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

SET THEORY

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

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

ACADEMIC PRESS

New York San Francisco London

A Subsidiary of Harcourt Brace Jovanovich, Publishers

1978

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

ACADEMIC PRESS, INC. 111 Fifth Avenue, New

York. New York 10003

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

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

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

PRINTED IN T H E UNITED STATES OF AMERICA

For. Paula, Pavel, and Susanna

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Aus dem Paradies, das Cantor uns geschaffen,

sol1 uns niemand vertreiben konnen.

David Hilbert

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CONTENTS Preface

xi

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

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

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

1 12 22 29 38 42 52 61 70 76

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

78 87 99

108 1 I4

126

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

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

ix

137 144 159

176 187 1 97

CONTENTS

X

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

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

216 229 244 261 283

PART I11 LARGE SETS

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

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

295 305 32 I 337 359 38 1

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

33. 34. 35. 36.

398 416 427 450 465 478

PART IV SETS OF REALS

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

493 509 527 537 550 563

HISTORICAL NOTES AND GUIDE TO THE BIBLIOGRAPHY

579

BIBLIOGRAPHY

5%

NOTATION

61 I

Index

615

PREFACE

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

xi

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PART I

Sets

CHAPTER 1

AXIOMATIC SET THEORY

1. AXIOMS OF SET THEORY

Axioms of ZermelwFraenkel:

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

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

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

=

uX,

the union of all

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

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

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

2

1. AXIOMATIC SET THEORY

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

Q

This principle, however, is false:

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

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

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

x=y

1.

3

AXIOMS OF SET THEORY

by means of connectives cpA4k

cpvbk

cp+*,

l c p 9

cpw$

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

vx cp,

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

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

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

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

C = {X :

.

~ 1 *% *

~p(x9

7

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

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

pn))?

...( pn)

3

-

V(X* 41,

...?

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

v = {x : x = x}

qm)

1.

4

AXIOMATIC SET THEORY

We define inclusion of classes CED

VX(XEC-XED)

++

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

E

CAx

# D}

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

u

{ S : S E C}

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

E

S and

{x : x E S}

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

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

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

X=Y

-

++

Vu(uEX-uEX)

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

=a

v x = b)

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

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

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

.

5

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

(1.1)

iff

a = c and

b=d

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

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

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

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

an)?an+l)

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

x

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

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

Yt+U E XA$(U,

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

p1,

..

. 3

..., p , ) ) pn))

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

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

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

1.

6

AXIOMATIC SET THEORY

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

X

n

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

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

Similarly, it follows that the empty class

0 = {u:u #

u)

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

3X(X = X )

(1.4)

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

n C n {x: x

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

E

n

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

would be a set.

Union For any X there exists a set Y =

u X:

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

(1.5)

Let us introduce the abbreviations (32 E

X )cp

for

3z(z

E

X

~ c p )

(Vz

X )cp

for

Vz(z

E

X

+ cp)

and E

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

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

=

u

{z: z

E

X }=

ux

the union of X . Now we can define

x

u Y=

{x,Y}, x

u Yu

z = (X u Y)u Z,

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

etc.

7

1. AXIOMS OF SET THEORY

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

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

VX 3Y Vu(u E

Y W U 5

X)

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

E

u

-b

z E X)

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

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

X

(1.6)

The notation {(x,y) :

. a * }

X

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

in (1.6) is justified because

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

xx

Y c PP(X u Y)

Further, we define

xx

Y xZ=(X x Y)xZ

and in general

x, x ... x

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

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

A--*Ax, E

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

X,,}

8

1. AXIOMATIC SET THEORY

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

G

X". It is

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

uu

ran(R) c

R,

uu

R.

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

(x, z ) ~f

and

implies

y =z

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

f:

XHY,

y =f,,

etc.

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

'Y

E P(X x

Y)

1.

9

AXIOMS OF SET THEORY

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

implies

x =y

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

f

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

x E X)

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

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

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

f- AX) = 1.

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

f-'(x)= y

x =f(y)

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

( x , z) E F

implies

y =z

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

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

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

X=U{Y:YEP} Let

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

= {y E

x :y = x )

(the equivalence class of x ) . The set

x/=

= {[XI :x

E

X}

1.

10

AXIOMATIC SET THEORY

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

XEJJ

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

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

s A (VX E s)[X

U {X} E

s]]

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

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

n

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

O=@, If n E N , let n

2={0,1},

1={0},

...

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

c--)

nEm

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

1.

11

AXIOMS OF SET THEORY

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

E

E A +n

+ 1 E A.

N is T-finite.

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

Replacement Schema

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

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

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

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

12

1.

AXIOMATIC SET THEORY

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

VX 3Y[U X E Y]

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

(1.9)

vx 3Y[P(X) E Y]

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

(1.10) If F is a function, then

VX 3Y(F[X] E Y)

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

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

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

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

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

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

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

2.

13

ORDINAL NUMBERS

If ( P , < ) and (Q, < ) are partially ordered sets and f: P + Q, then f is order-preserving if x a (/I is a limit). [Consider limn+wa,, where a,, = a, 1.1

+

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

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

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

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


1).

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

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

-

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

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

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

a

a

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

2.

21

ORDINAL NUMBERS

-

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

then

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

r[a

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

and hence y(u)

w'. Thus if

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

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

a2

UO.

E a,,

a3

E a2,

...

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

+

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



C f }=

{(ck : k

E

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

1. AXIOMATIC SET THEORY

36

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

bo=--,

1

b , =-

a0

+ a11 ’

a.

-

etc. Notice that (4.8)

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

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

b,; ii is the continued

1

a= a0

1

+ a,

+-a2 +1 ...

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

1

+p .

1

‘

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

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

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

u “N m

Seq=

n=O

4. REAL

37

NUMBERS

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

E

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

E

T.

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

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

E

N)

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

TF = {s E Seq : s cffor somef’E F}

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

. I ‘ and sequential trees without maximal nodes.

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

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

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

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

To= T,

(4.14)

T. =

f7

T,

T,+, = T,‘ if a > 0 is a limit

, a, >

. a *

> a, > ..'

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

a,,

, E a,

for all n E N

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

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

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

K,NS

= 2'8.

Proof:

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

'

'I,

= 2%

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

6.

43

CARDINAL ARITHMETIC

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

c A such that I X I = K

Similarly, [A] K for eoery infinite cardinal K~

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

< 2's Let l i= 2's

i K , ,

(6.12)

cf(2Na) > K,

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

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

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

K.

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

n

fl

i

i

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

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

1.

48

of length

K.

AXIOMATIC SET THEORY

By Lemma 3.4b, we have Cf

1 = Cf K

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

(6.13)

If

ti

is a regular cardinal, then

K

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

2" = KcfK

(6.14)

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

= ticfK(the gimel function).

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

(6.15) (a) If

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

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

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

and so (6.17) a p and lfK,Np < K, for all y < a, then: (a) if K, is regular or cf K, > K,, then K,Np = K,; (b) [f cf K, IK,,< K, , then K,KP = Kzf ". ProoL (i) has been already proved. (ii) K,Np Ih',Kp I (h'?)'~ = h'y. (iii) If K, is a successor cardinal, use the Hausdorff formula. If K, is a limit cardinal, then note that (6.19)

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

.

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

h',Na=

1

1%.

1

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

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

K,"'" for all a.

=p 7 = Nu+

50

1. AXIOMATIC

SET THEORY

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

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

Ktfl

=

K? . 2'l.

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

h't2= K,"

. 2N2.

A cardinal K is a strong limit cardinal if

2A< ti

A

Exercise 6.13. If

ti

is singular and is not a strong limit, then

Exercise 6.14. If

ti

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

ti w is regular and IS I 2 K , then { X E S : 1x1

a such that j3 E C (unbounded).

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

,

(7.14)

a

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

,

7.

57

FILTERS A N D IDEALS. CLOSED UNBOUNDED SETS

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

be the limit of

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

-.

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

K

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

E a is closed

unbounded.

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

ti

is

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

0,.

as follows: We let Po E C o be such Po > a, and for each 5 < y , let P, E C , be such that P, > sup{P, : I < t). Since ti is regular and y < ti, such a sequence (7.15) exists and its limit P is less than K. For each q < y, /? is the limit of a sequence ( P , : r] I 5 < y ) in C , , and so P E C,. Hence j? E C. H Let ( X u: a < K ) be a sequence of subsets of K. The diagonal intersection of X u ,a < ti, is defined as follows: (7.16) A{X,:a wl, this is clear: The sets { a : cf a = o}and { a : cf a = w l } are disjoint. If K = wl. the decomposition of w l into disjoint stationary sets uses the axiom ofchoice. It is an interesting problem (related to large cardinals) to show that the use of AC is necessary. In fact, every stationary S c K can be decomposed into K disjoint stationary sets. We shall prove this in Section 35. Here is a weaker statement:

Lemma 7.6. Every regular uncountable sets.

K

is the union of

K

disjoint stationary

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

(7.17)

{ a E W : a : 2 q}

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

f (a) = a::

FIGURE 7.2

The functionf is regressive; and so for every q < K, we find by Fodor's theorem a stationary subset S, of (7.17) and y, 2 q such that f ( a ) = yq on S,. If y, # yq., then S, n S,. = 0, and since K is regular, we have l{S,: q < K } [ = I{Y,:q 1, it is necessary that K' > K whenever 1 2 Cf K.]

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

'

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

'

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

8.

63

SINGULAR CARDINALS

'lleorem 23 (Silver).

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

,

= K,+ for all

2'"

K, < K ,

then 2" = K +

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

is stationary in cf

K,

K

: 2"- = K:}

then 2" = K + .

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

K""

= K+.

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

S = { a < cf K : cf K,

=w

and 2'O < K,}

1. AXIOMATIC SET THEORY

64

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

: ~ = K:

by the assumption.

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

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

9=

n

A,

a cpl(4 > %(a) > . * * which is a contradiction. Hence the relation (P < $ is well-founded and by Theorem 5, we can define the rank ll(~llof (P in this relation (called the norm of cp) such that

'

%(co

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

%I,

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

~ , + l l ~ l l .

68

1. AXIOMATIC SET THEORY

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

[ A I I [I

+

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

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

cp < s $

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

iff

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

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

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

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

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

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

and so -9=

u

{

S

F

+

IIIc/II < IIVII}

69

8. SINGULAR CARDINALS

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

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

9,= {g E 9: 3s E S o , S 4 I,, (Va E S) g(a) u 1 > . * .> u, > -.., a < K, of elements of B. The K,-chain condition is called the countable chain Condition (c.c.c.). Lemma 17.6. I f B is an infinite complete Boolean algebra, then sat(B) is a regular uncountable cardinal. Proof: Let K = sat(B). It is clear that K is uncountable. Let us assume that K is singular; we shall obtain a contradiction by constructing a partition of size K. For u E B, u # 0.let sat(u) denote sat(B,). Let us call u E B stable if sat(u) = sat@) for every nonzero u I u. The set S of stable elements is dense in B; otherwise, there would be a descending sequence uo > u 1 > u 2 > ... with decreasing cardinals sat(uo)> sat(ul) > .... Let T be a maximal incompatible subset of S. Thus T is a partition of B, and 1 T 1 < ti. First we show that sup{sat(u):u E T ) = K. For every regular 1 < K such that 1> I T 1, consider a partition W of B of size 1.Then at least one u E T is partitioned by W into 1 pieces. Thus we consider two cases: Case I . There is u E T such that sat(u) = K. Since cf K < K, there is a partition W of u of size cf K : W = {u, : a < cf K } . Let K,, a < cf K , be an increasing sequence with limit K. For each a, sat(u,) = sat(u) = K and so let W, be a partition of u, of size K , . Then UaccfW, is a partition of u of size ti. Case If. For all u E T , sat(u) < K , but sup{sat(u): u E T} = K. Again, let K, + K , a < cf K . For each a < cf ti (by induction), we find u, E T , distinct from all u p , p < a, which admits a partition W, of size K,. Then W, is a pairwise disjoint subset of B of size K. A partially ordered set ( P , < ) is k--saturated (satisfies the ti-chain condition) if there is no set W G P of pairwise incompatible elements such that 1 W 1 = K . It can be easily observed that if P and Q are partially ordered sets and P is dense in Q, then sat(P) = sat(Q). In particular, if B is a Boolean algebra and C is its completion, then sat(B) = sat(C). Moreover, if a partially ordered set P is

Ua sat(P) and is a limit of regular cardinals, thus a cardinal in 91[G]. In particular, if sat(P) = K,, then all cardinals and confinalities are preserved. We shall now show that the notion of forcing ( P , < ) defined in (19.1) satisfies the countable chain condition. The following lemma is a slightly more general statement: Lemma 19.3. Let S be an arbitrary set and let C he at most countable. Let P he the set of al1,functions p whose domain is afinite subset ofS, with values in C. Let P be partially ordered by inverse inclusion. Then every incompatible subset W of P is at most countable.

