of numbers).

Thus

ary relational

x ~ A is analytical

Clearly, by means

the projective

a set is analytical

of projection

of course,

relations

and, in particular,

sets are just those sets A for which

the

sets (O, l)-

in some ~. iff it can be obtained

on function/number

coordinates,

from a recursive

relation

and by taking unions,

in

-61-

intersections, and complements.

(This is because projection corresponds to

existential quantification, complementation to negation, etc.) Since there are only m recursive relationals, there are only ~ analytical relationals and sets.

However, there are 2

projective relationals and sets, of

course. Recall that the arithmetical sets (i.e. the sets obtained by quantification on number variables of recursive relationals only, together with the formation of unions, intersections, and complements) are just those which can be defined in elementary arithmetic.

This is pretty obvious from the definition.

Similarly, the analytical

sets are just those definable in elementary analysis (i.e. where there are variables ranging over real numbers). below.)

Again, this is a pretty obvious fact.

(See Theorem i

These observations should explain our nomenclature somewhat.

In ~7] , a hierarchy is defined on the arithmetical relationals - the arithmetic hierarchy, due to Kleene.

We shall define an analogous hierarchy on the

analytical relationals - the analytical hierarchy, again due to Kleene.

(Actually,

this is also described in []7], but not discussed to the same extend as here.

Thus

we shall give the con~lete definition.) The Analytica!Hierarchy A formula ~ is said to be in prenex form if all the quantifiers in ~ occur at the beginning.

The matrix of a prenex formula is the quantifier free part of 4.

Let ~ be a formula in prenex form.

For each quantifier Q in 4, write Q0 if Q

binds a number variable and Q1 if it binds a function variable. sequence of superscripted quantifiers is called the prefix of 4"

The resulting The reduced prefix

of ~ is obtained from the prefix of ~ by deleting all quantifiers with superscript O. Two quantifiers are of the same type if their superscripts are equal. of the same kind if they are either both universal or both existential.

They are

A block of

quantifiers is a sequence of quantifiers of the same kind. A prefix is ~ blocks starting with

(resp. H~) if its reduction consists of n alternating (in kind) 31 (resp V ~ ).

(The superscript I in ~ ,

etc., refers to the

-62-

type of quantifiers under consideration.

The arithmetical hierarchy is the hierarchy

corresponding to the analytic hierarchy, defined below, but with superscripts 0 instead of i everywhere.) A relational is ~in

(reap. Hi) if it is equivalent to a formula in prenex form

whose prefix is ZIn (reap. H~n).

(Note that we identify the relational by the formula

which defines it, and that we have not bothered to formalise our language, etc.) A set is 11n (reap. E~n) if the relation x ~ A is lln (reap. ~).n A relational (or set) is A I if it is both ~In and H I. n n We often let InI (reap ~n' A~n) denote the collection of all such sets. We define ~In(B), EIn(B), A~n(B) analogously, allowing B to figure in formulas as a parameter (and thus replacing 'recursive' by 'B-recursive', in effect). A relational (or set) is E1 ~n

(reap. H:n) if it is ~n(B) (reap. ~n(B)) for some B.

A relational (or set) is Al if it is both ~n and H~ ~n" ~n ~

Clearly, a relational R

will be Al iff it is An(B) for some B, since we can recursively combine two functions _n to give one function (in a one-one manner). The following are proved in we indicate).

[iq] (with

the exception of Theorem

4, whose proof

Theorem 1 shows that we have defined hierarchies as the subtitle above

suggests. Theorem 1 (Kleene) Every analytical relation is ~n or N~n for some n, and conversely.

Every

projective relation is ~In or H~ for some n, and conversely. ~n Theorem 2 (Kleene) Every ~In (reap. ~In) relational is equivalent to a relational of the form t

3~I V~2"'" Qn~n Qn x R(~I"''' ~n' ~[ ' x) where R is recursive (reap. recursive in the I

reap. B) and where Qn is 3 if n is odd and ~ if n is even, and where Qn is of the opposite kind to Qn"

Dually for N~ (and ~).

Theorem 3 (Normal Form Theorem; Kleene) There is, for each m, n, a primitive recursive predicate Tm, n, such that when-

-63-

ever R ( ~ ,

t) is an (m, n)-ary relational, recursive in B, there are numbers e, e'

such that:

~tR(~i, t) ~-+ ~t Tm,n(e , ~ ( t ) , ~(t)) ~tR(~,

t) ~-+ ~ t ~ T m , n ( e ' , ~(t), ~(t)).

(This is a slight reformulation of the 'Normal Form Theorem' as stated in [ I ~ . ) Theorem 4 (Analytical Normal Form Theorem; Kleene) Every ~ ( B )

relational is equivalent to a relational of the form !

~el W~2"'" Qn~n Qn t R(71(t) .....~ m (t),

~(t), ~(t)), where R is primitive recursive.

Dually for N~(B). (This follows easily from Theorems 2 and 3.) Theorem 5 (Hierarchy Theorem) For every n > O, there is a

n relational which is not ~ . ~n

Dually for ~I and n

F~n" Theorem 5 tells us that our "hierarchy" of analytical/projective sets is not trivial. Characterisation of ~

Sets in Terms of Well-Orderings.

We obtain an important characterisation of H~11 relationals in terms of wellfoundedness.

First some definitions.

Define a partial order on Seq by s .<* t i f f

s =is0,...,

Sm~ and

t = ~ tO ,..., tn ~ and m < n and sO = to & ... & sm = tm . A set C ~ Seq is called a bar if

~

~t(~(t) e C).

If C is a set of sequence numbers, we set S(C) = {u e Seql ( ~ v e Seq)(v ~<* u -~ v

o)

We define a total order
C is a bar iff S(C) is well-ordered by
This is the

-64-

Proof: Suppose C is not a bar.

Then

~

~t(~(t) ~ C).

For such an e,

~(0) >KB ~(i) >EB ~(2) >KB .... so S(C) is not well-ordered by <EB" Conversely,

assume
a bar by constructing,

inductively,

We show that C is not

a function ~ such that for every t, ~(t) ~ C.

Let w0 >KB wl >KB w2 >KB" .. be an infinite descending chain in S(C). definition, i ~ ~0"

(w°) 0 ~ (wl) 0 ~ ...

Let ~(0) = (w~0) 0.

(w~n)n+l,

Hence for some number ~0' (wl)0 is constant for every

If e(n) = (w~n) n is defined, then again the sequence

(w~n+l)n+ I .... is eventually constant from some (w~n+l)n+ I onwards.

~(n + I) = (w~n+l)n+ I. (w~t-l)t_l~* Let ~ e

w ~t-l. ~.

For every t, ~(t) = ~(w~0)0,. Hence ~(t) ~ C.

Let

, (w~t-l) t i~ = ~(w~t-l) 0

QED.

We say ~ codes the binary relation R iff R = {I~(~u, v~) = O}.

For such ~, we let R

denote the binary relation coded by ~.

we denote it instead by $ .

We say e codes a well-order iff ~

(Likewise for a total order.)

In case R

is transitive,

is a well-order.

If ~ is a countable ordinal and the well-order coded

by ~ has type ~, we say ~ codes ~. c m2.

Then, by

Clearly, every countable ordinal is coded by some

If ~ codes a well-order, we let II~II be the order-type of ~ .

(Thus ~ codes

II~ II) Theorem 7 (Kleene-Brouwer) Let P be an (m, n)-ary relational.

Then P is 9 1 (B) iff for some F: Nm, n ~> 60

recursive in ~, P(~i) +-~ F(~I) codes a well-order. I

Proof: Suppose P is El(B).

By Theorem 4, there is a recursive R such that

P(~i) ~-+ V~ ~t R(~(t), ~(t), ~(t)). Then, clearly, P(~i) +-~ C ~

Let C~t = {~(t) IR(~(t), ~(t), ~(t))}.

is a bar ++ S(C~ ) is well-ordered by
Define

y ~ ~2 by ?(n) =

I O, if n = ~u, v~ & u, v E Seq & u, v ~ S ( C ~ )

& u <•B v.

L I, otherwise. Then, clearly, y = F(PI) for some mapping F: N

m,n

÷

00 ~ which is recursive in

B, and P(TJ~) +-+ F(~i) codes a well-order. For the converse, it suffices to show that the relation "~ codes a well-order"

-65I

is H I .

Now, " ~codes

descending chains".

a well-order" +-+ "~< The clause ".<

is a total order having no infinite

is a total order" here is arithmetic.

The second

clause can be written in the form ~8[~n(~(~B(n which is E Il .

+ I), B(n)~) = O) +

~m(B(m + I) = B(m))],

QED.

Corollary 8 The relation "a codes a well-order" Proof: If it were El' then every ~

is fll but not

set would be El' contrary to Theorem 5.

Let R, S be binary relations on ~. y ~

~ such that

~u~v[R(u,

R ~< S +-+ otp(R) .< otp(S)

QED.

We say R .< S iff there is a one-one map

v) +-+ S(y(u), ¥(v))].

Thus if R and S are well-orderings,

(as ordinals).

Define a binary relation ~Ii, since

I"

~

on

by

~(~, B) ~-+ 3 ¥ [ ~ u ~ v [ ~ ( ~ u ,

~(~, B) ~+ R

~< R~.

The relation ~

is

v~) = O ~-+ B(~y(u), y(v)>) = O .&. y(u) = y(v) ÷

u = v]. I

The next result gives a necessary and sufficient condition for a ~I relational to be, in fact, A I ~I" Theorem 9 I

Let P be any $I (m, n)-ary relational.

Let F be recursive in B such that i

P(~)

+-+ F(~i) codes a well-order.

Then P is ~i iff sup{IIF(~t)ll: P(21)} <~I.

Proof: Suppose sup{llF(~i)II: P(~i)} <~I" = II~011"

Then for some ~0 s m~' sup{llF(~i)ll: P(~i)}

By our remarks above, concerning the relation ~ , it suffices to show

that P(2I) +-> RF(~I ) ~< R~0. Now, if RF(~I ) .< R~0, then since ~0 codes a well-order, F(~i) codes a wellorder, so P(Z[).

thus IIF<~)II-<

Conversely,

II~oll,

Now suppose P is ~II.

if P(21), then F(g() codes a well-order,

i.e. RF(~L ) .< Reo. If

sup(llF(~)II:P<21)}

+-+ (~i) [P(Pl) ^ II~ll ~< IIF(~)II~ which is ~ corollary 8.

and

Hence sup{llF(~l)II: P(~i)} <~I"

= ~I' then ['e codes a well-order]

(by our remarks on ~ ), contrary to QED.

-66-

sup{ll~II:~

For any B E we, let 6~ =

is recursive in B}.

By a proof similar to

the above, we have: Theorem iO Let P be any (O, n)-ary H~(B) relational (i.e. an n-dry H~(B) relation), and let F be recrusive in B such that P(?I) ~-+ F ( ~ ) iff sup{[IF(~)II: p ( ~ ) } < ~ .

codes a well-order.

Then P is ~ ( B )

(Note that since there are no function variables in

~ , F(~i) is always B-recursive.) Topological Characterisation of A~ Sets. Let X be a topological space.

Recall that the Borel alsebra on X is the small-

est o-field of subsets of X containing the open sets.

A subset of X is Borel if it

belongs to the Borel algebra on X. Theorem II The Borel subsets of N

m,n

are just the A I (m, n)-ary relationals. I

Proof: We show first that the A 1 (m, n)-ary relationals constitute a o-field of subsets ~I of Nm, n containing all open sets, whence every Borel set is ~ . By the choice of our topology, every open subset of N certainly ~ .

And clearly, ~

Pi' i < m, are all &~.

is closed under complementation.

We show that P = Ui<mP i is A~.

so that Pi is A~(Yi ). recursive in y.

is recursive in some ~, and hence Finally, suppose

For each i < m, pick Yi

It is easy to find a function Y so that each Yi is

Hence each Pi is A~(Y).

By Theorem 2, for each i < e there are

!

(m+l, n+l)-ary relationals Ri, Ri, recursive in Y, such that P'1(~i) ~-+ 3~ ~t R i ( % ~i , t) T

+-~ ~

~t Ri( %

2i , t).

Thus by theorem 3, for each i < m there are ~(i), 4'(i) < e such that P'1(?/) ~-+ ~ ~-+ ~ Hence P(~l) ++

"~t~Tm+l,n (~(i)' ~(t), ~(t), y(t)) ~t Tm+ l,n(~ '(i), ~(t), ~(t), ~(t)).

~i ~

~t~Tm+l,n(~(i) , ~(t), ~(t), y(t))

+-~ ~i ~

3t Tm+l,n(~'(i), ~(t), ~(t), ~(t)). 1 i , By the first equivalence P is ~i(~, y) and by the second it is HI(~ , y). I P is ~I"

Hence

-67-

Conversely, suppose P is A~. F: Nm, n + ~

By theorem 9, this means that for some continuous

and soma countable ordinal ~, P(~)~-+ IIF(gL)II Z $.

the equivalence of F: N

÷ ~

m,n

being continuous and being recursive in some 8.)

I t = {~ e m~l~ codes a well-order and II~[I Z ~}. so then is P, as F is continuous. proof is by induction on ~. I 0 = {~ e ~ I ~ =

~ - On<

Then P = F-I"I~.

But if I t is Borel,

Hence, it suffices to show that I T is Borel.

The initial step is easy.

Let

The

We have:

codes a well-order & II~ll = O}

{al ~n(~a(@n,

=

(We have here used

n~)

{el~n,

=

O)}

n>) = 0},

which is Borel, since each set { ~ l a ~ n , need some notation.

n~) = O} is open.

For the successor step we

For ~ ~ m~, n g m, let e(n) be such that if e codes a well-order,

then so does ~(n), and 0, if ~ = ~u, v ~ &

u ~

v & n $~ n & v <

n

i, otherwise.

Then, ~ + l

= {~ltl~ll

=

~ ~+ l}

n<j lll Cn)ll

Now, the maps fn: ~ ÷ ( n ) is I~+ I = ~ n <

are recursive, & hence continuous, so if It is Borel, so

fn-l"{~II1~ll ~ ~} = ~ n <

fn-l"l~.

Finally, suppose ~ < ~I and lim(~).

= {~I ( V n ) ~

Then I

= {~iIl~II ~ ~}

< n)(ll~(n)ll ~ ~)} = ~ n < [jg< {~IIl~(n)IT E ~} = ~n<~[J$<~fn-l"l ~ , so

I~ is Borel if It is Borel for all ~ < ~.

The proof is eomplete.[

I

An Well Orderings of the Continuum. Theorem 12 Suppose there is a h I well-ordering < of ~ n (i)

of type ~I' some n > O.

Then :

< N (m2) 2 is a A * non-measurable subset of (~2) (with respect to the product n

measure on m2 of the usual measure on 2) (ii) There is a A I subset of the reals which is not Lebesgue measurable. n Proof:

(ii) clearly follows from (i), of course.

We prove (i).

-68-

Define

~(~,

B) =

$ O, if ~ < B

t

i, otherwise

Suppose < were measurable Fubini's

under the product measure,

~, on e2.

Then, by

theorem,

B})JdB

=fodg =0. Similarly,

~ ( > ) = 0 and ~ ( = ) = O.

Thus ~ ( ( e 2 ) 2 )

= O, w h i c h i s

absurd.

QED.

Theorem 12 gives one example of the effect upon the reals of the existence of a k*n well-ordering concerning

of ee of type e I.

We give one more example

(there are many others)

the perfect subsets of ee.

Recall that a subset Y of a topological closed, and contains no isolated points.

space X is per fest iff it is uncountable,

Clearly,

if Y ~ e

is perfect,

then IYI = e

Solovay has proved that it is consistent with ZF + "the axiom of m-dependent choices"

that every uncountable

one can construct an uncountable

subset of ee has a perfect subset. subset of e

In ZFC, however,

with no perfect subset.

Theorem 13 If there is a A n~ well-ordering I

n

< of ee of type % ,

then there is an uncountable

subset of ee which has no perfect subset.(n > 0).

Proof: Define ~ by ~ ¢ ~ +-+ ~ codes a well-order Thus ~ is R l n

& (~B)[B

< ~ + ~(II~II = II~II)J•

And ~ is clearly uncountable.

Now, any perfect subset of~" is closed, hence Borel, hence A~. the well-orders

By theorem 9,

coded by all the members of such a subset are bounded above by

a countable ordinal.

But by definition o f ~ ,

at most one member of~#.

any such well-order

is coded by

The result is proved.[

We shall prove below that V = L implies the existence of a A ~ well-ordering 2

of

-69-

~m of type

~.

Hence, in particular, there is an uncountable H

subset of

2

with no

1

perfect subset.

With a little more effort, this can be improved to a ~I set.

Absoluteness and the Analytic Hierarchy Theorem 14 Let P be an (m, n)-ary relational. (i)

If P is U-absolute for all transitive e-models of ZFC (including inner models)

(resp. containing B), then P is ~2I (resp. ~I2 (8)). (ii) If P is absolute for all transitive c-models of ZFC (resp. containing 8), then % P is A~ (resp. A2(8)). Proof: We prove the result without the 8, and for m = n = I only. entirely similar.

Let ~

The general case is

be the first-order predicate language with a binary

relation E (to denote membership) and constants ~ ) ~ <

.

The proof proceeds in

several stages. Stage I. R

Let B(e, ¢) - "~ is a sentence of

i n t e r p r e t s E".

<

which is true in <m, Re, ~ ) k < > when

l

Then B i s A1°

To see this, let y~(¢) = I O, if B(e, #) [ I, otherwise. Then B(e, ¢) *-+

3B[B = ye & 8(¢) = O]

VB[~ =

*~

y~ +

8(~)

= 0],

so it suffices to show that the relational 8 = Ye is arithmetical (in e, 8). NOW, 8 = Y~ ++

~¢[¢ is not a sentence ÷ 8(~) = lJ

~W[B(F=~) &

~[BO<

=o~

=~]

m ~) = O ~-+ ~ ( ~ k , ~ )

= O]

& ~[~

is a sentence ÷ ( 8 ( ~ )

= O ~-+ 8(e) = I)] & . . .

& ~[

-~v+ is a sentence ÷ ( 8 ( 3 v ~ )

= O ~-+ ( 3 ~ 8 ( ~ ( v / ~ ) ) = O)],

which is clearly arithmetical. Stage II.

There is a A I relational Ko(c¢, i, j) such that whenever <m,]{~> is well-

founded and extensional and ~ : < ~ , ~ > =~ with A transitive, then K0(~ , i, j) *-+ ~otCi) =

j.

-70-

Let T. be a term of set theory which represents j. 3 e(i) = j +-+ ~ v 0 = Tj[~e(i)],

Then:

as A is transitive

++ <~, Re> ~ v 0 = rj [i], by isomorphism ~-+ <~, R e , ([)k<~ > ~ ~ = T O which is A~ by stage I. Stage III.

There is a A~ relational Kl(e , i, y) such that whenever <~, Re>, ~ , A are

as above, then Kl(e , i, y) +~ he(i) = y. As above, ~e(i) = y ++ [<~, Re, (~)~< > ~ ~ ~ ~w] & ~j~,

Re, (~)~< > ~ <~ (j), Tj> ~ ~],

I which is A I again. Stage IV.

For each ~, let WF(e) ~ R e is a well-founded

relation on ~ × ~".

Then

I WF is ~l" This was proved in establishing We may now prove the theorem.

theorem 7. Suppose first that P(~, n) is U-absolute.

be a theorem of ZFC such that P is U-absolute

for all transitive models of o.

Let a Then:

P(y, n) *-+ ~AEA is transitive & IAI = ~ & ~ o & y, n s A & ~ P(y, n)]. So: P~,

n) +-+ 3e[WF(e)

& <~, R >

~ ~ & (~i,

(z (i) = ¥ & ~e(j) = n & <~, R , By the above results, this is [~.

j ~ ~) ([)k<e > ~ P(i, j)]

This proves

(i).

For (ii), note that if P is D-absolute for all transitive models of ZFC, then ~P is U-absolute for such models, whence, by (i), ~P is ~ . Constructibility

QED.

and the Continuum

Theorem 15 (i)

e n~

is ~ .

(ii)


(iii) If V = L, then
(i) e s L q ~

is A~.

is U-absolute for transitive models of ZFC, so the result

follows from Theorem 14.

-71-

(ii)

Similarly.

(iil) If V = L, then I ~2" QED.

~(~
-~(~
Corollary 16 If V = L, then there is a A l well-ordering of the continuum of type ~ .I 2 i We have seen that the Borel sets are just the &~ sets. We show below that these sets ar~ to a large extent, absolute for transitive models of ZFC, and in particular for L. Theorem 17 (Mostowski) All ~ l ( g )

and E (g) r e l a t i o n a l s 1

are a b s o l u t e f o r t r a n s i t i v e

models of ZFC

1

containing 8.

Hence all ~I relationals are absolute for 'function-absolute' models of

ZFC. !

Proof: For ~I(B) relationals, this follows from Theorem 7, since the notion of a well! ordering is absolute for transitive models of ZFC. The result for 11(8) relationals follows by t a k i n g n e g a t i o n s .

I

Theorem 18 (Shoenfield) All 12(B) and H2(B) relationals are absolute for transitive models of ZFC containing B and all countable ordinals. !

l

Proof: We prove the result for ~2(8) relationals, since H2(8) then follows by taking negations.

Let P ( ~ )

such that P ( ~ )

1 be 12(8).

~-+ ( 3 ~ ) ~ ( ~ , ~-+ ( ~ ) ( ~

By theorems 2 and 7, there is a 8-recursive F

~ ) codes a well-order]. < Wl)(~f

~ ~g)(Vm,

n)~(~,

~)

codes a linear-

order & F ( % ~i )({m, n~) = O ÷ f(m) E f(n)]. Since F is 8-recursive, this can be written in the form P(~i) ~+ ( ~

< ~l)(3~)(~f

E w~)(Vk)Q(~i,

~[k, f~k, ~), where (by theorem 17),

Q is absolute for all transitive models of ZFC containing B. The ~ in the above can always be taken to exceed w, of course, so we can amend Q slightly to obtain P ( ~ )

+-+ ( ~

< Wl)( ~f s w ~ ) ( ~ ) R ( ~ ,

f~k, k), where R

is still absolute for all transitive models of ZFC containing 8. function contraction.)

(This is just

-72-

Let Z~,~

P(~)

~

= {u[(~n

( ~

< w)(u ¢ ~n) & ( ~ k

< wl)[
e dom(u))R(~,

~ > is not well-founded].

ulk, ~)}.

Clearly,

But Z ~ , ~

is absolute

for all transitive models of ZFC containing B, ~ and ~ .

Hence P is absolute

for all transitive models of ZFC containing B and all countable ordinals.

Corollary

19 I

(i) (ii)

QED.

!

111 ~2 and ~2 s u b s e t s

of w are constructible.

1

w

I f L ~ww i s H2, t h e n

w ~ L and L c o n t a i n s

all

Borel sets.

1

(iii) Proof:

I f ~ i s a ~2 s t a t e m e n t (i) (ii)

Immediate,

(of analysis)

such t h a t ZFC + V = L ~ ~, t h e n ZFC ~ ~.

from t h e t h e o r e m . w 1 w i s g2 and t h a t ,

Suppose L q

in fact,

we ~ L.

Then t h e s t a t e m e n t

!

( ~ a ) (a ~ L n ww)iS ~ Z2 s t a t e m e n t

L, by t h e t h e o r e m . (3~

But t h i s

which is true.

is absurd,

~ L ) ( a { L q w w ) by a b s o l u t e n e s s .

Hence i t must be t r u e

since L ~ (~)(a The l a s t

part

in

4 L q ww) means

f o l l o w s by t h e o r e m 17,

of course.

(iii) are

This is really

1 o f t h e o r e m 17, and h o l d s b e c a u s e ~2 s t a t e m e n t s

a corollary

(by t h e o r e m 17) U - a b s o l u t e

f o r L. I

Remark Theorem 18 i s t h e b e s t p o s s i b l e Con(ZFC + " t h e r e

here,

1

i s a A3 n o n - c o n s t r u c t i b l e

prove that

this

result

cardinal.

Solovay's

is result

(realIy)

true

for Solovay has proved that set of numbers").

In section

i f we assume t h e e x i s t e n c e

appeared in [22].

Con(ZFC) ÷ 5 , we s h a l l

of a very large

Chapter 6

THE JENSEN HIERARCHY OF CONSTRUCTIBLE SETS

The deeper results concerning the constructible universe require a study of the levels in the constructible hierarchy, and for this it is convenient to modify the definition of the hierarchy somewhat.

The study of the "fine structure of the

constructible hierarchy" was begun by Ronald Jensen, who first of all investigated the fine structure of the usual constructible hierarchy, which we defined in section i.

However, although this study could be carried out, it was unnecessarily

complicated, because the usual constructible hierarchy has some deficiencies from this point of view.

He was thus led to define a more convenient hierarchy, which

we shall describe below. Recall that we previously defined the constructible hierarchy by inductively defining sets L~, ~ E On, by setting L 0 = ~ , set of all x ~

L% = ~ <%L , if lim(~), and L +i= the

L~ such that for some E-formula ~ and some al, ..., an E L~,

x = {z ~ L~ I L~I= ~[z . .al, . . as the class L = ~

L o

, a~}.

We then defined the constructible universe

Now, the important facts concerning this definition which

we used when studying L in section I are, firstly, that the construction is (in a strong way) ~l-definable, and thus has certain absoluteness properties, and, secondly, that L + 1 contains all and only those subsets of L

which are L -definable

(and which, therefore, must be in L if L is to be a model of ZFC).

But if, indeed,

these are the only conditions which we require (and loosely speaking they are), then it is clear that our above definition is unnecessarily restrictive.

For instance,

there are many simply definable functions on sets under which L must be closed, but which increase rank - and these functions will lead out of the sets L . instance, unless lim(~), L

For

will not be closed under the formation of ordered pairs.

Since this function plays a central role in even the most elementary parts of set theory, we see that this defect becomes quite important (though not unavoidable) when we try to study the fine structure of L rather than L itself.

So, following

Jensen, we define a new hierarchy of "constructible sets", which is sufficiently

-74-

like the L-hierarchy to preserve the two properties mentioned above, but which has the extra property that each level in the hierarchy is closed under ordered pairs, etc.

More precisely, using the rudimentary functions defined in chapter 2, we

define a hierarchy <J

I e E On> (the Jensen hierarchy) such that each J

under the rudimentary functions, L = U e s O n J for this hierarchy.

In most cases, J

extension of L , and in fact, if
e

is closed

, and these two properties above hold

will be a "constructibly inessential" I ~ E On> denotes the familiar rank-hierarchy,

the precise relationship between the J- and the L-hierarchies is easily seen to be J0 = L0 = ~ and Lw+~ = V~+e

N Jl+~ for all ~.

Hence we have Je = L~ iff we = e.

In this chapter the Jensen hierarchy is defined and its elementary properties discussed°

In chapter 7 we shall investigate the fine structure of the Jensen

hierarchy.

A corresponding theory may also be developed for the L-hierarchy, the

only difference being that some akward complications arise becuase of the above mentioned defects in this definition. On Rudimentary Functions In chapter 2, we defined the concept of a rudimentary function, and proved some simple facts about general rudimentary functions. In order to define and discuss the Jensen hierarchy, we must carry this study further. Firstly, we observe that although rud functions increase rank, they do so only by a finite amount.

More precisely:

Lemma 1 (Finite Rank Property) Suppose f: V n ÷ V is rud. Then there is a pc ~ such that for all Xl, ..., Xn, rank (f(xl, ..o, Xn)) < max (rank (xl) , ..., rank (Xn)) + p. Proof: By induction on a rud definition of f. I Note that by virtue of lemma I, the rud functions are not the same as the ~0 ZF functions, although rud relations ar___~ejust the ~0 ZF relations (lemma 2.4). For example, the function f(e) = ~ + ~ is infinite amount it cannot be rud.

~ Z F but, since it increases rank by an 0

-75-

Next we show that relatively

rudimentary

functions

can be simply expressed

in

terms of rud functions. Lemma 2 If f is rud in A

-= V, there are rud functions

expressible

(in a u n i f o r m way w i t h respect

composition

of gl' °''' gn and the function h(x)

Proof: By induction on the

gl' "''' gn such that f is

to the rud definition

(rud) definition

= A

n

of f) as a

Xo

of f. I

Recall that a set X is said to be rud closed if for all rud functions f, f"X n

c X.

A structure M = <M, A> is said to be rud closed if for all functions are rud in A, f"M n

f which

-~ M.

We say a structure M = <M, A> is amenable

if u s M ÷ A

~ u E M.

Lemma 3 A structure M = <M, A> is rud closed iff the set M is rud closed and M is amenable. Proof: By lemma 2° I Lemmm 4 Let A transitive, Proof:

~_ Vo

If f is rud in A, then fIM n is uniformly

rud closed M = <M, A

n

[0(<M, A

~ M>) for all

M>.

By len~nas 2.4 and 2. I The next result will be of considerable

use to us later on.

Lemma 5 Every rud function is a composition

of some of the following rud functions:

F0(x, y) = {x, y} Fl(X, y) = x - y F2(x , y) = x × y F3(x, y) = {

I z ~ x

^ ~ y}

-76-

F#(x,

y) = { I z s x

^ ~ y}

F s(x, y) = Ux F 6 (x, y) = dora(x) F7(x,

y)=

s

t',

x2

Fs(x, y) = (x"{z} I z ~ y}o Proof:

Let

~ denote

composition.

the class of all functions

t#(u)

¢(Xl,

- x. E x., z j

Write Fx(Y)

Xn>

I x I, "-., x n ~ u

~ l:¢LXl . . . . .

on ¢, we show that for all ¢, t~ s ~ .

then be proved using ¢(x)

"'

F

by

8

..o, Xn) , set

: {<x I . . . . .

By induction

(a)

from F 0, ..

We must show that f rud ÷ f ~ ~ o

For each g-formula

will

obtainable

(The required

Xn] }. result

this fact.)

I ~< i < j <~ n.

for F3(x , y)°

Then t¢(u)

= u

i-I x F~-i(F#(s

n

u 2, un-J))

so t¢

(b)

Let ~l(X) . . . . .

Cp(~)

be such that t¢ . . . . t s ~ . 1 ' CP

Let ~(~) be any propositional Since x - y, x

u y (=

combination

O{x,

y}), x

of ¢i'

"''' Cp"

~ y (= x -(x-y))

e ~ , we clearly have

t¢ e ( . (e)

Let 7(Y, ~) be such that t~ s ~ . Clearly,

either case, (d)

~(x)

t 3y~(U)

= dom(t~(u))

and t Vy~(U)

Hence (e)

-

-@y~(y, ~) or

Vy~(y,

= u n - dom(u n - t~(u)).

~). So in

t¢ s ~ .

E x. = X.o z j

Let 0(y, ~) - y s x. +-+ y e x.. i j ~ ~[~j

Let ~(~)

iff ( V y

0

e ~ .

But look,

E

t (u) = u n n

¢(~) - x i s xi,

By (a), (b), t

t Vy~(U

u (0u)),

so t

s ~ , by (c).

I ~ j < i ~< n.

Let ~(y, z, ~X) - y g z

,\ y = x i

A

z = x°. J

By (a),

(b),

(d), t~ s e •

But

-77-

0(~) +-+

~ Y ~ z~(y,

Hence,

z, ~),

so by (c), t 0 ~ ~ .

for any s-formula

If f : V n + V, define prove by induction implies (a)

0, t 0 e ~

o

f*: V ÷ V by f*(u)

on the rud definition

the required

= f"u m.

Using our above result,

we

This easily

of f, that f rud ÷ f* ~ ~ .

result.

f(x) = x.. i

f*(u) (b)

= f"u n = u = u - (u-u) e ~ .

f(x) = x. - x.. I J f*(u)

= f"u n = {x-y

I x, y ~ u}.

Let F(u) = t0(u o ( 0 u ) ) Then f*(u) (c)

= F8(F(u),

f~ ( 0 u

Let 0(z, x, y) ~ z e x - y. x u 2) = {

u 2) e ~ , since

I x, y E u

^

z e x-y}.

t0 e ~ .

f(x) = {x i, xj}. f*(u)

= f"u n = {{x, y}

I x, y e u} = ~ u 2 ~ ~ . -+

(d)

f(~) = h(gl(~) . . . . .

gk(x)).

k

k

Let G(u) = i=~l g~(u), H(u) K(u)

= un

U

G(u)

be an e-formula

O H(u).

By hypothesis,

equivalent

to the formula

~z I o.. Clearly, (e)

f*(u)

= h*(i~ 1 g*(u))

~ Z k ( Z 1 = gl(~)^

= FS(([t0(K(u)) ] ~

f(y, x) = ~ z e y g ( Z ,

= h*(G(u)),

G, H, K ~ ~ .

... a zk = gk(1)

[H(u)

x un]),

and

By lemma 2.4, let 0(Y, ~)

A

Y = h(z I . . . . .

Zk)).

u n) e ~ .

~). -+

Let G(u) = { [ ( ~ v e y) [z c g(v, ~)] But f*(u)

As above G e ~ •

= F8(G(u) , u n+l) a ~ .

Hence

f rud ÷ f* e ~ , for all f.

Finally, _

let f be rud.

-+

Set f() above, ~* ~ ~ . f(x) =

a ~ e u}.

We show that f e ~ .

-+

= f(z), ~(y) Let P(~)

U O{{f(~)}}

=

= ~ in all other

= {<x>}.

O UfY{<x>}}

Thus P E d . =

cases.

Thus ~ is rud.

But look,

L)OFs(T*(P(~))

, P(~))

E ~.

I

So by the

-78-

As an immediate corollary of lenmms 2 and 5 we have: Lemma 6 Let A

~ V and define F 9 by Fg(X , y) = A

n x.

Every function rud in A may be

expressed as a composition of some of the (rud in A) functions F0, ..., F 9. We shall make immediate use of lermna 6 in investigating the logical complexity of the predicates

~ ~n for suitable M. M fixed arithmetisation of our language.

We assume, once and for all, that we have a

Le~mna 7 ~0 M

M is uniformly ~ for transitive, rud closed M = <M, A>. ~

Proof: Let

~*

be the language consisting of:

(i)

variables wi, i s ~.

(ii)

function symbols (binary) f0' "''' fg"

We shall assume we have fixed arithmetisation of the reader understands what is meant by a "term" of

~ *.

~ *.

We also assume that

Henceforth, let

M = <M, A> be arbitrary, transitive, and rud closed. We first define precisely how

~*

is to be interpreted in M.

Let Q be the set of functions p mapping a finite subset of {w. I i ~ ~} into i

M.

We may clearly assume Q is rud.

T of Vbl ~ ,

Let C be the (rud) function which to each term

~ * assigns the set of all component terms of T, including variables. be the rud predicate defining the set {wili s ~}.

P(u, g, v) +-~ [dom(g) = ~

A (Vx

Let

Let P be the predicate

c u)~[x ~ Vbl ÷ x c dora(v) A

g(x) = v(x)] ^

9 i~0( Vt0, t I ~ u)Ix = fi(t0, t l) ÷ g(x) = Fi(g(t0) , g(tl))]J. Thus P is rud in A. We may now define the interpretation of a term T of

* at a "point" p s Q

by: Y = IITII~ P ]

~

"r is an

~*-term"

Hence the function f(T, p) =

^

O g C

A ~g[P(C(r), g, @)

11"cll M[p], if r is an ~ , otherwise

~*-term

~\ g(T) = y] . and p g Q

-79-

iS(uniformly) ~

(for transitive, rud closed M).

Since M is rud closed we can use the above result to define I= ~0 as an MM predicate. Let

s Fml~0o

By lemma 2.4,

I1 ~ II ~

is rud in A.

Hence the function F

defined by

O, otherwise is rud in A.

So, by lemma 6, we may assume F = IIT II M, where T is a term of ~ *,

under the above interpretation (i.e. with F. interpreting f. for each i). I

l

In fact,

we may clearly pick a recursive function o mapping Fml~0 into the terms of ~ * so that whenever ~ s Fml~0, II # II~[~] ~-+ II~(~)II~[ ~] = I. this implies that ~ 0 As

an

is (uniformly) ~

But by our above result,

(for transitive, rud closed M). I

i~nediate consequence of this result, we have

Lermna 8 Let n >~ I.

Then ~M n is uniformly -~IYMfor transitive, rud closed ~M = <M, A>.

We conclude this section with a few miscellaneous results of use later.

The

first two are technical, and will often be used without mention. Lemma 9 Let M = <M, A> be rud closed. P such that R(x) +~

If R ~ M is In(M) ,~

there is a ~0(M)~ relation

-IxI ~ X 2 ~ X 3 ~., QnXnP(X, x I .... , Xn).

Proof: Suppose R(x) +-+ ~M ~ a ~0-formula.

~ ~ ~÷ v o.o Qn Vn ~(v, v I . . . . . -$n)Ix], where ~ is 1 2 3 m Using the rud functions <-, ..., ->, (-)O' o - ~ ' ( -)m-l' we

can easily obtain, via lermna 2.3, a ~0-formula ~ such that R(X) ++ ~M

~vl~v2°°" Qn Vn @(v, v I . . . . .

vn)~X].

Then

R(x) +-+ ~x I ~ x 2 ..o Qn Xn[~M ~[x, x I ..... Xn]], as required. Lemma I0 Let M = <M, A> be rud closed. P

M such that R is yM({p}). ~n

If R

G M is In(M), there is a single element

-80-

Proof:

Z

If R is

({Pl . . . . .

p }), then R is also

n

I

..., p >}).

n

n

The following result is a slight variation of lemma 2.10 and is just as important for us as was that result. Lemma II Let M~ be transitive

and rud closed.

the axiom of extensionality W is transitive° Proof:

and is rud closed.

Let f : M ÷ M be rud in A.

Since M is transitive,

M satisfies

M, so does <X, A

X~l

~

closed.

Let X ~ E l

Hence,

~ X>.

in particular,

M.

Then <X, A

Let ~ : <X, A

~

X> satisfies

N X> ~ <W, B>, where

Then for all z c X, ~(f(z))

the axiom of extensionality.

Similarly,

by lemma 4, <X, A

= f(~(~)).

Hence as

~ X> is rud

z ~ X + f(z) a X for f : M ÷ M rud in A.

induction on the (rud in A) definition of f, ~(f(z))

= f(~(z))

By

for each ~ s X.I

The Jensen Hierarchy Let X be a set.

The rudimentary

closure of X is the smallest set Y

~ X such

that Y is rud closed. Lemma 12 If U is transitive,

so is its rud closure.

Proof: Let W be the rud closure of U. composition, W = {f(x)

Since rud functions are closed under

we clearly have

I x ~ U

a f is rud}.

An easy induction on the rud definition of

any rud f shows that x ~ U ÷ TC (f(~)) For U transitive, importance

let rud(U)

~

W.

Hence W is transitive. I

= the rud closure of U

~ {U}.

Of crucial

is:

Lemma 13 Let U be transitive. Proof:

Clearly,

~(U)

@(U) Q [o(U Let X ~

~(U)

X s rud(U)

~

Then ~0(U

0 {U}) = N ~0 (U o

~(U)

~ rud(U)

= Def(U).

~ {U}) = Def(U),so

@(U) {U~.

it suffices

to show that

~ rud(U). Then, exactly as in the proof of lemma 2.4,

(by induction on the ~0 definition of X).

Now let

-81-

X c

~U)

n rud(U).

Then X is a

~0(rud(U)) subset of U.

may in fact assume that X is ~0rud(U)(u

u {U}).

By lemma 2.3, we

But X G U ~{U}

m rud(U)

and U o {U}, rud(U) are transitive, so X is actually ~0 U ~ {U}(u d {U}) =

X0(u ~ {u}). I Also very relevant is: Lemma 14 There is a rud function S such that whenever U is transitive, S(U) is transitive, U U {U} ~

S(U) and Une w sn(u) = rud(U).

Proof: Set S(U) = (U u {U})

u

( $ Fi"(U d {U}) ). i=0

The result follows by lemma 5.

Lemma 15 There is a rud function ~9 such that whenever r is a well-ordering of u, W~(r, u) is an end-extension of r which well-orders ~(u). Proof: Define iu, j?, j~ by:iU(x) = the least i Z 8 such that ( ~Xl, x 2 ~ u)[Fi(Xl, x 2) = x] j~(x) = the r-least x I e u such that ( ~ x 2 ~ u)[FiU(x ) (Xl, x 2) = x] .u

.U

]2(x) = the r-least x 2 e u such that FiU(x ) (Jl(X), x2) = x. -U

Clearly, i

,

.U Jl'

.U

J2 are rud functions of u, x, r.

Define W~(r, u) = {<x, y> I x, y e u

a xry} u

{<x, y> I X E u

A y ~ u} d

{<x, y> I x ~ u

a y ~ U

[iu(x) = iU(y)

j~(y)

^~[iU(x) < iU(y)] v

,U

~ ~Jl(x)r Jy(Y)

.u A J2(x)r j2(y))]]]}.

,U

V (Jl(x) =

I

The Jensen hierarchy, <J C~ I ~ ~ On>, is defined as follows: 30= J~+1 = rud(J~) J~ =

U <~ J , if lim(~)o

-82-

Len~na 16 (i)

Each J

is transitive.

(ii)

~ ~ B + J

(iii)

rank( J ) = On

Proof:

(i) By lemma 12.

JB ~ J

= ~.

(ii) Immediate. (iii) By induction: rank(J +i) = rank(rud(J )) = rank(J ) + ~,(By the finite rank property of rud functions, this last step is easily verified.) To facilitate our handling of the hierarchy, we "stratify" defining an auxiliary hierarchy <S

the J 's by

[ a ~ On> as follows:

s + I = s(s) SX =

Ua<~g

, if lim(%).

Clearly, the J ' s

are just the limit points of this sequence,

Lemma 17 (i)

Each S

(ii)

~ ~ B + S

(ill) J Proof:

is transitive. ~

= ~M<~

SB S

= S

(i)

By lemma 14.

(ii)

Immediate.

(iii)

By induction: J +i = r u d ( J Onem ~ n ( s ~ ) = O n ~

) = ~n~

S~e+n = S ~ + e =

~n(Ja) = Sm(e+l)" [

Lemma 18 <Sv I ~ < ~ >

is uniformly ~

Proof: Set ~(f) ~ "f is a function" (~

for all e. ^ dora(f) s On

~ dom(f))[(succ(v)

[lim(~) ÷ f(v) = ~ e v

A f(O) = ~

+ f(v) = S(f(v-l))) a f(a)]]"

A

In fact:

-83-

~J

Clearly, y = S

*-+

# is uniformly IO e.

~f(#(f)

the existential

^ y = f(~)).

And by definition, Thus it suffices

quantifier here can be restricted

show that whenever y < me, then <S ~.

For ~ = 0 it is trivial.

In other words, we must

This is proved by induction on

For limit ~ the induction step is i ~ e d i a t e . I ~ < T> e JB"

I ~ < ~>

~

I v < ~8> e Je.

Lemma 19 I ~ < ~> is uniformly ~

Proof: By an easy induction, J

= S

<~

for all a. I ~ < ~> is uniformly [~e for all ~.

, the result follows by lemma 18.

Lemma 20 There are well-orderings (i)

~

< ~ 1

(ii)

<

(iii) < <

of the S

such that:

+ <

_~ < ; I 2 is an end-extension 2

v+l

<

~

of < ; J I ~ < ~c> is uniformly %1 ~ for all ~.

Proof: We use lemma 15. <~+I
Set < = WO

0

= ~, and by induction:

(
~vel <~ ' if lim(%).

(i) and (ii) are im~aediate and (iii) is proved like lermna 18. Lemma 21 There we well-orderings

(i) (ii)

<j~ of the J

eI < ~

÷ <j ~ <j ; 2 el ~2 is an end-extension

<j

of <

~+i

(iii)

<<JB

(iv)

<j

; J~

I ~ < e> is uniformlyj [ ~ is uniformly lie;

;

Thus

{<~m(JB), ~B + m> I m < n} e J~,

is rud c l o s e d . ) . ~

<J~

So

Then, by our above remarks,

So by lenmm 13, < S

I ~ < ~B> is ~ .

for all n < ~, <S v I ~ < ~B + n> = < S as J

to J .

I ~ < Y> ~ J •

assume ~ = B + I and that T < mB ÷ <S it is clear that <S~

to show that for any e, ~ < me,

such that:

Since

-84-

(v)

the function pr (x) = {z [ z < j

J x} is uniformly Zl a.

("pr" stands for "predecessors" of course.) Proof: Set <j

= <

that x <

(i)-(iii) are immediate by ler~na 20. y +-+ ~ ( x

J~

<

v

y = pr (x) *-> ~v[x ~ S

y).

For (iv)

note sin~ply

Finally, for (v), note that

~ y = {z I z <

x}] (and that <

s J ), and use

lemma 20. I At this point, we could prove that the condensation lemma holds for the Jensen hierarchy, by an argument as in chapter 2, using p.r. function theory.

However,

there is a more direct proof which we can give (although it should be noticed that in essence the two proofs are really quite similar, and work for the same reasons). The proof we give is Jensen's. Theorem 22 (Condensation Lemma) Let X -
then there is a unique ~ and a unique B ~ e such that

(i)

g : X ~ JB;

(ii)

~(~) ~ ~ for all ~ ~ ~ q X ;

(iii) ~(x) Ej x for all x c X ; (iv)

if Y

~ X is transitive, then ~ Y

Proof: Let X ~ I L

J .

=~.

Then by lemma II, let v : X ~ W, where W is transitive.

prove b y i n d u c t i o n

on a that

W = JB f o r

~ = ~"(X

Assume, therefore, that whenever ~ < a and X~ ~ ~

We

~ a).

J , the unique isomorphism

1 of X~ onto a transitive set W~ yields W~ = J~,,(X~ q ~). J Note that, as < J Claim I: For all ~ c X

] ~ < ~> is ~ q ~, ~(J ) = J (~).

To see this, note first that for ~ c X A E ~i (X f~ J ). X -<El J , A

Since J

n X # ~.

hypothesis, ~' : X

~ X, A g ~ ~(X).

But A ~ J , so A

N J~ ~ J~,"(X n J

E-end extension of J , so ~ maps X N J

n ~, X

q J~ -<~1

J~"

For, let

So, if A ~ ~, then as

N (X n J ) # ~ . ]

Hence by induction

N ~) for some unique ~'.

But look, J

is an

isomorphically onto a transitive set also.

-85-

In other words, ~' = ~IX ~"(X ~

~ J , and ~" X

~ J~ = Jv,,(X ~ ~).

J ) = J~"(X ~ v) = J~(~)' by the definition of v, as claimed.

For ~ < e, define rudx(J ~) = the rud closure of X Claim 2: X = ~ s X

For the converse, let x ~ X.

function f, ~j ( 3 ~ ) ( 3 p (~

~ (J

~ {J~}).

~e rudx(Jv)"

To establish this claim, note that as X ~ ~I obvious.

So, ~(J ) =

e X N ~)( ~p e J

Then x s J

e Jv)(x = f(p, J )): ~ X)(x = f(p, J )).

J , X is rud closed, so = ~<~

~ is

rud(J ), so for some rud

But X ~ ~I J~' so

In other words, x E ~ s X

~

rudx(J).

Hence claim 2. Claim 3: For ~ E X ~ ,

~" rudx(J v) = rud(J (v)).

To see this, let ~ ~ X ~ ~o rud function f and some p s J ~(x) = f(~(p), J (~)). This proves

~

.

