Axiom of Constructibility: A Guide for the Mathematician

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

617 Keith J. Devlin

The Axiom of Constructibility: A Guide for the Mathematician I

II

Springer-Verlag Berlin Heidelberg NewYork 1977

Author Keith J. Devlin Department of Mathematics University of Lancaster Lancaster/England

Library of Congress Cataloging in Publication Data Devli% Keith J The axiom of constructibility. (Lecture notes in mathematics ; 617) Bibliography: p. Includes index. i. Axiom of eonstruetibility. I. Title. II- Series: Lecture notes in mathematics (Berlin) ~ 617. QA3oL°8 no, 617 [QA248] 510'.8s [511'.3] 77-17119

AMS Subject Classifications (1970): 02K15, 02K25, 02K05, 04-01, 04A30, 20A10, 20K35, 54D15 ISBN 3-540-08520-3 ISBN 0-38?-08520-3

Springer-Verlag Berlin Heidelberg NewYork Springer-Verlag NewYork Heidelberg Berlin

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 1977 Printed in Germany Printing and binding: Beltz Offsetdruck, Hemsbach/Bergstr. 2140/3140-543210

PREFACE

Consider the following four theorems of pure mathematics.

The ~ h - - B a n a c h

Theorem of /~nalysis:

If F is a bounded linear functional

defined on a subspaee M of a Banach space B, there is an extension of F to a linear functional G on B such that

The Nielsen-Schreier

llG~l= llF~l .

Theorem of Group Theory:

If G is a free group and

is a subgroup of G, then H is a free group.

The Tychonoff Product Theorem of General Topology~

The product of any

family of compact t o p o l o g i c a l spaces i s compact. The Zermelo l~ell-0rdering Theorem of Set Theory:

~Very set can be well-

ordered.

The above theorems have

two

things in common. Firstly they are all

fundamental results in contemporary mathematics.

Secondly, none of them can be proved

without the aid of some powerful set theoretic~l ass~a~ption:

in this case the ~xiom

of Choice.

Now, there is nothing wrong about assuming the Axiom of Choice. But let us be sure about one thing: we are making an assumption here. ~e are saying, in effect, that when we speak of *'set theory", the Axiom of Choice is one of the basic properties of sets which we intend to use. This is a perfectly reasonable assumption to make, as most pure mathematicians would agree. Horeover (and here we are at a distinct advantage over those who first advocated the use of the Axiom of Choice), we know for sure that assuming the Axiom of Choice does not lead to a contradiction with our other (more fundamental)

assumptions about sets.

In Chapter I of this book we describe four classic open problems of mathematics, as above one from Analysis,

one fromAlgebra,

one from General Topology, and

IV

one f r o m S e t T h e o r y . S i n c e we c a l l i% s h o u l d be o b v i o u s t h a t Indeed,

it

"problems" rather

they are not quite

c a n be shown t h a t

of any of these

these

problems.

than "theorems",

however,

t h e same a s o u r f o u r s t a t e m e n t s

above.

a s s u m i n g t h e a x i o m o f c h o i c e d o e s n o t l e a d %o a s o l u t i o n

But by making a fuzf~her assumption about sets,

we a r e a b l e

t o s o l v e e a c h o f t h e s e p r o b l e m s ( a n d many more p r o b l e m s known %o be u n s o l v a b l e without

such an assumption).

This assumption

The Axiom o f C o n s t r u c t i b i l i t y axiom, closely It

is an axiom of set

b o u n d u p w i t h w h a t we mean b y " s e t " .

i s known n o t t o c o n t r a d i c t

already

i s t h e Axiom Of C o n s t r u c t i b i l i t y .

indicated,

its

t h e more b a s i c

It

theory.

implies

assumptions

assumption leads to the solution

It

is a natural

the axiom of choice.

about sets.

And a s we h a v e

o f many p r o b l e m s known t o be r

unsolvable

f r o m t h e Axiom o f C h o i c e a l o n e .

axiom is eventually situation

accepted

is net unlike

that

as a basic involving

Time a l o n e w i l l

tell

whether or not this

assumption in mathematics.

Currently,

t h e Axiom o f C h o i c e some s i x t y

years

the ago.

The axiom is being applied more and more, and what is more it tends %o decide problems in the "correct" direction. And one can provide persuasive arguments which justify the adoption of the axiom. (Again as with the Axiom of Choice in the past, there are also argaments against its adoption.) However, since the axiom is being applied in different areas of pure mathematics, it is a proposition of interest to the mathematician at large regardless of the final outcome concerning its "validity".

Until

recently

the notion

only by the mathematical considerable particular,

acquaintance the notions

logician°

of formal languages,

theory.

in areas

of set theory,

who do n o t p o s s e s s t h a t we h a v e w r i t t e n whilst

all

logic,

satisfaction,

study requires logic

model theory,

t h e a x i o m h a s become o f i n t e r e s t

of these prerequisites

this

this

any kind of in-depth

extensively

--

a in

and a good

B u t w i t h t h e g r o w i n g u s e o f t h e Axiom o f C o n s t r u c t i b i l i t y

short

it would be very nice

mathematical

Indeed,

s e t was s t u d i e d

with the ideas and methods of mathematical

deal of pure set outside

of a constructible

account.

from logic.

Our b a s i c

is for this

premise in writing

if everyone had at least

is almost certainly

It

to mathematicians

a basic

not the case.

audience

has been that,

knowledge of elementary

We t h e r e f o r e

a s s u m e no

p r i o r knowledge of m a t h e m a t i c a l l o g i c . i s d e s i g n e d so t h a t t h i s else.)

(The one e x c e p t i o n i s C h a p t e r V, b u t t h e book

c h a p t e r c a n he t o t a l l y

ignored without affecting

anything

S i n c e i t would c l e a r l y he f a r t o o g r e a t a t a s k t o d e v e l o p t h i s m a t e r i a l t o a

l e v e l a d e q u a t e f o r a n y t h i n g a p p r o a c h i n g a c o m p r e h e n s i v e trea~nen% o f e o n s t r u c t i b i l i t y , we c h o o s e i n s t e a d %o c u t some c o r n e r s and a r r i v e a t t h e r e q u i r e d d e f i n i t i o n s

very

q u i c k l y . I n o t h e r w o r d s , we p r e s e n t h e r e a d e s c r i p t i o n of s e t t h e o r y and t h e Axiom of Constructibili%y, not the theory itself. Admittedly this approach may prove annoying to logicians -- but they do not need to read this account, being well equipped to consult a more mathematical account.

The book i s d i v i d e d up a s f o l l o w s . I n C h a p t e r I we d i s c u s s some w e l l known p r o b l e m s o f p u r e m a t h e m a t i c s . S i n c e e a c h o f t h e s e p r o b l e m s i s u n s o l v a h l e on t h e b a s i s o f t h e c u r r e n t s y s t e m of s e t t h e o r y , h u t can be s o l v e d i f one a s s u m e s t h e ,ixiom of Constructibility, as i l l u s t r a t i o n s

t h e y p r o v i d e b o t h a m o t i v a t i o n f o r c o n s i d e r i n g t h e axiom, a s w e l l of i t s a p p l i c a t i o n .

I n C h a p t e r I1 we g i v e a b r i e f

t h e o r y . T h i s forms t h e b a s i s o f our d e s c r i p t i o n of c o n s t r u c t i b i l i t y C h a p t e r IV a p p l i e s t h e Axiom o f C o n s t r u c t i b i l i t y considered in Chapter I.

Chapter V is different

some knowledge o f l o g i c i s a s s u m e d .(At l e a s t , i o n a p r i o r knowledge of l o g i c i s r e q u i r e d .

in Chapter III.

in order to solve the problems from t h e r e s t

for a full

of t h i s book i n t h a t

a p p r e c i a t i o n of o u r d i s c u s s -

The r e a d e r may be a b l e t o g a i n some i d e a

o f what i s g o i n g on w i t h o u t s u c h knowledge. We c e r t a i n l y as p o s s i b l e . )

account of s e t

t r y %o k e e p t h i n g s a s s i m p l e

I n C h a p t e r V we t r y t o e x p l a i n j u s t how i t i s t h a t t h e Axiom o f

Constructibility

e n a b l e s one %o answer q u e s t i o n s of m a t h e m a t i c s of t h e k i n d c o n s i d e r e d

in the previous chapters.

I n o r d e r %o i l l u s t r a t e

o u r d e s c r i p t i o n we p r e s e n t a f u r t h e r

a p p l i c a t i o n of t h e axiom, t h i s t i m e i n Measure T h e o r y . (We t h e r e b y p r o v i d e some c o n s o l a t i o n f o r measure t h e o r i s t s in Chapter I . )

who may h a v e f e l t

The hook i s s t r u c t u r e d

left

out by our choice of problems

on t h e a s s u m p t i o n t h a t many r e a d e r s w i l l n o t

w i s h t o go i n t o t h e s u b j e c t m a t t e r of C h a p t e r V v e r y t h o r o u g h l y , i f a t a l l .

It is to be hoped that mathematicians may wish to use the Axiom of Constructihility. For this reason the proofs in Chapter lV are given in some detail, except

VI

that

i n e a c h c a s e we s t a t e

without proof a very general combinatorial

i s a c o n s e q u e n c e o f t h e Axiom o f C o n s t r u c t i b i l i t y , order to prove the desired result.

The a d v a n t a g e o f t h i s

may u s e t h e p r o o f a s a model f o r o t h e r p r o o f s , of time investigating

Finally this

and t h e n u s e t h i s

principle principle

approach is that

in

the reader

without having to spend a great

t h e Axiom o f C o n s t r u c t i b i l i t y

which

deal

itself.

a word a b o u t o u r u s e o f t h e p h r a s e " p u r e m a t h e m a t i c s " . I n ~ r i t i n g

book i t has been c o n v e n i e n t to r e s t r i c t

mathematics other than set theory". "mathematician" in our title.

A similar

the meaning of t h i s remark applies

phrase to "pure

t o o u r u s e o f t h e word

CONTENTS

Preface

Four Famous Problems

Chapter I.

1

1. A Problem i n Real A n a l y s i s

1

2. A Problem in A l g e b r a

2

3. A Problem i n G e n e r a l Topology

5

4. A Problem in Set Theory

6

~at

Chapter If.

i s S e t Theory,?

11

1. S e t Theory a s a Framework for ~athematics

11

2. Set Theory Under t h e Microscope

13

3. A Language f o r S e t Theory

14

4. The Set-Theoretic Hierarchy

18

5. The Axiom of Choice

25

The Axiom of C o n s t r u c t i b i l i t y

27

1. The C o n s t r u c t i b l e L i e r a r c h y

28

2. The Axiom of Constructibility

30

Chapter III.

3. The Generalised Continuum l~pothesis 4. H i s t o r i c a l

33 ~emarks

34

A p p l i c a t i o n s of V= L i n M a t h e m a t i c s

Chapter IV.

1. C o m b i n a t o r i a l P r i n c i p l e s

from V= L

36 36

2. The S o u s l i n IToblem

39

3. The \~itehead Problem

~

4. Collectionwise I ~ u s d o r f f Spaces

58

5.

6~

F u r t h e r Remarks

VIII Chapter V

A Problem in Measure The ery

65

I. Lktensions of Lebesgue Measure

65

2. The Measure Problem

66

3. A Theorem in Model Theory

72

4. The Condensation Lemma

76

5. Solution to the Measure Problem

78

6. ~[istorical ]lemark

80

Appendix I

~ i e m s for Set Theory,

81

Appendix II

Independence Froofs in Set, Theo,ry

86

Glossary of key Terms

92

Special Symbols

94

Suggested Further Reading

96

Chapter I

FOUR FAMOUS PROBLENS

In order

both to motivate

and to illustrate

its

of pure mathematics, and one from set solved

we g i v e h e r e a b r i e f

one f r o m a n a l y s i s ,

theory.

on t h e b a s i s

solvable

use,

the consideration

set

account

o f f o u r w e l l known p r o b l e m s

one from algebra,

These problems all

of the usual

o f t h e ~ixiom o f C o n s t r u c t i b i l i t y ,

one from ~eneral

have one thing

theoretical

topology,

i n con~non: t h e y c a n n o t b e

assumptions

(axioms),

but they are

i f we a s s u m e t h e i~xiom o f C o n s t r u c t i b i l i t y .

1. A P r o b l e m i n R e a l A n a l y s i s l 1

Let X be an infinite set, < a linear ordering of X. We may define a topology on X by taking as an open basis all intervals (a,b) = { x e X

la < x < b

} for a , b e X

with a < b. A classic theorem of Cantor says that if X has no largest member and no smallest member, and if the above topology on X is both connected and separable, then X is (considered as an ordered topological space) homeomorphic to the real line, ~, (considered as an ordered topological space). The basic idea behind the proof is to take a countable dense subset of X (by separability), prove that this set is isomorphic to the rationals, @, and then show that X must be isomorphic to the Dedekind completion of the dense subset, and hence isomorphic to ~, the Dedekind completion of Q. Use is made of the fact that the connectedness of X is equivalent to the two facts (a) that for each pair a,b of elements of X with a < b

there is a third element,c, of X with

a < c
It

is not unnatural

the best

possible.

m~d s t i l l

obtain

to ask if

Can we, f o r the conclusion

the above characterisation

instance, that

question

of the real

weaken any of the conditions

such an X will is quite

natural.

be i s o m o r p h i c

to2

Let us say that

line

on X ( a n d

is < )

? From t h i s

standpoint,

the followin~

a linearly

1. S t r i c t l y of interest sentence.)

speaking, this is not a problem of real analysis itself. But it is clearly to any real analyst. (We make no a p o l o g y f o r a n y a m b i g u i t y i n t h i s l a s t

ordered set X satisfies pairwise disjoint

the countable chain condition (c.c.c.)

open i n t e r v a l s

if every collection

of

i s c o u n t a b l e . (The r e a d e r s h o u l d n o t w o r r y a b o u t

where " c h a i n s " g e t i n t o t h e a c t .

There a r e good h i s t o r i c a l

reasons for using the

word " c h a i ~ ' h e r e , a s w e l l as some, n o t so e v e ~ ¢ h e l m i n g , m a t h e m a t i c a l r e a s o n s . ) Clearly,

if X is separable, then X will satisfy

the c.c.c.

So i t i s n o t u n r e a s o n a b l e

t o pose t h e f o l l o w i n g q u e s t i o n :

Let X be an infinite set, < a linear ordering of X. Suppose that under the ordering<, X has no largest member and no snmllest member. Regard X as a topological space as above. If X is connected and satisfies the c.c.c., does it follow that

X~ ~ ?

This q u e s t i o n ~ r a s f i r s t

r a i s e d by M. S o u s t i n i n 1920. i t

a s The S o u s l i n P r o b l e m . C u r i o u s l y e n o u ~ , very simple to pose, it resisted y e a r s . Of c o u r s e ( ? ) ,

soon became known

a l t h o u g h t h e q u e s t i o n i s so b a s i c and so

numerous a t t e m p t s a t s o l u t i o n o v e r t h e n e x t f o r t y

i n v i e w o f t h e i m p o r t a n t r o l e p l a y e d by t h e f a c t t h a t t h e

r e a l s have a c o u n t a b l e d e n s e s u b s e t , one would e x p e c t t h e S o u s l i n Problem t o have a n e g a t i v e a n s w e r . But no c o u n t e r e x a m p l e was f o r t h c o m i n g . We s h a l l see why i n t h e ensuing chapters. 2. A i~r0blem i n A l g e b r a

We c o n s i d e r now a famous p r o b l e m o f group t h e o r y . As a f i r s t establish

step,

l e t us

t h e c o n v e n t i o n t h a t "group" w i l l always mean " a b e l i a n g r o u p " .

Let G, A, B be groups. We say that G is an extension of A b y B subgroup of G (written

A~G)

iff A is a

and G/A ~ B. (Thus B describes, in a sense, the

m~nuer in which G extends A.)

Given groups G and A with G an extension of A, there is a unique (up to isomorphism) group B such that G is an extension of A by B: namely the group G/A. The extension problem (for abelian groups) asks the followin~ converse question. Given groups A and B, determine the extensions of A by B. ~ e r e

is always at least

one such: namely, the direct sum A @ B . Dut there may be more than one. For instance, let Z be the group of integers, 2Z the subgroup of the even integers, and 2 the unique group of order 2. Now, Z/2Z ~ 2, so Z is an extension of 2Z by 2. But ~ ~ 2~ • 2 since Z is torsion free and 2Z • ~ has torsion. The following solution to the extension problem is due to Baer.

L e t G, G' be e x t e n s i o n s o f A by B. Thus G/A ~ B, G'/A ~ B. L e t I

G

~B

and

G

I

~ ~B

3

be t h e

canonical

projections.

We w r i t e

G

G ~ G~ i f f

there

is an isomorphism

V ~ G'

which makes the following diagram co~mute:

G _ ~ _

G'

(As usual, 1A denotes the identity morphism on A into whatever extension of A is being considered.)

It

is

easily

seen that

~

is an equivalence

relation

on t h e

set

of all

J

extensions clearly

the

o f A b y B. The r e l a t i o n correct

notion

G ~ G

is

stronger

o f " s a m e n e s s " when we a r e

than

isomorphism,

considering

and is

extensions

of A

b y B.

R. Baer proved, in 1949, that the set of equivalence classes of extensions of A by B under the above equivalence relation itself forms a group. We denote this group by ]kt(B,A). It is the ~roup of extensions o f A b ~ B .

It is outside the scope

of this book to describe the group operation involved in ~t(B,A), and indeed we shall not need to know it. ~ a t

we do need to know is that the identity element of

the group Ext(B,A) is the equivalence class of the direct sum A ~ B .

We should also mention that as a result of work by Schreier it is possible to give a description of the members of ,~xt(B,A) in terms of A and B.

Let us recall now that a group G will be free iff there is a set ~gili ~ I} of elements of G such that every non-zero element of G has a unique representation of the form

nlgil +

. • . +

nkgik

'

where il,...,i k are distinct members of I and nl,...,nk are non-zero integers.

We then s a y { g i I i ¢ I}

is a basis for G. A basis is thus the same as a linearly

independent generating set. We relate the notion of a free group to the extension

problem as follows.

2.1 Theorem Let H < G .

Proof: Let

If G/H is free, then G = II ~ N for some N < G ,

{H + k lkeE}

of G generated b y E .

N ~ G/H .

be a basis for (]/H , where }(zG. Let N =G,

the subgroup

It is easily checked that G = I{ @ N . o

2.2 Corollary. If B is free, then }~t(B,A) = O.

Proof: Let A ~ G ,

G/A ~ B. Since G/A is free, 2.1 gives G ~ A @ B . (~ore precisely,

the proof of 2.1 shows that G ~ A

The a b o v e r e s u l t the following

standard

@ B in the sense defined above.)

has a converse.

:in o r d e r

to obtain

the converse,

~e recall

theorem.

2.3 Theorem ( N i e l s e n - S c h r e i e r ) If G is free and H ~ G ,

L e t B be a g i v e n g r o u p .

I:roof:

L e t F be t h e f r e e

by the set

then H is free. m

If

(~xt(B,A) = 0 f o r

group on B (i.e.

all

g r o u p s A, t h e n B i s

the unique group which is freely

free.

generated

B). Let F

~

,B

extend the identity f~mctien on B. Set A = Ker(~). so by hypothesis,

Then 2 is an extension of A by B,

F ~ A ~ B . Hence there is an embeddin~

B

So b y 2.3, B i s f r e e .

)F.

Now , b y 2 . 2 , i f G i s f r e e , 1951, i f t h i s

then ~t(G,Z)

s t a t e m e n t h a s a v a l i d c o n v e r s e . I n o t h e r w o r d s , does t h e p r o p e r t y t h a t

Ext(G,Z) = 0 c h a r a c t e r i s e

t h e f r e e g r o u p s G ? D e f i n i n g a W-group t o be any ~roup G

f o r which Ext(G,Z) = 0, t h i s r e d u c e s t , recently,

= O. J . H. C. ~ l i t e h e a d a s k e d , i n

the only result

showing t h a t e v e r y W - g r o u p i s f r e e . U n t i l

o f n o t e on t h i s problem was t h e f o l l o w i n g , p r o v e d i n 1951:

2 . 5 Theorem ( S t e i n ) 9 ~ e r y c o u n t a b l e W-group i s f r e e .

We shall return to the %~hitehead Problem in later chapters.

3. A Pro b ! e m ! n

G e n e r a l Topolog7

L e t (X,~) be a t o p o l o g i c a l s p a c e . The f o l l o w i n g s e p a r a t i o n p r o p e r t i e s which X may satisfy are well kno~rn. X is T O if, whenever x , y e X and x ~ y there is an open set U which contains exactly one of x, y. X is T 1 if, whenever x , y ~ X and x ~ y there are open sets U, V such that xeU, y~U

and x C V , y e V .

X is T 2 (Hausdorff) if, whenever x,y e X and x ~ y there are disjoint open sets U, V such that x ~ U

and y e V.

X is re~galar if, whenever x e X and A = X open sets U, V such that x ~ U

and A ~ V .

is closed and x @ A ,

there are disjoint

X is T 3 if it is regular and T I.

X is normal if, whenever A, B are disjoint closed subsets of X there are disjoint open sets U, V such that A ~ U

and B ~ V . X is T~ if it is normal and T I.

It i s immediate t h a t T ~ - - * T 3 - - ~ T 2 - - ~ T 1 - - * T o . None o f t h e s e i m p l i c a t i o n s i s

reversible.

The f o l l o w i n ~ g e n e r a l i s a t i o n

of the Hausdorff property occurs in the literat-

u r e . We s a y a s u b s e t Y of X i s d i s c r e t e

i f e v e r y p o i n t of X has a n e i g h b o r h o o d which

c o n t a i n s a t most one p o i n t o f Y. (Thus i f X i s T1, a s u b s e t Y o f X w i l l be d i s c r e t e i f f Y h a s no l i m i t p o i n t s i n X, w h i c h s a y s t h a t t h e e l e m e n t s o f Y a r e s p a c e d w e l l a p a r t from each o t h e r ° ) A s p a c e i s c o l l e c t i o n w i s e I t a u s d o r f f i f , whenever Y i s a discrete yoU

Y

set,

t h e r e i s a f a m i l y { U ~y e Y } of p a i r w i s e d i s j o i n t

for all yeY.

(~¢e c a l l

open s e t s s u c h t h a t

such a f a m i l y a s e p a r a t i o n o f Y.)

Now, i t i s n o t h a r d t o show t h a t t h e t t a u s d o r f f p r o p e r t y does n o t i n g e n e r a l imply t h e c o l l e c t i o n w i s e ~ u s d o r f f

p r o p e r t y . I n d e e d , t h e r e a r e T~ s p a c e s w h i c h a r e

n o t e o l l e c t i o n w i s e I I a u s d o r f f . So we a s k what e x t r a c o n d i t i o n s on a s p a c e a r e r e q u i r e d i n o r d e r t o y i e l d t h e c o n c l u s i o n t h a t i t be c o l l e c t i o n w i s e t l a u s d o r f f ? C o n s i d e r a t i o n s o u t s i d e o f our p r e s e n t s c o p e l e a d t o t h e f o l l o w i n g p r e c i s e q u e s t i o n (which i s f a i r l y satisfy

the first

close to the best possible). axiom o f c o u n t a b i l i t y

of x has a countable b a s i s .

mentioning that this question first G e n e r a l Topology

--

i f f o r each p o i n t x the neighborhood system

The q u e s t i o n now i s :

c o l l e c t i o n w i s e t t a u s d o r f f 7 ~te i n v e s t i g a t e

of

Recall t h a t a space i s s a i d to

the no~l

Is every first

c o u n t a b l e T~ s p a c e

t h i s p r o b l e m i n I V . ~ . L e t us f i n i s h by

a r o s e o u t o f r e s e a r c h on a v e r y famous p r o b l e m Moore s p a c e problem (which i s s t i l l

p r o b l e m d e a l s w i t h t h e m e t r i z a t i o n problem ( i . e .

o p e n ) . This

which t o p o l o g i c a l spaces are m e t r i c

s p a c e s ? ) . Roughly s p e a k i n g , what t h e normal ~!oore s p a c e problem a s k s i s w h e t h e r every first

c o u n t a b l e T~ s p a c e i s c o l l e c t i o n w i s e n o r m a l .

~. A Problem i n S e t T h e o ~

The o l d e s t o f our f o u r p r o b l e m s - - t h e continuum p r o b l e m - - d a t e s back t o C a n t o r . The q u e s t i o n r a i s e d h e r e i s : How many r e a l numbers a r e t h e r e ? I n o r d e r t o make t h i s p r e c i s e we r e q u i r e some e l e m e n t a r y n o t i o n s from s e t t h e o r y .

~ n d a m e n t a l i n m a t h e m a t i c s i s t h e n o t i o n o f c o u n t i n g . !md i t t h a t our r e a d e r i s f a m i l i a r w i t h ( a t l e a s t

i s t o be e x p e c t e d

some o f ! ) t h e n a t u r a l nvmbers 0 , 1 , 2 , . . . .

Using t h e n a t u r a l n v ~ b e r s we may " c o u n t " t h e "number" o f e l e m e n t s i n any f i n i t e But what a b o u t i n f i n i t e

sets ? ~ell,

set.

why n o t e x t e n d t h e n a t u r a l number s y s t e m i n t o t h e

transfinite

? ~y

not indeed!

which commences with of any set.

By d o i n g t h i s

the natural

~Waat a r e t h e o r d i n a l

the question:

numbers,

we o b t a i n

the ordinal

and which is adequate

n u m b e r s ? We a n s w e r t h i s

What a r e t h e n a t u r a l

number system, to "count"

question

the elements

by first

answering

numbers ?