Proof: Let W be an incompatible subset of P; that is, if p and q are distinct elements of W ,then p ( x ) # q ( x )for some x E dom(p) n dom(q). We shall show that there is a countable subset A E S such that dom(p) E A for all p E W . Then it follows that I WI IK O . We construct a sequence A , E A C_ ... C A , c *.. of subsets of S , and a sequence W, c W, c ... c W, E ... of subsets of W as follows: First we pick

19.

INDEPENDENCE OF THE CONTINUUM HYPOTHESIS A N D AC

179

q E W and Ipt W, = (4)and A0 = dom(q). Given W, and A,, we choose for each p E P with dom(p) E A , , some q E W whose restriction to A, is p (if such 4 exists). Then we let W,,, be W, along with these chosen q's, and A , , = {dom(q): q E W,, Finally, we let A = uFz0A , . It is obvious (by induction) that each W, and each A, is at most countable. We complete the proof by showing that W = uFE0W,. If q E W , then there is n such that dom(q) n A = dom(q) n A,. Thus if p = q [ A , , there exists q' E W,, such that q' I A, = p . Since dom(q') 5 A , , I G A , it follows that q and 4' are compatible; however, both are elements of W and thus q = q'. Hence q E K+i.

,

u ,

Now the proof of Lemma 19.1 is almost complete. Since P is K,-saturated, it preserves cardinals, and 9I[G] has at least ti subsets of o (namely the generic reals a,, u < K). It remains to show that 931[G] 1 2N0I K . This is a consequence of the following general estimate of the number of subsets of a cardinal I in a generic extension.

Lemma 19.4. Let I he a cardinal in 9131. I f G is an 931-generic ultrajlter on B, then Proof: Every subset A c I in I1)1[G] has a name A E 91'; every such A determines a function a- lldi E ,411 from I into B. Different subsets correspond to different functions, and thus the number of all subsets of 1 in 91[G] is not greater than the number of all functions from I into B in 311.

In our case, 1 P I = ti; and since P satisfies the countable chain condition, it follows that I r.0. P I ItiNo (see also Exercise 17.22); this is because every u E r.0. P is the sum of countably many elements of P. Considering that KNo = K, we have I BIN0= ti, and so (2No)'mrG1 = ti. Exercise 19.1. If P is defined as in (19.1) and if K is arbitrary, then (2Ho)m[CI = (K'~)'~. [(2Xo)WlCl = ((2No)no)mlcl 2 (KNO)Wl"l 2 (KHO)W.]

Adding Subsets of Regular Cardinals The proof of independence of the continuum hypothesis (Lemma 19.1) consisted of adding a large number of generic subsets of o.This construction is easily generalized to arbitrary regular cardinals. (For what happens at singular cardinals, see Exercises 19.3 and 19.4.) Let ti be a regular cardinal (in the ground model 9 31)and assume that 2'" = ti. (We shall discuss the case 2'" > ti in the exercises.) Let ( P , < ) be defined as follows: P is the set of all functions p such that: (19.7) (i) dom(p) c ti and I dom(p) 1 < K ; (ii) ran(p) c (0, 1). A condition p is stronger than q iff p

3

q.

3.

180

FORCING AND GENERIC MODELS

Let G be a set of conditions generic over 91 and let f = f i s a function from K into (0,l}, and

X

(19.8)

=

u G.

As before,

{a < K :f(a) = l }

is a subset of K and X # 1' 11. We shall show that !lland '111[G]have the same cardinals (and cofinalities). To start, since [PI = K < =~ K, we have sat(P) I IPI+ = K', and by Lemma 19.2 all regular cardinals I > K remain regular cardinals in the extension. We shall now show that all the regular cardinals I 5 K are preserved. In fact, we shall show that for every a < K, every a-sequence of ordinal numbers in '111[G]is a set in 1 ' 11. (In particular, P'lRIG1(o!) = P"(a) for all a < K.) Then it clearly follows that if I I K is a regular cardinal in 1 ' 11, then A is a regular cardinal in !ll[G]. Let I be an infinite cardinal. We say that a notion of forcing P is I-closed if for every a I 1,if po 2 p1 2 ... 2 p c 2 *..

(< < a) is a descending sequence of conditions, then there is p E P stronger than all p e , 5 < a. Let us call P A-distributive if the intersection of I open dense subsets of P is open dense. (If P = B - {0},B a complete Boolean algebra, then by Lemma 17.7 this definition is equivalent to I-distributivity of B.) Note that if B = r.o.(P), then B is I-distributive if and only if P is I-distributive. Lemma 19.5. r f P is I-closed, then it is I-distributive. Proof. Let ( D , : a < I } be a collection of open dense sets. The intersection D= a D, is clearly open; to show that D is dense, let p E P be arbitrary. By induction on o! < I, we construct a descending I-sequence of conditions p 2 po 2 p1 2 . .. . We let p a be a condition stronger than all p c , < a, and such that p a E D , . Finally, we let q be a condition stronger than all para < I.Clearly, qED.

na
1 remains a cardinal in the extension. [The condition in (b) is satisfied, e.g., if G C H holds and cf 1 2 K.] Proof: Let P be the set of all functions p such that: (19.12) (i) dom(p) E ti and Idom(p)) < ti, (ii) ran(p) G 1, and let p c q iff p 3 4.

19.

183

INDEPENDENCE OF THE CONTINUUM HYPOTHESIS AND AC

u

Let G be a generic set of conditions and letf= G. Clearly,fis a function, and it maps ti onto 1;this is because for all a < K and all /3 < 1,the sets

D, = {p : a E dom(p)},

ED = { p : fl E ran(p)}

are dense. ( P , < ) is a-closed for every a < K and therefore all cardinals Iti are preserved. If A 1 in Lemma 19.9, then not only 1 is collapsed but "'1 instance :

as well. For

Exercise 19.5. In (19.12), let K = K1 and I, = K,. Then in W [ G ] there is a one-to-one function g : K? + K,. [If X is a countable subset of K,, let g ( X ) = the least a such thatf[(a + w ) - a] = X (wheref= G is the collapsing function). Use the fact that X E 91.1

u

If K is singular in (19.12),then (as was the case in adding a subset of collapsed to cf K. One of many ways to verify it:

ti) K

is

Exercise 19.6. In (19.12), let K = K,. Then in !UI[G], there is a one-to-one function g from I, into w . [Let f = U G, and let g(a) = the least n such that the function f I (a,+ - w,) is eventually constantly equal to a.]

,

In the above examples of forcing, either when adding a subset of ti or collapsing a cardinal to ti, we assumed that 2'" = ti. This is because if 2'" > ti, then either notion of forcing collapses 2'" to K. For instance: Exercise 19.7. Let ( P , < ) be the notion of forcing adding a subset of w , via (19.7). and let (Q,< ) be the notion of forcing that adds a mapping of Sl onto 2"O via (19.12). Then r.o.(P)= r.o.(Q). [Let Q'= ( q E Q : dom(q) is an initial segment of w , ; Q'is dense in Q.Show that P has a dense set P' isomorphic to Q': use the fact that every p E P has 2O ' mutually incompatible extensions.] [Another way to show that (P. < ) from (19.7) adjoins a G, and for every g E "{O, l}, let F ( g ) = one-to-one mapping of 2"O into K,: Letf= least a such that f ( a + n) = g ( n ) for all n.]

u

,

Thus adding a subset of w via (19.7) makes the continuum hypothesis true, and we have a forcing proof of the consistency of 2N0= K,.

Independence of the Axiom of Choice If the ground model satisfies the axiom of choice, then so does the generic extension. However, we can still use the method of forcing to construct a model in which AC fails; namely, we find a suitable submodel of the generic model, a model 91 such that 91 c 91 c 91[G]. I shall describe a general method of construction of models without choice in Section 21. In the present section I shall only prove the basic result.

3.

184

FORCING AND GENERIC MODELS

Thorem 45 (Cohen). There is a model of ZF in which the real numbers cannot be well-ordered. Thus the axiom of choice is independent of the axioms of ZF.

Before we construct a model without choice, we shall prove an easy but useful lemma on automorphisms of Boolean-valued models. Let B be a complete Boolean algebra and let n be an automorphism of B. We define, by induction on p ( x ) , an automorphism of the Boolean-valued universe I/”, and denote it also by n: (19.13)

.(a)=

(i) 0; if n(y) has been defined for all y E dom(x), let (ii) dom(nx) = n[dom(x)], and (nx)(ny) = n(x(y)) for all n(y) E dom(nx).

Clearly, n is a one-to-one function of V” onto itself, and n ( i )= 2 for every x. Lemma 19.10. Let cp(x,, . . ., x,) be aformula. l f n is an automorphism ofB, then for all xl, ..., x, E V”,

---.

(19.14)

JJcp(nx1,

nxn)JJ= n(JJcp(x1,

xn)Jl)

Proof: (a) If cp is an atomic formula, (19.14) is proved by induction (as in the definition of IIx E y(J,( J x= ~ 1 1 ) . For instance, IInx

ZYII

= =

c c . .1 (1

t s dom(ny)

z E dom(y)

(Ibx = tll = KZII

. (ny)(t)) . (ny)(nz))

(b) For other formulas, the proof is by induction on the complexity of 4).

In practice, (19.14) is used as follows: Let (P, c )be a separative partially ordered set. If n is an automorphism of P , then n extends to a unique automorphism of the complete Boolean algebra B = r.0. P :

44 = c M P ) : P 5 4

(19.15)

Then (19.14) takes this form: (19.16)

where xl,

p

..., x,

~p(x1,- . * , xn)

1‘ E

iff

np

~ ( n x , ,. . ., nx,)

V”.

Exercise 19.8. If G is an !Ill-generic ultrafilter on B and n an automorphism of B (in !IN), then H = n[G] is !IN-generic and !IN[H] = !Ill[G].

19.

INDEPENDENCEOF THE CONTINUUM HYPOTHESIS AND AC

185

Now, let the ground model 91 be the constructible universe: 911 = L We first extend 911 by adding countably many Cohen generic reals: Let P be the set of all functions p such that: (19.17)

(i) dom(p) is a finite subset of w x w, (ii) ran(p) E (0, l},

and let p < q iff p 2 q. [Note that any one-to-one mapping between (0 x w and w induces an isomorphism between this notion of forcing and the one that adds one Cohen generic real (when the conditions are functions with dom(p) E o).] Let G be a generic set of conditions. For each i E w, let a, = {n E (0 : ( 3 p E G ) p(i, n ) = 1)

(19.18) and let (19.19) Let u, , i (19.20)

A = {ai : i E

w}

w, and A be the canonical names for aiand A :

dom(ui) = {.' : n a,(.') =

( 19.21)

E

E

w)

C { p E P : p(i, n ) = I}

(n E w )

dom(A) = {ui : i E w } A ( u i )= 1

Lemma 19.11. I f i

(i

E

w)

# j , then every p forces ui # u j : /(a,= U j l ( = 0

ProoJ For every p there exists q =) p and n

E

w such that q(i, n)= 1 and

q ( j , n) = 0.

From now on, let us work in 'X)I[G].Let 9 be the class of all sets hereditarily ordinal-definable over A : 9 = HOD(A) As we have shown in Section 15, 9 is a transitive model of ZF. Since the elements of A are sets of integers, it is clear that A E 9.We shall show that the set of all real numbers cannot be well-ordered in the model 9 (in fact that A cannot be well-ordered). If R" is well-orderable in 9,then there is a one-to-one mappingfE 9 of Pm(o)into the ordinals. The following lemma shows that this is not true: Lemma 19.12. (In "m[G]) There is no one-to-onefunctionf: A + Ord, ordinaldefinable over A.