Suppose first that x g rudx(J ).

Q X, x = f(p, J ).

But p e J

Then for some

By lemma II and claim I,

~ X so ~(p) ~ J (~).

Conversely, suppose y ~ rud(J (~)).

Hence ~(x) ~ rud(J (~)). Then y s rud(~(J )), by

claim I, so for some rud function f and some p s ~(J ), y = f(p, ~(J )). #(J ) = ~"(Ju ~ X), so for some q ~ J ~(J )) = #(f(q, J )) ~ ~"rudx(J )o

~ X, p = ~(q) and we have y = f(~(q),

Hence

~ , and claim 3 is proved.

By claims 2 and 3, we have W = ~"X = ~"( ~ u s X ~ ~ rudx(J~)) = ~ " r u d x ( J ~) = = ~"(x

~ ~).

Now,

rud(J (~)) =

rud ( J )

= JB' where

Chapter 7 THE FINE STRUCTURE OF THE JENSEN HIERARCHY As mentioned in chapter 6, a theory similar to the one following can be developed for the usual L-hierarchy, if desired. Central in our discussion will be the concept of a "uniformising function" for a relation, which is a sort of "choice function" for a given relation. Specifically, a function r is said to uniformise a relation R iff dom(r) = dom(R) and for all ~, ~yR(y,x)~+R(r(x),x). ~ ~ Let M = <M,A>

n ~ I.

We say M is Z -uniformisable iff every In(M) relation

on M is uniformised by a Z (M) function. n ~

A few moments reflection will reveal that

Z -uniformisability is a very strong condition to demand of an arbitrary structure n M, since in the more obvious cases, the definition of a uniformising function for a given relation would appear to increase the logical complexity by one or more quantifier switches. Z -uniformisable. n

However, it will turn out that for all ~, all n k I, J

For n = i, this will be easy to prove, but for n > I,

indirect approach will be necessary.

is a more

We shall outline the approach required after

we dispose of some of the more easy results.

First, Zl-uniformisability.

Zl(J ) well-ordering of each J~ gives us this with little effort.

The

In fact, we have

a much stronger result, of importance in applications of Zl-uniformisability. Let F be a class of structures M = <M,A>, n ~ I.

Say F is unifor!n~ly Z n-

uniformisable if, whenever ~ is a parameter free Zn-formula such that

II~ II~is

a relation on M for each M e F, there is a parameter free Zn-fOrmula ~ such that for each M ~ F,

II~ [I~ is a function uniformising

II~ II~"

Theorem I <J ,A> is gl-uniformisable.

In fact, the class of all <J~,A> is uniformly

Zl-uniformisable.

<J ,A> Proof : Let ~ be a El-formula w.p.f. J tion on J .

By c o n t r a c t i o n

such that

of quantifiers,

II~(y,~)ll

is a z I rela-

we c a n , i n a u n i f o r m w a y , f i n d

-87-

a E 0 formula ~ (w.p.f. J ) such that ~<j ,A>~(y~)'e-~z.~(z,y,~). g(~) = the <j-least w such that ~<j ,A>~((w)0,(w)l,~ ).

Define g by:

Then g is (uniformly)

E I ( < J ,A>), since w = g(~)4-~ ~<j ,A>~((%~)0 ,(w) l,~) ^ Set r(~) -- (g(~))l" II ~ (Y,~) If <Ja'A>. Remark.

~t[t = pr (w) A (~w' e t) -~ $((w')0,(W')l,~)]

Then r is (uniformly)

El(<J ,A>) , and clearly uniformises

I

We call the above construction the canonical Sl-uniformisation

Observe that if R(y,~) is a ll(<Ja,A>) predicate,

procedure.

then the canonical gl-uniformisation

of R is a function whose l I (<J ,A>) definition involves only those parameters which occur in the definition of R. Let us take a little time off to examine the above construction more closely, Suppose R(y,~) is a given E I relation, say R(y,~) ~-* 3zP(z,y,~), where P is S 0. obtain the gl uniformisation of R, we first obtain a gl uniformisation relation {<w,~>Ie((w)0,(W)l,~)},

of the E 0

and then simply pick out the requisite component of

the result as our required function. result is also gl"

To

And since <j is a E I well order of J

the

However, returning now to the notation of theorem I, we see that,

if we try to extend this procedure to the case n > i, we cannot conclude that the function g is En, the problem being the last conjunct in the explicit definition of g.

Let ~{(w,~) denote the predicate

II -n~((w)0,(W)l,~)II <J ,A>.

was no problem, since, if ~(w,x) is E0, so is (~w e t)~(w,x). -~ ~(w,~) is Sn_ I, and we cannot automatically

For n = I, there However , for n > I,

conclude that (~w e t)P(w,~) is !:n-l;

though it is certainly En, of course, so that the resulting uniformisation original gn relation will turn out to be In+l" uniformisabity of all J

of the

Thus, in order to establish the I n

for n > I, we must adopt a more subtle approach.

Closely connected with En-Uniformisability

is the notion of a "gn skolem

function". Let M = <M,A> be transitive and rud closed. ~

By a E - - n

skolem function for M we ~

. . . . . . . . . .

mean a gn(M) function h with dom(h) ~ ~ × M, such that for some p e M, h is sM({p}), ~

and whenever P e EM({x,p}) for some x e M, then 3yP(y) + (~i e 0~)P(h(i,x)).

(With

-88-

h,p as above, we say that p is a good parameter

for h.)

functions need not be (and in general~re not) total~ skolem functions appear. Lena

Note that E

n

skolem

As far as existence of E

n

is concerned, we can get away with slightly less than might first

In fact:

2 Let M = <M,A> be transitive

dora(h) ~-- ~ x M.

and rud closed.

Let h be a EM({p})

Suppose that whenever P ~ ~M({x})

~yP(y) * (~i c m)P(h(i,x)).

function with

for some x c M, then

Then M has a E skolem function. ~ n

Proof: Set h(i,x) = h(i,<x,p>).

It is easily seen that h is a E

n

skolem function

for M. Note that in the above, if h is actually EMn' then h = h.

This is used in establish-

ing the following result: Lemma 3 If <J ,A> is amenable, E l skolem function h

A

then it has a E l skolem function,

for <J ,A> which is uniformly

in fact, there is a

El <J~'A> for all amenable

<J ,A>. Proof: Let <~ili < ~> be a recursive enumeration Let <J ,A> be amenable. be the canonical

~ By lemma 6.8, ~<j

El-uniformisation

of Fml~ I.

. . ,A> is (unlformly)

of the ~i

El

<J

~'

A>

.

_ Let h - h ,A

<j ,A> El ~ relation {l~ <j

o o A>#i(Y,X)}.

(By Theorem 1 and the ensuing remark, h is thus uniformly El <J~'A> for amenable <J ,A>.).

By the remark following

lemma 2, it is clear that h is a E1 skolem

function for <J ,A>. | We refer to h ,A

as the canonical E l skolem function for (amenable)

<J ,A>.

(We shall have occasion to examine the above definition more closely in Chapter 13.) By a similar argument, we have: Lemma 4 If <J ,A> is amenable

and I -uniformisable, n

it has a E

n

skolem function.

The following lemmas indicate our reason for using the word "skolem" here.

-89-

Lemma 5 Let M be transitive Then whenever

and rud closed,

x e M, x s h"(m

and let h be a E

x {x}) -~E

skolem

n

function

for M.

M. n

Proof:

Set X = h"(m × {x}). that P ~ X # ~.

Clearly,

x s X.

Let p be a good parameter

P c E~({y I .... , ym}).

By definition

Yl = h(Jl'X) .... 'Ym = h(Jm'X)" P e E~({p,x}).

Let P e E~(X), P # ~. n

Hence,

We must

show

for h, and pick yl,...,y m ~ X with

of X, there are Jl ..... Jm E ~ such that

Since h is ~ ( { p } ) ,

P # ~ ÷ ~yP(y) ÷

it follows

(~i g w)P(h(i,x))

that

~ (~y g X)P(y). I

Lemma 6 Let M be transitive

and rud closed,

If X ~ M is closed under ordered pairs,

and let h be a E

skolem function

n

then X ~ h"(~ × X ) ~ E

for M.

M. n

Proof:

Set Y = h"(m x X).

By lemma 5, X ~ Y. Let P e E~(Y) -

that P N Y + @. Pe

Let p be a good parameter

Yl = h(Jl'Xl) ' .... Ym = h(Jm'Xm)" as h is g~({p}),

(~i e ~)P(h(i,x)) Cgrollary

P # ~.

We must show

'

for h, and pick yl,...,ym

e Y with

" e m and x I ..... x m ~ X such that Pick Jl ..... 3m

g ({YI' .... Ym })"

But clearly,

n

Let x = <x I ,. ..,Xm>.By

P is then g~({p,x}),

÷ (Hy e Y)P(y).

assumption,

so P + ~ + ~yP(y)

x g X. ÷

|

7

Let M, h be as above. ordered pairs.

Let X ~ M and suppose h"(~

Then X ~ h " ( ~

x X) ~ E

x X) is closed under

M. n

Proof:

Let Y = H"(~ × X).

Clearly,

Y = h"(~ x y), so the result

follows by the

lemma. I Lemma

8 (G~del)

There ~-i ~ a

is a bijection ~ : On 2 +-+ On such that ~(~,B) J is uniformly E l a for all ~.

Proof: Define

a well-order

>- ~,B for all ~,B, and

<* of On2by

(~,B) <* (y,~) +-+ [max(~,B)

< max(T,~)]

V [max(~,B)

= max(T,6)

A

-90-

Let ~: <* ~ On.

It is easily

J seen ~hat ~-I ~we is ~i ~ (uniformly).

Lemma 9 There Proof:

is a EI(J ) map of us onto

Let Q = {~I~(0,~)

= ~}.

(m~) 2 for all ~.

Thus Q is closed and unbounded

Q = {~I~ : ~2 ~_+ ~}, so us ~ Q ÷ us = s.) We prove ~.

Assume

may assume there

+ I.

If B = O, then ms = m g Q, so we are done by Case Then clearly,

there is a EI(J ~) map j : us +-+ mB.

is a EI(J B) map of mB onto into mE.

(mB) 2, so there is certainly

Then f is EI(J ~) and f maps

By hypothesis,

a EI(J B) map g of

= g(<j(~),j(T)>).

(m~) 2 one-one

f-l(v),if

into wE.

Now ran(f)

= ran(g)a

Case 3: lim(~) ~ ms ~ < ~.

and is ZI(J ). into ~y.

Q.

v E ran(f)

otherwise

onto we see that h is El(J ) and h : m~---~

(m~)2.

In this case let <~,T> = - i ( ~ ) .

Set c = {zlz <* }

(g J ). s

Pick y < ~ with ~,~ < my. And by assumption,

into my.

So, setting

is a Zl(J

) map of (m~) 2 one-one

f~[,~>)

there

Thus ~ c

Then ~

-i

[ms

maps

g(-i(~))>),

onto d = g"(g"c) 2.

f-l(0),

mapping

(my) 2 one-one

I,~ < ms, we see that f

But d s Js' so we can define

if e E d

h(0) = , onto h = us---,

otherwise

(m~) 2.

The lermna is proved.] Using

this lemma, we may now establish

the following

onto ~ ,

is a Zl(J s) map of us

is a map g ~ J

= g(
c one-one

El(Js) map h on us by

Clearly,

Js' so if

= <0,0>,

one-one

Hence we

(for v e ms)

h(~)

Thus ~,~

I.

Then g 8 rud(J E) = J , so for ~,y s ms, define f(<~,~>)

we define h by

on

Then ~ -IFs suffices.

B ~ I.

(~E) 2 one-one

the lemma by induction

it is true for all ~ < ~.

Case I: us e Q. Case 2: ~ =B

in On. (Clearly,

important

result:

a

-91-

Theorem I0 There is a EI(J ~) map of ms onto J for all s. onto ~ ~ (~s) 2 be I~ S({p}),s where p e J

Proof: Let f : ~

is the <j-least element of

Js for which such an f exists.

Define f0 fl by demanding that f(~) = onto By induction, define f : ~ ~ (~s) n thus: n fn+1 (~) ~ . Hence each fn is E ~ ( { p } ) . Let

for all ~ ~ ms. f0 = i d I ~ ;

h = hs, the canonical E I skolem function for J . s Claim I: X is closed under ordered pairs.

Set X = h " ( ~

x(~s x{p})).

To see this, let yl,y 2 E X, say Yl = h(Jl'<~l'P>)' Y2 = h(J2'<~2'P>)" Let Js • " <~i,~2 > = f2(T). Then {} is a E l ({T,p}) predicate, so by definltlon of h, s X, as claimed. So by corollary 7, X ~ E I J ~. ~s ~ X ,

By the condensation lermna, let ~: X ~ JB' B ~ s.

Since

we clearly have B = e here.

Claim 2: For all i e ~, x s X, ~(h(i,x)) = h(i,~(x)). To see this, observe first that as h is EJs l , there is a rud function H such that y = h(i,x) ~-+ (~t ~ J )[H(t,i,x,y) = I]. y = h(i,x) E X (if defined). (~t s X)[H(t,i,x,y) = ~ .

Now let i ~ ~, x ~ X.

Since X ~ EiJs,

Thus, by the above, since x, y e X ~ E I J

By le~na210~therefore,

,

(~t ~ X)[H(~(t),i,~(x),~(y))

Since ~"X = J , this can be rewritten as (~t s Js)[H(t,i,~(x),~(y))

= ~.

= I].

Thus

~(y) = h(i,~(x)), as claimed. Now, f ~ (~s) 3, so as ~ e E~s({~(p)}).

= id~,

~"f = f.

And by isomorphism, ~"f is

So as ~(p) ~ jp, the choice of p shows that ~(p) = p.

So, by claim 2, if i ~ m, ~ e ~s, ~(h(i,<~,p>)) = h(i,), which is to say ~X

= id~X.

Thus X = J~.

Now define h : ( ~ ) 3 ÷ j

by setting l y, if (~t ~ ST) [H(t ,i ,<~ ,p> ,y) = I]

h(i,~, ~) =

~otherwise

Then h is EI(J ), and clearly h"(~s) 3 = h"(m x (~e x {p})) = X = J~. Therefore, h = f3 is as required by the theorem. I Observe that in lemma 9 and theorem i0, the maps constructed generally have parameters in their definitions.

Note also that, being total, these maps are in fact

Al(Js)An important concept in our study will be that of the 'projectu~'

of a given

ord-

inal. Let s,n k 0.

n The E_n-proSectum of s, p s ' is the largest p ~ s such that <J p ,A> is

amenable for all A e En(J ~) n ~ J

). P

(For some intuitive explanation as to the motivation of this definition, we refer the

-92-

reader to

[~].)

We shall also require an obvious generalisation of the notion of the En~projectum of an ordinal. Let <J ,A> be amenable.

n The E_rl-proiectum of <J ,A>, p ,A, is the largest P ~ ~ such

that <J p ,B> is amenable for all B c 2n(<J ,A>)~ ~ { ~ ) We shall make s t r o n g use of the En-projectura i n p r o v i n g t h a t e v e r y Ja i s En-Uniformisable, all n ~ i.

Since most of the following le~m~as are directed towards this

g o a l , i t i s worth i n d i c a t i n g b r i e f l y E

n

our s t r a t e g y .

(The reason f o r i n t r o d u c i n g the

projectum is that it provides just what is required in order to follow this

strategy.) We already know that <J6,A> is El-uniformisable attempt to "reduce" En(J ~) predicates which is itself En(J ).

for all <J~,A>.

to E l(<J£n,A >) predicates

What we shall do is for some A ~ J p n

To carry out this reduction, we need to have at our

disposal a En(J ~) map of a subset (at least) of J n onto J . p

Thus, what we shall do

is to simultaneously prove, by induction on n,~, the followingtwo propositions: (P I) J

is En+l-uniformisable

(P 2) There is a En(J ) map of a subset of wp n onto J . The proof of (P I) goes roughly as follows. Let f : =_ ~pn onto j

be ~n(J ).

Let R be a En+l(J~) predicate on J .

Now, f-i is a En(J ~) relation, so by assuming

En_Uni formis ability ' f-I can be "shrunk" to a En(J ~) map of J

into ~P n •

This

reduces R to a En+ l(J ) predicate R' on Jpn" c~ Now find a En(J ~) predicate A c J R' by a El(<Jpn,A>) uniformising

n such that R' is in fact EI(<Jon,A>).

Uniformise

function, and then reverse the procedure to recover a En+l(J ~)

function for R.

There is one doubtful point in the above outline.

we in fact find a set A as required.

Can

That we can has to be proved as we proceed, so

we shall in fact simultaneously prove three propositions,

(P I), (P 2), and a propo-

sition (P 3) to be formulated precisely later. Lemma II Let n ~ I, and assume J

is E -uniformisable. n

Let ¥ ~ ~ be the least ordinal such

-93-

that ~(my) ~ ~n(Ja) ~ Ja" Proof: By lemma 33, J

Then there is a En(J ) map of a subset of my onto Ja"

has a E skolem function, h. n

assume p i s

the <j-least

Let a ~ m y ,

a s Zn(J ), a ~ Ja"

that a ~ EJa({q}). n that h i s

We may

e l e m e n t o f J a f o r w h i c h s u c h an h e x i s t s . Let q be the <j-least element of Ja such

Define h by h(i,x) = h(i,<x,p,q>).

a E skolem function n

Set X = h".m( x Jy).

for J

~

and t h a t

is

It is easily seen a good p a r a m e t e r

for

Now, there is a El (JY) map g : my onto ~ Jy, so h o g Wis"

a En(J ) map of a subset of ~y onto X. X=J

J Let h be E ~({p}).

Hence it suffices to show that

.

Clearly, X ~ Z

J~"

Let ~ : X ~ JB' B ~ ~. Then ~IJy = id~Jy, so in particular, EJB Also ~"a is ({~(q)}). But look, this implies that a = ~"a ~ JB+I" n n

~"a = a.

Hence we must have 8 = ~ ( a n d h e r e we h a v e u s e d o u r h y p o t h e s i s t h a t ~ m y ) G En(J a) J J~). Thus, in particular, a = ~"a is Ena({~(q)}), so by the choice of q, we see J . a Zn~ ({~ (p)}) E nthat ~(q) = q. Again, it is easy to see that h' = ~ = h o ~ -i ~s skolem f u n c t i o n

for J , so by choice of p, ~(p)

defined by the same ~

n

Lemma ii p l a y s

a direct

But t h e n h , h '

formula (with parameter p) in J , so h = h'.

immediately that ~ ~ h = - i ~ ~(i,x) = h(i,x).

= p.

= ~, of course.

are both It follows

So for i e ~, x ~ J , ~ ~ h(i,x)= Y

Thus ~IX = id[X, and X = J • part

in the proof

o f (P 1 ) - ( P

3).

The n e x t

lena,

however,

is only used during the proof of the lemma which follows it, and may, on first sight, appear somewhat uninspiring. Lermna 12 Let <J~,A> be amenable, p = P~,A'I

If B ~ J p

is EI(<J~,A>), then ZI(<Jp,B>)

Z (<J ,A>). 2 Proof: Case I. There is a ZI(<J~,A>) map of some y < mp cofinally into ~ . Let g be such a map, and let B be Z0(<J ,A>) such that B(x) +-+ ~zB(z,x) for each x e J . p

Define B' by B'() +-> (~zs Sg(~B(z,x),

is AI(<J ,A>).

And since B(x) +-+ ( ~

for ~ e y, x s J . P

Thus B'

~ y)B'(<~,x>), Zl(<Jp,B>) ~- ZI(<Jp,B'>).

-94-

Thus, we need only prove that 7~l(<Jp,B'>

~E2(<J

,A>)o

It clearly suffices

prove that 10(<Jp,B'>)

~ ~2(<J ,A>).

Let R be E0(<Jp,B'>).

Thus R is rud in B' and some parameter p ~ Jp.

p, <J ,B'> is amenable p

to

By choice of

so by lenm~as&2 and63, there is a E0(J@) predicate P and

functions fl ..... fm+k' rud in parameter p, such that R(~) +-+ P(~,fl (~)'''''fm (~)' B' ~ fm+l(~) ..... B' ~ fm+k~)). = B' q fm+k(~)

Hence R(x) +-+ ~Yl ..... @Yk[Yl = B' q fm+l(~) .... ^Yk

^P(~,fl(~) ..... fm (~)' Yl ..... Yk )]"

Now P is certainly E0(J ), and

fl'''''fm+k are rud in parameter p, so it suffices to show that the function b(u) = B' q u is E2(<J ,A>). ~x[x E y +-+ x s u ^B'(x)], Case 2.

It is in fact Hl(<J ,A>), because: y = b(u) +-+

and B' is &l(<J ,A>).

Otherwise.

As before, we must show that 10(<Jp,B>) _~ E2(<J ,A>).

Again as before, this reduces,

by the amenability of <J ,B>, to proving that the function b(u) = B N u is P E2(<J ,A> ) on J . Now, we clearly have P y = b(u)~+ (~x ~ y)(x ~ u ~ B(x)) ~ (Vx ~ u)(B(x) ÷ x e y). Now, the second conjunct here is H l(<J ,A>). Xl(<J ,A>) , which is sufficient. (<J ,A>).

We show that the first conjunct is

It reduces to showing that (~x c y)B(x) is

But look, we know that Case i fails to hold, so this is proved just as

in the proof of le~ma O.16(iv).

(The failure of Case i gives a "replacement axiom"

for E l (<J ,A>) predicates - which is just what is required:)

]

The next lemma is the key step involved in proving, by induction, the as yet unformulated

(P 3).

Lemma 13 Let <J ,A> be amenable, O ~p onto J .

i = O~,A"

Suppose there is a E (<J ,A>) map of a subset of i

Then there is a B ~ J , B ~ EI(<J ,A>), such that ~n(<Jp,B>) = p

~(Jp) N E n + l(<J ,A>) for all n >~ I. Proof: Let u ~ p ,

onto and let f : u ~ J

~l<J~'A>({p}).

be EI(<J ,A>).

Pick p e J

such that f is

Let <~ili < ~> be a recursive enumeration of Fml~ I,

-95-

El ~ B = {li c m ~ x c Jp A ~ <j ,A>~iIx,P)}'

Se%

Now, <J ,A> is amenable,

and hence rud closed, so by lemma6B, B ~ El(<J

,A>).

And

of course B ~-J . P Commencing with lemma 12, an easy induction shows that for all n >. i, Zn(<Jo,B >) CEn+l(<j

,A>).

For the converse,

let R(~) be a En+l(<J

sake of argument,

that n is even.

Let P be a El(<J

c Jp, R(~) ~-+~yl~Y2...~ynP(~,x~). By choice of f, any x ~ J

,A>) relation on Jp, n >~ I.

Assume,

for the

,A>) relation such that, for

Define P by P(~,~) ++ [~,~ E Jp ^ P(f(~),~)].

is ll<Je'A>({p,~})

is rud in B and some parameter 9 < rap.

for some ~ < mP, so by definition of B,

In particular,

P is AI(<JQ,B>).

Again, D = dom(f) is rud in B and some parameter T < mp, SO D is also 81(<Jp,B>). But for

~ c Jp, R(x) -~ +-+ (~z I E D) (~z 2 c D) ... (~z n ~ D)P(~,x), ~ ~' which is thus

En(<J0 ,B>). We are now ready to formulate

(P 3) and prove our promised uniformisation

theorem.

The proof will indicate even more clearly than before the motivation for the projectum. Let e, n >~ O.

A E

"-~

master code for J

c~

is a set A -~J n, A ~ En(J ~), such that whenPc~

ever m >. I, Z (<J n,A>) = ~(J n) n En+m(J~) •

Pc~

m

Theorem 14. Let ~, n >. O. (P i) J

is E

P~

(Uniformlsation Theorem) Then: n+l

-uniformisable.

(P 2) There is a En(J ) map of a subset of ~pn onto J .

(Unless n = O, when the map

is EI(J ).) (P 3) J

has a E

n

master code.

Proof: We prove the theorem

(for all n) by induction on ~.

For ~ = 0, it is trivial.

So assume ~ > 0 and that (P I)-(P 3) hold (for all n) for all ~ < ~. prove

(P I)-(P 3) at ~ by induction on n.

Case i : n = 0. (P I) is already proved

(Theorem I)

We

-96-

(P 2) p0 = 5, so (P 2) is already proved (Theorem i0) (P 3 ) Case

Since

2 : n = m + 1 , m ~ O.

0

pa = a, Let

A = ~ is

n p = Pa for

a 20 m a s t e r

code

for

J

.

convenience,

We first prove that p is the least ordinal such that some En(J ~) function maps a subset of up onto J . To this end, let 6 be the least such ordinal.

Suppose first that 6 < p.

Then

B = {~ ~ ~6 I ~ ~ f($)} is a ~n(J ) subset of J , so by definition of p, <Jp,B> is P amenable.

Thus, as 6 < p, B = B ~ 6

~ J

~ J .

E f(~) +-+ ~ e B ~-> ~ ~ f(~), which is absurd.

So B = f(~) for some ~ g ~6, whence Hence p ~ 6.

definition of p, this means that for some ~n(J ) set B ~ J 6 , Since <JI,B~ must be amenable, 6 > I.

Suppose p < 6.

By

<J6 'B> is not amenable.

If 6 = y + I, then since there is a Zl(J )

map of ~y onto m6, there is a En(J ~) map of a subset of ~y onto J , contrary to the choice of 6.

Hence lim(6).

T < 6 with B n J T ~ J6"

It follows, since <J6,B> is not amenable, that there is

By induction hypothesis, J~ is En-Uniformisable.

T < 6, lemma II implies that ~(~T) n Zn(J ~) ~ J~onto

JT' so this implies

for some B < ~, B Q J such that B Q J then B n J ~

~(j )

O Zn(J ~) ~ J~.

is J6-definable.

is Zr(J6).

= (B Q J ) O J

So as

But there is h e J , h :~T

In particular, B Q J

8 J .

Hence

Let 6 be the least such, and let r be least

r By definition, <J r, B Q J T > is amenable, so if r < P6' P6 r e Jo~ ~ J6' contrary to the choice of 6. Hence ~ ~ pB.

r By induction hypothesis, there is a Er(J ~) map g from a subset of ~p$ onto J . since B ~ J~ ~ J6+ I and B Q J~ ~ J~, 6 + I > 6, or 6 ~ 6. map g' from a subset of m#~r onto m6.

And

Hence there is a Er(J 6)

Then f o g' is a En(J ~) map of a subset of

r ~P6 onto Je.

r But we have established that P6 ~ r < ~' so this contradicts the

choice of 6.

Hence 6 = p.

(P 2) follows i~mnediately from the above result of course. We turn now to (P 3).

By induction hypothesis, let A be a Em master code for J .

Set q = pm for convenience. By the above, let f be a En(J ) map of a subset of up onto J~.

By choice of A,

f' = f p (f-l,,jq) is a El(<Jq,A>) map of a subset of up onto Jq . n = Pq,A" I it is clear that ~ = p~

By choice of A,

Finally, of course, <J n ,A> is amenable.

So, we may

-97-

apply len~na 13 to <Jn'A> to obtain a EI(<J ,A>) set B ~ Jo such that Zr(<Jp,B>) = ~(Jo) ~ Er+l(<Jn,A>) Er(<Jo,B>) = ~(Jo)

for all r ~ I.

NZn+r(J ~) for all r ~ i.

Finally we prove (P I).

J

n

function uniformising R.

f-I

Since f is Zn(J ), so is f-l.

Set r = f = r o f-i

function which uniformises R.

Let R(y,~)

Then R is Zn+l(J ~), and hence ZI(<J0,B>).

is En-Uniformisable , by induction hypothesis,

function uniformising

master code for J .

Define, with f as above, a relation R on J0 by

R(y,x)~ +-+ [y,x~ e J0 ^ R(f(y),f(~))]. Let r be a ZI(<J0,B>)

Hence B is a Zn master code for J~.

Let B be, as above, a Z

be a Zn+l(J ~) relation on J .

By choice of A, B ~ En(J ) and

But

so we can let f be a Zn(J ~) It is clear that r is a ~n+l(J )

The proof is complete. I

The above results give us two (intuitive) equivalent formulations of theZn-projectum: Theorem 15 Let ~,n ~ O.

Let 6 be the least ordinal such that some Zn(J a) function maps a

subset of ~6 onto J .

Let y be the least ordinal such that @(~y) O En(J~) ~ J~.

n

Then 6 = y = p • Proof: That ~ = 0~ was actually proved during the proof of Theorem 14. know that J 6 < y.

is Z -uniformisable, n

Now by definition,

Z = {$ I ~ ~ f(~)}"

lemma Ii tells us that 6 ~ y.

let u ~ ,

Then Z ~ ~

and let f : u onto ---~ J

Since we now Assume

be Zn(J ).

Let

and Z g Zn(J ), so by definition of

y, Z s J . Thus Z = f(~) for some $, so ~ e f(~) +-+ ~ ~ f(~), which is absurd. Hence 6 = y.

QED

Finally, in this chapter, we define some specific master codes which have the advantage of being preserved under condensation arguments. some elementary results concerning embeddings of

To do this, we require

J 's.

Lermma 16 Let M, M be transitive, with <M,A> rud closed.

Suppose

be rud in A and let f be rud in A by the same definition. ~(f(x)) = f(~(~)).

~: <M,A>-<

<M,A>.

Let

Then for all ~ e M,

-98-

Proof: Clearly, ~(x q A) = ~(x) ~ A for all x e M. <M,~>

is E 0

, SO ~(g(x)) = g(~(~)).

Also, if g is rud, then g ~

The result follows by lemma 6.2 and

induction on the definitions of ~ , f. Lena

17

Suppose ~ : J~-<~0 J~"

Then for any ~ < ~ ,

Proof: By induction on ~ < ~ .

~(S~) = S (~).

Clearly, ~(S 0) = ~(~) = ~ = S O = S~(0).

Also, by

lemma 16, ~(S +i) = ~(S(S )) = S(~(S )), so by induction, ~(Sv+ I) = S(S (.~)) = S (~)+i = S (~+i). Let X = ~(S%).

Finally, suppose lim(%) and that v < ~ -> ~(S ) = S (~).

Since S k is transitive,

JN ~ (~y ~ S%)(~x g y)(x e S%).

as ~ : J--:

J , J ~ (~y s X)(~x ~ y)(x ~ X). Hence X is transitive. E0 lim(%), so % = ~$ for some ~, and S k = J~. By le~na 6.18, therefore, <S

S~ El "

I ~ < %> is (uniformly)

definition of this sequence, Then since ~ : J N - < ~ 0 J X ~ ~.

But X e J

limit ordinal ~.

Using the uniform, parameter-free,

So Now,

El

let ~ be the sentence which says ~x~)(x ~ S ).

and S% ~ J~, whence

is transitive,

(~S%)

: Sk-< X, and S% I= ~,

so this can only happen if X = S~ for some

Again, ~()~) = ~(On ~ S l)

=

On A S

=

T

,

SO

~(S l) = S (~).|

Lemma 18 Suppose ~:<J~,A>-<~0 <J ,A> and that sup(~"~)

= ~.

Proof: Suppose ~ is E 0 and ~<j ,A> ~y~(y, ~(~)). ~<j ,A>(~y ~ S (~))~(y, ~(X~).

Then ~ : <J ~ ,A>-<~I <J ,A>. Then for some ~ < ~ ,

we have

So as ~ is E0-elementary, l=<j_,~>(~y~ c S )

(y, ~), by lemma 17. Lemma 19 Suppose <J_,A>~ is amenable and ~: J ~ - < ~ 0 J A ~ J

such

that

~:

a

Thus A is

x

,~t>-~

~

Proof: Let A = U

<Ja,A>

<J

is

amenable.

n (A n S ( , ~ ) ) .

<J ,A> a n d <Jc~,A> i s E0~

_ ~(A ~ S ). the unique For

let

cofinally.

amenable.

(Since <J_,A> is amenable, this is quite sound.)

A ~ J

such

x ~ Ja"

But <J~,A> i s

Then there is a unique

that

Pick

amenable,

"~(A c~ S ) = A f'l S "J < mE w i t h

x ~ S (v).

for

all

v < wE.

T h u s x clA =

s o A c~S,~ e J_a and s o A c~ S ( ~ ) e

Ja.

So,

-99-

as Ju is rud closed, x ~ A

= x ~(A ~S

(v)) g J .

By a similar trick, if x e Js, then ~(x n A) = ~(x) % A . ~<j5,~> #[~].

Let u e J~ be transitive with ~ e u.

is E0-elementary , <~(n), A ~ ~(u)>

1= ~[~(~)].

Let ~ be E 0 and suppose

Thus ~ ~[~].

So, as

But ~(u) is transitive, so

b < j ,A> ¢ [~ (~)]" n The standard codes, A , and the standard parameters, p~, are defined by induction as follows.

Set A 0 = pa0 = @.

code for J .

n are defined, and that A~ is a gn master Suppose A~, p~

n+l By theorem 14 there is a El(<Jpn, A >) map of a subset of ~o~ onto

J n, so we may define P~ n+l = the <j-least p e J n such that every x ~ Jpn is El-definable in <J u,-A n>~ from Pe P~ ~ P~ parameters in {p} O Jpn+l. Let <~ili < m> be a recursive enumeration of Fml~ I, and set -El n > ~"l[ x, n+l~ A n+l = {li g ~ ^ x e Jpn+l ^ ~<jpn,A~ P~ J} Then An+la is a En+l master code for J~.

For, clearly, A TM~ g J n+lP~

<J^n~e, An>~ is amenable, so A n+l ~ is E l(<Jpn, A >), and hence En+l(J ). show that if m ~ I, then Em(<Jpn+l, An+l~) ~ = ~(Jpn+l ) ~ Em+n+l(J~).

And So, we must For this, it

clearly suffices to show that Em(Jpn+l' A~ +I>) =~(jp~+l) ~ Em+ l (<J n, A~>).

But

look, <Jpn, A~ > is amenable, pn+l = p 1n,An , and there is a El(<Jpn, A~>) map of a n+l P~ ~ subset of mp~ onto J n, so this follows from lemma 13. P~ As we have already indicated, the main point of these definitions is that the standard codes and standard parameters are preserved under condensation arguments. To prove this, we require an obvious generalisation of the two concepts for the case n = i. Let <Jp,B> be amenable, and set p = P~,B"

Let

P~,B = the <j-least p ¢ J~ such that every x e JB is El-definable in <J ,B> from parameters in {p} ~ J . P A1 6,B = {[i e ~ A X ¢ Jp ^ ~ p , B >

~i Ex' PB,B ]}"

1 Then AB, B is a E 1 master code for <JB,B >, and we clearly have:

-i00-

°

n+l = pl n+l = I n+l = AI n n" ps,A ~ n n, PB Pnn~B,An'B A8 p~,A~

Pp

Lemma 20 Let ~ > I, m,n ~ 0,

Let <J~,A> be amenable,

(i)

There is a unique

(2)

There

and let ~: <J_p,A>~Em<Jp~,A n> .~

Then:

e ~ p such that ~ = p~ and A = A~"

is a unique ~ ~ 7, ~ : J --<Em+n J , such that for all i ~ n: (a)

~ i z(p~)

i = p~

(b)

(~Dpi) :<J~,.

and

A_l>~< ~ ~

< J .p~ ,

m+n-I

A~ >.

a

[We call J~ ~ J the canonical extension of <J~,~> ~ <J n' An>'] Proof:

By induction

n - I.

on no

For n = 0 the result is trivial, -

so assume it holds

n

Let m >~ 0, and let z:<J~,A>-< E

for

-

<J n' A >, where <Jp,A> is amenable. 0~ to show that there is a unique <J~,B> such that m

It clearly

suffices

= 0~,~, A = A~,~,

and a unique ~ -~z such that ~(p~,~)

= pn and such that

~: <J~'B>-~E

<J n-l, An-l> (For then, by induction hypothesis, there is m+l P~ ~ " I n a unique ~ such that B = @n-i and B = A n-l ,~ whence ~ = 0~,~ = 0~ and

% = A -I - = A n etc.) B,B E' Let X = {x ~ J n_ 1 I x is El-definable

Pc~

ran(m) Thus

in <J n-i' An-l> ~

Po~

from parameters

in

O {pn} }. X-< ~ <Jpn-l'

~: <J~'B>-<~ ~i

A n-l> ~ '

Let @-l : <X, An-I ~ ~ X > ~ <J~,B>.

<J n-i An-l>

Pa.

' O:

and set A = A n N J ~ .

"

We show that ~ = ~ -- "

Then 7:

Thus,

already

Define P ~< O n by ~05 = sup~)<~z(~)

<J-,A>-

~ cofinally,

so by lemma 18,

~

: <J~,A>-<EI<Js,A >. y e X ~ J~.

P

We show that ran(T)

Then for some i e m and some x E ran(~)

such that ~<~ ~n-l>~.[, d n i A i A(i, )., Hence

ran(z)

= X nJ~.

pn]. ~

its transitivis ation.

This means

'

ran(~) c X ~ J~.

y is the unique y e J n-i

Pc~

Thus y is the unique y e J n- such that p 1

~ut look, m: < J ~ , A > - < Z # ~ , A > = X f~J~. P

Clearly,

, so as x e ran(m),

that m -I is " the unique

But X ~ J~ is an E-initial P

y ~ ran(T)

collapse

also.

of X N J~ onto

segment of X and 3 -I is the

Let

-I01-

unique collapse of X onto its transitivisation, J~, so - l ~ x collapse of X ~ J~p onto its transitivisation. ~, therefore, ~ J ~

= ~, so ~ m ~.

nJ~ is the unique

Thus ~-I~x ~ J~P = - I .

By choice of

Next we show that ~: <J-'B>'~-P ~m+l<JP~n-~'~ An-l>a ;

<Jn-l' An-l~ ~ -" If m = 0 we are already done, so assume m > O. m+l P~ Let y be Em+l-definable in <J n - l , A.n-l> from parameters in ran(n) O {pn}. We must i.e. that X - ~ E

show that y ~ X.

Let ~ be a Em+ 1 formula such that y is the unique y g J n-i such

that ~< j n-l, An-l>~[y' +x, pj,n

where +x e ran(~).

Let ~(v,v,q)÷ -

...z m ~(~,v,~,q), where ~ is Z l if m is even and ~i if m is odd.

~ZI~... Let h be the

canonical E l skolem function for <J n-i ,An-i>, and set h() ~ h(i, <x,pn>). Suppose z e Jpn ~ X. z, pn .

So as

Then h(z) is 0~El-definable in <Jpn-l' An-l> from the parameters

z,p~n ~ X_ ~ • h(z) s X also.

~[~<~)'~' P~] i.

that the condition ~<J n-l'A~-l>

So it suffices to show

satisfied b~ some ~ ~ X OJ~n.

n Pa By definition of pc, every x e Jpn-I is El-definable in <J n-l, A n-l> from parameters in {pn} 0 J n, so we clearly have h"J

P~

P~

= J n-l"

P

Hence, it suffices to show that

the condition jpn ) ... (-zm ¢ Jpn)~ ~<j n-i ,An_l>~h(z), h(z), ~, p:]] Pa n-l> is satisfied by some z e X n Jpn' But A na is a El master code for <Jon-l_~, A , (*) (~z I s Jpn ) (Wz 2

n,An> (I, pn) ~

so (*) is a Lm Pa

unary predicate on J Pan-

By assumption, ~:<JD,A >

-<~ <J n, An>, so as ran(~) = X GNu, X N J~ -4Em <J0 n, An>.

hm Pe

-c~

So, as

+x, p~n e ran(n) V {p ~ }, (*) is satisfied by some z e X ~ J~ g X a J n, and we are P~ done.

Thus ~ is Em+l-elementary. =

We

prove

that

-

-

Claim: A = {li s ~ ^ x e J_ ^ ~<JB_.~>+i[~' To see this, observe that as ~: <J~,A>~Em And by definition of

and

=

~]~.

<J n,An>,a A() ~-+ An(). Pa

A~, A~() ~ ~E13 n_l,An-l>$i['lt(x), P~I" Pa c~

Finally, as ~: <J~,B>-<EI <Jpn-l,A n-I a > and ~(x), Pan e ran(~) -

El ,An_l>~i[~(x ) ~<J n-I P~

we have

E1 phi iff ~ <j~,~>~i[x,P].

-102-

The claim follows immediately. We show that

~ = p~,~.

By construction of <J~,B>, every x ~ J~ is El-definable

in <J~,B> from parameters in J~ u {5}.

Thus if h is the canonical E l skolem

for <J~,B>, J~ = h"(~0 x (J~ x {p})).

function

a subset of ~

onto J~.

Z = {$ e mPI~ ~ f(~)}"

This implies ~ >. p!B,B"

'

For suppose not.

Z is thus a EI(<J~,B>) subset of J~ ~ J

is amenable, whence Z = Z n ~ e f(~) +-~ ~ ~ f($)

Hence there is a EI(<J~,B>) map f of

~ J I

P~,~ ~

a contradiction.

C e (~(J~) ~ EI(<J~,B> ).

J~.

Let

i

, so <Jpl _, Z>

Thus Z = f(~) for some ~ ~ ~ , whence

We show that ~ ~< p ! _

8,B

also.

As we remarked above, every x ¢ J~ is El-definable in

<J~,B> from parameters in J_P u {~}, so clearly C e E~J~ '~> (J~ O {p}). C is therefore rud in A.

By the claim,

But <J~,A> is amenable, so <J~,C> must be amenable also.

Since C was arbitrary, this implies that p ~< p ! _ B,B" Next we show that ~ = p! B,B"

Hence

D = pl _ B,B"

Since every x ~ J~ is El-definable from parameters in

J~ o {~} in <J~,B>, it suffices to show that ~ is the <j-least such. not, and let p' < j ~ also have this property. some x e J~.

Let

Well suppose

In particular, p = h(i,<x,p'>) for

Now, h is the canonical S 1 skolem function for <J~,B>, h is the

n-I ,An-l>" canonical E l skolem function for <Jpn-l, An-i> ~ , and ~: <J~,B> -~Ll <J p~ c~ Hence p~n = ~(}) = ~(h(i, <x, p'>)) = h(i, <~(x), ~(p')>). X f] J n. P~

Note that ~(x) = ~(x)

By choice of pn, every y E J n-I is El-definable in <Jpn_ I , An-l> ~ from P~

a~

parameters in J n O {pn}. So by the above, every y E J~n-i is El-definable in P~ <Jpn 1 ,An-l> from parameters in Jp~n <){~(p')}" But p' < jp, so ~(p') < j ~ ) = pn, so this contradicts the choice of pn.u Hence ~ = ~,~. Since ~ = pl - and ~ = p~,~, the claim implies immediately that A = A~,~. ~,B

This

completes the existence proof for the lemma, and we turn to the question of uniqueness.

First we show that there is at most one <J~,B> such that p = pl~,B_ and

= A~,B . I_

For let <Js0,B0> and <JB "BI> be two such. Set P0 = pIB0 ,B0 and

Pl = pl For each j g ~ and each ~ s J ~I 'BI " P' (x)

~ ~ >~j[<~> p0] iffA(<j ~ > ) i f f > El Cj[<~>, Pl]. <JB0~B~ ' , <J~l,]~l>

-103-

Let h.l be the canonical E l skolem function for <JBi, Bi>, i = O, I. hi(<J,x>) = hi( j, <x, pi >), i = O, I.

Then JB. = h'"J-'l P each i. i enumerates all of Fml~ I, so if x, y E J~, then by (x):

Set

But <~jlJ e ~ )

h0(x) E h0(Y) iff hl(x) c hl(Y), h0(x) = h0(Y) iff hl(x) = hi(Y), and B0(h0(x)) iff Bl(hl(x)). by ~(h0(x)) = hl(X).

Hence we may define an isomorphism 0:< J~0'B0> =~ <JBI 'BI>

Since JB0, JBI and transitive, ~ must be the identiy, so

B 0 = B I and B 0 = B I. such that Finally, suppose that there are ~0' 01 ~ ~' ~i: <J~,B>-< Z 1 <Jon-i , A n-l\ ~ ~' n i = O, i. Let h, h be, as before, the canonical Z I skolem functions ~i(~) = p~, of <J~,B>, <J n_l,A n-l>, respectively. Then for x e J~, j g ~0, we have P~ ~0 o h(j, <x, ~>) = h(j,<~ (x), pn>) = ~i o h(j, <x, p>). So, as h"(m × (J~ x {~}))

: J~' ~0 = ~I" I

Chapter 8

THE COMBINATORIAL PRINCIPLES []~

In this chapter we shall establish, for later use.

Let K be an infinite

in L, a powerful

cardinal.

By

combinatorial

principle

[]<, we shall mean the following

principle: There is a sequence
< K+ ^

lim(%) > such that

:-

(i)

C% is closed and unbounded in %;

(ii)

cf(~) < < + I C% I < K ;

(iii)

if y is a limit point of C%, then Cy = y n C%.

Note that by (ii) and (iii), if cf(l) = ~, then otp(Cl) = K. We shall prove that, in a very strong way, V = L -> ( V K )

[3 . K

First some definitions. Let e ~< ~8, lim(e).

We say ~ is regular at 8 iff there is no Js-definable

of a bounded subset of ~ cofinally into ~.

We say ~ is In-regular at S iff there is

no [n(Js) map of a bounded subset of e coflnally into ~. B iff ~ is In-regular at ~ for all n. obviously

map

Clearly,

If V = L and e is singular,

~ is a regular at then there are

8, n such that e is not In-regular at 8.

Let ~B >~ ~ > I.

We say that 8 is s-minlmal

iff for some p e Js, we have for

all X, p e X-< J 8 ^ The following

~ rh X e On ÷ X = J~.

lermna is extemely useful, and is clearly related to the concept of

~-minimal i ty. Lemma I Let 8 ~< y, X is transitive

c j . Y

There is a smallest N -< J

(and hence N

L and IXI < 8, then N

N

such that X c

N and N ~

O

Y O O e On).

Furthermore,

if e is regular in the sense of

e E O.

Proof: Let N O be the (canonical)

skolem h=ll of X in Jy, and set ~0 = sup(N 0 n

e).

-105-

By induction, let Nn+ 1 be the (canonical) skolem hull of Nn set Sn+ I = sup(Nn+ 1 ~

8).

Let N = ~ n < m N n .

u an in J y and

The lemma follows at once.

There are many variations of lemma i, some of which we shall use without particular mention. proof.

All variations have the same u-iteration idea behind their

The same idea also gives our next lemma.

Lemma 2 Let s > ~ and suppose that for some ~6 >~ s, s is regular at B and 6 is sminimal.

Then cf(~) = ~.

In fact, there is a 11(J6+i ) map of ~ onto a cofinal

subset of ~, so s is not ll-regular at B + I. PrOof: Pick p e J6 so that p ~ X-< JB ^ s ~ X e On -~ X = J6"

11 skolemfunction

Let h be the canonical

for JB+I' and let H be a IOJ6+I predicate such that

y = h(i, x) +~+ ( ~z c J6+l ) H(z, y, i, x), Define hj, j ¢ ~, by y = hi(i, x) +-+ y, x ~ ~ 6 + j

^ (~z

c S ~+j) H(z, y, i, x).

Note that each hj e J6+l' and that the sequence is [l(J6+l )" Define ~i' Xi' i e ~, thus: s 0 = m, X i = hi"(~ x (Jsi x {p})), si+ I = sup(s (~ X i) . Set X = ~ i c ~ Xi' ~ = suPie~ ~i" Then X = h"(~ × (J~ x {p})) " ~ I J6 ~ J6+l"

But p e Y and ~ = s ~ Y

J6+1"

e On.