The number 0 we define to be the empty set, ~. The number 1 we define to be the set {0 } (i.e. the set with precisely one element, that element being the natural number 0). The number 2 we define to be the set [0,i} . Proceeding inductively, we define the number n+l to be the set {0,i,

. . . ,n } . Notice that the number n is

always a set with exactly n elements, those elements being precisely the numbers smaller than n. To obtain the ordinal numbers we continue the definition into the transfinite. The first infinite ordinal, denoted by ~ , {0,i,

. . . ,n, . . . . . . . . .

of all natural numbers. The second infinite ordinal,

} ~+i,

0,I, . . . ,n, . . . . . . . . . In general, the next ordinal number after ~ w i l l

is the set

is the set

,~ } .

he the set ~ ~ ~ }

. And when we

have defined the sequence of ordinals

this sequence having no last member, the "next" ordinal number will be the set <0,i,

. . . ,n, . . . ,~,~+I, . . . ,~, . . .

of all ordinals constructed so far.

In general that

we u s e l o w e r c a s e G r e e k l e t t e r s

by our definition

of ordinal

number, if

(i.e. be smaller than ~ ), written

~reover,

~ , ~ are ordinal

~ < ~

ordinal numbers are totally ordered by ~

to denote ordinal

, just in case

numbers. Notice

m~abers, ~ will ~ ~ ~

precede

. Thus the

. Indeed they are well-ordered by

~

.

regarded as the set of all smaller ordinal numbers, each ordinal number is

itself well-ordered by E .

Now, e a c h o r d i n a l namely

~ . It

number has associated

can be shown that

with

every well-ordered

it set

a canonical

well-ordering

P can he put into

an order-

:

preserving, written

one-one correspondence with a unique ordinal,

otp(P).

In this

way t h e o r d i n a l

elements in any well-ordered

set

(otp(P)

called

the order-type

o f P,

n u m b e r s c a n he u s e d t o ' b o u n t " t h e n u m b e r o f being the answer for

the well-ordered

set P).

But by Z e r m e l o ' s W e l l - 0 r d e r i n g

Theorem ( w h i c h i s a c o n s e q u e n c e o f t h e Axiom o f C h o i c e ) ,

every set

Hence we may u s e t h e o r d i n a l

c a n be w e l l - o r d e r e d .

elements of any set.

The p r o b l e m h e r e i s t h a t

upon the well-ordering fact

that

it

answer will Different

the result

of our counting depends

c h o s e n . Now i n t h e c a s e o f f i n i t e

s e t s we a r e u s e d t o t h e

does not matter

i n w h i c h o r d e r we c o u n t t h e e l e m e n t s o f t h a t

a l w a y s be t h e s a m e . B u t f o r

well-orderings

o f t h e same s e t

of counting the elements of that

set.

this

+~

well-ordering

when t h a t

numbers for size

. . . . . .

this

this set,~o~has

; the

i s no l o n g e r t h e c a s e .

can l e a d to d i f f e r e n t

the

,I~3,5,

order-type

, which is the second ordinal

ordinals

sets

set

results

to the process

the set,~,

order-type

~

of natural . B u t we c a n

the set ~ as follows: ~0,2~,

Under

infinite

For example, consider

numbers. Under the u s u a l w e l l - o r d e r i n g , also well-order

numbers to "count" the

of the

constructed

s e t h a s no l a r g e s t

"counting"

of well-ordered

infinite

sets.

. . . . . . same

~

set

~

by taking

• is the

ordinal

the set of all

number

previous

member. T h u s , a l t h o u g h we c a n u s e t h e o r d i n a l

sets,

they are really

Fortunately,

only suited

however, using the ordinal

for measuring the n u m b e r s we may

o b t a i n a number s y s t e m which i s a b l e to " c o u n t " the elements of an a r b i t r a r y

set.

We make u s e o f t h e f a c t

by

that

the ordinal

G i v e n a s e t X, we d e f i n e w h i c h may be p u t i n t o of X is the least usually

be a c a r d i n a l

of all

~.

it

~+l,

ordinal

~+2,

definition

IXI f o r that

.

number

the cardinality

o f X. The c a r d i n a l i t y

number which e q u a l s

e

of X is

some s e t X i s c a l l e d

an ordinal

number ~ w i l l

c a n n o t be p u t i n t o o n e - o n e c o r r e s p o n d e n c e w i t h a n y

number. Clearly,

But

t o be t h e l e a s t

well-orderings

is immediate from this

number i f f

ordinal

So t o o i s into

order-type

munber. It

cardinality

o n e - o n e c o r r e s p o n d e n c e w i t h X. ~ q u i v a l e n t l y ,

d e n o t e d b y I X I . Any o r d i n a l

a cardinal

smaller

its

numbers are themselves well-ordered

etc.

0,1,2,

. . . ,n,

are not cardinal

o n e - o n e c o r r e s p o n d e n c e w i t h ~ . The f i r s t

• . • are all

cardinal

numbers, since

cardinal

after

m~nbers.

e a c h may be p u t

¢o i s d e n o t e d b y

~1'

the next by w 2 ' and so on. Thus for any ordinal m,nber ~ , cardinal after ~

. We usually reserve

~,k, ~

~

denotes the ~

th

to denote cardinals. Also, since we

sometimes have occasion to consider a cardinal number both as a cardinal and as an ordinal, it is convenient to have a notation which indicates the usage. On occasions where the distinction is important we use the Hebrew letter "aleph", with o~

considered as a cardinal. (We write

~ 0 instead of

S~ denoting

co .)

With t h e c a r d i n a l numbers we now have a good s y s t e m f o r m e a s u r i n g t h e s i z e o f a s e t X. L~ery s e t X w i l l have a s s o c i a t e d w i t h i t a unique c a r d i n a l number, i t s cardinality. most

A s e t X w i l l be c o u n t a b l e i f f

~0 )°

numbers, ~ , such that

Now, a c l a s s i c a l

result

i s u n c o u n t a b l e . Hence ~ 1 I~l =

~

cardinality

of Cantor t e l l s

i s a t most

~

(i.e.

at

us t h a t t h e s e t o f a l l r e a l

~ ~1" We know t h a t t h e r e i s a unique o r d i n a l

. So what is this ~

here was I, i.e. that II~I= ~I"

its

? Cantor himself conjectured that the

But the problem resisted all attempts at solution

over the years: it is known as the continuum problem.

I t i s p o s s i b l e t o r e p h r a s e t h e continuum p r o b l e m s l i g h t l y . then

~ =

{ ~ 1 ~ < K ] , so we may c o n s i d e r t h e s e t of a l l

denoting this

s e t by

2 . We d e f i n e t h e c a r d i n a l 2

we d e f i n e c a r d i n a l e x p o n e n t i a t i e n by s e t t i n g representations,

we s e e t h a t ~ 1

which s o l v e s t h e e q u a t i o n 2~

=

~'

2~

=

~ is a cardinal,

f u n c t i o n s from

K i n t o 2,

I 2 t . (Mere g e n e r a l l y ,

K ~ . ) By c o n s i d e r i n g b i n a r y

= 2S° . So t h e continuum p r o b l e m a s k s f o r t h e

= ~

. The continuum h y p o t h e s i s ' CH, i s t h e a s s e r t i o n

~1 ° The g e n e r a l i s e d continuum h y p o t h e s i s , GCII, i s t h e s t a t e m e n t t h a t f o r a l l = ~+I

" Denoting the next cardinal after ~ by

the alternative form

that

2So

K

t o be

If

(~K)(2 K =

K ranges over all

infinite

We s h a l l r e t u r n l a t e r

~+, we may write the GCH in

K+), it being understood from the notation here

cardinals.

t o t h e continuum p r o b l e m . I n t h e meantime, w h i l s t we

a r e c o n s i d e r i n g o r d i n a l numbers, l e t us i n t r o d u c e one o r two f u r t h e r n o t i o n s which we s h a l l r e q u i r e . If ~

Firstly,

n o t e t h a t our o r d i n a l numbers f a l l

is an ordinal number, then

~ =

{0,1,2,

i n t o two c a t e g o r i e s .

. . . ,~, . . .}

= ~Ti~<

~}.

10

Now, either i% is the case that ~ has a largest member, =

fO,l,2,

.

.

.

,V,

•

.

.

,~}

,

or else ~ has no largest member, = {0,1,2,

In the former c a s e , ~

. . . ,~,

. . . }

.

is the first ordinal number after

~ (i.e. ~ =

~+i ), and we

say ~ is a successor ordinal. In the latter case, for every ordinal ~ < ~ an ordinal strictly between ~ and ~

(in particular,

there is

~+l is such an ordinal), and

we say ~ is a limit ordinal. For example, all positive integers are successor ordinals, whilst ~ is a limit ordinal. It is easily seen that, in fact, every infinite cardinal number is a limit ordinal.

For readers observations. into

totally

The o r d i n a l

the transfinite.

ordinal,

numbers are an intuitive

ordinal,

results

The p o i n t

takes place.

We t o u c h o n t h i s

just

painted

, being points

a digression seems b e t t e r

as with the natural

ordinals

to deal with.

means o f t h e f o l l o w i n g

a limit

ordinal.

difference

in turn,

t h a n no p i c t u r e

ordinals

o f much g r e a t e r

At a n y r a t e ,

unimportant.)

have

magnitude

we h a v e

at all.

and p r o o f s by i n d u c t i o n

(induction

details

on t h e o r d i n a l

numbers

we now h a v e t h e new c a s e o f l i m i t

on

~

of ordinal

for arbitrary

here,

but the very definition

Further

and

it would constitute

the picture

F o r e x a m p l e , we may d e f i n e a d d i t i o n

'next ordinal',

last

w h i c h do n o t a t a l l

but unfortunately

here.

numbers, except that

induction

numbers, then another

b e i n g a t t h e same t i m e b o t h a c c u r a t e

in IV.I,

t o be p r e c i s e

numbers

a copy of the natural

the presence of this

at which a collection

point

u s make t h e f o l l o w i n g of the natural

copy of t h e n a t u r a l

( T h e r e i s some a m b i g u i t y o f n o t a t i o n

is used to denote this

continuation

be many l i m i t

We may c a r r y o u t d e f i n i t i o n s just

numbers, let

Unfortunately,

picture

there will

of

too great

~

is,

in this

the characer

far

then another

a n d s o o n , ad i n f i n i t u m .

phrase ad infinitum misleading.

with ordinal

As we p r o c e e d u p w a r d s we m e e t ,

numbers, then a limit limit

unfamiliar

Y< @ since

n u m b e r s by

~ ):

t h e s y m b o l ' +1 '

of ordinal

addition

makes

c a n be f o u n d i n a n y e l e m e n t a r y t e x t .

Chapter II

%a~AT IS SET TI~ORY ?

1. S e t T h e o r y a s a Framework f o r ~ ! a t h e m a t i c s

The u s e o f e l e m e n t a r y s e t prevalent

in all

theoretical

branches of pure mathematics,

notions

and r e a s o n i n g

is,

a n d n e e d s no e l a b o r a t i o n

of course, here.

But

for the most part, scant attention is paid to the set theory itself. For instance, although a group is defined to be a set together with an operation (i.e. function) defined on that set, the precise nature of that set is never considered. This is, of course, because it is totally irrelevant (in group theory) what the set actually consisSs oft what matters is the hehaviour of the ~roup operation defined on the set. In other words, the structure is what counts, not the material used to support that structure. By and large the same is true for any part of algebra. In analysis the situation is a little different, since it is not irrelevant to ask the question: "What are the real numbers?" It is really a matter of opinion as to whether the construction of the real line belongs t o the field of analysis, of course, since a quite common (and entirely reasonable) approach is to take the real number system as basic. Another common approach is to take as basic, say, the integers, and then to construct the rationals, and thence the real numbers as Dedekind cuts. Or, taking this process one step further we may commence with just the natural numbers. But this is about as far as any analyst would bother to go. But there is no real reason to stop here. As we saw in I.~, the natural numbers can themselves be defined as certain sets. Indeed, in order to construct the natural numbers we need only make one basic existence assumption: namely that nothin~ exists~ So, we see that it is possible to construct the entire real number system starting with nothing. The assumption that the natural numbers exist has been dispensed with. Thus, not only does set theory provide us with a useful framework for discussing structures, be they algebraic or analytic, it also answers any relevant questions of existence (modulo the assumption that nethin~ exists>.

12 'i~ne a b o v e d i s c u s s i o n may c o n s t r u c t will

even the natural

doubtless

the natural took this

strike

the reader

as a little

natural

were precisely

numbers. But wait.

assumption was that operation:

there

~ilst

it

reduce

problems to operations

collection

of the empty set, now) f o r

instance,

own r i g h t .

of sets

is

itself

we may c o n s t r u c t

existence.

~at

of a function

ordered

pair

as a set

o f two o b j e c t s

Well,

verified

is perfectly ically°

that

acceptable.)

H e n c e we o b t a i n

of generality

standpoint

inductively

by (xl,o..,Xn)

single

and an n-dry

set,

which are n-tuples.) other

concepts us

with

various

operations

pairs

and notions

and

to define

2 as

the only raw existence ourselves that

theory

entire

is all

G i v e n t h e one b a s i c

collect-

about.

We

operation

that

only from the existence

required

other

one b a s i c

in mathematics°

concepts

So much

of mathematics

(or interpretation,

? For

if you

i s w e l l k n o w n . And we may d e f i n e

the

~x~,~x,y~}.

as particular

the functions

is

the function within

set

of the sets

operation

the n-tuple

concept,

set-theoret-

are unary!

Indeed,

(Xl,...,Xn)

is

Thus, not only does set also

from a

more than a

h o w e v e r , we may d e f i n e

it

loss

defined

which is only defined

we r e q u i r e ,

w h i c h we n e e d .

definition

we may w i t h o u t

, and is thus nothing

a function

theory.

so this

may be defined

(In fact,

of mathematics

because

just

x = x' & y = y',

sets.

= ((xl,...,Xn_l),Xn)

the existence

nothing),

x and y by

they are,

Besides

that

we d i d a l l o w

the definition

pair

function

(i.e.

we m a y r e g a r d

of the sets

Thus the ordered

all

true

and starting

iff

of mathematics

provide

on s e t s °

= (x',y')

functions

assume that

set-theoretical

(x,y)

empty set

This is what set

of ordered

(x,y) = is easily

(sets),

a set,

all

(sets)

of nothin~

how d i d we d e f i n e

we w e r e t h e n a b l e

certainly

about the various

what about flmctions?

prefer)

is

of objects

(set)

any given

in its

Well,

we

the number 1 the set whose only

i s a s e t w i t h no e l e m e n t s ,

given any collection

existence

The i d e a t h a t

0 a n d 1. And s o o n . I n t h e e n d we o b t a i n e d

ion as an object

(It

of the

O. H a v i n g 0 a n d 1 a v a i l a b l e ,

the set whose elements

(for

suspicious.

t o be t h e n u m b e r 0 . We t h e n t o o k f o r set

some e l a b o r a t i o n .

numbers by assuming only the existence

n . ~ n b e r s ~ We a s s u m e d t h e e x i s t e n c e

member was t h i s

all

would seem to require

provides

on s e t s all

the

theory

us with

the

13

To f i n i s h

this

section,

let

us point out that

intended to convey anything particularly k i n d of l e v e l

it

is possible

ment. But, hopefully, us with a unified,

to raise

o u r above d i s c u s s i o n

is not

deep o r p r o f o u n d . I n d e e d , i f t a k e n on t h a t

all

sorts

of q u e s t i o n s c o n c e r n i n g our d e v e l o p -

we managed t o c o n v e y t h e g e n e r a l i d e a o f how s e t t h e o r y p r o v i d e s

f u n d a m e n t a l f r a m e w o r k f o r m a t h e m a t i c s , so t h a t a n y b r a n c h o f ( p u r e )

m a t h e m a t i c s c a n be r e g a r d e d a s s e t t h e o r y r e s t r i c t e d

to particular

overenthusiastic

s t a t e m e n t , maybe we s h o u l d s a y

that

r e a d e r be c a r r i e d

in most instances

of mathematics in this

there

away b y t h i s

is absolutely

way, so i t w i l l

m a t h e m a t i c s t h a n books on s e t t h e o r y . to all ical

c o n c e r n e d when a p r o b l e m i n ,

approach for its

last

no p o i n t i n a c t u a l l y

still

sets.

(Lest the

regarding a branch

be n e c e s s a r y t o r e a d o t h e r books i n

I n d e e d , i t v e r y o f t e n comes a s a g r e a t

say, algebra,

actually

suprise

does r e q u i r e a s e t - t h e o r e t -

s~ution.)

2. S e t Theory. Under t h e }~icroscope

H a v i n g s e e n how s e t t h e o r y p r o v i d e s u s w i t h a ~ m i f i e d framework f o r a l l pure mathematics,

of

it is now high time that we examine this rather powerful set

concept. Since everythinz in mathematics is now a set, just what is a set ? ~e have said that, besides assuming the existence of the empty set, there is only one basic notion in set theory: The ability to regard any collection of objects as an object in its own right. O.K., fine. But what determines a collection ? In the case of finite collections there is (in theory) no problem: we may determine a finite collection by listing precisely what the elements of the set are to be. But in the case of arbitrary collections this is no longer possible. The only possibility is %0 give some property which characterises the elements of the collection concerned. Thus, if P is some property, the "collection" of all those x for which P(x) holds is a well-defined collection. As usual, we shall denote by

{ x l r(x) } the collection of all x for which P(x) is true.

14

But the above discussion has only postponed the day of reckoning. No sooner have we defined what we mean by "collection" than we are faced with the question: ~hat is a property ? A simple answer would be to allow any "property" which can be described in the ] ~ l i s h language. But this is much too vague, since it will include such "collections" as the set of all bicycles or the set of all mathematicians , neither of which should be included in a mathematical theory of sets. Indeed, our definition should only include collections of genuine mathenmtical objects. On the other hand, we do not want to exclude any bona fide mathematical collections. So, a not unreasonable approach would be to define a language which will describe those and only those collections we require for mathematics. The reader who has not seen anything of this nature before may well be suprised to discover that this is indeed possible. He will be even more suprised at the simplicity of the language.

3- A Language for Set T h e o ~

The notion of a formal la ~ua e is familiar to all computer users. Programming languages are such. That is, they have a fixed (and highly restricted) vocabulary, and fixed (and very rigid) rules of grammar. Nevertheless, despite the apparent

inflexibility

o f a progranuning l a n g u a g e ,

for expressing

highly

complex procedures.

it

i s w e l l known t h a t

So i t

it

is usually

is with the language of set

we d e f i n e b e l o w .

We commence by listing the basic symbols (words) of our language.

Variables

:

Vn

(n = 0,i,2,...)

Logical connectives

Quantifier

Brackets

:

symbols :

:

(

,

)

A

~

~

,

;

V

3

,

;

--t

;

~ --~

,

~

}

ade~late theory,

15

Set-theoretic

symbols :

The v a r i a b l e s sets.

E

a r e t o be u s e d t o d e n o t e t h e o b j e c t s

S i n c e we do n o t w i s h t o p l a c e a n y p r i o r

may a t a n y one t i m e d i s c u s s , e v e n t h o u g h we s h a l l

The l o g i c a l

connectives

selves

denotes

iff.

rather

on t h e number o f s e t s we

an infinite

collection

of variables,

a b l e to use them a l l .

have the following meanings:

variety),

i.e.

-1

denotes not,

^ d e n o t e s and ,

--, d e n o t e s i m p l i e s

v , and

Had we w i s h e d , we c o u l d o f c o u r s e h a v e u s e d t h e s e w o r d s t h e m -

than introduce

s y m b o l s f o r t h e m , and t h i s

w i t h p r o g r a r a m i n g l a n g u a g e s much c l e a r e r , the usual practice

of our discussion,

restriction

we a l l o w o u r s e l v e s

n e v e r be ( t h e o r e t i c a l l y )

d e n o t e s o__rr ( t h e n o n - e x c l u s i v e ----

=

,

w o u l d h a v e made t h e a n a l o g y

h u t on t h e w h o l e i t

and u s e t h e s y m b o l s . ( F o r one t h i n g ,

seems b e t t e r

sentences

to adopt

are actually

easier

t o " r e a d " when t h e s y m b o l s a r e u s e d . )

The q u a n t i f i e r exists

.

.

.

such

that

The b r a c k e t s a "clause"

symbols read:

for

V

read for all

, and f o r

~

read there

.

serve as punctuation

symbols, denoting

the start

and f i n i s h

of

or "sentence".

Finally

we h a v e t h e two s y m b o l s w h i c h make t h e l a n g u a g e a s e t - t h e o r e t i c

l a n g u a g e . We w a n t o u r l a n g u a g e t o be a d e q u a t e f o r d e s c r i b i n g Taking it

a s known t h a t

see that

what is required

concept of set

theory.

apparatus"

already

expressing

just

we may d e s c r i b e is that

Suprisingly

described,

all

any mathematical

any such concept in set-theoretic

olur l a n g u a g e i s s t r o n g enough, we r e q u i r e

it

suffices is that

t e r m s , we

enough to describe

that,

besides

concept.

any

the "standard

o u r l a n g u a g e be c a p a b l e o f

two f u n d a m e n t a l n o t i o n s : x is an element of y and

We i n t r o d u c e

t h e symbol

x is equal to y. ~ t o d e n o t e i s an e l e m e n t o f and = t o d e n o t e i s e q u a l t o .

16

That then is our language.

It is called the language of set t h e o ~ ,

for short. Despite its rudimentary nature,

or LST

it is powerful enough to express all the

concepts of mathematics which we use. Before we can indicate how this may be "proved", we must of course say what the rules of grammar are to be.

As b a s i c

formulas

(or basic (v nev

We t h e n b u i l d

clauses) )

,

up the formulas

we a l l o w a l l

expressions

of the form

( v n = Vm)

(i.e.

the clauses

and sentences)

by applying

the following rules: if

T

, T

are formulas,

(~)

so are

, (~-,)

, (~),

(~v)

,

(~) if T

is a formula,

then so are

(~Vn~) , (3Vn~)

The meanings of the above rules should be quite clear in view of our previous discussion.

Now, t h e r e a d e r believing

that

mathematics. concepts,

tbis

language

Nevertheless,

unreadable.

B u t we a r e n o t it

is

give a proof

concepts

notion

as follows.

computing,

let

in order

of expressing

all

Of c o u r s e ,

for all

true.

as a typical

since,

we u s e i n m a t h e m a t i c s We f i r s t

concepts

this

to successively we t h e n c e

treat

be i n c r e d i b l y

complex, unreadable.)

i n LST. What i s

important

on t h e o n e h a n d t h e c o l l e c t i o n

of all

defined

collection, undefined

software.

by saying

that

whereas the collectcollection.

For readers

as a regular

So w h a t unfamiliar

w h a t we do i s u s e o u r

more and more complex notions. it

of

cannot

is a vague,

point

define

but the most basic

~ f e l l , we c l e a r l y

develop an extensive

us elaborate

we n e e d i n

computer program is virtually

7md how do we p r o v e t h i s ~ sense,

have trouble

the concepts

the concept will

i n LST i s a p r e c i s e l y

has been defined,

kind of thing may well

in a c t u a l l y ~

in any rigorous

ion of all

language

is

always possible.

expressible

with

is capable

this

(Just

interested

properties

we do i s

with this

t h e f o r m u l a o f LST w h i c h d e f i n e s

and totally

is that

unfamiliar

part

Once a new

of the language.

In

17

a remarkably ics

speedy manner,

this

process

leads

us to the usual

"language"

of mathemat-

(by which we mean the "la~raage"which mathematicians use when writing up proofs).

At which point we can safely say that our task has been accomplished.

In order described set

h

starts

theory.

concepts

to illustrate off.

L e t u s e x p a n d LST t o i n c l u d e

let

define

we j u s t

notions

o u r new c o n c e p t

of from

available.

(Vvo((Voeh)-'(Vo~V2)))

:

h = uv2

u s s e e how t h e p r o c e s s

some o f t h e f a m i l i a r

~r~emember, w h a t we m u s t do i s a t e a c h s t a g e

which are already

~ v2

the above outline,

(Vvo((%eh) ~(av3((Vo'~V:;)" (~)~v2-~'~).')

:

v I = tVo~

ffv2((v2~h)~(v 2 = %)))

v2 = { % , h }

v 2 = YoUr 1

:

:

v 2 = (v0,Vl)

v 0 is an ordered

(vv)((v3ev2)~ ((v3 = re)v (v) = h))))

v2 =

:

U~Vo,V 1]

(Yv3((v3ev2)~-~ ((v 3 = [ v 0 J ) v (v 3 =[v0,vl}))))

pair

v 0 is a function

:

~Vl(~V2(v 0 = (vi,v2))))

(vh((vic Vo) ~

(h i. an ordered pair))) ^

(VvlVV2Vv3((( (h,v2)~ %)~ ( (h,v3)~ v0))~(v 2 = v3))).

etc.

Hopefully, by now we have convinced the reader that LST will indeed prove adequate for our needs. We will certainly have convinced him of the pointlessness of ever a c t n a l l y ~ L S T

in order to do mathematics. So why have we introduced this

18

p o w e r f u l y e t cumbersome l a n g u a g e ? W e l l , t h e p o i n t i s , defined,

the collection

of properties

s i n c e LST i s p r e c i s e l y

d e s c r i b a b l e w i t h i n LST i s a w e l l d e f i n e d

collection. It is immediate that any property which is describable in LST will be a bona fide mathematical property. And by our above discussion, LST envelopes all of the expressive power of the everyday "language of mathematics~ We shall thus be able to define a collection to be any "collection" defined by a property expressible in LST.

~. The Set-Theoretic Hierarchy

At this point it is very tempting to argue as follows.

(i)

The b a s i c p r e m i s e o f s e t t h e o r y i s t h a t g i v e n a n y c o l l e c t i o n

of o b j e c t s

(sets), we may regard the entire collection as an object (set) in its own right.