Proof: Assume that f: A 3 Ord is one-to-one and ordinal-definable over A. Then there is a finite sequence s = (xo, . . . , x r > in A such that f is ordinal-

3.

186

FORCING AND GENERIC MODELS

definable from s and A. Sincefis one-to-one, it is easy to see that every a E A is ordinal-definable from s and A. In particular, pick some a E A that is not among the x i , i 5 k. Since a E OD[s, A] (in W[G]), there is a formula cp such that (19.22)

W[G] k a is the unique set such that q(a, al, .. ., a,,, s, A )

for some ordinals al, . . ., a,, . We shall show that (19.22) is impossible. Let a be the canonical name for a, and let xo, . . . ,xk be the canonical names for xo , . . . , xk (see 19.20); let s be the obvious canonical name for the sequence (xo , . . . , x k ) . We shall show the following: (19.23) For every po that forces q(u. Oi,, . . . ,Oi, , s,A ) there exists band q Ipo that forces a # b and cp(b,Oil, ..., din, s, A ) . Let po 11 q(a, Oil, ..., i,,, s, A). Let i, io, ..., i k be such that a = a , , xo = uior. . . , xk = u i k .Let us pick j E w such that j # i, and that for all m E w, ( j , m) # dom(p,). See Fig. 19.1.

‘0

i

‘k

J

FIGURE 19.1

Now let n be the permutation of w that interchanges i and j , and nx = x otherwise. This permutation induces an automorphism of P: for every p E P , (19.24)

dom(np) = {(nx, m ): (x, m) E dom(p)} (.p)(.x,

m ) = P ( X , m)

( x E dom(p))

In turn, n induces an automorphism of B, and of 91’.It is easy to see (cf. 19.20, 19.21) that (19.25)

n(a,)= a j ,

n(a,) = a,

n(uj)= a, n(A)= A

(all x # i, j )

n(s)= s

and

Since ( j , m ) # dom(po), for all m,it follows that (i, m ) # dom(np,), for all m,and thus po and npo are compatible. Let q = po u np,. Now, on the one hand we have PO 11~

( a ih,, 9

* * *

9

hn

9

S, A )

20.

187

MORE GENERIC MODELS

and on the other hand, since nti = 4, ns = s and nA = A, we have np,

ti,, ..., G", s, A )

Ik q+j,

Hence q ~kcp(ai,...) and cp(aj,...)

and by Lemma 19.11,4 Ik ai # a j . Thus we have proved (19.23),which contradicts (19.22). W 20. MORE GENERIC MODELS

Products of Partially Ordered Sets Let (Pi, < ) and (P,, < ) be two partially ordered sets. The product ( P , ti is inaccessible and each I Pi I < 1,then I W I < 1. Proof: If p = ( p i : i E I) and q = (qi : i E I) are incompatible in P, then for some i E s ( p ) n s(q), pi and qi are incompatible in Pi, and in particular pi # q i . Thus we can regard elements of Was functions whose domain is a subset of I of size < ti, with values in the P i , and show that if W consists of pairwise incompatible functions then I W 1 has the required bound. This we do precisely as in Lemma 19.8 (or 19.3): We construct A,

A, c

... & A,

G

...

and

W,

E

u

W, E ... G W, E . . *

(a
a .We let p > a = {p'" * P E P ) p-' is the Easton product of P ,, K E A and K > 1. Moreover, P is (isomorphic to) the product Ps' x P>'. First we notice that P>' is 1-closed: If C I P'A consists of pairwise compatible conditions and ICI I 1, then p = U C is a condition in P"; (20.11) holds for all regular y > 1, and holds trivially for y II because if (ti, a, p) E dom(p), then K > 1. Furthermore, P' satisfies the L+-chaincondition: If W c P s a is incompatible, then I W I I1.The proof given in Lemma 19.8 works in this case as well because Idom(p)( < I for each p E P" (and because the GCH holds). Thus P = P>' x Ps' where P>' is bclosed and sat(PsA)' x G" and 9 l [ G ] = 911[G>'][G5'] = 9Il[G"'][G'']. By Lemma 20.5, f is in 9ll[G"] and so K is not a regular 1 IK cardinal in 'J)l[G"']].However, this is a contradiction since sat(P") I ' and hence K is regular in 9l[G"]. It remains to prove that (2'rmG1= F(1), for each 1~ A. Again, we regard P as a product P>' x P" and G = G>' x Gs'. By Lemma 20.5, every subset of 1 in 9)1[G] is in 91[Gs'] and we have (2'rWc1= (2')mc'Al. However, an easy computation shows that IPS'( = F(1); and since sat(P") = A + , we have lr.o.(Psa)l = F(1) and hence (2'r)'mlGl5 F ( I ) . O n the other hand, we have exhibited F(1) subsets of 1for each 1E A, and so 9ll[G] k 2A= F(1). H

20.

1%

MORE GENERIC MODELS

Exercise 20.7. The singular cardinals hypothesis holds in Easton's model.

then

[If K is singular then every$ c f ( K ) -+ K is in 91 = 91[Gscf"],and so if F(cf K ) < K ( K ~ ~ =' (K" ~ ')'I ~ ~I;(2K)* ~ Ilr.o.(Ps cf ' ) I K = (F(cf K))" = K+.]

Forcing with a Class of Conditions We shall now show how to generalize the preceding construction to prove Theorem 46 in full generality, when the function F is defined for all regular cardinals (in 91).This generalization involves forcing with a proper class of conditions. Although it is possible to give a general method of forcing with a class, we shall concentrate only on the particular example. Thus let 911 be a transitive model of ZFC + GCH. Moreover, we assume that 91 has a well-ordering of the universe (e.g., if 91satisfies I/ = L). Let F be a function (in 91) defined on all regular cardinals and having the properties (20.8)(i)-(iii). We define a class P of forcing conditions as follows: P is the class of all functions p with values 0 and 1, whose domain consists of triples (K, a, fi) where K is a regular cardinal, a < K and fi < F(K),and such that for every regular cardinal y, (20.11) I {(K, a, f i ) E dom(p) : K 5 Y ) I < Y (and p is stronger than q iff p 2 4). As before, we define P" and P" for every regular cardinal I,. Note that P" is a set. To define the Boolean-valued model 91' and the forcing relation, we use the fact that P is the Easton product of P,, K a regular cardinal. For each regular A, we let B A = r.o.(P"). If A < p then the inclusion P" s P'" defines an obvious embedding of BAinto B,; thus we arrange the definition of the B A so that BAis a complete subalgebra of B, whenever I, < p. Then we let B= B A . B is a proper class; otherwise it has all the features of a complete Boolean algebra. In particular, X exists for every set X c B. Also, P is dense in B. T o define 911', we cannot quite use the inductive definition (18.8) since B is 91IBi;the formal definition of 91' not a set. However, we simply let 91' = does not present any problem. Similarly, todefine IIx E ylJ and Ilx = yll, we first and I 5 p, then IIx E yllB1= IIx E yllB, and so we let notice that if x, y E IJxE yll = JIxE ~ ~ where 1 1A is ~ such~ that x, y E %IBi. The same for IIx = y(1. As for the forcing relation in general, we cannot define llqll unless cp is X does not generally exist if X s B is a class. restricted; this is because However, we can still define p I1 q using the formulas from Lemma 18.8. Now, we call G c P generic over 1' 11 if (i) p c q and p E G implies q E G,(ii) p , q E G implies p u q E G,and (iii) if D is a class in 91 and D is dense in P, then D n G # 0. The question of existence of a generic filter can be settled in a more or less the same way as in case when P is a set. One possible way is to assume that 1 ' 11is

uA

1

uA

1

3.

1%

FORCING AND GENERIC MODELS

a countable transitive model. Then there are only countably many classes in W and G exists. Another possible way is to use the canonical generic ultrafilter. It is the class G in ’IllB defined by G($)= p for all p E P (here we need the assumption that 3ll is a class in %IB). Thus let G be an W-generic filter on P.For every regular 1,G A= G n P 5 is generic on P“. If x E W B A and 15 p, then iG1(x) = iG,(x),and so we define i c ( x ) = ic,(x) where 1 is such that x E WB1. Then we let W [ G ]= ic[Wm”];it follows that W[G] = W[G,]. Using genericity of G and properties of the forcing relation, we get

’

uA

Forcing Theorem :

(20.16)

9131[C] lk cp(xl,

where x . . . , x,

E

. . ., x,)

iff

9NB are names for x

(3p E G)p 1 cp(xl, . . ., x,)

. . ., x,

.

The formula (20.16) is proved first for atomic formulas and then by induction on cp; in the induction step involving the quantifiers, we use the fact that G intersects every dense class of W. We shall now show that W [ G ]is a model of ZFC. The proofs of all axioms of ZFC except power set and replacement go through as when we forced with a set. It is no surprise that the power set and replacement axioms present problems. It is easy to generalize the constructions of Section 19 either to get a class of forcing conditions adding a proper class of Cohen reals, or a class of conditions collapsing Ord onto o.The present proof of the power set and replacement axioms uses the fact that for every regular 1 (or at least for arbitrarily large regular A), P = P I x P where P, is 1-closed and P is a set and sat(P,) 5 1’. (Here P = P” x P “.) Power set. Let 1 be a regular cardinal. Lemma 20.5 remains true even when applied to P’’ x P“. It does not matter that each D, in (20.14) is a class. The “sequence” of classes (D, : a < A ) can be defined (e.g., as a class of pairs { ( p , a ) : p E D,)) and since P>’is 1-closed, the intersection D, is dense, and there exists p E G n P” such that p E D, for all a c 1.The rest of the proof of Lemma 20.5 remains unchanged, and thus we have proved that every subset of L in %l[G]is in 9lI[G,]. Since P 5’ is a set, it follows that the power set axiom holds in W[G]. Replacement. To show that the axioms of replacement hold in Yt[C], we combine the proof for ordinary generic extension with Lemma 20.5. It suffices to prove that if in YJ[G], q(a. u ) defines a function K : Ord -+9l[G], then K[A] is a set in !Ul[C]for every regular cardinal 1. Without loss of generality, let us assume that for every p E P,

nacA

(20.17) p

IF

for every a there is a unique u such that cp(a, v).

Let 1be a regular cardinal, and let us consider again P = P>’x P “, and G = (G n P>’)x G1. As in Lemma 20.5, let us define, for each a < 1,a class

21.

197

SYMMETRIC SUBMODELS OF GENERIC MODELS

D, c P>': p E D, iff there is a maximal incompatible W E P S a and a family {a',".;: q E W } such that for each q E W , (20.18) As in Lemma 20.5, each D,, a < I, is open dense; since P" is I-closed, D, is dense and there exists p E G n P >' such that p E D, for all a < I. We pick (in W )for each a < 1 a maximal incompatible W, E P s and a family {a',".;: q E W,} such that (20.18) holds for each q E W,. Now if we let S = {at; : a < I and q E W,}, then it follows that

nu
I P " ( A ) ( .The set P of forcing

21.

209

SYMMETRIC SUBMODELS OF GENERIC MODELS

conditions consists of functions p with values 0, 1 such that 1 dom(p) 1 < K and dom(p) c ( A x K ) x K ; as usual, p < q iff p =I q. Let G be an 9bgeneric filter on P. For each a E A and each < K, we let


xo} is infinite. Then pick xl, x 2 , . . . similarly.]

An antichain in T is a set A G T such that any two distinct elements x, y of A are incomparable, i.e., neither x < y nor y < x. Let us call a tree T a Suslin tree if:

(22.2)

(i) the height of T is w , ; (ii) every branch in T is at most countable; (iii) every antichain in T is at most countable.

Lemma 22.1. There exists a Suslin line if and only ifthere exists a Suslin tree. Proof. (a) Let S be a Suslin line. We shall construct a Suslin tree, The tree will consist of closed (nondegenerate) intervals on the Suslin line S. The partial ordering of T is by inverse inclusion : If I, J E T, then I 5 J if and only if I 3 J.