Let Y = X

n

J6"

Then Y -< J6' as

Thus, Y -- J6' so ~ = s n Y = s q J8 = s.

The sequence <s i I i e ~> is clearly II(JB+I) , so as s = ~ = suPiE ~ si, it suffices to show that ~. < s for all i c ~. i definition, s 0 -- ~ < ~.

Suppose ~i < s.

We prove this by induction on i.

Now, h i N J6 is Js-deflnable, and since

s I. < s ~< 6, there is a function in J6 mapping s i onto J~i definable map of ~ionto X i.

By

Thus there is a J6-

But si+ 1 = sup(s ~ X i) , so there is a J6-definable map

of ~i onto a cofinal subset of ~. . l+l

Since ~ is regular at 6, this means ~i+l < ~" I

-106-

Theorem 3 Assume V = L.

Let < be an infinite +

and let S

be the closed

+ consisting of all ordinals ~ such that ~ < ~ < K , ~

unbounded subset of R and ~ < ~ ÷

cardinal,

I~I J~ ~ K.

Then there is a set E

~

S and a sequence


1%

= ~, 8 S>,

such that: + in K ;

(i)

E is stationary

(ii)

C% is closed and unbounded

in %;

(iii) If y is a limit point of C%, then y E S, y ~ E, and C

= y ~ C%. Y

Proof:

Let E = {~ s S I ( ~B > e)(e is regular at B and B is e-minimal)}. We show that E is stationary

in ~+

Let C c <+ be closed and unbounded +

Let X be the smallest X ~ J regular,

+

~ = X ~

~(C) = C n ~, ~(S)

+

~ < .

Well,

by definition

Let

= S ~ e.

closed and unbounded that e s Eo

++ such that ~ X and X n K

in e, so ~ ~ So

= ~(S) is

~ e C, so it will suffice

to prove

regular at B. And

C>), then we clearly have


I I s S> by cases.

For this we require

some

Let B(e) = the least B ~ ~ such that

Let n(~) = the least n such that ~ is not In-regular

b e any u-sequence

For the remaining n(e)

cofinal

at ~(e).

in ~.

cases, we assume ~ does not fall under Case I.

= I and 8 = ~(~) is a successor

Let B = T + Io

~0(JB)

) = ~,

c~ < ~ . I

Let C

~< T.)

in ~+, S Q e

j$ ~ j ++,~ is certainly

If ~ s S, then ~ is not regular•

is not regular at B.

Case 2.

% (K +

s On * Z = JB' so B is s-minimal.

We define the sequence

Case I.

Then ~ < ~ , and, clearly,

Similarly,

as ~(<+) = ~ and - I :

~ On.

+

Since S is closed and unbounded

of X, if p = ~(<S,

p s Z-<J B a Z ~

notation°

~: X ~ J~.

in K +

+ . Since K ~s

Let u ~ y predicate

fi(~) = q +-+ ( ~ z

We show that cf(~)

= ~.

ordinal. (Note that since lim(~), we have

< ~ and let f: u ÷ ~ be Z (J$) and cofinal 1 such that f(~) = q ~ +

( ~ z s JB) F(z, n, ~).

~ S~T+.)I F(z, q, ~).

Then fi c J8 and fi

in ~. Define

Let F be a fi' i c ~, by

=- JT' so fi is JT-

-107-

definable.

But ~ is regular

suPiem ~i = sup(f"7) Let C

= e.

at T, so ~i = sup(fi"Y)

Hence

be any ~-sequence

For the remaining

cf(~)

of E.

in ~.

under Case 2, so we have already =

Case 2 will be required

By choice

@(~y)

of B, there is a ~n(JB)

into ~ such that f ~ J~o

Since ~ ~S,

Note

that by

taken care of

later.

n Then PB = <"

6 = B(e), n = n(~).

Since p~ is the least y such that n PB ~ ~"

clearly,

cases, we assume e does not fall under Case 2.

The fact that E

Claim A: Let e sS,

But,

= m.

cofinal

virtue of le~mna 2, every e s E falls all members

< e"

N ~n(JB)

~ JB' and < is a cardinal,

map f of a bounded

it is easy to obtain,

subset

using

of ~ cofinally

f, a ~n(J~)

subset

n of ~ not in J~, so O~ = ~o Case 3.

n(e)

= i & lim(B(~))

or else n(e)

> i.

n-I Set B = B(~), n = n(~), ~ = @(e) = O@

n-I = AB .

A = A(~)

(So p = B and A =

if n = I). Note that,

if n > i, then e is ~

Let h be the canonical

~I skolem

n-i

-regular

function

at B, so in all cases,

e ~< P.

for <J , A>. P

Using claim A, let p = p(~) = the <j-least

p g J

such that h"(~ x (< x {p})) P

=J

. P Let h have the

(uniform)

~i definition

y = h(i, x) +-+ ( 3 z

~ J )H(z, y, i, x), 0

where H is ~~0<JP ' A> For T < p, set y = h (i, x) +-~ y, x ¢ J <J~, A ~ JT > is amenable,

hT is its canonical

Define a map g from a subset g(m~ + i) =

~ J )H(z, y, i, x). T

~Yl skolem

function.

I h(i, <~, p>), if h(i, <~, p>)

Let r I' F 2 be the uniformly AF2(~)

~ (~z

of K onto ~ by

otherwise

= my + i + Fl(~) = i

T

undefined.

~I functions

= v.

such that

E

Thus,

if

-108-

Set G(z, I(<JP' A>)

T, ~) +-+ ~, T < ~ a H(z,

(with parameters

p and,

r, rl(.) ,
if ~ < p, e as well)

p>),

Thus G is (uniformly)

(for amenable

<Jp, A> with

p e Jp). Clearly,

g(%)) = T +-+ ( ~z c Jp) G(z,

For X

Jp, set ~x = sup(~

=

Define reeursion

functions

T, u), a uniform El definition

of g.

c~ X).

k: @ + K, £: @ -~ ~, m: @ ÷ p for some

@ .< < by simultaneous

as follows.

k(%)) = the least r ~ dom(g) £(%)) = ~x~' where

X

= hm(v)

~ + i), where

such that g(T) > £(%)) and T!

I£(%))I Jg(r)

~ <.

(~ × (< × {p})).

m(O)

= max(<,

m(%)

= sup%)<% m(~)), if < p (otherwise

such that p e Jn+ I.

n is the least ordinal undefined),

for lim(l)o

such that g o k(~) < y and A ~ Jm(~)

m(%) + i) = the least y > m(~)

£(~), m(%)) e hy"(~0 x (< × {p})) Suppose m(%)) < p is defined.

and ( -~z a Jy) G(z,

Then £(9) < ~.

For suppose ~x

e J

Y

and

g o k(w), k(9)).

= ~"

Then hm(%) )

%) maps < cofinally

But hm(%) ) is [l(<Jm(9)'

we can find T < p such that A r~ Jm(v)

amenable, contrary

into ~.

to r < p Z 8.

Hence

It follows

the recursion

down when m(%)) = p.

m(%)) < p + m(~ + I) < p, so this means limit ordinal

@ ~< <.

g o k is monotone;

(b)

k is monotone°

(c) U,<% (d)

is JT-definable, is defined.

Clearly,

that k, £, m are all defined on the same

and in fact £(~) < g o k(v) < £(%) + I) for all %).

(By (~), g o k(%) + I) < £(9),

and [£(%))[Jgok(%)+l)

~< K, so by

of k, k(v) ~< k(~ + I).)

£, m are normal°

(For £, note

hm(%))"(to x (< x {p})) =

that if lim(%)~

X% = h t.,"(~ × (< x {p})) mkAJ

[J <% X .)

~ < %) ÷ m(~) e X%).

Claim B.

hm(9)

We have:

(a)

definition

e Jr' where

So, as <Jp, A> is

that, as ~ e S and g is onto, k(9)

also

only breaks

A ~ Jm(~) >).

sups< 0 £(~)

Suppose not.

= ~o

Then sup~
< P, for if not,

then

=

-109-

~v<0

Xv = hp"(~ × (K x {p})) = Jp, so supv<0 ~(~) = ~.

But look, to say that

SUPv<8 m(v) < p is to say that the recursion never broke down below <, i.e. that 0 = ~.

Let ~ = SUPv< ~ £(v).

Since ~ < ~ s S, there is y < a such that I~I JY ~ <.

Since g is onto, we may pick ~ < ~ with y < g(w) < a.

Now, r s dom(g), g(~) > ~(~),

and I~(~)I Jg(~) ~ ~, so by definition of k, k(r) g ~.

So, as k is monotone, k(~) = w.

So, by (a) above, £(T + i) > g O k(r) = g(r) > ~ = supv< ~ £(v), which is absurd, proving claim B=

Hence, set C

= {~(v) Iv < ~}, a closed unbounded subset of a.

This completes the definition of
I a ~ S>.

We must verify condition

(iii)

of the theorem. Let ~ < a be a limit point of C . % < e with ~ = ~(%)o

Then a must fall under Case 3 above.

Since ~ is a limit point of Ca, ~ = ~(%) = supv< % ~(v).

definition of k and inequality

(a) above, we see that ~ c S.

Pick By the

Hence C-- is defined.

We show that C-- = ~ n C . Note that since lim(%), the definition of m ensures that <Jm(%), A q Jm(%)> is amenable.

We shall use this fact without mention in what follows.

Claim C.

~

X%.

Since ~ = £(%) = sup~<% E(v) = sup~<% ~x ' it suffices to show that v < X -~ ~X

--

X%.

required,

But look, v < % ÷ ~xp c X%, and since a ~ S we have J~ IJm(%) ~ K, SO as K = X%-~y ~ < % "*" e v

Let ~

-i

~ : <X%, A r~ X%> ='~ <J~, ~>.

~I

~

X%, a s

v

Thus ~: <J~, ~>'<~I <Jm(%)' A (I Jm(%)> and

~: <Jp--, A>-o Let ~: J~-<~ 7.20)°

J6 be the canonical extension of <J~, ~> ~

(If n > l~Ithen in fact ~: J~ "<~n-I JB') Thus, in particular, ~ = p~-I , ~ = A~-I . By claim C, ~ 7

Set ~ = - I (p).

= id[7, so ~(7) = e.

Let ~ be the canonical ~i skolem function for <J~, ~ . = -i

A> (by lemma I0 <Jp'

Clearly,

o hm(~) o ~ and ~ has the canonical ~i definition y = ~(i

~(z, y, i, x), where H = ~-l"Ho

x) +-~ ( ~ z

~ J~)

-Ii0-

The following diagram sums up the situation:

k=A

i

n

Claim D.

p[ =
As in claim A, we must show that

~(<) ~ [n(J~) ~ J~.

Define a map g from a

subset of < into ~ by T = ~(9) ++ ( ~ z c Jm(%)) G(z, T, n). Then g is A <J--, %> Jm(x) > ~i <Jm(%)' (p, ~)o But g _~ ~ x ~, so -I,,~ = g, so g is ~I P (~' ~)" nn-I Since 7 = p I and ~ = A-p ' this means that ~ is ~n(J~). But by definition of m, k"% ~- dora(g) and g~(k"~) = gI(k'hj. v < % ÷ ~ in -~.

< g o k(~) < ~

~+i

So as, ~(~) < g o k(v) < ~(~ + I) for all v,

, so g o k"% is cofinal in ~

x~

= ~, whence g is cofinal

Now, "~: J~-
regular".

Hence g ~ J~o

So, as ~ g S, this easily implies

~D(~) ~ ~n(j~) ~ j~, as

required° Claim E,

[ = $(~).

If ~ ~ ~ < ~, then ~ is regular at ~.

For if not, J ~ w o u l d

contain a cofinal

map of a bounded subset of ~ into ~, whence, as ~: J ~ y $ @ , ~ w o u ~ c o n t a i n map of a bounded subset of ~ into ~, which is not true. of

a cofinal

But look, ~ is a ~n(J~) map

a subset of K cofinally into ~, so claim E is proven.

Claim F.

n = n(~).

By claim E and the existence of g, n(~) ~ n. already.

Suppose n > I.

So, if n = i, claim F is proved

We must show that n(~) B n, i.e. that ~ is In_l-regular at

-IIi-

~.

Well, suppose not.

cofinally into ~.

Then there is a ~n_l(J~) map of a bounded subset of

Since ~j~ "~ is regular", this map cannot lie in J~.

So, as

~" ~ S, we conclude that e(<) ~ ~n-I (J~) ~ J~" Then, aS in claim A, 0~ -I = <. n-I by claim C, 0~ = p >~ ~ > ~, which proves the point and the claim. Claim G.

But

~- = p(~--). n--I

By claims D, E, F and the fact that ~ = P~

, P(~--) = the <j-least p' such that

~'(~ × (m × {p'})) = J--. But look, ~"(m x (K × {~})) = ~"(m × (~ × { --l(p)})) = P #-I ,,ha(%) ,,(~ × (K × {p})) = - I "X% = J~. Hence p(~) ~<j P. Again, let p' = ~(p(~)). Pick i ~ ,

~ e ~, with p = h(i, ).

h"(~ × (~ × {p})) ~ h " ( ~

x (K × {p'})).

Then p = h(i, <~, p'>), so

Hence h"(~ × (< x {p'})) = Jo" whence

p ~<j p'.

Thus ~ ~<j p(~), provin~ the claim.

Claim H.

~ is defined from ~ exactly as g was defined from ~.

Define g' from ~ as g was defined from e.

Since ~ is the canonical ~I skolem

function for <J~, A>, claims D, E, F, G show that this means that g' ( ~ T(e ~-) +-+ ~(i, <~, ~>) = r(E ~). ~: <JF' A>'
So for ~0 e K, T e ~, since

<Jm(%)' A ~ Jm(%)> and ~rI7 = id[7, we have g'(~) = T +-+h(Fl(~),

) = T +-~ hm(%)(Fl(~), ~(~) = r.

+ i) =

) = T ~-+(~Z C Jm(%)) G(z, T, ~) *~+

Hence ~ = g', proving claim H.

Using the above results, a straightforward

induction proof shows that if we

define k, ~, m from ~ as k, ~, m were defined from ~, then for all ~ < %, we have ~(k(~)) = k(~), ~(~(~)) = %(~), ~(~(~)) = re(v).

EFor example, if T < % and lira(T),

then as m(T) e X%, the m-induction step here is ~(m(T)) = ~T(suPi< T m(i)) = the least e X% such that ~ > re(i) for all i < T = m(T)o]. ~

= id~,

But ~ < % ÷ %(~) ~ ~, so as

~ < % -+ ~(~) = 9~(~J). Also, the k, ~, m recursion must terminate at %.

Thus C--~ = {~(~)

I ~ < %} = {~(~)

It r e m a i n s o n l y t o p r o v e t h a t

I ~ < %} = 7 n Ca, as required.

~" ~ Eo

For this,

it

suffices

t o show t h a t ~-

cannot fall under Case 2.

By claims E and F, if ~ did fall under Case 2 we would

have n = I and succ

Thus ~ = ~

~:

(~).

J~'<~l Jm(%)' so this is absurd.

n-I

= ~

0

= ~ is a successor ordinal.

The proof is complete. |

But

-112-

Theorem 4 Assume V = L.
Let K, S be as above.

There is a set E ~ K + and a sequence

~ S> such that: +

(i)

E is stationary in K ;

(ii)

C% is closed and unbounded in % ;

(iii) If cf(%) < K, then ICxI < K; (iv)

If y is a limit point of C%, then y s S, y ~ E, and C = Y O C%. Y

Proof: Let E, be as in Theorem 3. So, if ~ is regular, we are done.

Thus, for each a E S, otp(C)

Assume, therefore, that ~ = cf(K) < K,

shall replace by a new sequence as follows. satisfies Case I or Case 2, set C*a = Ca.

Otherwise,

normal sequence of limit ordinals cofinal in K. consider.

~< K.

If

We a

let be a

There are two cases to

Set y~- = K for convenience, i

Case A.

YT < °tp(Ca) ~< YT+I for some T < K. Set C*a = {8 e C a I otp(~ ~ Ca) >~ yr}.

Case B.

otp(C a) = yl for some % Z ~ with lira(X). Set C*a = {8 e Ca [ ( ~

If

< %)Eotp(B • Ca) = y ~ } .

Ica[ = K, then otp(Ca) = ~, so by Case B, [C~I = ~ < K.

Thus to show that

satisfies the theorem, we need only prove that if ~ is a limit point of C*a' then ~

= ~ ~C*.

Now, if ~ is a limit point of C*a' it is a limit point of Ca, whence a ~ Ca = O--.a Suppose 0~ satisfies Case A at r. otp(~ ~ Ca ) > YT"

Since a ~ C'a, we must have

And of course, otp(~ o C a ) ~< otp(C a) ~< YT+I"

YT < otp(C~) ~< YT+I' SO ~ satisfies Case A alsoo

a -'- {B E C--a [ otp(8 a C-~-) >~ yT} = {8 ~ ~" ~Ca

]

Hence

Therefore, otp(~ ~

~C a) >_. yT} =

{B g ~- ~ C a I otp(~ ~ C a ) >~ yT} = ~ h {8 E C a I otp(B N C a) >~ YT } = ~ ~ C*.a Now suppose a satisfies Case B.

Then, since ~ is a limit point of C*a' we must have

otp(~ ~ Ca) = y% for some limit ~ < ~. Hence 0~ = {8 ~ C~- I ( ~ v

Thus, otp(C~) = y% and ~ satisfies Case B also.

< %)[otp(B n G~) = yv]} = ~ ~ {8 ~ C a I ( 3u < %)

[otp(B ~ G a) = yu]} = ~ ~ C a.

-113-

Let < be any infinite cardinal, and let S be as above. all half-open intervals [~0' rl) ~ S = ~ . +

Let ~ be the set of

[T0' rl ) such that Y1 e S and Y0 is least such that

(Since S is closed, T0 is always a successor ordinal.)

Clearly,

Theorem 5 Assume V = L.

Let <, S be as before, and let

~ be as above.

Let I s ~ .

There is a sequence such that: (i)

ClI is closed and unbounded in I;

(ii)

If of(%) < <, then IC~l < <;

I then y e I and C YI = ¥ ~ C (iii) If y is a limit point of C%, Proof: Let <~

.

I v ~ 8> be the monotone enumeration of all limit ordinals in

I U {sup(l)}.

For each v Z 6, let ~9 be the set of all possible sequences

=
I T Z ~>. By induction on ~ Z ~, we prove the proposition T P(v) £ ~vC ~ ~ and for each r < v, each ~ s C , there is ~' E ~)C such that ~ ~'.

P(0) is trivial, since i 0 = T O + ~ (where I = [TO, TI).) as

Succ(T0).

Assume P(v).

Let

~ =
I r ~ 9> E

~ o

We can extend

~ to

IT ' in N~+IC by setting

~' = , where CIg+I = I~+l - %~ (so

CI~+I is just an ~-sequence). Assume lim(~) and ( ~ n such that

g E

~'.

< ~) e(~).

~

Let T < ~,

~ ~ C .

We seek a ~'

in C

Let 0 = ef(v), and let <~. I i Z 0> be a normal sequence i

such that ~0 = T and ~9 = 9" ~0 ~ ~I

Hence P(~ + I).

"'° and

Define a sequence < ~i I i Z 0>such that

as follows. Set ~0 = ~ ° Let ~i+l e ~ i + l I be such that ~ i ~ ~i+l by hypothesis. For limit i Z 0, set ~* = ~j<[ i ]' and extend ~ * i to

~i e C

~i8 C

by setting

= ~*i ~{<~j

~i

I ~ < i>}

Thus P(~)

i By P(6), there is a sequence

~ = e C~, which satisfies the

theorem.~ The next result proves that V = L + (~<)

~

K

in a very strong way.

Theorem 6 Assume V = L, and let < be any infinite cardinal.

Then ~ <

holds.

In fact,

-114-

there

is a set E -c <+ and a sequence


I % < <+ A lim(~)>

such that:

+

(i)

E is stationary

in < ;

(ii)

C A is closed and unbounded

(iii)

If cf(%)

(iv)

If y is a limit point of CA, then y ~ E and C

< K, then

in ~;

IC%I < <; = y c~ C%. T

Proof:

Let S, ~ be as above.

Let E,
I % ¢ S> be as before

also.

For each I E ~ +

let be as in Theorem ~ I for some unique

5.

I I ~ J , so set C% = C%.

If lim(%) Then
and ~ ¢ ~

- S, then

I % < <+ A lim(%)>

clearly as required. | Remark In chapter

Skolem theorem above will

12 we s h a l l

of model

use

theory,

~:

to establish

in L.

However,

b e o f u s e t o us i n t h e n e x t

chapter.

a strong

the strong

v!

form of the Lowenhelm-

form of

O K established

is

,

Chapter 9

THE GENERALISED

Recalling

Chapter

regular cardinal. cardinality Hypothesis

<.

SOUSLIN HYPOTHESIS

FOR SUCCESSOR CARDINALS

3, we make the following definitions.

A Souslin <-tree is a normal

(<, <)-tree with no antichain of

(Such a tree can clearly have no <-branches

for <, SH(<),

is the assertion

by an extension of the earlier argument,

The first step, therefore,

either.)

The Souslin

that there is no Souslin <-tree.

chapter 3 we have already shown that if V = L, then SH(ml) we shall prove,

Let < be an uncountable

is to prove a general

fails.

Thus,

in

In this chapter,

that V = L +

( ~ < ) ~ SH(<+).

form of the combinatorial

principle

of chapter 3. Suppose < is an uncountable by S

O<(E)

we mean the following

regular cardinal.

assertion:

If E ~ < is stationary

There is a sequence

<S I~ ~ E> such that

= ~ and for every X ~ <, the set {~ e EIX ~ ~ = S } is stationary

of chapter 3 is just

in m, then

in <.

Thus,

0~i(~i).

Theorem I Assume V = L. set stationary Proof:

Let < be any uncountable

in <.

B~ i n d u c t i o n

Then

~

<

<S , C > a s t h e < j - l e a s t

C

c ~, C ~ is closed and unbounded

S

= C

either

in a, and T s C

succ(~) or else

Suppose <S I~ e E> is not as required.

lim(a)

S i n c e <Sc~l~ e E> i s J K + - d e f i n a b l e

N

~J<+, 0

pair

~ E + S

Set ~

~ y ~ S , with

a s above e x i s t s .

in < and ~ e C ~ E

-> S h e

~ S .

Define submodels

~ < <, as follows: ~ < E On and E ~ N.

Ng+ 1 = the smallest N - < J<+ such that N n < e On and N ~

S ,

Let <S, C> be the <j-least pair of

f r o m E, so i s <S, C>.

= the smallest N -~ J + such that N <

Nl =

such that

b u t no s u c h p a i r

subsets of < such that C is closed and unbounded

N

and let E = < be any

(E) holds.

on a E E, d e f i n e

= ~ if

regular cardinal,


u{N

} ~ N.

, if lim(%). n <.

Since < is regular,

i s a n o r m a l s e q u e n c e c o f i n a l

in <, whence Z = {~ I~, = ~} is closed and unbounded

in <.

Thus E ~ Z ~ C ~ ~.

-116-

Let ~ = ~

e E ~ C.

~(<S, C>) = KS n ~ ,

Let ~: N C n~>

z-l: j s _ < j<+, <S ~ ,

~ Js"

and ~(<S

C ~>

<S ~ ~, C ~ ~> = <S , C >.

cardinal

the choice of <S, C>.

application

<.

Now,

of ~ +(~+), <

Im ~ ~ q E>.

Since

in ~, and y g C' q E ÷ S' ~ y + S . Y S q ~ = S .

Hence

But ~ E C ~ E, so this

QED.

in L, of a Souslin <+-tree for an arbitrary

along the lines of theorem 3.8 will not work, because

are any cofinal branches,

generalised

n(<) = ~, and

if < > ~, it is easy to see that a straightforward

limit stages m in the construction,

the form S .

Im e E>) = <S

In particular,

We turn now to the construction, infinite

= id~,

is the <j-least pair <S'~ C'> such that S',

C' ~ ~, C' is closed and unbounded

contradicts

Then z ~

at

unless cf(m) = ~, we will not be sure if there

let alone the ones we require to kill any T~e antichain of

It is T h e o r e m 8.6 w h i c h

(together with the strong form of

above) enables us to overcome

~ w h i c h we

this difficulty.

Theorem 2 Assume V = L. Proof:

Let < be any infinite

+ Then there is a Souslin < -tree.

cardinal.

Let E ~ <+, be as in T h e o r e m 8.6. <Sale e E> satisfy O<+(E)

for this E.

We construct

induction on the levels so that for each ~ < <+ The elements

a Souslin <-tree, T, by

T~m is a normal ~

(m, <+)-tree.

of T will be members of <, and we shall have m
set T O = {O}. ordinals

,

By T h e o r e m I, let

If T m is defined,

as extensions

T~a is defined.

of each member of Tm.

We must decide w h i c h

extend in order to obtain T . m T~e is a normal

T + 1 is obtained by appointing Suppose,

finally,

two new that lira(a) and

(if there are any:) a-branches

By induction hypothesis,

(m, <+)-tree of course.

We

of T~a to

we are assuming

that

We first show how we shall attempt to

associate with each p o i n t x e T~m an a-branch b e of T~a with x e bm. ~ x x Let x e T~m be arbitrary.

Let be the monotone enumeration

Let ~(x) be the least ~ such that x e T~y of elements x P~(x)

of T~~

= the least

.

Define a sequence

as follows: (ordinalwise)

y e T~(x)

such that x ~
of Cm.


~< ~ < %>

-117x x = the least y E JY~+I such that p~ ~T y" P~+I x pq = the unique y E Tyq such that for all ~ < q, x P~ ~T y' if it exists (otherwise undefined). x . If the above definition breaks down (i.e. if p~ ~s undefined for some limit ordinal q < ~) for some x, then our entire construction of T breaks down. (Fortunately, however, once we have finished describing the construction of 7' we shall be able to verify that this state of affairs does not arise~) Otherwise, and let us assume, for the time being, that this is the case here, we set bex = {y ~ ~ e l ( ~ We define T each b~x' x s T~e. ~

Otherwise, let T

e

an element of Se.

< %)(Y ~r P~)}' an a-branch of T~e with x E b .x e

as follows.

If e ~ E

If e ~ E and S

let T

consist of one-point extensions of

is not a maximal antichain of T ~ ,

do likewise

consist of one-point extensions of each b ~ such that x lies above x (Since Se is a maximal antichain of T~e, T~e+l will still be normal-

see the proof of theorem 3.8.

This is because, for each x ~ T~e, there is x' s S ~

such that x and x' are T-comparable, whence, letting x" be the T-maximum of x, x' , the one-point extension of the ~-branch b~,, will be an extension of x on T .) x This completes the definition of T

in this case.

We set T~ = ~ < < + T ~~ e .

Provid-

ing the definition of the p-sequence did not break down at any stage, T will clearly be a normal (<+, <+)-tree.

(In fact, it is easily seen that for all ~ < K+,

Irel Z lel, so r has width K+ in " a very strong way.) some p-sequence did break down.

Well, suppose the definition of

Let e be the least limit ordinal such that for some

x e T~e, the sequence was not well defined, where ~, , etc. are as above. defined.

x was not D

Now, since lim(q), yq is a limit point of Ca, so Yn # E and

Cyq = yq ~ C
Let n < X be the least (limit) ordinal such that p

= {y~l~ < ~}.

Thus, if we define from Yn as

~(x) ~ ~ < X> was described above to be defined from a, then for all ~ < n,

x q~x = P~"

But look, determined the y -branch b Yq, and since x

yq # E, we know this branch had an extension on Tyq, by construction. sequence has an extension on Ty . have a contradiction.

Hence the

But then p~ is defined, and we

-118-

Hence T is a well-defined, Suppose

otherwise.

C = {~ e K I T ~

ha~

normal

Let A be a maximal

(K +, ~+)-tree.

We show that T is Souslin.

antichain of T of cardinality

domain a subset of ~ & A Q e

is a maximal

<+.

Set

antichain of T~a}.

~

Clearly,

C is closed and unbounded

s C ~ E such that A Q ~ = S . element extended since

IAI =

in <+.

But T

some member of A ha.

Hence, by choice of <S

was constructed

]~ c E>, there is

(in this case) so that every

Thus, picking x E A with ht(x)

+ ), there must be y E A q a such that y
that A is an antichain of T, and so we are done.~

~ ~ (possible

This contradicts

the fact

Chapter i0

INEFFABLE CARDINALS AND THE GENERALISED KUREPA HYPOTHESIS

For this chapter and the next, the reader will find it helpful if he has some acquaintance with the notion of a weakly compact cardinal. consult [8] .)

(For full details,

In the meantime, we shall content ourselves with a brief review of

the definitions. Let ~ be any uncountable cardinal. ~n>l~l <...< e n < K}. homoseneous

For n s ~, we write [
If f: [Kin ÷ p for some cardinal p, we say X ~ m is

for f if If"[x]nl = I.

n If X is some ordinal, we write K ÷ (X)p tO mean

that whenever f: [K]n ÷ ~ there is a set X ~ K of order-type % which is homogeneous for f.

(These notations are due to Erd~s.) We say K is w e ~ ! y

compact iff K ÷ (K)~.

is equivalent to the compactness languages.)

(The name arises because this property

theorem for certain infinitary first-order

Weakly compact cardinals are always inaccessible,

then the first inaccessible. inaccessibles beneath K. weak compactness.

but are much larger

In fact, if K is weakly compact, there are

There are many equivalent formulations of the notion of

One of these is that K will be weakly compact iff < is

inaccessible and every (K, ~)-tree has a K-branch. i b le. Another equivalence is that K will be weakly compact iff it is]~l-indescriba It will be recalled that K iS~In-indescribable

iff whenever ~(UI"'''

Un' W) is a

first-order formula of the language of set theory with the additional unary predicate symbols ~1 .... , Un, W and W c V --

(QU n ~ (~N

I

V K)~VK, ~ VX)(~U2G

is such that ( ~ U

~ VK)(~U I

2

~ VK)(~U

~ V~) .0. 3

s, U ..... Un, W> ~ ~(U ..... Un, W)~, then there is % < < such that I I VX)(~U

3

~ V X) ... ( q U n ~ vx)[

k ~(~ ..... Un, ~)~" The following slightly stronger notion is due to Jensen and Kunen. (I

We call

I. To be precise, the notion of ineffability itself is due to Jensen, who defined the concept in terms of the property expressed in Theorem i. Kunen later obtained the characterisation of ineffability in terms of partition properties, which we have here adopted as our definition.

-120-

a cardinal K ineffable

iff it is regular,

uncountable,

there is a stationary set X ~ < which is homogeneous cardinal

is weakly compact,

if m is ineffable,

then it i s ~ - i n d e s c r i b a b l e ,

exceed the first ~2-1ndescrxbable

cardinal which isT[l-indescribable n following equivalence Theorem I

of ineffability

f: [<]2 ÷ 2,

Clearly, every ineffable

In fact, Kunen has shown that

whence it is easily shown that there

(It can also be shown that ineffable cardinal,

for all n.)

and in fact exceed the first

For our present purposes,

the

is the important one.

(Kumen)

Let < be a regular uncountable
for f.

The converse is not true.

are < weakly compact cardinals beneath <. cardinals

and whenever

I ~ < <> is such that A

{~ ~ < I A

= A ~}

cardinal.

Then < is ineffable

iff whenever

c ~ for all ~ < <, there is A =_ < such that

is stationary in ~.

Proof:~+-~Let f: [<]2 ÷ 2. For ~ < K, define f : ~ + 2 by f (~) = f(v, ~), ~ s e. Identifying

~D(~) with ~2, our hypothesis

X = {~ e K I f

= ~I~} is stationary.

Then ~[X is regressive ~=

We may assume X ~ 2 = ~ of course.

on X, so by theorem 3.1 there is v 0 < 2 such that

{~ e X I ~(~) = V0 } is stationary.

f(~, ~) = f~(~) = ( ~ ) ( ~ ) {--~) Let
gives an ~ E <2 such that

= ~(v) = ~0' so ~f is homogeneous

I ~ < <> be such that f

graphic ordering on the f 's°

But if ~, ~ ~ ~ , v < ~, then

~ ~2 for all e < <.

Let <* be the lexico-

Define h: [K] 2 + 2 by setting, I 0, if f

h(~,

for f.

for ~ < B < <,

<* fB

~) = I, if f~ <* f .

By ineffability,

there is a stationary

e, ~ e X & ~ < ~ ÷ f~ <* f~. follows. (~8

For each 9, let ~

e X)(8 >~ ~

v

Define a sequence < ~

I ~ < <> of members of X as

be the least ordinal in X such t h a t e

~ f vI~ = f s ~ ) .

can always be found, of course.) ~

set X c < such that, say,

>~ ~ and

(Since ~, S e X & ~ < B ~ f~ <* fB' such an ~ Note that if lim(v) and <e

n

I D < v> is defined,

will in fact be the first member of X not less than supq
C = {y e K I v < Y -> ~9 < Y}, a closed unbounded subset of K. so is -~= X ~ C ~ {v g K I lim(~))}.

Thus, v e ~

-> ~

Let

Since X is stationary,

= ~), by our above remark.

-121-

It follows that for any ~ e Y , ( ~ 8

e Y )(8 ~ e ÷ f

therefore, we have f e <2 and {~ e < I fe = f ~ }

= fs~).

~ ,

Setting f = ~ e ~

which is stationary.

f~'

The

lemma is proved. I Recalling chapter 4, we call the following assertion the Kurepa Hypothesis for i, KH(<), for K any uncountable regular cardinal:

There is a family

~ ~ ~(<) such

+ that I~ I = K

and for all e < K, {X ~ e I X e ~ } has cardinality at most Ie I + ~.

Thus, in chapter 4, we proved that KH(~ I) holds in L.

In this chapter we shall prove

that if V = L, then KH(<) holds iff < is not ineffable.

Since KH(<) can be formulated

in terms of (<, <)-trees, much as in chapter 4 for the case < = ~I' this gives an equivalence, in L, of ineffability in terms of the non-existence of "very fat" (<, ~)-trees.

(In the next chapter, we shall give an equivalence, in L, of weak

compactness in terms of the non-existence of "very thin" (<, K)-trees, namely Souslin <-trees.).

Half of this equivalence is already true in ZFCo

From now Theorem 2

on, < will denote an arbitrary, uncountable regular cardinal.

(Jensen; Kunen)

If K is ineffable, then -I KH(<). PrOof: Assume < is ineffable, and let < < ÷ ]{x ~ ~ I x e ~}I ~< I~I

~ c + ~.

~(<) be such that We show that I ~I ~< K.

For each e >~ ~, let enumerate {x N ~ I x c ~ }, and set R e = ( I T ~ f~}. there

is

R ~

<2 s u c h

each 9 < K, set f Let f e ~

.

that

Then R - ~ 2

so (since I(~21 = Ic~I) by theorem I

X = {c~ ~ < I R f3 ~2 = R } i s

= R"{~}.

We show that ~ ~ { f

stationary

in

K.

For

1 9 < K}-

If f ~ fv for any ~ < <, then since < is regular we can find a

closed unbounded set A c K such that ~ e A ÷ ( ~ X is stationary, there is ~ ~ X r~A. rl R"{v} = R " f v }

< ~)(f ~ ~ # f

For such an ~, v < ~ ÷ f ~

(~ ~).

Since

# f ~ ~ = l}

= {T I T E f~} = fe.~ Hence f r~ e ~ {f~ I ~ < e} =

{x ~ ~ I x e ~ }, contrary to f e ~ .

Hence f = f for some ~ < <, as required. l)

The theorem is proved. I There remains the proof, that, if V = L and K is not ineffable, then KH(<) holds.

-122-

Now, this can he done by means of an argument generslising that of theorems 4.3 and 4.4.

However, it is possible to establish a much stronger form of KH(K) from the

same assumptions here, so that is what we shall do.

And, as in chapter 4 for the

case KH(~I) , we shall find it more convenient to "reduce" KH(<) (in its strong form) to a generalised version of (a strong form of)

~+

for <.

The following Strong form of Kurepa hypothesis is due to Chang.

Let <, % be

uncountable cardinals, with < regular and % ~ <. Say KH(K, %) iff there is + ~ ~(~) such that I~I = < and for all infinite x s ~%(<), I{f ~ x I f s ~}I ~

Ixl.

Clearly KH(<, <) ÷ KH(<). fails°

(Here, we use ~D%(<) to denote {x ~ < I Ixl < %}.) In particular, therefore, if < is uneffable, then KH(K, <)

We shall prove that if V = L, then for all regular uncountable < and all

uncountable % < m, KH(~, %) holds, and if ~ is, in addition, not ineffable, then KH(<, ~) holds.

In order to do this, we shall define strong forms of a generalisation

of the combinatorial principle

¢ + of Chapter 4.

The following principle is known as such that S

~

~ +°

There is a sequence <S

I ~ < K>

~(~) and IS~I ~ l~I + ~ for all ~ < K, and whenever X ~ < there is

an unbounded set B ~ < such that e = sup(B n e) implies X n ~, B ~

s S .

By generalising the proofs of theorems 4.3 and 4.4 one can prove that V = L & + not ineffable *ZFC ~ < +ZFC KH(m)" However, we shall here consider a two-cardinal version of

0

which applies to the principles KH(<, %).

+ The following principle is known as ¢ ~ , % . <Sx I x ~ ~ ( K ) >

such that Sx ~

~(~x)

There is a sequence

and ISxl ~ Ixl for each x ~

~3%(<), and

whenever X ~ < there is an unbounded set B ~ K such that for any x s ~)~(<), if =~x

is a limit point of B ~ x, then X n ~, B ~ e s S .

Theorem 3

(Jensen)

For any regular uncountable cardinal < and any uncountable cardina] % ~ <,

~+(~, %) ÷ KH(~, %). Proof: Recall that H< is a model of ZF-.

Let <Sx I x s ~ % ( < ) > satisfy

For each x ~ ~I(K), let M x be the smallest M ~ H ~

0+(<, %).

such that x o {x} ~ M and

-123-

( ~ a ~ Ox)(S Set

CM).

x~

~ = {f ~ < I ( W x

satisfies KH(~,

e @%(<))(f

~x

s M )}. x

Clearly,

if

I ~I

~ <+ then

X), so in order to prove KH(K, X) w e shall assume

I~I ~

N

and work for a contradiction. Let
I v < <> enumerate

all unbounded members

set X ~ < such that for each e e X, lim(~) ~.

(Since < is regular,

B Q x, then X n a, B ~ a e S j ~

and if v < a, then f

(~x

e

Let
~%(<))

I ~ < ~}, an unbounded c~ B

Let x E ~ % ( K ) . x

is

subset of <.

imraediate.

Since f

the

So, as x c Mx,

{~ E X I ~ is a limit Set in eg+l

< <)(f ~ f ).

We

is finite,

suppose

f ~ x is

then since infinite.

Let

f ~ x o ~ E M ÷ f ~ x s M . x x

same way that

f was

f n x ~ B ~ M . x

Now, B i s

But clearly, defined

a limit

point

of

f ~ B is ZF--definable

f r o m X a n d B.

Hence

QED.

(Jensen)

Assume V = L. cardinals Proof:

is a limit point of

Then f q x = (f ~ x q B) U (f n x - B), so

f ~ x (and f ~ B), so X q B, B n ~ e Sx~ B ~ M x.

Theorem 4

procedure.)

~ ev+l is unbounded

If f ~ x

Otherwise,

limit point of f N x.

as f ~ x - B m u s t b e f i n i t e ,

f ~ B E M . x

is unbounded in

by showing that f e ~ .

this

from X ~ B and B Q $ in

~

Thus a ~ ~ B v < a~+l .

We prove that f q x ~ M . x

and M ~ ZF-, x

B be the greatest

~x

+ 1 < a ~+I for each ~ < ~, we see that ( ~

obtain our contradiction

x ~ M

[if e =

I v < ~> enumerate

point of B}, and for ~ < <, set B v = min(B - ~ ) .

but f ~ + i

Pick a closed unbounded

such a set is easily obtained b y an inductive

Let B n K be unbounded and such that

f = {~

of ~ .

< <,

~+(<,

For each x ~ Set S

x

For all uncountable

= ~ if x s

B qx,

< and all uncountable

let Mx be the smallest M -
I x ~

(~) and S = 60 x ~%(K)>

obtain a contradiction. unbounded

cardinals

~) holds.

~(~),

prove that <Sx

regular

~(L)x) n M

satisfies

~+(<,

x

if x s ~).

Suppose not.

Let X _oK be the
set B c K for which

then X ~ ~, B n ~ e M ~ .

( ~x

e

~(~))

[if

6D%(~) -

@

(K). t0

We

We shall

such that there is no

~ = Ox

is a limit point of

(Note that X is L<+-definable

from %.)

-124-

By induction on ~ < K, define submodels N O ~ N I ~

... - ~ L K + as

... ~ N

follows. N O = the smallest N ~ L K +

such that % g N n < s K.

N + I = the smallest N ~ L K + N6 =

~

<~

~ {N}

~ N and N o ~ ~ ~.

N , if lim(~).

Thus ~ < T < K
such that N

÷

N v ~ N T -< LK+.

Set a

I v < <> is a normal sequence in <.

o (~) = a v and ~v (X) = X ~ ~ .

= N v ~ <, each ~ < K.

Clearly,

For each ~ < K, let ~ : N v ~ L B ( v ) .

Set B = {B(~)

Clearly,

[ ~ < <} an unbounded subset of K.

obtain the required c o n t r a d i c t i o n b y showing that B satisfies

0+(~,

We

%) for X,

c o n t r a r y to the choice of X. So, let x ~ = Ux

~(<)

be arbitrary, but fixed from n o w on, and suppose that

is a limit point of B ~ x.

We m u s t show that X ~ ~, B ~ a s M . x

Note that since we clearly have B(~) < ~ + I

< B(~ + I) for each ~ < K, and

since ~ is a limit point of B, we m u s t have a = ~

for some limit ordinal n. n

from %, - In : LB(n) -~ L + ,

Now, X is L K + - d e f i n a b l e ~I(x

Q a n) = X.

Thus, X n a n

is L $ ( n ) - d e f i n a b l e

Again, B ~ a m is ZF--definable

-I n (%) = ~, and

from %.

from LB(n) and % in e x a c t l y the same way that B

was d e f i n e d from L + and % above. So, as % e M x and M x ~ ZF-,

[B(~) e Mx] ÷

IX N a , B ~ ~n ~ M j ,

and we are

r e d u c e d to p r o v i n g that B(~) ~ M . x To avoid confusion b e t w e e n ordinals and sequences of ordinals, B* denote <8(v)

[ v < K>, and for any T < K, set B*~T = <~(v) -I o a~

For ~ < ~ < K, set a T = o
Thus a ~T : L B (~) -< L B (~)"

I ~ < T < K> and for any i < ~, set a*~i =
from n o w on let

I v < T>. Let ~* denote

I ~ < T < i>. "~T

The p r o o f that B(n) e M

n o w consists of our e s t a b l i s h i n g a series of claims,

x

which we shall number for r e f e r e n c e as we proceed. (I)

v

e

n n M

x

÷

C~

Let ~ e n ~ M o x

~

B(9), B*~(v + i), a*I(~ + I) c M X . Since e

that ~ < T and ~(T) E x ~ M . x

n

= ~ is a limit point of B q x, w e can find T < n such Clearly,

~* ~ T and o* ~ T are ZF--definable

from

-125-

%, $(T) and ~

(= the largest cardinal of LB(~)) in the same way that B* and o* were

d e f i n e d f r o m X, ~+ and K.

(Warning:

It

i s n o t t h e c a s e t h a t t h e c o l l a p s i n g maps

corresponding to the ~ 's are the same in this situation. when the c o l l a p s i n g maps f o r t h e X, B(~), a

situation

t h e e l e m e n t a r y e m b e d d i n g s c o r r e s p o n d i n g to t h e o obtained.

This is all

X, ~(~) ~ M

easiiy

~T

's,

What is the case is that

a r e combined a s b e f o r e t o g i v e then just

t h e s e q u e n c e o*~r i s

c h e c k e d by i n d u c t i o n on t h e c o n s t r u c t i o n . )

and M x ~ ZF-, ~ * ~ , ~ * ~

e M .

x

But ~ ~ M .

x

Hence as

Hence

x

B*~(~ + i) = (~*~) ~ (~ + I) s Mx, a*~(~ + i) = (a*~) ~(v + i) ~ Mx, and ~(~) = (~*~r)(~) e Mx.

And as ~

is the largest cardinal of LB(~) , ~

~ Mx.

This

proves (1) o Let ~: M x ~ L~. ~* = ~ c ~ n M

Set ~ = ~"(n Q Mx).

~(B*~(~ + i)), x Since ~ is a collapsing isomorphism, the following are

~(~*~(~ + I)).

By (I), set ~* = ~ E ~ n M

x

easily

checked:

(a)

~ is an ordinal and ~* is an ~-sequence of ordinals.

(b)

~* is a system of maps indexed by the set {<~, 3> l ~ < T < ~}.

(d)

~ TE = $* ~

(e)

<(L~(~))~<~,

= ~(~

)o

(~ T)Vis

Note that <(LB(~))~
a directed elementary system.

(o T)Vis

transitive direct limit . of the system (e).

a directed elementary system with Let <, (~v)~<~>be

a direct limit

Define an elementary e ~ e d d i n g h: ~<~LB(~) , s> by letting

range over ~ in the following diagram:

LB(-l(v))

L~(~) Hence is well-founded.

.0

~

LB(D)

~



We may thus take U to be the unique transitive

-126-

limit of (e), so that U = L~ for some ~. For each v < ~, ~ -I ~

~ < ~(~) such that o ~ -I : L~ (v)- < L +, so there is ~v

~-l(v)

<

[= "~ is the largest cardinal". e~(v) v o -1 n o h: ~ - <

Similarly,since

L<+, let ~ < 8 be such that ~ L ~ " 7 is the largest

cardinal". Clearly, (2)

~

=

for each v < ~:

~(e _i(9)),'~ ( ~ )

= 7, and h(~) = ~ .

Also, we shall show that: (3)

~ < ~ ÷L~ To prove

s ~ o

~

= id~

(3), it suffices to show that v < T < ~ ÷ O

Then ~ = ~(~) for some ~ g ~ -I

~VT(~) = ~(q ~-~ (v),

(4)

o

- i (~) (~))

(~)

~ M . x

Now, o

~

= idlg .

-i (v),

Let

-~(~) (¢) = ~, so

= ~(~) = ~, which proves (3).

7 = supv< ~ ~v" This follows directly from (2) and (3).

(5)

z e @ ( ~ D ) ~ M x ÷ ~(z) e

@(~)

Note first that as ~ = e

=

A L~.

[Jx = supv< n e v "

and x ~ M x, e

= supve~n M

~ x

Let ~ e z ~ M

x

.

Then ~ g ~ , so by the above, ~ ~ ~

r~

(2), ~(g) e g~(V) and ~(v) < ~o But ~(z) = {~(~) (6)

By (4), ~ ( v )

V

for some v ¢ n ~ M

x

.

By

~< 7, so ~(¢) e 7.

I $ ~ z ~ Mx} , so (5) follows immediately.

If z e L ~ o L d and z is a bounded subset of ~, then h(z) = ~-l(z). By (4), z ~ ~

sufficiently

for all sufficiently

large v < ~.

large ~ < ~.