( n ) In order to obtain an adequate and meaningful set theory, we allow all (and only these) properties which can be expressed in LST to determine collections.

( n i ) By (1) and (II), given any property P which is expressible in LST, the collection

{ xlP(x)]

is a set.

The above argument certainly leads to a well-defined theory of sets. Moreover, all the sets we need in mathematics will be there. The only trouble is that the theory is inconsistent: Indeed, the Inconsistency is easily derived. Consider the property P(x) : x ~ x

fo~la

. (i.e. x is not an element of x) I) can be expressed in LST ( the

(~(Vo~Vo)) does i t ). So, assumi~ ( h i ) above, t x I r ( x ) ~ i s a set. Call

this set y. Since y is a set in its own right, it must be the case that either l(y) is true or that P(y) is false. But look, if }'(y) holds, then y must fulfill the definition of P, so it must be the case that y 4 Y, whence y cannot satisfy the property which defines y, whence P(y) is false. Thus, if l'(y) is true then it is false. Hence it must be the case that P(y) is false. Thus y does not satisfy ~, which

19

means that it must be the case that y e y ,

which means that y must satisfy the property

which defines y, which means that P(y) is true. We have thus arrived at a contradiction.

So w h a t h a s gone xcrong ? The a n s w e r i s t h a t from (I)

and ( n )

to (nI).

property

P considered

We i n t e r p r e t e d

a b o v e i s so b a s i c

(n)

that

we h a v e b e e n c a r e l e s s

correctly.

there

(Indeed,

in going

tl~e d e s t r u c t i v e

i s no a v o i d i n g h a v i n g t h i s

property

around for the definition of sets.) But we did not interpret (I) correctly. Let us reconsider

(I),

e m p h a s i s i n g t h e p o i n t w h e r e we w e n t w r o n g . ( I )

GIVEN a n y c o l l e c t i o n The p o i n t

is,

disposal.

We c a n n o t i n c l u d e

in (I).

of sets,

we may r e g a r d

we c a n o n l y f o r m a new s e t

the entire

out of sets

this

collection

which are already

sets which are not there!

L e t u s s e e why o u r i g n o r i n g

says:

Thus t h e c r u c i a l

at our w o r d GIVEN

one w o r d l e d t o t h e c a t a s t r o p h e

Let P be any property expressible in LST. (e.g. let F(x) say x ~ x . )

as a set.

it

did.

We wish to

consider the collection of all those x for which P(x) holds. Lhlt what are all those x ? Certainly, until we have formed the collection Ix IP(x)}, this collection will not he available to be considered as one of the all those x. Indeed, there may be many other "sets" which will not be available mltil afterwards. So how do we escape from what is beginning to look like a hopeless situation? ~'/ell, the answer is already there in our basic premise (I) above, lU~undamental to set theory is the notion of a hierarchy of sets. We build sets. They are not handed to us on a silver p~atter.

But no, already there is another possible snag. How do we build sets? Well, you say, we use the properties expressible in the language LST. If P is a property expressible in LST, then at any stage in our inductive process of building sets we may f o r m a s a new s e t t h e c o l l e c t i o n the property

of all

those x which are available

P. But l o o k , w h a t do we mean b y s a y i n g t h a t

an x satisfies

which satisfy P 7 In order

to know whether x satisfies P we must check whether P(x) is true or not. But F may well contain in its definition quantifiers of the form for all

z ...

, or

there is a z such that ...

20

Thus, in order for the statement "P(x) is true" to be at all meaningful, must it not be the case that the universe of all sets is already a well defined collection~ (For how do we know what "for all z ..." means if we do not know what all the sets are?) There are two ways out of this dilemma, lie consider first the classical one of E%.

Zermelo.( I"

The basic idea is as follows. We take as basic the iterative buildin~ofoLsets approach at which we most painfully arrived above, but ignore for the moment the problem about actually defining sets by means of properties expressible in LST. This is achieved by introducing the notion of the unrestricted power set operation. If x is any set, we denote by

~(x) the collection of a1__~isubsets of x. Thus:

@(x)= ~yly~x

} .

For the time being we make no comment as to just what sets are in use of the word "unrestricted" above.

~(x). ~Tence our

In order to obtain our universe of sets now,

we commence with the empty set, ~, and iteratej a d i n f i n i t u m 9 the power set operation. In order to make this more precise~ we need first to introduce the ordinal number system.

Before we can have a hierarchy, we must have some means of indexing the hierarchy. For a hierarchy of sets we require the ordinal m ~ b e r s . So we shall take this system as given. This can be avoided, but only at the loss of some degree of clarity. So our approach is similar to that of the analyst who takes as basic the natural numbers and proceeds from there to construct the real line. In an appendix we see how to eliminate our assumption of the ordinal number system from the development, replacing the assumption by a system of axioms from which everything can be defined. And this is akin to the elimination of the assumption of the system of natural numbers by introducing %he Peano axioms. The reason why few authors commence with the Peano axioms is, of course, that it is far clearer %o work with the intuitive picture of the natural numbers than with a bunch of abstract axioms.

1. I t i s t h e s e c o n d s o l u t i o n which g i v e s us t h e ,Lxiom of C o n s t r u e t i b i l i t y .

21

And t h e same i s t r u e h e r e .

The o r d i n a l

number s y s t e m i s j u s t

as intuitive

a s i s the

natural number system. Indeed, it simply makes precise the natural idea of "counting", only with the ordinal numbers we allow the counting to extend into the transfinite. its to the structure of the ordinal n~nnher system, the description given in 1.4 will suffice. The class of all ordinal numbers will be denoted by On. The class On is not a set in the sense of our universe of sets, of course, just as the set of all natural numbers is not a natural nlumber. (A better analob'ry would be that the set of natural numbers is not a finite set~) This is because our hierarchy o f sets should continue ad infinitum, in the sense appropriate to set theory. (This is quite different from the "infinite" nature of the natural numbers. Althou~h in a sense the natural numbers continue "for ever", in the sense of set theory they constitute quite a small set, and are thus fairly insignificant when compared with the universe of all sets.)

We define now the set-theoretic hierarchy. The ~ t h l e v e l will be denoted By Va . (As usual, a

of this hierarchy

denotes an arbitrary ordinal number.)

To commence, we start with the empty set. Thus:

V0 = ~ •

Next we form all possible subsets of the sets we have so far. Thus:

vI = [ ~ }

Thus, although ~0 had no members, already V 1 has a member. Next we repeat the process, fermin~ all possible sets from what is now available. This gives:

v 2 = ~ ~, ~ }

}

Repeating the process, we obtain next:

v3 = [ ~,

~f~,

t~.~},

{¢,~}j

;).

22

In general, for every positive integer n, V

n

will have 2n-I elements, as is

readily checked. Hence V~ is already sufficiently large

to make enumerating its

members a pointless exercise, so let us now write down the general rule. If V

n

is

defined, we set:

~+i

=

e(Vn)

So what do we do when we have defined V

for every natural n~mnber n ? The n

answer to this question is inherent in the notion of a limit ordinal. At stage ~

we

just collect to~ether everything we have so far. Thus:

V~

=

4in< V n.

Notice that, although we have not defined any new sets at stage ~ , we have nevertheless made some progress. ~

is quite a different collection from each of the

previous collections V n. Indeed, V~

is infinite, whereas each V n is finite. We now

continue as before, setting:

v~+ i In general now, if V~

and i f

=

~(v~)

.

is defined, we set

is a limit ordinal and we have defined V O

for all ~
It is not hard to see that for all ~ ,

In particular, V~n

On=

o~ ,

so V~+ 1 contains all subsets of oo . Thus, each V

n

is finite, V ~

is cotuntably infinite,

and Vo)+l is uncountable. Thus, despite the rather slow start, once the hierarchy gets

23

going it grows at a phenomenal rate. (Indeed, its growth is exponential, in a very precise sense.)

It is also easily seen that if ~ < ~ , then V

~

V~. Thus our hierarchy is

cumulative.

Now, the

idea

behind the above construction was that the sets of mathematics

should be precisely those collections which we can construct by iterating the process of building new sets out of old ones. Thus, if we denote by V the collection of all sets, we shall require:

(Pi)

v = U~Cn

v~

.

The principle (PI) can be taken as a definition of set. We call the collection V of all sets the universe of sets.

But what has happened to our language LST in all of this2 Can we now dispense with it altogether? Well no. So far we have not said very much about the power set operation. All we know is that

~(x) is to consist of all subsets of x. Certainly,

since we commence with just the empty set, V will contain nothing other than genuine, mathematical sets. But will it contain all the sets we need? In order to ensure that it does, we now place some restriction on the operation of the sets which

@ , in that we specify some

~ (x) must contain. (This restriction on ~

must, of necessity,

come after the event, so as to speak, as we indicated earlier, hut our development is not circular.) We thus formulate our second principle.

(P2) Principle of Subset Selection.

If x is a set and P is a property which is

expressible in LST, then { y ~ x i P(y) }

is a set.

The above principle requires some comment. Firstly, the reader may wonder why we do not formulate (P2) with V~

in the place of x. Certainly we have indicated

24

as much in our previous discussion. The point is the t~¢o versions are equivalent, so we have chosen to give the version more conmlonly used. ~[oreover~ formulated as above, the principle is clearly a partial description of the power set opera~ion.

Our second comment concerns the meaning of the ~,rase "expressible in LST" in (P2). By definition, if the property P involved is expressible in LST there must be some formula of LST which describes P. The intention is that this should enable us to define any of the sets we require in mathematics. Now, when we define sets in mathematics, we usually refer to other sets in the process. (To take a trivial example, when we define the set of all irrational real numbers we refer to %he set of all real mnnbers.) Thus, in the description of P we should be able to refer to any sets we please. This we allow. In other words, there is no restriction placed upon the way we use the variables in the LST formula describing P in order to denote specific sets. Hence, the variables involved in this description ~all into three distinct categories. Some of the variables will be used i n conjunction with quantiflets, and hence are internal to the formula. Others will be used to denote specific sets. And one further variable will behave as a genuine variable. This will he the variable corresponding to the y in P(y): the set variable which will, according to the definition of o u r set, range over the set { y e x IP(y)} o

It light

is probably instructive

o f o u r new d e f i n i t i o n

x the collection

to re-examine our original

of sets.

So l e t

y = {zexlP(z)} w i l l

positive is.

result

For let

a least

that

to a contradiction

y @ x . And now we c a n s a y j u s t

x be a n y s e t .

Then t h e r e

s u c h c~ . S i n c e no new s e t s

be of the form

P be t h e p r o p e r t y

he a set a c c o r d i n g

aFply the argument which led previously

contradiction

i s an o r d i n a l

appear at limit

x ¢ x. For each set

to ( P 2 ) . B u t now i f we we j u s t

end up w i t h t h e

what the collection ~

such that

levels

in the

[ x I I'(x) }

x e V~. T h e r e i s t h u s

in the hierarchy,

~ will

~ + l for some ~ . Then x c ~ hut x ¢ V~ . Since x c V~ , we have ye

x

--~

yEV~

y6

x

~

y~x.

fiance, as x ~ V~, we have

.

25 Thus, in particular, x ~ x. But x was arbitrary. Hence every set satisfies the property P. In other words { xlP(x)}

= V. Not only does the property P not lead to a contrad-

iction, it is universally true!

That then is set theory as developed by Zermelo. In h o n o ~ of Zermelo and Fraenkel, who first made the above procedure rigorous (which we have not bothered to do here), we denote this system of set theory by ZF. Thus, ZF is that system of set theory which has for its basic principles the two principles (PI) and (1>2) above. (For a more accurate account of what the ZF system entails, the reader should consult Appendix I.)

5- The Axiom o f Choice

With the system ZF we new have a servicable, natural set theory, which corresponds to our intuition about sets, and which is adequate for the development of mathematics from set theory. Well, almost, but not quite. Since the early part of this century the Axiom of Choice (AC, for short) has been regarded as a justifiable assumption about sets. This principle has several formulations. The most basic is the following.

L e t x be a s e t f from x to sets

(of sets,

such that

f r o m e a c h member o f x . )

every set

prove that

a statement

in Appendix II.) exist,

it

guarantee of Choice,

then x has a choice

about sets

thought

by introducing

o r AC f o r

we k n o ~ f o r

of non-empty sets

Nonetheless,

is generally this

for each a in x, f(a)

(To b e p r e c i s e ,

prove that

A choice ~ a.

Now, i n ZF one c a n p r o v e t h a t

element of x is non-empty, not possible.

naturally).

short:

set.

although

if x is finite x is

function.

infinite

Just

in a certain

we c a n n o t p r o v e t h a t

an additional

and each this

is

i n t h e s y s t e m ZF one c a n n o t

has a choice

t o be a r e a s o n a b l e

for x is a function

(Thus f chooses an element

But if

sure that

is not provable

function

assumption

how o n e may

system is

choice that

axiom to our system.

indicated

functions

always

t h e y d o . So we T h i s i s t h e Axio.__~m

26

AC : Every s e t o f non-empty s e t s has a c h o i c e f ~ m c t i o n .

AC has v a r i o u s o t h e r e q u i v a l e n t f o r m u l a t i o n s . Cne o f t h e s e i s t h a t t h e C a r t e s i a n p r o d u c t o f any f a m i l y o f n o n - e m p t y s e t s i s n o n - e m p t y . A n o t h e r i s Z e r m e l o ' s Well-0rdering Principle:

EVery s e t can be w e l l - o r d e r e d . S t i l l

known Z o r n ' s Lewma: I f P i s a p a r t i a l l y

another is the well

o r d e r e d s e t such t h a t e v e r y t o t a l l y

ordered

s u b s e t o f P h a s an u p p e r bound i n P , t h e n P h a s a maximal e l e m e n t . ( I n each c a s e t h e e q u i v a l e n c e w i t h AC i s p r o v a b l e i n t h e s y s t e m ZF.)

~e d e n o t e by ZFC t h e s y s t e m ZF t o g e t h e r w i t h t h e Axiom o f C h o i c e . To be precise,

our s y s t e m ZFC i s what i s commonly r e f e r r e d

to as Zermelo-Fraenkel Set

Theory., b u t t h e meanings o f t h e n o t a t i o n s ZF and ZFC a r e now s t a n d a r d i n s e t t h e o r y , d e s p i t e the m i s l e a d i n g a s p e c t of t h e s e a b b r e v i a t i o n s .

Chapter III

TH~ AXIOM OF CONSTRUCTIBILITY

Using techniques described in Appendix II, it has been proved that the four problems of pure mathematics described in Chapter I cannot be solved on the basis of the system ZFC. In view of the fundamental nature of these questions, this surely indicates a serious

deficiency in our set theory. So what went "wrong" ? Well, we

surely cannot abandon the principle of subset selection, (P2), since this is the one principle which is indispensible to us if we are to develop a set theory which is adequate for mathematics. So we must re-examine (PI). The reason why ZFC is not adequate for the solution of our four problems must lie in the fact that we took as a basic, undefined notion, the unrestricted power set operation,

~

. So let us go

back to the point where we felt it necessary to introduce this idea. Our argument then was that before we could introduce the principle of subset selection we must have at our disposal all the sets. But is this the case ? What if we stay even closer to our "iterative building of sets" idea than before ? Indeed, let us regard the original idea of sets being describable collections as basic. Let us define our hierarchy of sets as follows. We shall start with the empty set, and at limit levels we shall collect everything together as before. But in proceeding from stage ~ t o stage ~ + l we shall introduce just those subsets of what is available which we can describe using LST. "But look," you say, "we had to abandon this kind of approach before, because until we have all the sets available we do not know whether any statement expressed in LST is true or not." Well, if by "true" you mean really true (i.e. in the universe V), then this is correct. But consider the following alternative. At stage ~ we h a v e constructed some sets. This collection of sets can be regarded as a partial universe. (Indeed, it i_~ssuch!) And relative to this partial universe, any statement expressible in LST (which refers explicitly only to sets in this partial universe) will either be true or false. Hence, if at stage ~+l we say to ourselves "Ne shall build all sets we can using LST properties, referring Q n ! ~ t o what is available right now , "

then we can indeed proceed without trouble. In other words,

we shall construct our new hierarchy almost as if we ~ r e

actually carrying out the

construction step by step, at each stage having only those sets available which we

28

have constructed so far, and introducing new sets only when we can construct them using properties expressible in LST. This reasoning leads us to the definition of the constructihle hierarchy.

I. The Constructible Hierarchy

By analogy with the Zermelo hierarchy, V~, ~ e On, we define the constructible hierarchy of sets, L=, ~ e On. We shall commence as before with the empty set, and at limit stages we shall just collect everything together that we have so far, again as before. But in passing from L~

to L + l we shall not take "all" subsets of L~

(whatever that means), but only those collections from L~ which are, in a precise sense to be outlined

Suppose other v

n

things,

will

~

below, describable

i n t e r m s o f L~.

i s some f o r m u l a o f LST. Then

quantifiers

range over all

of the form ~v

sets.

or

n

More p r e c i s e l y ,

~ will ~v

n

in general

contain,

. And i n g e n e r a l ,

amongst

the variable

t h e i n t e n d e d domain o f t h e q u a n t i f i e r

is V. Suppose now that M is some set which contains all of the sets which any variables in ~ may denote. By ~

M

we shall mean the formula

restricted to N. Of course, in a strict sense What is different is the way we interpret

M

T

M

~ with all quantifiers

is exactly the same fo~nula as T •

and

T

. A quantifier

Vv

in ~

is

n

read as "for all sets Vn" , whereas a quantifier

~v

in

cpM is read as "for all sets

n

v

n

in M". Notice that there need be no connection between the validity of T

validity of q~". For instance (and in terms of ZF set theory), let ~

and the

be the sentence

3 v 0 ( v 0 is an uncountable cardinal) (expressed in LST, of course). Now,

so

~V~I

is false. But since

that in q~VcOI , the quantifier

a)1 is the first uncountable cardinal, and

c~1 E V~z ,

~V~k is true, as is T itself. The point is

~ v 0 only farces over V@I , which contains no uncountable

cardinal. (The reader may he disturbed by the fact that this example is constructed in ZF, a system which we appear to have abandoned. He should contain his concern for a while, for we have a pleasant suprise in store for him.)

29 G i v e n a n y s e t M now, we d e n o t e b y Def(M) t h e c o l l e c t i o n w h i c h c a n be d e f i n e d that all

M contains definable

definition

by a formula

any sets subsets

of the form

which variables

o f M. N o t i c e

only refer

that

in

of all

subsets

of ~

~ ", w h e r e T i s a f o r m u l a o f LST s u c h T

since

denote.

We c a l l

the formulas

Def(M) t h e f a m i l y

TM i n v o l v e d

( i n a n y m a n n e r ) t o m e m b e r s o f N, we o n l y r e q u i r e

of

in this the availability

of M in order to define Def(M).

We d e f i n e

L0 =~

the constructible

L~ =

;

U~<~L~ , i f ot i s a l i m i t

Comparing this

elements),

as follows:

;

L~+ 1 = D e f ( L ~ )

. In fact,

hierarchy

since

ordinal

with the Zermelo hierarchy,

we may e x p l i c i t l y

we h a v e Ln = Vn f o r

define

all

natural

we c l e a r l y

any finite

set

h a v e L~ ~ V< f o r a l l

(by enumerating

numbers n, whence also

L~ = V~. B u t

L~+ I ~ V~+I, since V~+ I is uncountable w h e r e a s L~. I will be countable. remarks presuppose reader

to supress

It L

is

for

easily

n On = [ ~ ~ < ~ ]

resembles

a while

~<~

implies

L~c

--oc . T h u s t h e c o n s t r u c t i b l e

with

is quite

slow.

Indeed,

about

arithmetic

hierarchy

for

convenience

later

L

L~ .

=

U~on

on, let

us define

any ordinal

~

of growth of the

any infinite

s h o u l d h a v e no d i f f i c u l t y

for

grows in a manner which

But the rate

equality.)

For notational

this.)

L~, a n d t h a t

• ( T h i s i s b e c a u s e LST h a s o n l y c o u n t a b l y

cardinal

(These

i n t h e s y s t e m ZF. Once a g a i n we a s k t h e

a n y u n e a s e h e may f e e l

seen that

hierarchy

h a v e I L~I = I ~ l

of this

we a r e w o r k i n g

the growth of the Zermelo hierarchy.

constructible

familiar

that

its

ordinal

~

we s h a l l

many f o r m u l a s .

Readers

in supplying

a proof

,

30

We c a l l

the collection

L the constructible

universe

(of sets).

2 . The Axiom o f C o n s t r u c t i b i l i t y

The a x i o m o f c o n s t r u c t i b i l i t y all

sets.

define

Or, t o p u t i t

a set

another

B u t a s we s h a l l

suggests

the universe

that

the use of the word "axiom" here

introduced

is

in fact

the symbol V to denote

of constructible

sets.

the axiom of constructibility, V--L

Thus, what V = L says is the empty set

L contains

quite

we

the axiom

is unnecessary. ai~propriate.

The

to our next definition.

We h a v e a l r e a d y

of denoting

the collection

t h e way we h a v e i n t r o d u c e d

see in a moment, our terminology

same r e m a r k a p p l i e s

that

w a y , w h e n we a s s u m e t h e a x i o m o f c o n s t r u c t i b i l i t y

t o be a n e l e m e n t o f L. Of c o u r s e ,

of constructibili%y

L denotes

is the assumption

and successively

So we h a v e on h a n d a c o n v e n i e n t

And way

.

that

set

of sets.

namely:

in order

f o r m new s e t s

So i f we w o r k i n c o n s % r u c t i b l e

the universe

theory.,

to obtain

all

sets

we s t a r t

off with

from old by means of the Def operator. instead

of taking

as our basic

assumptions

t h e two a x i o m s :

(P1)

v=

(P2)

Principle

we t a k e

U~onV~

;

of subset

selection

,

t h e two a s s u m p t i o n s :

(L1)

V =

U~0n

(I2)

Principle

L~

(i.e.

of subset

v = L)

selection.

L e t u s a s s u m e now t h e d e f i n i t i o n us.

This provides

for each set

us with a well-defined

x there

o f x . (More on t h i s well-defined

will point

power set

surely

of set which constructible universe

be a s e t w h o s e e l e m e n t s

in a moment.) ~ence,

operator.

of sets.

within

There is thus nothing

set

Now, w i t h i n are precisely

our universe to prevent

theory this

gives

universe

the subsets

there

will

be a

us from carrying

31

out our previous definition of the Zermelo hierarchy, V~

( ~ ~ On) . (Only now we

are working within a fixed universe of sets, of course.) we find that

U~eon

V~

= v . In fact, for all ~

And what do we find ? Well,

, L ~ ~ V . (Equality can only

occur very infrequently.) I n other words, the principle (PI) is still valid! The only difference between what (P1) meant before and what it now means lies in the fact that whereas the power set operator is unrestricted in ZF set theory, in eonstruetihle set theory it is ~overned by a much more precisely defined universe of stes. Thus, constructible set theory is not an alternative to ZF set theory, it is an extension of it. In other words, we may regard the principle V = L as an axiom of set theory which we assume in addition to the ZF axioms. This illustrates both our use of the word "axiom' and our abbreviation for the axiom itself.

At this point we feel we owe the reader some comments concerning the assumption made above that in constructible set theory, each set has a well-defined power set. We were not claiming that this could be proved from the axiom of constructibili%y. But i t

is a very reasonable

And i n t h e l i g h t constructibility, with constructible

role.

a s s u m p t i o n t o make when d e v e l o p i n g a n y t h e o r y

o f o u r comments a b o v e c o n c e r n i n g t h e n a t u r e we c a n d e m o n s t r a t e t h a t set

theory

it

o f t h e axiom o f

i s no w o r s e t o a s s u m e i t

t h a n w i t h ZF s e t

theory,

of sets.

where it

in conjunction

plays a fundamental

We did not explicitly list the axiom of power set existence as a basic

principle of ZF set theory, a l t h o ~ h

it certainly is so, because the principle (F1)

already conveys the essential structure of the system. But what is the situation when we append the system ZF by the additional assumption V = L ? Could this not invalidate the axiom of power sets (and thus lead to an inconsistent theory)? The answer is no. indeed, assuming the axiom of constructibility as an additional axiom of set theory on top of the principles of ZF is totally harmless, in view of the follcwing classical result of GSdel:

2 . 1 Theorem If

t h e s y s t e m ZF + V= L i s

inconsistent,

t h e n s o i s t h e s y s t e m ZF.

32

T h u s , t h e axiom o f c o n s t r u c t i b i l i t y I t c a n n o t l e a d u s i n t o a n y more t r o u b l e from the statement

t h a n c o u l d a l r e a d y t h e ZF s y s t e m . (Note a l s o

of 2.1 the convenient notation we now have for constructible

theory: ZF + V = L .)

set

At this point, readers whose pet subjects require the Axiom of

Choice may be asking themselves They do not need to worry. Indeed,

i s t h e r e f o r u s t o a s s u m e i f we w a n t t o .

in constructible

if they could live with ZF + V = L instead of ZFC.

They may safely use AC in conjunction with ZF + V = L.

set theory we do not need to assume AC as an axiom: it is a

theorem:

2 . 2 Theorem ( i n t h e s y s t e m ZF + V= L ) ~ery

set of non-empty sets has a choice function.

2 . 2 was p r o v e d by GSdel. P r o o f s of b e t h 2 . 1 a n d 2 . 2 may be f o u n d i n t h e more r i g o r o u s development of c o n s t r u c t i b i l i t y

theory presented in our article

As the basic set theory underlying mathematics the system ZFC , Zermelo-Fraenkel additional

[De]

.

it is now customary to take

set theolV . ~%enever a result is proved using the

assumption V = L, the statement of the result must then be prefaced by

the sentence:

't~ss~mne V = L" . (This is what is customarily done when a result is

proved using the GCH.