4. SOME APPLICATIONS OF

218

FORCING

The collection T of intervals is constructed by induction on a < w I . We let I, = [a,, b,] be arbitrary (such that a, < b,). Having constructed I,, /3 < a, we consider the countable set C = {(I, : /3 < a} u {b, : /3 < a} of endpoints of the intervals I,, /3 < a. Since S is a Suslin line, C is not dense in S and so there exists an interval [a, b] disjoint from C;we pick some such [a,, bJ = I,. The set T = {I, : a < q}is uncountable and partially ordered by 2. If a < /3, then either I, 3 I, or I, is disjoint from I,. It follows that for each a, {I E T : I 3 I,} is well-ordered by 2 and thus T is a tree. We shall show that T has no uncountable branches and no uncountable antichains. Then it is immediate that the height of T is at most 0 , ;and since every level is an antichain and T is uncountable, we have height(T) = wl. If I, J E Tare incomparable, then they are disjoint intervals of S; and since S satisfies the countable chain condition, every antichain in T is at most countable. To show that T has no uncountable branch, we note first that if b is a branch of length wlr then the left endpoints of the intervals I E b form an increasing sequence {x, : a < ol} of points of S. It is clear that the intervals (x,, x,+ a < wl, form a disjoint uncountable collection of open intervals in S, contrary to the assumption that S satisfies the C.C.C. (b) We shall now assume that there exists a Suslin tree and construct a Suslin line. The idea is to let S consist of branches of the tree. However, since the set S so constructed might not generally be dense, we shall first normalize the tree. Let a be an ordinal number, a I wl. A normal a-tree is a tree T with the following properties: (22.3)

(i) (ii) (iii) (iv)

height(T) = a ; T has a unique least point (the root); each level of T is at most countable; if x is not maximal in T, then there are infinitely many y > x at the next level (immediate successors of x); (v) for each x E T there is some y > x at each greater level (less than a);

(vi) if

/3 < a is a limit ordinal and

{z : z

< x} = {z : z < y}, then x

x, y are both at level

/3 and if

= y.

See the following exercise for justification of conditions (ii) and (vi): Exercise 223. If T is a normal a-tree, then T is isomorphic to a tree T whose elements are 8-sequences (0 < a), ordered by extension; i f f G sand s E T, then t E T,and the 8th level of T is the set {t E T : dom(t) = 8).

A normal Suslin tree is a Suslin tree that satisfies the normality conditions (22.3). We shall show that given a Suslin tree, we can obtain a normal Suslin tree. Let T be a Suslin tree. T has height wl, and each level of T is countable. We first discard all points x E T such that T, = {y E T : y 2 x} is at most count-

22.

SUSLIN'S PROBLEM

219

able, and let T, = {.x E T : T, is uncountable). Note that if x E TI and a > o(x), then I T.1= S1for some y > x at level a. Hence TI satisfies condition (v). Next, we consider only branching points of the tree, i.e., points that have at least two immediate successors. For each x E T,, there are uncountably many branching points z > x in 7''(otherwise, all branching points z > x would be below a certain level and then there would exist an uncountable branch). Thus T2 = {thebranching points of TI}is a Suslin tree with properties (i), (iii), (v), and (iv*), where (iv*) stands for: each x has at least two immediate successors. Next we satisfy property (vi): We consider all chains C in T2 with the property that C = { z : z < x} = { z : z < y) for some x # y. For every such C we add to T2 an extra node a, and stipulate that z < a, for all z E C , and a, < x for each .Y such that x > z for all z E C . The resulting tree T3 satisfies (i), (iii), (iv*), (v), and (vi). To get property (iv), let T4 consist of all x E T, at limit levels of T,. The tree T4 satisfies (i), (iii), (iv), and (v); and then T5 c T4 satisfying ( i i ) as well is easily obtained. Now we can finish the proof of Lemma 22.1. Let T be a normal Suslin tree. The line S will consist of all branches of T (which are all countable). Each x E T has countably many immediate successors, and we order these successors as rational numbers. Then we order the elements of S lexicographically: If a is the least level where two branches a, b E S differ, then a is a successor ordinal and the points a, E a and b, E b at level a are both successors of the same point at level a - 1 ; we let a < b or b < a according to whether a, < b, or b, < a,. It is easy to see that S is linearly ordered and dense. If (a, b ) is an interval in S, then one can find x E T such that I , c (a, b), where I , is the interval I , = {c E S : x E c). And if I , and I , are disjoint, then x and yare incomparable points of T . Thus every disjoint collection of open intervals of S must be at most countable, and so S satisfies the countable chain condition. The line S is not separable: If C is a countable set of branches of T, let a be a countable ordinal bigger than the length of any branch b E C . Then if x is any point at level a, the interval 1,does not contain any h E C , and so C is not dense in S. Before going further, let me make the following observation: If T is a normal tu,-tree and if T has a branch of length ol, then T has an uncountable antichain. For, if h is an uncountable branch, pick for each x E b a successor z , of x such that z , $ h. Then A = ( z , : x E b ) is an antichain. Thus in order that a normal (]),-tree T be Suslin, it suffices that T has no uncountable antichain. Suslin lines have often been used in point set topology, mostly to construct nonseparable examples of various types of spaces; it turns out that since existence of Suslin lines cannot be proved in ZFC, the same is true for many of these topological examples: Their existence cannot be proved in ZFC. If T is a tree, let P , denote the partially ordered set that is obtained from ( T , I ) by reversing the order. Note that any two elements of P I . are either

220

4.

SOME APPLICATIONS OF FORCING

comparable or incompatible; they are incompatible just in case they are incomparable as points of the tree T . Thus if T is a Suslin tree, then P T satisfies the countable chain condition. Lemma 22.2. If T is a normal Suslin tree, then P , is K,-distriburive. Corollary. If T is a normal Suslin tree, then B c.c.c., atomless, complete Boolean algebra.

= r.0.

P , is an KO-distributive,

ProoJ Let D , , n = 0, 1,2, ...,be open dense subsets of P , .We shall prove that (-)7."=,, D, is dense in PT , First I claim that if D G P , is open dense, then there is a < o,such that D contains all levels of T above a. To prove this, let A be a maximal incompatible subset of D. A is an antichain in T and hence countable. Thus pick a < w 1 such that all a E A are below level a. Now if x E T is at level >a, there is d E D such that a 5 , d (by density), and by maximality of A, d is comparable with some a E A . Hence a I d (draw a picture), and consequently a 5 , x a1 such that each t E is comparable with some a E A n etc. If a. < a l < a2 < ... a, < ... is constructed in this way and if a = limn a,, then A n T, is a maximal antichain in T,. W a1 > a.

x2,

x,

Now we are ready to construct a Suslin tree, using element (besides 0 )is Lemma 22.5. Lemma 22.8. If

0.Again,

the key

0 holds, then there exists a s u s m tree.

Pro08 We construct a normal Suslin tree ( T , < T ) by induction on levels. To facilitate the use of 0,we let points of T be countable ordinals, T = wl, and in fact each T, (the first a levels of T ) is an initial segment of wl. We construct T, , a < w l rsuch that each T, is a normal a-tree and such that 5 extends T, whenever p > a. TI consists of one point. If a is a limit ordinal, then (T,, < T ) is the union of the trees ( T p , < T ) , /? < a. If a is a successor < T ) obtained by adjoining ordinal, then (T,,,, < T ) is an extension of infinitely many immediate successors to each x at the top level of T,. It remains to describe the construction of T,, if a is a limit ordinal. By 0, we are given a sequence S,, a < wl, with property (22.15). IfS, happens to be a maximal antichain in ( T , , < T), then we use Lemma 22.5 and find an extension (T,, < T ) of T, such that S, is maximal in T,, 1. Otherwise, we let T,+ be any extension of T, that is a normal (a + 1)-tree. (In either case, we let the set T,+ be an initial segment of countable ordinals.) T, is a normal Suslin tree. It We shall now show that the tree T = suffices to verify that T has no uncountable antichain. If A c T (= w is a maximal antichain in T, then by Lemma 22.7, A n T, is a maximal antichain in T,, for a closed unbounded set of a’s. It follows easily from the construction that for a closed unbounded set of a’s, T, = a. Thus using the diamond principle, we find a limit ordinal a such that A n a = S, and A n a is a maximal antichain in T,. However, we constructed T,+l in such a way that A n a is maximal in T , + l , and therefore in T. It follows that A = A n a and so A is countable. W

(x,

u, K, for all K < 2'O.I [Proof of Corollary: cf The proof uses almost disjoint subsets of w. If A,, A2 are subsets of w then we say that A, and A 2 are almost disjoint if A , n A2 is finite.

4. SOME APPLICATIONS OF

242

FORCING

Lemma 23.9. There exists a family of 2O ' pairwise almost disjoint subsets of w. Proof: Let S be the set of all finite (0,1)-sequences: S = uF=f=o "(0,1). For every f: w + {0, l}, let A, E S be the set A, = {s E S : s c f } = {f n : n E w } .Clearly, A, n A, is finite i f f # g; thus {A, :f E "(0, 1)) is a family of 2O ' almost disjoint subsets of the countable set S, and the lemma follows.

I

Let K < 2,O and let us assume that MA, holds. We fix a family {A, : a c K ) of almost disjoint subsets of o. Lemma 23.10. For every X c K there exists A (23.31)

a

E

X

iff

Cw

such that for all a < K ,

A, n A is infinite

Granted Lemma 23.10, the mapping that sends each X E K to some A E o such that (23.31) holds is one-to-one, and maps P ( K )into P(w).Hence 2" I 2'O. Proof. Let X c K. We let (PI c )be the following notion of forcing: A condition is a function p from a subset of w into (0, 1) such that:

(23.32)

(i) dom(p) n A, is finite for every a E X; (ii) { n : p ( n ) = 1) is finite.

The set P is partially ordered by inverse inclusion: p 5 q iff p extends q. We first show that P satisfies the countable chain condition. If p and q are incompatible, then { n : p ( n ) = 1) # { n : q(n) = 1) and since there are only countably many finite subsets of w, it follows that P satisfies the C.C.C. For each p E K - X, let D, = ( p E P : A, G dom(p)}. Any q E P can be extended to some p E D,: Simply let p ( n ) = 0 for all n E A, - dom(p). Since A, is almost disjoint from all A,, a E X,p has property (i) in (23.32) and hence is a condition. Thus each D, is dense. For each a E X and each k E w. let El,k = { p E P : { n E A , : p ( n ) = 1 ) has size at least k} It is easy to see that each Ea,k is dense in P. Let 9 be the collection of all D, (p E K - X )and all E,.,(a E X,k E w ) . By MA,, there exists a $generic filter G on P. Note thatf= G is a function on a subset of w. We let

u

A

= { n : f ( n )=

1) = { n : p(n) = 1 for some p

E

G)

If OL E X,then A n A, is infinite because for each k there is some p E G n E l . k , If p E K - X, then A n A, is finite because for some p E G, A, c dom(p) and { n : p ( n ) = 1 ) is finite. Theorem 53. MA,, implies that if P i , i ing the countable chain condition, then condition.

E

Hi

I, are partially ordered sets, all satigyE,

P i also satisjes the countable chain

23.

243

MARTIN’S AXIOM AND ITERATED FORCING

We first observe that it is enough to show that (assuming MA,,), the product of two partially ordered sets preserves the C.C.C.

ni

Lemma 23.1 1. fj t, Pi does not satisfy the c.c.c., then there is a finite S such that Pi does not satisjj the C.C.C.

ni

C

f

Proof: Let W be an uncountable pairwise incompatible subset of the product P . For each p E P , the support of p, s ( p ) = { i E I : p i # 1) is finite. By the A-lemma (Lemma 22.6), there exists an uncountable Z c W,and a finite set S c I such that s ( p ) n s(q) = S for any distinct elements p , q E Z. Then {p 1 S : p E Z} is an uncountable incompatible subset of Pi.

flies

If ( P , < ) is a partially ordered set, we say that P satisfies the strong countable chain condition if every uncountable W E P has an uncountable subset Z such that the elements of Z are pairwise compatible.