Thus by (3), ~ "z = z for all

Hence for all such v < ~, z s L~(v) and ~ (z) = z.

(This

follows from the definition of ~.) Again, by (2) we have ~-l(z)_~ e - I

for all sufficiently

large v < ~, whence

(v) o

- i (v), n

(~-l(z)) = ~-l(z)

h(z) = o

o -I

~-i(v), n

for such ~ .

o ~-l(z)

v

Thus, by definition of h,

= o

o ~-l(z) = ~-l(z), as required.

-l(~), n

-127-

(7)

If z ~ ~

~ L 6 and z c ~, then h(z) = - l ( z ) .

By (2), h(z) c_ ~r]' and by (4) and (i), ~-l(z) ~ ~q . that ~ O

<~

= D

g ~ M ~ , we see that h(z) = U sq~ M h(z) ~ e o = (by definition of ~) x -i x h(z) ~ '~ o ~(e -l(v)) = (by (2)) O <~ h(z) n ~ - l ( ~ ) = (by (6)) ~ h(z ~

U

<~

(8)

-i

(z) ~ - I ( ~

) = ~J <~

Similarly,

~-l(z) = ~

~-l(z) ~ x

-I

(z ~ ~ ).

The result follows by (6).

~(~ ) f~M x ~ LB(~). We have x e ~>(~ ) n M x and x is cofinal in ~q .

of ~ , B(q). (9)

Also,

Ixl < ~. < ~ n and ~ is a

Thus if x ~ LB(q) , then LB(~) = "~ q is singular" contrary to the definition

cardinal.

Hence x ~ L8(~) , which proves (8).

8-< 6. Suppose, on the contrary, that 6 ~< ~.

Let z ~

~(z) s ~(~-) C~L~.

Thus ~ ( z ) ~

h(~(z)) = -l(~(z))

= z.

here.

n M x _c LB(q) , contrary to (8).

(10)

So, recalling from above

Hence

~(~q)

~ and ~(z) E L 6 ~ L~.

In other words,

6D(~ ) f~ M x.

By (5),

So by (7),

z ~ ran(h) c LB(n).

But z was arbitrary

~(q) ~ M X. By (9), ~ E L~.

And of course L 6 ~= Mx, so L 6 = ZF-.

But clearly, ~*, 5" are

-

ZF -definable from B, ~ in the same way that ~*, ~ course, ~ is ZF--definable

were defined from K +, K.

from ~ as the largest cardinal of ~ .

It follows that - i ( ~ , )

and - i ( ~ , )

an end-extension of ~ q

and similarly for - I (5*) and o*~q.

since ~ L -i

are defined, and that, in particular,

is the unique transitive limit of <(L~(~)) <~, ( ~ v ) is the unique transitive

Hence B , a

And of ~ L 6.

-I(~*)

If - I (~*) = ~ ,

is

then

, we see that

limit of <(LB(v)) , so ~

-I

(~) = B(q),

(F) giving B(~) E ran(~ -I) = Mx, as required. and so B(q) is ZF--definable ran(~ -I (~*)) such that ~L x kJ { x } ~

M

B(n) s M o x

x

so ~ s M . x

from - I ( ~ , )

Otherwise, we must have B(n) = ~-I(~*)(D), and ~ as the unique element y of

"~ is the largest cardinal". Y -I Also, of course, ~ (~*) E M . x

But ~ = sup(x) and Hence, again we see that

This proves (IO).

But (iO) is what we set out to prove, so the theorem is established. I

-128-

In view of Theorems 3 and 4, we see that all of the principles KH(K, ~) for < K hold in L.

The situation regarding the principles KH(K, ~) is described

entirely by TheoremS 2, 3 and our next result. Theorem 5

(Jensen)

Assume V = L.

For all uncountable regular cardinals K,

~+(~, ~) holds just in

case ~ is not ineffable. Proof: Suppose K is ineffable.

Then by Theorem 2, -~KH(K), so -IKH(K, ~) of course,

whence by theorem 3, --lO+(K, K). Conversely,

suppose K is not ineffable:

argument very similar to the above.

Then we prove

~ +(K, K) by an

In fact the similarity here is so strong,

that we shall content ourselves with a brief description of the differences only. To begin with, let be the
a but

for any Ag

K, {a C K I A

= a N A} i s n o t s t a t i o n a r y

i n K.

Such a s e q u e n c e

exists by virtue of theorem I, of course~ For each x ~ ~)K(K), let M

x

be the smallest M -~L

such that

x ~ {x} ~ {A Ox } c M. Define <S

x

J x ~ ~

K

the models Mx, of course).

(K)> now exactly as before The verification

that

(using the new definition of

~+(~, K) holds proceeds as before.

The only difference is in the verification of claim (8), where it will be recalled we made heavy use of the fact that ~ < K in the previous case. the non-ineffability M . x

of K, and the correspondingly

Now we use, instead,

different choice of the models

We proceed, then, as follows.

(8)'

~(~n) ~ M x ~ L6(~). Suppose ( 8 ) '

fails.

In particular,

is LK+-definable,

Aa f~ Li3(rl). n so e Nn, whence

= aq() ~ LB(q).

therefore,

Thus X = {y E aq [ A y

Also,

= y ~A

} c LB(n)" q

Suppose is

stationary

[=e6(n )

"X i s

stationary

i n K", s o X* r e a l l y

is

in a

rl "

.

stationary

Then, setting i n ~.

X

,

-i = c~ ( X ) , I=. "X* q h +

But, setting

A = ~l(Aa

) ~ we

-129-

have X* = {y e < ] A y

= y ~ A}, since this is true in L<+ by elementary equivalence.

This contradicts the choice of
7

[ y < <>.

C g LBCD)_ _ such that I=LB(n)I"C is closed and unbounded in

and y c C ÷ A

~ y Q A s ". Let C* = a -I(C) A = ~- (A__) (as before). Then C* D N is closed and unbounded in K, A ! K , and y s C* ÷ A ~ Y N A, since all of this holds Y in L + by elementary equivalence. e

But C* O a n = on(C*) = C, which is unbounded in

(since this is true in L~,~),~ ~ so as C* is closed, ~

E C*.

Hence A

+ e n N A. n

But ~

~ A = on(A) = A

, so we have a contradiction.

This proves (8)'.

D The reader is left to check for himself that the rest of the proof goes through exactly as before. [

Chapter ii

WEAKLY COMPACT CARDINALS AND THE GENERALISED SOUSLIN HYPOTHESIS

In this chapter we shall prove that if V = L, then for any uncountable regular cardinal K, SH(K) iff ~ is weakly compact.

Now, we know already in ZFC that if < is

weakly compact, then every (K, ~)-tree has a x-branch, so if ~ is weakly compact, then SH(m) holds.

Hence, all we must prove is that if V = L and K is an uncountable

regular cardinal which is not weakly compact, then there is a Souslin K-tree.

If K

is a successor cardinal here, then this is already known by virtue of Chapter 9. Hence, in proving the above result, we need only consider inaccessible ~.

So, for the

rest of this chapter, we shall be assuming V = L and that ~ is an arbitrary inaccessible cardinal. K-tree.

Assuming K is not weakly compact, we shall then construct a Souslin

The idea behind the construction is much as in chapter 9.

Indeed, for such

K, we could establish a result for K which is very similar to the principle %+.

(In fact, Jensen does this in [11]).

~]% for

However, this would involve us in more

work than we need, so we shall content ourselves with proving the following characterisations of weak compactness in L: Theorem I (Jensen) Assum V = L.

Let K be an uncountable regular cardinal.

The following are

equivalent. (i)

K is weakly compact.

(ii)

If ~(B, D) is a first-order formula of the language of set theory with the

additional unary predicates B, D, and if B ~ K is such that ( ~ D then for some 6 < K, ( ~ D ~ (iii) If E = K

= ~)(~JK ~(B, D)),

~)(kj B ~(B ~ 6, D)).

is stationary in K, then for some ~ < K, E Q ~ is stationary in ~.

Proof: The proof that (i) ~

(ii) is entirely similar to the proof that ~l-indescrib-

ability (in its usual form) characterises weak compactness.

(In fact, (ii)

is clearly just a "constructibility version" of of Hl-zndescrlbabllzty.)

And

I the implication (i) ÷ (iii) holds in ZFC, being a simple application of NIindescribability.

We are left with the implication ~(i) * ~(iii).

Assume,

-131-

therefore,

that K is not weakly compact.

We must find a stationary

and, for each limit ordinal a < ~, a closed unbounded set C ~ a C ~E

= ~.

such that

Now, if < is a successor cardinal, Theorem 8.6 at once supplies us

with such an E and such sets C.

There remains the case K inaccessible.

Since < is not weakly compact, we can, by the above equivalence

(i) +-+ (ii),

find a formula ¢(B, D) of the language of set theory with the additional predicates (~B

B, D, and a set B ~ ~, such that ( ~ D

< <)(~D

set E ~

~ B)@j@

unary

~ <)(~JK ¢(B, D)) but

~ ¢(B ~ B, D)).

Let E be the set of all limit cardinals a < < such that for some B > a: (i)

a is regular at B;

(ii)

B is a-minimal;

(iii) B h a

~ JB & ( ~ D

~ @(~) ~ JB)(~ja ¢(B n a, D)).

We show that E is stationary in K. Let X be the smallest X - < J regular,~=Xn~ ~(B) = B o a .

~ ~.

Let C ~ K be closed and unbounded

+ such that s X and X Q K e On.

Let ~: X ~ J B .

Then @ < K, 7~a = i d ~ ,

Since C ~ a is unbounded

in a, a ~ C.

prove that C Q E + @, to show that a ~ E.

ill K.

Since ~ is

~(K) = ~, ~(C) = C Q a,

It thus suffices,

in order to

In fact, it is easily seen that a is

regular at B and B is a-minimal by means of the simple argument used in the proof of the corresponding (~D

~)(~j

part of theorem 8.3.

¢(B, D)), so as v

-I

: JB-< JK+,

K

to say ( ~ D Case I.

~ @(a)

OJB)(~ja

~J8

~(B N a, D)), giving

(~D

(iii), note that ~j<+ ca)(~ja --

(iii).

¢(Bsa

'

D)), which is

Hence E is stationary

cf(a) = ~.

Let C be any 0~-sequence

of successor ordinals cofinal in a.

closed and unbounded in ~ and disjoint Case 2.

For condition

Clearly C is

from E, which disposes of Case I.

a is not a limit cardinal.

Let k < a be maximal such that X = 0 or else k is a limit cardinal. C = a - (%+I).

Clearly,

as E consists only of limit cardinals,

unbounded subset of a disjoint Case 3.

a is inaccessible.

from E, so Case 2 is proved.

Set

C is a closed

in K.

-132-

Since ~ is regular, we may define submodels X x

= the smallest X - ~ J

+ such that {B N ~ } ~ { ~ I n

~ J +, ~ < ~, by setting: < ~} ~ X and ~

Thus, <X I~ < ~> is a strictly increasing elementary in particular, that C q E

<~ I~ < ~> is a normal sequence

= ~.

Let ~: X

~ JB"

Thus ~ r ~

= id~

such that ~j~ - ~ ( B

Hence ~j + ( ~ D

c

~)~=j~- ~ ( B

~ ~)[~j

other words,

chain continuous

in ~.

-~(B

N~,

, ~(~) = ~ D).

at limits, and We show

e E.

and, ~(B N ~) = B q ~v"

Now,

And any such D must of course lie in J +.

f3 ~, D)], so as ~, B N ~, ¢ X

~ ~, D)].

e ~.

Let C = {~ I~ < ~}.

Assume not, and suppose that ~ < ~ is such that ~

there is D c ~

~][~ ( ~ D

= ~ qX

Hence ~j~ ( ~ D

if y >~ B, there is D ~ ~(~ ) n J

~

)[~j~

~<J+, -n~(B q ~

, D)].

such that _kJa -, ~(B ~ v ,

In

D).

Thus,

Y by definition of E, since a

~ E, there must be a y such that a

regular at y and ~ is ~ -minimal. look, w-l: JB -<Ja+'

By lemma 8.2, ~

v

< 7 < B and a

is not regular at y + I.

is But

so lim(B) , whence y + 1 < B, so this means that there is a map

~ JB of a bounded subset of ~

onto an unbounded

subset of ~ .

map of a bounded subset of ~ cofinally into ~, which is absurd.

Hence ~-I(~) is a Hence C q E = ~, as

required. Case 4.

Otherwise.

Thus ~ is a singular limit ordinal such that cf(~) > ~. Theorem 3.8, let 8 = 8(~), n = n(~).

Using the notation of

(Thus B is the least ordinal such that ~ is not

regular at B and n is the least integer such that ~ is not Zn-regular P = {y ~ ~[[y >. ~i] & [y is a singular

limit ordinal]

at B.)

Let

& In(y) = n] & Ilia(8) -> lim(8(y))]}

We may now make the crucial claim for our argument: Crucial Claim.

There is a closed unbounded

set C _~ ~ such that whenever y is a limit

point of C, then y e P. Leaving the proof of the crucial claim until later, let us see how this enables us to dispose of case 4.

Let C consist of all the limit points of C.

cf(~) > m, C is closed and unbounded in ~. y is a limit point of C, so y ~ P.

We show that C q E = ~.

Let y e C.

Then

Now, by the argument of Case 2 of Theorem 8.3,

cf(~) > m implies that n > i or else lira(8). or else lim(B(y)).

Since

Hence as y ¢ P, we must have n(y) > i

But by le~mna 8.2, if ~ ~ E then n(~) = I and succ(B(~)).

Hence

-133-

y~E. We turn now to the proof of the crucial claim. We proceed much as in Case 3 of Theorem 8.3, and use the notation established there. Set p = p(~) = p~-l, A = A(~) = A n-I 6 . In-regular

at 6, p~ ~ ~ ~ p.

x e J?is If-definable

Since a is In_l-regular

In particular,

we may let p~J~ be such that every

in <Jg' A> from parameters

in e ~ p } .

Let h be the canonical [i skolem function for <Jp, A>. [l(<Jp, A>) definition y = h(i, x) ~-+ ( ~ z For r < p, set y = hr(i, x) ~-+ y, x ~ J

at 6 but not

Let h have the (uniformly)

e Jp)H(z, y, i, x), where H is [0 <Jp' A>

& ( ~z e Jr)H(z, y, i, x).

<Jr' A ~ Jr> is amenable, h r is its canonical

Thus if

[I skolem function.

Define a map g from a subset of ~ onto ~ by: h(i, <~, p>)~if h(i, <~, p>) ~ g(~

+ i) otherwise undefined.

As in Theorem 8.3, there is a (uniformly) G such that g(~) = r ~-+ ( ~z ~ Jp)G(z, Claim A.

[i <Jp' A>

(p; possibly e as well) predicate

r, v).

There is a cardinal y < ~ such that ~ n h " ( ~

x (y x {p})) is unbounded

in ~.

By choice of P, A, etc., there is T < ~ and a [ (<JP, A>) map f such that f"T 1 Pick i 0 e ~, ~0 s is a cofinal subset of a. Pick q so that f is [I <Jp, A>({q}). with q = h(i 0, )" such that y > T, ~0" course.

Since ~ is a limit cardinal, we can find a cardinal y <

We show that f"r ~ h"(~ x (y x {p})), which proves the claim of

Let X = h"(~ × (~ × {p})).

that X ~ I ~

<7p, A>.

It thus suffices

But this is immediate, If-pairing

Since {q} U (r + I) ~ X, it suffices to show to show that X is closed under ordered pairs.

since y is a cardinal,

and hence is closed under the Godel

function ~: On 2 +-+ On.

By claim A, we can find a cardinal y < a as such that sup(g"y) = e.

Fix

from now on as the least such. We define functions ~: e + y, ~: 0 ÷ e, m: e ÷ P for some limit ordinal e ,< by means of a simultaneous

recursion

(which breaks down when any of ~, ~ , m is not

-134-

defined)

as follows:

~(v) = the least r ~ dom(g)

- y such that i < v + ~(i)

< T and g(r) >

~(~) and

m('0) ~ h"(~ x (g(r) x {p})). ~(v) = sup(6 A X )

if < 6 (otherwise

undefined)

where X

'

'

and q(v) = max(y m(O) = max(y

if < p (otherwise

m(~ + i) = the least

(b) ~(~) < go~O0

Again,

x {p}))

(a) ~, go~ are monotone;

If

undefined),

~(~),

and A ~ Jm(~)

go~(~),

a J~ and

and

~(~), m(~)

(--]z ~ J )G(z,

s

go~(~),

~(~)).

[, m are normal.

then m(~) must have been defined,

is defined,

so is m(9 + I), and thus so is

n h m(~+l) "(w × (Jn x {p}))),

~(~ + I) = sup(e

for lim(~).

< ~(~ + i).

[(~) is defined,

if [(v)

such that p s J . r

n < p such that ~ > m(~),

h "(m x (go~(~) Clearly,

m ~ )

+ I, supi<~g=~(i))

+ I, r), where "[ is least

m(%) = supv<%m(~),

= h . ."(~0 x (Jq(~) x {p})) ~

hm(~+l)

is Jr-definable

for some T < p .< B).

ordinal

@ .< y, and e is determined

where

and hence ~(~)

is defined.

[(v + I) (this last because

~ = max(y

+ I, go$(~))

and

Hence ~, ~ , m are all defined

by ~ taking

the value

6 or else

m taking

on a limit the value

p (at some limit stage). Claim B.

sups<@ ~(~) = 6.

Suppose -~ = sups<@ [(v) < ~. X

=

~

<@X

= h"(~ x (J-~ x {p})).

[l<Jp, A>({p; =

possibly

Then sup~
6}), g"Y ~ X .

< p, for if not,

then

6 > ~f and p ~ X-~f[l <Jp, A>, so as g is

But sup(g"y)

= 6, so 6 = sup(X fl 6) = supu
6 < 6, a contradiction. We define,

by recursion,

@ ~< @ such that sup(t"@)

a normal

t: @ ÷ ~ for some limit ordinal

= 8.

t(O) = the least ~ < @ such that t(%) = sup~<%t(v),

function

(if ~ < p)6 e X

if < @ (otherwise

If n = I, t(v + I) = t(~) + I.

undefined)

and [(~) > ~I" for lim(l).

-135-

If n > i, t(~ + i) = the least ~ > t(~) + I such that J

Nh*"(~ x (Xt(~) x {p*}))

--~XT ' where p* = p~-2 ' A* = ABn-2 ' p, = P6n-2 , and h* is the canonical El skolem function for <J ,, A*>. P We show that t(~ + I) is defined whenever t(v) is. is when n > i.)

(Clearly, the only problem

This will show, of course, that lim(6) and sup(t"e) = e.

Y = ~ q h*(~ × (Xt(v) x {p*})).

We must show that Y is bounded in e.

Xt(.o) = hm(t(~))"(~ x (Jn(t(~)) x {p})).

Clearly, h ( t ( ~ 0 )

Let

Now,

e JB' so we see that JB

contains a map of a subset of ~(t(v)) < e onto (~ × (Xt(~) x {p*})).

But h* is

~n_l(JB), so there is a ~n_l(JB) map of a subset of n(t(~)) < e onto Y.

Since ~ is

~n_l-regular at B, this implies that Y is bounded in ~. Set C = { ~ot(~)l~ < 8}, a closed unbounded subset of ~. We show that if ~ is a limit point of C, then ~ e P. = ~(~), where lim(~), of course.

Suppose, in fact,

Let ~-I: <X~, A ~ X>> =~ <JD-' ~>"

~: <J_, ~>~0 P ~ > ~ ~I <Jm(%) ' A N Jm(%) >' so ~: <J--' P

<JP '

A>

Let ~: J~

the canonical extension of this Z0-embedding given by lemma 7.20. above, we see that sup(god"X) ~I~ -- ida7 and ~(~) = ~.

= ~, whence ~ c X%.

In particular,

Let p = ~-i (p), and set ~ = t -I (~).

Then JB be

~I

By inequality

(b)

therefore,

(For the latter here,

note that ~(%) = 7 s C = ~ot"e.) Claim C.

~ = 6(~).

Let ~ .< ~ < ~.

Then e is regular at ~.

For otherwise, J~ would contain a map

of a bounded subset of ~ cofinally into -~, whence, as ~: J ~ l

J6' ~(~) would be a

map of a bounded subset of ~ cofinally into ~, lying in JB" contrary to the choice of B. Define a map g from a subset of ~ into ~ by g(v) = T +-~ ( ~Z S Jm(x))G(z, Thus g is ~l<Jm(x) ' A ~Jm(X)> g c ~ x e, ~-i,,~ = ~. means that g is ~n(J~).

(p; possibly ~ also).

Hence g is ~I <j_, p A> ({p, e}).

Set g = g ~ (e x ~).

Since

Since ~ = p~-l, ~ = A~-I, this

But by definition of m, ~"% ~ dom(g) and g~(~"%) =

But v < % ÷ ~(~) < go~(~) < ~(~ + I) (see above) so g ~ " E ~(k) = ~.

~, ~)-

gI(~"%).

is a cofinal subset of

Hence g maps a bounded subset of ~ (namely ~"% c y < ~) cofinally into ~.

-136-

This shows that ~ is not ~n-regular at ~, which completes the proof of claim C. Claim D.

n = n(~).

We saw during the proof of claim C that ~ is not Zn-regular at ~. claim C itself, n(~) ~ n.

If n = I, then we are done.

that ~ is Zn_l-regular at 6. ~.

P ~

Let x E J~, be such that f is

P

~l<J~-*, A >({~,, x}). p*, ~*(x).

By lemma 7.20, = ~~IJ--,. P

<J ,, A*> and ~*(p*) = p*.

LI

Define f' in <Jp*, A*> by the same definition in parameters

Since f g ~

x ~, ~"f = f, so f ~ f'.

Let u = dom(f).

Since u is a

m # p , u ~ J-* (by amenability) and ~*(u) = u. %n_l(j~ ) set bounded in ---, P predicates "f is a function ~' and "dom(f) c=u" a r e ~ l < J ~ is a function with dom(f') 9 u. ~*(x) ~ X k a n d

We must show

Let f be any ~n_l(J~) map of a bounded subset of ~ into

n-2 , ~, = A~-2 , --* Set --* p = p~ p = ~ n-2 , ~

~*: <J--,, A*> ~ v

Suppose n > I.

Hence, by

' ~*>({p*, x, u}), and so f'

Hence f = f' , and f is [I

<Jp,, A~>

dom(f) is bounded in ~ = sup~<~(sup(~ ~ Xt(v))).

of t, f is bounded in ~.

Now, the

, ({p , ~*(x)}).

Thus, by definition

Hence ~ is [n-I regular at ~.

That ~ E P now follows immediately.

For, by the proof of claim C (i.e. by the

existence of g), ~ is a singular limit ordinal; by the definition of t, ~ ~ ~I; by claim D, n(~) = n; and by claim C, lim(6) + lim(6(~)). lim(B), then since ~: J ~ [ l

JB' lira(B) also.)

(This last is because if

This proves the crucial claim, and

hence the theorem. We can now fill in the remaining gap in our study of SH(~) in L. Theorem 2 (Jensen) Assume V = L.

Let K be an inaccessible cardinal which is not weakly compact.

Then there is a Souslin <-tree. Proof: We construct a Souslin ~-tree much as in Theorem 9.2.

Thus, the elements of

our tree, %, are members of K, and we have ~ < B + ~ < 6. T

The construction

proceeds by induction on the levels so that for each ~ < K, T [~ is a normal (~, <)-tree.

Set T O = CO}.

If T

is defined we shall define T + I as in

Theorem 9.2 (viz. by appointing two successors to each member of T~ in some canonical manner).

But

Thus, as before, the construction hinges upon the limit

-137-

stages.

For these we proceed as before, only things are a little easier in some

respects. By Theorem I, let E ~ < be stationary in K but such that for no limit ~ < < is E Qa

stationary in a.

Let <Sa Is < <> satisfy

a < <, lim(a), and T~a is defined.

0K(E).

If ~ ~ E, let T

~

(Theorem 9.1).

Suppose

consist of one-point extensions

(chosen from < in some canonical manner) of each a-branch of T ~.

Since < is

~

inaccessible

and IT~al < < (as T ~

is, by hypothesis,

a normal (a, <)-tree),

IT +If < ~, and so T~(a + I) is an (a + I, <)-tree. Otherwise we consider S a. define T b

as above.

Otherwise,

of T~e such that x s b X

~

is not a maximal antichain of T ~ ,

for each x e T~a we pick (if possible)

and b X

any trouble, of course.)

If S

N S

# ~.

then we

an s-branch

(This last requirement will not cause

X

To obtain T , we appoint a one-point extension to each

bx, x ~ T ~ ,

in a canonical manner.

construction

is whether T ~(a + i) is normal in each of the above three cases.

we know E ~ ~ is not stationary such that C Q E = ~. construction,

Well,

in e, so we can find a closed unbounded set C ~ a

We may assume y E C ÷ lim(T) here of course.

for each y E C, all T-branches

induction on y g C we can construct, x.

The only point we must worry about in the

Then, by

of T~y were extended on T . Y

Hence, by

for each x s T~a, an s-branch of T~a containing

Hence T ~(a + I) is always normal for limit a < <. Set r = ~ < K T ~ ,

in Theorem 9.2.

a normal

(<, <)-tree.

The proof is complete. |

That T is Souslin is proved exactly as

Chapter 12

THE GAP-I TWO CARDINALS THEOREM

For the whole of this section, we shall let ~ first-order

language with a (distinguished)

theorem for such languages infinite

~-structure,

such that originally

~ ~ ~

and I ~ I

= ~-

The fol~owing generalisation

(E2~).

to an ~-structure

that the Lowenheim-Skolem

~k[<~,

K> + <%, k>].

+

++

<< , ~> * <% "two-cardinal"

, %>.

of this concept was

Let us say an ~-structure

~

the interpretation

We write ÷ <<', ~'> if every ~ - s t r u c t u r e

example,

The Lowenheim-Skolem

of course, and says that if 0~ is any

= K and IU~I = X, where U ~ denotes

arily equivalent

~

unary predicate U.

countable

and if ~ is any infinite cardinal then there is an ~ - s t r u c t u r e

suggested by Vaught

iff I~I U inn.

is well-known,

denote an arbitrary,

of type <~', %'>.

has type of the predicate

of type is element-

Thus, it is easily seen, for

theorem stated above is equivalent

to the assertion

Now, it is easily seen that for no <, % do we have

+ + But what about << , <> + <% , %>?

analogue of the Lowenheim-Skolem

In other words, do we get a

theorem if we just consider pairs of

+ cardinals of the form (K , ~)?

For obvious reasons,

the statement

÷ <X +, %>] is known as the "Gap-i Two Cardinals Theorem'. this problem,

÷ <~B+y, ~B>]. theorem)

let us discuss briefly various other results on this topic in general.

Namely, he showed that if GCH be assumed, Since we clearly have

~e VB ~

+~, ~ >

<~ +~+~ , ~ > ÷ <~ +~, ~ > for any ~, 6, this completely solves the problem

selves with finite gaps.

, %> is false for all ~, %.

so, for example,

problem for the appropriate n.

to increase a ++ the statement

And by using the downward Lowenheim-Skolem-Tarski

theorem as above, the problem of decreasing

a finite gap can be reduced to a Gap-n

Thus the Gap-n Theorems play a central role in the

study of two cardinals models of ~ . such studies,

Hence we need only concern our-

It is easily seen that it is not possible

finite gap in a two cardinals theorem, + <%

then

in a very strong form,

(by the downward LSwenheim-Skolem-Tarski

when there is an infinite gap on the left hand side.

+++

K>

Before we go on to consider

In [2~ , Vaught proved the "Infinite Gap Two Cardinals Theorem" assuming GCH.

~K ~%[<~+,

To make life easier, we generally assume GCH in

since model theory becomes much more fun when there are saturated

-139-

structures,

etc. available.

Now, even without GCH, Vaught proved long ago that for

+

any

K ,

+ <~i'~>"

Later Chang proved that if GCH holds then for all

+ regular

k,<~l,m> + .

Thus, only the result

<~i,~> +

remains to be proved with regards to the Gap-1 Theorem. Jensen under the assumption due to Silver. Silver.)

V = L.

for singular

This was proved by

We shall give here a proof of this result,

(Jensen's proof was much more complicated than the later one of

Since this proof is really just a generalisation

of both that of Vaught

and that of Chang, we c~nmence with an outline of these results. First we give a sketch proof of

<<+,<> + <~i,~>.

We use "structure" to

mean "infinite structure" throughout. Recall that if T is a consistent Z of ~-formulas

set of ~ - s e n t e n c e s ,

a l-t[~e of T is a set

with free variable x (only) which is consistent with T. A model 6~

of T omits Z iff there is no a e 6Y such that well-kno~rn, and is proved by a Henkin-style

~

Z[a]. The following lemma is

diagonalisation

argument.

Lemma i. Suppose Z has the property that whenever%0 (x) is an ~ - f o r m u l a T ~(x) model of

+ o(x) for all o c Z, then T ~ T

which omits Z.

such that

Then there is a countable

~Bx~(x).

l

Lemma 2. (Keisler) +

Let

K

be a cardinal,

~an

countable, there is a countable

Proof:

~-structure ~ ~

Let T = Th(~ ,b)U {b -

T

b =

Ug = U %.

enumerate

< c l n < ~ ) , where c is a new constant.

is consistent

to find a countable model We use lemma 1.

Let

such that

~'l is

~

•

Let

n

Z = {U(x)}U {x ~ b nlb n EU ~ } . so

~ ~ ~

OZ= <<+ ,<, <,''" >.

We may assume

<~,b>,

~ ,

<< ,<>. If ~ -

of type

Suppose

~*

Clearly,

Z U T

is finitely satisfiable

and • is a 1-type of T. of

T

which omits

~ (x,yI ,..-,yn,z) is an

in

It clearly suffices

Z. G-formula

such that

-140-

T ~0(x,~,c) see

+ E(x), where ~ = bil,...,bin.

at once that

T~

(Vt)(3z>

By a simple cc~pactness argument, we

t)q0(br,~,z)

for any

r < ~.

+

Suppose that and

b = ~

T

~ -~ B x ~ (x,b,c), however.

enumerates ~ , and

T ~ ~t)(B

~

T~(~t)(~z

has type



z > t)g0(br,b,z).

Then, since

>t)(3x)q0(x,~,z).

Th~

,b)

is complete

Using the facts that

and cf(K +) > K, this implies that for some b s U~ , r

A contradiction, so we are done.

I

Theorem 3 (Vaught) +

For all Proof:

K, ~ <~l,~>.

By lemma 2 and a simple elementary chain argument.

Assume GCH frc~ now on.

|

We prove that if k is regular and uncountable, then

+ <~I ,~> ÷ <~ ,k>. all

K

(Together with Theorem 3, this will prove ÷ <~+,X>

of course. )

for

Let ~ be an arbitrary uncountable regular cardinal until further

notice. If 6~ is an ~ - s t r u c t u r e ,

~

denotes the language ~ with the members of 6%.

added as constants which denote themselves in Let ~Z be an ~-structure.

We say 6Z is saturated if whenever Z(V 0) is a set of

fewer than 16~I formulas of ~ , w h i c h simultaneously satisfied in ~ . whenever

"v 0 eU" e Z(v 0 ).

~Z .

is finitely satisfiable in ~ , then Z(v 0) is

We say 0Z is U-saturated if the above holds

The following lemmas are standard.

Lemma 4 (i)

If

~ is an ~ - s t r u c t u r e ,

there is a saturated structure

~-

~

of cardinality

k. (ii) If

~,

~ are saturated

Z-structures of cardinality X and

06 - ~ , then

Lemma 5 Let ~

be an <~i,~> ~-structure,

and let ~

be a saturated

~-structure of

-141-

cardinality ~ such that ~ cardinality % such that

~ ~.

~

Then there is a saturated ~-structure ~' of

~',

~+~',u

=u

, ~=~'.l

Also easy, is the following lemma, which depends upon U-saturation quite strongly. Lemma 6 Let

OLbe a U-saturated ~-structure of cardinality %.

saturated ~-structure

~of

cardinality % such that

Then there is a

0~ ~ ~ and U ~ = U !

I

Finally we have the key lermna for our proof. Lepta 7 Assume ~ has a binary predicate letter E. are just those ~-structures ~ in which E

Let T O be the ~-theory whose models

is extensional and, whenever H is a

there is x ~ U ~ such that H = {y[yE~x}.

finite subset of U

Let < ~ [~ < ~ < ~+> be

an elementary tower of U-saturated models of T

of cardinality % such that 0 ~ < 0-[~ is U-saturated.

< ~' < 8 * U ~I~ = U01~'. Then ~ =

Proof: Let ~(v0) be a set of less than % ~ - f o r m u l a s finitely satisfaible in ~ . which are

~

-formulas.

that ~L ~ ~ Ez ].

containing "v 0 e U" which is

For each ~ < B, let ~ Since ~

is U-saturated, there is z

Similarly there is a w e g U

{xlxE ~ w } S U ~ and ExE ~ w

consist of those ~ s

÷ ~ ~ l~xjj

~ U ~ such

such that {z I~ ~ ~ < B} The sets { x l x E ~ w

}, ~ < B,

form a collection with the f.i.p., since any finite number of them will contain z

for ~ sufficiently large.

that t e ~


}.

So, as ~-60 is U-saturated, there is t ~ U

Clearly,

such

~[~ ~Et]. I

Theorem 8 (Chang) +

Proof: Much as in Theorem 3, we construct a proper elementary tower < ~ I ~ saturated ~-structures of cardinality % such that ~ < B < %+ + [ ~ ~

~ ~B & U~

= U~B] and such that ~ 0 ~ ~

<el, m> ~-structure. stages require lemma 5.

< %+> of ~ ~B &

where INS01 = ~ and ~ is a given

At limit stages we use both lemmas 7 and 6.

Successor

The initial stage is easy using lemma 4 (i), since we

-142-

need only ensure that IL~°I = J ~ 0 j. generality in assuming that ~ 0 ~ TO"

And of course, there is no loss of Then

~ =~

<~+~

is as required.~

+ Finally we prove that if < is singular, then <ml' ~> ÷ << ' <> holds if V = L. More precisely, we show that if GCH + D

<

be assumed, then this result holds.

assume this from now on, and let < be a singular cardinal, ~ = cf(~) < R.

So

We shall

work relative to a fixed (but arbitrary) sequence G : U ~ K of regular cardinals such that G(O) = O, G(1) > e, ~ < ~ < ~ ÷ G(a) < G(B), and sup(G"~) = <. <S I~ < K+ & lim(~)> be such that: then S S

(i) S

K

.

is a closed subset of e; (ii) if cf(~) > w

is unbounded in ~; (iii) IS I < ~; and (iv) y s S

is just the set of limit points of C Since K is singular,

÷ S

= YnS

.

IS [ < K for all ~.]

Note that ~ e S + lim(a). y ~ is special if

~ = 0

< >> is an elementary tower of saturated structures such that I ~

each a.

A mapping r: 0L ÷ ~ is a rankin~ of ~

r(x) = the least a such that x e ~ + I then it is special.

[Clearly,

in ~, where realises

Let 0 ~ b e an ~-structure of cardinality K. <~el~

By ~J<, let

~

, where

I = G(a) for

iff there is such a tower for which

for all x e ~ .

Note that if ~

Similarly UTSpecial and U-ranking.

has a ranking,

The following lemma is easily

proved using our lelmmas on saturation: Lemma 9 (i)

Let ~ b e

(ii)

Let 0i, ~

is f: ~ ~

~

(iii) Let ~ ---~ . (iv)

an ~-structure.

that 4 ~

(~,

such that

~ ~ ~.

be special ~-structures with rankings r, s (reap), 0-[ - ~ .

There

such that s(f(x)) = r(x) for all x e ~ . he U-special with U-ranking r and let ~

be an ~-strueture of type <ml' e>"

with U-ranking r,

(~,

~

he special with ranking s,

There is f: ~['< ~ such that ran(f) =- U ~ and s(f(x)) = r(x) for all x s ~

Let ~

Let

There is a special structure

~ - ~

, ~+£',U ~,

~'

Let

~ be a U-special

There is a special structure

=U

•

~-structure

with ranking r

such

,andr=r'.l

be U-special

~-structures with U-ranking r, r' (reap.). We write: ~I' r ) ~ ( g t ' , r ' ) iff ~ ' < ~Y~' & r ~_r ' & U = U eL' r) coy(01', r') iff ~ ~< ~' & r~U ~L = r'~U ~ & U = U & (r(x) < y ÷ r'(x) < y) &

(r(x)

>~ y +

r(x)

= r'(x))

-143-

(01, r) c~ (61', r') iff (~y)[(O],

r) 0~y (Ol', r')].

Theorem i0 (Jensen) +
Proof:

60>

+

<<

(Silver)

,

<>

Let

as before.

.

~ be a given <Wl, ~> i-structure.

We construct a sequence < ( ~

We may assume that ~ ~ TO,

, r )[a < <+> such that < ~

Is < <+> +

is a proper elementary tower of U-special structures such that a < B < < + U

= U

, and r

~L = U < < + ~ e

is a U-ranking of ~

, each e, where ~ ~ ~ . Thus ~ 0 The construction is carried out so as to

will be as required.

preserve the following conditions: (A) a < E ÷ ( ~ ,

r ) oc (~B'

(B) ~ e S B ÷ (0~ , r ) <

rE);

(O~B, rE) ;

(C) if ~ = S (G(B)), then x e ~[ 7

- ~ 7

÷ r (x) >. B. c~

7

By lemma 9 (i), let (0l 0 , r0) be arbitrary so that ~ 0 successor stages in the induction are trivial. and (C) arise.) Case i.

S

By lermma 9 (iv),

(In particular, no new cases of (B)

is eofinal in ~. = ~ ,cS r ,.

Note that r~ is a function. by (B), r E _c rB,.]

[For suppose B, B' e S , B < E'.

We show that r~ is in fact a U-ranking of

let ~ B = <{x ~ O~ Ir (x) < B} .... >. elementary submodel of ~ (since S~ = U y a S

of cardinality G(B).

Then B ~ S B,, so For each B < ~,

~B is a U-saturated

Well, by (C) at earlier stages

Sy), we see that only the first G(B) terms in <(~y17 c S > give

For each ~ < 6, set

(Again, since S~ = ~ y a S each

0~.

We must prove that each

elements of rank less than B, so certainly

induction,

~.

Assume, then, that ~ < <+, lim(~), and <(0qB, rE)IB < ~> is defined.

Set 0 ~ = D ,¢S 0"~,, r

otp(S~)).

-

~B,[

I ~BI

~E,~ = {x ~ ~ I r

= G(B).

Now set 6 = min(G(E),

(x) < B}.

Thus

~B = ~ < 6

Sy, we only need (B) at earlier stages for this.)

~ B,~" By

is a U-saturated structure of cardinality G(E) and < ~E,[I[ < 6>

is an elementary tower such that [I < [2 < 6 ÷ U~B'[I = U ~E'[2. is a U-saturated structure of cardinality G(B). ing of ~[ , and hence that ~

is U-special.

Hence by lermma 7,

This proves that r

Clearly,

is a U-rank-

(A), (B), and (C) are preserved.

-144-

Case 2.

Otherwise.

(Hence cf(e) = ~.)

Let <s In < ~> be cofinal in s with s 0 = sup(So). n G(~) > otp(Se).

Let ~ be least such that

Since (~I, r) =# (~', r') & ~ < ~' < ~ ÷ (~, r)= ,(~', r'), we

can pick a monotone sequence <~nln < m> of ordinals such that ~0 = O, ~I > ~' and ~n < ~' such that (O~si, rei)=~i+l(~ei+l,

rsi+l).

Set ~ e = ~ i < ~ e • x ~ ~si •

(= ~s'<sO~e')" For x ~ ~Ls, let i(x) be the least i such that i Set rs (x) = max (#i (x) , rsi(x)), each x. We show that the function rs is

a U-ranklng of ~L e.

Define ~B, B < ~ as before.

Suppose first that ~i < B ~< ~i+l"

Then ~B = {x s ~silre.(x) < B}, which is a U-saturated model of cardinality G(B), I by induction. EThe equality here hold because of two facts. Firstly, no new elements x such that rs(x) < B appear after ~ei.

Secondly, if j < i and x e ~ j ,

rej(x) < ~,

then re.(x) < B (even if rs.(x) # rsi(x)). ] Now suppose that B >~ suPi<~ i. Then i j ~B = Ui<m ~ , i ' where ~ B,i = <{x ~ 01silrsi(x) < B} ....>, and l e n a 7 applies (by induction hypothesis).

Hence r is a U-ranklng of 0] and ~ is U-special• s s' s

We

must check (A), (B), and (C). For (B), note that -~is transitive and (~e 0, re0) "~ (0~e, rs) holds by definition of re . For (C), note that (C) holds for y = s 0 because S = S O{s 0} and becuase, for i > O, ~i > ~" It remains to check (A). By the s s0 transitivity of =, it suffices to show (~si, rsi)=~i(~ s, re). Suppose first that i(x) = i.

Then re(x) = rsi(x).

since (0~sj,. re ) ~ i ( ~ s i , j re(x) = rej(x) = rei(x).

Suppose next that i(x) = j < i but rei(x) >~ ~i"

rsi) , rsj (x) >~ ~i' so rej(x) = rsl(x),, whence In either case, (C) is proved. |

Then

Chapter 13 GAP-1 MORASSES In chapter 12, we showed that if V = L, then V K V k [ < < + , < > + ], where V = L was only required for the case I singular (and even then all that was really required was ~

and GCH, not the full power of V = L).

What about the Gap-2

Theorem ? Well, Silver ([19]) has shown that, for example, <~4,~2 > ~ <~3,~i > cannot be proved in ZFC + GCH, so some assumption stronger than GCH is required even for regular cardinals.

V = L is a natural assumption to try, and Jensen has shown that

in fact V = L does imply the Gap-2 Theorem.

We shall give his proof of this in the

next chapter, but first we must develop an important combinatorial property of L to make this proof work. This property is quite a general one, and is not specifically concerned with model theory. Nevertheless, the motivation for its development is the Gap-2 Theorem, and this does give a convenient starting point. Let us therefore consider the Gap-2 Theorem for a few moments. Suppose we were to try to generalise the method of Chang outlined in the previous chapter. Well, to make the construction work, it was necessary to use U-saturated structures (or strictly countable structures in the case of the Keisler-Vaught proof of the first case), and it is easily seen that if ~ is U-saturated, then I~I

= IUOZI (and in the Keisler-Vaught

case countability was important),so there is no chance of extending the elementary tower from ~+ to ~++ (or to ~2 in the Keisler-Vaught proof) to obtain a gap-2 structure,even if we were to make clever use of the fact that we were c(~nmencin6 with a gap-2 structure and not just a gap-1 structure.So,we must adopt a somewhat different approach if we wish to prove the Gap-2 theorem in L.But some similarity with the Keisler-Vaught/Chang method would seem to be inevitable.For,consider the reasoning behind their approach.They wish to construct a certain model of type . So they construct approximations to such a model of type <~,l> so that these approximations get "better" successively along a ~+-chain. In other words, the elementary chain construction can be regarded as a process of successive apyroximation to the desired structure. Viewed in this light, and bearing in mind the requirements needed to make the Keisler-Vaught/Chang proofs go through, we see that in order to

-146-

prove the Gap-2 theroem we ought perhaps be able to determine cardinality

%++ by means of a process of successive

each approximation

being of cardinality

%.

a structure of

approximation

of length %+,

Now, if one just uses an elementary chain,

then there is no hope, since all one could ever determine by such a process would be a structure of cardinality system,

%+.

But if one were to construct some sort of directed

then there may be some chance.

For example,

and suppose we appoint immediate successors canonical way.

to each cofinal branch of ~ in some

Then we can regard the levels of T as being approximations

"top level" of our extended countable,

suppose T is a Kurepa tree,

tree,

and the tree structure

There are ~I levels of T, each of which is itself gives a precise description of the way in

which the top level is uniquely determined by the approximation first try at proving the Gap-2 theorem might be to well-order tree in some manner,

procedure.

Thus, a

the levels of a Kurepa

and then proceed up the tree, level by level, attaching

countable homogeneous

structure

(say) a

to each point of the tree, and then take the direct

limit (in some manner) of the "topped" attached

to the

structure.

Thus, if ~

and ~

are structures

to points ~, B of T and ~ and B are on the some level of ~, with ~ preceeding

in the well order of the level of e, ~, we would demand if ~
: ~

~

~

.

(for example)

~

~

~

B

, and

(say) that there is an elementary

Other requirements

would then be necessary.

the e m b e d d i n g s o B , e
system

(i.e. ~ B y o ~

We would want

=~¥

for

e
%B

= Oy6~

precede y, d in the well-orderings ~.

Given such a set-up,

of the levels, then we would want

the "top-level"

structures would be determined

entirely by the lower system as limits of directed elementary level would consist of a proper elementary union would be as required. reader,

tower of structures

systems,

and the top

of length ~2, whose

Now, in case we have created too much enthusiasm in the

let us at this point state that it is easy to see flaws in the above outline.

A Kurepa tree just is not good enough.

However,

hopefully,

convinced the reader of its feasibility

for a short while.

the speed of our outline And it is the picture he

had in mind at that time which we shall work on, since chances are it resembled the structure we really require more than it did a Kurepa tree.

-147-

Our strategy then is to approximate a structure of cardinality K + by means of a K-sequence of structures of cardinality less than <. we shall assume K = ~I here.

To fix ideas once and for all,

We shall thus describe how to approximate a structure

of cardinality ~2 by means of an ml-sequence of countable structures.

Everything we

do in this chapter generalises to arbitrary regular, uncountable K in place of ~I in a straightforward manner.

To begin with, let us consider a very simple "structure" of

cardinality w2, namely the ordinal w 2.

For notational convenience, let us first of

all identify ~2 with the ordinal interval

E~I, m2 ) , and consider this.

tried to approximate [~I' w2) by intervals

[~, B) where e < B < ~I"

Suppose we

Then we could

illustrate the approximation procedure thus:

•

,

•

o

•

°

4

~

°

The idea is that, to make an intelligible diagram, we "flip over" the intervals concerned, as illustrated.

(Since each interval [e, B) is countable, we can assume

they are all disjoint, so that this "flipping" really is just diagramatical.)

Note

that we have not described how the interval [~I' w2) is actually determined by the smaller intervals [~, B), ~ < B < ~I"

What we want then is some way of determining,

for each point of [wl, w2), which points of the various

[~, B)'s approximate it.

for each 6 such that w I < 6 < w 2, the interval E~I, 6) has cardinality ~I' so can indeed be defined as the limit of an Wl-chain of countable intervals certain embeddings.

E~, B) under

Thus, in the following diagram the line descending from 6 is

intended to convey the "path" which we must follow in order to determine

[ml' 6).

Now,

-148-

(More precisely, if ~ preceedes ~' on this line then there must be an embedding ~$$, of [~, ~) into [~', B'), as shown, whence, providing the OBB,'s commute along this path, ~9 I, 6) will be the limit of the ~-system along it.

0 ....... °

Now, for each ~ ~ [Wl, ~2 ) , we can carry out the above procedure.

The only

difficulty is to ensure that the directed systems for all the different 6's all coalesce nicely below 91, because this part of the structure must consist of at most 91 countable intervals, for reasons already indicated. of a morass.

This brings us to the notion

More precisely, we are going to describe an el-morass, since the more

general notion of a K-morass, for < regular and uncountable, is entirely similar.