Instead of saying: "In the system of set theory ZFC + GC}[

.... ", one assumes ZFC as basic and just writes "Assuming G~I ...".

To sum up t h e n , we a r r i v e d o b t a i n a good d e f i n i t i o n to exist

if there

unrestrictive describing

a t the n o t i o n of c o n s t r u c t i b i l i t y

o f se__~%. Our a p p r o a c h was t h a t a c o l l e c t i o n

i s ' s o m e way o f d e s c r i b i n g

intuitionwas

still

sound, and t h a t

all

removed t h e w e a k n e s s i n h e r e n t i n ZF t h a t totally

undescribed.

we o b s e r v e d t h a t

t h a t had i n e f f e c t

in a fairly

to any o t h e r s e t s

H a v i n g o b t a i n e d t h e n o t i o n of t h e c o n s t r u c t i b l e

a n d h a v i n g d e f i n e d t h e axiom of c o n s t r u c t i h i l i t y ,

to

c a n o n l y be s a i d

i%. ( T h i s b e i n g made p r e c i s e

m a n n e r , a l l o w a n c e b e i n g made f o r t h e r e f e r e n c e

a collection.)

by t r y i n g

in

hierarchy~

our p r e v i o u s ,

ZF

h a p p e n e d was t h a t we h a d

t h e power s e t o p e r a t i o n

is taken as almost

S i n c e m o s t t h e o r e m s i n m a t h e m a t i c s c a n be p r o v e d w i t h o u t a n y

33

great

u s e b e i n g made o f t h e f u n d a m e n t a l

nature

of what constitutes

a set

(except

for

recourse to AC now and then), we decided to retai~ the "simpler" system ZFC as basic, but we could now call o n the axiom of eonstructibility to help us out in those cases where ZFC is not enough. Thus, if a theorem can only be proved using V = L, then we know that at the heart of this lies a fundamental question about the nature of sets. This approach does not preclude our adopting the system ZF + V = L as basic (thereby retaining the intuition of our original development). It would just be the case that for the most part it doesn't matter one way or the other (whence any approach will suffice). But for the four problems introduced in Chapter I it matters a great deal what our set theory is. This is the subject matter of the rest of this chapter and of the next.

3. The Generalised Continuum Uypothesis.

Recall

that

t h e GCH r e s i s t e d

GCH i s t h e a s s e r t i o n

all

case of the CH ( 2 ~ =

attempts

at proof

~(2

~ =

~+).

over the years.

As we m e n t i o n e d

Indeed,

in 1.~,

even the special

~1) went unsolved. Then in 1965, P. J. Cohen gave a rigorous

proof of the fact that neither the CH nor the G ~

could be settled on the basis of

the system ZFC alone. (In Appendix II we give some indication of how such a result may be obtained.) This highlighted an old result of G~del:

3.I Theorem Assume V = L. Then C~}[ is true.

tt

is not possible

be found in [De] this

reduces

to give a rigorous

.) But let

%o p r o v i n g

that

us see what is there

as there

proof

of this

involved.

are at most

are characteristic

result

Consider

~1 subsets functions

of

here.

( S u c h may

t h e c a s e o f CH. Now, ~

• (There are clearly

as many subsets

of ~

o f s u c h , w h e n c e t h e two

sets ~(t0) and

co 2 have the same cardinality.) Consider now the constr~iction of the

constructible hierarchy. At stage c0+l some subsets of co will appear. Perhaps by makin~ reference to some of these we can describe more

subsets of ~ to come in at

34

stage

¢~ +2. Then we may be a b l e t o d e s c r ~ e s t i l l

on. Thus, u n l i k e subsets

the situation

o f co a p p e a r t o g e t h e r

w i t h t h e Zermelo h i e r a r c h y in ~+1~

o~ may k e e p a p p e a r i n g , a s o u r f a c i l i t y and t h i s stage

is a very fundamental fact,

in the constructible for describing it

~1

levels

c a n be shown t h a t

have m e n t i o n e d e a r l i e r , ~L=I = I~l

A similar

of the constructible that

it

is quite

• Thus we c o n c l u d e t h a t

hierarchy.

w h e r e , by d e f i n i t i o n , hierarchy

Thus

all

new s u b s e t s increases.

thi~ process

subsets

easy to prove that

of But,

stops ~t

of ~

will

appear

~(~) c

L~.

B u t we

f o r any i n f i n i t e

2 = I~Sl=l~(o~)l < - ILeal = ~ 1 '

ordinal

p r o v i n g CH.

a r g u m e n t p r o v e s GCH a s s u m i n g V = L.

In V . ~ we say a little more about the above,

sets

t o go i n t o L~+ 3 , and s o

new c o l l e c t i o n s

c51. I n o t h e r w o r d s , i f we asmune V = L, t h e n a l l

in the first

o¢

more s e t s

of natural

numbers occur after

stage

in that we indicate w h y no n e w

c~1.

T h u s , by a s s u m i n g V = L we may s o l v e one o f t h e f o u r p r o b l e m s o f C h a p t e r I . I n t h e n e x t c h a p t e r we c o n s i d e r

~. ~ s t o r i c a l

appeared in his

classic

of the constructible

axiom of c o n s t r u c t i b i l i t y of mathematics).

s h o u l d be t a k e n a s a b a s i c

universe

as a tool

give the first

GBdel's approach to set

of these,

a d o p t V = L a s an a x i o m . I f , in an iterative

that

the

(and thence

in order to establish i n ZFC o f GCH.

and when we add on 3 . 1 a s w e l l

t h e o r y was t h a t

there

is,

of sets which the Zermelo-Fraenkel axioms attempt

approach, then there

and f i r s t

formulated

axiom of s e t t h e o r y

i n Z~ o f AC and t h e n o n - r e f u t a b i l i t y

( 2 . 1 and 2 . 2 t a k e n t o g e t h e r we g e t t h e s e c o n d . )

i s due t o K. G S d e l ,

V = L. B u t i t was n o t G S d e l ' s i n t e n t i o n

R a t h e r he u s e d c o n s t r u c t i b i l i t y

both the non-refutability

I f we a d o p t t h i s

hierarchy

m o n o g r a p h [G8 ] . I n h i s m o n o g r a p h , GSdel a l s o

the axiom of constructibility,

t h a t we b u i l d

remaining problems.

~emarks

The d e f i n i t i o n

a "fixed"

the three

i s o f c o u r s e no a p r i o r i

to describe.

r e a s o n why we s h o u l d

on t h e o t h e r h a n d , we t h i n k o f s e t s m a n n e r , t h e n we may c e r t a i n l y

i n some s e n s e ,

insist

as being something ~ l a t we p r o c e e d

35

in the manner which led us to the axiom of constructibility. c o u r s e we t a k e ,

in order to construct we m u s t u s e q u i t e on i t s

which leads

f r o m t h e s e a x i o m s . ~nd when we do t h i s

either

the Zermelo hierarchy

a few of our axioms. But t h i s

head,

so as to speak,

to our final

Had i t

not been for

ultimate

does not have any bearing

V = L as a useful

like

tool

on t h e s e t t h e o r y

set theory msee

unprovable in Zermelo-Fraenkel set chain of results

theory.

about constructibility,

This stimulation

Of c o u r s e ,

AC some s i x t y in several

years

areas

then applications

of pure mathematics (as .

) , known t o be

and o p e n e d t h e way f o r v a r i o u s

theory, tell

author

at least

and t o t h e w r i t i n g the outcome ( j u s t

certainly

merits

areas

people to of pure

to reconsider of this

the

book i n

a s was t h e c a s e w i t h

a g o ) . But s i n c e t h e axiom of c o n s t r u c t i b i l i t y

o f m a t h e m a t i c s now, i t

and which

B u t i n 1967, R. B.

to solve problems in different

only time ~i]l

non-refutability

set theoretical,

involved.

a result

ten years,

S u b s e q u e n t work of J e n s e n l e d t o a whole

led the present

o f V = L a s an axiom o f s e t

particular.

our

on t h e i n t u i t i o n

For until

t h e comment on p a g e ( i v )

apply the axiom of constructibility mathematics.

of standing

to establish

AC a n d GCH, w h i c h a r e h i g h l y

Jensen proved from the axiom of constructibility opposed to abstract

process

see that

hierarchy

some r e m a r k a b l e d e v e l o p m e n t s o v e r t h e l a s t

one m i g h t e x p e c t t o d e p e n d h e a v i l y

status

or the constructible

would undoubtedly have remained unchallenged.

had been to statements

we s l m l l

choice.)

t h e GSdel a p p r o a c h o f r e g a r d i n g results

whichever

we a r e g o i n g t o end up w i t h a s y s t e m o f a x i o m s a n d p r o c e e d t o d e v e l o p

our set theory rigorously

analysis

(Of c o u r s e ,

attention.

has applications

AFFLICATIC}:S OF V = L IN .~V~T]~LiTICS

Chapter IV

It is to be expected that proofs involving the axiom of constructibility have a quite different flavour from proofs in ZFC. Indeed, we only need to use V = L in the first place because the "usual ~' methods fail. The same is true of proofs involving AC. A good illu~ration is the standard proof of the ~[ahn-Banach theorem of analysis.

The proof falls into two distinct parts: one part being clearly of an

analytic nature

(more accurately,

it has the flavour of Banach space theory),

other part, ~le application of Zorn's le~T1a, being set theoretical course~

it does not take the pure mathematician

in ~ature. Of

long to master the use of AC. Dut

with V = L we are already at a much deeper level, and many applications can really only be fully understood with a solid background of) mathematical

logic. Fortunately

investigation

of constructibility,

combinatorial

consequences

application

Principles

B e f o r e we c a n s t a t e set

of V = L

in set theory and (parts

this is not always the case, however.

In his

Jensen established a collection of purely

of V = L, and for many problems a straightforward

of one of these consequences

i. Combinatorial

the

is all that is required.

from V = L.

the combinatorial

principles

we r e q u i r e

some n o t i o n s

from

theory.

The infinite cardinals the re6ular cardinals. A of ~ < ~

fall under t~o categories,

These are defined as follows.

is unbounded in ~

if there is no ~< ~

there is a Y ~ A with i >

the singular cardinals and

If ~ is a limit ordinal,

with A ~

~

(equivalently,

~ ). A function f from an ordinal

~

cofinal if f is order preserving and the range of f is unbounded in ~ ity of o( , c f ( ~ ,

is the least

6

example,

cf(~) = ~

, cf(~+~) = ~

for each

into ~

is

. The cofinal-

such that there is a cofinal map f : ~ - - ~

in~nediate that cf(~(~ is a limit ordinal,

. It is

and indeed is an infinite cardinal.

, cf(~w) = to

a subset

For

. (For this last example take f:e-*,0~

37

defined by f(n)

= ~

n

. ) An i n f i n i t e

(in which case cf(a)< ~ ), ~

cardinal

~ is regular

if

cf(K) = K . Otherwise

is singular.

It can be shown that every successor cardinal (i.e. a cardinal of the form + K

for some cardinal < ) is regular. Thus,

are ~ + I

and

cardinals) ~

'~I' ~2'

.... are all regular, and so

~ ~z1+I. The limit cardinals (i.e. cardinals which are not successor and ,~a~ are singular.

~ 0 is a regular limit cardinal. It cannot be

proved that there are any other regular limit cardinals, though it is most unlikely that this could be disproved. (Assuming V = L certainly would not help to disprove it.) Hence, any limit cardinal other than

~ 0 which we meet is likely to be singular.

Another definition of singularity is in terms of cardinal addition. An infinite cardinal K such that

is singular iff there is a cardinal ~ < K and cardinals K = ~XK=

~=

~n < ~

~ n"

Let ~ be any uncountable regular cardinal. A set C c_ K K

if C c_ K

are closed and ILnbolmded in

is also closed and unbounded in <

K

K-complete filter in

~(<). Indeed,

~t for all eL< k , where k< K , then

. ~le indicate the proof for the case

leave the reader to generalise this to any

We a r e g i v e n c l o s e d

~

A = 2, and

and u n b o u n d e d s u b s e t s A a n d B o f t h e u n c o u n t a b l e

closure first. Let f:~ - - ~ A ~ B , where ~ < K

regular

. We verify

is a limit ordinal and f is order preser-

A and supf[~] e B so supf[~] e A n B .

be given. We must find a @ ~ A(~B, ~ > ~

We check unboundedness now.

. We define a function f:o0 --~ K

by induction. Since A is unbounded we can find an ordinal f(0)e A, f(0) > ~ B is unbounded we can now find an ordinal f(1)6 B with f ( 1 ) > f ( 0 ) . can always find

C

)~ < K •

. We wish to prove that A n B is closed and unbounded in ~

ving. Then supf[~] e

C.

every set of ordinals has a suprenmm of course. )

The closed unbounded subsets of K generate a

Let ~ K

is closed if, whenever

is a limit ordinal and f:6---* C is order preserving, then sup~<6f(o( ) ~

(Since the ordinals are well-ordered,

cardinal

(~ < A )

, in which case cf(<) is the least )~ for which such K ~ < K

can be found. For instance,

6 <

K < K

f(2n÷2) ~ A with f(2n+2) > f ( 2 n + l ) ,

. Since

In general we

a n d t h e n f(2n+3) e B with f ( ~ ÷ 5 )

38

>f(2n+2).

Let

@ = supf[~].

Clearly,

~ = SUPn< f ( 2 n ) = SUPn< f ( 2 n + l ) . But f ( 2 n ) e

A for all n and f(2n+l)e B for all n. Thus

~ e

A and

~ e B. Hence $ s A ~ B

and we

are done.

The r e a d e r m i g h t l i k e t o c h e c k t h a t w h e n e v e r C ~ ~ i s c l o s e d and unbounded ( K an uncountable regular

cardinal),

t h e n C' = ~ ~ e

C ~ ~ is a limit

ordinal

}

i s a l s o c l o s e d a n d unbounded.

A subset E of a regular uncountable if E ~ C

cardinal

<

~ ~ for all closed and unbounded sets C ~ ~

closed unbounded set will be stationary.

The converse

is said to be stationary

. By our above result,

every

is easily disproved.

(i~emove

one limit point from a closed unbounded set.) Indeed, we can find disjoint stationary subsets of K . ~owever, the set [~eKl~ ~ }

a stationary set is certainly ~mbounded

(in ~ ), since

is closed and unbounded for any ~ < K • For example, E and F are

disjoint stationary subsets of

~2' where

E= { ~ 2 1 c f ( ~ ) = ~

,

F={~21cf~)=

We are now ready to introduce one of Jensen's

~i~

combinatorial

principles.

The

one we consider is both the simplest and the most widely used so far. (These two aspects may, of course, be related.) After Jensen, we denote this principle by (i.e. diamond).

: Let [S~
a

be an uncountable

such that each S~

stationary set E s <

regular cardinal.

is a subset of ~

such that X n ~

= S~

Then there is a sequence

and whenever X ~ K

for all

~ e E.

(The symbol <~ is sometimes used to denote the specific when

b< =

e l , with ~ K

there is a

instance of this principle

then being used to denote the version for any other

Assuming V = L, Jensen proved we know that there are at least

~

. So let us see what O

K + many subsets of ~( . ~

K

says. ~ell now

says that there are

•

39

K

many sets S g ~ such that on a stationary set S~

the initial part X ~ ~

predicts the correct value of

of any set X ~ K • Since any stationary set is unbounded,

we see that in fact the ~ ' s

give arbitrarily good approximations to X, and that

with E the appropriate stationary set, X = ~ EE(X n ~ ) = ~ E

S~ . Now, readers

familiar with GCH will observe at once that if we assume GCH (a weaker assumption than V = L), we can define a sequence whenever X ~ K %~hat O

, then for each ~ < K

[S~}~<~ ~ere

such that S ~

is a

~
, ~

for all ~ < ~ ~

and

, with X n ~

= S~

.

says is that we can find such a sequence with %he property that on a

stationary set of

~'s the ~

here can he taken to be ~ itself. This turns out to

be much stronger than the GCH result. Indeed, Jensen has proved that

0

does not

fell~g from GCH.

Suprisingly, (i.e.

very

large)

the fact set

of

that

every set X gets

~ ' s i n t h e m a n n e r Xn ~

W o r k i n g i n t h e s y s t e m ZFC, D e v l i n h a s shown t h a t of a sequence infinite that

ordinal

Xn~

with S ~

such that

~ f o r w h i c h X~ ~

is required.)

arbitrarily

set

{S<~
Thus the essence

well,

but that

Indeed,

of

~

in

whenever Xg K

application

it

that

0

on a s t a t i o n a r y is

OK i s e q u i v a l e n t

is not that

we c a n a p p r o x i m a t e

in our first

= S~

= S~ . ( I n m a n y a p p l i c a t i o n s

= S~ . T h i s d o e s n o t m e a n , h o w e v e r , in O.

approximated

at

irrelevant. to the existence

there of

is at ~

least

this

we c a n a p p r o x i n m t e

is all any set

leas% once in the manner

we n e v e r make u s e o f t h e s t a t i o n a r y

o f ~ b e l o w we make e s s e n t i a l

use of the

fact that the approximations are good on a stationary set.

2. The S o u s l i n

Problem

'de s o l v e

the Souslin

Froblem by proving

the following

theorem:

2.1 Theorem Assume V = L. Then there is a linearly ordered set (Y,~) such that: (i) (ii)

~

one

is a dense linear ordering without end points;

Y satisfies the c.c.c. (i.e. there is no uncountable family of disjoint

40

open intervals); (iii)

(Y,~)

is Dedekind complete

glb)

a

(Y,~)

(i.e.

every bounded subset

of Y has a lub and

;

is not isomorphic

We f i r s t

observe

that

to the real

it

is

line,

sufficient

E .

to prove the following

result:

2 . 2 Lemma Assume V = L. Then there (i)

is a dense linear

(ii) (iii)

Every closed

ordering

without

X is not separable

(i.e. X

claiming

that

(i)

at once that

and (iii)

X fails

of 2.1.

is maximal. Set K = X -

Y satisfies

2.1(i)-(iii).

Hence X is contrary

isomorphic

2.1.

o f X. I t

the c.c.c,

failed

to satisfy So l e t

~-(a ,b ) . Clearly, 1~1 1 1

s o I~ i s u n c o u n t a b l e .

Suppose it to a subspace

failed of ~

let

{(ai,bi)

(X,~) be as in Y satisfies It

lev~na we may assx~ne

K is closed

2.1(iv).

implies

follows

I i ~ I } be an

This violates

to satisfy

. But this

.

2.1(it).

o f X. By Z o r n ' s

and nowhere

2.2(ii).

Hence

Thus (Y,~) ~1~ .

that

X is

separable,

to 2.2(iii).

lie a s s u m e V = L f r o m now o n . We c o n s t r u c t The c o n s t r u c t i o n which is set

;

is immediate that

as well.

open intervals

I 3 ~ K

countable

To s e e t h i s ,

this

{a.tie 1

;

2.2 implies

family

d e n s e i n X. B u t

(X,~) such that:

dense subset)

uncountable family

of disjoint

of X is

Suppose it

to satisfy

set

has no countable

Let (Y,~) be the Dedekind completion

conditions

ordered

end points

and nowhere dense subset

We a r e t h u s 2.2.

is a linearly

is by induction.

important,

~1 of all

not what those

countable

extension

to

following

discussion,

ordinals.

Clearly,

it

points

are.

We d e f i n e

w+~ , then the extension c( a n d ~ w i l l

is

an ordered the ordering

So we s h a l l first

of that

set

ordering

always denote limit

to

on t h e p o i n t s

take

the ordering

(X,~) as in 2.2.

for

on ~ ,

(o+t6+~0

ordinals

the set

less

of X X the

then its

, etc. than

In the cO1 .

41

By induction on o( we define an ordering <~* on ~ s o that: (i) < * is a dense linear ordering without end points;

(ii) if=<~, then <. *

(In connection so <* will c4

with

(ii),

recall

a l w a y s be a s u b s e t

To commence we take

=

that

<~(~). any binary

of the Cartesian

relation product

is a set

of ordered

c~ x c~ . )

< ~* to be any ordering of ~

isomorphic to ~, the

rational line. And if ~ is a limit of limit ordinals we shall set

<~ = ~<@<:

which by (ii) will satisfy (i) and (ii). And when we are through we shall set ~<

~I < ~

that < *

pairs,

,

<* =

to obtain our desired ordering of ~ I (i.e. X). It will be immediate

satisfies 2.2(i). So we must define < *

from

* in each case to ensure

(somehow) that both 2.2(ii) and 2.2(iii) will hold.

Ensuring that 2.2(iii) will hold is easy. It suffices that we arrange that for no ~ < ~ if A ~

is

~ ( = {Plg<~

~ ~I) dense in (~i,<*). Why? Well, since cf(~l) =

~ l is co~mtable, then A H ~

for some

~<~l'

~l'

so if A were dense in (~ir<*),

so would ~ be dense in (~it<*)° So how do we arrange that no ~ will be dense in ~i,<*)? Suppose we are about to define <*~+~ from < ~ . Let D#

be a proper Dedekind

cut in (~,<~) (i.e. a proper initial segment of (~,<~) with no lub). Since (~,<~) is a countable, densely linearly ordered set, it will he isomorphic to Q, so there is no difficulty in finding such a D@. %qe define <~+~ by ordering the ordinals

~+n

(n e~) isomorphically to Q and placing this entire copy of Q where the "bole" at the top of D~

Thus ~

is. Thus, for each n e ~

will no% be dense in

we shall have

~ +~o, and hence certainly not in

~I"

~e now show that by choosing the cuts ])~ carefully now, we can also ensure that 2.2(ii) will hold. The idea is to apply the Baire category theorem to avoid certain closed and nowhere dense sets. tie first make precise what is mean% here.

Let K be a closed and nowhere dense subset of ~. Let 0 be a proper Dedekind

42

c u t i n Q. We s a y D a v o i d s K i f

there

are rationals

a < b with a~D
and E a , b ] s K = ~.

2 . 3 Len~a L e t Kn, n = 0 , 1 , 2 , . . .

be c l o s e d and n o w h e r e d e n s e s u b s e t s

o f ~. Then t h e r e

is a proper Dedekind cut which avoids each K . n

Proof:

This is really

of the sets irrational

just

a special

case of the ~ire

category

K i n l~ a r e c l o s e d a n d n o w h e r e d e n s e s u b s e t s n x which is not in the closure

of~,

of any of the sets

t h e o r e m . The c l o s u r e s so there

is an

K . Take D t o be t h e n

cut defined by x.

S i n c e ( ~ , < ~ ) H ~, 2 . 3 w i l l

apply to this

then is which closed and nowhere dense subsets w h e r e we make u s e o f V = L. I n f a c t ,

By ~ set

~E~ll

sets

that

[S~}~lbe

a sequence such that

A~

= S,~.~ i s s t a t i o n a r y

<~+~

we a r e g o i n g t o a v o i d .

in

S~ c_ ~

The d e f i n i t i o n

of the ordered

is as required

by 2.2.

This is

principle

and whenever A c

el

~

the

~1 o

now, we u s e 2 . 3 t o p i c k De

S~, ~ ~% , w h i c h h a p p e n t o b e c l o s e d

it

of ~

What we m u s t s p e c i f y

w h a t we u s e i s t h e c o m b i n a t o r i a l

, let

In order to define

ordering.

to avoid all

and n o w h e r e d e n s e s u b s e t s

set thus completed,

In view of our earlier

it

those

of ~ .

remains only to prove

r e m a r k s we n e e d o n l y c h e c k

2.2(ii) now.

L e t K -~ ~1 be c l o s e d a n d n o w h e r e d e n s e i n

~1 (under <*).

~e m u s t p r o v e

that K is countable.

The i d e a i s t o f i n d a c o u n t a b l e w h e r e we make u s e o f ~>. By d e f i n i n g picking

any

~ ~ C for which Kr~

ordinal

a suitable

= Sw , we s h a l l

a

for which ~ = K~ ~

closed unbounded set be a b l e t o f i n d

. This is

C _c ~ 1 a n d

such an

~

.

.

43

Define functions f, g from

e I to ~i so that for each ~e~l

(~(~),~(~)~ ~

and

el × ~i to

Define functions h, k from

=

-

K,

~.

w I so that whenever ~ < * T

,

h(~,~), k(~,~) ~ (~,~)* ;

Set g(~))~

~

k(9,T)

K whenever KO(?,T)* ¢ ¢.

~

I c~ i s a l i m i t

C ={~c01

, and for

h(~,r) ~

all

~,T<

~

ordinal

, h ( ¢ , T ) < oC

and for all

~<~

, f(~)<~

and

a n d k(~),T) < ~ .

2 . ~ Lemma (i) (ii)

C is

closed

~ e C --~

and unbounded in ~is

a limit

(iii) ~ E C - - ~ K n ~

Proof:

is

~I;

ordinal;

closed

and nowhere dense in ~ .

It is an instructive exercise to prove part (i). (Set theorists will see this

immediately. For the non-expert, notice that closure is trivial, and to prove the unboundedness of C use the fact that the limit of any monotone increasing sequence from

~1 is a countable

o f C. F o r p a r t closed in

~

in

~

(iii),

limit if

(under

<~),

ordinal.)

And p a r t

~ c C, t h e n s i n c e and since

(ii)

~ is closed

~ is closed

is

included

in the definition

u n d e r f a n d g , K~ ~ w i l l

under h, K n ~

will

be

be n o w h e r e d e n s e

.

By

~ , fix now some ordinal ~ ~ C such that K n ~

= Sa .

2.5 Lemma Let q > ~ . with

Proof:

~ <*

•

Then (i) K n ~ <*

By i n d u c t i o n

T

and

on ~ . For

= Ko~

, and (ii) if ~ e q -

K there are

(~,~)*oK = ~ .