Lemma 23.12. If P and Q are partially ordered sets such that P satisfies the strong C.C.C. and Q satisfies the c.c.c., then P x Q satisfies the C.C.C. Proof: Let W be an uncountable subset of P x Q. If the set

W,

= {p E

P : ( p , q ) E W for some q E Q}

is at most countable, then there exists some p such that W, = ( 4 E

Q : (P,4 ) E

W)

is uncountable. Since Q satisfies the c.c.c., there are ql, q2 E Wpthat are compatible. Then (p, q l ) , (p, q 2 ) are compatible elements of W. If W, is uncountable, then since P satisfies the strong c.c.c., there exists an uncountable Z G W, such that any two elements of Z are compatible. For each p E Z pick q p E Q such that ( p , q p ) E W. If q p , = qp2for some pl, p 2 E Z, then ( p , , q p , ) ,( p 2 , q p , )are compatible elements of W. Otherwise, the set {qp: p E Z} is uncountable, and hence some q p , and qp2 are compatible. Then (pl, q p , ) , ( p 2 , q p 2 )are compatible. W Exercise 23.10. If P and Q satisfy the strong c.c.c., then P x Q satisfies the strong C.C.C. Exercise 23.1 1 . I f P the strong C.C.C.

=

nit,Pi and ifall Pi.i

E

I , satisfy the strong c.c.c., then P satisfies

In particular, if P is the notion of forcing that adjoins K generic reals ( 1 9 4 then P satisfies the strong C.C.C.(compare with Exercise 23.4). Theorem 53 will now follow from this lemma:

Lemma 23.13. MA,, implies that every partially ordered set that satisfies the satisfies the strong C.C.C.

C.C.C.

Proof: Let P be a partially ordered set that satisfies the C.C.C. and let W = {wa: a < wl} be an uncountable subset of P . We will use MA,, to find a filter G such that Z = G n W is uncountable.

4. SOME APPLICATIONS OF FORCING

244

First I claim that there is some p o E W such that every p I p o is compatible with uncountably many w , . Otherwise, for each a < o there is p > a and some u, I w, which is incompatible with all w , , y 2 /3; then we can construct an of pairwise incompatible elements. ol-sequence {u,, : i < ol} For each a < olrlet

D, = { p I po : p I w, for some y 2 a } By the above claim, each D, is dense below po . By MA,,. there exists a filter G on P such that p o E G and G n D, # 0 for all a < wl. It follows that G n W is uncountable.

H

24. SOME COMBINATORIAL PROBLEMS

In this section I shall discuss various combinatorial problems that are related to the topics covered in the two preceding sections. I shall prove several theorems dealing with trees, almost disjoint sets and functions, and ultrafilters, and state several related results without proof. Trees

First let us consider Aronszajn trees (see Section 22), that is, trees of height o1such that each level is at most countable. The Aronszajn tree constructed in

Lemma 22.3 has the following property: Its nodes (elements of T) are one-toone functions from some countable ordinal into a fixed countable set, and x I y just in case x c y . Let us call an Aronszajn tree that can be represented this way a special Aronszajn tree. [Special Aronszajn trees (and their generalization to greater heights) have application in model theory; this is because " there is a special Aronszajn tree" can be expressed as: " a certain sentence has a model of type (Kl, KO)."] Now let us consider a stronger property: An Aronszajn tree T is regular if there exists a function f: T -P o such that (24.1)

f ( x ) #f(y) whenever x and y are comparable

Exercise 24.1. A regular Aronszajn tree is special. [If x E T is at level a, represent x by the (a 1)-sequence ( f ( x o ) , j ( x , ) , j ( x e ) , . . . , j ( x ) ) , where xo < x, < ... < xc < ... are the predecessors of x.]

+

. . .,

Exercise 24.2. If T is a special Aronszajn tree, then the set of all successor nodes of T is a regular Aronszajn tree.

Lemma 24.1. r f T is an Aronszajn tree, then the following are equivalent: (i) T is regular (a) T is the union of countably many antichains.

24.

245

SOME COMBINATORIAL PROBLEMS

(iii) there exists a function R from T into the set Q of all rationals, such that x x, > ... > x,, > ....) Exercise 24.15. If ( P ,

If 2N0= K,,then there exists a p-point. (b) More generally, if Martin’s axiom holds, then there exists a p-point.

Theorem 57 (a)

258

4. SOME APPLICATIONS OF FORCING

In the proof of Theorem 57 we shall use functions from o to o.Let us say that a function h: u -+ u majorizes a functionf: o -, w if h(n) > f ( n )for all n. A family of functions X majorizes all functions f: u -+ w if for every f there is h E X such that h majorizesf: An easy diagonalization shows that no countable family majorizes all functionsf: w -, o.We can use MA to extend this observation: Lemma 24.10. (a) If .#' is countable, then X does not majorize all functionsf: o o. (b) If Martin's Axiom holds and I X I < 2'O, then X does not majorize all functions. -+

Proof: (a) If cR= {h,};=o, let f be a function such that f ( n ) I h,(n) for all k = 0, ..., n. ( b ) Let P be the collection of all finite sequences p = (p(O), . . .,p(n - 1)) of natural numbers; p < q iff p extends 4. The notion of forcing (PI < ) satisfies the C.C.C.

For each n, let D , = { p E P : n E dom(p)}; each D , is dense in P . For each h E .R, let Eh = {p E P : p(n) 2 h(n) for some n ) ; again, each E h is dense in P . Since I H 1 < 2'O, there exists a filter G on P that intersects each D,and each E,,. Then f = G is a function from o to o that is not majorized by any h e x .

u

The key element of the proof of Theorem 57 is the following lemma: Lemma 24.11. Lpt Y be a family of subsets of o with the finite intersection property and let us assume that (a) Y is countable, or more generally, that (b) MA holds and ( Y I < 2K0.kt A. 2 A , 2 . * * 2 A, 2 be a decreasing sequence of elements of Y. Then there exists Z G w such that: (i) 9 u {Z}has thefinite intersection property; (ii) Z c A, mod.f., for all n E o. Proof: We may assume that if X,Y E Y,then X n Y E Y. For each X E 9, let h,: w -+ o be some function such that h d n ) E X n A,. B y Lemma 24.10, the family {h, : X E 9 }does not majorize all functions; and so there is a functionf: o o such that for every X E Y, -+

(24.24)

f ( n ) 2 h,(n)

for some n

Now we let

u {k~A,:kIf(n)} m

(24.25)

Z=

n=O

It is readily verified that Z - A,, is finite for each n, and by (24.24),Z n X # 0 for every X E 9.

24.

SOME COMBINATORIALPROBLEMS

259

Proof of Theorem 57. Let .d,, a < 2no, be an enumeration of all decreasing sequences {A,},“=, of subsets of w. We construct, by induction on a < 2N0,a chain of families Yo E . . . E Yuc .. * of subsets of w, such that each 9,is closed under finite intersections and 19, I < 2N0for all a. We let 9, = {X C w : o - X is finite}. If a is a limit ordinal, we let 9, = 9,. Having constructed 9,, we construct 9,+1 as follows: Let .d,= {A,},“=, be a decreasing sequence of subsets of w. If some A, is disjoint from some X E g U ,then we let Y,, = 9. Otherwise, the family 9 = teu u { A , : n E w } has the finite intersection property and by Lemma 24.1 1, there exists Z c w such that Z c A, m0d.f. for all n, and 9‘ = te u {Z}has the finite intersection property. Then we let 9,+ consist of all finite intersections X , n ... n X, of elements of 9“. Finally, we let Y = (9, : a < 2’O}, and let D be any ultrafilter such that D 2 3.We claim that D is a p-point: if A, 2 A , 2 ... 2 A, 2 .. . is any decreasing sequence ofelements of D, then {A&“=, = .d, for some a < 2”O and we have 2 E Y,, such that Z c A, m0d.f. for all n.

up hs(rs,)> ..., a contradiction.

Degrees of Constructibility The relation between two sets of ordinals

(26.22) x is constructible from y , i.e., x

E

yy]

is a transitive relation (and reflexive). Thus one may be tempted to investigate the partial ordering of the corresponding equivalence classes, the degrees of consrructihilify. One might even expect a certain similarity between degrees of constructibility and degrees of recursive unsolvability (Turing degrees) investigated in recursion theory. Unlike the case for Turing degrees, there has been no systematic study of degrees of constructibility. There are a few facts known and a few results that I should mention. Theorem 64 provides an example of a minimal degree. Using a similar technique, models have been obtained where the degrees of constructibility form various finite patterns. Not much is known about, however, how, for instance, the initial segments of degrees look in general. (One result worth mentioning is a model in which the nonconstructible degrees have order-type w*, i.e., the order-type of negative integers.) Any two degrees have a least upper bound. (This is easy: If A and Bare sets of ordinals, let C = r[A x B] where r is the canonical mapping of Ord2 onto Ord; the degree of C is the 1.u.b. of the degrees of A and B.) The following theorem of Hajek, stated here without proof, shows that the same is not true for lower bounds: I t is consistent that there are two r e d s X , Y, Cohen-generic over L such that the constructibility degrees of X and of Y have no greatest lower bound. It follows that 4x1 n YY] is not a model of ZFC (and it is unclear whether it is even a model of ZF). The situation is similar for suprema and infima of countable sequences of degrees. For instance, if d o , d , , ..., d , , ... is an increasing (or decreasing) sequence of Ldegrees, then its supremum (infimum) may or may not exist.

4.

292

SOME APPLICATIONS OF FORCING

A related problem is the structure of transitive models of ZFC (containing all ordinals). Above, I mentioned an example of two models %Il and %I2 of ZFC such that (xn, n (xn2 is not a model of ZFC.Similarly, there is an example 3 Iu1, 3 . * . 2 91" 3 ... of models such that of a decreasing sequence 91, is not a model. Or consider this:

n;=o

Exercise 26.5. I f lllo c 111, c ... c 111, c ... is a strictly increasing sequence of models 111, is not a model of ZF. (containing all ordinals), then U = [If U n V, E U for all a, then there is n such that U n V, E 9.R" for arbitrarily large a ; hence U G Zn, .]

u."=o

A somewhat better picture can be obtained when generic extensions are used. For instance: Exercise 26.6. If Bo 2 B , 1 .. . z B, 2 ... is a sequence in 111 of complete Boolean algebras and if G c B is generic over !Ill, then m

W [ G n B,]

91 = n=O

is a model of ZF. [31 is transitive and closed under Godel operations. To show that for each a, V. n 9l E W, note that for every n, V. n 31 E 91[G n B,]:

n (V. n W[G, 00

V, n 9 l =

n

B,])

k=n

where G, = G n B, .]

Even when we have a sequence of models Iu1[G n BJ as in Exercise 26.6, the intersection need not satisfy the axiom of choice. I shall conclude this discussion with the following lemma, which shows that in some cases the intersection of a decreasing sequence of models of ZFC is a model of ZFC:

Lemma 26.6. Let K be a cardinal and let B be a K-distributive complete Boolean algebra. Let Bo 2 B 2 . * * 2 B, 2 * . .,a < K , be a descending K-sequence of comB, . lf G is an 9J-generic ultrajilter on plete subalgebras of B, and let B, = B, and G , = G n B, for all a I K , then

nu 0. is an atomless measure on S , there exists Z c S such that p ( Z ) = 4. More generally, given Z o E S , there exists Z c Zo such that p ( 2 ) = fp(Z,). [Construct a sequence S = So 2 S I z ... z S , 2 ..., a < w , , such that p(S,) 2 i,

Exercise 27.3. If p

27.

297

THE MEASURE PROBLEM

and if p ( S , ) > i,then i 5 p(&+ ,) .= p ( S J ; ifa is a limit ordinal, let S, exists ct < w , such that p(S,) = 4.3

=

S,. There

We shall eventually prove various strong consequences of existence of a nontrivial a-additive measure and establish the relationship between the measure problem and large cardinals. Our starting point is the following theorem which shows that if a measure exists, then there exists at least a weakly inaccessible cardinal.