It

turns out to he a little easier to think not of intervals [~, ~), [91 , 92 ) but of certain closed subsets of such intervals.

This is because, having defined a morass,

we shall construct one in L, and to do so we shall consider p.r. closed ordinals only. (It is not really surprising that there will be a morass in L of course, since our basic notion is clearly tied up with the ideas of the condensation lemma to some extent.) We commence with the formal definition of the (91-)morass.

Unfortunately, the

reader cannot really avoid sitting down and drawing several diagrams for himself to illustrate each axiom as we give it.

We shall give a couple of instances of such

diagrams to start the ball rolling however, immediately following the completion of

-149-

the definition. Let ~

be a set of ordered pairs of primitive recursive closed ordinals

< ~ < ~2 such that whenever , e ~ ,

set sO = {~ ~ ~

+ lI(~-)[<~,

a < a' ÷ ~ < ~'.

,> ~ 2 ] ~ ;

S = S O U SI; S

= {~ e SII e ~ } ,

for a e sO;

a

= that unique ordinal a e S O such that e ~ ,

Let -9 be a tree on S 1 such that D -9 T ÷ a Let {~

Iv-~}

< a • T

be a commutative system of maps ~

Then, the structure ~ =

<S,~,-9,

for ~ e SI.

:(v + I) ÷ (~ + I).

> is an (~I' l)-morass if the

(~)~T

following axioms are satisfied. (MO)

(a) S

is closed in sup(S ), all ~ g S0;

(b) ~I = max(S0) = sup(S° ~ ~I ); ~2 = sup(S~l)" (MI)

If ~-~ T, then ~ into

Sa

N

T so

= id~,

~ T(~ ) = at, ~

(v) = ~, and ~ T

maps S ~ ~ v

that:

T

(i)

if y is initial in S~ , then ~ r(y) is initial in Su ; T

(ii)

if y immediately succeeds ~ in S~ ~ v, then ~ ~ ) in

Sa

~mm-succeeds ~ T (~)

; T

(iii) if ~ is a limit point in Sa

N ~, then ~ T(~) is a limit point in S~ .

T is closed in ~ .

(M3)

{a I ~ - - ~ }

(M4)

~ not maximal in S~

÷ {a I~-~ ~} unbounded in ~ . T

(MS)

{a I~ --~} unbounded in a t ÷ •

(M6)

~ a limit point of So_ & ~ - ~ T 7 a limit point of Sa_- & ~ - ~ T T

(M7)

~U~

~

& i = sup~<--~(~) = sup~<~(~)

÷ ( ~'

+ ~

I & ~XI~

&a e ~ v e S ~ { a

~ s)(~

T' ~ T).

That cor~letes the definition. Suppose that for each ~ ~ S I there is U

~ ~ x ~ such that:

= ~ .

]~_-3 ~ _ ~ r ( ~ ) }

-150-

(i)

~ s S~T~

T ÷ U

= U T n(v × ~);

(ii) U = ~ E s ~ I U ~ is such that enumerates all bounded subsets of w . 2 Let ~ ~*

be as above, and consider the structure j~* = < ~ ,

(U)

is a universal ml-morass iff for each v-~ T, ~ T " • ÷~0

~.
We say ~

UT~

The following diagram is intended to illustrate the notation•

so ~e~ nT T

%

0

] so

Note that we use vertical lines to denote the tree-e.

Comparing this with our

earlier diagram, we see that an initial segment S~I N T of S~I for T ~ S~I, is determined as the direct limit of the system < ( S ~

N ~)~-@T'

(~)~

~T>.

axioms (M4) and (145), we see that this really will determine .... all of S~I ~ T.~ now clear why (M2) should be necessary. procedures Sml.

[By It is

We clearly want the various approximation

to fit together smoothly as we progress from left to right along, say,

In diagramatic form, we could illustrate

(M2) thus:

-151-

W

(M2) asserts that the line between ~ and ~ is a -~-line, ~-IV = ~-~. TT UV

and that

(Note that the second conclusion here cannot even be asserted until

the first is known, since ~

is only ever defined for ~

(M3) and (M4) are natural enough,

though

~ .)

(M4) has the strange-seeming

consequence

that if ~ is a successor point of S O , then IS I = I, so that the approximations S~I do not "get better" monotonically which,

together with

(M5), ensures,

as we proceed up S O .

in particular,

to

And of course it is (M4)

that every initial part of S~I is

entirely determined by the morass below ~I" (M6) and (MT) are "continuity" systems

fit together.

importance

axioms for the way the various approximation

We leave the reader to puzzle out the diagrams.

of (M6) and (MT) will become apparent when we use the morass,

The in the next

chapter. So far, we have not said why we bothered morass.

Certainly

next chapter,

to introduce

the concept of a universal

for the gap-2 theorem which we shall prove using a morass in the

all that we require is a morass.

universal morass is required. easy, but instructive,

However,

for several other uses, a

(One of these uses occurs in [2].)

exercise to deduce the combinational

existence of a universal <-morass.

Finally,

it illustrates

Also, it is an

principle<

from the

the fact that one can

-152-

"carry along" quite a lot of extra structure

on a morass,

if required.

(When we

prove that there is a morass in L, we shall in fact prove there is a universal one~) Before we commence with our proof that there is a morass in L, let us indicate how the above notion generalises.

In the above, we determine

cardinality ~2 by means of ~I countable structures. is one cardinallty higher than the approximating or, more precisely, visualising

an (~I' l)-morass.

an (~I' 2)-morass,

~3 is so determined.

~

Since the structure determined

procedure, we call it a Ga_~_7~ morass,

The reader should now have no trouble in

i.e. a gap-2 ~1-morass, where a structure

Of course, one would require three dimensions

such a structure,

since it would resemble a "morass of morasses"

such a structure,

one would have, instead of

However,

More surprisingly,

of cardinality

to illustrate

so as to speak.

In

~ as above, a set ~ + of ordered triples,

and instead of S O , S I a ~iple of sets S O , S I , S 2. for all n g ~.

structure of

Similarly,

there are gap-n morasses

there are also morasses with infinite gaps.

the complexity of even gap-2 morasses

is so great as to warrant a separate

treatment of their own (though such a treatment would be uniform for all countable gaps greater than I, at least).

In the next chapter we shall use a gap-i morass

prove the gap-2 theorem in L (note the different meanings gap-n morass, one can prove, in an analogous, gap-(n + i) theorem.

Henceforth,

of 'gap' here).

though more complex manner,

we shall leave all generalisations

to

Using a the

to higher gaps

(in either sense of the word "gap") to the readers imagination. From now on, we shall assume V = L. Our strategy is to formulate a certain combinatorial a universal

(~i' l)-morass with the aid of ~ .

(in L).

is not intended to be particularly

~

principle, ~ , and construct

We shall then show that ~ intuitive.

It simply embodies enough

of the fine structure of L to enable one to construct a morass, morasses

are concerned,

For any structure all ~0-formulas

and so, as far as

it is the "hear~' of the matter~ of the form <Jp, A>, we write X- iff X ~ Jp and for

~(v0, Vl) with parameters

~<X,AnX>

is true

(~)(~

from X,

> ~)~(~' JB ) iff ~<j ,A> ( ~ ) ( ~ > ~)~(~' JB )" O In this case, we say X is a ~-submodel of <Jp, A>. Clearly, if lim(p), then

-153-

X-~Q <Jp, A> implies X~<~I <Jp, A>. cofinal in p, then X ~ Q

Conversely,

if X-<~0 <Jp, A> and X ~ p is

<Jp, A>.

In order to formulate ~ , we need to first of all define the domain and pairing predicate of our intended morass. Set

~ = {<~, v>l~ < v < ~02 & , is p.m. closed & ~ = mlJv & b J

"~ is the

largest cardinal"}. Define S O , S I, S, S , ev as usual.

Note that S~ ~, is uniformly ~ ' ( { ~ M

}),

~ ~ S I. For ~ ~ S I, set: B(~) = the least B >. v such that ~ is not regular at B; n(~) = the least n >. 1 such that v is not In-regular at ~; p(~)

n(~)-I • = p~(~)

(Note that p(v) >. ~, by choice of ~, n, p.)

~n(~)-l"

A(v) = "~B(~)

: There is a sequence such that, for each ~ ~ S I, C~ is an unbounded subset of ~ such that: (I) <Jr' C > is amenable; (2) Every x ~ J

is ~I - d e f i n a b l e

in <J),

C > from parameters

(3) < C~ 1~ E S~ ~ > is uniformly ~I <Jp(~), A(~)>. M (4) Suppose ~ .< v, ~ c_ J~, and ~: <Jr' ~ > ~ Q <J~' c~>. ~(~)

i n c~.);

Then 7 E S I, ~ = C~, and

= ~v' and there is a ~ ~ ~ such that ~: <Jp(~), A(~)>-4~I <Jp(v)' A(~)>.

Furthermore,

0 is uniquely

determined

by ~ ' - .

(5) Suppose ~ is a limit point of So~- and ~: <J~, C--> -~Q <Jr' C >. v X = s u p v , < ~ ~ ( ' 0 ' ) , t h e n X ~ Sc~ a n d C)~ = % c~ Cu. M A s s u m i n g ~) , we d e f i n e --

a m o r a s s on S as f o l l o w s .

For v, ~ E SI, set ~ _ ~ ~ +_~ ~

By

If

< ~

& (~)[~:

<J~, C ~ - < Q

<j

, C > &

~(4), the map ~ in the above is unique, so we may denote it by ~^

For 7 _ - ~ ,

define ~ :

It is immediate

(~ + I) ÷ (~ + I) by ~-- = (~-~ ~7) o{<~, ~-->}.

that -~ is a tree and the systems of maps {~v~I~-.~ ~},

.

-154-

{~F-q~}

are commutative.

SetJl

<s,~

=

,

~

,

(~)7~,

(MO)

This is easily checked.

(MI)

By ~(4)

~v(a-~) = a .

>

We show that J~ is an (~I' l)-morass.

The rest is easy to check.

In particular,

(MI) (ill)

holds because of our definition o f ~ Q . (M2)

This is easy to see, using

~(3) & (4).

(M3)

The union of an increasing chain of Q-elementary submodels is clearly a Qelementary submodel, so (M3) follows from ~(4).

(M4)

Suppose ~ is not maximal in Sa that ~-- > y.

Claim i.

and let y < ~

be given.

We seek a ~ ~ ~ such

We require the following, very general:

For any ~ s S a ,

~jp (~) "~v is regular".

If 0(~) = ~, then this holds because ~ s S~ , of course. NOW, as P(~) ,< B('), kj0(v ) "~ is a regular cardinal"•

Suppose p(~) > ~.

Since ~ ~ S

a regular cardinal in J ~ " , by absoluteness considerations.

is , I=j p(~) Hence, ~jp(~) "a is a

regular cardinal", since, in~'ZFC"+ V = L, if < is regular and I=j then % is regular.

QED.

Now, by choice of +, we can pick • ~ S regular".

By

"% is regular",

~(3) ' C v 8 J0(T) .

, T > +.

By claim i, l=jp(T ) "a

is

Also, the canonical I] -skolem function for

<J , C+> is an element of Jp(T)' and does not lead out of JO(T).

But look, using

this skolem function, there is a canonical way of constructing, within JO(T)' X~Q

<J , C > such that y ~ a

y E ~

N X s ~ .

-~ +. (M5)

~ X s On.

So, as ~

is regular in Jp(T)'

In the real world, if we now set J~ % X, then

~(4) implies that

But by construction, 0~ > y, so (M4) is proven. Let {%1-v-~ v} be unbounded in ~ Let x c J .

<J~, C > from ~.

% ~ " J^ ~

•

We show that J - ~ d - ~ _ ~

By ~Q(2), there are ~ g e

~-~ "j--.

^

such that x is ~l-definable in

By hypothesis, pick ~ - ~ ~ so that ~ s o~-.~ Since

<J, %>,x ~ ~-. ~ "J-,,

as required•

-I (M6)

Let ~ be a limit point of ScL_, ~ - ~ + , ^ -e % and ~ % = +~ .

% --supT<~ ~ v ( r ) .

We show that

-155-

Since r a n ( ~ )

~ J%, ~

: <J~, C~>~y

<Jx' C

by choice of %, this embedding is cofinal. But by ~(5), C

~ % = C%.

Q X>, by ~$absoluteness.

Hence ~

: <J~, C~>~.

Hence, by uniqueness, ~ %

(MT) Let ~- be a limit point of Sc~ , ~ - ~ v

But

^

and ~

= ~X"

= sup=--p~"~-. Let ~ s {~ I~ ---~n ~ - ( r ) }

V

for all

"r ~ Sc~- ~ ~'.

F o r e a c h "r e Sc~-

S e t ~)' = s u p { n ( T ) l - r

We s e e k '~' s S

13 v ,

iet

such that

v ~ x)' ~ u .

unique

n a Sc~ s u c h t h a t

~lCr) b e t h a t

e Sa_ n'~'}.

T h u s ~)' e Sa ,

a s Sa i s

closed.

~c ~ q ~

)(-c).

We show t h a t

"O

~'--~.

By (M2), if r, ~' ~ S0~_~ ~, r < ~', then n~,~(~) = ~ ,r ,q(~,)~J and ~ (~),~,o(y) = ~n(~,),~ (~,)~J (T). Thus, we may define functions ~0' ~I by:

V

Clearly, ~I o ~0 : J~-<~0 J

cofinally, so ~I ° ~0 : J~-~Q J "

Let x e J--. Then for some ~ s S

n~-, x e J

= ~I ° =0(x) (by com~utativity considerations). Let C = ~-I"(CIv n ran(~l)).

and by (M2), ~ -

~^

=

(x)

Hence =I ~ =0 = ~ "

Thus ~I: <Jr'' C>~<~0 <J , C > cofinally. ^

~(4), therefore, C = C ,, whence ~' ~ ~ and ~I = ~'~" ~0 = ~v'"

(x)

Similarly, ~_-~'

By

and

This proves (M7).

Hence .~ is an (~I' l)-morass. Let U ~ (i)

2

×

~2

It remains to expand ~

be the canonical ~ 2 - p r e d i c a t e

to a universal morass.

such that

U"{~} ~ ~;

(ii) enumerates all bounded subsets of ~2; ~T

(iii) v < T -~ U {v} <j U"{T}. For each ~ s S

, set U

= U ~(v × ~).

1

~ S~I and ~-~ ~, then ~--~ ~ J~-<~l J~' so U Clearly,

<S, ~ , -~, ( ~ v ) ~ _ ~ ,

We turn now to the proof of ~ . ~1 -skolem functions.

Thus U

is ~ , by choice of U. J induces a ZI~ set U~ ~ × v.

If

(U)~eSI> is a universal (~I' l)-morass.

Before we commence, we must choose specific

So, let <JP" A> be arbitrary.

Define a function h = hp'A as

-156-

follows.

Let <~i li < w> be a recursive enumeration of all formulas of the form

~i(x) - ~y~i(y, x), where ~i is ~0"

Fix <~il i < m> once and for all:

For i e ~,

x ~ Jp, define rp,A(i, x) _~ the <j-least v ~ Jp such that l=<jp, A>~i ((v)0' (V)l, x)Define hp,A(i, x) ~_ (rp,A(i , x)) I,

Thus, for i g ~, x g Jp, y ~ Jp,

y = h(i, x) *-+ ~v[v is the <j-least v ~ J

P

such that ~<j ,A>~i((v)0, p

(v) I, x)

& (v) I = y] *-+ ~ v ~ w [ w (~v' g w)-7~i((v') 0 ,

= {zlz <j v} & ~<j ,A>Ti((v)0, (v)l, x) & o (V')l, x) & (v) l = y]. Let O be a 10 -formula such that for all

and all v E J , w = {zlz <j v} +-~ l=j ~te(t, w, v). Then, v y = h(i, x) ~+ ~ v ~ w ~t[~<j ,A>[O(t, w, v) & y = (v) & ~i((v)0, p I (~v' g w)-~i((v')0, (v')l, x)]]. Define H by o,A H(z,y , i, x) +-+ ~<Jp,A>[e((z)2,

(V)l, x) &

(z)l, (z) 0) & y = ((z)0) l & ~i(((z)0) 0, ((z)0) I, x)

(Vv' ~ (z)~)~Ti((v') 0, (v' I , x)] Then H p,A is Lv<J 0 p ,A> and y = hp,A(i, x) ~-+ ( ~z ~ Jp)H(z, y, i, x). q < 0, then H n ' A q J n y = h ,A(i, x). on J . P

Moreover, if

= Hp'A ~(Jq)q' so for x, y ~ Jq, y = hq, A qjq(i, x) ++

Also, of course, if <Jp, A> is amenable, then hp ,A is ~]Jp,A> predicate

But, even if <J , A> is not amenable, we still have: P

Claim 2.

Let q .< 0, p s JP"

Then h"(~0 × (Jn × {P}))-.

Suppose first that z = h(i, ), some y E Jq. definable in <Jp, A> from y and po such that ~<Jp,A>~i((v)0,

We show that z is El-

Clearly, z = (v) I where v = the <j-least v c Jp

(V) l, ).

But then z is as stated.

Now set X = h"(~ x (Jn x {p})), and let Zl,..., zm E X be given such that ~x~i(x, z ..... Zn~). By our remark above, we can find Yl ..... Ym e J <Jp ,A> I q such that, for each j, zj is Zl-definable in <Jp, A> from y~, p. Hence we can find j = j(i) so that, setting y = (E Jq), (*) ~<<Jo,A> ~x[~i(x , z I ..... Zm) +-~ ~j(x, )]. Now, ~<j ,A> ~x~i(x , Zl, "'" , Zm) , by hypothesis " O So, ~<j ,A>~X%j(x, ), by (*). o So, ~<j ,A> ~x~j((x) 0' (X) l' )" O Hence rp,A(j, ) is defined, whence so is h(j, ); and clearly, ~<j o

,A

>% (h(j j

'

)

).

-157-

Thus, ~<j ,A>~j(h(j, ), z I ..... Zm), by (*), P giving ( ~ x e X)[~<Jp,A>~i(x , z I ..... Zm) ]. This proves The point about all of this is the following. consider [I skolem functions

for amenable <J P' A>.

to consider arbitrary structures <Jp,A> if <J ZI

<J

A>.

p'

the claim.

Earlier it was only necessary

to

For our present purposes, we want

Of course, h

will not necessarily be

p,A

A> is not amenable, but this will not matter.

Claim 2 is the

P'

important point, together with the fact that the definition of hp, A is uniform over all <Jp, A>. We may now get down to business. Let ~ e S I. A = A(~).

For convenience,

we set a = a , B = S(~), n = n(v), p = p(v),

This convention will be retained throughout.

Le=~na I n

pB,
Since a is the largest cardinal

in J , there is a [n(JB) map f such that f"a

is cofinal in ~, by choice of B, n. follows

that

n PB "< x).

(For if

f = f q(~ x ~) e J n _cJB.) PB g"x) = J13"

Now, f o r

each

Since ~JB ''~ is regular",

f ~ JB"

This means that there is a [n(JB) map g such that

"r < %

there

is

fT e Jx) s u c h t h a t

f,r:

ONTO c~ 4"r; and i f

we take fT to be the <j-least such, then fT will be El-definable J~.

Using f, we can therefore

g~

construct

S 1 into

two d i s j o i n t

sets

from a, T in

a ~n(Js) map ~ so that ~"a = ~.

is a ~n(J~) map of a subset of ~ onto JB' so pnS ~< a .

Partition

It

n Pf3 > x), t h e n by a m e n a b i l i t y ,

P , R by s e t t i n g

Then

QED.

P = {~) e s l l n = 1 & g = Y + 1

for some y}, R = S 1 - P. We shall take for Cv a canonical p.r. coding as a subset of ~ of the set E

~ ~ x J

~

to be defined below.

Setting

subset of v, so it will suffice to verify The construction

E 0 = { ( x ) Ix E E }, E 0 w i l l ~ 0 ~

with E

Case I.

~ e P.

in place of C .

takes place in two entirely disjoint

whether v c P or ~ e R. Let B = Y + I.

Thus y = y(~).

b e an u n b o u n d e d

cases, depending on

-158-

Note that as lim(v), v g T.

Also, of course, p = B, A = ¢.

Lemma 2 There is p ~ J J

such that every x ~ J

Y

T

is J -definable from parameters in Y

u{p}.

Proof: By lemma 1, t h e r e i s p ~ JB such t h a t e v e r y x ~ J~3 i s Z l - d e f i n a b l e i n Jig from parameters in J a o { p } .

Since J~ = rud(J ), there is a rud. function f and an Y

element p ~ Jy such t h a t p = f ( J T , p). lemma.

Let x ~ Jy be arbitrary.

We show t h a t p i s as claimed i n the

Then for some Z0 formula ~ and some z E Ja'

x is the unique x ~ J~ such that ( ~ y

~ J~)[~jBC(y, ~, p, x)].

Since

JB = ~m<wSm(jT )' we can find m < ~ such that ( ~y ~ sm(JT)) [~Sm j )¢(y, ~, f(Jy, p), x)~.

~(~

statement p ~ J

(with parameters x, z, p) in J . ¥

u{p}.

Every x s J

T

Hence x is J -definable from z, Y

QED.

Let p = p(~) = the <j-least p e J (i)

But this can be written as a first-order-

¥

such that:

is J -definable from parameters ¥

(ii) e is ~I -definable

in J

Y

in ~ u{p};

from p.

This is possible by lemma 2 (for (i)) and the fact that ~ < ~ ~ T (for (ii)). For each m < ~, set Xm+ I = {x c J I x is ~~m+l-definable T ~{p}}.

in J

from parameters in

Thus Xm+l-
Note that

p e Xm+ I for all m, and hence ~ e Xm+ I for all m. Lemma 3 For all m < m, Xm+ I N J~ is transitive. Proof: For each ~ < ~, let f~ be the <j-least map f~: ~ ON~+O ~.

Since ~ is the

largest cardinal in J , f~ e Jv is always defined, and is El-definable

in JT

from ~, ~ (and hence from p, ~). Suppose $ c Xm+ I q ~. definition,

~ ~ Xm+ I.

is Xm+ I ~ J .

QED.

Then as ~, $ ~ Xm+l-<~m+l Jy, f~ c Xm+ I.

Thus ~ = f "~ ~ Xm+ I.

But by

Hence Xm+ 1 q ~ is transitive, whence so

-159-

Set J

= Xm+ I ~ J), all m < ~. Put ~ = ~. Thus < ~ m l m < ~> is non"Om+l 0 decreasing and cofinal in ~ (if not eventually equal to ~, which we shall later see it is, in fact, not~). Let ~m+l: Xm+l ~ JY

, all m < ~, and put y m+l

= ~. 0

By definition of JYm+l' there is a JYm+l-definable map of ~ onto ~m+l' so as is regular at y, Ym+l < ~ for all m.

But clearly, ~m+l ~ Ym+l for all m, so

SUPm<~y m = ~. Se~ Pm+l = ~ m + l (p)' e a c h m < ~, and set P0 = o. Let E~ = {Im < ~}" Lemma 4 The sequences <E I~ s P>, have the following properties. (0)

E 0 is an unbounded subset of ~ and E

(I)

<J , E > is amenable.

(2)

Every x s J

(3)

<E717 s So

(4)

Suppose ~ ~ v, E ~ J ~ ,

~ J .

is ~ -definable in <J , E > from parameters in o. ~ ~ nP>

is uniformly ~I ands:

<J ,A> P

<J~, E > ~ Q

<J , E >.

Then ~ s P, ~ = E~, and

~(o~) = ~ .

Furthermore, there is a unique ~ _~ o such that ~: Jp (~) ~ . JP (~) I and ~(p(~)) = p(~). (Note that, as V, ~ ~ P here, p(~) = B(~), p(v) = ~(~),

and A(~) = A(~) = ~.

Note also that, by (2), CT(and hence ~) is uniquely

determined by ~ . ) (5)

Suppose ~ is a limit point of So_ and ~: <J~, E~>-. % = sup~,<~ ~(~'), then X g S

Proof:

~ P and E% = E

If

~ JX"

(0) Since S U P m < ~ m = ~ and m < ~ + Ym < ~' (O) is clearly true.

(i) Since E 0 is an unbounded m-sequence in ~, <J , E > is trivially amenable. (2) Now, the monotone enumeration of E 0 is clearly y<Jv 'E~ > f: ~ + ~ defined by f(m) = Ym is ~<J~'E~>. -I

In other words, the map

Since ~ s S~, for each m < ~, the

<j-least map ~m: ~ ONTO Jf(m) lies in J , and is El-definable in J~ from ~, f(m). <J,Ev> ONTO Hence, using f and <~ Im < ~> we can construct a ~ ({o}) map g: ~ -~ J • m ' -I But look ~ = Y0 = first element of E~, so o is El -definable in <J , E >.

Hence g

-160-

is in fact ~ <Jv'Ev> . -I (3) Clear

(2) follows

from the definitions.

(Note that ~ c S~

of S~ .) v (4) Suppose v, E, ~ are as in (4). And,

We m u s t

N v ~ B(~) < B(v), by definition

show that ~ E P, E = E~, and ~ ( ~ )

since p(~) = 8(~) = y(~) + i and p = B = y + i (and hence Jp(~)

and Jp = rud(Jy)),

we must show that there

J Y and ~(p(~))

: Jy(~) ~ 6':

immediately.

JP(~)-<~l

= p.

its range.

M N J

~ ~ o such that

(For then, o must extend uniquely

desire

In other words,

to a

t o f i n d a ~ : Jy(~)---< J y , we c o r m e n c e by

set

M = {x ~ JyI x is Jy-definable Claim 3.

= rud(Jy(~))

Jp ")

B e a r i n g i n mind o u r e v e n t u a l defining

is a unique

= ~.

from parameters

in ran(~)

~ {p}.}.

= ran(o). V

Firstly, Hence,

note that the monotone

as r a n ( ~ ) ~ l

enumeration

of E

<Jr' Ev >' m < ~ ÷ Ym' Pm ~ ran(~).

is -~ <Jv'Ev> and has type ~. -i In particular,

= Y0 e ran(~) ~ M. Clearly,

ran(U) ~ M ~ Jr"

For the converse,

Since x s M, we can find m < ~ such that, from z, p in J . Y

m = id~J

vm

x is ~l-definable so x ~ ran(~),

and cofinal

in v.

, we see that x is ~m-definable in Jv from Ym' ÷z, Pm"

proving

nJv

for some ~ ~ ran(O),

We can assume m is chosen large enough

is nonmdecreasing ~l~J

let x s M

Thus,

be arbitrary.

x is ~m-definable

for z, x s Jvm' since

as ~ I :

J Ym ~ _~m J

from z, Pm in J

Ym

But Ym' ÷z, Pm e r a n ( ~ ) ~ l

.

So, as ¥m < v' J,

the claim.

~

w

Let ~: M ~ J--, and set ~ = ~-I. = y complete (i)

~"~

Let p = ~(p),

~ = ~(~).

of claims.

=~.

This holds by claim 3, since M (ii)

Thus ~: J---<J • Y Y

the proof of (4) by means of a series

G b-

and Y

Q J

is an g-initial

= v • M and sup(v riM) = sup(ran(tO)

section

of M.

= v.

This holds by (i), claim 3, and the fact that m < ~ + Ym s ran(O). (iii)

~ < y + v < y & 6(~)

= v.

We

-161-

Suppose ~ < y. o(~)

>. v.

But ~(~)

(iv)

is defined.

< ~, of course,

+ J have v = ( I ~ l ) Y. which means ~(~)

Thus ~(~)

But ~ ~ M - < J

Since ran(~)

so v < ]..

= M,

(ii) implies

that

Since 7 < 8 and ~ s S , we clearly

, so this implies

,~ e M.

So, by

(ii), ~(~) = v,

= ~.

~ is regular

at ~.

Suppose not,

and let ~ be a monotone

J--definable Y

map of a subset

of some

~

0 < ~ cofinally

into ~.

Let f be defined

in J

by the c5: J---~ Jy image of the Y

Y definition

of ~.

a J -definable Y Suppose

Set 0 = ~(~).

Thus ~ < ~, by

map of 0 < n cofinally

therefore,

that v = y.

(i).

If ~ < y, then by

into ~, contrary

Thus ~ j _ ( ~ )

to ~ being

(iii),

regular

( ~ ~ < 0) (f(~) >~ ~).

f is

at ¥.

Applying ~,

~j

(~)(~ ~ < 8 ) ( f ( ~ ) >, ~). Set f' = f N (v x 0). Thus, f' is J -definable, y whether or not ~0 < y. We show that f' maps a subset of 0 cofinally into ~, contrary to ¥ = y(,0). ~< ~ + d(~)

For

~ < ~, we clearly have f ( ~ ( ~ ) )

< ~, we see that

Let ~ < ~ be given. we can find = ~(~(~)) (v)

By

~< ~ + f'(~(~))

(ii), we can find

~ < ~ such that ~(~) > ~ . > ~(t)

Every x e J-- is J~-definable Y

Let x g J~.

Thus ~(x)

p for some z 0 = <60> c m , (i)

~j~f V z 3 y ~ y '

(ii)

~jy

i< ~ such that ~(2)

Then f' (~(~))

Z< ~,

and f' ( ~ ( ~ ) )

s J • i (i

is J -definable Y

from z0,

such that:

z, p) + ( ~ 6 1 ..... ~6m)(Z . .<61' . .

z such that ~j

Hence

from

[y' = y ++ ~(y', z, p)];

But O(x), p ¢ M - < Jy. z

Thus

I

y

~(o(x),

z, p).

Zl ~ M.

By choice

*---

i

=h

Next, note that as a is Zl-definable

is J -definable y

from

of Zl, Zl ~<J z0' so as ~

, we see that z

Hence x is J_-definable Y

I

--

Now, ~: M ~ J--, ~(~) = ~, = Y

--

i

Then z

-~

= <~i > for some 61 ~ ~. = z

' 6m>)];

~(y, z , p). 0

--

~(p) = p, so, setting ~(Zl) (y, Zl, p).

Since

in e O {p}; ~ is ~l-definable

Let ~ be a formula

-+

~j_

> ~.

is defined

e J , so by choice of p, ~ ( x )

~-+ y ~ Jy & ~j

Let z I be the <j-least

~

and has value 6(~(~)).

such.

¥

z0 ~ J , z

is defined

from parameters

some m.

~z ~y~#(y,

(iii) y = ~ ( x )

p.

< v, so as

> ~, as required.

in J~; and p is the <j-least

~(x),

= ~(~(~))

--in

= <6 > e ~ I

and y = x ++ y g J-- & Y

__

from 61, p, which are members from p in J.f, applying

of ~ O { ~ } -

~ immediately

gives

-162-

a ~l-definiti°n of ~ f r o m ~ Finally,

in J--.

suppose that ~'

<j ~ h a s t h e a f o r e m e n t i o n e d

derive a contradiction with the definition of p° particular, be able to define p f r o m p '

properties

also.

We

Now, by choice of ~', we must, in

and some ~ ~ ~,in J--. Hence, applying ~, we Y

see that p is J -definable from p' = 6(~') and 7 = 6(7) ~ ~. Y

But look, every x ~ J

Y

is J -definaBle from parameters in ~ ~ {p}, so we see that we can define every x s J from p a r a m e t e r s

in a 0 {p'}

in J

, by r e p l a c i n g

o f a ~ { p } , by t h e J - t e r m w h i c h d e f i n e s Y

p in a J -definition

p f r o m ~5, p , .

from p' in J~, applying O tells us that a is El - d e f i n a b l e

Again,

as ~ is

from p'

Y

o f x f r o m members ~i-definable

in J Y .

~' <j p, so p' <j p, whence all of this contradicts the choice of p.

But

Hence ~ is as

claimed in (v). (vi)

~ is not El-regular at ~ + I. For each m < ~, let ~m be the supremum of all ordinals in ~ which are

definable in JT from parameters in ~ u~{p}. = SUPm< Tm. (vii)

But by (v),

Since is clearly ~l(J~+l), we are done.

~ s S I, and moreover, ~-~ = 7. We have O: J ~ Q

(~

By (iv), rm < ~' each m.

m+l-

J , and by (i), ~(~) = 6 ( 7 )

ONTO > ~)($ ~ ~ V ( ~f ~ J~)(f: ~ ÷ ¢)).

= ~.

Since v g S , ~j

Thus ] = j _ ( ~ ) ( ~

(V~)

~ > ~)($ ~ ~ V ( ~ f

g Jr)

ONTO (f: ~ ÷ $)).

Hence ~j_ "¢ < ~ ÷ ~ is countable". By an entirely similar argument, J~ But if ~j_ "~ is countable", then, ~ j _ ( ~ ~ 7 ) ( [ ~ [ = ]~I) . Since ~ ~ S , ~ = ~i v ONTO by our last result in J~, ~J7 ( ~ ) ( ~ ~ > ~)( ~ f ~ J~)(f: ~ ÷ ~), whence, crossing by ~ , we see that ~jv "all ordinals are countable " . Hence ~j_ "~ is uncountable". _ J_ ~ J_ Collecting all our J~ information together, we see that ~ = ~i v = [largest cardinalJ v. to J

This proves (vii).

(viii) B(~) = Y + i, n(~) = i. (ix)

~ ~ P.

(x)

p = p(~).

(By (iv), (vi))

(By (viii)). (By (v), since B(~) = ~ + 1 and 7 = a-~, by (viii), (vii), reap.).

Finally, since ~: J-- ~ J

it follows from (vii) - (x) that ~"E-- = E .

is an m-sequence, whence so is E, and so o " E = E . V

But E

Thus, by (i), E = E--. This

-163-

proves

the first part of (4).

of (4), uniqueness (5) Since E

following

Since y(~-) = ~, ~ already satisfies immediately

is an U-sequence,

from (v).

the requirements

This completes

(4).

--

11

it is clear that with ~, <~, I ~s in (5), <~ E ~ = Ev,

so
~ ¢ R.

Note that in this case,

lim(p) is assured.

Set p = p(v) = the <j-least p ~ J

such that: P

(i)

Every x s Jp is El-definable

(ii) ~ is ~I -definable This is possible because

in <Jp, A> from parameters

in ~ D { p } ;

in J P from p.

(for (i)) ' plp,A = 06n ~ a (lemma I) and

(for (ii)), a < v ~ p.

Let K = K(~) = the least < < p such that ~, p e JK+I" Set h = hp,A, and for T < p, set h We retain these conventions p

T

= h

T ,A qJ

throughout

T

(e.g. h will always be hp

for ,A

= p(~), A = A(~)). For K < T < p, set X

= h "(~ x (J T

T

x {p})).

T

By choice of p, we clearly have ~

X ~<~
<JT' A Q JT >"

By claim 2, X ~ v

~

L

1

= J . •

P

Lermna 5 For each Proof:

T,

X

T

~

J

is transitive.

Exactly as in lemma 3. Set

J~

= X T Q

J

.

Thus

<~ IK < T < p> is non-decreasing

and supK<~
=

~.

T Let rrr:

<X.r, A Q X.r > ="~ <JY ' AyT >.

S e t p~ = ~ r ( p ) .

T

Lemma 6 YT < v, and moreover, Proof:

Firstly, is f ~ J

<JyT ~ AyT>

E Jr.

notice that as <Jp, A> is amenable, p

such that f: ~ ~-+ X . T

B = { ~ s ~If(t)

e A}.

Hence X T s J p and there

Set E = {< ~, n > s ~21f(~)

Thus E, B E J . p

= p, this is, of course,

h T ~ J0 .

trivially

We claim that,

true.

Suppose

c f(q)},

in fact, E, B s J .

~ < p.

Then, placing

If

-164-

ourselves in Jp, we see that E, B ~ ~, ~ < v, and

v

is a regular cardinal, so

that E, B s Jr"

But if this is true J , then it really is true.

<~, E, B> ~ J .

But <e, E, B> is a well-founded,

its transitivisation must exist in J . ation is just <J

extensional structure, so

But by construction,

, ~, Ay >, so <Jy , A YT

Hence

> g J .

this transitivis-

QED.

Y~

T

Le~na 7 (i)

The sequence <<J

, A >I< < T < O> is [?Jp,A>({p}). YT YT (ii) The sequence is ~]Jp,A>({p}).

Proof: Directly from the proof of lemma 6. Set p< = y< = v< = ~ A Set E

= ~.

Clearly, v T

7 T for all T, so sup<~r
= {>IK ~ T < p}. Y

Len~na 8 E

is a []J0'A>({p}) predicate on J~.

Hence the monotone enumeration of E 0~ is

~]Jp,A> ({p }). Proof: Clearly, <~TI< .< T < p> is ~]Jp,A>({p}).

The lemma follows at once by lemma

7.1 Lemma 9 <J , E > is amenable. Proof: Let ~ < ~ be given.

Pick T < p, T > <, with ~

~ 6.

It is easily seen that if

we define ~ from <JYT' AYT>' PT' ~T in the same way that E <Jp, A>, p, ~, then ~I,,~ is an initial segment of E . z-l"E = E.

But ~ I ~

Also, ~ is cofinal in J~ , so in fact E = E

T

T

6, <JYT' AYe >' p~' v ¢ a

Jr' so E e Jr"

E

QED.

~ J~ = ~ ~ J~

¢

J~.

Let x s J~, y s J .

Hence, as 6 < vT,

The following are equivalent:

(a) x is Zl-definable from y, p in <Jp, A>; (b) x is Zl-definable from y in <Jr' E~ >"

~ J~ . ~

Lemma I0

was defined from

T

= id~,

so

But, by lemma

-165-+

Proof:

(a) ÷ (b).

Let x be as in (a).

Pick i e m so that x = hp,A(i, ).

Clearly, z = x ~-+ (~yr, Ay , p~, ~T)[z e J~r & z = hy~,Ay~(i, <~, p >)J. Thus x is as in (b). (

-> (a).

Let x be as in (b).

Pick i ~ m so that x = h

Y~ is cofinal in ~, z = x ~-+ ( ~ ) ~

< T < p & z = hyr,E~jy

lemma 7, is ~<J I p ,A> ({p}). <E OJ

,E~

J~ < r < p> is ~<Jp,A>({p}). I

(i, ).

Since

(i, <~>)].

By

And by lemmas 7 and 8,

Hence x is as in (a).

QED.

Lemma ii The sequences <E J~ e R>, have the following properties. (0) E 0 is an unbounded subset of ~ and E

~ J .

(I) <J , E~> is amenable. (2) Every x ~ J (3) <ETJ7 E S

is ~ I-definable in <J , E > from parameters in ~. n v ~ R> is uniformly ~]JP 'A>

(4) Suppose ~ ~ ~, E ~ J~, and ~: <J~, E > ~ Q ~(~) ~:

= ~ .

Furthermore,

<J,

m>.

Then ~ g R, E = E~, and

there is a unique ~ m o s u c h

that

<JP(~)' A ( - ~ ) > ~ I <JP(~)' A(~)> and ~(p(~)) = p(~).

(Note that by (2), ~ (and

hence ~) is uniquely determined by ~ . ) (5) Suppose ~ is a limit point of S~_ andS: then % ~ S ~ Proof:

<J~, E ~ > ~ Q

<J , E~>.

If X = supa"~,

and E k = E~ ~ J%.

(0) E 0 = {yTJK ~ r < p}, which we know to be unbounded in v.

(1) This is just lermna 9. (2) By lemma iO and the choice of p. (3) This is clear from the definitions.

~ s~

(Note that by definition of S ,

n~ ÷ p(~) < p(~).)

(4) Let ~, E, ~ be as in (4), and prepare for a long and tedious journey. We proceed in a manner analogous to that in lemma 4. Set M = {x e J Jx is ~ -definable P I

in <J , A> from parameters in P

(ran(~) ~ J ) O {p}}. Claim 4.

M ~J

= ran(W).

(cf. Claim 3 in lenma 4.)

-166-

Let x s M q J .

Since x s M, there are y s ran(~) A J

definable in <Jp, A> from y, p. definable in <J , E > from y.

such that x is

But x ~ J~ and y ~ J , so by lemma i0, x is But y ~ r a n ( ~ ) ~ l l

Conversely, suppose x g ran(~).

<J , E >.

l-

Hence x ~ ran(~).

Then, as ran(O) ~ J , (2) above tells us that

there are ~ ~ ~ such that x is ~ l-definable from ~ in <J , E >.

Let Y0 = <~> s J .

Let ~ be a ~0-formula such that: (i)

~<J~,E~> ~ y ~ t ~ z ~ z '

(ii) z = x +÷ z s J Set K = {YIY ~ J

& ~<~

[~(t, y, z') ~+ z = z']

& ~<Jv,E~> ~t~(t, Y0' z)

~ >~t~(t, y, x)}.

K is a ~]J~'Ev>({x}) relation on J . x s ran~)-
Since ~ = y~ = the least element of E$,

Since Y0 s K, K ~ ~.

<J , E >, K q ran(o) # ~.

So, as

Let Yl s K ~ ran(~).

Then Yl ~ ran(~) O J~

and, clearly, x is ll-definable in <J , E > from Yl by the formula ~. I0, x is ~l-definable in <Jp, A > from YI' p" M~

<JP' A>.

So, by lemma

But look, Y I' p g (ran((Y) A J ) o {p}

Hence x ~ M n J , and the claim is proven.

1 Let ~: <M, A q M> ~ <J~, A>, and set 6 = ~-I . Let ~ = ~(p), ~ = ~(~).

Thuse:

<J~, A>-fZI <Jp,

(Recall that ~ is ~I -definable from p in J P .)

A>o

As in lemma

4, we must establish a whole series of claims. (i)

&~J~ = ~. By claim 4, since M A J ~

(ii)

~ <

p "+ ~

<

p

Suppose ~ < 0.

is an E-initial segment of M.

__

& £Y(~#) = ~. Then ~(~) is defined, and hence less than p of course.

and claim 4, ~ ~< ~(~) < p, so ~ < p. Suppose not.

Thus v < ~.

Set ~ = ~(~).

We must show that ~ = ~.

Since ~ is the largest cardinal in J , and ~JB ''~ is

regular" and v < p .< ~, we see that ~j = (~+)J~. <Jp, A> and ~ = ~(~) ~ M.

By (i)

Hence ~ s M.

But ~ = ~

~ ran(cs) = M A J.~ c_ M-<~I

Thus, by (i) and claim 4, ~(~) = ~.

But

~(~) =-~ and ~ < ~, so we have a contradiction. (iii)

-~

~

S1

, and moreover, c¢- = ~.

We have~Y: J---~Q J , and by (i), (Y(~) = ~(~) = ~ . ~j ( ~ ) ( - ] ~

> ~)(~ >. e V ( ~ f

~ J~)(f: m

ON+TO

Since ,~ ~ S

~)). Thus, ~ j _ ( ~ ) ( ~

> ~)(~ >. ~ v

-167-

( ~f ~ J~)(f: e

ONTO ÷ ~)). Hence ~j_ "g < e + ~ is countable".

argument, ~ j v ( ~

>. 7)(I$ I = 171).

By an entirely similar

Suppose i=jq "7 is countable".

Then, by our last

ONTO result, ~ j _ ( ~ ) ( ~ > ~ ) ( ~ f e J~)(f: w ÷ ~), whence, crossing to Jv b y e , we v j see that ~j "all ordinals are countable", contrary to ~ = ~i v. Hence ~j_ "~ is uncountable", J cardinal] -~.

which implies, by our earlier remarks, that ~ = ~ This proves

Now, ~: <J--, ~>-< P ! = A~ (iv)

[largest

(iii).

<JP,

A>, so by lerm~a 7.20, there is -~ such that ~ = p~-i

, and a unique ~ -~y such that

" J~n

J8"

~ is In_l-regular at ~. Suppose not.

~.

=

Let f be a ~n_l(J~) map of a subset of some p < v cofinally into

By (iii), we may assume p = ~ here.

Then, using (iii) again, we can construct,

from f, a ~n_l(J~) map ¥ such that f"~ = v. which codes a well-order of ~ of type v.

Hence we can find a ~n_l(J~) set A ~

We show that A ~ J~.

If v = 8, then thereis nothing to worry about, of course. way of contradiction, ?'.

Hence v < 0.

v < 8.

So assume v < 8.

By

suppose A ~ J~.

[Why?

Then ~j_ "there is a map of a subset of ~ onto 6 Well, if O = B, then this holds because we are assuming

And, if O < B, then since ~ = 0~-i, ~j~ ,,~ is a regular cardinal", so by our

result which we claim to imply ~ < 0, we see that ~j~ "v < p", whence v < 0, as we said.]

So, by (ii), v < p and ~(~) = v.

is a map of a subset of ~ onto ~". is absurd.

Hence A ~ J~.

But ~"" J ~ ' ~ n

But ~J8 "v is regular", since ~ = B(v), so this

O.K., then, A g ~)(~) f~

least one definition, means that ~

J8 and ~(~) = ~, so ~J8 "there

n-I

In_l(J~)and

A ~ J~, which, by at

-n-i ~< ~. But 0~ = P >~ "~ > ~, so at last we have

our desired contradiction to prove (iv). (v)

v is not In-regular at ~.

By definition of M, every x ~ J-- is [ -definable in <J--, ~> from parameters in O 1 0 --~ ~{p}. Thus J--p = h~pZ'(~, × (~ x {p})). Hence there is a El(<Jp, A>) map of a subset of _~ cofinally into _v.

Since p = p~-l, ~ = El,

(vi)

7 = 8(~), n = n(7).

(By (iv), (v)).

(vii)

-v ~ R.

this map_ is in fact ~n(JT).

(By (vi), since, as $: J~-<~l JB' we have lim(~).)

-168-

(vii)

7 = P(V), A = A(~).

(ix)

~ = p(~).

n-I .) (By (vi), since 7 = ~ n-I , A = A ~

Now, by definition of M, every x s J--pis El-definable in <J~, ~> from parameters in ~ u {p}.

JP and ~(~) = ~, d(~)

And as 6: J ~ l

f r o m p in J~p, since the same is true for ~, p, p. suffices to show that ~ is the least such.

=

p, we see that ~ is 11 -definable

So, by virtue of (vi) - (viii), it

Well, suppose p' <j p is another one.

particular, then, p = h~p,~(i, <~, p'>) for some i s ~, ~ ~ ~. ~

~

setting p' =~(p'), ~(~) e ~.

Applying 6, and

~

~

we see that p = hp,A(i, <~(6), p'>).

In

_

Since ~(~) = e, we have

Hence, every x ~ J

is E -definable in <Jp, A> from parameters in ~ d {p'}, P 1 since we can replace p by hp,A(i, <~(~), p'>) in any such definition of x from

parameters in ~ o{p}.

Also, as ~ is El-definable fromp'

he ~l-definable from p' in Jp. contradicts the choice of p. (x)

6(K(~)) = ~(~).

But as ~' <j p, we have p' <j p, so all of this Hence p is <j-least, which is what we required.

(By (iii), (ix), and the fact that ~(~) = ~, ~(~) = p.)

We can now speedily dispose of (4) of the lemma. since

,

in J~, we see that ~ will

First, note that ~ is unique

-- A> from parameters as we observed above, every x ~ J~ is El -definable in <Jp,

in ~ ~ {~}.

Secondly, note that by (iii) & (vii) - (x) above, ~"E-- is an initial

segment of E .

But look, ~"m--v = ~ " E ~

= E~ q ran(O) =~"~.

Hence E--v = E, and (4) is

completed. (5)

Let v be a limit point of Sc~_, and suppose~: = sup~"~.