~=~

we u s e t h e c l o s u r e

of

~ u n d e r f a n d g . And

44

the induction step at limit stages is trivial. We p r o v e i t f o r dense i n that

~

So assume now t h e r e s u l t

@+c~ . By t h e i n d u c t i o n h y p o t h e s i s , S~ = K n ~

, so by c o n s t r u c t i o n ,

[~¢,~'3"~ K n ~

= ¢ and a l l

D~

a v o i d s S~ . Thus t h e r e a r e

the ordinals

~+n ( n o ~ ) l i e

holds for 9 .

i s c l o s e d and nowhere ~¢<* ~" i n ~ such i n (~,TO*. L e t

9,~,9',T'<~ arise from applying part (ii) of the induction hypothesis (i.e. part (ii) of the le,~a) to ~,:¢',

respectively.

Then (p,~')*(~ K ~ ~ = ~, so as o( i s c l o s e d

under the function k, (~,T')*n K = ¢. But since aii the ordinals

~+n (n~)

lie in

i*

(~,~) , ( i ) and ( i i ) f o l l o w i m m e d i a t e l y a t ~ +co. The p r o o f i s c o m p l e t e . Q

By ( i ) o f 2.~ now, we have K = ~ @ ( K n ~ ) = K n ~ . Hence K ~

. In particular,

K i s c o u n t a b l e . The p r o o f o f 2 . 2 i s now c o m p l e t e .

3. The i~hitehead Problem

Ass1~ing V = L, we prove that if G is an abeliau group for which ~t(G,Z)

= O,

then G is free. Just as in 1.2, we use the word "group" to mean "abelian group". We con,hence with some preliminaries from group theory and homological algebra.

Let G B

~G

~ ) B be a group e p i m o r p h i s m . A s p l i t

such t h a t

~o~

of 9

= 1B . Any such ~ must c l e a r l y be a monomorphism.

L e t G be an e x t e n s i o n o f A by B. tee s a y G i s a s p l i t t h e c a n o n i c a l p r o j e c t i o n G ~ > B has a s p l i t . of ~

, it

i s a homomorphism

extension just in case

In this case, if B 9, G is a split

i s e a s i l y s e e n t h a t G = A @ Im(~), so G " s p l i t s "

i n t o th~ two components

A and (up t o isomorphism) B.

I t f o l l o w s from t h e above t h a t E~t(G,Z) = 0 i f f e v e r y e x t e n s i o n o f Z by G is a split

e x t e n s i o n . We may a l s o c h a r a c t e r i s e

of short exact sequences.

the p r o p e r t y hkt(G,Z) = 0 in terms

45

A sequence (of morphisms) A1 ~ ) A 2

~

•

•

~.4)

An

~An+ 1

is exact if Im(gi) : Ker(gi+i) for aii i : l,...,n-l.

For example, the sequence 0 is exact iff ~

A short

be exact

canonical

s e q u e n c e i s one o f t h e f o r m

iff

)A

~, G ~ , B

G is an extension

~ 0.

of Im(x) by B , in which case

~ will

be t h e

exact sequence 0

if

C &- B course,

"0

~ 'B

projection.

A short

splits

>G

is onto.

0 It will

~

is one-to-one. And the sequence G

is exact iff ~

>A

there

is a

B

, A

V,A

~ • B

with~o~=

w i t h ~ ° ~ = 1 C. ( P r o v i n g t h i s this

is a split

a t once i m p l i e s extension

that

~ >C

-0

1A. T h i s w i l l

be t h e c a s e i f f

is an easy exercise

we h a v e j u s t

on e x a c t n e s s . )

there

is a

And o f

f o u n d a n o t h e r way o f s a y i n g t h a t

B

o f A b y C.

Thus, b~t(G,Z) = 0 iff 0

every short ,~

>H

exact sequence •

G

, 0

splits.

So far we have succeeded in expressing some remarkably trivial facts in a very impressive looking language. But we at once achieve the payoff, since we may now apply the following fundamental result from homological algebra.

46

3.1 Lemna Let O

~A

× .B

~0

~ *C

be exact. Let G be any group. Then there is an exact sequence 0

~Hom(C,G)

~ ~Hom(B,G)

¢ ~Hom(A,G) ~

,~t(C,G)

,~t(B,G) ,Ext(A,G)

where, in particular,

k*(~) = ~ ° ~

and

F*(~) = T ° P

• • 0 ,

- m

A proof of 3.1 may be found in almost any book on homological algebra. It is not difficult once one has a characterisation of hkt(H,K) in terms of equivalence classes of short exact sequences. But the proof is very tedious, and at any rate outside the scope of this book, so we leave it to the reader to investigate the matter himself.

We a r e now i n a p o s i t i o n

to start

our proof.

We s h a l l

call

a group G a

W-group if Ext(G,Z) = O. By 1.2.2, if G is free, then G is a W-group. It is our task to prove the converse. We first of all need three facts about W-sroups.

3.2

Lena Let G be a W-group. If fI <~ G, then I! is a W-group.

~'roof: Let 0

.H

~G

~-G/H

~0

be exact. By 3.1 there is an exact sequence ~t(G,~)

Since

G

is

a

•

~t(~,Z)

W~group,

~t(~,z)

=

~t(~,z)

= o,

o.

Hence, by e x a c t n e s s ,

a n d we a r e d o n e . O

-~

O.

47 3-3 Lemma If G is a W-group, then G is torsion free.

Proof: Suppose G has torsion. Then for some g ~ G , Then Let

Z



is finite, say of order n.

~ Z/nZ . But by 5.2, < g > will be a W-group. Hence Z/nZ is a W-group.

~ )Z/nZ

be the canonical projection. Clearly, Ker(~) ~ ~. Thus there is an

exact sequence

,Z

0

,Z

~ "~In~

,0.

Since Ext(Z/nZ,Z) = O, this sequence must split. Hence there is an embeddi~

~/~

, ~.

But Z/nZ has torsion and Z is torsion free, so this is impossible.

3.4 Lemma Let H ~ G . there is a

Proof:

H

Suppose that G is a W-group but that G/H is not a W-group. Then

~Z

which does not extend to a

G

~)Z.

• G/H

)0

Let 0

,H

~, G

be exact. By 3.I there is an exact sequence

~o~(G,Z) ~ >~om(H,Z) e >~t(G/~,Z) ~ ,~t(G,Z), where 1~(~) = ~ O l H =

~ restricted

t o H, f o r a l l

~ e Hom(G,Z).

Now, b y h y p o t h e s i s on G/H, E x t ( G / H , Z ) ¢ 0. 2Lnd by h y p o t h e s i s on G,

~t(G,z) = 0.

I~(I~) ¢ ~o~(~,Z).

~ence any

~euce, ~.(~)

G

~ .

= Ker(T) = ~t(G/~,Z) ¢

T h i s means t h a t

In other words,

~

there is a

0. Hence ~ - ~

Ker(e) ¢ Hom(H,Z). ~ ¢ 1~(r)

with

is not the restriction to H of any

G

for

~ ~Z. Thus

is as required.

Suppose now t h a t we a r e g i v e n a W-group, G, a n d we w i s h t o show t h a t G i s free.

How m i g h t we p r o c e e d ? One way i s t o t r y t o c o n s t r u c t ,

f o r G. A u s e f u l t o o l t h e n would be t h e f o l l o w i n g r e s u l t ,

explicitly,

a direct

a basis

consequence of

48

the proof of 1.2.1.

3.3 Lemm Let H ~ G. If G/H and H are both free, then G is free and any basis of H can be extended to a basis of G.

We might try to use 3.5 as follows. Suppose we can represent G as the union

o f an i n c r e a s i n g ,

continuous chain

( I n c r e a s i n ~ h e r e means t h a t limit ordinal ~
{G~]w<~

o f s u b g r o u p s , f o r some l i m i t o r d i n a l T •

v<~* i m p l i e s G# ~

, G~ =

~<

G , . C o n t i n u o u s means t h a t f o r any

G~ . ) Then we c o u l d t r y f i r s t

to find a basis

for GO, then extend it to a basis for GI, then extend that to a basis for G2, and so on. 3.6 provides us with a sufficient condition for this process to work:

3.6 h e r a Let G =

~ w ~ r G ~ , where T

is a limit ordinal and {G~]~< r

is an increasing,

continuous chain of subgroups of G. Suppose G O is free and that G~+I/G ~ all ~ < T

is free for

• Then G is free, and so is each group G/Gw.

}Toof: 2y i n d u c t i o n , we c o n s t r u c t an i n c r e a s i n g ( u n d e r i n c l u s i o n )

sequence {X~]~ T

o f s e t s such t h a t X~ i s a b a s i s f o r G~. I f ~ < T i s a l i m i t o r d i n a l we s h a l l have X~ =

~w< X~ . At s u c c e s s o r s t a g e s we o b t a i n X~+1 from X~ by 3 . 5 . C l e a r l y , i f we

set X = ~<

X~ , X w i l l be a b a s i s f o r G. Moreover, f o r any ~ < T , t h e s e t

{G~+x is a basis for G/G~

I x ~X-X~

}

.

If G is a torsion free group, we shall say that a subgroup H of G is pure if G/H is torsion free. Using 3.6, we obtain a useful condition for a group to be free for the case that the group concerned is countable.

3-7 Lemma ( i ~ o n t r y a g i n ' s C r i t e r i o n ) Let G be a countable, torsion free group. If every finitely generated

49

subgroup o f G i s c o n t a i n e d i n a f i n i t e l y

~roof: Let G = ~ g n I n ~ j

~ e n e r a t e d pure subgroup o f G, t h e n G i s f r e e .

. Define a sequence [ G }

of subgroups of G as f o l l o w s .

L e t GO = 0. Suppose Gn has b e e n d e f i n e d and i s f i n i t e l y is finitely

g e n e r a t e d . Then < G n , g n >

g e n e r a t e d , so by a s s u m p t i o n t h e r e i s a f i n i t e l y

g e n e r a t e d pure subgroup

o f G c o n t a i n i n g t h i s g r o u p . L e t Gn+1 be such a g r o u p .

Clearly, G =

~ n < ~ Gn" For each n , Gn i s p u r e i n G, so G/Gn i s t o r s i o n

f r e e . Hence Gn+I/G n i s t o r s i o n f r e e f o r e a c h n . But Gn+1 i s f i n i t e l y n , whence so i s Gn+l/Gn . But i t

is a standard result

g e n e r a t e d t o r s i o n f r e e group i s f r e e .

We s h a l l p r o v e our main r e s u l t

g e n e r a t e d , each

o f group t h e o r y t h a t any f i n i t e l y

So, a s GO i s f r e e ,

that (ass~ing

3,6 t e l l s

us t h a t G i s f r e e . ~

V = L) e v e r y W-group i s f r e e

by i n d u c t i o n on t h e o r d e r o f G. F o r t h e c a s e o f c o u n t a b l e groups we s h a l l u s e 3 . 7 . For t h e g e n e r a l c a s e we u s e a g e n e r a l i s a t i o n

o f 3 . 7 due t o S . U. Chase.

Suppose now that we are given a group B and we wish to construct, explicitly, an extension C of Z by B. As domain for our extension it is natural to take the

C a r t e s i a n p r o d u c t B ~ Z. We must t h e r e f o r e d e f i n e a group o p e r a t i o n on t h i s s e t . Since we want C to be an extension of Z we shall require ~

to embed ~ in C, where

: z ~ B xz is aefiueaby =(n) = (0,n). And since C should extend Z ~ , demand that ~ :3 × Z - ~ B

is a group homomorphism, where

we shah

~((b,n)) = b. This will

i m p l y t h a t Ker(~) = Im(~), s o t h e s e q u e n c e

0

~Z ~ ~ C ~ , B

~0

w i l l i n d e e d be e x a c t .

We s a y C i s a ( B , Z ) - ~ r o u p i f f : ( i ) C has domain B x Z ; (ii) (iii)

~ : ~ - ~ C i s a group morphism, where ~ : C - - ~ B i s a group morphism, where

I n t h e c o n t e x t o f ( B , Z ) - g r o u p s , t h e symbols ~ ,

~

~ ( n ) = (O,n) ; ~((b,n))

= b.

w i l l always have t h e above m e a n i n g s .

50

One example o f a ( B , Z ) - g r o u p i s t h e e x t e r n a l shall

see this

direct

su~ B • Z , h u t a s we

is not the only example.

3.8 Lemma L e t B1 he a W-group, B0<~ B1. L e t CO be a t o r s i o n there is a torsion

( B 0 , ~ ) - g r o u p. Then

f r e e ( B 1 , Z ) - g r o u p C1 such t h a t C 0 ~ C 1.

P r o o f : S i n c e B0 <~ B I , B0 i s a W-group. L e t T : BOb ~--~C 0

free

by T((h,n))

=

~(h)

B 0 3 - ~ CO

+ (0,n)

C0 - - ~ B 0. D e f i n e

split

. I% i s e a s i l y

checked that

group i s o m o r p h i s m . Moreover, ~ 0 o T ( ( b , n ) )

= h and T ( ( 0 , n ) )

= (0,n).

assmme, w i t h o u t l o s s o f g e n e r a l i t y ,

T is the identity

here,

Bo e Z . Then C 1 ffiB1 @ Z

that

T

is a

Hence we may

and t h a t

CO =

is as required.

5 - 9 Lemma L e t B1 be a W-group, B0 ~ B1. Suppose t h a t B1/B 0 i s n o t a W-group. L e t CO he a t o r s i o n

free

( B 0 , Z ) - g r o u p , and suppose t h a t

Then t h e r e i s a t o r s i o n extend to a split

f r e e ( B 1 , Z ) - g r o u p C1 s u c h t h a t CO <~ C1 and

~((b,n)) =

now have

0.

~0 does n o t

X = IB0 @ Z

and C O = B 0 e ~ , where we define

• ( b ) + (0,n). I n particular, s i n c e

~o(h) =

T((h,0)) , we shall

%(b) = ( h , 0 ) .

By 3 . ~ , l e t

B0 ~ ~ Z

have no e x t e n s i o n t o B1. S e t

r~

define a function

%~:C0----*C 1

by

"('((h,n)) = (b,n+~(h)). Clearly,

C 0 ~ xo B

o f C1 ~ - - ~ B 1.

Proof: As in 3.8 we may assume that by

B0 ~ . _ ~ CO s p l i t s

~

i s a g r o u p morphism. We p r o v e t h a t :

~1 = B1 ~ ~ ' and

51

(*)

B l U e C1 s u c h t h a t t h e restriction of ~ t o B0 is ~ o~ .

T h e r e i s no

In order to prove (*), we assume the contrary, i.e.that there were such a ~.

Define

9:CI---~ ~

by

O((b,n)) = n ,

and define

~:B 1

, Z by

~=

Then ~

and ~

~o~

are group morphisms and for

•

any beB 0 ,

~(b) = e o ~ ( b ) = O-rO~o(b)= o o~((b,O))= ~ence

~

O((b,~(b)))=

'~(b) •

extends ~ on B 1 , which is impossible. This proves (*).

Now define a function f:C1---+BI× Z by

(b,n)

, if b~B 0

(h,n-V(b))

, i f b E B O.

f((h,n)) N

Clearly, f is a bijection. Let C 1 b e t h e g r o u p on B l X Z i n d u c e d f r o m C1 b y f .

f ~

= IBo× ~ , C 0 ~

Suppose

to B 0 is

C 1 . Amd clearly, C 1 i s a t o r s i o n

BI ~--~--IC 1

~0 " Set

~ = f-I o ~I " Thus B 1

~(h) = £ - i o ~1(b ) = f-1 This contradicts

(*)

The f o l l o w i n g

direct

proofs

were to split

of this

result,

~ ~1

(Bl,~)-group.

, where the restriction

of

and for b~Bo,

~o(b) = f-1((h,O) ) = (b,~(b)) = ~((b,O)) = ~O~o(b)-

. H e n c e C1 i s

resldt

CI~-~'B1

free

Since

as required.

V1

was p r o v e d b y K. S t e i n

i n 1 9 5 1 . T h e r e a r e much m o r e

but the one given here has the advantage

that

it

will

52

generalise

to the uncountable

that,

whereas

the countable

carry

through

the proof

case.

The o n l y d i f f e r e n c e

case is provable

in the uncountable

b e t w e e n t h e two c a s e s

i n ZFC, we n e e d V = L i n o r d e r

is

to

case.

3.10 Theorem Every countable

Proof:

W-group is

Let G be a countable

free.

W - g r o u p . By 3.3, G i s t o r s i o n

free.

So b y 5 . 7 ,

in order

to prove that G is free i% suffices to prove that G satisfies Pontryagin's criterion. We a s s u m e it d o e s n o t and w o r k f o r a c o n t a d i c t i o n .

L e t B0 b e a f i n i t e l y finitely

generated

generated

subgroup of G which is not

B is a (countable)

s u b g r o u p o f G. And b y i t s therefore,

be finitely

union of a strictly

increasing

i n g w i t h B0 . N o t i c e

that

group.

By i n d u c t i o n

-

thereby

set

of generators

o n n we c o n s t r u c t

The i d e a

contadicting

definition,

generated.

c h a i n {B n }

e a c h Cn i s a t o r s i o n

(B,Z)-gronp.

very

free

of finitely

the fact

that

generated

a strictly

subgroups,

of B implies

increasing

that

commenc-

B/B 0 i s a t o r s i o n

s e q u e n c e { Cn }

a n d C ffi U n <

the Cn'S so that

by 3.2,

B as a

f o r B0 .

(Bn,Z)-group

is to construct

B i s p u r e i n G. S i n c e

H e n c e we c a n r e p r e s e n t

a s B 0 ~ B, t h e d e f i n i t i o n

Let S be a finite

such that

in any

pure subgroup of 6. Let

B = { g ~ G I ( 3 k a ~ ) ( k - O & k.geBo)}

B0 ~ B, B c a n n o t ,

contained

C

of groups

Cn i s a t o r s i o n

n ~B

does not

free

split,

B ~ G i s a W - g r o u p . lee n e e d t o know t h e

following fact.

(*)

If

C is a torsion

the values

Proof of (*): determined

it

takes

Let b~B.

by the values

o n S. B u t C i s

free

torsion

group and

B 9-~

C

then

is uniquely

determined

by

on S.

Pick n>O of free,

~

w i t h n . b e B 0. S i n c e S g e n e r a t e s

on S. T h u s n . ~ ( b )

so if

n.c = n.~(b),

is

determined

then c = ~(b)

B0 ,

e(n.b)

by the values . Hence

~(b)

is of is

53

uniquely determined by the values ~ takes on S.

QED(*).

Fix now some enumeration { gn ~ of all functions

g:S--~S x Z with

~.g = IS.

Set C O = B 0 ~ Z . Suppose now that Cn has been defined. If gn extends to a split

~

of

to a split

Cn X ~ ) B n , l e t of

Cn+ 1 b e a n e x t e n s i o n

Cn+l~--~+l.

(Since ~+l/~

o f Cn s u c h t h a t

is a torsion

s o b y 3 . 9 s u c h a Cn+ 1 c a n b e f o u n d . N o t i c e a l s o

that

group,

by ( * ) ,

if

~ does not extend it

is not a W-~onp,

there

is such a ~ ,

it will be unique.) Otherwise use 3.8 to let Cn+ 1 be any (Bn+l,~)-greup extending Cn. That completes the construction.

We finish by showing that that

B

Let

~ ~C

C

,B does not split.

were to split ~ . For some n,

Suppose, on the contrary,

gn must be the restriction of ~ to S.

~n be the restriction of ~ to B n. Then ~ n splits

Cn )rn •B n . But ~n is the

unique extension of gn to a split of ~n" Hence, by construction of Cn+l, not extend to a split of ~n+l"

Cn+l~-~-~JBn+l . But this is absurd, since

~)n does

~n+l splits

The proof is complete. ~3

Assuming V = L, we shall prove by induction on the infinite cardinal ~ that i f G i s a W-group of o r d e r initial

case

~

, then G is free.

~ -- ~ O" We c o n s i d e r

work e q u a l l y w e l l f o r t h e i n d u c t i o n bother

to look at

cardinal all that

less

has nothing

less

than

K will

in

cardinal

leave the case where cardinal),

t o do w i t h V = L. R a t h e r , suffices,

together

K , s o we s h a l l

~ is a singular

we h a v e p r o v e d t h e r e s u l t at

~ . It

turns

with the fact

that

any set

paper [Sh] .

for out

of

each of which has cardinality

c a s e t h u s h a s n o t h i n g t o do w i t h V = L, we s h a l l

Shelah's

the

GCIt ( w h i c h h o l d s b e c a u s e we a r e

be a u n i o n o f f e w e r t h a n ~ s e t s

~ . Since this

A proof appears

step at any regular

a limit

establishes

~ 1 " The a r g u m e n t we u s e w i l l

t h a n ~ , a n d we w i s h t o p r o v e t h e r e s u l t

a s s u m i n g V -- L f o r o u r p r o o f ) cardinality

n e x t t h e c a s e K --

This will

(and hence necessarily

cardinals this

these cases.

The a b o v e r e s u l t

omit it.

not

54

Let us call a group G ~-free

if every subgroup of G of order less than ~ is

free. By 3.10 and 3.2, we have at once:

3-11 L e n a L~ery W-group i s

~l-free.

[]

Now, a g r o u p G w i l l be t o r s i o n

free iff

every finitely

generated subgroup is

free. Viewed in this light, we see that the property of a group being

~-free is a

natural generalisation of the property of being torsion free. In particular, if G is K-free, it is torsion free. Accordingly, we define a subgroup H of a to be K-pure if G/H is K-free.

~-free group G

Generalising the Pontryagin criterion to the

~i

case we now have:

Chase's Condition:

G is an

~l-free

of G is contained in a countable

The f o l l o w i n g r e s u l t

group and every countable subgroup

~ l - p u r e subgroup of G.

is trivial.

3 . 1 2 Lemme L e t G b e a group of o r d e r the union of an increasing, w i t h AO = 0 a n d , f o r a l l

~1" Then G s a t i s f i e s

continuous sequence

~ < w 1 , A~+ 1 i s

Chase's condition iff

~ A~}~<~I

~l-pure

G is

of countable free groups

in G .

Now, unlike in the case of the Pontryagin criterion, we cannot prove that any group which satisfies Chase's condition is free. However, we have a partial result in this direction:

3 . 1 3 Le-n.~ L e t G be a g r o u p o f o r d e r ~A~]~I

~1 w h i c h s a t i s f i e s

be a s i n 3 . 1 2 . Then G i s f r e e i f f

C -= "~1 s u c h t h a t Ay

is

Sl-pure

Chase's condition,

and let

t h e r e i s a c l o s e d a n d unbounded s e t

in G for all

~C.

55

Proof: (~--) L e t ~ } ~ < ~

be the monotone enumeration of such a set C. Set ~p

For

each 9

,

A9

= A~

,

each ~ .

is ~l-pure in G, so G / ~

3.6 (applied to the sequence [ ~ } ~ < ~

is ~l-free, so A~+I/A~

is free. So by

) we see that G is free.

(--~) Let X be a basis for G. By an easy induction we can define a closed and unbounded set C _c ~I and an increasing sequence that for each ~

C,

Xv

{ Xp~C

of subsets of X such

is a basis of A~ . Then, for each ~ ff C, the set

is a basis for G/A~ . Hence G/Ap is free for all ~ e C, and hence is certainly free.

Thus e a c h A~, ~ ~ C, i s

~l-

~l-Pure.

In order to complete our proof that ( a s s u m i n g V = L ) , we n e e d a g e n e r a l i s a t i o n

e v e r y W-group of o r d e r

~l is free

of the combinatorial principle

~

.

By a slight generalisation of the argument used to es±~blish ~ from the a s s u m p t i o n V = L, one may p r o v e t h e f o l l o w i n g ( a p p a r e n t l y )

Let K be an uncountable regular cardinal. Let E ~

stronger result.

be stationary.

Then there is a sequence ~Sa]a~ E such that S ~ ~ for each ~ e E, and whenever X ~ all

~

This result itself,

there is a stationary set E ' ~ E such that X n ~ = S~ for

I

~

E

.

i s a l s o due t o J o n s o n . S h e l a h h a s %hough i t c l e a r l y

Either by an entirely above r e s u l t ,

implies

proved that it

does n o t f o l l o w from

~ .

analogous argument, or else as a corollary

we may o b t a i n t h e f o l l o w i n g c o n s e q u e n c e of V = L, t a i l o r

present needs. (We give only the

of the

made f o r o u r

~i case, since this is all we need now.)

56

3.1~ Lemma Assume V = L. Let B be the union of a strictly increasing, continuous sequence {B~}v<,I

of countable sets. Let E ~ w I be stationary. Then there is a sequence

{g~] 9 e E such that

gg:B~---*B~ Z

that h[B~] g B~ x ~ for all to B 9

for all 9 ~ E, and whenever

9 g E, there is a ~

h:B--*B x Z is such

E such that the restriction of h

is g~ .

3.15 Lemma Assume V = L. Let B be a W-group of order

~1" Let { B ~ ; < ~, be a strictly

increasing, continuous sequence of countable (free) subgroups of B with union B. Then there is a closed unbounded set C ~ ~ 1 such that B~+I/B p ~

is free for all

C.

P r o o f : Suppose n o t . Then t h e s e t E is stationary in ~1"

Let

= ~e

~1 I B~+I/B ~ i s n e t f r e e }

{ g ~ ] ~ c E be as in 3.14.

We d e f i n e , by i n d u c t i o n ,

a strictly

of t o r s i o n f r e e groups such t h a t each C# torsion free (B,Z)-group, and

C

~ ,B

increasing,

c o n t i n u o u s sequence { C ~ J ~ %

i s a ( B y , Z ) - g r c u p , C = U~<%C# i s a does not split, thereby contradicting the

fact that B is a W-group.

Set C O = B 0 e ~ . Suppose now that we have defined C~ is a limit ordinal we set C 9

= ~T<

for a l l T < p

. If

CT . If ~ = T + I , there are two cases to

consider.