Theorem 66 (Ulam). Ifthere is a a-additive nontrivial measure on S , then either there exists a two-valued measure on S and I S I 2 the least (strongly)inaccessible cardinal, or there exists an atomless measure on 2'O and 2'O 2 the least weakly inaccessible cardinal. Theorem 66 will be proved in a sequence of lemmas, which will also provide additional information on the measure problem and introduce basic notions and methods of the theory of large cardinals. First we make the following observation. Let K be the least cardinal that carries a nontrivial a-additive two-valued measure. Clearly, K is uncountable and is also the least cardinal that has a nonprincipal countably complete ultrafilter. And we observe that such an ultrafilter is in fact K-complete:

Lemma 27.1. Let K be the least cardinal with the property that there is a nonprincipal a-complete ultrafilter over K , and let U be such an ultrafilter. Then U is K-complerr. Pro05 Let U be a a-complete ultrafilter over K, and let us assume that U is not ri-complete. Then there exists a partition {X, : a < y} of K such that y < K and X , 4 U for all a < y. We shall now use this partition to construct a nonprincipal a-complete ultrafilter over y. thus contradicting the choice of K as the least cardinal that carries such an ultrafilter. Let f be the mapping of K onto y defined as follows:

f ( x )= a

x E X,

iff

(x E

K)

Now, any mapping of K onto y induces a a-complete ultrafilter over y; we define D E P ( y ) thus: (27.6)

ZED

iff

~ - , ( Z ) UE

(We leave it to the reader to verify this.) Moreover, our mappingfis such that D is nonprincipal: Assume that { a ) E D for some a < y. Then X, E U , contrary to our assumption on X,. Thus y carries a a-complete nonprincipal ultrafilter. We are now ready to define the central notion of this chapter. An uncountable cardinal K is mra.surable if there exists a K-complete nonprincipal ultrafilter U over K .

5.

298

MEASURABLE CARDlNALS

By Lemma 27.1, the least cardinal that carries a nontrivial two-valued a-additive measure is measurable. Note that if U is a K-complete nonprincipal ultrafilter over K, then every set X E U has cardinality K because every set of smaller size is the union of less than K singletons. For similar reasons, K is a regular cardinal because if K is singular, then it is the union of less than K small sets. The next lemma gives a first link of the measure problem with large cardinals. Lemma 27.2. Every measurable cardinal is inaccessible.

ProoJ We have just given an argument why a measurable cardinal is regular. Let me show that measurable cardinals are strong limit cardinals. Let K be measurable, and let us assume that there exists A < K such that 2L 2 K ; we shall reach a contradiction. Let S be a set of functionsf: 1 + {0, 1) such that I S 1 = K , and let U be a K-complete nonprincipal ultrafilter over S. For each a < 1,let X,be that one of the two sets { f S ~: f ( a )= 0), { f S~: f ( a )= 1) which is in U,and let &,be0or 1 accordingly. Since U is K-complete, the set X = X,is in U . However, X has at most one element, viz. the functionf that has the valuesf(cr) = E , . A contradict ion.

n,
0, there are only finitely many sets X E W of measure 2 l/n. A a-complete nonprincipal ideal I over S is called a-saturated if it satisfies (27.8). More generally, a K-complete nonprincipal ideal I over S is tc-saturated if it satisfies (27.8)(i) and if every disjoint family W c P ( S ) of sets not in I has size < K. Let me make a small digression: Let I be a K-complete nonprincipal ideal over S, and let B be the quotient Boolean algebra B = P ( S ) / I . Note that B is K-saturated (cf. (17.15)) just in case there is no W c P ( S ) of size K such that X 4 I for each X E W and X n Y E I whenever X and Y are distinct elements of w. Exercise 27.4. Let S, 1, and B be as above. Show that B is K-saturated if and only if I is K-saturated.

27.

299

THE MEASURE PROBLEM

[If W c P ( S ) - I has size K , and X n Y E I whenever X # Y E W , let W = { X , : tl < K ) , and let Z , = X, X , ; the family (2, : a < K } is disjoint, and Z , $ I for each tl.]

Up..

The following lemma is an analog of Lemma 27.1 : Lemma 27.3.

(a) Lut K he the least cardinal that carries a nontrivial a-additive measure and let ,u he such a measure on K . Then the ideal I , of null sets is K-complete. (b) Let K he the least cardinal with the property that there is a o-complete a-saturated ideal over K, and let I he such an ideal. Then I is K-complete. Proof. (a) Let us assume that I, is not K-complete. There exists a collection of nuli sets { X , : ~1 < $ such that y < K and that their union X has positive measure. We may assume without loss of generality that the sets X a , a < y, are pairwise disjoint; let m = p ( X ) . Let f be the following mapping of X onto y:

f(x)=cc

iff

XEX,

(XE

X)

The mapping f induces a measure v on y : (27.9) The measure v is o-additive and is nontrivial since v{a} = p ( X a ) = 0 for each a E y. This contradicts the choice of K as the least cardinal that carries a measure. (b) The proof is similar. We define an ideal J over y by: 2 E J ifff- l(Z) E I. The induced ideal J is a-complete and maturated. Let {ri : i (27.10)

E

I} be a collection of nonnegative real numbers. We define

1 ri = sup{ 1 ri : E is a finite subset of I } icl

isE

Note that if the sum (27.10) is finite (i.e., is not a), then at most countably many terms ri are nonzero. Let K be an uncountable cardinal. A measure p on S is called K-additive if for every y < K and for every disjoint collection X a , a < y, of subsets of S, we have (27.11) Exercise 27.5. Let p be a two-valued measure, and let U be the ultrafilter of all sets of measure one. The measure p is K-additive if and only if U is K-complete.

If p is a K-additive measure, then the ideal I, of null sets is K-complete. The converse is also true, and we get a better analog of Lemma 27.1 for real-valued measures :

5.

300

MEASURABLE CARDINALS

Lemma 27.4. Let p be a measure on S, and let I , be the ideal of null sets. If I , is K-complete, then p is u-additive.

Proof. Let y < K, and let X u ,a < y, be disjoint subsets of S. Since the X, are disjoint, at most countably many of them have positive measure. Thus let us write

{Xu: a < y } = (Yn : n = 0, 1, 2, . . .} u { Z , : a < y } where each Z , has measure 0. Then we have

Now first p is a-additive, and we have

and secondly I, is K-complete and

c.

P(Xd Thus PCU. Xu)= Corollary. Let K be the least cardinal that carries a nontrivial a-additive measure and let p be such a measure. Then p is K-additive.

An uncountable cardinal K is real-valued measurable if there exists a nontrivial u-additive measure p on K. By the corollary above, the least cardinal that carries a nontrivial a-additive measure is real-valued measurable. We shall show that if a real-valued measurable cardinal K is not measurable, then K I2'O. Note that if p is a nontrivial u-additive measure on K, then every set of size < K has measure 0, and moreover K cannot be the union of less than K sets of size < K. Thus a real-valued measurable cardinal is regular. We shall show that it is weakly inaccessible. We shall first prove the first claim made in the preceding paragraph. Lemma 27.5.

(a) If there exists an atomless nontrivial a-additive measure, then there exists a nontrivial a-additive measure on some K I2'O. (b) If I is a a-complete a-saturated ideal over S, then either there exists Z G S such that I ( Z = {X c Z : X E I } is a prime ideal, or there exists a acomplete maturated ideal over some K I2'O. Proof: (a) Let p be a measure on S. We construct a tree T of subsets of S, partially ordered by reverse inclusion. The 0th level of T is {S}.Each level of T

27.

30 1

THE MEASURE PROBLEM

consists of pairwise disjoint subsets of S of positive measure. Each X E T has two immediate successors: We choose two sets Y, Z of positive measure such that Y u Z = X and Y n Z = 0. If a is a limit ordinal, then the ath level consists of all intersections X = X, such that each X, is on the 5th level of T and such that X has positive measure. We observe that every branch of T has countable length: If {X, : 5 < a} is a branch in T , then the set { 6 : 4 < a}, where 6 = X, - X,+ is a disjoint collection of sets of positive measure. Consequently, T has height at most wl. Similarly, each level of T is at most countable, and it follows that T has at most O 2' branches. Let { b , : a < K}, K I 2'O, be an enumeration ofall branches b = { X , : 5 < y} such that X, is nonempty; for each a < K, let Z , = {X : X E b,}. The collection { Z , : a < K } is a partition of S into K sets of measure 0. Using what is by now a standard argument, we induce a measure v on K as follows: Letfbe the mapping of S onto K defined by

n, K. Let d (the diagonalfunction) be the function on

(28.20)

d ( a )= a

K

defined by

(a < K )

Since U is K-complete every bounded subset of K has measure 0 and so for every y < K, we have d(u) > y for almost all a . Hence [dl > y for all y < K and thus [ d ] 2 K. However, we clearly have [d] < j ( ~and ) it follows that j ( ~>) K. Exercise 28.6. Let U be a nonprincipal a-complete ultrafilter over S and let A be the largest cardinal such that U is I-complete. Then j ( A ) > A. [Let { X , : a < A} be a partition of S into sets of measure 0; letfbe a function on S such that /(x) = a if x E X , . Then [f]2 A.]

We have thus proved that if there is a measurable cardinal, then there is an elementary embedding j of the universe in a transitive model M such thatj is not the identity mapping; let us call j a nontrivial elementary embedding of the uniuerse. We shall use this to prove the following theorem : Theorem 67 (Scott).

If V = ,!I then there are no measurable cardinals.

Proof. Let us assume V = L and that measurable cardinals exist; let K be the least measurable cardinal. Let U be a nonprincipal K-complete ultrafilter over K and let j : V + M be the corresponding natural embedding. As we have shown, J(K) > K. Since V = L, the only transitive model containing all ordinals is the universe itself: V = M = L. Since j is an elementary embedding and K is the least measurable cardinal, we have M 1 j ( K ) is the least measurable cardinal; and hence, j ( K ) is the least measurable cardinal. This is impossible since j(K) > K. If there exists a measurable cardinal, then we obtain a nontrivial elementary embedding of the universe. Let me show that conversely, if j : V + M is a nontrivial elementary embedding then there exists a measurable cardinal. (A word of caution: “There exists a nontrivial elementary embedding of V ” or even “ jis an elementary embedding of V” is not a statement expressible in the language of ZF.) Lemma 28.5. l f j is a nontrivial elementary embedding of the universe, then there exists a measurable cardinal.

Proof. Let j : V -+ M be a nontrivial elementary embedding. Notice that there exists an ordinal a such that j ( a ) # a ; otherwise, we would have rank(jx) = rank(x) for all x, and then we could prove by induction on rank that j(x) = x for all x. Thus let K be the least ordinal number such that j ( K ) # K (and hence j ( ~>) K). It is clear that j ( n ) = n for all n andj(o) = o since 0, n + 1, and o are

5.

312

MEASURABLE CARDINALS

absolute notions and j is elementary. Hence K > o.We shall show that measurable cardinal. Let D be the collection of subsets of K defined as follows: (28.21 )

XED

K

is a

KE~(X) (XGK)

iff

Since K < j ( K ) , i.e., K E ~ ( K ) we , have K E D ; also 0 4 D because j ( 0 )= 0. Using the fact that j(X n Y) = j ( X ) n j ( Y) and that j ( X ) c j ( Y) whenever X c Y , we see that D is a filter: If K E j ( X ) and K E j ( Y), then K E j(X n Y);if X E Y and K E ~ ( X )then , K € j ( Y ) .Similarly,j ( K - X ) = j ( ~- )j ( X ) and thus D is an ultrafilter. D is a nonprincipal ultrafilter: For every a < K, we have j({a})= {j(a)}= {a}, and so K 4 j({a})and we have {a}4 D. We shall now show that D is K-complete. Let y < K and let X = (X,: a < y) be a sequence of subsets of K such that K E ~ ( X ,for ) each a < y. We shall show that X, E D.In M (and thus in V), j ( X ) is a sequence of lengthj(y) of subsets ofj(K); for each a < y, the j(a)th term of j ( X ) is j ( X , ) . Since j(a) = a for all a < y and j ( y ) = y, it follows that j ( X ) = ( j ( X , ) : a .c 7). Hence if X = Om K. Exercise 28.9. Ifj: V + M is a nontrivial elementary embedding, if K is the least ordinal moved, and if 1 = lim{K, j ( K ) , j(j(K)), . . .}, then there exists A c 1 such that A $ M. [Assuming that M contains all bounded subsets of 1, the above proof that G M goes through.]