<J~, E ~ > ~ Q

<J , E >.

Note that as Sc~ is clearly E~J~-'E~>({~}), and as S ~

Set is

E~J~'E~>({~ }) by the same definition (and, by (4), (7(~) = a ), X is a limit point

of

Sc~ , s o i n ,¢

fact,

X ~ Sa . -,)

Set ~ = p(~), A = A(~), p = p(~). ~: <J--, ~>-< P GG)

show that

Set

n = sup~"~,~=A

be the unique map

Note that if e = m--, ~ =

P

so ~ ~ r a n ( & ) .

E3, = E ) cl J x "

By (4) above, let ~ e ~

<J , A> such that ~(p) = p. ~1

= ~,

We m u s t

~J

n

.

x~ Note that

fact 6"J--~ J .

then a,

p,

~ c J

T]

,

and i n

By E -absoluteness, 6: <JT' 7>-<~ <Jn' i>, so as this embedding is 0 ~0 • ~ coflnal,cY: <JT' ~>-~Z 1 <Jn' A> ' cofinally. p

n

Set X = Xn (= hn(e x (J~ x {p})) = hn,E(m × (Ja × {P}))')

-169-

Claim 5.

ran(~) ~ X. -~ Jp , so if x ~ ran(~),

Now, ran(~) parameters

in ~ u {p}.

Hence ran(~) n h"(~ x ( j

since h f~ (Jn) 3 = h n and ~, p c Jn' Claim 6.

v

then x is El -definable

ran(~)

x {p})).

~-h n "(~ x ( J

in <J p, A> from

But ran(~)

x {p})).

c J . n

Hence,

QED in claim 5.

= %.

n

By claim 5, (*) E e ran(~) there is ~ e ran(~)

f~

~ ~ ÷ ~ c X q ~ = ~ . n

such that

t< E•

Let

By (*), ~ e ~n .

~e i. Hence

By choice of ~, Lg v •

Thus

n Now suppose Now, ~ = sup @ " 7

$ s ~n"

(and lim(~)),

E = hT(i,

).

~"~

=(Y"~ -~%. Thus

Then E ~ X, so for some i e ~, z ~ J , E = hn(i,

Thus

so we can find • < p such that,

~ e X r ~ v = ~.

Hence

~ ~ %, proving

By lemma 8, $ ( ~ ) v

).

setting • = ~(~),

= u~.

So, as ~-r < 7,

.< %.

claim 6 is proven.

Set y = Yn' A Note that,

¥

= A

y~

, ~ = ~ . n

Thus,

~: <X, A q X> ~ <J , A > and ~(p) = pn. = Y

as <J~' A>', T-I: <Jy, Ay>-<10

by claim 6, ~-l~l

<JP' A>.

Also, note that

= ida%.

By le~na 7.20,

there is ~ such that y = p~-l, Ay = A ~ -I, and a unique

~

_= ~-i such that ~: J~-<~n JB" Claim 7.

~ = B(1); n = n(1);

Leaving result

¥

the proof of claim 7 for a moment,

(5) follows

~"E l is an initial

from all of this• segment

E o ~ J%, and we are done. resemble

% s R; y = p(%); A

somewhat

of E .

the verification

of

p~ = p(1);

let us observe

= ~-IIJl = idIJ%.

~[J

(4) above.

= ~-l~jy; ~-l~jx_ = id~J%; ~-l"j Y

These

We establish

(ii)

= X; X q v = X. Y

just collect

together

various

% < 7 ÷ ~ < n .< P & ~(1) = v.

= ~.

how the required

Hence,

E l = $"E% =

now with the proof of claim 7, which will

before.

(i)

~(<(%))

Well, by claim 7, it is easy to see that

But ~ J l

We commence

= A(%);

remarks

made

above.

a chain of results,

as

-170-

Suppose % < y.

Then by (i), ~(%) >~ o, so as ~(%) = v-l(1) < n, ~ < n ,< P.

since v < p, v = h(i, ) for some y ~ J = h (i, ) e X, (ili)

Thus u(v) is defined.

(by definition of p).

Now,

Since v, y, p e jn,

By (i), ~(v) = l, i.e. ~(1) = v.

~ is In_l-regular at -~. This is proved in exactly the same way that we proved (iv) in the verification

of (4). (iv)

~ is not In-regular at ~. By claim 5 and the definition of ~, X = sup(X ~ %) = supEhq,~(~ × ( J × {p}))~%].

Applying ~, ~ = supEh

A (~ x ( J x {p })) n ~]. Hence there is a [l(<J¥, Ay>) map Y' T n-i of ~ cofinally into X. But y = p~ , Ay = An-I so this map is [n(J~). (v)

~ = B(X), n = n(%).

(vi)

% e R

(vii)

y = p(l), A

(By (iii),

(iv).)

(By (v), since, as ~: J---
(By (v).)

Y (viii) p

n

= p(%).

By definition, X = h ,~(m x (J~ x {p})). Thus every x e Jy is If-definable

Applying n, Jy = hy,Ay(~ x ( J

in <Jy, Ay> from parameters

in e O {P~}.

× {pn})).

The rest

of the argument is now as in the proof of (ix) in the verification of (4). ~

(ix)

~(<(X)) = <.

(By (viii), since ~(e) = ~.)

That completes the proof of claim 7, and hence completes the proof of lem~a ii.I Finally, it is easy to see that lemmas 4 and II together imply ~ .

(Really, it

just requires the observation that the sets P and R are so simply defined, that clause (3) of the two len~nas combines to give one definability statement.)

Chapter 14

THE GAP-2 TWO CARDINALS THEOREM.

We shall assume V = L throughout this chapter. ~K ~I[<~ ++, ~> + ].

(JENSEN)

Our aim is to prove that

The basic strategy is as follows.

We shall be given a

++

<<

,

K>-model ~L.

We shall then construct, by induction, a <%++, %>-model

~ ~0"L .

The construction will differ from the gap-I construction only in that, instead of constructing an elementary X+-chain of models, we shall construct a directed elementary system of models following a k+ -morass.

(We shall call such a system a complex.)

We commence by describing the construction of a model complex. From now on, ~ Let ~

denotes an arbitrary l+-morass, fixed once and for all.

he as before, and suppose ~

T be a theory in ~

contains a binary predicate symbol -=.

such that T ~ "-=~ linearly orders the universe".

Let

For any set X

and any point a, X-= a denotes the set of sentences {x~= alx s X}. An ]~-complex for T is a structure ~ that

writing MT for ~vs

(CI) (T s S I ÷ M r ~ T) (C2) • ~ S 1 ÷ e

e M T

&

The maps O~--TT:

M

<(MT)TeSI , (eT)TESI , (~TT)T_~T> such

and M* for ~
(T s S I - S%+ + ]M I = %).

& M T

~ M*< T

(C3) u ~ s l n T ÷ M u - < M ~ (C4)

SaT

=

T

(& M

e . T

+ M , by (C2)).

~ MT, T--~ T, form a commutative system.

(C7) ~'u < om-->£Y-r TT [~ V = id[M ~, any ~ --2 T. (C8) T a limit in-~ ÷ M

= UT

T ~T

cy- "M--. TT

T

We show how, in certain circumstances, one may construct an ~-complex.

At

various points in the construction, which we shall denote by the symbol (~), we must assume the existence of a certain model.

In any specific case, it must therefore be

verified that at each such point, this assumption is justified.

In particular, we

shall eventually show this is the case for a certain situation we will develop in order to prove the promised two-cardinal theorem.

For convenience, we assume ~

is a

-172-

skolem language

and T is a skolem theory.

To commence

the construction,

with an arbitrary

model M

of T of cardinality

0

choice of M 0 may be somewhat Case I.

~ minimal

If • =

by induction

on • c S I, we start

%, though in specific

set

M

there

is

Otherwise,

M >M*

such

that

proceed for

as follows.

s o m e x c M, M ~ M * . . ~ x .

= the

skolem

hull

o f Me ~ { x }

Case 2.

-

(C3)

~ a limit point

Consider

direct

limit.

still

the directed Define

By commutativity

sufficiently Then x c ~ by

hold,

whilst

system <(M )

a map j :

= id~M*.

Otherwise,

Similarly

Case 3.

then M

=~--

T'r

back"

~ x-~ ~ .

~ immediately

that

T = ~--

for

T ..~%

"M--.~

>.

-

have

(cs)

arisen.

all

(O'v),~ ~

Let <M,

r --~'r,

> be a

~) ~ S 1 (3 r ,

as follows.

~-- "(Sc~ - ~ 7 ) . TT

Let x c M *T.

So as ~-- ~-- = id~--,

%_

TT

Then by

(e~)

T

= M.

T

T

Pick -T

[If ~v-~ ~'T' just pick T with ~--T > ~ "

v c SaT n

T, so ~ is not maximal

(C8), which holds

for any ÷x c M*T' whence

T

T

(c4)

j: M*T "<M°

For T -~ T, setC~--

in S

at ~ by hypothesis,

high in --~ to achieve x so'~"M--.v ]

Then M * - ~ M, so set M

By a "pulling x s M*,

in-~.

(and ~) sufficiently

by e

of

Hence v = ~ T (~) for some -T --~, -~ E S I ~ 7 .

and x = O ~ ( x ) .

T

no new cases

such a j can be defined

high in -~ to obtain x c ~

define e T c M

= x.

, (¢Y~)~_~_~

M* ~ M such ~C

considerations,

~T~" (S~n 7).

O--T o tY--~v-l(x).

J

~

commutes

(M4), ~ is a limit point

can pick v

set

•

For some ~ < T, x ~ M . By (M5), SaT n r = UT_~r

such

in--~.

= ~-D U (~), the following

S~

in M and

T

(el)

For

T

T

Clearly,

the

in~.

T

M, x ,

instances

less cavalier.

Q S I, set M~ = M 0.

(~) Assume

which proceeds

, so we

Set j (x) =

Choose M so that

TT

=CY--. T

By eom~nutativity,

for any "r-~ ~.

argument

of the kind just used to define j, we see that if

It is now easy to check

succeeds ~ i n - ~ .

(CI) - (C8).

There are three

suheases.

-173-

Case 3.1.

r minimal in S~ .

Then ~ is minimal in S~_, and M* = ~ . T

T

T

(~) Assume there is M >-M* and ~: M*~-< M such that M ~ M * < ~Y(e 7) and ~e

rM*

M* M* = i d ~ P r ~ ( @ 7 ) , where Pr (x) = {z ~ M* IN* ~ z ~ x }

(~)

Set ~T = ~ ( e ~ ) ' ~-- = ~.

for any x ~ M*.

be the skolem hull of Me~ ~ {e~} in M, and put

and let M

(~I) - (C8) are easily checked.

TT

Case 3.2.

T immediately succeeds ~ in Se .

Then • immediately succeeds ~ = ~ - i

(n) in So_, and Me = M , M-% = M--. T ~ T T

(~) Assume there is M ~ M * (= M ) ands: M * ~ M such that M ~ M * ~ T D T T M* ~Pr~(t)

M*

= idtPr~(t), where t = ~ T n ( e T )

Set ~

= o(t), and let M T

~-- = ~ - -

TT

~(t) and

(so t ~ Mn = M*).

be the skolem hull of M* ~ {e } in M, and put T

T

T

.

~

(CI) - (C5) are easily checked, whilst no new case of (C8) arises.

We check

(C6), which then implies (C7) by induction. To verify (C6), by induction we need only consider ~ E Sc~_ ~ ~.

And again by

T

induction, M~T I= M ~ Thus ~ ( ~

e~.

Hence as ~ n ~ M T ) : ~-T--<M , we see that M

"M~) = id~(~n"M~n), which yields ~ T ~M--n= ~ ° U ~ n ~

~U~"M--~t.

= ~ n I M ~ n ' as

required. For future use, we note that for any y e Me (*) M n ~ ~ [ ~ , ~ ( ~ ) ]

and x s

, any 2-formula ~,

iff M T ~ [ ~ , ~ ( ~ ) 3 .

(Apply ~ to the LHS and note that, as e-- - ~ n M ~ , which holds by (C2) and induction n M* M* hypothesis, an application ofonn-- yields en ~ ~ t, so M*n -c Pr~(t).) Case 3.3.

T a limit point in S ST

Then T is a limit point in Sc~ T. Case 3.3.1.

There are two subcases.

% = sup <~iT(~) < T.

(~) Assume there is M ~ M * and ~: M*-< M such that M ~ M * ~ ~(e%) and M* M* T T ~IPr T(e~) = id~Pr=T (fX).

-174-

Set T

~o m ~

= o~%),

let M r be the skolem hull of M*T O{e~} in M, and put~--rr =

•

The proof that (el) - (C8) hold is similar to, hut easier than, in Case 3.2. Note that for any y e

k

(*) M%

~,

, x ~

~=~[~,

~YTI(x)] iff M r

Case 3.3.2.

sup.<~ ~ ( . )

%

,

a fo=ula ,


= ~.

F o r each . e Sa_ Q T, let q(.) = the least q (in-~)

such that .-~ O-q~TT(.).

T

(Note that

a s ~rT.r ( . )

i s n o t m a x i m a l i n Sa

, there

is

no p o s s i b i l i t y

that

q = "rrT.r ( . )

T

Clearly,
here.)

(.)

I" ~ Sa_ ~ ~> is non-decreasing.

By (M4)

it is in fact

T

strictly increasing. s S

Set ~ = sup.aso~_~ ~ an(.).

(~T ~, there is a .' e S

C~-T

By (M3), a aS and in fact, whenever

such ~hat ~--~ v'-9 ~-- (.).

C%

So by (M7) and the fact

TT

that T irm~ediately succeeds ~- in -~ , we see that ~ = ~ . T

We shall define M

T

as a

diagonal limit of the models M I(v), ~ c Sa_ ~ ~-. T

For., n(.,

y)

y s S_

c3 T w i t h

= ey,n(T)(.)

q s S

= the

. ~< y ,

set i

unique

I I

such that , =9 n -e -- (~).

//

q(~,Y)

Then:

/

.-~n(.)-~n(.,

(i)

Y)-~

i

~(.)

i

c M*n(", y) (ii) ~Yq(.),q(.. y) "M* q(.) -Since ~T--T is cofinal on T in T and the ~ (.)'s are cofinal in ~T' it is easily

%,

seen that

T

(iii) M*T = O"sSa_ a 7<~n("), ~ T ( " ) "M*n(")" T

Claim.

If.,

y s S N -T, . .< y, -> y E M* x E ~T then ~n (.)' ÷ ' T

Mq(.) ~ * ~ ,

~

÷ ,q(.)(~)] iff Mq(y) ~ %[~I("),q(" • ~) (Y)' ~ , n

÷)] (~) (x

Assuming this for the time being, we define MT, etc. as follows. Let X, h be arbitrary with X c3M* = ~ and h: (~T - M~T*) +-+ X. X DM*. T

Define

a set - -

mappingS:

M-- ÷ M by T

Let M be the set

-175-

h(x), if x ~ M~m ] ~CY~(x), where ~ ~ S~_ t ~

O'(x)

& x s M--~& ~ = ~--m~(~)' otherwise.

By (C6) and induction, this clearly defines a mapping.

We ma~e M into a structure and

o into an elementary embedding by demanding that for all ~-formulas ~, and all --> ~ ~ S ~r, ye ~-T

(iv)

M ~ $~(.)

M*

n (~)'

÷x ~

~-"'~'

_ (v)(y), ~Y(X)] iff Mn(~) ~ *[~, ~ ,~(~)(~)]. ~TT

By (iii), the y's take care of the whole of M*, and the x's take care of X. Hence all of M is covered by this definition. between different choices of v. that o~: M--<M,

By the claim, there is no conflict

Hence this does define a unique structure on M such

By an argument as in Case 2, there is j: M*-<M. T

that j = id~M*, so M*m ~ M "

Set M

= M,o~Tr = ~ ,

and @m =eft- (@~).

limit point of S~ , a simple "pulling back" argument shows that M verification of (CI) - (C8) is now straightforward,

Pick X now so Since r is a ~ M*-~ e • The

(C6) holding by definition, of

course. remains the proof of the claim, which is by induction on y e S _ ~ ~. If r = ~ S _ , the result is trivially true. Suppose ~ immediately succeeds ~ in Se_. There

T

T

Then, by induction, it suffices to prove the claim for this one pair ~, ~.

Now,

->

fy~,n(~)(~), y s M n(~)" O ~ q(~),n(v,T) : MO(~) -< Mn(~,Y)' and cY n('o),o(~,Y)°cr~,~(~)= C7 ,q(~,y), so for any $, we will clearly have Mn(~) ~ ~[~, ~

,n(~)~)] iff Mn(v,y) ~ $[~D(~),n(~,y)(~), CY ,n (~ ,y) (x)] .

Now, n(Y) immediately succeeds y in-~and n(y) immediately succeeds n(~, Y) in S . Hence Case 3.2 applies to n(y)% (~)

Note that, as ~ ~ M* (ii) above gives ~ (,~)'


So by (*) of 3.2, we obtain

M (~,y) ~ +[%(~),~(~,y)(~),

EY ,~(~,~)&)] iff Mq(y) ~ #[c~ (v),~(~,y)(y),~y,~(y)(x)].

v>

The claim follows from these two equivalences. in S~ . Let X = sup

~ (v). ~
3.3.1. applies to q(y).

Finally, suppose ~ is a limit point

Either X = ~(y) or else X < ~(~) and so Case

So, either by identity or else by (*) of 3.3.1., respectively, ~

-~

we see that for all y s MX, x ~ (v)

<

,

M (y) ~ $ ~ , cY¥,~(~f)(x)] iff M 1 ~ Sly, C¢ ,l(x)~.

Let ~ = supv< eq(~).

-+

Let ~ be the unique ~ ~ S

such that Y-4 ~

X.

Then

-176-

immediately succeeds y in -~; for if there were ~ such that y ~ ~ ~ , e S

~ y we would have ~

~ ~,

then for

contrary to the choice of ~ and n ~ S .

So~

T

the definition of %, we see that Case 3.3.2. itself applies to q, whence~

recalling

by induction, for y s M ~q(~), ~ s Scr_ ~ Y' ~x g My , we have (by (iv)) T

Mq(~) ~ ~[y, ~

,q(~)(x)] iff M~n ~ ~ [ ~

Applying ~Y~%: M ~

(~),

y,~(v)

to the RHS we obtain, by commutativity,

LHS iff M% ~ +[~q(~),~ . (v)(y),
O'y,q(y)(1)]

which is precisely what we wanted. This completes our "construction" of the ~ - c o m p l e x ~ . ++

We prove that, if V = L, then from now on.

~

~%[<<++, ~> + <%

, l>].

So, assume V = L

We consider the case ~ = ~ first.

Let 0-[ be a given <<++, <> ~-structure. Let H ++ be the set of all sets + hereditarily of cardinality ~ < . We may assume ~ has domain H<++ and U = <. ++

Let -~ be a well-order of H<++ of type < assume that

0~ =
If X is a set, I X

such that x ~ y + x - ~ y .

We may

e,...>.

denotes the language ~ with the elements of X added as

constants. • -+

Let

y, x) he a set of formulas. (~

-> ~ ~y~(y, ~) for {

Assume ~

~ ~#(y, ~)I(~ = ~I ~

9 ->

...

We write

->

-~

~-

->

-+

~y~(y, x) for { rye(y, x ) I ~ }

^ ~ n )& ~I ....

,

~n ~ ~}"

is recursively arithmetised in ~ from now on.

Similarly

A set I(Y) of

~i~i-

-+

formulas is M-arlthmetic, any set M, iff there are u e M and an arithmetic set (y, I) of ~-formulas such that ~(y)

1'(u, ~).

An ~-structure M is arithmetically U-saturated iff whenever ~(x) is an M-arithmetic set of£M-formulas with M ~ ( ~x E U)~(x), there is u e U M such that MI = ~(u).

The following is easily established.

and

-177-

Lemma i (i)

Let ~I0 be any ~-structure. 1

(ii)

There is a countable arith. U-sat, structure

0

Let B < ml'

and l e t

< ~ l a I ~ < B> be an e l e m e n t a r y t o w e r o f c o u n t a b l e ,

U-sat. ~-models with the same U.

arith.

Then ~ = 0a
Define a quantifier Q ("there exist -~-cofinally many") by

Qx~ (x) ++ (Vz) ( ~I > z)~ (~). ++

Since ~ has type <<

, K>, we clearly have

(.) ~ ~ (q:) ( ~x ~ u)~ (~, :) ++ ( ~ Define Qx~(x) ÷ ~ as For any

~ u) (~)¢ (~, :).

~X~ (x) was defined,

~[-structure

M, Th(M) denotes the complete ~M-theory of <M, (m)mgM>.

L ermna 2 Let M be any ~-structure, ~(~) any set of ~M-formulas. iff Con(Th(M) + M ~ x

Then M ~ Q~(~)

+ ~(x)),

Proof: (+) M ~ Qx$(x) ÷ ( ~ m E M ) ( ~ z ~ M)[M ~ $(~) ~ m -~] ÷ M-~ ~ + {$ (~)} is finitely satisfiable in PI ÷ Con(rh(M) + M - ~

+ {~ ~)}).

(÷) M ~ ~ Q ~ ( ~ ) * ( ~m c M)[M ~ (~x)(~(~) ÷ - ~ +-~Con(Th(M) + { m ~ x }

~m)]

+ {$(~)}).

I

Lena 3 Let M - ~ be countable and~rith. U-sat. formulas. =

Let ~(x) be an M-arith set of £ M -

Then M ~ Qx Z÷ (x)+ iff there is a countable arith. U-sat. M' ~ M such that and, for some c e

,

~ M ~ C + ~(c).

Proof: (÷) By lemma 2 . (2) L e t c be new c o n s t a n t s .

S e t M = M U {c}.

T h e n , by l e n a

T O = Th(M) + M - ~ C + ~(c) is a consistent ~_-theory.

2,

It clearly suffices to

find a model M' of T O such that whenever H(x) is an~-arith, set of ~ - f o r m u l a s with M' ~ ( 3 x ~ U)H(x), there is u e UM such that M' ~ H(u).

-178-

Let enumerate all M-arith.sets

of (such) ~ - f o r m u l a s .

We

I

define, by induction on i ~ ~, certain consistent extensions T. ~ T such that z 0 i < j < m ÷ T. c_ T.. l ]

To make the construction work, we shall ensure that

÷

T i = Th(M) + M -~c + X i, where X i is M-arith, X 0 is M-arith and r 0 = Th(M) + M ~ X i are defined. Otherwise,

for each i.

+ ~(~) = Th(M) + M - ~

If -~Con(T i + ( ~x s U)Hi(x)),

+ X 0.

->

Let Xi (x), Hi (x, ~) be

--

-~

÷

So by (*),

÷

(~x ~ U)(Qx)[ Xi(x ) +Hi(x' x)].

MF

Then

c U)Hi(x, -> x))), so by lemma 2, M ~ Qx - > I X i(x) ->

+ ( Bx e U)Hi(x , x ) ] , i . e . M ~ (Qx)( ~x e U)[-X i(x) + + H i(x, x)]o ÷

Suppose Ti,

such that X i = Xi(c), Hi(x) = Hi(x' %).

Con(Th(M) + M -~x-> + (Xi(x) -÷ + (~x

Thus

set Ti+ I = T i, Xi+ I = X i.

Con(Th(M) + M ~ ÷c + X i + ( ~x E U) Hi(x)).

M-arith. sets of ~M-formulas

Set X 0 = ~(~).

By arith. U-sat. there is thus a u s UM

such that M ~ Qx Ix - i (x) ÷ + ~i (u, ÷x)].

So, by lemma 2, Con(Th(M) + M~= ÷x + Xi(~)

+ Hi (u, ÷x)), i.e. Con(Th(M) + M~= ÷c + X i + Hi(u)).

Set Ti+l = T.I + H.(u),l

Xi+ 1 = X i + H i(u). Let T = ~.

T..

i<~

Let M' by any countable model of T.

Clearly, M' is as required.I

l

Define another "cofinally many" quantifier by Q , x , ( ~ ) ++ '~[Iz ++

Since 6~ has type <~

I .< < +

~x(~_= ~ ~ I~1 .<

< ^ ,(x)~].

, K>, we c l e a r l y have

(**) O~ F (Q'x)( ~x e u)#(x, u) ~+ ( ~ x E U)(Q*x)@(x, u). Exactly as before, we may prove: nellmRa

4

Let M ~ fi~-be countable and arith. U-sat. M

-formulas.

Let %(x) be an M-arith. set of

The following are equivalent:-

(i)

~ F Q*x~(x)

(ii)

Con(Th(M) + ~ M ( X )

+ ~(x)) !

(iii) There is a countable arith. U-sat. M' ~ M such that U M

e E M',

M

~I~M(C)

where~M(X)

+

[(c),

= {IxI ~< <} o {z _cxlz e M A M

the Q-set M-~ x. I

= UM and, for some

F Iz[ ~< K}, the

Q*-set corresponding

to

-179-

For an~ ~-formula ~(~), let ~(x) be the ~-formula

Let ES(x) = {$(x)J~(~) is an G-formula}, an arith, set of ~-formulas. If M -= 6~e and t g M, then clearly, M ~ ES(t) iff prM(t) -< M.

(Hence "ES" stands

for "elementary submodel".) L ennna 5 Let M -= 0-L, M ~ ES(t), a ~ prM(t). M ~ (~w Proof:

-~t)(Q~)(~

For any ~M-formula 9, "2

÷

÷

g U)[~(a, w, u) ~-+ ~(a, v, ~)].

e~(U ~) e [HK++] ~+, so ~)~(Un) c H ++, and in fact Furthermore, M ~

~

~ ( U n) is an ~-definable point in 0~.

~z(e(U ~) ~ z).

So as Pal(t) ~ M, prM~(t) ~

~z(e(U ~) ~ z). Now, M z ~ ,

~ z ( ~ U ~) ~ z).

M ~ ( ~z ~ t)(~(U n) ~= z), whence ~)M(u~) e Pal(t).

so Hence

By choice of ~ ,

therefore, ~)M(u~) = prM(t). +

+

Let w -~t. b ~ prM(t).

+

Set b = { ~

M

•

Then b ~

Set ~(b, a, ~) ~ ( ~ ] ~ U)[<]> ~ b ++ +(a, ÷v, ~)].

M ~ (Qv)~(b, ÷ a, ~).

Well suppose not.

Then M ~

b, a ~ priM(t) -< M, there is such a z in Pal(t). M ~ ~[~(b,

+

~(a, w, u)}.

a, ~) ÷-,~ ~ t].

Let ~(x) be the

~z ~ [ ~ ( b ,

), so We must show that

a, ~) ÷-~v÷ ~_ z].

So, in particular,

But w ~ t and yet M ~ ~(b, a, w)'

I

~-formula ~(x) -- ( V a _~ Pr(x))(la I ,< K ÷ a s Pr(x)).

ES*(x) = ES(x) + {O(x)},

So as

Let

an arith, set of ~-formulas.

We are now ready to prove the two fundamental lermnas for our proof. Lennna 6 Let M -_- ~

be countable and arith. U-sat.

There is a countable, arith. U-sat.

M' >-M such that UM'= U M and for some x c M', M' ~ M -~x + ES*(x). Proof: By lemma 3, it suffices to prove M ~ QxES*(x).

But thus clearly holds for 0 ~

Lenuna 7 Let M - ~

be countable and arith. U-sat.

Let t ¢ M be such that M ~ ES*(t).

Then there is a countable arith. U-sat. M' > M such that U M' = U M, and a map

-180-

o'; M - ~ M' such that ~[prM(t) = id[prM(t) and M' ~ M-~

~(t).

Proof: As a trivial consequence of lermna 4, we see that there is a countable, arith. U-sat. M > M

such that ~

= I#M, and an a E M such that M ~ M ( a ) .

Since M ~ lal ~< ~, let f ~ M be that element of M such that (recalling that U ~ = K) M ~ "f is the ~ - l e a s t <-enumeration of a". Then, ignoring f for a few moments, we have: (I) z ~ M + ~

z ~ a.

For let z e M.

Then {z} g M, so as M ~ M ( a ) ,

d = (aaPr(t)) M.

M ~ {z} c a, so M ~ z s a.

Let

Then

(2) M ~ d ~ t. For as M - < M ,

M ~ ~(t) ^ Id I ~< K ^d ~Pr(t).

Since M ~ Idl ~< ~, let g e M be such that M ~ "g is the -~-least ~-enumeration of d". Let H(t', a') be the set of all ~M-formulas ( ~ u E U)[~(d, t, a, ~) ~+ +(d, t', a', u)], where ~ is an ~-formula.

Thus H is M-arith.

Since t, a _=~t (by (I)), lem~a 7 yields M ~ (Qt', a')H(t', a'). is a countable arith. U-sat. M' >-M such that (3) M' ~ N ~ t ' ,

= ~

So, by lemma 3 there

(= UM) and for some t', a' s M',

a' + H(t', a').

By definition of H, M' ~ la'l -< K, so let f' c M' be such that M' ~ "f' is the ~ - l e a s t ~-enumeration of a'" By (3), for all ~-formulas ~, we have ->

(4) M' ~ ( ~ u

->

~ U)[#(g, t, f, u) ~-+ ~(g, t', f', u)].

It follows that for any u c U, (5) M ~

f(u) ~= t implies M' ~ f(u) = f'(u).

For suppose u E U and M ~ f(u) -= t. ~ f(u) -- g(u').

Then M ~ f(u) ~ d.

So for some u' s U,

So by (4), M' ~ f'(u) = g(u') = f(u).

Also, for any u ~ U, (6) M ~ f(u) --="t implies M' ~ M ~ f ' ( u ) . For let u ~ U, M ~ f(u) _~t.

By (4), M' ~ f'(u) ~t'.

But by (3), M' ~ M ~ t ' .

Hence M' ~ M -=f'(u). A particular consequence of (4) is that for all ~-formulas ~ and all u E U,

-181-

(7) M' ~ ~(f(u)) ~+ ~(f'(u)). Finally, if x E M, then by (I), M ~ x s a, whence for some u e U, x = f(u).

Thus we

may define a map ~: M ÷ M' by setting O(x) = f'(u) where u s U is such that x = f(u). By (7), ~: M "~M'.

By (6), M' ~ M"~ oit).

By (5),

We are now ready to prove our theorem.

Let j ~ b e an ~I -morass.

an ~ - c o m p l e x of models of the theory of ~ . arith. U-sat.

~[Pr~i(t) = id~Pr~(t).

I

We construct

By lemma I, let M 0 ~ 67~ be countable and

We construct the J~-complex as above, with M 0 at the start, so that,

in addition: (C9)

r E

S1 -

S~I ~ M

T

M M0 is countable arith. U-sat• and U r = U .

(CI0) T ~ SI÷ M T ~ ES~(eT ) (so by (C3), ~ ~ S I n T Case I.

Lemma 6 takes care of this case, so we must only check that U

induction, UM~ = UM0. Case 2.

By lermna 8, U MT = U M~

•

Hence UM~ = U

M0

MT

=

uM0.

By

.

All we need to do is check that (C9), (CIO) are preserved.

Now, any M - a r i t h . T

M . T

÷ M*T ~ ES*(c )).

(CIO) is trivial.

o f ~ M - f o r m u l a s c o n t a i n s o n l y f i n i t e l y many c o n s t a n t s f r o m T So, by a "pulling back" argument as used to define j in Case 2, we can reduce the

question

set.

t o one c o n c e r n i n g N-- f o r some T--.~ T. T

The r e s u l t

f o l l o w s by i n d u c t i o n .

Hence (c9) i s a l s o p r e s e r v e d .

Similarly,

UN~ = UN'r = UN0 f o r some ~ - . - ~ .

Case 3.1.

Since M* ~ ES*~--) by induction hypothesis, lemma 7 yields all we require T T

here. Case 3.2.

Since M*T ~ E S * ~

( e ~ ) ) by induction hypothesis, we have all we require in

len~na 7. Case 3.3.1.

By lemma 7 again.

Case 3.3.2.

By (iv) of 3.3.2 and an argument as in Case 2 above, (C9) and (CIO) are

preserved here. Hence the theory of ~ h a s Let ~ =

[J

M .

Then ~

~ ~

an /[-complex ~ = <(MT) TeS~, (~T) T¢S% ( ~ T ) ~ _ ~ T >" M0 U = U , and (by (C3)), l~I = °~2, so ~ i s as

++ required.

This proves <<

' ~> ÷ <m2' w>.

If X is regular and uncountable, <~++, K> ÷ <X ++, %> is proved in a similar fashion, using a X+-morass, with U-saturation replacing the notion of arith. U-

-182-

saturation. If % is singular, a similar argument with U-special structures in place of Usaturated structures, using

~% to code up the construction of the U-special models

in terms of U-saturated models (as in Chapter 12), will give <<++, <> * <%++, %>.

Chapter 15

SMALL LARGE CARDINALS

We consider the following types of cardinal: inaccessible, Mahlo, weakly compact, and ineffable, and show that the existence of such is consistent with V = L. We shall not, however, discuss these properties in any other way. First, let us observe that since the predicate "v

is a cardinal" is ~i' every 0

cardinal must be a cardinal in the sense of L. Recall that K is inaccessible iff < is regular and ( ~ e ,

% < <)(8 % < K).

Theorem 1 If K is inaccessible, then ~L "K is inaccessible~ Proof: The predicate "v 0 is inaccessible" is~] IRecall that ~ is Mahlo iff every closed unbounded subset of ~ contains an inaccessible cardinal. Theorem 2 If K is Mahlo, then ~L ''~ is Mahlo". Proof: Suppose C s L, C ~ K, and that ~L "C is closed and unbounded in R".

Then, in

the real world, C is closed and unbounded in K, so there is an inaccessible cardinal % in C.

" ". By Theorem I, J=L "~ ~ C & ~ is inaccesslble

QED.

Weakly compact cardinals have been discussed in chapter ii, of course.

The

results there obtained gain importance by virtue of: Theorem 3 (Keisler) If ~ is weakly compact, then ~L ''~ is weakly compact~ Proof: A standard equivalence of the notion of weak compactness for ~ is the statement that for every structure of the form , where T ~ V K, there is a transitive structure <X, s, T'> such that V -< <X, s, T'>.

K

g X and

(This equivalence is due to Keisler, and can be

established in several ways.)

Using this fact, we shall establish our theorem

-184-

by proving that with ~ weakly compact, ~L SH(<). ~L "T is a normal (~, K)-treeo

(K, <)-tree".

Then, in the real world, T is a normal

Since < is weakly compact,

T has a <-branch,

identify b with its canonical enumeration. the set of predecessors b~

Let T s L be such that

We shall

Now, for each ~ < ~, b ~

of the element b(~) of b.

s L, being definable

say b.

is just

Hence for each e < ~,

from T and b(~), of course.

We show that b s L,

which proves the required result,

since then ~L "T is not Souslin".

~ ( ~ ) ( b ~

~ There are just two points to note here.

Firstly,

s L<).

as K is inaccessible,

V

W ZFC and L K

cardinal and each b ~

~ ZFC.

Secondly,

Clearly,

as R is a

K

s L, then each b ~

~ L .]

Now, we can easily code up

the two predicates L , b in the above structure as a single unary predicate. Hence, by Keisler's extension property

(above), there is a transitive

set X

such that V K s X and -~<X, c, Y, Z>, where L< ~ Y ~ X and b ~ Z ~X.

In particular,

<X, s, Y, Z> ~ ( ~ ) ( Z [ ~

this implies <X, c, Y, Z> ~ " Z ~ clearly have Y = L

here,

Y

easily seen that Z ~ transitive,

b s L

Since K s X,

But look, X is transitive,

some y (in fact for y = sup(X ~ On).).

= b.

~ L.

g Y".

E Y).

Hence <X, c, Ly, Z> I= "b c L y

v!

so we And it is

Thus, as X is

QEDo

Y The critical

fact used in the above proof is worthy of special mention.

Stating it in general terms, it says: Lerm~a 4

(Keis ler)

Suppose < is weakly compact. Xc

If X ~ K and for all e < K, X ~ ~ s L, then

L. The proof differs hardly at all from the argument used in theorem 3, of course. Finally, we prove a result which bears upon Chapter I0.

Theorem 5

(Jensen; Kunen)

If K is ineffable,

then ~L "K is ineffable".

Proof: We use Theorem i0.I.

Let
I e < <> ~ L be such that A ~ c-- ~ for all e < <.

We seek a set A ~ ~ such that A s L and I=L "{~ ~ ~ I A Q ~ = A } is stationary".

-185-

Now,

in the real w o r l d

there is a set A ~ K such that

X = {e s K I A ~ ~ = A } is stationary A ~ ~ c L, since A

~

So by len~a 4, A s L. predicate

in K.

But look, for each ~ < K,

= A B n ~, where we pick any B s X such that B > ~ here. Thus X e L also.

"X is stationary"

is ]]I' ~ L

But as X is stationary

"X is stationary".

and the

The theorem is proved.~

Chapter 16

MEDIUM LARGE CARDINALS

We show that the existence is consistent with V = L.

The proof can easily be extended to the case where

÷ (a)< 2 for all a < L • I

[Recall that ~ ÷ (~)<w means that for all f: [~]<~ ÷ 2 2

there is X ~ ~ of order-type ÷ f(ol ) = f(a ), where 2 Theorem I

~ such that for each n < w, °l' °2 g Ix] n ÷

[<]<~ =

~

n<~

[KJ n.

We then say that X is homogeneous

Then ~e K ÷ (~)<2"

Suppose f e L, f: [<j<~ ÷ 2o homogeneous & (~n

for f.

Notice

< lol)(Va0,

We seek X ~ <, X ~ L, X infinite, which is

that (~<]<~)e = [<]<~o

~i s [o]n)(f(o0)

= f(ol ))}"

ZFC that f has an infinite homogeneous Now, since f does <~,

for f.]

(Silver)

Assume < ÷ (~)<~. Proof:

of a cardinal K such that < + (~)<m 2

the same definition

~

={~ ~ < I Iol <

Clearly,

set iff < ~ , ~

have an infinite homogeneous

~ > is not well-founded.

Hence,

Set

But look, <~,-~ > s

it is provable

in

> is not well-founded.

set in the real world, L, and in fact ~

in L as in V (namely the above, L-absolute

has

definition.)

as L ~ ZFC, the theorem follows from the following lermma.

Lemma 2 Suppose is a partially ordered structure

in L.

Then is well-

founded iff ~L " is we ll-founded. Proof:

Suppose is well-founded. chains from A. founded".

-g-decreasing

Hence there can be no such chains in L, so I=L" is well-

Conversely,

suppose I=L" is well-founded".

L, we define, by induction, of .)

Then there are no infinite

in

an order preserving map r: A ÷ On (i.e. a ranking

The existence of such an r clearly implies

is well-founded.

Then, working

that ~A, --~ > really

QEDo

Remark Theorem I ties in closely with a result in

[6] .

There it is proved

that,if

-187-

V = L, then K + (W)<~ is equivalent order structure

~'~ = of countable

such that no element of X is further details of this result, reader

to

E6].

to the assertion

~% -definable

(Fr(~, ~)) that for every first

language there is an infinite set X ~ from any other elements

of X.

For

and the property Fr(<, %) in general, we refer the

Cha?terl7

LARGE LARGE CARDINALS

In this chapter we shall study the effect of very large cardinals upon the constructible universe.

By "very large cardinals" we mean, for example, measurable <w

cardinals, Ramsey cardinals, or cardinals ~ such that K

+

(~i) 2

.

Since all

measurable cardinals are Ramsey, we shall consider Ramsey cardinals specifically.

All

<m of the proofs can be adopted to the weaker notion < ÷ (~i)2 , but the details are a little more akward.

We refer the reader to []~] on this matter.

And for general

information concerning measurable cardinals, Ramsey cardinals, etc., we refer him to [~.

That every measurable cardinal is Ramsey is proved, in effect, in chapter 19.

Preliminaries Suppose <X, <> is an ordered set. x is increasing if x --

Let ~ denote <x0,... , Xn> throughout.

Say

<...< x . 0

n

We shall be concerned with structures of the form

~=

where
linearly orders a subset of A (called the field of
We shall always assume 6"t is a Skolem structure, although in general

we will not explicitly mention the Skolem functions.

Thus, if ~

is the language for

0-[, ~ will be countable and for each ~-formula ~(v0,... , v n) there will be f# s such that ~ For ~

~ V l . . - V V n [ ~ V 0 # ( v 0, v I ..... v n) ÷ #(f#(v I ..... Vn), v I ..... Vn)].

such a~structure and _a ~ A n+l , define the type of a as the set

{#(v 0 ..... v n) ~ ~ 4, ~

I0~

~[~]}.

Thus for each ~(v 0 ..... Vn) ~ ~, exactly one of

is in the type of ~. A subset H ~ A is 6-~-homogeneous (or a set of indiscernables for ~]~) if

(i)

H ~ f i e l d of
(ii)

H is infinite;

(iii) for each n e m, all increasing n-tuples from H have the same type. Recall that if X is a set, % a cardinal, [X] <~ = {Y ~ XIIY 1 < ~}.

~ ] % = {Y ~ XIIYI = %} and

In the following, the set X is said to he homogeneous for

-189-

f.

For K, %, ~ cardinals, write K + (%)~m iff (~f:

[K] <~ ÷ ~ ) ( ~ X

c ~]%)(~a,

We say K is a Ramsey cardinal iff ( ~

b c [X]<~)(la I = ]b] ÷ f(a) = f(b)). < K)(K ÷ (K)~).

It is proved in ~

that <

is Ramsey iff < ÷ (K)~ ~. Lemma 1 If K is Ramsey, then K is inaccessible. Proof: Let K be a Ramsey cardinal. (i)

Suppose < is singular, and let % = cf(K) < K.

ing sequence of members of K with U<.K

= K.

Let ( < ) <% be a strictly increas-

Define f: [<] 2 ÷ 2 by setting

f({a, B}) = 0 iff ~ and ~ are separated by K for some y < %, and putting Y f({a, ~}) = 1 otherwise.

Extend f trivially to have domain [K] <~.

have no homogeneous set of power K, contrary to < being Ramsey. (ii) Suppose now % < < ~ 2 % for some %.

Clearly, f can

Hence < is regular.

Then we may identify < with a subset of ~2,

and hence define f: ~ ] 2 ÷ % by setting f({a, B}) = the least y c % s.t. a(y) + ~(y). Clearly, if f is constant on IX] 2 for some X ~ <, then IXI ~ 2, contrary to K being Ramsey. | Lemma 2 (Rowbottom) Let K be a Ramsey cardinal, IField(
~=

a given structure with

Then there is H G A of power K homogeneous for ~ .

be the language for ~ .

f: [B] <m ÷ ~ ( ~ )

Let B ~Field(
IBI = <.

Define

by setting f(~) = type of ~, where ~ is the increasing

enumeration of {ala e ~} c [B] <~.

By lenmna I, I ~ ( ~ ) I

is Ramsey, f has a homogeneous set of power K.

= 2~

< K.

Thus as K

Clearly, such a set will be

~ -homogeneous. I For later use, we require some concepts from recursion theory.

Let A, B ~ ~.

Recall that A is many-one reducible to B iff there is a recursive function # such that ( V n

~ ~)(n s A +-+ ~(n) e B).

Let A, B ~ ~. (~n

A is

I~(B)iff

c ~)(n e A +-~ R(n, B)).

there is a l~-predicate R(n, x) such that

Similarly for A being,S(B).

A is AI(B)m iff A is both

-190-

Lemma 3 Let B be AIm(C)" There is a [1-predicate S 1 (X, Y) such that (I) ( V D

=~)(D

= B K-+ SI(D, C)).

And there is a]T1m-predlcate S2(X, Y) such that (2) ( ~ D ~ ~)(D = B +-+ S2(D, C)). Proof: Let R I (n, X), R 2 (n, X) be, respectively, that ( ~ n

a Jim-predicate , a lqlm-predicate such

~ ~)(n ~ B K-+ Ri(n ' C)), i = I, 2.

(kln g ~)(n e D ÷ Rl(n , C)) & ( ~ n predicate,

SI(D, C).

Then D = B iff 1 This is a Ira-

~ m)(R2(n, C) ÷ n E D).

To obtain $2, simply interchange R 1 and R 2 in the above.|

Lemma 4 (Shoenfield) Let A, B, C = ~, where A is [In(B) and B is AIm(C)" Then:(i)

If n < m, A is AIm(C)

(ii) If n >. m, A is [In(C). Similarly for ~ in place of [. Proof: Let S(r, X) be a [In-predicate such that ( V r

g ~)(r c A K-+ S(r, B)).

By lemma

3, ( I ) , (*) ( ~ r

~ ~)(r £ A ++ ( ~ D g ~)(SI(D, C) & S(r, D))).

The RHS of the above equivalence is easily seen to be a [Is-predicate of r, C, where s = max(m, n).

If m > n, then, in addition,

(**) ( Vr e ~)(r e A +-+ ( V D c_ m)(S I(D, C) ÷ S(r, D))) and here

the RHS is aTFlm-predicate of r, C. |

Corollary 5 (Transitivity of AIn) Let A, B, C c ~ . --

If A is AI(B) and B is AIn(C) then A is AI(c). n

'

In particular,

n

if A, B =_ m, B is A n,I and A is many-one reducible to B, then A is Al.n ~ For the rest of this chapter~ we assume that there is at least one Ramsey cardinal.

BRC denotes the statement "There exists a Ramsey cardinal".

-191-

The Main Theorems We shall prove three theorems.

In this section we shall simply state these

results and derive some corollaries. Theorem 6 (Solovay) There is a non-constructible A~ set O~ ~ ~ such that every construetible subset of ~ is many-one reducible to 0 ~. Corollary 7 Every constructible subset of m is A~. Proof: By corollary 5 above. | Theorem 8 (Silver) If <, X are cardinals,

~ < < ~ X, then ~ .

Theorem 8 was first proved by Gaifman, under the assumption that there is a measurable cardinal. Since, by theorem 8, L is the union of the elementary chain L i ~

L 2 < . . . , we

have Corollary 9 If < is an uncountable cardinal, then ~ . | Corollary I0 (Metatheorem) ZFC +

NRC ~ "If X > e is a cardinal and x s Ll, then ~LX(x) ~-+ ~L(x)".

Corollar~ Ii (Metatheorem) If ~ is a function from ordinals to sets defined in ZFC and if ~ I " ' ' '

~m < ~ ' 1

then fL(el,..., ~m ) is countable. Proof: As ~ ..... ~m ~ L~ I' 9 gives fL(~ ,.... ~m ) = fL~l(~l, .... ~m ) s L L ~L are countahle. For example, ~i'

.

l

As another example, we have

Corollary 12 (Rowbottom) If ~ < ml' then V~ ~ L is countable.

In particular, ~(~) ~ L is countable.

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Proof: By corollary ii, as V L = V

f~ L and @L(~) = ~(0~) FI L.

Corollary 13 Let ~(v0) be D-absolute for L, and suppose there is a cardinal % > ~ such that #(%)o

Then for some ~ < ~I' ~L(~).

Proof: Assume a recursive g6del numbering of the formulas of set theory.

Define

(in ZFC) a function ~: m ÷ On by setting (for each formula ~(v0)) the least uncountable cardinal s.t. ~; if this exists. ~(~)

t O;

otherwise.

By corollary Ii, for each ~, L ( ~ ) for the given ~, ~(~) > O. ~L ~ ( ~ ) "

is a countable ordinal.

Hence ~(~(~)).