Suppose first that T e E let C~ be an extension of C x (Since T e E, B~/B v B~

and

B~---~Br

× Z

splits

C

x~ ~Bv . Then we

such that gT does not extend to a split of C~ ~

is not free, so by 3.10 it is not a W-group. ?md as B ~ < s

is a W-group. So 3.9 guarantees the existence of such a

Otherwise we use 3.8 to let C~

C~ .)

be any extension of C~.

• By . B,

57

Clearly, to split

C

C = U~<~C~

~ ~ B.

o f 9 t o B~ . ( S i n c e

is a torsion free (B,Z)-group.

Then we could find a r ~ E

9 splits

X

3.16

gr

of

is absurd, since the restriction

CT+lX~--X~B+I of 9

to B+I

B

e ~C

were

equal to the restriction

, ~ o 9 = 1B, so 9 [ B ~ ] ~ B ~ x ~

Thus C + I was d e f i n e d so t h a t no s p l i t But t h i s

with

Suppose

for all

agrees with splits

9<~1.

)

~ on B r •

~+1"

Lemma Assume V = L. If G is a W-group of order

~i' then G satisfies Chase's

condition.

P r o o f : Suppose n o t .

Now, G i s

81-free

by 3.11,

so i t must be t h e c a s e t h e n %hat

t h e r e i s a c o u n t a b l e s u b g r o u p B0 o f G s u c h t h a t w h e n e v e r C 4 G B0 ~ C, t h e n C i s n o t

~l-pure.

is countable with

P i c k s u c h a B0. Then, w h e n e v e r C 4 G i s c o u n t a b l e

and B0 g C, we c a n f i n d a c o u n t a b l e C ' ~

G w i t h C ~ C'

So b y i n d u c t i o n we c a n d e f i n e a s t r i c t l y

increasing,

such that C'/C is not free.

continuous sequence [B~}~<~I

of c o u n t a b l e s u b g r o u p s of G, c o - , - e n c i n g w i t h B0, s u c h t h a t B t + I / B 9 each ~<~1. order

S e t B = D 9<

is not free for

B~ . S i n c e B ~ G, B i s a W-group. And c l e a r l y ,

B has

~ 1" Thus b y 3 . 1 5 %here i s a c l o s e d unbounded s e t C ~ ~ 1 s u c h t h a t Bg+I/B ~

is free for all

~ ~ C. But t h i s

i s a b s u r d s i n c e f o r no ~

i s Bg+I/B 9 f r e e .

U

At last we may prove:

3 . 1 7 Theorem Assume V = L. I f G i s a W-group o f o r d e r

P r o o f : By 3 . 1 6 a n d 3 . 1 2 we may r e p r e s e n t continuous sequence is

{A~]9<w1

~1' then G is free.

G as t h e u n i o n of a s t r i c t l y

increasing,

of c o u n t a b l e f r e e g r o u p s s u c h t h a t A0 = 0 and A~+1

El-Pure in G for each ~.

(*) For ali ~ < e l '

P r o o f of ( * ) : I f A9

%

is

is

81-Pure in G

~l-pure

iff

l~+I/i9 is free.

i n G, t h e n G/A~ i s

~l-free,

so A~+I/A9 i s f r e e .

58 Conversely, Suppose A~+I/A ~ is free. Let 7>~ . By the Third Isomorphism Theorem of elementary group theory,

(41A ~)I(A~+ilA~) ~ ArlA~+ I

•

Now, A~+ i is ~l-pure in G, so AT/A~+ 1 is free. Hence, as A~+I/A ~ is also free, 3.5 tells us that Az/A ~ Hence A~

is

is free. But T > w

~l-Pure in G .

is arbitrary here. Hence G/A~

is

~l-free.

QED(*).

By 3.15 now, t h e r e i s a c l o s e d and unbounded s e t C ~ m l such t h a t Ag+I/A 9 is free for all free.

~ e

C. By ( * ) , A v

is

~l-pure

for all

9 ~ C. ~enee by 3 . 1 3 , G i s

Q

~. C o l l e c t i o n w i s e H a u s d o r f f Spaces

I t w i l l be sho~rn t h a t i f we assume V = L, e v e r y f i r s t

c o m l t a b l e T~ s p a c e i s

c o l l e c t i o n w i s e H a u s d o r f f . The p r o o f i s by i n d u c t i o n on t h e c a r d i n a l i t y

of the discrete

set involved.

L e t (X,~) be a t o p o l o g i c a l s p a c e , ~ an i n f i n i t e collectionwise Hausdorff

(K-C~H) i f e v e r y d i s c r e t e

cardinal.

We s a y X i s

K ....

s u b s e t o f X of c a r d i n a l i t y

<

has a s e p a r a t i o n .

The p r o o f t h a t ,

a s s u m i n g V = L, e v e r y f i r s t

w i s e H a n s d o r f f p r o c e e d s by e s t a b l i s h i n g a =

~0 t h e r e s u l t

c o u n t a b l e T~ s p a c e i s c o l l e c t i o n -

~-CWH f o r a l l

~

by i n d u c t i o n on K - For

i s p r o v a b l e i n ZFC a l o n e . The i n d u c t i o n s t e p a t

t h e i n d u c t i o n s t e p a t any u n c o u n t a b l e r e g u l a r c a r d i n a l .

~1 i s t y p i c a l

And t h e i n d u c t i o n s t e p a t

s i n g u l a r c a r d i n a l s r e q u i r e s o n l y t h e GCH, and d e p e n d s upon t h e s i n g u l a r i t y So, j u s t as w i t h t h e WVaitehead P r o b l e m , we g i v e o n l y t h e c a s e s ~ =

~.i Theorem Let X be any T 3 space. Then X is ~0-CWH.

of

~ 0 and

of ~ . K -- }~1"

59

Proof:

Let Y ~ X be discrete, IYl = ~0" Notice that any discrete set is closed,

and that any subset of a discrete set is discrete, and hence closed.

Enumerate Y as ~yn} • For each n, set Y

ffi ~Yn+l,Yn+2, • . •

Now, Y0 S Y0 and Y0 is a closed subset of X, so by r e g u l a r i t y disjoint

there are

open s e t s U0,¥ 0 such t h a t Y0 EU0 and Y0 ¢ V 0. P r o c e e d i n g i n d u c t i v e l y ,

each n we can gind disjoint open sets Un+l,Vn+ 1 such that Y n + l ~ U + l

and

for

, Y+I~Vn+I

,

Vn÷l

The sets U ,

n = 0,1,2, ...

clearly separate Y.

We turn next to the case ~ = that if V = L there is a sequence whenever f : ~ l

~i" Here we need V = L. Now, Jensen has proved

~f~<el

such that f~: ~ - - ~

for each ~ , and

~ ~ there is a stationary set E ~ ~1 such that

ft~ = f~ for all

e E. (The symbol f~A denotes the restriction of f to A for any set A ~ dom(f).) Indeed, this is equivalent to the principle

$~

. For our present problem we need

a somewhat stronger result:

%.2 Lemma Assume V = L. L e t AS b e , f o r each f : ~ 1 - - ~ co , a s t a t i o n a r y such t h a t w h e n e v e r f : o ) 1

>co

and g : ~ l - - - ~ c 6

A ( ~ ( c ~ + l ) . Then t h e r e i s a s e q u e n c e [ f ~ < ~ l whenever f : ~ f

~ ~

, the set

and f ~

= g~

s u b s e t of

, t h e n Af(~(c~+l) =

such t h a t f~ : ~ - - - , c o + l f o r a l l

{o( E oJ 1 I f P ~

= f~

is a stationary

c~ , and

s u b s e t of Af.~S

U s i n g ~ . 2 , we p r o v e :

%-3 Theorem Assume V = L. L e t X be any f i r s t

o~1,

c o u n t a b l e T% s p a c e . Then X i s

~I-C~t.

60

Proof:

Assume the theorem is false. Let Y be a discrete subset of X of cardinality

~i having no separation. We may without loss of generality assume that X is an ordinal and that Y is the ordinal

~i o For each x ~ X, let {Nn(x)}

enumerate a countable

base for the neighborhood system of x.

If f:~--*o~ , where k ~< * I ' and if ~ < ) ~ ,

we set

W(f,~) = U~< Nf(~)(~).

If H-~o~ 1 , f:~

a~, ~ ~< OOl, and c~ ~ ~ , we set

~(n,f,~) =

U , ~ H ~ ~ Nf(~)(~).

For f:~l---~ now, set

Af = {~ ~ ~I I Notice that

if f : ~ 1

ft~

' ~

and g : ~ 1 - - ~

A f n ( = + 1) = Ag~ ( ~+ 1) .

implies

= g~

CLAIM: F o r e a c h f:~l--*~,

, then

Af i s s t a t i o n a r y .

P r o o f o f c l a i m : Suppose n o t . L e t C _~ ~ 1 be a c l o s e d and unbounded s e t w h i c h i s disjoint

from Af (some f ) . o( ~

Let {~<%

C

Thus, implies

c(

be t h e monotone e n u m e r a t i o n of C. By ~ . 1 , X i s

each

~) < a~1 we c a n f i n d a f a m i l y

--av,

such t h a t x ~ ~

Moreover, s i n c e

( c l o s u r e W ( f , ~ ) ) N 0~1 =

~

. We ~ay c l e a r l y

(closure W(f,~))

of p a i r w i s e d i s j o i n t assume t h a t ~

N ~1 = c~

f o r each

.

~0-ClfH, so f o r

open s e t s Ux, x ~ c ~ + 1 -

Nf(x)(X) , each x~

~1"

~ < o~1, we may assume a l s o

61

that for x > ~

,

U x f~ closure W(f,c(#) = @ . It is in~nediate that

separation of Y, which is a contradiction.

By t h e c l a i m , l e t { f ~ * < ~ l

~.,~

~

(iv) ~ _~ ~

be a s i n ~ . 2 now.

&

K~--

~I

el' so that:

K~ ;

=~;

~ ~

.

In order to carry out the definition to

is a

co I ;

(ii) ~'< ~ - - ~ H ~ -c H&

(i~i)

3"~

~FO(CLAIM)

By induction, we define sets Hr, K , ~ <

(i) Hr, K v _c

U

defined by letting

we n e e d a p a r t i a l

function

from ~ 1

9 (~) be the least element of

(eios~

~(%,~))

n ( ~I - ~) '

if there is such an element, with ~ (~) undefined otherwise.

To cow,hence we set H 0 = X 0 = @ . And if

K V =

Suppose now that

U~
is a limit ordinal we set

.

~ ' = ~ + 1 and we have defined H , K

. There are two cases to

consider.

Case I.

f:~--~o~ , ~(~) is defined, and for all ~ < c ~

~(~) <~

, if 9 (~) is defined then

•

~.Tow, ~ c _ ~ u

~,

so

~(~,~) _- ~(~,%,~)

~ ~(~,%,~)

.

Henc e ,

closure W(f~,~) = closure ~([L,f~,~) 0 closure W(K~,f~,~).

62

Since

v(~)

i s defined, either

f,,~) (or both). ~f

~(~) ~ c l o s u r e W(~,f~,~) or

(~) e c~Los~, w(Z~,

~ (~) e c~osure W(~,,¢,,~), we sst

H

= I~u[,r-K

r].

On the other hand, if O(v() ~ c l o s u r e W(H~,f~,~) ( i n which c a s e we must hare ~ ( ~ ) ~

clos~e W(K~,f~,~) ), we set Hr Kv

Case2. Otherwise. H

Set

= H~

;

That completes the definition. The only point we need to check is that

H r ~ K~

= ~

for all ~

. Now, the only thing which could conceivably go wrong

here is that for distinct ~ ,

into H + 1

and

we must have

~(~)< J

•

in Case I,

~)(~) = ~)(~') and we put (say)

~ (c~)

~)(~t) into K % 1 . But look, if ~ < o~/ , then for ~ to be in Case 1

( V ~ < ~ ) ~ if v(~) is defined then ~ ( ~ ) < oc'] , so in particular

B~t by definition, ~(~') >~ C( /

, so

we ~ u s t

have

~(~) ~ ~(~')

#

In

o t h e r w o r d s , n o t h i n g can go wrong.

Set

= 13 Clearly,

H and K a r e d i s j o i n t .

~

,

K=

B e i n g s u b s e t s of

U~X~. ~1'

they are discrete,

and hence

63

closed. By normalitywe

P i c k some f u n c t i o n

can find disjoint open sets U, V such that H K U and K K V.

f: ~1

,~

so that

E Z

Since H u K =

ml

' this

~

~f(~)(~)

definition

is

~

sound.

V .

Let

E= { ~ e ~ l I ft~ = f ~ } .

By assumption, E is a stationary subset of

Let

C = {o(~ coI

Af .

I oc is a limit ordinal & (V~<~)[ if ~)(~) is defined then

9 (~)< c( ]~.

It is not hard to prove that C is closed and unbounded in

point

cx e E • C . S i n c e

c~ e A f ,

(closure

(olosure W ( & , ~ ) ) n ~1 ~ ~ " Henoe

U~l. Hence there is a

W(f,c~)) N ~0 1 ~

~

9 (~) ~

(.closure W(H~,f~,~)) ~ K + l

(closure W(K~,f~,~)) n H + 1 • Hence either

else

~ e E,

~ (~) i s defined. Thus, rising now

that ~ ~ C, we see that Case 1 applied in the definition of ~ + l

means that either

. So a s

or

else

the f a c t

and K + I. This

9 (~) 6

9 (~) 6 (.closure W(H,f,~I)) ~ K

or

?(c~) £ (closure W(K,f,051) ) N H . Bug N(H,f,el) c U and W(K,f,el) c V .

Hence either

@(c~) E (closure U ) N K

and K _c V , U ~ V

or else

9 (o~) E (closure V ) ~ H °

= ~ , and U and V are open. Thus (closure U ) ~ K

= ~ . We have thus arrived at a contradiction.

But H_C U

-- (closure V)(IH =

The theorem is proved. [3

84

5. Further

Remarks

The t h r e e combinatorial to carry

enumeration

through

are other

applications

construction

such as

is used to help carry of just

mentioned

we h a v e made u s e o f ,

our proofs

might lead

of set

theory.)

a detailed

highly

of infinite

t h e GCH s o l v e s

And a f i n a l

this

remark concerning

to the Souslin

result

the ~hitehead

Axiom o f C o n s t r u c t i b i l i t y

Our a c c o u n t

i s due t o W. F l e i s s n e r °

for

taking

a solution

but rather

outside

set

Souslin's

the origin

of the three of each of our

GCH a s a n a x i o m

more information

This is not the case. in Appendix II,

Problem;

Problem nor the C~

theory.

we g i v e

of the problems needs

that

outlined

of our three

the major breakthrough

i s due t o K. K u n e n ° The s o l u t i o n S. Shelah.

, the simplest

leading

The v e r s i o n

about Using

R. B.

S. Shelah has shown

problem.

results.

P r o b l e m u s i n g V = L i s due t o R. B. J e n s e n ,

which constituted

chapter

GCH w o u l d b e e n o u g h t o s o l v e

of the type of procedure

neither

~

reason

a set,

solve

a simple

every case V = L

In the next

numbers is required.

t h e GCH d o e s n o t

down. T h e r e

f r o m GCH. B u t t h e n a t u r e

that

i s no i n t u i t i v e

cardinal

to isolate

enabled us

manner.

does not follow to suspect

break

in almost

induction.

the principle

of what constitutes

complex varieties

solution

that

but

I n w h i c h c a s e i t w o u l d n o t be t h a t

Jensen has shown that that

the reader

(Although there

analysis

the arithmetic

from the proof,

made u s e o f

These principles

is not possible

w h y V ffi L w o r k s i n t h i s

principles

problems.

~

kind.

given all

which would normally

through a complicated

We h a v e a l r e a d y

three

of a similar

of V = L where it

principle

some i n d i c a t i o n

principles

an inductive

applications

combinatorial

o f V = L w h i c h we h a v e j u s t

The o r i g i n a l

and indeed it

to applications

is

of the

o f t h e p r o o f we g a v e h e r e

of the Whitehead Problem using V = L is due to

owes much t o a n a c c o u n t b y P . F k l o f .

The r e s u l t

on CWH s p a c e s

Chapter V

A P R O B L ~ IN MFASURE Tt~-DRY

This chapter prove a result

is

intended

to serve

in measure theory.

investigate

the notion

and thereby

we s h a l l

In order

1..Extensions

of a constructible achieve

additive

function

~

from

~

X. ( T h i s d e f i n i t i o n

Our s t a r t i n g

a little

have need to

more thoroughly

than before,

t o g i v e some i n d i c a t i o n

o f how t h e

point

that

such an extension

there

in

interval

why d i d we n o t status

[0,1 ~ , so there

is the following to a measure

~

to construct

[0,1 3 which vanishes the value

1 on t h e w h o l e

of course,

but includes

i s no g r e a t

Is there

on all

let

us remark that

? The

in Lebesgue measurability

u s e o f t h e Axiom o f C h o i c e .

the assumption

o f t h e Axiom o f C h o i c e i s s t r i c t l y

a set

which is not Lebesgue measurable.

of reals

problem in Chapter

t h e same a s t h e p r o b l e m s

for

of [0,1]

of

of [0,1 3 shows that

considered

the ether

problems.

It

is

So

I ? Well, because

there,

t h e p r o b l e m we n e e d t o d e v e l o p t h e t h e o r y

to a degree not required

an extension

subsets

subset

loss.)

invariant.

the measure extension

is not quite

to investigate

defined

example of a necessary that

include

question.

is a non Lebesgue measurable

we h a v e a f u r t h e r

in order

interval

, and which takes

cannot be translation

known f r o m w o r k o f ~ necessary

~

unit

o f a s e t X we m e a n a c o u n t a b l y

L e b e s g u e m e a s u r e on R ,

W h i l s t we a r e o n t h e s u b j e c t , on [0,1]

, of subsets

the closed

excludes

Lebesgue measure on [0,I ] proof

into

a n d on a n y p o i n t s

L e b e s g u e m e a s u r e on t h e u n i t

order

o u r p r o o f we s h a l l

of Lebesgue Measure.

on the empty set

its

set

our second goal --

By a m e a s u r e on a ~ - a l g e b r a , ~

"usnal"

to obtain

On t h e o n e h a n d we s h a l l

Constructibilityworks.

Axiom o f

space,

two p u r p o s e s .

and moreover,

of constructibility

in

86

A classical

result

o f Ulam t e l l s

extension

of Lebesgue measure on [0,1]

interval.

(We p r o v e t h i s

answers our basic uncountable

fact

question.

But it

set X for which there

D e p e n d i n g on y o u r v i e w p o i n t , measure theory, interest~

below.)

It

this

question interest

raises ~

on a l l

subsets

on a l l

in measure theory. is

of this certainly any

of X ?

importance

But either

impossible

is no

Is there

subsets

of fundamental

way i t

in Zermelo-Fraenkel

in is of set

(We shall see why in a little while.) However,

it is certainly possible that one could obtain a negative assumptions.

then there

a new p r o b l e m .

defined

is either

solution

theory with just the ~xiom of Choice.

the C~holds,

H e n c e t h e Axiom o f C o n s t r u c t i b i l i t y

immediately

a positive

if

to a measure defined

is a measure

or of peripheral

i s known t h a t

us that

solution from these

To date no one has been able to do this. What we can (and shall) do is

obtain a negative

solution from the .Axiom of Constructibility.

2. The Measure Problem

We investigate measure

the problem:

does there exist an uncountable

set X and a

~ defined on all subsets of X ?

First let us observe that it is enough to consider cardinal numbers ~ instead of arbitrary sets X. For, given any pair X , ~ f:X~-,K, of

K

then f induces from

. By

a

K

k

we shall henceforth mean a measure defined on

~

is

cardinal

~-additive

sets of measure zero, then

any measure is

Suppose

~ = IXi and if

.

cardinal. We shall say that

definition,

K

~ b e a measure on an uncountable

ion of fewer than

if we let

in the obvious way a measure defined on all subsets

measure o_~n a cardinal

all subsets of

Let

~

as required,

K . Let

~

be any uncountable

iff, whenever Av, ~ < ~ , is a collect~ A

has measure zero. Thus, by

~l-additive.

~ is a measure on the uncountable

cardinal

K

. There is clearly a

67

largest cardinal since

K

X such that

p is

X-additive.

is the union of its one-element

is a set A c K

subsets, )x ~< K . By definition of ~ , there

of positive measure which is the union of

measure zero:

A =

D e f i n e a map f : A - ~

~.A~

)~

disjoint sets of

.

X by s e t t i n g f(a)

For B ~ k

By our remark above, ~ ~ '~1" And

=

iff

a e A~ .

now, s e t

~(B) = It is easily seen that on a cardinal

K

o is a

M f-1~B~ )

A-additive measure on X • Hence, defining a measure

to be stron~ if it is

K-additive,

we see that we have proved

the following lemma.

2.1 Lemma Suppose that for some uncountable all subsets of

cardinal ~ there is a measure defined on

K • Then there is an uncountable

strong measure defined on all subsets of

cardinal ~

k . u

Thus, in order to show that on no uncountable (defined on all sets),

~ such that there is a

cardinal

is there a measure

it suffices to show that on no uncountable

cardinal

is there

a strong measure. As a first step we prove that if there is a strong measure on the uncountable

cardinal

~ , then

~ is an inaccessible

So what is an inaccessible

cardinal

cardinal.

? Well, we know already what it means

for a cardinal to be regular. And we also know what it means for a cardinal to be a limit cardinal.

The first infinite cardinal, S0, is both regular and a limit cardinal.

But are there any other cardinals which are both regular and a limit cardinal is a consequence

? It

of the GSdel Incompleteness

Theorem that a positive answer to this

question is not possible. But considerations

outside the scope of this book lead to

the conclusion that a negative answer is extremely unlikely the possibility that Zermelo-Fraenkel

(indeed, as unlikely as

Set Theory is inconsistent).

It is also known

that if a negative solution were to be obtained assuming the Axiom of Constructibility,

68 then a negative just

answer would already

follow from Zermelo-Fraenkel Set Theory with

t h e Axiom o f C h o i c e . R o u g h l y s p e a k i n g ,

an inaccessible

cardinal

and certainly much bigger than ~I' ~2'''" ~n'"'' ~ " ' "

F i x ~ a s t r o n g m e a s u r e on a n u n c o u n t a b l e

cardinal

~'""

is

"very large",

etc.

K f r o m now o n .

2.2 Lemma If f < ~ Proof:

Since

~

, then ~ I ~ < ~ }

has measure zero.

is strong° D

2 . 3 Lemma K is regular.

Proof:

Suppose not.

By 2 . 2 , tion.

Then t h e r e

~(Kp) = 0

for all

i s a ~ < K and o r d i n a l s

~

. So a s

~

is strong,

~v< ~ , ~<~

we g e t

~ such that

~ ( K ) = 0, a c o n t r a d i c -

[]

The p r o o f t h a t

We s a y t h a t f:B--* ~

our measure

Let f:A--~ < K

such that

bounded if

for

incompressible g:B-, K

f(})<

f is constant

assume t h a t

will

for all

~ E B, t h e r e

As a f i r s t

step

longer.

some a u x i l i a r y

B of positive

We s a y t h a t

if A has positive

there

is a

f is nowhere

h a s m e a s u r e z e r o . We s a y t h a t

f i s n o w h e r e b o u n d e d a n d w h e n e v e r B~A h a s p o s i t i v e for all

measure

definitions.

f is almost bounded if

has measure zero.

{~AIf(~)_<~}

i s a B' c

measure and

i n o u r p r o o f we show t h a t we may

~ i s n o r m a l . We r e q u i r e

{~6 AIf(~)>~

take a little

whenever B a K has positive

, w h e r e A ~ ~ . We s a y t h a t

and g(~) < f ( ~ )

only important

~

on B ' .

each ~
cardinal

~t i s n o r m a l i f f ,

is such that

such that

K is a limit

f is

measure and

~ ~ B, t h e n g i s a l m o s t b o u n d e d . T h e s e c o n c e p t s a r e measure, of course,

for

if A has measure zero,

then

69

any map f is simultaneously almost bounded, nowhere bounded, and incompressible.

2./I Lenmm Let f:A--~K

be nowhere bounded. Then we can write A as the disjoint union

of sets B and C such that:

(1) frB i s i n c o m p r e s s i b l e ; (2) t h e r e iS a g:C-* K such that g(~)< f(~) for all ~ e C and such that g is nowhere bounded.

Proof: Using Zorn's lenmm we obtain a maximal family

=

{(Ci,gi) I i~If }

such t h a t : (1) Ci-CA has p o s i t i v e measure; (2) g i : C i - - ~ K i s nowhere bounded; (3) t e Ci ~ (~) i f i , j

gi(~)
elements of K, t h e n C. ~ C. = ~o 1 j

By (1) and ( ~ ) , K must be c o u n t a b l e . ( I f K were u n c o u n t a b l e , t h e n f o r some p o s i t i v e n ~ b e r n , Ci would have measure g r e a t e r t h a n 1 / n f o r u n c o u n t a h l y many i , c o n t r a r y to the measure b e i n g f i n i t e . )

c=

Ui~Kci

We s e t

,

~=

Ui~i.

Since l~ is countable, g is nowhere bounded. And clearly~ B

=

A

-

~ ~ C-~ g(1)< f(~). Set

C. If B has measure zero, f~B is trivially incompressible. If B has positive

measure, the maximality of ~

implies that fiB is incompressible.

2.5 Len~a There is an incompressible function f: ~ -~ ~.

Proof: We define a sequence of sets {A n} and functions hn:An -~ ~ To commence we set A 0 = K

by induction on n.

and let h 0 be the identity function on z . (By 2.2, h 0 is

nowhere bounded. ) Suppose now that n = k + I and that we have defined A k and h k with bk:Ak--* ~ nowhere bounded. We apply 2.~ to Ak, h k to obtain a set Ak+ l_c A k and a map

70

hk+l:Ak÷l--+ ~ which is nowhere c o n s t a n t , such that hk+l(#)
&An

Then h 0 ( ~ ) > h l ( ~ ) > . . .