Let me now consider ultrapowers and the corresponding natural embeddings j,: V + Ult. To introduce the following lemma, let us observe that if j: V -,M and if K is the least ordinal moved, thenj(x) = x for every x E V,, and j(X) n V, = X for every X G V,. Hence VF+ = V,, (and P ( K=)P ( K ) ) .

Lemma 28.9. Let U be a nonprincipal K-complete ultrafilter over M = Ult,(V) and let j = j, be the natural embedding of V in M .

K,

let

(a) " M c M , i.e., every K-sequence (a, : a < K ) of elements of M is itself a member of M . (b) u M . (c) 2" I ( 2 " ) M < j ( K ) < (2")+. (d) I f I i s a limit ordinal and ifcf I = K, then j(I)> lim,+Aj(a); ifcf 1 # K, then j ( A ) = lim,+Aj(a). (e) If I > K is a strong limit cardinal and cf I # K, then j ( I ) = I .

z

Proof. (a) Let (aC: ( < K ) be a K-sequence of elements of M. For each 5 < K, let ge be a function that represents a t , and let h be a function that represents K : [SCI = at.

[hl = K

We shall construct a function F such that [ F ] = (at : 5 < K). We let, for each a < K, F(a) = ( s t ( a ) : 5 < h ( 4 ) Since for each a, F(a) is an h(a)-sequence, [ F ] is a K-sequence. Let 5 < K ; we want to show that the 5th term of [F] is a t . Since [h] > 5, we have 5 < h(a) for almost all a ; and for each a such that ( < h(a), the 5th term of F(a) is gda). But [ce]= 5 and [ g C ]= a t , and we are done. (b) Assume that U E M, and let us consider the mapping e of " K onto&) defined by e( f )= [ f .Since " K E M and U E M, the mapping e is in M . It follows that M 1 J ~ ( KI) 5 2". This is a contradiction since K < j(K) a n d j ( ~is) inaccessible in M. (c) 2" I (2")M holds because P M ( ~=)P ( K )and M c V; (2")M is less than j ( K ) since j ( K ) is inaccessible in M; finally, we have Ij ( K ) I = 2" and hence j ( K ) < (2")+. (d) If cf I = K, let 1 = lim,.+KI, and letf(a) = 1, for all a c K . Then [f]>

5.

316

MEASURABLE CARDINALS

j(1,) for all a < K and [ f ] c j(1). If cf I > K, then for everyf:

K + 1 there exists c 1such that [f]<j(a). If cf I = y < K, let I = Iim,+, 1,;for everyf: K + I there exists (by K-completeness) v < y such that [f]< j ( I , ) . (e) For every a c 12, the ordinals below a are represented by functions f: K + a ; hence I j ( a ) I I I ' a 1 < 1;by (d) we have j(1)= lim,,+A j ( a ) = 1. H

a

Remark. Note that in (e) it suffices to assume that cf I # cardinals a < I.

K

and a' c I for all

Normal Measures We defined the notion of a normalfilter in (7.18); a filter is normal if it is closed under diagonal intersections. If D is a nonprincipal complete ultrafilter over K and if D is normal (as a filter), then we call D a normal (two-valued) measure on K. Lemma 28.10. Let D be a nonprincipal K-complete ultrajilter over following are equivalent :

K.

Then the

(i) D is normal. (ii) Iff is such that f(a) c a for almost all a, then there is y c K such that f ( a ) = yfor almost all a. (iii) I n the ultrapower UltD(V), K =

where d is the diagonal function. (iv) For every X C K , X E D if and only

[d]

if^ €jD(X).

Thus D is normal just in case the diagonal function is the least nonconstant function (i-e., the least function that is not constant almost everywhere). Proof: (i) + (ii): Let us recall that the diagonal intersection of {X, : y < K} is the set

A X , = { ~ < K : ~0EX,) Y t i + , then the set

{ a < K : 2" > a + } has measure one, for every normal measure on K . Consequently, if2" = a+ for all cardinals a < K , then 2" = K + .

-= K : tl is a cardinal} is closed unbounded and

(As for the notation, note that { a hence has a normal measure one.)

Proof: Let D be a normal measure on K , and let M almost all a, then, since [d], = K, we have

M k

=

Ult,( V ) . If 2" = a+ for

2'=K+

In other words, there is a one-to-one mapping in M between P"(K) and ( K + ) ~ = K + because ~ K + (in fact, (K')" However, P"(K) = P(ti) and ( K + ) 5 P M ( t i )= P(ti)),and so 2" = t i + . Similarly, we have: Exercise 28.15. If D is a normal measure on K and { a : 2" I a + + } E D,then 2" 5 K + + . More generally, if D < K and {K,: 2NmI K.+B} E D, then 2" 5 K.+B. [Iffis such thatf(K,) = K.+P for all a c K , then [ f I D = (K,+jc,#' I KK+B.] Exercise 28.16. I f D is a normal measure on 2" < K K + K . [If{(.) = %+,. then [/I = ( K K + K ) M . ]

K

and { a : 2"< K,+,}

E

D, then

A similar method can be used for singular cardinals of measurable cofinality. The following results (which have been partly superseded by the more general theorems of Silver and Galvin-Hajnal; see Section 8) illustrate the method.

.

5.

320

MEASURABLE CARDINALS

Lemma 28.14. Let K be a measurable cardinal, let D be a normal measure on K and let j : V -+ M be the corresponding elementary embedding. Let I > K be a strong limit cardinal of cojnality K . Then 2A< j ( A ) . Proof. We recall that by Lemma 28.9(d), since cf A = K, we have j ( A ) > A. We shall show that (28.27)

2' = I" I

(A")"

I( A j ' K ) ) M c j ( A )

The first equality in (28.27) holds because I is a strong limit. We have I" I (I")" because ("I)" = "I.As for the last inequality, we have

M C j ( A ) is a strong limit cardinal and since I < j ( I ) and j ( K ) < j(A), we have M k

A]'") < j ( I ) .

W

For applications of the lemma, see Exercises 28.18-28.20 below. K = cf 1 is measurable. If 2" = a' for all a < 1,then 2" = 1+. [Let D be a normal measure on K ; let 1= [ e l D .For almost all a, e(a). I (e(a))+,and SO Ult C 1"2 1 '. Thus 2" = 1" 5 (1")"2 (A')" 2 A+.]

Exercise 28.1 7. Let 1 be a singular cardinal such that

The result of this exercise is a special case of Theorem 23. In fact, Silver's original proof was motivated by measurable cardinals and is similar to the present method. I shall present the ultrapower proof in Chapter 6. Exercise 28.18. Let K be measurable and let

2" < K ( 2 . ) + . [Use Lemma 28.14: j(1) = (N~(,+.,)" 2

1 = K,+, be a strong limit. Then

Nj(K)+j,K);

j ( K ) + j ( ~ m ; if rn = K, then it is clearly false. Thus we always assume 2 I rn < K and 1 I K . If n = 1, then (29.2) holds just in case either K > A, or K = 1 and cf K > rn. We shall concentrate on the nontrivial case: n 2 2. We start with two negative partition relations

(AL

Lemma 29.2. For all K and 1, 2" % (4: I n other words, there is a partition of 2" into K pieces that does not have an injnite homogeneous set.

Proof. In fact, our partition has no homogeneous set of size 2 3. Let S = "(0, 1) and let F: [S]' + K be defined by F ( { J g } ) = the least a < K such

323

29. INFINITARY COMBINATORICS

that f ( a ) # g(a). I f J g, h are distinct elements of S, it is impossible to have F ( { J sf) = W J 4 )= F ( { g , h}). Lemma 29.3. For every K, (Thus the obvious generalization of Ramsey’s theorem, namely K 1-, (K1)i, is false.) To construct a partition of [2“12that violates the partition property, let us consider the linearly ordered set ( P , < ) where P = “(0,l}, a n d f < g if and only if f ( a ) < g(a) where a is the least a such that f ( a ) # g(a) (the lexicographic ordering of P). Lemma 29.4. The lexicographically ordered set “(0, 1) has no increasing or decreasing K -sequence. +

Proof: Assume that W = {f,: a < K + } c “2 is such thatf, A, and therefore we can find a 1 < ... C(k < a k + ]< ... < C(2k in H such that F ( ( a l , ..., a,})= F({ak+l, a2k}). It follows that H is homogeneous for F : If a1 < ... < a,, and PI < ... < /I,, are two sequences in H, we choose a sequence y 1 < < y,,in H such that both x,, < y1 and p,, < yl. Then

-=

.-.?

The combinatorial methods introduced in this section will now be employed to obtain a result on measurable cardinals considerably stronger than Scott's theorem in the preceding section. Let me first make a few observations about models with definable Skolem function. Let 91 be a model of a language Y such that 91 2 K and let I G K be a set of indiscernibles for PI. Let us assume that the model VI has definable Skolem

29.

331

INFINITARY COMBINATORICS

functions; i.e., for every formula* q ( u . u l , . . .,0), there exists an n-ary function h, in PI such that: (i) lg, is dejnable in PI, i.e., there is a formula $ such that

iff

y = h(xl, ..., x,)

PI 1 $[y, xl,

..., x,,]

for all y , xl, . . ., x,, E A ; and (ii) lg, is a Skolemfunctionfor cp (see Section 10). Let B E PI be the closure of I under all functions in Y and all the functions h,,,, cp a formula. B is an elementary submodel of PI, and in fact is the least elemen-

tary submodel of PI that includes the set I; let us call B the Skolem hull of I and let us say that I generates '8. We augment the language of PI by adding function symbols for all the Skolem functions and call Skolem terms the terms built up from variables and constant symbols (which are 0-ary functions) by applications of functions in Y and the Skolem functions. Since B is an elementary submodel of PI, the interpretation of each Skolem term t is the same in B as in PI. For every element x E B there is a Skolem term r and indiscernibles y 1 < ... < y,, (elements of I) such that (29.14) x = t'"[yl, Y,] = t " [ y l , ..., Y,] a * . ,

Now if $ is a formula of the augmented language, i.e., if $ also contains the Skolem terms, it still does not distinguish between the indiscernibles: If a l < ... < a, and p1 < ... < p, are two sequences in I, then $ ( a l , ..., a,,) holds (either in '21 or in B) if and only if $(/II, . . ., p,) holds.

Theorem 71 (Rowbottom). If K is a Ramsey cardinal, then the set of all constructible reals is countable. More generally, i f A is an injnite cardinal less than K , then I P"(A)l = A. ProoJ Let K be a Ramsey cardinal and let A < K. Since K is inaccessible, we have PL(A)c L, . Consider the model (29.15)

PI = ( L , E, P"(A),

PI is a model of the language V = {E, Q,c a } z 5 A where Q is a one-place predicate (interpreted in PI as P(A) n L) and c,, a I A, are constant symbols (interpreted as ordinals less than or equal to A). Since ti is Ramsey, there exists a set I of size ti of indiscernibles for PI. The model PI has definable Skolem functions: Since K is inaccessible, LKis a model of ZFC + V = L and therefore has a definable well-ordering (see also Exercise 13.14). Thus let B G L, be the elementary submodel of PI generated by the set I. Every element x E B is expressible as x = t ( y l , ..., y,,) where t is a Skolem term and y 1 < ... < y, are elements of I.

5.

332

MEASURABLE CARDINALS

We shall now show that the set S = P L ( I ) n 9 has at most I elements. Since S is the interpretation in 9 of the one-place predicate Q, it suffices to show that there are at most I elements x E b such that b C Q(x). Let t be a Skolem term. Let us consider the truth value of the formula (29.16)

t ( a l , ...,a,)=

[(Ply..., Pn)

for a sequence of indiscernibles a l < ... < a , < PI -= ... < P,. The formula (29.16) is either true for all increasing sequences in I or false for all increasing sequences in I. If (29.16) is true, then it is true for any two sequences al 0.1

It has been proved among other things that no K,, n < w, is a Jonsson cardinal and that the least Jonsson cardinal is either weakly inaccessible or has cofinality w . While it is consistent (relative to ZFC + there exists a measurable cardinal) that all Jonsson cardinals are Ramsey cardinals, there is also a model in which a Rowbottom (and hence Jonsson) cardinal exists and has cofinality w. One of the still open questions is whether K,,can be J6nsson. Somewhat more is known about Rowbottom cardinals: Exercise 29.13. I f

cf

K

ti

is a Rowbottom cardinal, then either

ti

is weakly inaccessible or

= Q.