Hence ~L[~(~) > O & ~(D(~))].

Now, we know that

By D-absoluteness, ~L(~(~)). In other words, ~ L ( L ( # ) ) ,

Thus

and we

are done. | Corollary 14 (i)

ZFC +

~RC ~ ( H a < ~l)(e is inaccessible in L)

(ii)

ZFC +

3RC ~ ( ~

< Wl)(~ is weakly compact in L)

(iii) ZFC +

~RC ~ ( ~

< ~I)( ~ ~ ÷ (m)2)"

Proof: By corollary 13 and the results of chapters 15 and 16. | Note that the ordinals of a model of set theory form a linearly ordered subset of the model.

Thus, models of set theory are included in the class of models we

discussed initially. Theorem 15 (Silver) If % > ~ is a cardinal, then there is a set of indiscernables for which contains all uncountable cardinals in %. Corollary 16 The class of all uncountable cardinals is a class of indiscernables for . Proof: By theorems 9 and 15. ] Corollary 17 Let ~(v 0) be D-absolute for L.

If there is a cardinal % > ~ such that ~(%),

-193-

then cL(~I). Proof: By D-absoluteness, ¢(~) ÷ cL(~).

By corollary 16, cL(~I).

I

Corollary 18 (i)

toI is inaccessible in L.

(ii)

cdI is weakly compact in L.

(iii) If ~ = ~I' then ~L ~ ÷ (w)2 ~.

I

We are now ready to prove theorems 6, 8, and 15. Proof of the Main Theorems Let ZFL be the theory ZFC + V = L.

Recall that if

model of ZFC, then there is an isomorphism If

~

ZFL, then A' = L

c~

for some ~ = t ~

01= is a well-founded

~ : ~-~ ~ , where A' is transitive. > w.

A' is called the transitive collapse

of A, ~ the collapsing isomorphism. 0~ Suppose for

6-[ = is a model of ZFL, and H c On

~L (under the E-ordering of On

holding in ~

in ~ ) .

on increasing tuples from H.

then N is ~-homogeneous

is a set of indiscernables

Let F~(H) denote the set of formulas Note that if H ~ B and

~ = -<~,

and F@(H) = F0~(H).

It is worth recalling, at this point, our earlier convention that we assume without explicit mention that all structures have a complete set of Skolem functions. A set of formulas is an Ehrenfeucht-Mostowski F~(H) for some model

(E-M) set if it is of the form

O~l of ZFL and some 6~L-homogeneous set H ~ On ~.

Let ~ be an E-M set, ~ an infinite ordinal.

A (~, ~)-model is a pair (~L, H)

such that : (i) (ii)

0% ~ ZFL 01 H c On is ~ -homogeneous

(iii) H generates

0l (via the Skolem functions of ~ )

(iv)

~ = F~(H)

(v)

The ordinal ordering of ~ type ~.

(henceforth written as <~) well-orders H with order-

-194-

If (i), (ii), (iv), (v) hold but (iii) fails, then by replacing 0~ by (the substructure of

~ generated by H), (i) - (v) will all hold for

by virtue of ~I being a Skolem structure,

~H

~H

Say (~I' HI) and ( ~ 2 ' H2) are

isomorphic if there is f: ~I I =~ ~ 2 such that f: ~ . (See, for example,

(because,

~).

Let ( H I , HI), (~2' H2) be (~, ~)-models.

result is well-known.

~H

The following

[18]).

Theorem (Ehrenfeucht-Mostowski) If ~ is an E-M set and ~ is an infinite ordinal, then there is a (~, ~)-model. Furthermore, this model is unique up to isomorphism. I Note that if ( ~, H) is a (~, ~)-model and H' is an infinite subset of H of order-type ~(~ ~), then ( ~ H ' , If

H') is a (~, ~)-model.

6~ is a well-founded model of ZFL then we know 0~1~ for some

= U ~ > O.

Thus well-founded

(~, ~)-models are seen to occupy a special place.

Call an E-M set ~ well-founded if for all infinite ordinals e, the (~, ~)-model is well-founded

(using the above theorem of Ehrenfeucht-Mostowski).

Ler~a 19 Let ~ be an E-M set.

The following are equivalent.

(i)

~ is well-founded.

(ii)

For some e ~ ml' the (~, ~)-model is well-founded.

(iii) For all countable ordinals e (~ ~), the (~, ~)-model is well-founded. Proof:

(i) ÷ (ii).

Immediate, by definition.

(ii) ÷ (iii).

Let ~ ~ ~I be such that the (~, ~)-model is well-founded.

By

our remark above, if m ~ B < ~ , the (~, B)-model is (up to isomorphism) a I submodel of the (~, e)-model, and hence is well-founded. (iii) ÷ (i). Assume ~ is not well-founded. model,

(~,

H), is not well-founded.

members of ~

such that ( ~ i

Let (an)n< ~ be an infinite sequence of

with an+l~ ~ an for all n.

Skolem term f. of

~

Then for some ~ ~ ~, the (~, ~)-

Each a i is of the form fi(h) for some

and some tuple h from H.

g ~)(a i ~ ~I~H').

Let H' ~ H, IH'I = ~

Let g be the order-type of .

, be Thus

-195-

B < ml and ( ~ I H ' , well-founded.

Since each a i s ~ H ' ,

it is not

Thus m(i) + -~(iii). |

It , is well-founded transitive

H') is the (@, B)-model.

and ~ is an infinite ordinal, denote the unique

(¢, ~) model by M(~, ~).

Then, M(~, ~) = for some y = Y Call an E-M set ~ cofinal if ~ contains all formulas of the form

(absolutely~) Uy > ~.

f(v 0 ..... Vn_ I) E On ÷ f(v 0 .... , Vn_ I) < v n where f is a function symbol of the

(Skolem)

language.

Le~na 20 Let # be an E-M set. (i)

If ~ is cofinal,

then for every limit ordinal ~, if( ~, H) is the (~, ~)-model,

then is cofinal in . (ii) If for some limit ordinal e the (~, e)-model

( ~ , H) is such that is

cofinal in , then ~ is cofinal. Proof: Ohvious.| An E-M set ~ is remarkable

if ~ is well-founded

and cofinal,

and for any f, if

f(v0, .... Vn_ I, Vn, .... Vn+ m) < v n is in ~, then so is f(v 0 ..... Vn_ I, Vn,-.., Vn+ m) = f(v 0 ..... Vn_ I, Vn+m+ I ..... Vn+2m+l). This last condition is the remarkable

condition on #.

It is due to Rowbottom.

Lemma 21 (Rowbottom) There is a remarkable E-M set. Proof: Let K be a Ramsey cardinal. And we know On L< = <. Assume H = ( h ) ~ < < ,

By lemma I, < is inaccessible,

so ~ ZFL.

Using lemma 2, let H ~ < he LK-homogeneous

where ~ < S < K ÷ h a < h B.

Assume that h

of power K.

is as small as

possible. Now, H is well-ordered =
e> ~

.

of order type K.

Then if ~ = FL<(H),

19 and 20, ~ is well-founded

and cofinal.

Thus H is cofinal in K.

Set

( ~, H) is the (~, ~)-model.

By lemmas

We show ~ satisfies the remarkable

condition. Assume

f(v 0 ..... Vn_l, v n .... , Vn+m) < v n is in ~.

Partition H into i, ~0'

-196-

~I"''' d

d_~,... (~ < K), where _c has length n and each_~d has length m + i.

= (h , h + I ....

h +m).

Since H is ~-homogeneous,

Thus

one of the following must occur.

-'60

(i)

For all ~ < ~ < ~, f(!, d_~) = f(!, ~ ) .

(ii)

For all a < B < <, f(!, d_~) < f(!, ~ ) .

(iii) For all ~ < B < <, f(!, d_~) > f(!, ~ ) . We must show only (i) can be the case. ordering.

Assume

< B < K ÷ e minimally,

h

(ii) holds.

< e B. ~ e .

Let e

Clearly,

(iii) cannot hold as < is a well-

= f(c, d~)

~ < <.

Then, by (ii),

Clearly, {e I~ < <} is ~-homogeneous. That is, f(!, h , h +i,... , h + m) ~ h .

contained the negation of this formula.

Hence

As h~ was chosen But we assumed that

(ii) cannot hold. J

From now on, let ~ be a remarkable E-M set, % an uncountable (M(~, %), H) the unique transitive

(~, %)-model.

And let ( h ) ~ < %

cardinal, enumerate H in

increasing order. Lemma 22 H is cofinal in On M(~'%). Proof: By lemma 20. | Lemma 23 Let O < a = ~ N ~M(~, Proof:

%).

< %, and put H' = {h~JB < ~}.

Let N = M(~, X)~H'.

Thus

Then On N = {x ~ onM(¢'%)Ix < ha}.

(N, H') is the (¢, ~)-model.

As ~ is remarkable,

On N _c {x ~ onM(~'X)Ix < h~}.

For the reverse inclusion, suppose x E On M(~'%),

x<

h .

Then x = f ~ ,

h), where ~ h

course, x may be independent of h).

H' is cofinal in 0n N, so

is increasing and h = (h ,...) (of Thus f(k, h) < h .

By the remarkable

condition on ~, whenever h' is an increasing sequence from H of length that of h and ~ < h'

then f(k, h) = f(k, h')

such an h' with h' < h .

onM(~, %) = %.

we may choose

Then h' is an increasing sequence from H', so

x = f(~, h') s N, giving x e On N. J Lemma 24

As ~ is a limit ordinal

-197-

Proof:

By absoluteness, Assume y > %. a =

Ua

Thus

On M(~'%) Thus

% ~ O ~ M(~'%).

and % < h a.

lOn N] < %.

As H ~ On M(~'%), T ~ ~ -

= y for some y s On.

By lemma 22 there is a < % such that

Take H', N as in lemma 23.

Now,

IN I = JH'[ = ]a[ < %.

But by lemma 23, On N = {x s yIx < h } e %, which is absurd.

Hence y = %. Corollary

25

M(~,

J

%) = .

Proof of theorem

8

Let K, % be uncountable and = M(~, M(~,

%)~H' ~ M ( ~ ,

On M(~'%)

%). %).

Thus M(~,

cardinals,

But by lemma 23, On M(~'%)[H' X)~H'

=
K'

~>.

Therefore,

Proof of theorem

an initial

~> = M(~,

segment

%)~H' ~ M(~

of '

%)

in the next proof.

15

We know = M(~,

%), so H is a set of indiscernables

must show that if K < % is an uncountable = M(~,

K) = M(~,

= {X ~ %IX < h } = h K

%)IH',

(as h K

~ %).

Hence

E H), where

= F({mn+lln

Let 0~ = {~(Vl,..., theorem 8 (in particular,

then ~ s H.

Now, we have

S H.

QED.

some preliminaries. (~, m )-model.

By the last proof,

This shows

the remarkable

L~ v n) j9 ~(Wl ..... ~n)} corollary

recursive

for
s>.

E-M set ~ is unique.

(where n ranges over ~).

9), ~(v I ,..., Vn ) ~ 0 ~ *-+ L ( ~ I , . . . ,

g~del numbering

Thus

Then, by ~n)"

from now on.

26 The predicate

We

K

H is the set of indiscernables

< m}).

a fixed,

K = h

~

Let be the transitive ( ~ n e m)(mn+l

cardinal,

for .

so by lemma 23, K = On L K = {X ~ onM(~,%) IX < hK }

The proof of theorem 6 requires

Lena

K'

<)

and let H' = {h J~ < K}.

forms
25, = M(~,

QED.

We use the above notation

Assume

By corollary

Let H = {hal~ < %} be as above,

•

= .

K < %.

"~ is a remarkable

E-M set" is a ~ l - p r e d i c a t e

on

~(~).

Then

-198-

Proof: By definition,

"~ is a remarkable E-M set" ~-+ "~ is a well-founded,

cofinal

E-M set satisfying the remarkable condition". That "~ is a cofinal E-M set" is A~ is easily seen.

And "~ satisfies the

remarkable condition" is of the form (~p

~ ~)(Vq

~ ~)[,~(p, q) ~ p ~ ~ ÷ q

~

~]

0 where ~ is a recursive binary relation, so this is ~[I" We show "~ is well-founded" lemma 19 (iii). ~-+ ( ~ R

in ~TI.

Let W = {2nln < ~},

We use the equivalent formulation given by Then "~ is well-founded"

c-W2)[R well-orders W ÷ (if < ~ , 5 >

well-orders

is the (~, <•, R>)-model,

then S

m)].

Thus "~ is well-founded"

is of the form (~B) r~(B) ÷ ~ ( B ,

form (~B)[Z~(B)v

~)], which is essentially

~(B,

~)], hence of the

(VB)~I(B,

~)].

This is

Proof of theorem 6 We know the remarkable E-M set is unique. E-M set" is]~ I.

And by lemma 26, "~ is a remarkable

Thus there is a~21 predicate P such that P(A) ~-+ A = 0 ~ (in fact,

P is precisely the predicate just stated).

Hence:-

(a) ( ~ n

e ~)[n E 0~ +-~ (~A)[P(A)

^ n E A]], which means 0 ~ is 113

(b) ( ~ n

~ m)[n e 0~ ~-+ (~A)[P(A) ÷ n E A]], which means 0~ is ~ .

Hence O ~ is a A~ set of integers. Now, the formula "v 0 is a remarkable E-M set" is D-absolute for L, as is easily seen.

Thus, if 0~ e L, we could relativise our previous proofs to L and prove, for

example, theorem 15 in , which is absurd. Finally, suppose A ~ ~, A E L. (O~, ~l)-model.

Hence O % #

Then A e L0~I.

L.

Let (L~I, H) be the transitive

Then A = fe~l(h) for some f, some h from Hm.

set theory designating n, for each n c m. ~n

L e t ~ n be the term of

Then, n E A ++ ~ n

e f(h_) +-+

E f(v ..... Vm_l) ] s O . Thus n e A +-+ ~(n) c O ~ , where # is recursive (since 0 our g~del numbering is recursive). In other words, if A s ~L(~), A is many-one reducible to 0~.

QED.

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That completes our discussion on this topic. mentioned at the outset, ~ ÷ (~i) 2

Of the three large cardinals

i8, of course, the weakest, logically speaking.

It is not however the weakest property which implies the results we have obtained (e.g. the existence of O~). Fr(~, ~i).

An even weaker notion which gives the same results is

This matter is discussed in detail in [ 6 ] .

Another property which also

implies the above results is the existence of an elementary embedding f: L some ~, 6 such that f(~) > ~ for some ~ < I~I.

~ L~ for

However, since neither of these

properties are (strictly speaking) large cardinal properties, they do not fall within the scope of this chapter.

Now, in the case of the former property, the relevant

proofs are all given in [A].

But for the second property, we do not know of any

suitable reference to give the reader.

We shall therefore give the proof here.

It

is due to Silver, though the result itself was first proved by Kunen using an alternative argument.

Before we commence, let us observe that by an easy application

of the condensation lemma, the existence of an elementary embedding of the above kind follows from the existence of a Jonsson cardinal.

(A cardinal ~ is said to be Jonsson

if every first-order structure of cardinality ~, with a countable language, has a non-trivial elementary substructure of cardinality <. in detail in [ 5].)

Such cardinals are discussed

This gives us another "cardinal property" which implies the

existence of 0 ~. Lemma 27 (Silver) Suppose that for some ~, 6 there is f: L ~ ordinal y < I~I.

L 6 such that f(y) > ¥ for some

Then there is an embedding i: L ~ L

such that i(~) > ~ for some ~,

and such that there is a stationary (proper) class U of ordinals ~ for which

i(~) = ~.

Proof: Let y be the first ordinal moved by f, and put D = {X ~ yIX E L

& y E f(X)}

Then an easy argument shows that D is an ultrafilter in L (= L ~ ~(y) =

~L(y), as

y

< lel.)

~ ~(¥).

we shall use D to form an ultrapower of

L. Define an equivalence relation ~D on (YL) ~ L by gl~D g2 iff {~ g ylgl(~) = g2(~)} ~ D

-

Let A be the set of equivalence

200

-

classes.

Let g denote the equivalence

class

of g. Define a binary relation E on A by gl E g2 iff {~ g ¥Igl(~)

c g2(~)} ~ D.

Then the canonical embedding e of into defined by e(x) = (x)a
is well-founded. Suppose

(~n

~ V = L. y + i ~ B.

~ ~)(gn+l E gn ).

Let O =

Ue > ~ be s.t.

~ n ( g n c ne).

Take ~ with

[B[ = [YI and

Then ~ , some 6 < [yl +. n new =

(as 7 + i is a transitive

subset of B), so {~Ign+l(~)

by elementary substructure

Now, y + I ~ L 6

s gn(~)} = {~If~l,(~)

e fn(a)}) for each n s ~.

f({~If'n+l (~) ~ f~(~)}) = {~l(f( f'n+l ))(~) ~ (f(f'))(~)}, n e (f(fn))(y), which is absurd.

But,

so for all n

This proves the claim.

Thus is a well-founded model of ZFC + V = L of "order-type" ~ Let i = (a) i(y) > y. ~((eonstant

some ~ o e. Now,

e f~(~)},

and isomorphism.

So we must have 7 g f({~If~+l(~)

(f(fn+l))(y)

Note that

On.

Hence

~ (~ is the collapsing map). Thus i: L ~ L.

We show i is as required.

(constant function ¥)/D >E (identity function on y)/D.

So, as

function y)/D) = ~(e(¥)) = i(y), and (identity function on y)/D has at

least ¥ E-predecessors,

we have i(y) > y.

(b) i fixes each member of some stationary class U of ordinals. be any cardinal.

Clearly,

To see this, let Z

if ~(x) ~ i(%), then x = z for some z c ¥%.

Thus if

cf(%) > [YI, we see that ~(x) s i(%) implies x = z for some z c Y~, where ~ < ~, whence

~(x) e i(~).

Thus if i is a cardinal with cf(%) > IY[, i(%) = ~<%i(n).

For n < %, [i(n)[ = [ ~o e(n) l = le(n)[

= [{~IX e L & x(~) s ~ a.e.} i

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~< l{xlx S L & x ~ Yq}I

= I (Y~)LI ~< Iemax(IyI +,I nl +) I = max(IyI +, I~I +) .< max(cf(%),

Hence i(X) = X.

%)

Thus U = {% E Cardslcf(X)

A set X is said to be U-indiscernable (i)

X o U ~ On.

(ii)

U is a stationary class.

(iii) For each formula #(v I ,. .., v n, Vn+l,.

> IyI} will be as required.

I

if

., Vn+ m) of set theory, and for all

S-,...,l Bn e ,U ~ (~X), if al,..., am, a'l,..., a'm e X are s.t. al <...< am and a

v

t

<. .<

I

"

then

am~

~ ¢(BI,. .. , Bn, a ] ,..., am) iff ~ ~(B ..... Bn, a',..., am).' 1 1 Example Let i, U be as in lemma 27. {y, i(y)} is U-indiscernable. of U and of

Let y he the first ordinal moved by i.

Then

This is because i: L ~ L and i does not affect elements

~{y, i(¥)} = y, this latter by choice of y.

We shall make use for this example later on. For Z ~ L , elements of Z.

let Defn(Z) be the class of all elements definable in from Defn(Z) is called the Skolem hull of Z in .

Lem~na 28 Suppose that X is U-indiscernable

and IXI ~ 2.

let y(a, U) be the least ordinal ~ a in Defn(~ o U ) .

Let s 0 = ~X. Then:

(i)

For each ~ ~ X, ~ # Defn(a o U ) ,

so y(a, U) > a.

(ii)

For each a ~ X, y(a, U) exceeds any element of X.

(iii) For all a s X, y(e, U) = Y(~0' U), = y(U), say.

For each ~ e X,

-202-

Proof:

(i) Suppose ~ ~ Defn(a

uU).

Then there is a term • of set theory s.t.

L

T (BI"'" ~ <...< 1

~

Bn' ~ I ' ' " '

6m)' where

BI,...,

~n a U and 61,...,

6m ~ e with

.

m

Hence, ~

~vl... ~Vm(Vl

<'''<

As X is U-indiscernable

Vm < ~ & a = r(Bl,...,

01

Thus there are 8 ,..., ~ with 1 m

<...<

there is e

~ a 1

--TL(~

which

1

s.t. ~ 0

Bm" V l " ' ' '

Vm))"

< a 0 s.~.

so as IXl >~ 2, by U-indiscernability

~ X and (hence) 1

, .... Bn, 0 .... , 8m) 1

is absurd.

(ii) Suppose y(~,

U) ~< ~', where e' E X.

term

U) = oL(B

~ s.t. y(~,

<...<

I

0m

~cL(BI''''' Bn' 01,..., Ore)"

But ~ , .... ~n' e ..... 6 a U u (f~X), 1 1 m

1

Vm))"

and s 0 ~ X,

~Vl''" ~Vm(V 1 < " ..< Vm < a0 & ~0 = ~ (gl''°''

~

~0 =

Bn' Vl'''''

~

m

are all in ~.

~ ~v...~

I

..... Bn, 6 ..... 6m) , where g ,..., B E U and I i n

So, using

3 v m [v I <...<

(i),

v m < ~ & ~ < ~(B

Now, ~ < T(e, U) .< e', so ~ < ~'. by ~ , the second member 1

As y(e, U) ~ Defn(a u U), there is a

i .....

Bn, v .... , v m) .< e'].

As e, ~' s X we can replace ~ by e0 and ~'

of X, to obtain,

for some e I <...< 0

m

< ~ , 0

~0 < ~L(BI .... , 6n, el .... ' om) ~< ~I" But by U-indiscernability,

I

<

~L(B

I

the first

inequality

gives

..... B n, O ..... 8 m) I

so by the second

inequality,

~ I

< ~ , which I

(iii) Note that ~ ~ X -> e >. e0 + Defn(~0

is absurd.

u U) ~ Defn(~

oU).

By (ii), as

X, Y(~0' U) > e for any ~ s X. Thus

for e e X, ~f(~ , U) > e and Y(~0' U) e Defn(~ 0

UU).

So by definition

~(~, u), ~(~, u) .< ~(~ , u). Suppose

that,

0 in fact, y(~, U) < Y(~0' U).

Take terms

~(~, u) = oL(~ ..... ~n' ~ ..... ~m ) I

I

Y(~0" U) = rL(8 ', .... B', 6', .... $') l K 1 r

o,

• s.t.

of

-203-

where 8i, ~'i e U, 81 <...< 6

m

< ~, 6' <...< 6 ' < ~ . 1 r 0

~ "''I ~ V m [ Vl <...< V m <

~

&

< ~(S I .......... !

So,

v m) < ~(~

....

S'~" 6'1 ''''~

'

!

AS Bi, B i ~ U and 6. e ~

= ()X, U - i n d i s c e r n a b i l i t y

allows us to replace ~ by

0 in the above, obtaining O 0 0

<...< 0 1

<<;L(B

Thus > ~

1

< ~ m

s.t. 0

..... B n, ~ ..... ~ m ) < ~L(B', .... B' 5' ..... ~'r). 1 I K~ 1

c;L(B ,..., B n, g ,..., 8 m ) is an ordinal I 1 and < y(e

0

, U), contrary to the definition

in Defn(e 0 ~ of 7(~0,

U) which

is both

U). I

0

L e m m a 29 Let X be U-indiscernable,

IXI h 2.

Let U* = {K ~ U]]< ~ U I = <}.

T h e n for

each ~ e X there is f : L ~ L which fixes each element of ~ u U* and sends e into y(~). Proof:

Fix ~ ~ X.

Now, Defn(~ o U) ~ L and so is a w e l l - f o u n d e d

Collapse Defn(~ o U) to a transitive stationary,

model of ZFC + V = L.

model by the i s o m o r p h i s m h .

As U is

e

the image of Defn(e o U) under h a is just L.

Let f

= h ~-I .

Then

fe: L ~ Defn(~ U U) ~ L, so fe: L ~ L. Now, a is a transitive

subset of Defn(~ o U) so 8 e ~ ÷ he(B)

= B.

Thus

f (B) = B for all 8 ~ ~. As y(U)

is the least ordinal > ~ in Defn(e

above shows that f (e) = T(U). Finally,

let < E U*.

h (<) = {h (6)16 ~ Defn(~ of U*,

lha(~)I ~ ~.

(i.e. h ( y ( U ) )

Then,

o U) and ~ # Defn(~ t) U), the result

= e.)

from the definition

of the collapsing map,

o U) & ~ ~ <} ~ {h (6)J6 E U & 6 ~ <}.

Thus by definition

But as h~ is the collapsing map, h~(<) ~ K; whence h~(~) = ~.

Thus < s U* ÷ f (<) = <. I e

Lemma 30 Let X, U, U* be as above.

Then X' = X

O{y(U)}

Proof: Let ~(v0,... , Vn+m) be a formula of set theory.

is U*-indiscernable. We show that for

' I ~ X' , ~ <...< 6m+ I, 8 ,..., 8n ~ U' O ~ , ~ ,.-., 6m+ I, 6' ,.--, 6m+ 1 0 1 1 1 6' <...< 6' we have I m+l'

-204-

~L(BI .... , Bn, 61 .... , 6m+l ) ~-+ ~L(BI,..., Bn' 6'i,..., 6'm+l). Case I. 6m+ I

= 6v m+l = y(U).

the X, U situation.

Now, y(U) ~ Defn(a 0 U U), so y(U) is a constant term in

Thus as 6 ,..., 6m, 6' ,.. . 1

1

'

6' must be in X, the U-indiscernabilm

ity of X gives the result. 6 1

Case 2. 6re+l, m+l < y(U).

Then all 6 i ,

6[ are in X, so the result is known already 1

(note that U* ~ U). Case 3. 6m+ I < 6'm+l = ¥(U) .

By cases I and 2 we may assume 6 I = 6'1 ,..., 6 m = 6'.m

That is, we must verify that ~L(BI'''" ~L(BI,...,

Bn, 61,..o, 6m, 6 ~ i ) .

Bn, 61,..., 6m' 6m+l) ~-+

By lemma 29, the LHS above holds iff (where

= 6m+ I) ~L(f (8 I) .... , fq(Bn) , f (6]) ..... fq(6m) , fq(6m+l)) i.e. iff ~L(B I ..... B n, 6 I ..... 6m, y(U)).

(This is because 6m+ I a X.)

Hence

cL(B 1 ' .... 8n' 61' .... 6m' 6m+l) +-+ %L(B 1 ..... ~n' 61 ..... 6m' 6m+l)" ' | Theorem 31 (Kunen; Silver) The hypothesis of lemma 27 implies the existence of a closed, cofinal class of indiscernables Proof:

for L s.t. L is generated by this class.

(Silver) Let i, U be as in lemma 27.

Then as we observed after proving lemma

27, if T is the first ordinal moved by i, {y, i(y)} is U-indiscernable. So put X 2 = {¥, i(y)}, U 2 = U. By len~na 30, X

= X

U {y(U)} is U3-indiscernable, where 2 U 3 = U~ = {K g U211K N U21 = K} ~ U 2. In general, therefore, we define 3

Xa+ I = X~ u{y(Ua)}

,

Ua+ I = U~,

,

a ~ On.

X6 = a<~ ~ Xa

,

U~ = d~d U

,

~ =

Then for each a, X a is U -indiscernable. is a cofinal class of indiscernables

U6 e On.

Since ~Q U a = ~, it follows that X = ~ X

for .

By the usual arguments, Defn(X) ~ L under the collapsing isomorphism g. g"X is a cofinal class of indiscernables

Now

for , and every element of L is

definable from elements of g"X (in ). To show that g"X is closed in L, we need only show that X is closed in Defm(X).

-205-

This follows from lemma 28.~ Clearly, by virtue of the above, all of our earlier results hold with the hypothesis of lemma 27 in place of the assumption this topic.

~RC.

At this point we shall leave

Chapter 18

RELATIVE CONSTRUCTIBILITY

We define, by analogy with the definition of the constructible hierarchy, a slightly extended hierarchy as follows. If X is a set and A is a class, we let D e ~ ( X )

denote the set of all subsets of

X which are definable in the structure <X, s, A ~ X> by some formula of the <X, E, A ~ X>-language with the one free variable v 0. We define the hierarchy of sets eonstructible from A, for A an arbitrary given class, inductively, thus: Lo[A ] = L +I[A ] = DefA(L ~A]) L6[A ] = We set L[A] =

~g0ne

~

<6L [A],

if lim(~)o

[A], the universe of sets constructible from A.

It is easily

seen that the obvious analogue of lemma I.i holds in the present situation. the above hierarchy resembles the constructible hierarchy in many ways. sketch briefly some of the similarities.

In fact,

We shall

However, we shall not be as rigorous as we

were in chapter 2, rather we shall just indicate the main points.

This will result

in our present treatment having a somewhat different flavour from that of chapter 2. We leave it to the reader to check for himself how everything we shall do can be recast in the chapter 2 mould~

From now on, we shall assume A is a set.

It is easily seen that L~A] is an inner model of ZF.

In fact just as L can be

characterised as the ~-smallest inner model of ZF, so can L~A] be characterised as the ~-smallest inner model M of ZF such that A n M s M. The L[A] construction is, in a very strong sense, absolute.

In particular, if

M is a "reasonable" transitive set and A e M, then for each ordinal ~ g M, (L [AJ)M = L [A].

By "reasonable" we mean that, firstly, the notion Def A should be

absolute for M, and, secondly, for each ordinal ~ e M there is an m-sequence in M which satisfies the inductive conditions of the L[A]-construction.

(We saw in

chapter 2 that neither of these requirements is particularly stringent.)

Now, these

-207-

requirements can be formulated as a first-order sentence and which is valid in each infinite L [A]. shall a s s ~ e

~0 which is provable in ZF,

The formula x = L [A] is ~ZF and we I '

that the ~0 above contains clauses asserting this fact.

assume that ~0 asserts also that there is no largest ordinal.

Finally, let us

We fix o 0 from now on.

Note, however, that ~0 is still a theorem of ZF, and is true in L [A] whenever lim(a). We have, therefore: Theorem I

(i)

Let M be a transitive model of o . Then: 0 If A E M and e E M, then (L [A]) M = e [A] ;

(ii) If A q M e M and ~ £ M, then ( L [ A n M])M = L [A ~ Proof:

(i) holds by definition of o0, of course. we have (Le[A n HI) M = e [A n M]. by induction on ~. e [A], so ee[A ] ~ M. thesis, A n e DefA(L

M]

= e [A].

For (ii), note first that by (i),

The equality e [A ~ M ]

Assume true for ~.

In particular then, (Le[A n M]) M =

Hence A n L EA] = A ~ M N L [A].

[A] = (A A M )

~ L [A e M].

[A n M]) = L + I [ A n M ] ,

= L [A] is proved

So by induction hypo-

Hence e +I[A ] = D e ~ ( L

using the induction hypothesis

[AJ) =

again.

Limit

stages in the induction are trivial, so (ii) is proved. | Applying Theorem I to M = L[A], we obtain: Theorem 2 For every ~, (L [A n L[A]])L[A]

,[A]

= L [A].

Hence (L[A ~ L[A]])L[A]

= L[A] and

v ° L[A n L[A]]' I By a procedure analogous to that for L, we can define a well-order,

L[A].

In fact, this can be done uniformly for all sets A.

construct a ElZF formula WO (v0,

Vl,

is A~ P

well-orders L[A].

~(x, y, ~, A) ~-> W0(x, y, A) ~ y E L [A].

Assume ~0 has conjunctions

More precisely, we can

V2) such that for any set A, WO is absolute for

L[A] and {<x, y>IwoL[A] (x, y, A ~ L [ A ] ) define~by


With WO as above,

Then it is easily seen that

asserting this fact from now on.

It will still

be true that ~0 is a theorem of ZF and that ~0 is true in L [A] whenever lim(~). Suppose now that M is a transitive model of ~0 and that A c M, y = sup(On n M). Then for all x, y, ~ E M, ~(x, y, ~, A) +-~ ~M(x, y, ~, A), as is easily

-208-

seen.

We can thus sharpen our absoluteness

statement

if x, y s e [A], then WO(x, y, A) ~-+ woM(x, y, A). < y.

for WO to obtain:

For suppose y c e [A], some

Then WO(x, y, A) +-+ ~(x, y, ~, A) +-+ ~M(x, y, ~, A) ++ woM(x, y, A).

wringing

So,

the maximum possible out of all this, we state:

Theorem 3 Let M be a transitive model of a

0

with A ~ M g M, and let y = sup(On q M).

For all x, y ~ ey[A], WO(x, y, A) ++ woM(x, y, A n M ) . I As we have stated, we shall not go into a great deal of detail in our present study.

But we clearly cannot fail to prove a "condensation

situation,

lemma" in the present

even if we do not obtain the sharpest version.

Theroem 4 (A Condensation Lemma) Suppose lim(e).

If M ~ L

EA], then there is a unique T ~ ~ and a unique map

defined on M such that, setting B = ~"(M n A), ~: <M, s, A ~ M> ~ < L [ B ] , =

Proof:

s, B>.

Since a 0 is valid in L [AI, this is not hard to prove, using our earlier results.|

Remark

We ought really be a little more precise concerning

"L [A]", etc.

our use of the symbols

We often use this (and the above proof furnishes

to denote the structure
EA]>.

This is because

a convenient

example)

the whole idea behind

is that they should be as similar to L as possible, whilst still

allowing us to refer to A; hence we at all times allow reference to A (at least as a unary predicate). We leave it to the reader to verify the following fact: Suppose A ~ K, some cardinal K. way.

Then, above <, the L[A]-hierarchy

For example,

the condensation

entirely similar,

and in particular

Loosely speaking,

this is because,

hierarchies

and the L-hierarchy behave in the same

lemmas for the two hierarchies

above <

do not lead out of the hierarchies

are

concerned.

if we are only concerned with the behaviour

above K, then when we carry out a condensation

argument for the LEA ] -

hierarchy, we can always include all of ~ when we take our elementary thereby ensuring that B = A in theorem 4.

of the

submodels,

Using this fact, it is easily seen that:

-209-

Theorem 5 Suppose V = L[A] where A ~ ~, ~ an infinite cardinal.

Then for all cardinals

~ ~, 2 ~ = %+. Proof: By induction on ~, IL JAIl = I~I for all infinite a. lemma, if ~ ~ ~ then

But by the condensation

|

e(%) ~ Lx+[A ].

+ Clearly, theorem 5 is the best possible in general, since if %

is smaller than

~, then we could find an A ~ < which codes %++ subsets of %, so that, in L[A],2 % = %++o If < is a successor cardinal, however, we can slightly strengthen theorem 5 to give: Theorem 6

(Hajnal)

Suppose V = L[A], where A ~ ~+ , ~ an infinite cardinal. Proof: Since ILK+[A]J = ~+, it suffices to show that Let x ~ ~.

Then 2~ = ~+

e(<) ~ L +[A].

Let % be a cardinal such that x s L%[A].

Since K + is regular,

we can find N ~ L x ~ A ] such that (< + I) ~ {x} c N and ~+ ~ N s such an N, define N O = the smallest N O ~ L % [ ~

o

(To obtain

such that (K + I) u { x } G N O

and inductively define Nn+: = the smallest Nn+ 1 ~ L%[A] such t~at Nn U sup(N n ~ K +) ~ N n + I, whence N = ~ n < ~ Nn is clearly as required.) Collapsing N we have~: %
and ~ = r+ q N = ~(~ + )• x ~ L [ A N 6].

s, A ~ 6>, where y = ~ " ( N ~ % )

T

But ~(x) = x, since ~ 6

= ida6 and 6 > <.

But look y, ~ < ~+, so Ly[A ~ 6] ~ L +[A].

Ly[A ~ 6] ~ L +[A].

Hence x ~ L +[A].

In particular,

QEDo

We shall use the same trick as above in order to adopt the proof of the case LEA] where A ~ L [ A ] . 1

Recall that

<S

~ a and, whenever X ~

I ~ < m > such that each S

Hence

~ in L to

~ asserts the existence of a sequence {a ~ ~

I X ~ ~ = S } is

stationary in ~I" Theorem 7 Suppose V = L[A], where A c ~I"

Then 0 holds.

Proof: By induction, for each limit ordinal a < ~I' let <Sa, Ca> be the < L [ A n ~ least pair of subsets of ~ lying in L[A ~ d unbounded in ~ and y e C a ÷ y Q S a % Sy.

-

such that C a is closed and

If no such pair <S , Ca>

exists, or

-210-

if succ(~), set Sa = C

=~.

Suppose now that <S~ I a < 601> is not as required by 0.

Let <S, C> be the
(Clearly, any countable N ~ L602[A] will be as required, so lets take

the smallest for definiteness.) y = ~"(N ~ 2 )

As above, pick N - < L ~ 2 [ A ] such that

and 6 = N Q ~I"

Let ~: ~ , where Note that ~(~i ) = 6.

Now, <<S , Ca> I e < ml > and

<S, C> are L~ [A]-definable (using the predicate A), so lie in N. And clearly, 2 ~(<<S a, Ca> I a < 601> = <<S , Ca> I ~ < ~> and ~(<S, C>) = <S ~ ~, C ~ 6>. But look, since ~-I: e [A ~ < L~ [A] and ~-I(<s a 6, C ~ 6 > ) = <S, C>, we see that Y 2 <S n ~, C ~ ~> is the = <$6, C~>.

In particular, S 0 ~ = $6.

this contradicts the choice of <S, C>. Note

Hence

But ~ = sup(C ~ 6) e C, so

QED.

In the above proof, we had to be slightly more careful than for L, since we had

to allow for the fact that when we carry out the eventual condensation argument

for

e

[A], we come down to some L [A n 6] and not to e [A]. This problem would not, of ~2 ~ course, arise if A were a bounded subset of 601' for it would then be the case that 6 > sup(A) in the above, whence A ~ ~ = A.

Corollary 8 Suppose V = L[A], where A ~ 601" Then there is a Souslin tree. Proof: By theorem 3.8. I Recall that S

~ + asserts the existence of a sequence <S~ I a < 601> such that each

is a countable subset of

~(~) and whenever X ~ 601 there is an unbounded set B ~ 601

such that for each limit point ~ of B, X h a ,

B ~ ~ ~ S .

By virtue of our above note,

it should be clear that if V = L[A], where A ~ ~ < ~I for some ~, then

~ + holds.

However, by exercising a hit of extra care, we can prove a stronger result, promised in chapter 4.

First we require a rather cute little lemma.

Lemma 9 Suppose V = L[A], where A _c

60 1 •

If L [IA

~ a] < 601 for all a < 601' then ml is

-211-

inaccessible in L[A ~ s] for all ~ < ~I" Proof: Suppose that for some ~ < ~I' ~I is not inaccessible in L[A ~ s]. must be O < ~I such that ~I = Io+IL[A ~ e] f e e

EA~, f: m +-+ O.

~: N ~ L¥[A ~ y j ,

Now, 0 is countable, so there is

Let N be the smallest N ~ L

where y = N ~ ~I"

[AJ such that f c N.

Let # = max(s, y) < ~i"

Let

(Clearly, the transitivisation of N will

ha~e this form.) The., in L[A ~ ¥], ~ is countable, so FL[A countable".

Then there

~] ~

Then clearly, m I = ~i L[A a BJ.

~s QED.

Theorem I0 Suppose V = L[AJ, where A ~ ~ . 1

Then

~

holds.

Proof: We define a function ~: ml ÷ ~I by cases, depending upon the nature of A. Case Io There is s < ~I such that ~I Let e0 be the least such ~. that ~0' s s M. Case 2. ( ~ e

For each s < ~I' let M s be the smallest M ~ L I[A] such

Set ~(s) = ~i ~ M

< ~i)(~i L[A ~ ]

Set 6(s) = ~ L[A ~ s ]

1

.

Clearly, ~(s) e ~I-

~ ~i)o

By lemma 9, el is inaccessible in LEA ~ s ]

so 6(s) ~ el

Set s 0 = m in this case. That completes the definition of ~: w I + ~I"

For each s < ~I' we set

= max(s, s 0). Set Ss = satisfies

0+.

e(s) ~ e~(~)[A ~ ~], each s < ~I"

We show that <Ss I ~ < Wl>

Well suppose not, and let X _c ~I be the
there is no B as in

0+.

By induction on 0 < ~ < ~i' define submodels

~A] as follows: 2 N I = the smallest N -
Nv

L

N + 1 = the smallest N -
Clearly, <su I ~ < ~i > is a normal sequence in ~I"

] for each u, and set B = {B(u)

We show that B satisfies

Let s be a limit point of Bo

I 0 < v < ml }, an unbounded

~ + for X, contrary to the choice of X.

Since s

< B(v) < ~ + I

for each ~, we clearly

-212-

have ~ =

for some limit ordinal n.

Claim I

L6(~)[A ~ ~] is a model of ZF-.

Suppose first that Case I holds. V = L[A], L ira] = H i , so L L~(~)[A ~ ( ~ ) j

also.

We must show that X ~ a, B Q a e S , i.e. that

Let ~s: Ms ~ e6(a)[A q ~(~)].

[A] is a model of ZF-.

Now, as

Hence so is M , and hence

But look, g < 6(e), so A n ~ c L~(a)[A n 6(a)].

Thus, working

inside the ZF- model ed(a)[A ~ 6(a)j we can carry out the proof that: "e~(a)[A q ~] is an inner model of ZF-".

In other words, L~(a)[A n ~] really is a model of ZF-.

Suppose now that Case 2 holds. ZF--model.

Thus, as a is infinite, ~ = a.

And in L[A O e] it holds that e

~3

[ A n a] = H

~3

Now, L[A n a] is a

, which is a model of ZF-.

Thus once again we see that L~(a)[A n ~] (really) is a model of ZF-, and the claim is proved. Now, the function ~ is clearly L

[A]-definable. 2 It follows that X is Lm2 [A]-definable.

<S

Hence so is the sequence

I a < ~0 >. Thus, by applying 1 ~ -i : LB(n) fA _ a] -< Lc0 [A] we see that X p~ a is L~(O) [A n a]-definable. Again, since 2 on-l: LB(n)[A n a] =~ N -< L [A], it is easily checked that B q ~ is ZF--definable 2 from B(n), A q ~ in the same way that B was defined from m2, Ao /x

Thus, by virtue of claim I, in order to prove that X ~ s, B q ~ s L~(a)[A n s] , it suffices to prove the following: Claim 2

B(n) < ~(a).

Suppose first that case i holds.

Now, a < ~i' so by choice of a0'

< ~i L~A (~ a0] • Hence ~L~I [A n a0]"s is countable". But look, a

-< L [A], so it is easily checked that co 1 M~ n LmI[A na0] ~ L I[A (~s0] , and it is also easily seen that ~S

0

c M

a

: (MS (~ L I[A n~0]) ~ L~(a)[A ~a0].

• countable". FL6(a)[A n a0] "c~ i s ,

Hence as a s M ,

It follows that ~L

we may conclude that

r. ~ "~ is countable". ~((x) L~ ? aJ

But (noting that ~ > ~0 ), o:': LS(~)[A n ~] ~< L w [A] and o~ (~) = ~I' so 2 ~LB(q)[A q ~] "a is uncountable". Hence 6(a) > B(q). Suppose now that case 2 applies.

Since on : L S ( n ) [ A n

~o 2

[A] and

-213-

o~i(~)

Ll[i

B(q) ~< ~2

i. B(q)[A n ~]"~ ~]o

is the only uncountable cardinal".

But ~(~) = ~3 L[A ~ ~].

It follows that

Hence B(q) < ~(~), and claim 2 is proved.

QED. Corollary Ii Suppose V = L[A], where A g ~i"

Then KH holds.

Proof: By theorem 4.4. Now, we stated in chapter 4 that Silver proved the consistency of ZFC + ~ K H assuming the consistency of ZFC + "there is an inaccessible

cardinal".

result shows that his assumption of a large cardinal here was necessary

by

The next (by virtue of

the Godel IInd Incompleteness Theorem). Theorem 12 Assume

(Solovay) ~KH.

Then m2 is inaccessible

Proof: Suppose ~2 is accessible in L.

in L.

Now, ~2 must be regular in L, since regularity

is a ~I property valid for ~2 in the real world.

Hence, as L I= GCH, it follows

that ~2 = ~0+)L for some e < ~2 such that I=L "0 is a cardinal". e < ~2 we c~n find A ~ I that m 2 = ~2 LLA] .

such that ~i = ~I L[A] and 181L[A]

Now, by corollary Ii, there is

" N is a Kurepa family".

Now, as

= ~I"

It follows

~ s L[A] such that I=L[A]

But since ml and ~2 are the same for L[A] and the

real world, this means that

~

really is a Kurepa family, whence KH is proved.

QED. The above results indicate to some extent how L[A] and L "agree" from "the top" + of A upwards, and to some extent near "the top" of A.

However, since a set A ~ K

can code up a lot of information concerning <, it is clear that L and L[A] need not at all agree below K.

For instance, we could have an A ~ ~2 which codes up m2 subsets

of ~, whence I=L[A] 2~ = ~2"

Thus, as A gets larger, so one would expect the

"disagreement" between L and LLA ] to become worse.

It is therefore perhaps a little

suprising to discover that there is a very large set A, which affects L quite considerably

(i.e, which makes V = L fail in a spectacular manner),

behaves very much like L.

such that

We discuss this phenomenon in the next chapter.

L[AJ

Chapter

19

THE INNER MODEL e [p]

R e c a l l that a cardinal there is a <-complete, ultrafilter

< is measurable

non-principal

u on ~ is normal

iff < is regular and uncountable,

ultrafilter

on < (i.e. in

iff it is <-complete,

~(<)).

non-principal,

and

An

and such that for

every f: < ÷ < such that {~ E
It is a standard

a normal ultrafilter.

For a proof of this result,

measurable

cardinals

the following

in general,

information

p will denote

concerning

For our present purposes,

concerning measurable

an aribtrary

and ~ will denote a (fixed) normal ultrafilter

cardinal carries

and for further details

we refer the reader to [8].

lemma gives the crucial

For the w h o l e of this chapter,

result that every measurable

cardinals.

(fixed) measurable

cardinal,

on p.

L e m m a 1 (Rowbottom) Suppose % < p and f: If"[x]nl Proof:

[~]<~ -+ %.

Then there is X E ~ such that for each n < ~,

= I.

Since ~ is p-complete, f:

it suffices

to prove that w h e n e v e r n ~ ~ and % < p and

[E]n ÷ I then there is X 6 u such that

induction on n. Let % < P, f:

If"[x]nl

The case n = I is trivial.

[p]n+l ÷ %.

f (~) = f ( < ~ > ~ O )

For each ~ E P, define

some g(e) ~ %.

(a set we considered,

We prove this by

the result is true for n.

f : [pin ÷ ~ by

for each ~ e p, there is X

e ~

for each ~ e [pin. such that

Now, g: p ÷ %, so as ~ is p-complete we can

pick S E ~ such that g"S = {y}, some y E ~.

Claim

Suppose

if e < rain(a) and f (a) = O otherwise,

By induction hypothesis, f "Ix ] n = {g(~)},

= I.

it may be recalled,

Set r = {~ c p J ( ~

< B)(~ e X )}

in chapter 3~).

T ~ ~. D e f i n e h on p - T by setting h(B) = least a s.t. B ~ X .

all 8 e p - T.

Thus,

if T ~ ~, then by the normality

that h"U = {~} for some e < p. This proves

the claim.

But then U n X

Clearly, h(8) < B for

of ~, there will be U ~ u such

= #, which is absurd,

since U, X

~ ~.

-215-

Set X = (S f~ T). required.