= @. Suppose o t h e r w i s e , and l e t ~ l i e i n t h i s

, so [ ~ ( ~ ) ~

But t h e o r d i n a l s a r e w e l l - o r d e r e d ,

is a strictly so t h i s

intersection.

d e c r e a s i n g sequence of o r d i n a l s .

is impossible.

By t h e a b o v e ,

: is a disjoint

~i(~ -

u n i o n and we may d e f i n e h : ~ - ~ K h ( } ) ffi hn(~ ) iff

An+i) by

~e~

-~+i"

I t i s e a s i l y checked t h a t h i s i n c o m p r e s s i b l e .

Let h: ~--~

he i n c o m p r e s s i b l e now. D e f i n e a f u n c t i o n

~ :~(~)-~[0,1]

by

I t i s e a s i l y c h e c k e d t h a t # i s a s t r o n g measure on ~ . We p r o v e t h a t 9 i s n o r m a l . Suppose A h a s p o s i t i v e

~ - m e a s u r e and t h a t g : A - ~ < i s such ÷~hag g(~) < ~

e A. L e t B = h - l [ A ] . Then B h a s p o s i t i v e I f £ e B, f ( r )

/ s k Then

~ - m e a s u r e . L e t f = goh. Thus f : B - ~ ~ .

= g ( h ( r ) ) < h ( ~ ) . So a s h i s i n c o m p r e s s i b l e t h e r e i s a

[~e BIf(~) ~ ~ ]

has p o s i t i v e

for all

A < ~ f o r which

~ - m e a s u r e . S i n c e ~ i s s t r o n g and ~ < ~ t h e r e i s a

such that D ffi {Te B~f(~) = ~ } has positive ~-measure. Let E = {?eAlg(T ) =A}. T e D~+g(h(T)) = f *-~ h(T)e E. Thus D = h-l[E]. Hence E has positive v-measure°

But E = A a n d g i s c o n s t a n t on E, so we a r e done,

We t h u s s e e t h a t w i t h o u t l o s s o f g e n e r a l i t y we may assume t h a t i n f a c t t h e measure ~ i s n o r m a l .

2 . 6 Le-w,~ L e t A have p o s i t i v e m e a s u r e . L e t h : A ~

be s u c h t h a t h ( ~ ) <

Then h i s a l m o s t bounded. Proof: Let

E = { ~
C l e a r l y t E must be c o u n t a b l e .

h a s p o s i t i v e measure } .

for all

~EA.

71

Let

B =

{~AI

h(T) $ E } .

B must have measure zero. For otherwise, by normality, there is a of positive measure such that h(T) = A for

T ~ B', giving

X< ~ and a B'~ B

A ~ E, contrary to the

definition o f B. Since

K is regular,

A0 = sup(E)<~

. But

{r~Alh(r)

>

~o} ~ B.

Thus h is almost hounded.

2.7 Lemma For almost all ~ e ~ , ~

is a regular cardinal.

Proof: Suppose not. Let E = {~ Icf(oc)< ~}. Thus E has positive measure. Hence, by normality, there is a k < K For each

< ~

such that E 1 =

o~e El, pick a mapping h : ~ - ~

a map g~'E1-~ ~

{ c~ I cf(~) = ~} has positive measure. such that sup(h~[~]) = ~ . Define for each

by g~(~) = h ( ~ ) . then for all ~ ~ E i, g~(~) = h~(~) < ~

.

Applyir~ 2.6 to g#, we see that there is a set N~ of measure zero and an ordinal

T,<~

such that g~(~).
re~lar, T < K. ~ t

if ~ i - N ~ .

~2 = ~i -

U~<~.

Set

~ = sup{~l~<X}

Since ~ i~ strop,

= sup{gi(~) ~ < ~ } ~ sup Thus E 2 c { ~ [ c ~ < Y L

Since

• Since ~ is

~(S 2) > 0. Now, for ~ ~ ~ ,

{~ ~ < ~ }

~ T

~t(E2)>O , this contradicts 2.2.

2.8 Theorem is inaccessible.

Froof: Ne know that ~ is regular. So if ~ is not inaccessible, there must be a cardinal k such that

K = X+, the least cardinal greater than ~ . Then { ~ t ~ i s regular } ~ {~ I~% ~} ,

so by 2.2 {~ J~ is regular} has measure zero, contrary to 2.7.

Notice that since any inaccessible cardinal is greater than

MI' 2.8 provides

us with a proof of Ulam's theorem that, assuming CH, there is no extension of Lebes~ue measure to a measure defined on all sets of reals.

72

3- A Theorem i n Model Theory, (1"

A p r e d i c a t e language c o n s i s t s of t h e f o l l o w i n g : logical connectives quantifiers variables

brackets

V

^ (and), ~

v (or),

all),

3

-~ ( n o t ) ,

---,(implies);

(there exists);

Vo,Vl,V2,... ;

( , ) ;

(possibly) n-ary predicate symbols ~I' ~2' "'" (each n).

The formulas of the language are built up as follows: I. If Xl,...,x n are variables, then P~(Xl,...,Xn) is a formula; 2.

~,

If

,

are

fo~.ias,

so are

~^~), (~.,), (~), ~,);

3. I f 9 is a formula, then so are (~Vn~), (3Vn~). The meanings of the above operations are s~f-evident.

If a variable v

occurs in a n

formula ~ within the scope of a quantifier v n in ~

~ v n or BVn, we say that occurrence of

is bound; otherwise the occurrence is free. (Thus a free variable is one

for which we could "substitute" any value we chose from the values available. And a hound variable is one which is an integral part of the meaning of the formula in which it occurs.)

Let S be any predicate language. An S-structure is a structure of the form 1 1 2 2 ~I = (A,RI,R2,...,RI,R2,...).

If 9 is a formula of S and if Xl,...,x m are the free

variables of 9 , and if al,...,a m are elements of A, we say that ~

is satisfied

i_~nOl at the point (al,...,am) iff ~ is a true assertion about the structure when R n interprets pn and x i

Ol

denotes a. ; we write i

l

~I

~ 9

[a i .... , a ]

in this case. (The notion of satisfaction can be rigourously defined , but for our purposes the obvious intuitive meaning is adequate.)

1. Although t h i s s e c t i o n i s , i n a s t r i c t s e n s e , s e l f - c o n t a i n e d , i t may r e q u i r e some p r i o r knowledge of t h e m a t e r i a l f o r a p r o p e r u n d e r s t a n d i n g . The m a t e r i a l c o v e r e d i s a l s o r e q u i r e d f o r s e c t i o n 4.

73

F o r example, l e t S be t h e p r e d i c a t e language f o r group t h e o r y . This h a s o n l y one b i n a r y p r e d i c a t e ,

d e n o t i n g e q u a l i t y o f two g r o u p e l e m e n t s , and one t e r n a r y

p r e d i c a t e d e n o t i n g x . y = z . Any g r o u p w i l l be an S - s t r u c t u r e ,

The f a c t t h a t a g r o u p

i s a b e l i a n can be e x p r e s s e d i n any o f t h e f o l l o w i n g f o r m s : (i)

G b

(ii)

f o r all g , h i n G,

Vx~y~z(x.y = z - ~ y . x G~

(iii) for all g,h,k in G,

Vz(x.y = z-ey.x

= z) = z)

G ~ (x.y = z--~y.x = z)

[g,h] [g,h,k] o

(For reasons of clarity, one rarely retains the formalism of the definitions for very long. Thus, in the above we write x.y = z instead of P(x,ytz) for the appropriate P. Likewise we use x,y,z etc. instead of Vo,Vl,V2, .... )

I f we h a v e ~ ¢ o S - s t r u c t u r e s

ur__~e o f Z

, and w r i t e

0~ ~ Z

~ , Z , we s a y t h a t ~ i s an e l e m e n t a r y s u h s t r u c ~ -

, i f 0 ~ i s a s u b s t r u c t u r e of ~ in the obvious s e n s e ,

and for all formulas 9 of S with free variables (say) Xl,...,x n and all elements al~..o,a n of 0-[ , ~

9

iff

[al,.--,a n ]

( L o o s e l y speaking, ~ t ~ ~

~

means that a l t h o u g h

b ~ [al,...,an] ~

-

c o n t a i n s more elements than gZ ,

0~ and ~ looke very much alike with regards to their internal structure. For example, i f G and H a r e g r o u p s and G ~ H , t h e n G w i l l be a b e l i a n i f f H i s a b e l i a n . )

The n o t i o n o f e l e m e n t a r y s u b s t r u c t u r e was i n t r o d u c e d by T g r s k i , who a l s o showed t h a t t h e c o n s t r u c t i o n o f e l e m e n t a r y s u b s t r u c t u r e s can be r e d u c e d t o a p u r e l y algebraic construction.

More p r e c i s e l y :

3.1 Lense L e t S be a p r e d i c a t e l a n g u a g e . L e t ~ h e there are finitary

functions fl,f2,..,

an $ - s t r u c t l L r e , ~ = ( A , . . . ) .

Then

d e f i n e d on A i n t o A ( c a l l e d Skolem f u n c t i o n s ) co

such that for any set X c A, if B = n__Ulfn[X=] and if ~ is the structure ~

relativised

%0 B, then X c ~ _< ~[ ~ [3

A p r o o f o f 3.1 can be f o u n d i n any e l e m e n t a r y t e x t on m a t h e m a t i c a l l o g i c . As

74

an i l l u s t r a t i o n

o f t h e use o f 5 . 1 we p r o v e t h e f o l l o w i n g c l a s s i c a l

t h e o r e m o f model

theory.

5 . 2 Theorem (The L~wenheim-Skolem-Tarski Theorem) L e t S be a p r e d i c a t e l a n g u a g e , ~%an S - s t r u c t u r e o f i n f i n i t e L e t X~ a b e an i n f i n i t e Then t h e r e i s a ~ %

Proof: Let { % } if ~

cardinal,

cardinality

and l e t X be a s u b s e t o f ~ o f e a r d i n a l i t y

such t h a t X ~ ~

and

l~=

~ .

k .

k .

be a sequence of skolem functions for ~ . Let B = n~Ulfn[X] . By 5.1,

is the relativisation of 0~ to B, then ~ 6 q .

Since we clearly have IBI = k ,

we are done.

U s i n g 5 . 1 , we s h a l l a l s o p r o v e t h e f o l l o w i n g r a t h e r s p e c i a l i s e d v e r s i o n o f t h e above t h e o r e m . This w i l l be i n s t r u m e n t a l i n our s o l v i n g t h e measure problem from

the Axiom of Constructihility.

5-3 Theorem L e t K be an u n c o u n t a b l e c a r d i n a l which c a r r i e s

a s t r o n g m e a s u r e , L e t S be any

p r e d i c a t e l a n g u a g e which c o n t a i n s a u n a r y p r e d i c a t e symbol U, and l e t

o~= (A,~, he any S - s t r u c t u r e

such t h a t

)

IAJ ~ K and IUI < ~ =

such that IBI = ~

...

(B,UnB,

and I U ~ B } % ~0"

. Then t h e r e i s a ~ ~ 0-~,

... ),

Q

The p r o o f w i l l t a k e some t i m e . L e t us remark t h a t t h e d i f f i c u l t y

i s t o keep

B l a r g e w h i l s t e n s u r i n g t h a t U ~ B i s s m a l l . I n o r d e r t o do t h i s we must r e l y h e a v i l y upon t h e f a c t t h a t ~ c a r r i e s a s t r o n g m e a s u r e .

To commence t h e p r o o f , n o t i c e t h a t we may assume t h a t A i s i t s e l f number ( j u s t t r a n s f e r

%he s t r u c t u r e

from A %o i t s

a cardinal

c a r d i n a l number), and h e n c e t h a t

K ~ A . We f i x some s e q u e n c e { f n } o f Skolem f u n c t i o n s f o r

Suppose fn is k(n)-~,7. We define a k ( n ) - a ~ f u n c t i o n f

07. i n a c c o r d a n c e w i t h 5 . 1 .

from Kk(n) %o U by:

75

if this is an element

=

of

an arbitrary element o f U, otherwise. In order

to prove 3.3 it

such that

I fn[X ] 1 ~

clearly

~0 for

all

suffices n.

form the domain of the required if necessary, order i,j

all

then for

is

irrelevant),

We a r e t h u s r e d u c e d go p r o v i n g t h e f o l l o w i n g . strictly

increasing

n-tuples

~ of cardinality

n

on X

fn

extra

will

functions

are commutative (i.e.

and that

K

if x. = x. for 1 j

the some

- 1,

,Xj_l,Xj+l,---

,Xk(n) )-

If X ~ ~ , leg x[n]denote

the set

of

f r o m X. L e t { f n } b e a s e q u e n c e o f f u n c t i o n s

Sk(n)l ~ ~ ,

fn: w h e r e ~, < K • T h e n t h e r e

f

some m w i t h k ( m ) - - k ( n )

f n ( X l , . . -,Xk(n) ) = f m ( X l , . . .

all

X of

Now, b y i n t r o d u c i n g

the functions

in which the arguments appear j ~k(n),

a subset

(For then the range of the functions

substructure.)

we m a y a s s u m e t h a t

with 1 ~i<

to find

is a subset

X of ~ of cardinality

~ such that

I f n [ x ~ ( n ) ] ] I-< ~o f o r a l l n. (Notice that we have, in effect, replaced the set U by its cardinality, )~ , for convenience.)

Since the intersection of countably many sets of measure one has

measure one (and thus has cardinality ~ ), we see finally that it suffices to prove

*,he f o l l o w i n g -

3 . / t Len~ua Let K be an uncountable Let f:

~n]_~

<

cardinal

K . Then there

and let ~ he a normal strong

is a set

measure on ~ .

D c ~ of measure one such that

If[Dtn]] I ~< ~fl"

Proof-" By induction on n° Case I: n = i.

Leg E =

co.table. Let N = implies that

{ ~ I f-l[~}]

{ ~ ~f(~)¢ E ~ .

~ N ) = O. Thus D =

Case 2: n = k+1.

has positive measure } ° Clearly, E i,

Since f(~)< ~ < ~ ~ - N suffices.

For each ~ , define a map h : ~ k ] _ ~

f(.~,Xl,...,Xk) h~(x I

for all ~, the normality of

,Xk ) L 0 , otherwise.

by:

, if ~< x I ,

n

76

By t h e

induction

hypothesis

we c a n f i n d

a set

D~ o f m e a s u r e o n e s u c h t h a t

and ,_ In~[ Di[k] ]l~g 0. set S~ = h~[D~ k]] . Thus IsJ ~ o n, define gn: ~ - - ~

"

K -

~0nt°~S~ . For each

s~:

~ t

D~

by

gn(~) = s~(n). By case I there is a set N

Set

F =

~ -

of measure zero and a countable set E ~ k such that

n

n

n=~_ONn

,

E = n=0 E n

Then ~ F ) = I and E is countable. And for all n,

~ F ~

£n(~) ~E.

Thus,

S~ ={s<(n)InE~}

~.eF-.

= {gn(~)In~-=

E.

Now set

We p r o v e t h a t

f[D In]]

c E . L e t x e D In3 a n d l e t

such that x ={*}u y . By definition of D, y e u ~ ~ D~F, It

thus

s o S~-cE. H e n c e f ( x ) e E . suffices

t o show t h a t

measure.

I

~

Then g(Y) <~

for

measure such that

all

V~

g is constant

y e D" But D %

1)11

Then

D

oc<~

such that

, so by normality on

But

by

g(V) = t h e l e a s t ~1

y e D f k ] be

. Thus, f(x) = h ~ ( y ) ~ S . .

~(D) = 1 . W e l l s u p p o s e n o t .

Define g:D ~

and let

Thus f[D~n~]~E.

Dj = { r e F l r C N

has positive

ct = r a i n ( x ) ,

there

r @ D~.

is a set

II

I

D~ D

of positive

, s a y w i t h v a l u e ~ 0 • Thenp

---)'C¢ D %

.

has measure one, so this is impossible. The proof is complete.

E]

3 . ~ completes the proof of 3 . 3 .

~ . The Condensation Lem~a

The m o s t s i g n i f i c a n t is the condensation

property.

property

of the constructible

hierarchy,

L~ , R e On,

77

~ . l Theorem ( C o n d e n s a t i o n Le~na) L e t K be an i n f i n i t e

cardinal.

Suppose (N, ~ , = ) ~

u n i q u e o r d i n a l ~ and a u n i q u e s t r u c t u r a l

isomorphism ~ such t h a t

~ : (N, ~ ,:) ~ (L~, ~ Moreover,

~(~)

(LK, e , = ) . Then t h e r e i s a

,=).

if @ is an ordinal such that {~ l~<~]~ N, then for all ~ < ~ ,

= ~ , a n d for all ~ N

such that x ~ ~ , ~ (x) = x.

The p r o o f o f %.1 i s o u t s i d e t h e scope o f t h i s book. I t can be f o u n d i n [ D e ] . Coupled w i t h 3 . 2 , ~.1 t e l l s

us t h a t no m a t t e r how l a r g e t h e c a r d i n a l

always a c o u n t a b l e o r d i n a l V such t h a t t h e s t r u c t u r e ( L a , ~ , = ) . Now, t h e Axiom o f C o n s t r u c t i b i l i t y union of a l l

the partial

why i t

it

says t h a t the u n i v e r s e is j u s t the

i f a n y t h i n g e x c i t i n g h a p p e n s " h i g h up" i n t h e u n i v e r s e LV. This i s

i s o f t e n p o s s i b l e t o c a r r y o u t c o n s t r u c t i o n s by t r a n s f i n i t e

t h e Axiom o f C o n s t r u c t i b i l i t y .

the

(L~, ~ ,=) l o o k s j u s t l i k e

i s a l r e a d y m i r r o r e d w i t h i n some c o u n t a b l e p a r t i a l

IV.1 i s a r e l a t i v e l y

there is

u n i v e r s e s L~. Hence, assuming V = L p r o v i d e s us w i t h an

extremely uniform u n i v e r s e of s e t s ; universe,

~ is,

induction using

Indeed, the combinatorial principle

~

introduced in

e a s y , d i r e c t c o n s e q u e n c e o f t h e c o n d e n s a t i o n le~ma. (One d e f i n e s

~ s e q u e n c e by a t r a n s f i n i t e

i n d u c t i o n , u s i n g the c o n d e n s a t i o n l e p t a to prove t h a t

t h e s e q u e n c e does what we r e q u i r e o f i t .

Any r e s u l t

by a d i r e c t a p p e a l t o t h e c o n d e n s a t i o n lemma i t s e l f .

proved using

0 c o u l d be p r o v e d

However, t h i s u s u a l l y r e q u i r e s

some competence i n t e c h n i q u e s o f M a t h e m a t i c a l L o g i c , so i n p r a c t i c e

the non-special-

i s t w i l l p r o b a b l y f i n d i t e a s i e r t o a p p l y ~ a s we d i d i n , s a y , IV°2. For a s i m i l a r r e a s o n we do n o t g i v e h e r e t h e d e r i v a t i o n o f ~ . I t can be f o u n d i n [ D e ] . )

U s i n g ~ . 1 , we can o b t a i n a v e r y q u i c k p r o o f o f t h e GCH from V = L. F i r s t we n e e d a lemma which i s i t s e l f

rather illuminating.

we p r o c e e d up t h e c o n s t r u c t i b l e

We have a l r e a d y m e n t i o n e d t h a t a s

h i e r a r c h y from, s a y , s t a g e ~ ,

new s u b s e t s o f ~ w i l l

c o n t i n u e t o a p p e a r (which i s n o t t h e c a s e w i t h t h e c u m u l a t i v e h i e r a r c h y , The f o l l o w i n g leDmm t e l l s

us t h a t t h i s

of c o u r s e ) .

c e a s e s a s soon a s i t p o s s i b l y c a n .

~o2 Lemma Assume V = L. Let ~ be an infinite cardinal. Let x c ~ . Then x ~ L K +

.

78

Proof: Since V = [ J L~, we can find a cardinal k

such that x e L x. By 3.2 we can find

an elementary submodel

(N, ~ ,=) ~ (L~, ~ such that

~ N

{~l~<~}u{x}

and

~:(~, ~,=)

INl=

,=)

~ . By ~.i, l e t

~ (L~, ~

,=).

Now, I L~ I = INI = ~ . But it is easily proved (by transfinite any infinite ordinal ~ ,

IL~l = 191

and x c_ K = { c ~ l ~ < ~}C-N, xeL~+

, as required.

. Hence

so by ~.i,

induction) that for

%" < ~+. Hence L r

~(x) = x. Thus x c

c

L~+ . But x e N

~[N] = L r , giving

[]

/I.5 Theorem Assume V = L. Then GCH h o l d s .

Proof: Since IL~I = ~ ]

for all infinite ~

, and since we know that 2 ~

for any infinite cardinal ~ , it suffices t o show that

= I~(~)I > K

~ (K) _c L~+ . But this is

immediate by %.2.

5. S o l u t i o n t o t h e Measure P r o b l e m .

5 . 1 Theorem Assume V = L. L e t ~ be a n y u n c o u n t a b l e c a r d i n a l .

measure on

~

Then t h e r e

i s no s t r o n g

.

Proof: Suppose otherwise. Let ~ be a normal strong measure on < .

Now, by ~.2 we have:

where ~

is the formula (W)(x_~ y -

P(x)),

and where L~I is the unary predicate which interprets P (so P(x) "means" x g L~I ), and where

o~ interprets the free variable y. To see that (*) is valid, we just ask

ourselves what 5~ says when interpreted in the structure (L~,L~ , ~ ,=). It says that L~, contains all subsets of ~ ; which we know to be true by ~°2. (The astute reader

79 may have n o t i c e d t h a t o u r l a s t us t h a t

~

sentence is not entirely

i s a t r u e s t a t e m e n t a b o u t m and L ~ .

i s t r u e when i n t e r p r e t e d

in the ~artial

Well, the point is that the truth

~,

and the collection

b y %.2, a l l

of all

Certainly,

~.2 tells

But why s h o u l d i% t h e n f o l l o w t h a t

u n i v e r s e L ~ w h i c h i s what (w) a c t u a l l y

asserts2 L~,

correct.

or falsity

of T

clearly

o n l y depends on

s u b s e t s o f ~ . No o t h e r s e t s a r e i n v o l v e d . But

of these sets are in the partial

u n i v e r s e L~. Hence ~ w i l l b e t r u e i n L<

iff it is true.)

By 5.5 %here is a set N of cardinality K such that I N n L ~ I k ~ % a n d (N,L~nN, e,=)-< (L~,L%,~ ,=).

Then, clearly, (N,~ ,=)-< (L~,~,=), so by ~.i, let

~:(N,~ ,=) =- (L~,~ ,=). Then,

l~{ = IL~ = {N[ = ~ . So ~

Let W = ~ [ L ~ N ]

. Thus

~

. (In fact it can be shown %hat ~ = ~

o)

IW~ ~ ~0" Clearly,

X:(N,L~,,nN, e ,=) ~ (L~,W,e ,=).

~,

(i)

(LK,L%,~,=)

~

So,

(2)

(N,L~,nN, e,=)

~ c~ ~

q~ 1o~]

•

So, apply/=g x ,

(3)

(L~,W,~,--) ~ ~ ~ I

Now, e v e r y p o s i t i v e

•

integer n is definable

in the language of set theory.

in the partial

u n i v e r s e L~ b y a s e n t e n c e

(The s e n t e n c e w h i c h d e s c r i b e s

explicitly

t h e way i n

which n is built up starting from the empty set suffices. Admittedly this sentence will b e rather long if n is large, hut Ixhat will n o t affect matters at all.) Hence, as (N,e ,--)~ (L~,e ,=), we see that N will contain all positive integers. Similarly, o~ will he an element of N. (]/I particular, this ~ustifies otur deduction of (2) from

(I) able.) (a)

~ t by ~.I, ~ (n~w,~,--)

see that this i~plies that ~(~) = ~

. Hence from (5)"

~ ~ .

What does (~) mean 7 It means %hat in the partial universe L~, all subsets of 60 lie in the set W. But %~> ~ , so the partial universe L r

contains al!l subsets of ¢o .

Hence (~) implies that W actually does contain all subsets of ~ . But W is countable, so this is impossible. The proof is complete. D

80

By v i r t u e

of our d i s c u s s i o n

in section

1, we s e e f r o m 5 . 1 t h a t we h a v e now

proved the following result:

5.2 T h e o r e m Assume V = L. Then for subsets

of X.

6. H i s t o r i c a l

there

a measure

defined

on all

t o t h e M e a s u r e P r o b l e m a s we h a v e p r e s e n t e d i t h e r e i s due t o

R. B. J o n s o n , a n d D. S c o t t . no u n c o u n t a b l e c a r d i n a l

h a d p r e v i o u s l y b e e n o b t a i n e d b y S. Ulam , D. H. F r e m l i n , (In particular,

carries

t o deduce t h e s o l u t i o n

F . Rowbottom. )

X is

Remark

R. M. S o l o v a y . Weaker r e s u l t s

Both s o l u t i o n s

set

Q

The s o l u t i o n

possible

no uncountable

Scott had proved that V = L implies that

a t w o - v a l u e d m e a s u r e d e f i n e d on a l l

subsets.

to the g e n e r a l measure problem from t h i s

It is

result.

a r e due t o S o l o v a y , t h e one g i v e n h e r e b e i n g b a s e d u p o n work o f

Appendix I

AXIOMS FOR SET THEORY

In Chapter II ~e developed the system ZF of set theory, and claimed that this system was adequate for almost all our purposes, at least if we assumed AC as well. But we did not spend much time analysing the definition of the system itself. We just relied upon an intuitive idea of what a set should be, and proceeded from there to construct the Zermelo hierarchy of sets. Since the ZF system remains with us even if we abandon the intuition behind it and opt for constructible set theory, it is worth while analysing just what assumptions the ZF system entails.