Thus X s ~.

We show that If"[x]n+l I = I, as

In fact, we show that f" IX] n+l = {¥}.

and let ~ be such that ~ = But n c S.

<~>'-~ ~.

Hence f(a) = g(n) = y.

Let

¢ye IX] n+l.

Since ~ ~ T, c~ e [xn]n.

ultrafilter on p".

Hence f(~) = fn (~) = g(~)"

QED.

Consider the inner model e[~] now, and set ~ = ~ ~L[~]. chapter 18, ~e[u]"V = e[~j".

Let n = min(~),

By the results of

Now, it is easily seen that ~e[~] ''~ is a normal

(Roughly speaking,

this is because this sentence is T~I.)

It

follows that in order to study the inner model L[U], it suffices to study all consequences of the axiom "V = L[X] where X is a normal ultrafilter on the measurable cardinal p".

Extending our notational conventions

simply as "V = L [~] " .

a little, we shall write this

Note that by our above remarks, we have essentially proved

that Con(ZFC + "there is a measurable cardinal") ÷ZF Con(ZFC + V = LEvi). Now, in view of" the dramatic effects the assumption V = L[~] has on L (see Chapter 18), one might expect that L[~] looks very different from L.

It is therefore

perhaps somewhat of a suprise to discover that L and L[~] have many similarities. shall prove that there are L[~]-analogues concerning L.

We

of all of the results of chapters 1-5

In fact one can go much further.

In [4], it is proved that all of

the results of chapter I0 carry over from L to e[~], whilst Solovay ([21]) has indicated that the results of chapter Ii (and some of the prerequisitary Section 2) also carry over to L[~]. Kunen's paper

[13].

In particular,

only normal ultrafilter on p.

results of

Other properties of L[~] are established

in

it is proved there that if V = L[U], then p is the

V = L[U] also implies several ill~minating results

concerning the large cardinals considered in chapters 16 and 17, as is shown in [6] and [13]. Before we commence our study of L[U ] , let us remind the reader that we often r i ,, as an abbreviation use "L ~LuJ

for the structure
If n is some mapping, Fix(~) denotes the set {x e dom(~)I~(x)

[U]>. = x}.

We first of all prove a condensation lemma for the L[u]-hierarchy. a subset of

Since ~ is

~(p), it will be observed that, by virtue of the trouble we had with

condensation lemmas in chapter 18 (where collapsing tended to lead out of the

-216-

hierarchy

concerned),

this result is somewhat

striking, even though it is easily

proved. Lemma 2 (Silver) Let K > p he a cardinal.

Suppose X -
[~

and X q p

s ~.

Then there is a

unique ~ and a unique y such that 9 .< y .< K and: (i)

~: X ~ e~[~]

(i.e. ~: <X, ~, u ~ X> =~ <ey[~],

(ii)

if Y % X is transitive,

then ~ Y

~, ~ q ey[~]>);

= idlY;

(iii) Fix (~) q p ~ ~. Proof: Let ~: X ~ M, where M is transitive. s p ~X,

~(e) .< ~.

(ii) is immediate.

Now, for all

So if Fix(n) A P # ~, then {e c pl~(~)

< ~} s ~, whence

by the normality of ~, ~ is constant on a set in ~, which is absurd since is one-one.

Hence

Let Z = Fix(~) n p . A qZ

c ~.

(iii) holds.

We are left with

Thus, by (iii), Z E ~.

(i).

Suppose A ~ X ~ ~.

But by choice of Z, A ~ Z -=~(A).

Then

Hence ~(A) e M q ~.

Thus

~"(X n ~ ) ~ M n ~ . Now suppose A g X ~ ~(p) - ~. IX q Pl = P, so ~(p) = p. the above. But look, ~L

Then P - A s X q ~.

Hence p - ~(A) = ~(p) - ~(A) = ~(p - A) ~ M q ~, by

It follows that ~(A) ~ M n 6)(0) - ~. [~] "V = e[~]",

as M is transitive,

Now, p is a cardinal and

M = Ly[~]

so by our last

Thus ~"(X ~ ~) = M q ~ .

result, ~M

for some unique ~.

"V = L[M q ~].

Thus,

QED.

Using the above result, we can prove GCH holds in L[~].

First a lemma.

Lemma 3 Let m ~ 5 < P < 0, and let T -= L [~] be countable. (i)

There is a set X ~ ~ of indiscernables

for the structure

<e0[~ ] , ~, ~ n eo[~ ] , (a)a.<~' (t)t~T>' (ii) For any ~ < 6, 16 qLs[~] ~(X ~ ( ~

+ I) o r ) I = le(6)

qe01~][(X u(~ + 1) uW)I

I~I ÷ ~.

(For A ~ e 0[~], e 8[~][A denotes the smallest N - < L Proof:

(i) follows

[~

such that A ~ N . )

from lemma I by an argument as in lermma 17.2

=

-217-

(ii) follows from (i) by an easy indiscernibility Remark.

In [4] is established

strengthens strongly.

an "L[~]-condensation

lemmas 2 and 3, and highlights For our present purposes,

argument. I

lemma" which considerably

the similarity of L and L[~] quite

however,

lemmas 2 and 3 are quite strong enough

as they stand. Theorem 4 (Silver) Assume V = L[~].

Then GCH holds.

Proof: Let K be an arbitrary infinite

cardinal.

depends upon the size of K relative Case I.

f~e6+l[~] I ~

of indiscernables %

<.

X ~ ~, N q O e ~. NOW, a ~ <

"a

[~]

to p.

to show that if a ~ ~ and a ~ e6+ I[~] - e 6[~], then

So let a, 6 be as above.

for the structure

6 is a cardinal.

~e

The proof

~ < p.

It clearly suffices

le(<)

We show that 2 ~ = K+ •

Let N

<e%[~],

E, ~ q e%[~],

(~)~.<~, {6}, {a}>, where

LX[~ ] ~(X O (K + i) O {6, a}).

So, let ~: N ~ L [~] ~ N, so ~(a) = z"a = a.

g L~+I[U]

By lemma 3 (i), let X g ~ be a set

L~[~] ".

by lemma 2. But I=L)~[~]

Hence ~ = 6.

Since X G N ~ p and

Set ~ = ~(6).

"a

e

L6+ 1 [~]

In particular,

- L 6[~]'', so

therefore, y > 6.

But

Y by lemma 3 (ii),

le(<) (~ N I = <.

Hence as ~(<) = <, J~(<) ~ ey[U] I = <.

So, as

y > 6, 16)(<) q Ls+l[~]l,< < , as required. Case 2.

< >~ p.

By induction on cardinals ~ >~ <, Ie%[u]l for all ~ >. <, ~(X) c L%+[~]. of the analogous the hypothesis Note.

result for L.

= X.

The proof of this is essentially

as far as properties

to show that

the same as the proof

(Since we are working above p, all of the time now,

of lemma 2 is trivially satisfied.)

It will be observed

Hence it suffices

|

from the above proof that L[U] can be handled just like L

of cardinals greater than O are concerned.

where we use lemmas 2 and 3 in a non-trivial way.

It is only below p

-218-

Theorem 5 (Jensen) Assume V = L[~]. Proof: By induction,

Then

~ holds.

for each limit ordinal o < ~I let <So, C > be the
pair of subsets of o such that C y s C ° ÷ S ~¥ ~ S~. S~ = C~ = ~.

is closed and unbounded in ~ and

If no such pair <S , C~> exists, or if succ(~), set

Then <S~I~ < ~i> is as required.

For, suppose not, and let

<S, C> be the are L p +[~]-definable

Note that <<S o, Co>Io < ~ 1 > and

(by means of the above definitions).

Let X ~ ~ be the .

Let N = ep+[~]~X.

But clearly, ~I q N is transitive.

[~].

By lemma 3 (ii),

(This is because,

is an L +[~J-definable map f : ~ +-~ ~.) O ~ N q p s ~, so let ~: N ~ L

for the structure I~ 1 ~ N I = m.

if ~ g ml ~ N, then there

Hence ~ = ~I q N s ~ . 1

Then ~ 6

Since X ~ N, --

= ida6, ~(~i ) = 6,

~(<<S , C >Io < ml>) = <<So, C >I~ < 6>, ~(<S, C>) = <S N 6, C ~6>. follows that 6 ~ C and S ~ ~ = S 6 (see chapter 3), a contradiction.

It I

Theorem 6 (Jensen, Devlin(indep.)) Assume V = L [~]. Proof:

Then

0 + holds.

(Devlin) Let U ~ ~ be the . lemma 3, I@(~) q M

for the structure

For each ~ < ~I' let Me = Lp+[~] ~(U ~ (~ + i)).

I ~< l~I + ~.

Let ~ : M

S~ = @(6) q L6(~) [~], each o < ~i.

~=L6(~)[~ ].

By

Set

We show that <So]~ < ~i > satisfies

0 +.

Well suppose not, and let X ~ m I be the
~+.

Note that <Solo < ~i > is Lp++[~]-definable,

Let W s ~ be the .

whence so is X.

for the structure

By induction on v-< ~I' define:

NO = L O + + [ ~ J ~

;

N + I = e p ++[~] ~(W ~ (o

+ I)) ;

N% = U
;

°0 = ml q NO ~+I

;

= ~I q N~+l;

e% = ml ~ N%.

-219-

Clearly, 9 < T < ~01 ÷ N -< N T ~ Lp++[~] and <~ 19 < ~i> is a normal sequence in ~I" For each 9, let ~-o: N each 9.

~ e~(.u)[~]

Set B = {e 19 < ml }.

0 + for each X.

Clearly, ~ ~e

= idle

We obtain a contradiction by showing that B satisfies

Suppose then that ~ is a limit point of B.

limit ordinal ~ < ml"

and f7 (~I) =

We must prove that X ~ ~, B ~

Then ~ = a n for some

c L~(~)[~].

Now, ~eB(n)[~ ] "e is uncountable" (since cr n-I(e) = ml )' but ~L~ (~)[~] ''~ is countable" (since ~ - I.- e~ (~) [~] ~ M

and ~ s M a--< e P +[~]).

Hence 6(n) < ~(~). Now, X is Lp++[~]-definable, so as cy~l: L6(~)[~ ] ~< Lp++[~] , we see t h a t X • is Ls(~)[~]-definable. that B ~ Claim.

Hence as B(B) < 6(~), X Q a e e6(a)[~], and it remains to show

~ e6(~)[~ ].

There is a set W' ~ ~ ~L6(~)[~ ] of indiscernables for the structure

Let us call an (~-t-definably) ordered subset Y of a first-order structure set of [n-indiscernables for ~ of ~e and all Y 0 " " '

iff for all In-fOrmulas ¢(v ..... Vn) of the language 0

Yn' Y 0' " " '

Yn' ~ y w i t h Y0 <'''< Yn and Y0'

~0L ¢(~0''''" ~n ) iff ~0L ~(~'0''''' ~'n).

<'''<

an e ++[~]-definable point of e ++[~]. p

Y'n'

For each n < ~, let Wn E ~ be the
set of [n-indiscernables for the structure <ep++[~], g, ~, (~)~<~i >. p

~ a

Hence each W

n

Thus each W

is an element of N . n

is

Set

W'n = °'n(Wn)" Then W'~ e LB(q) [~] is the lying in ~.

Set W' = ~ n n~<W'.

clearly,W ' is a set of indiscernables for the structure .

LB(~)

But look, the sequence <W'nln < ~> is clearly LB(q)+l[~]-definable

sentence above which characterises each W').n Hence W' ~ L6(n)+3[~ ]. W' e L6(a)[~].

But as ~ is O-complete, W' ~ ~.

Then, [~],

(by the

It follows that

The claim is proved.

Now, B was defined, in ZF- ' from W and L P ++[~].

Hence, by the above claim,

working entirely within e6(~)[~], we can define a set B' -~ ~ from W' and LB(~)[~ ], in the same way that B c_ ml was defined from W and LO++[~].

(In fact, we could

-220-

define a B' right up to ~ 6 ( e ) [ ~ !

but we shall "stop short" at ~ for convenience.)

This puts B' in L~(e)[~], so we are done if we can show that B' = B Q ~ . Now, ~ D is a collapsing isomorphism of N and ~

is one-one, we see that Fix(~n) s~.

B' = { ~ I v

< ~}.

onto L B

(n)

[~].

So, as p is normal

Hence Z = W ~ W ' Q Fix(~y ) s ~.

Suppose

We prove by induction on v < ~ that a'~ = a~. Limit stages in this

induction are trivial, and the initial step is a degenerate case of the successor step, so we shall just do that.

Suppose then that v < q and we have proved that for v

all ~ ~ ~, ~T = a'T"

We prove that ~ + I

Now, to obtain ~ + I ' av+l = ml ~ N~+I"

= ~+I"

we set N + I = LO++[~ ] I(W O(e +I)) and then put

Note that as ~ + i

< ~' we could just as well have set a~+l =

~ N + I here. !

Thus, to obtain av+l we would set N'v+l = eB(~) [U] ~(W' ~ (a~+]~ and then put

~v

v+l = a ~

N' ~+I'

So, what we must prove is that a QNv+ I = a Q N ' ~+i" Let ~ ¢ ~ Q N + I .

Then as ~ ~ N + I

there must be some ~ - t e r m t and some

elements Y0 <'''< Yn in W and some elements 8 0 <'''< 0m ~ ~v such that Lp++ [~] = t

(Y0""'

Yn' 0 0 " ' ' '

om)"

Now, ~, ~0''''' Om ~ ~ < ml so by choice of W we can assume that, in fact Y0"'''

7n ¢ Z here.

Hence, applying ~ ,

we obtain

teB (n) [~ =

(Y0 ..... Xn' eO ..... 0m)"

But by induction hypothesis, 00,... , 0m ~ a~(= av).

Hence, by definition, $ s N' ~+i"

Similarly, if ~ E a NN'v+ I , then ~ ~ a ~ N~+ I , which proves that a N N A N' v+l"

+1 =

QED.

Finally, we turn to the proof of our analogue of the main result of chapter 5. Now, as Silver remarks in the introduction to his paper [20] , if there is a measurable cardinal, then there can be no ~

well-ordering of the reals.

Thus, the

~

best analogue to

corollary 5.16 we could hope for is that V =

existence of a A~ well-ordering of the reals.

L[~3

implies the

We shall devote the rest of this

-221-

chapter to Silver's proof of this fact. We consider a fixed first-order language ~ with binary predicates =, e, unary predicates U, W constants a, b, and constants o for all ~ ~< 03. ~' denotes a Skolem expansion of ]~ , and T O denotes the usual axioms for Skolem expansions. A standard P 03-model, ~I, is a ~-structure with a transitative universe such that UCtis a limit ordinal and o0%= ~ for all ~ .< ~. where we here demand that

0~

We say a theory T in ~

Likewise a standard P '

~-model,

T O also. or ~ ' admits K iff there is a standard m-model of T

in which U denotes <. Let ~ L b e a standard

~' t0-model, I a set of formulas of ~ '

(I, ~)-set of indiscernables for ~

We call X a

iff:

(i)

X is a cofinal subset of U ~ a n d

otp(X) = ~;

(ii)

if x!<...< Xn are elements of X, then ~l~(X°l ..... Xn) iff ¢ ~ [;

O

(iii) if Y is an initial segment of X with no greatest members, then

01~Y Q U 0L is an

initial segment of U . Let P(T, [, ~) mean that either succ(~) or else there is a standard

~' ~-model

of T which has a (I, ~) set of indiscernables. Lemma 7 Suppose P(T, I, B) for some uncountable limit ordinal 6, or else that ( V ~ < ~I)P( T, [, ~)>

Then for all y, P(T, I, 7)-

Furthermore, T admits any cardinal

03.

Proof: Let y be a limit ordinal.

Extend

]~' by adding new, distinct constantaC 6,

6 < ¥.

Let T* be the theory T tJ {¢(C 6 .... , C6n)l + e I & 6 1 <'''< 6n < Y}" 1 By hypothesis, T* is finitely satisfiable, so T* has a model, ~ * . Let

be the ?'-reduct of ~ * . assume Claim. ~[.

~ = ~Y

Suppose C ~ Y ,

here.

Let Y =

{ c~~* 16

< 7}.

Since ~ Y - <

~ , we may

Note that as U(v 0) s [, Y -¢ U~.

otp(C) ,< ~, and that X is a (I, e)-set of indiscernables for

If g: C -~ X is order preserving, then there is a unique ~ ~ g such that

~:~c ~.

-222-

To see this, set g(t ~ ( x I .... , Xn)) = t 0%(g(xl) ..... g(Xn)) , for each term t of ' and each x I <...< x

in C.

This defines a one-one function g, since

n

tl(x ) = t2(X ) +-+ tl(~ ) = t2(v) g ~ ~+ tl(g(x)) = t2(g(x)7. Similarly, g is elementary. concerning ~ , (I)

~

This proves the claim.

There are three facts

Y which we now require.

is well-founded. Suppose not.

Then there is a descending u-chain, s a y ~ ,

countable C ~ Y such t h a t ~

~C.

under c

Pick a

By hypothesis, there is a standard w-model

and a (I, ~)-set X of indiscernables for ~ , where e > otp(C). By the claim, there is h: (2)

~

C ~0~.

But then h"J~is a decreasing w-chain under g, which is absurd.

Y is ~ - c o f i n a l Suppose not.

in U ~ Let ~ >. u, and let ~ be a standard u-model of T and X be a

(~, e)-set of indiscernables for ~ ,

by hypothesis.

Let z = tY~(yI ,. .., yn ) be an element of U Yl <'" "< Yn lie in Y. z' = t (xl,..., Xn). elements.

Let x I,..., xn be the first n elements of X Let Y' ~ Y have order-type u, with Y I " ' ' '

and set

Yn its first n

Now, given any x ~ X we can clearly find an order preserving map

g: Y' ÷ X such that g(yi ) = x i and x e ran(g). g(z) = z'.

exceeding every member of Y, where

And as gl(x) = ~-l(x) ~

Let g ~ g , g: ~ I Y ' ~< ~ .

Thus 0g z (by choice of z), we have x c z' ~ U . But

x ~ X was arbitrary, so z' exceeds every member of X, contrary to X being cofinal in

(3)

If Z is an initial segment of Y with no greatest element, then ~ [ Z q U ~

is an

initial segment of U ~. Suppose not.

Then some z I = tl(Y I ..... Yn' Yl ..... Ym) s U

e~-less than z 2 = t2(Yl,... , yn ) s U ~ ~ ~ Z , YI' <''" < Ym' are in Y - Z.

-

~ [ Z is

where Yl <. " .< Yn are in Z and

Let ~ ~ u + u, and let 0-[ be a standard u-model of T,

X a (I' ~)-set of indiscernables for ~ , order type ~ + u with Y I " " '

Yn' Y ~ " " '

by hypothesis.

Let Y' be a subset of Y of

Ym c yv and otp(Y' ~ Z) = u.

order-preserving map of Y' onto an initial segment of X.

Let ~

Let g be an

g, g: ~ ~ Y' ~ 01.

-223-

Let X' consist of the first ~ elements of X. X'~

g"Z).

Clearly, g(zl) ~ U ~ - 5fiX' (since

Similarly, ~(Zl) e ~(z2) ~ U ~ D 5-fiX'. This contradicts condition (iii)

in the definition of ([, a)-indiscernability of X for 6L. That proves the three facts about ~ ,

Y we require, and we may now prove the

theorem. By (I), let f: ~ above, ~

~ ~

= , where D is transitive.

and f"Y satisfy conditions (i) - (iii) in the definition of (~, ~)-

indiscernability.

Hence, a s ~

only prove that ~

is a standard m-model.

{~I~

~ ~

~ T, in order to establish P(T, ~, y) we need It clearly suffices to show that

~ ~} forms an initial segment of the ordinals of ~

suppose t ~(YI''''" yn) e~ ~o , where ~ ~ m and Yl < ' " < "t(~) < ~" s I, so with ~ ,

~, e as in (3), s i n c e ~

under ~

whence t~(Yl

.....

yn ) =

To see this

Yn are in Y.

Then

is a standard u-model, if

~b x I <...< x m are in X, then t (x I ..... Xn) = ~'°Lfor some e' < s.

~,

By (2) and (3)

~,,, Thus "t(~) =

~'$.

Hence P(T, ~, y). It remains to prove that if < > ~ is a cardinal, then T admits ~. (~, <)-set of indiscernables for a standard m-model ~ U ~is

of T.

Then U

Let X be a

= K.

For, as

an ordinal and X is cofinal in U ~, it suffices to show that each ~ E X has

cardinality less than <.

Now, if Y is a proper initial segment of X containing ~,

IYI < K, and by (iii) in the definition of (~, K)-indiscernability, ~ ~ 6 l ~ y . I ~ ' [ = ~, SO I B ~ Y 1 < <.

But

QED.

Lemma 8 Let < be a Ramsey cardinal. is a ~ such that ~ P ( T ,

I, ~).

Let T be a ~ - t h e o r y which admits <.

Then there

Hence T admits all cardinals % > m.

Proof: By lemma 7, it suffices to establish P(T, I, ~) for some I" Let

01 = be a standard m-model of T with U

= K, by assumption.

Let

X ~ K be a set of indiscernables for Ol of order-type ~.

Let ~ he the set of all

formulas satisfied in ~

Set

~is

on increasing sequences from X.

~ = ~X.

well-founded, so is isomorphic to a transitive structure.

Since It therefore

-224-

clearly suffices to show that X satisfies (i) - (iii) in the definition of (I, <)indiscernability for the structure ~ . collapse of ~

are as required.)

(For then the images of X and ~ under the

The only non-trivial part is (iii).

initial segment of X with no greatest member. segment of U ~.

Suppose

~Y,

where z I <''" < zn are

Choose zI <...< zna in X - Y for a < < so that a < B ÷ zne < zl" 8

indiscernability, w ~<

n U * is not an initial

Then for some Yl <'" "< Yn in Y and some terms t I, t 2 of ] I ,

~t tl(Yl,..., yn ) exceeds t 2~(Yl ''''' Yn' Zl .... , z n) E U ~ in X - Y.

~Y

Let Y be an

8÷wa~w

= t2(Y 1 .... ' Yn' z ] " ' "

Zn) < t l ( Y l " ' ' '

By

Yn ) and

B•

Thus tl(Yl,... , yn) has < predecessors, which is absurd.

[Note that we have

used the fact that Y has no greatest element in order to show that a + ~ ÷ w e ~ w B. For we must plck z 'I < . ..< z'n in Y greater than Yn to see that t2(Y],..., Yn' ,

Zl"'''

Zn) + t2(Yl ..... Yn' zl ..... Zn) (as the latter is not in

replace z'. by z~. and z i by z 8 by indiscernability.] l • i

~Y)

and then

I

From now on we shall assume ~ ' is arithmetised in ~ in some recursive manner. Now, if M is a transitive model of ZFC and T, I' ~ ¢ M, then P(T, I, ~) iff [P(T, I' ~ M .

The proof of this fact is essentially an "absoluteness" argument,

involving the absoluteness of semantical notions and of well-foundedness,

and the

effective way in which a structure with a (I, a)-set of indiscernables can he constructed from ~ and e if there is one at all. to the reader.

We leave the details of this proof

The proof shows, in fact, that we can replace ZFC in the above by

some finite fragment of ZFC, which we shall assume is axiomatised by the single sentence

~I

(which is thus a theorem of ZFC).

Hence, if M is transitive and

M ~I'

then for all T, I, ~ ~ M, P(T, I, ~) iff ~P(T, I, ~)]M Now, by lermna 7, P(T, I, ~i ) iff ( ~

< ~I)P(T, I, g)"

And every countable

ordinal ~ is in some countable transitive structure M such that P(T, I'

t° I)

o~.

Hence

iff

(*) for every countable transitive M such that T, I s M and M

I'

[~P(T,

I, e)]M.

It will be helpful to recast these observations in terms of well-foundedness of relations on ~.

If S ~ ~ and R is a well-founded, extensional relation on ~, we say

-225-

that x re~)resents S in <m, R> if S is the image of x under the collapse of <~, R>. Clearly, the predicate "x represents S in <m, R>" is A~. If ~ is an ~ -formula and R is a binary relation on ~, let (~)R be the result of replacing every occurence of e in ~

by R.

Translating (*) into these terms, we

get: P(T, ~, ~i) iff [for every well-founded relation R on ~ and every m, n ~ ~, if R is extensional and (~])R" and if m, n respectively represent T, ~ in <~, R>, then (V~P(m, n, ~))R]. of P(T, I, 91)-

Since "well-foundedness" is T[~, this gives us a~21 representation

In other words, we have proved:

Lemma 9 There is a ~ i formula ~ such that for any ]~ '-theory r and any set ~ of ~ 'formulas: P(T, I, ~i) iff <~, <, T, I> ~ ~. | Using Theorem 18.3, we shall obtain a characterisation of
for

some closed unbounded set C ~ p. We say a set M is acc_~table iff M is a transitive model of the sentence of chapter 18 such that

~ N M s M and

~ qM

= ~ qM.

~0

A few remarks are called

for here. Clearly, ~ c_ ~.

Moreover, if M is a transitive set such that P - Z ~ M when-

ever Z s M, then the conditions 3' ~ M such that

~' ~_ ~

~ ~ M s M and

and ( ~ Z

~ ~ M = ~ q M hold iff there is

E M)(Z ~ p -~ Z ~ ~' v (P - Z) E ~').

This

removes the ostensible dependence of acceptability on ~, and will facilitate our handling of acceptability in first-order logic. Lemma IO (i)

If x =--m, then x E L[~] iff there is an acceptable set M such that x ~ M and

~M O E L [

~ q MJ.

(ii) If x, y ~_m and x, y E L[D], then WO(x, y, ~) iff there is an acceptable set M such that x, y g M and ~M WO(x°, ~,

} qM).

-226-

Proof: U s i n g lemmas 18.1,

18.3, it is easy to reduce the p r o o f of (i) & (ii) above to

the proof of the following statement: (*) if x ~ 9 ,

x s L[~],

then there is an acceptable

set M such that for some

ordinal ~ ~ M, x s e [ ~ ~ M]. So let x ~ ~, x ~ L[~]. o f ~ 0 with x, ~, ~0 ~ N.
rg'

= <M',...>

C l a i m I. elements

= ~

~x

Pick ~0 with x g L 0 ~ ] °

Let ~

be a skolem expansion

(~) ~ >.

Let N be a transitive model of the structure

Let X g ~ be a set of indiscernables

I

Let

~I.

If ~ s X is the X - s u p r e m u m of ~ n X, then no element of M' lies between all of ~ ~ X

and ~.

Hence X is closed in M'.

Suppose not, and let w = t (Xl,..., xn, ~, ~I''''' where x

for ~ .

<...< x

n

< ~ < ~I <'''< ~

m

are elements

~(~) = t~(x] ..... Xn' ~' ~I .... , Bn), where

of X.

~m) lie b e t w e e n ~ ~ X and ~, For ~ ~ ~, B g X, set

~I ..... ~m are the first m elements

of X

above B. By an easy indiscernability is a regressive

argument,

function on X - ~.

that ~ is one-one on X - ~.

B n X < ~(~)

But another indiscernability

(For, let B'

contradicts C l a i m 2.

the normality

of ~, of course, w h i c h proves

= <M, e, {x}, { ~ ,

(Since x ~_~, h(x) = x of course.)

some B ~ ~ ~ M ' .

This

the claim.

Since Z e M'

By indiseernability,

some element of X above Xn, it must contain all elements

H e n c e ~' ~ . M n ~.

Since ~ is the

some final segment of X.

Z -- t (x ,..., x n) for some x <...< x in X. I I n

subset of 9.

shows

But w = t~(x ,..., x n,

note that as Z, X are in ~, Z fIX is eofinal in p

Now let h: J~' ~ ~

Hence

= t (xl, .... x n, B, B I ..... ~m ) > ~' > ~(~').)

If Z ~ M' ~ ~, then % includes

Firstly,

argument

B s X be such that B' < B.

X - s u p r e m u m of ~ q X there is 6 g X such that x n < ~ < ~. ~, ~ 1 ,.. ., a m) > ~, whence ~(B)

< B for all such B.

{~'},

Let Y = h"X.

(For let A e ~'.

of X above Xn, as claimed.

(~)~.<~>, where M is transitive. By claim i, Y is a closed unbounded T h e n A c M and so A = h(B)

By claim 2, B ~ C for some final segment C of X.

segment of Y, so as Y is closed and unbounded

since Z contains

in P, so is h(C).

for

h(C) is a final

But h(B) -~ h(C).

-227-

Hence A = h(B)

s ~.)

In fact, we have the stronger follows. Hence

result

For every S c p, S ~ M, either

that ~' = M c~ ~ •

To see this,

S E ~' or p - S s ~'

for S ~ p, S s M, S ~ ~' ~+ P - S g p' ~+ p - S g M r~ ~

S s M n ~

÷ S E p', giving

~' = M n ~.

In particular,

show that M n ~ = M Q ~, whence M will be acceptable, Well,

as ~ c p, we clearly have M n ~ ~ M ~ ~.

Then S ~ ~ ÷ S ~ ~' -> p - S e p' -~ P - S c ~ ÷ S E M G ~ -> S ~ M n ~ ,

whence M n ~ =

M ~ p.

x e e 0[~ ], so x g [L 0[~']]M = e ~[~'] Corollary

j'%'-~

÷ S ~ M n ~ .

therefore,

Thus

M n ~ E M.

We

of course. Suppose

then that S c p, S g M.

P - S e p ÷ S # ~.

Hence

Hence M is acceptable.

(since M ~ 0

f~

).

But look,

This proves

(*), as required.I

ii

@(~) rl e[~] Proof:

because

argue as

is a I~ subset

Let T 1 be the

~-theory

(i) Every member

of

6~(~).

whose

axioms

of b is a subset

are the following:

of U including

a closed unbounded

subset of

U. (2) W is a transitive ~z(z

(3)

~-U

a

^ z ~ W+

model

of

z ~ b vU-

c~0 such that a, b e W and z ~ b).

_=~o

(4) (a E L[B])W Let Tl(x)

= T

D{o

E aln e x} k) {o # al n ~ ~ - x}, each x 9 m.

By lermna i0, if x ~ 0% is a standard p which

w-model

then x ~ L[U]

of Tl(X) with U 0t = p.

is closed and unbounded

acceptable, x e L IV].

b

=

iff Tl(X)

complexity

every

subset

of

~s

Hence if T I (x) admits p, then

[, ml ).

For suppose

x _c ~.

By lemmas

7 and 8,

~, ~i ).

is recursively

as ( B[)P(T~,

~ (because

suppose

is trivial.)

p iff ( ~ ) P ( T I ( X ) ,

But since TI(X)

(To see this,

in p in the sense of ~% really is so) whence W

It is now easy to prove our corollary. Tl(x ) admits

Then b ~ ~

~ r~ W , and !emma iO applies.

The converse

admits P-

obtainable

Hence,

from T I,

this has the same logical

by lemma 9, we see that {x _~ wITl(x)

admits

-228I 0} is ~3"

But by the above, this set is just

~(~) ~ L ~ ] .

QED.

Theorem 12 There is a [3I formula ~ such that whenever x, y ~ ,

WO(x, y, ~) iff

<~, <, x, y> ~ ~. Proof: If P(T, I, ~i ), we say t embodies y i__n_n~ if t is a term of

]~', y c m,

woW(t, a, b) ~ t c o c I, and for each n s ~, n g y ~+ ,,o s t" ~ I" The importance of this notion arises from the following: Claim.

Suppose x, y ~ ~ and that P(TI(X), I, ml )"

Then WO(y, x, ~) iff there is a

term which embodies y in IIf 6L is a standard w-model with a (I, 0)-set of indiscernables (whence U~=

O) then as we saw above, W ~is acceptable, b ~[=

~ (~ W ~ , and (x s Lib]) W"

by lemma i0, WO(y, x, ~) iff [y e W ~ & ~ woW(y, x, b)]. indiscernables for ~ .

Hence

Let X be a (I, 0)-set of

We may assume, without loss of generality, that ~

= ~%~X.

Now, if y = t (xl,..., Xn), where x then ~

<...< x are elements of X, and if WO(y, x, ~), I n +W(y, x, b), so woW(t, a, b) e I, whence t embodies y in [.

Conversely, if t embodies y in ~, then for any x I <...< x n from X, y = t (xl,..., Xn), whence WO(y, x, ~).

This proves the claim.

It follows that for x, y ~ m , WO(y, x, U) iff ( 3 t e r m t of P ' ) ( ~ [ ) [P(rl(x), I, ~I) & (t embodies y in I)]-

Now, the statement "t embodies y in [" is

number-theoretlc, so the above equivalence provides the required [~ characterisation of WO(y, x, ~). [ Corollary 13 Assume V = L[~].

Then there is a A I well-ordering of ~ ) . 3

Proof: Since WO(x, y, ~) +-~ x # y 12. I

~-~WO(y, x, ~), this follows directly from theorem

Chapter 20

STRONGLY COMPACT CARDINALS AND RELATIVE CONSTRUCTIBILITY

It is not within the scope of this book to discuss the property of strong compactness.

All that we shall do is give the definition of strong compactness,

which is a generalisation

of the notion of measurability,

and show that existance of

such a cardinal has a dramatic effect on the constructibility A cardinal K is strongly compact iff it is uncountable, cardinal 4 h K, every K-complete

of the universe. regular,

and for each

filter on 4 can be extended to a R-complete

ultra-

filter on 4. (The name arises because if we restrict one obtains an obvious strengthening compactness

4 to be equal to K in the above,

of an alternative

then

definition of the weak

of K.)

Lemma i If < is strongly compact,

then K is measurable.

Proof: Let ~ = {x ~ KIIK -- X I < K}, a K-complete K-complete

ultrafilter

filter on K.

Extend D to a

~ on K.

Theorem 2 (Vopenka-Hib~cek) Assume there is a strongly compact cardinal, Proof:

Suppose V = LEA] for some set A. K-complete

Then V t L[AJ for any set A.

Set p = max(K,

filter ~ on 4 by p = {x ~ % 1 1 %

complete ultrafilter

<.

+ ITC(A) I), 4 = p •

- x I < %}.

Let ~ ~

Define a

be a K-

on %.

We shall define a class V%/~, a sort a "ultrapower"

of V.

(If V were a set,

it would in fact be the ultrapower of V by (4, ~).) For f, g g 4V, write f ~ g iff {~ ~ %If(~) = g(~)} ~ ~. an equivalence

"relation"

This clearly defines

on 4¥.

For each f s 4V, set f = {glg ~ f & rank(g) Since each f is a set, V%/~ is a well-defined V%/p by f E g iff {~ g 41f(~) ~ g(~)} e ~-

~ rank(f)}.

class.

Set V4/D = {flf e %V}.

Define a binary relation E on

Let V4/~ also denote the

vt

structure

,,

-230-

.

Then, by a very simple but tedious argument, we have, for any ~-formula

~v~/~(fI,

~n) iff

¢(v I ..... v n) . . . . .

{e e %l¢(fl(e),... , fn(e))} e p.

It follows

that if, for each x e V we define Cx: i -~ V by C (~) = x for all e e X, and define X x C: V -~ VX/~ by C(x) = Cx for each x, then CV /p(C(x!) ..... C(Xn)) iff ¢(x I ..... Xn), so C: V-< VX/U.

In particular, Vl/u ~ ZFC.

Note that although we have used standard

notation here, which we should strictly restrict for discussing sets, this is only a convenience,

and everything could be carried out formally within ZFC if required.

This will be true throughout this chapter. Claim i.

E is well-founded.

Suppose fn+l E fn' each n e ~. Un = {~ e ~[fn+l(~) ~ fn(~)} e ~. particular, U + ~.

Pick ~ s U.

Then for each n e m,

Since p is <-complete, U = ~ n <

Un E p.

In

Then fn+l(e) E fn(e) for all n s ~, which is absurd.

Hence the claim. Claim 2.

For any g ~ VX/~, {flf E g} is a set.

{~ e Xlf(~) ~ g(~)} c p.

Then

Thus, by picking f from f suitably, we may assume

f(~) ~ g(~) ~ {~} for all ~ g ~. a set.

Suppose f E g.

Thus {flf E g} ~

{flf s%[ran(g) k) {~}]}, which is

This proves claim 2. Now, E is clearly extensional,

"map"

so by claims I and 2 we can define a collapsing

~: VX/~ ~ M, where M is transitive. Let p = ~oC.

Thus p: V - < M .

Hence ~ ,< p(~) for all ~ ~ On.

Clearly, p: On-> OhM and ~ < ~ ÷ p(~) < p(B).

But M is transitive.

Hence ~ e On ÷ ~ ~ M, so

On ~ M . At this point, we break off our discussion of VX/p and define a somewhat modified "ultrapower",

(V%/~)~ of V.

In fact, the definition is exactly as before except that we take for the "domain" of (VX/p) ~ the class {flf ~ ~V & Iran(f) I < X}. for each x g V.

Note that C(x) s ( V % / ~

Let I denote the identity "map" on (VX/p)~ into VX/p.

It is easily seen that for any ~-formula iff {~ s X]~(fl(~), .... fn(~))} e p.

~(v I ..... Vn), ~(V%/~)~(~ I ..... fn )

[In fact the proof,

(by induction on 4) is

-231-

standard except for the case ~ =

~ X 9, when we know that

{~ ~ %I ~x%(x, fl(~),..., fn(~))} E ~.

But all we need to do here is observe that

Iran(fi) I < ~ for all i = I .... , n, so there are fewer than ~ n-tuples as (*-varies oVer ~, so we can define a function f: ~ -~ V so that Iran(f) I < ~ and {~ g %l~(f(~), fl(~) ..... fn(~))} ~ M, whence ~(V%/P)%(fl,..., fn ) of course.] It follows that C: V ~ (V%/u)~.

Hence I: (V%/M~-< VI/U.

(Vl/~)% is a well-founded (class) model of ZFC. model M ~.

Set pa =

~oC,

so p~: V-4 M ~.

Let

In particular,

{~ collapse it to a transitive

As before, On ~_M~o

Set ~ = [ o I ~ ~9-~, so that *: M~-< M.

The following commutative diagram sums

up the situation.

x

P

Claim 3.

(i)

p(A) = p~(A).

(ii)

P(O) = P~*(P)"

(iii) p(%) + p~(%). We shall assume claim 3 for the moment, and derive the required contradiction. Now, On ~ M ,

M~and

V = L[A], so M = LIP(A)] M = LIP(A)] and M R =

e[p~(A)j.

So

by (i) of claim 3, M = M ~. +

Again, X = p , so p(%) = [p

(p)+

]M and

p~(X)

=

[p~(p)+]M

, whence as M = M ,

(ii) of claim 3 implies that pW(%) = [p~(p)+J M e =

[p(p)+]M = p(X).

(iii) of claim 3.

First we must establish

Claim 4.

It remains to verify claim 3.

If f ~ %V and ITC(ran(f))I < %, then ~(f) = {~(f).

This contradicts

-232-

Note that

~(f) = {~(g) Ig E f} and

prove the claim by E-induction.

~(f) = { ~ ( g ) Ig E f & g E (V%/p)%~}.

Suppose g m f.

We

Then {~ e llg(e) e f(~)} g p, so we

can assume g(~) g f(~) u (~} for all ~ e %, whence ran(g) _= TC(ran(f)) k) {~}, and so in particular

ITC(ran(g))I < X, which implies g E (V%/p~.

The induction is therefore

trivial, and the claim is thus proved. Applying claim 4 to C

x

e %V we see that ifITC(x) I < %, then p(x) = p~(x).

ITC(A) I ,< p < % and TC(p) = p, so (i) and (ii) of claim 3 are immediate.

But

We are left

with the proof of (iii) of claim 3. Now, each element of p~(%) lles in MR, so is of the form f ~

%V with

Iran(f) I < 1.

But if

~(f)

~ p-~(l) =

(~(f) for some

~ (C(%)), then f E C(%), so

{~ E %If(~) E %} ~ p, and we can assume therefore that ran(f) _~ %. regular and Iran(f) I < x, we can find T < % with ran(f) ~ T. whence

Hence TC(ran(f)) c T,

IrC(ran(f))I < %. So by claim 4, ~(f) = ~ ~(f).

of p4~(l) is of the form

implies

~(i) e p(%). ~(i) =

In other words, we have shown that any element

~(f) where f E %V and Iran(f) I < %.

Let i = id ~(% × %).

above,

Since I is

Then i(~) e C%(~) for all ~ < %, so ~ E C(X), which

Assume now that p(%) = p~(%).

~(f) for some f E %V with

i ~ f, whence {~ c lli(~) = f(~)} c p. Iran(f) l = X, a contradiction.

Then

Iran(f) I < %.

~(i) E p~(%), so by the

so, as

~ is an isomorphism,

Thus {~ c %If(~) = ~} c p, which implies

Hence p(l) ~ p~(%), proving (iii) of claim 3.

QED.

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Measurable Cardinals and h 3I Well-Orderings.

Annals of Mathematics,

Vol. 94 [1971], pp 414 - 446. 21. Solovay, R.M.

The Fine Structure of L[~].

22. Solovay, R.M. & Jensen, R.B.

Handwritten notes.

Some Applications of Almost Disjoint Sets.

Math. Logic and Foundations of Set Theory (Proc. Colloq. Bar--Hillel. 23. Vaught, R.L.

in

at Jerusalem, 1968)~ed.

North Holland [1970], pp 84 - 104. A L~wenheim-Skolem Theorem for Cardinals Far Apart.

Theory of Models" (Proc. 1963 Berkeley Symposium). pp 390 - 401.

Berkeley [1972].

in "The

North Holland [1963]

-235-

GLOSSARY OF NOTATION Approximate

Symbol

P age

meaning

t

language of set theory

%

the complete satisfaction

On

the class of all ordinals

2

V

the class of all sets

3

P(x)

the power set of x

3

ordinals

4

i relation for

~L

(in general)

I

lim(~)

is a limit ordinal

4

succ(~)

is a successor ordinal

4 5

xn

the n-fold Cartesian product of X

X

a finite string of the form x0,...,

dom(R)

the domain of R

5

r a n (R)

the range of R

5

R~'Z

the restriction of R to Z

5

R"Z

ran(R~Z)

5

id

the identity

TC

transitive

V

the ~'th level in the cumulative hierarchy

6

sup

s up remum

8

xn

5

6

function

6

closure

C~

cardinals

Ixl x

Y )t < +

8

(in general)

the cardinality of x

8

{flf: x + y}

8

Ix~L

8 cardinal

to K

8

<

the successor

GCH

generalised

max

maximum member

cf(K)

the cofinality

K~

t h e weak p o w e r o f < by X

12

'f-x

the X-expansion of

14

continuum hypothesis

9

ii

of K

~.

-236-

0 x

the constant of ~ X denoting x

14

Con (T)

T is a consistent theory

14

HK

the set of all sets hereditary of cardinality less

ZF

+M

than

14

ZF minus the power set axiom

14

the relativisation of ~ to M

15

~n' lIn'~ZF, IIZFn' AZFn

17

(x) 0' (x) 1

the inverses to the pairing function

17

GRP

generalised reflection principle

21

~nM(N 1' ~M(N) Fml

AM(N) , etc.

22

the set of formulas of the considered language

Fml ~n, F m ~

22 22

In-elementary submodel

22

Def (X)

the set of all first-order definable subsets of X

25

L

the ~'th level in the constructible hierarchy

25

L

the constructible ~n~verse

25

V=L

the axiom of constructibility

29

p.r.

primitive recursive

29

rud

rudimentary

29

%R

the characteristic function of R

3O

the result of replacing x by t in

36

the canonical well-order of L

4O

the predecessors of x in a tree

44

the ~'th level of the tree T

44

the restriction of T to

44

"
(x/t)


T

UB<~T B

(K, X ) - t r e e

a tree of height K and width

44

Stt

the Souslin hypothesis

46 48

KIt

the Kurepa hypothesis

53 54 54

-237-

~,

HI 1 etc. (within Chapter 5) n' An'

62

II~IIMEp]

the interpretation of the term T in M at p

78

rud(U)

the rudimentary closure of U U{U}

8O 81 81

the ~'th level in the Jensen hierarchy

J

81 82

S <j

the canonical J-well-order of L

84

pr(x)

the <j-predecessors of x

84

h

the canonical El skolem function for <J ~, A>

88

c~,A n Pc~

the ~n projectum of

91

n P~,A

the ~n projeetum of <J~, A>

92

An

the standard ~n code for J

99

the standard parameter for A n

99

c~ n P~

104

~(~)

106

n(~)

106 the Souslin hypothesis for

sH(<)

115

0< (E)

115

[<]n

119 119

n

K ÷ (~)~ the Kurepa hypothesis for <

KH (K)

121

KH(~, ~)

122

<

122

÷

122

~K,X 122

-
Q-elementary submodel

152 153 156

Hp,A

186

[K] <~

< ÷ (~)<~ 2

186

-238-

193

F~(H) E-M set

Ehrenfeucht-Mostowski set

193

(~, e)-model

~x Defn(X)

193

the (elementary) submodel of Olgenerated by X

195

the skolem hull of X in L

201

DefA(x)

206

L[A]

206

L[A]

206 207


214

L [~] P

214

a measurable cardinal (in Chapter 14)

214

-239-

PAGE INDEX OF IMPORTANT DEFINITIONS absolute, 16

function, 5

amenable, 75

functional, 59

analytical hierarchy, 60 GCH, 9 analytical relational, 60 good parameter, 88 antichain, 44 inaccessible, 13 arithmetical hierarchy, 61 arithmetical set, 61

~l-indescribable, 119 n e-induction, 5, 6

Aronszajn tree, 45 ineffable, 120 branch, 44

inner model, 15

canonical extension, IOO

Jensen hierarchy, 81

canonical El-uniformisation 87 Kurepa tree, 53 cardinal, 8 choice, 3, 24

M-language, 21

class, 3

limit cardinal, 8

closed set, 8

limit ordinal, 4

cofinal set, 8 master code, 95 cofinality, Ii measurable cardinal, 214 collapsing lemma, 23 s-minimal, 104 ~[-complex, 171 morass, 149 constructible, 25 axiom of constructibility, 29

normal function, 13

condensation le~m~a, 34, 38

normal tree, 44

continuous function, 13

normal ultrafilter, 214

continuum, 59 ordinal, 4, 7 cumulative hierarchy, 6 parameter, 22 power, 8 domain, 5 power cardinal, 9 finite rank property, 74

primitive recursive function, 29

-240

primitive recursive closed, 33

standard codes, 99

projective relational, 60

standard parameters, 99

projeetum, 91

stationary set, 48

prolific tree, 46

strongly compact cardinal, 229 ~-structurc, 14

Ramsey cardinal, 189 successor cardinal, 8 range, 5 successor ordinal, 4 rank, 6 supremum, 8 recursion schema, 29 reflection principle, 19, 21

total relation, 5

regressive function, 48

transitive class, 4

regular cardinal, ii

transitive closure, 6

regular at ~, 104

transitive collapse, 23

In regular at ~, 104

transitivisation, 23

relation, 5

tree, 44

relational, 59

two cardinals theorem, 138

rudimentary closed, 33

type of a structure, 138

rudimentary function, 29 uniformisation, 86

Z~' 22

sequence, 6

uniformly

Shoenfield absoluteness lemma, 71

universal morass, 150

singular cardinal, Ii

~n-Skolem function, 87

weak power, 12 weakly compact cardinal, 119, 130

Souslin tree, 45 weakly inaccessible, 13 Souslin K-tree, 115