When we developed the ZF system, we took the ordinal numbers as basic, so let us continue for the time being to do this. So our first question is: Given the ordinal number system, what assumptions about sets are required in order to construct the Zermelo hierarehy?

Well, we certainly must be able to form the power set of a given set. So let us formulate this assumption as an axiom.

Power Set Axiom If x is a set, then there is a set consisting of all subsets of x.

This axiom facilitates our passing from V of V~ when ~

to V~+ 1 . What about the defintion

is a limit ordinal. We then set V~

=

~<~V~

.

So we certainly must be able to form the union of any set of sets:

Axiom of U n i o n If x is a set, there is a set whose members are precisely the members of the members of x (this set being denoted by

Then for ~ a limit

ordinal

we may s e t

~x).

V~ = ~ ) { V ~ < ~ } .

But wait a moment.

82

How do we know that ~V@ ~ }

is a set? The answer is we do not. Or, to put it

another way, we are making an assumption when we take this collection to be a set. (Admittedly it looks a perfectly reasonable assumption, but nonetheless it is an assumption.) ~e certainlyknow that

~ = ~ I ~ < ~} is a set. (We are assuming the

ordinal number system as basic at the moment.) And we obtain the collection V~ I~< ~

if we replace each element ~ of ~

by V~ . This leads to the Axiom of

Replacement. Roughly speaking, this says that if we have a set and we replace each element of this set by some other set, then the new collection is also a set. In order to make this more precise, we must say how the replacement process involved here is to be specified. (i.e. What rules of replacement are allowed?) Unless the original set is finite, we cannot explicitly say which elements are to be replaced by which new sets. We need some general description of the replacement procedure. Since we shall clearly only be interested in replacement procedures of a bona fide mathematical nature, the correct formulation of what is required should now be obvious. As allowable replacement procedures we take those and only those which are describable in our language LST. Formally now:

Axiom o f Replacement If x is a set and F is a function from x to sets which is definable in LST, then

iF(a) ~ a ~ x~ is a set.

Since the function F on c~ defined by F(~) = ¥~

is clearly (is this clear?)

definable by a formula of LST, the axiom of replacement now tells us that {V~ I ~ < ~ }

=[F(~)I~<~

isaset.

Given the ordinals then, the above axioms enable us to define the Zermelo hierarchy of sets. We now ask ourselves what principles are necessary in order to define the ordinals. It turns out that we do not need any more "non-obvious" ones (although some of them are so obvious as to be "non-obvious").

83

To start off with t we must assume there is a set with no members.

N u l l S e t Axiom There is

a set which has no members.

That provides us with the first ordinal, O. Next, if ~

is an ordinal, we

shall want to define the next ordinal, ~ + I , as the set ~

~}. This can be written

as

Since we already have an

~[~s~}

. And this can be written as ~ , ~ } ~ .

axiom of union, all that we need now is~

Pairin~ Axiom If x and y are sets there is a set whose members are exactly x and y (i.e. the doubleton set ~ x,y~ ).

Starting with the null set axiom now, the pairing axiom and the union axiom enable us to construct, inductively, all natural numbers. But what about the "next" ordinal number,~

2 We shall wish to define ~ to be the set of all natural numbers.

Since we need an axiom in order to do this, we may as well formulate this axiom in the most convenient form, namely:

Axiom of Infinity There is a set whose elements are all the natural numbers.

It turns out to be enough to assume that some infinite set exists, but the above formulation is adequate. It is this one axiom which takes us from the realm of finite sets into the infinite sets. Once the initial jump from the finite to the infinite has been accomplished, we need make no more raw existence assumptions about sets. (So the axioms of null set and infinity are the only ones which assert that a set simply exists, all other axioms which provide us with new sets being really descriptions of operation~ which we allow in forming new sets from old ones.)

84

We now have enough axioms to keep the construction of the ordinals and the Zermelo hierarchy going "for ever". For example, we obtain the second limit ordinal, ~+~

, by applying the axiom of replacement to the function which sends each n i n ~

to

~ + n , thereby obtaining the set ~

cO

o

{~+nln~

+nlnE

~,

whence we may obtain ~ + ~

as

c~.

In fact, we do not need to assume the axiom of subset selection now. This axiom is provable from the axioms of replacement, union, pairing, and null set. To see this,

suppose P is a property expressible

wish to prove that [y~ x lP(y)}

is a set.

F(y) = (Recall that

At t h i s now w r i t t e n

~F(y)ly~x~

point it

down a l l

adequate set theory,

if P is expressible

this

t h e a s s u m p t i o n s we n e e d i n o r d e r t o c o n s t r u c t

we h a v e

a sound and

s o we a r e d o n e . " U n f o r t u n a t e l y t h e r e a r e two p r i n c i p l e s of these has been overlooked because it

i s a t h e o r e m of l o g i c .

a s e t s h o u l d be d e t e r m i n e d by w h a t i t s t h e above f a c t

by t h e axiom o f u n i o n ,

is a set.

o b v i o u s . I f two s e t s a r e e q u a l , t h e n c l e a r l y problem there,

But t h e n

i n LST, so i s F. So by t h e

i s t e m p t i n g t o s a y t o o n e s e l f "O.K. t h a t d o e s i t ,

we h a v e o v e r l o o k e d . The f i r s t

s h o u l d a l s o be v a l i d :

which

i s so v e r y

t h e y w i l l h a v e t h e same e l e m e n t s . No

B u t we a r e d o i n g r e a l

members a r e .

se_~ t h e o r y h e r e ,

so

I n e t h e r words t h e c o n v e r s e o f

i f two s e t s h a v e e x a c t l y t h e same e l e m e n t s , them

they are the same set:

Axiom o f E x t e n s ! e n a l i %~ L e t x a n d y be s e t s .

We

~ {y}' i f r ( y ) [ , is ~p(y)

~ yJ = { y,y} .) Clearly,

=

x is a given set.

C o n s i d e r t h e f u n c t i o n F d e f i n e d on x b y

axiom o f r e p l a c e m e n t , { F ( y ) ~ y ~ x ~ i s a s e t . {yex~P(y)}

i n LST, a n d t h a t

t f x a n d y h a v e t h e same e l e m e n t s , t h e n x = y .

This is the axiom which tells us that we are talking about sets, i.e. collections of objects viewed as objects in ~ e i r

own right.

88 The l a s t

axiom i s c o n n e c t e d w i t h a p o i n t w h i c h t h e r e a d e r may have a l r e a d y

n o t i c e d . Even w i t h our axioms a s o u t l i n e d above, i n o r d e r t o c o n s t r u c t t h e o r d i n a l numbers and t h e Zermelo h i e r a r c h y we must be a b l e t o c a r r y o u t i n d u c t i o n a r g u m e n t s : b o t h c o n s t r u c t i o n by i n d u c t i o n and p r o o f by i n d u c t i o n . How can we be s u r e t h a t t h i s is

p o s s i b l e ? The answer i s , we n e e d a n o t h e r axiom. Comparing our p r e s e n t t a s k o f

a x i o m a t i s i n g s e t t h e o r y w i t h t h e f o r m u l a t i o n o f t h e Peano axioms f o r t h e n a t u r a l numbers ( a f a i r l y

good c o m p a r i s o n t o make, by t h e w a y ) , t h e axiom we n e e d h e r e w i l l

c o r r e s p o n d t o t h e " p r i n c i p l e o f m a t h e m a t i c a l i n d u c t i o n " o f t h e Peano axioms. What i t w i l l s a y i s t h a t t h e membership r e l a t i o n ,

~ , is well-founded.

Axiom o f F o u n d a t i o n I f a i s any non-empty s e t ( o f s e t s ) , minimal w i t h r e s p e c t t o t h e r e l a t i o n

t h e r e i s a member o f a which i s

~ .

Now a t l a s t we can s i t back and admire our h a n d i w o r k . Assuming o n l y t h e e i g h t axioms l i s t e d

above i t

i s p o s s i b l e to give a r i g o r o u s development of s e t t h e o r y ,

i n c l u d i n g t h e c o n s t r u c t i o n o f t h e o r d i n a l and c a r d i n a l number s y s t e m s , t h e Zermelo hierarchy,

and i n d e e d t h e c o n s t r u c t i b l e

h i e r a r c h y . And from t h i s p o i n t one i s a b l e

t o d e f i n e e v e r y t h i n g one r e q u i r e s f o r a n a l y s i s , theorist

would a d m i t t h a t t h e axioms do n o t form a p a r t i c u l a r l y

of principles as at all

a l g e b r a , e t c . Of c o u r s e , e v e r y s e t attractive

collection

on which t o b a s e m a t h e m a t i c s . But none o f t h e axioms can be r e g a r d e d

s u s p e c t ( c a n t h e y T ) . And t h e y were a r r i v e d a t by t a k i n g a f a i r l y

natural

n o t i o n o f what t h e s e t t h e o r e t i c u n i v e r s e s h o u l d l o o k and t h e n a n a l y s i n g What b a s i c assumptions lay behind this picture.

So, a t t r a c t i v e

o r n o t , t h e p h r a s e "ZF s e t t h e o r y "

i s nowadays t a k e n t o mean t h e s y s t e m o f s e t t h e o r y whose axioms a r e t h e e i g h t axioms listed

above.

Appendix II

INDEPENDENCE PROOFS IN SET THEORY

The r e a d e r w i l l c e r t a i n l y

be f a m i l i a r w i t h t h e t e c h n i q u e of p r o v i n g a t h e o r e m ,

starting from a set of axioms. But how can we prove that a certain statement is not provable? Indeed, with the theory ZF (or ZFC), could this be even conceivably possible, since ever3rthing we do in mathematics is, ultimately, done in ZF (or ZFC)2 It was one of the major mathematical achievements of this century when F. J. Cohen found a method whereby, working in ZFC one can prove that certain statements are neither provable nor refutable in the system ZFC. In particular, Cohen demonstrated that the continuum hypothesis could not be proved in ZFC. (Combined with Gadel's earlier result that the C!I follows from V = L,

an axiom which Gadel had shown

could not be refutable in ZFC, this established the undecidability of the CH in ZFC. In fact, Gadel's result can be dispensed with, since Cohen's method may also be used to establish the non-provability in ZFC of -~C}I.) Since then, Cohen's method has been greatly simplified, and used to demonstrate that many classical open problems in mathematics are, in fact, undecidable in ZFC. We present here a brief outline of this method.

Our starting point is the Zermelo hierarchy. Recall that we obtain %~+I from V~

by setting

L+I

= ~(V~). Now, as everyone Imows, associated with every subset

of a given set X is a particular function from X to the set

{ 0,I } , namely the

characteristic function of the set, which takes the value i on points in the set and the value 0 on points outside the set. Given a subset of X, we "know" its characteristic f11nction, and conversely, given any function from X into

{0,1 } we "know" the

set for which it is the characteristic function. Thus, in any set theoretical argument we could work with functions from sets into [ 0,1 } instead of with sets. (This would clearly be rather pointless in practice, but presents no great difficulties.) So suppose we define a "Zermelo hierarchy" of characteristic functions as follows:

87

vo

V~+1 V~

=

~;

= "[ f t f:V~--~{0,1} }; =

~ ~< V~ , if c< is a limit ordinal.

Then clearly, what we end up with is just a disguised version of the usual Zermelo hierarchy. Indeed, the two hierarchies correspond level by level in a natural way. Thus, it we set V the class V

= ~c4~On V~ ,

is a functional analogue of the universe, V.

But what is so special about restricting our functions to have only the two values 0 and I ? Well, if f is the characteristic function of the set x, then f(a) = I means that a is an element of x and f(a) = 0 means that a is not an element of x. And there are no further possibilities. Or are there ? Certainly, from our point of view there are no other eventualities. But are there any internal, mathematical reasons restricting us to just these two. Well, what is the role of the two values 0 and I in all of this ? A few moments reflection leads to the conclusion that they are truth values: I denotes truth and 0 falsity. The reason why w~efeel that the functions involved in the definition of V

should only take the values 0

and i is because normal logic only permits two possibilities, true or false. If f

x

denotes the characteristis function of x (regarded as a subset of some "big" set U), then for any a (in U), fx(a) = 0 means that the statement " a ~ x " is false, and fx(a) = I means that this statement is true. Let us now see what restrictions are placed on the values of the functions fx by the logic, ignoring for the moment any restrictions which arise from our intuition. Well, if we have two sets x and y , then for any a, the truth of the statement " a ~ x ny" is equivalent to that of both " a ~ x " and "a~y", so we must have f x o y ( a ) : fx(a).fy(a). Similarly, f x ~ y ( a ) =

~(fx(a),fy(a)).

Also, %_x(a) = I - f(a). Now, although we have so tar been

regardin~ the functions as still mapping into [ 0,i ) , it is possible now to see

just

what restrictions are placed upon the range space by the logic. It must be a boolean

88

algebra!

Although we just allow our characteristic functions to take two values,

the mathematics would be just the same if they took values in any boolean algebra. And of course, the set ~0,i~ is a boolean algebra with the operations a.b = a times b ;

avb = m a x ( a , b )

;

-a = 1 minus a. (IWnich connects in with our remarks above about the functions fx' of course.) In other words, what is important about the values our "characteristic functions" may take is that they behave in the same manner as truth values (remember truth tables?), and thus reflect the logic involved in the combination of sets and assertions about sets. In fact, when we come to look carefully into this point, we soon realise that what is required is not a boolean algebra but a complete boolean algebra (i.e. a boolean algebra in which every set has a sup and an inf.). This should become clear below.

So l e t ~

be a n y c o m p l e t e b o o l e a n a l g e b r a

now. We d e f i n e

the

~ -valued

universe of sets as follows:

V (]B) =

V (~)

U~V~

U ~

)

, if ~ is a limit ordinal.

On

"

An element of V (~) will be a J~-valued set • But just what is a

~3-valued

set? Well, if f is such an entity, then f provides us with a probability distribution f o r

"the set denoted by f" . That is,

if

f:Vy)--~]B

, then for any u eV (~),

if f(u) = I, then u is, with probability I, an element of f in V ~ ) .

And if f(u) = O,

then u is, with probabliity O, an element of f, which is the same as saying that

with probability

1, u i s i n t h e c o m p l e m e n t o f f i n V ( B ) . And i f

will be an element of f with probability b in V ~B).

f(u) = b, then u

So in V (~) it is not always

89 the case that partly

a s e t u i s e i t h e r i_~n a s e t f o r e l s e n o t i n f :

out. This state

of affairs

may s t r i k e

logic of the system is concerned all that the probability probability

it

c a n be p a r t l y

i n and

u~s a s odd, b u t a s f a r a s t h e i n t e r n a l

i s i n o r d e r . F o r e x a m p l e , t h e l o g i c demands

o f a s e t n o t b e i n g i n a n o t h e r i s t h e b o o l e a n complement o f t h e

of t h e f i r s t

s e t b e i n g i n t h e complement o f t h e s e c o n d . 2~d t h i s

turns

out to be t h e c a s e .

So what do we have so far? We have a "universe of sets". And for the elements of this "universe" we can answer basic membership questions (the answer being always an element o f the boolean algebra). But there is more t o set theory than saying what the truth value of a statement "a e x" is. For a start, what about statements of set equality~ Well, since two sets will be equ~l iff they have the same members and

we

can answer membership questions, we can define, in a natural manner, the probability, or truth value, of the statement denote by

"x = y" for sets x and y in our "universe". We

~ x = Y~I the probability that x = y holds in our universe. (Although the

precise definition of ~ x = yl~ as an element of the boolean algebra is quite natural, there are some technical complications involved, so we do not give it here.) But this is still net enough. In set theory we want to know the truth value of any assertion about sets. Now, any assertion about sets can be written in the language LST. So what we require here is a definition of the ~robabili%_~ or truth value , ~

, of any sentence ~

o f LST when interpreted in the boolean valued universe

V (B). This is defined by induction on the construction of ~

. If ~

is a basic

membership or equality statement we are done already. And we procced inductively now, setting:

tl~,~l

=

llW~(x)lt = (We may o b t a i n t h e d e f i n i t i o n ~

l~Ltv!t~l

;

i~f ~i~(~)ll for ~-~ ~ =

•

from the i d e n t i t y ~

90

and t h a t

for

~*~

from

~--~

Notice that left

t h e r e i s some d u p l i c a t i o n

of the e q u a l i t y

boolean operations

=_ ( ~ ) ~ ( ~ - - ~ ) . )

signs are logical

in~

. But t h e r e

of n o t a t i o n

symbols, and a l l

h e r e . All symbols to the

those to the right

i s no n e e d t o t r y t o a v o i d t h i s .

are

Indeed, it

e m p h a s i s e s t h e r e a s o n why a c o m p l e t e b o o l e a n a l g e b r a i s r e q u i r e d .

The o p e r a t i o n s

o f a b o o l e a n a l g e b r a b e h a v e e x a c t l y a s do t h e i r

counterparts.

At t h i s

canonical logical

p o i n t l e t u s w a r n t h e r e a d e r ~ l a t we h a v e v a s t l y

i n t h e above a c c o u n t , a n d t h a t he s h o u l d n o t c o n c e n t r a t e just

to outline

sacrificed

the method, not to present

for simplicity.

n e x t two, h i g h l y c r u c i a l

it.

on a n y d e t a i l s .

matters

Our aim i s

Hence a b s o l u t e a c c u r a c y h a s b e e n

(But n o t to too g r e a t an e x t e n t , parts

simplified

o f t h e d e v e l o p m e n t , we s h a l l

o f c o u r s e ! ) As t o t h e not attempt to indicate

how t h e v a r i o u s p r o o f s go: t h e y a r e h i g h l y t e c h n i c a l .

It turns out that all with probability

i n V~ )

(for any ~)

1. M o r e o v e r , i f we c a n deduce a s t a t e m e n t ~ from a s t a t e m e n t

means o f a l o g i c a l

argument, then

i n V(~) w i t h p r o b a b i l i t y o f ZFC w i l l

o f t h e a x i o m s o f ZFC a r e v a l i d

be v a l i d

i1~11~ I 1 ~ .

1. And, of c o u r s e ,

with probability

Hence a n y t h e o r e m o f ZFC w i l l be v a l i d it

follows that

~

the n e g a t i o n of any theorem

0.

Suppose now that we are given an assertion ~ wish to demonstrate that

~ by

about sets (e.g. CH), and we

is not decidable in the system ZFC. We assume that

has, in fact, not been decided! Which is the same as saying that we have been unable to calculate

II~I~ in the universe V(B)when

perhaps, by a careful choice of algebra

•

~

is the particular algebra G0,1) . But

, we can prove that in

V ~B), ll~il lies

strictly between 0 and i. Then we shall know at once that neither ~ nor its negation is provable in ZFC. And there is our methodl

As m i g h t be e x p e c t e d , t h e p r o b a b i l i t y

distributions

and t h e t r u t h v a l u e s o f

s t a t e m e n t s a b o u t s e t s i n a b o o l e a n v a l u e d u n i v e r s e depend h e a v i l y upon t h e a c t u a l

91

b o o l e a n a l g e b r a i n v o l v e d . E e n c e , i n o r d e r t o p r o v e t h e i n d e p e n d e n c e ( i n ZFC) o f a particular

statement, the first

be e x t r e m e l y d i f f i c u l t .

s t e p i s t o f i n d a s u i t a b l e b o o l e a n a l g e b r a . T h i s can

I n d e e d , most r e s u l t s

require the explicit

b o o l e a n a l g e b r a . The " s t a n d a r d examples" r a r e l y s u f f i c e .

c o n s t r u c t i o n of a

Then t h e r e i s t h e p r o b l e m

o f p r o v i n g t h a t t h e s t a t e m e n t i n q u e s t i o n has a t r u t h v a l u e s t r i c t l y i n t h e " u n i v e r s e " c h o s e n . This can a l s o p r o v e t r i c k y .

b e t w e e n 0 and 1

I n d e e d , on o c c a s i o n s what one

ends up d o i n g i s f i n d i n g two b o o l e a n a l g e b r a s , and p r o v i n g t h a t t h e s t a t e m e n t i n question is truewith other!

probability

( T h i s a l s o does t h e t r i c k ,

1 i n one u n i v e r s e and w i t h p r o b a b i l i t y

0 in the

of c o u r s e . )

Thus, although one is ultimately proving that some statement is not provable, what one actually does when applying the method outlined above is just construct a boolean algebra and prove a lemma about that algebra. All this can be done in good old ZFC. (Although, by virtue of our discussion on constructibility,

it would be

permissible to use V = L as well, and indeed this is sometimes necessary.) So even when he is proving some independence result, a set theorist is just doing set theory. ~

GLOSSARY OF KEY T E R ~

AC, 25

Ext, 3

axiom of choice, 25, 32

extension problem, 2

axiom of c o n s t r u c t i b i l i t y , 30 first countability axiom, 6 bound variable, 72

free group, 3 K-free group, 5~

cardinal, 8

free variable, 72

cardinality, 8 CE, 9, 33

C~H, 9, 33, 78

Chase's c o n d i t i o n , 5~

g e n e r a l i s e d continuum hypothesis, 9,

choice f u n c t i o n , 25, 32

33, 78

closed set, 37 cofinal, 36

i n a c c e s s i b l e c a r d i n a l , 67

cofinality, 36 collectionwise Hausdorff, 6, 58

L, 29

condensation lemma, 76

language of s e t theory, 1~

c o n s t r u c t i b l e h i e r a r c h y , 28

limit cardinal, 37

constructible set theory, 30

l i m i t o r d i n a l , 10

continuous sequence, ~8

LST, 16

continuum hypothesis, 9, 33 continuum problem, 6

measure, 65

6~H, 58

measure on a cardinal, 66 measure problem, 66, 78

Def, 29 d i s c r e t e s e t , ~5

n a t u r a l number, 7 normal measure, 68

elementary substructure, 73 exact sequence, A5

order type, 8 ordinal number, 7

93 Pontryagin's criterion, ~8

Zermelo-Fraenkel set theory, 26

power set, 20

ZF, 25

predicate language, 72

ZFC, 26

pure subgroup, ~8

Zorn's lemma, 26

r e g u l a r c a r d i n a l , 36

separation, 6 short sequence, ~5 s i n g u l a r c a r d i n a l , 36 skolem f u n c t i o n , 73 S o u s l i n problem, 2, 39 split, ~ s p l i t e x t e n s i o n , ~4 stationary set, 38 strong measu/'e, 67 S-structure, 72 subset selection, 23 successor c a r d i n a l , 37 successor ordinal, I0

unbo~mded set 36 universe of sets, 23

V, 23 V = L, 30

W-group, 5, 46 well-ordering principle, 26 Whitehead problem, 5, A~

SPECIAL SYmbOLS

SYMBOL

CoC,C. <}

G/E +

APPROX.~INING ( w h e r e p o s s i b l e )

countable

chain

condition

subgroup factor

group

group operation direct

sum

external

PAGE

direct

sum

subgroup generated

by A

group of integers G "---~H

group homomorphism

im(~)

image of

Ke,(~)

kernel

of

group of homomorphisms from G to H

1X f :X---~Y

f[A] f-l[A ]

identity on the set/group X function from X t o Y image o f f on A pre-image of f on A

~,~,r

ordinal numbers

~+i

first ordinal after cardinality

of X

first infinite ordinal set of all natural numbers infinite ~'th

Wo g~ K* (~,y)

cardinal

uncountable

numbers cardinal

8 as an ordinal

as a cardinal ~'th first

uncountable cardinal

ordered

pair

9 9

cardinal after

of x and y

as a cardinal

9 9 12

95

N

and

15

V

or

15

not

15

implies

15

iff

15

v

for all

15

3

there exists

13

power s e t of x

20

¢

empty set

20

v~

~('th level in Zermelo hierarchy

21

--1

~(*)

M

L~ cf

28 ~'th

level in constructible

cofinality

hierarchy

function

29 36 38 38

f~A

restriction satisfaction

of f t o A

38

symbol

72

elementary substructure

x~n~

all

increasing

n-tuples

73 from X

75

SUGGESTED FURTtIF~READING

For short accounts of set theory, parts

[Ba]

of m a t h e m a t i c a l l o g i c ,

see the various articles

in

development of constructibility

K. J . D e v l i n : C o n s t r u c t i b i l i t y

theory,

see

, in [Ba] above.

F o r t h e r e a d e r who w a n t s t o f i l l

i n t h e gap

we

Vhitehead Problem, the only source at present

[ Sh]

and o t h e r

K. J . Barwise ( E d ) : Ilandboo.k of H a t h e m a t i c a l Logic , N o r t h H o l l a n d (1977)

For a reasonably detailed

[De]

f o r m a l l a n g u a g e s , model t h e o r y ,

left

in our solution

is Shelah's

original

of the paper:

S. S h e l a h : A Compactness Theorem f o r S i n g u l a r C o r d i n a l s ~ F r e e A l g e b r a s ~ W h i t e h e a d P r o b l e m and T r a n s v e r s a l s , 21 ( I 9 7 5 ) , 319

As a m a t t e r o f h i s t o r i c a l

-

Israel

J o u r n a l of ~ I a t h e m a t i c s

3~9.

interest,

l e t us a l s o m e n t i o n t h e o r i g i n a l

mono-

g r a p h on c e n s t r u c t i b i l i t y :

[GB] K. GBdel: The C o n s i s t e n c y of t h e Axiom of Choice and o f t h e G e n e r a l i s e d Continuum H y p o t h e s i s . A n n a l s o f H a t h e m a t i c s S t u d i e s 3, P r i n c e t o n U n i v e r s i t y P r e s s (19~0)°