[To show that K = 1’ is not Rowbottom, letf, be a one-to-one mapping ofa onto 1, for each a, such that 1 2 a < ti. Let PI = (ti, I , No. To show that ti is not Rowbottom if K > cf K = 1 > No,let f be a nondecreasing function of ti onto I and use f to produce a counterexample.]

A related problem is Chang’s conjecture: Every model of type (N2,K,)has an elementary submodel of type (N,,No). Chang’s conjecture is consistent relative to the theory ZFC + “there exists a Ramsey cardinal.” We shall show in Section 32 that Chang’s conjecture has the strong consequences stated in Theorem 71. For the following exercise, we recall that functionsfand g on w 1 are almost disjoinr i f f @ ) # g(a) for all a 2 some a. . Exercise 29.24.* If there is a family 9 of K 2 almost disjoint functionsf: w , + w then Chang’s conjecture fails. [Consider a model 91 with the universe 9 u w 1 and the designated predicate wl. If (9 u E, B, ...) ,,) is an E-formula and 5 , < ... < t,,, then (M, E) k cp[cr, . . . , cE] if and only if cp E C.Thus I is a set of indiscernibles for (M, E). Now we let A be the Skolem hull of I in ( M , E ) . Since 91 = (A, E) is an elementary submodel of ( M , E), it follows that I is a set of indiscernibles for 91, Z(%, I) = C, and that f " " ( I ) = . R ( M . E )=( IA). Hence (VI, I) satisfies (a), (b), (c). For each E.M. set I: and each ordinal a, let us call the (C, a)-model the unique pair (VI, I) given by Lemma 30.3. The uniqueness proof of Lemma 30.3 easily extends to give the following:

,

,,

Lemma 30.4. Lat C be an E.M. set, let a Ip, and let j : a + p be orderpreserijiny. Then ,j can be extended to an elementary embedding of the (X,a)-model into the (1, ,!+model.

Proof. Extend j as in (30.13).

We shall eventually show that a measurable cardinal implies existence of an E.M. set Z having a certain syntactical property (remarkabilitp) and such that a)-model is well-founded. Let us investigate well-roundedness first. every (1,

5.

344

MEASURABLE CARDINALS

Lemma 30.5. The following are equivalent, for any given E.M. set Z: (i) For every ordinal a, the (C, a)-model is well-founded. (ii) For some ordinal a L ol, the (C, a)-model is well-founded. (iii) For every ordinal a < ul, the (C, a)-model is well-founded. ProoJ: (i) + (ii) is trivial. (ii) + (iii): If (PI, I) is the (C,a)-model and if /? Ia, let J be the initial segment of the first /? elements of I; let 9 = M " ( J ) . Clearly, (23,J ) is the (Z, /?)-model. Since a submodel of a well-founded model is well-founded, it follows that if p 5 a and the (Z, a)-model is well-founded, then the (X,j?)-model is also well-founded, and thus (ii) implies (iii). (iii) + (i): Let us assume that there is a limit ordinal a such that the (X,a)-model is not well-founded; let (PI, I) be the model. There is an infinite sequence a, ,a I, a 2 , . . . in PI such that a E a , , a E a etc. Each a,,is definable from I; that is, for each n there is a Skolem term t,, such that a, = t:[x l , .. ., x,J for some xl, , . . , x k nE I. Therefore there is a countable subset I , of I such that a,, E .#'"(I,) for all n E o.The order-type of I , is a countable ordinal p and (.#"(Io), I,) is the (C, /?)-model.This model is clearly non-well-founded since it contains all the a,,. Hence for some countable /?,the (C,/?)-model is not well-founded.

I shall now define remarkability. To avoid unpleasant surprises in the middle of a proof, let me from now on consider only (C,a)-models where a is an infinite limit ordinal. Let us say that a (Z, a)-model (PI, I) is unbounded if the set I is unbounded in the ordinals of 91, that is, if for every x E Ord" there is y E I such that x < y. Lemma 30.6. The following are equivalent, for any given E.M. set Z: (i) For all a, rhe (C, a)-model i s unbounded. (ii) For some a, the (C,a)-model is unbounded. (iii) For every Skolem term t ( v I ,..., on) the set Z contains the formula ( p , ( v l , ..., u,,+ ,) staring (30.16) i f t ( u l , . . ., u,,) is an ordinal, then

t(o,,

. . ., v,,) < u,+

ProoJ: (i) + (ii) is trivial. (ii) + (iii): Let (PI, I) be a (Z, a)-model, where a is a limit ordinal, and assume that I is unbounded in Or#'. To prove (iii), it suffices to show that for any t , (30.16) is true in PI for some increasing sequence x 1 < ... < x , + in I . Let t be a Skolem term. Let us choose x < * * . < x,, E I and let y = t"[x 1, . . ., x,J. If y # Orrf', then (30.16) is vacuously true; if y E Or#', then there exists x,+ E I such that y < x,+ and we have PI t= t [ x , , ..., x n ] < x,,+ (iii)+ (i): Let (PI, I) be a (Z, a)-model, where a is a limit ordinal, and assume (iii). To prove that I is unbounded in Or&', let y E O r 6 . There is a

30.

345

SILVER INDISCERNIBLES

Skolem term rand x 1 < ... < x, E I such that y = t " [ x , , . . ., xn]. Now ifx,, any element of I greater than x,, (iii) implies that y < x,+ 1.

is

Thus we say that Z is unbounded if it contains the formulas (30.16) for all Skolem terms t. Let a be a limit ordinal, a > w, and let (PI, I )be the (C, a)-model. For each 5 < a, let i, denote the 5th element of I. We say that (PI, I) is remarkable i f it is unbounded and if every ordinal x of PI less than i,,, is in .f'({i, : n E o}).

Lemma 30.7. The following are equivalent for any given unbounded E.M. set C : (i) For all a > w, the (C, a)-model is remurkable. (ii) For some a > w, the (C, a)-model is remarkable. (iii) For every Skolem term t(x,, . . ., x,, y,, . . ., y,), the set I; contains the formula cp(x,, ..., x,, y , , . ... y,,, z , , ..., z,) stating (30.17) ift(xl, . . . , x,,

. . ., y,) is an ordinal and

J'~,

t(x1, . . . . X , , J ' ~ ?

...,yn)


is impossible since that would mean that y,, > ys whenever a < j. Thus (7. : a < K ) is an increasing sequence of ordinals. I claim that J = {yrn: a < K} is a set of indiscernibles for L,. This is so because for any formula cp, the truth value of cp(y,,, . . ., y a k ) in L, does not depend on the choice of y z , < ... < y., in J because by the definition of the urn, the truth value of

a

cp(r(x1,

..., x,,

u,,),

..., t ( x , , ..., X,,

Urnk))

does not depend on the choice of a1 < . . * < a,, . Hence {y. : a < K } is a set of indiscernibles for L , . Since i,, is the first member of u,, , it follows by (30.26) that y,, < i,, . Now if A = H ( J ) and n is the transitive collapse of A, then, as we proved in the first paragraph, n[A] = L, and K = n[J] is a set of indiscernibles for L, of order-type K such that . f L i ( K ) = LI . However, n(yJ I y,,> < i,, , and so the wth member of K is smaller than i,,, contrary to our assumption. Hence (L,, l) is remarkable. This completes the proof of Theorem 72. Absoluteness of 0‘

The set 0’ is not only unique if it exists, but is the unique well-founded remarkable E.M. set in every transitive model of ZF that contains it. And conversely, if 91 is a transitive model of ZF containing all ordinals and X E 911 is such that 911 k (Cis a well-founded remarkable E.M. set), then X = O x . This follows from the fact that C = 0’ can be expressed as a ll property and from the Levy-Shoenfield absoluteness lemma.

,

350

5.

MEASURABLE CARDINALS

Lemma 30.14. There is a ll, formula @(C) such that @(C) holds i f and only if C is a well-founded remarkable E.M. set. Corollary. The property ’‘ C is a well-founded remarkable E.M. set ” is absolute for euery transitive model u1 of Z F containing all ordinals. Hence YJI k (0* exists) ifand only if0” E “,in which case (O‘)IIR = 0”. The corollary is a consequence of Theorem 36’ (the parametric version of the Levy-Shoenfield absoluteness lemma). Although the theorem is stated for Z E: w, it is applicable here since a set of formulas Z can be replaced by a subset of o,the set of all Godel numbers of cp E C.

Proof: A set of formulas C is a well-founded remarkable E.M. set if: (30.28) (a) C is an E.M. set; (b) X is remarkable; (c) for every a, the (C,a)-model is well-founded. Since (a) does not readily admit a nice formulation, we shall first replace it by another condition. Let 2 be the language {E, cI, c 2 , . . ., c,, . . .} where c,. n < w, are constant symbols. For every formula cp(ul, . . ., u,) let @ be the sentence cp(cI,. . ., c,) of 2.For each set of formulas Z, let 2 be the set containing (i) all @ for cp E C,(ii) the sentence “c, is an ordinal and c , < c2,” and (iii) the sentence “cp(ci,, ..., ci,)++q(cj,, ..., cj,)” for every cp E C and any i , < ... < i n , j , < ... < j,, (iv) all axioms of ZFC + I/ = L. Let (30.28) (a’) 2 is consistent. Clearly if Z is an E.M.set, then 2 is consistent for we simply interpret the constants c,, n < o,as some Silver indiscernibles. Conversely, if 2 is consistent, then 2 has a model and that model provides us with a (C,a)-model (with indiscernibles c, , n < w ) and the proof of Lemma 30.3 goes through. Therefore if C satisfies (a’), (b), and (c), we proceed as in the proof of Theorem 72 and prove that C = 0’. Thus C is a well-founded remarkable E.M. set if and only if it satisfies (30.28)(af),(b), (c). Now (a’) and (b) are syntactical conditions, and so the conjunction of (a‘)and (b) is a Ao-property over HF (= Vw),the collection of all hereditarily finite sets. We shall now show that (c) can be written as a ll, property of C. If C satisfies (a’) and (b), then for every limit ordinal a there is a unique (up to isomorphism) (C,a)-model and we can find one ((A, E), I) such that I = a and that y. Let U, be the class of all limit cardinals K with cf K > y; by transfinite induction we define a sequence of classes Uo 3 U 1 3 . . * 3 U, ... as follows: (30.34)

U z + l = { ~ ~ U) u , : , n ~= J K} U, = (? U,

(2 a limit)

,< A

(That is, consists of fixed points of the increasing enumeration of U z . ) Each U, is nonempty, and in fact a proper class. To see this, verify, by induction on a, that each U, is a proper class and is 6-closed, for each 6 with cf 6 > y ; that is, whenever ( K :~ 5 < 6) is an increasing sequence in U,, then lims+a K; E U,. Hence each U, is nonempty, and we choose a cardinal K E U,,,,. Thus K is such that cf K > y and K is the Kth element ofeach U,, a < ol. We E). shall find a set of K 1 indiscernibles for (k, Since j : L + L is elementary and j(K) = K, it is clear that the mapping i = j 1 L, is an elementary embedding of (LK, E) into ( L K ,E). We shall use i

354

5.

MEASURABLE CARDINALS

and the sets U, n K, a < w l , to produce indiscernibles y,, u < o l ,for L,. Let X, = U, n K for each u < ol, and recall that y is the least ordinal moved by i . For each a < w l , we let M, = .#"*'(y u X,) (30.35) M, is an elementary submodel of L,. If n, is the transitive collapse of M,, then because IX,I = K, we have nu[M,] = L, . Thus if we denote i , = n; then i, is an elementary embedding of L, in L,.Let (30.36)

Yo

= &(Y)

Lemma 30.16. (a) The ordinal y, is the least ordinal > y in M , . (b) l f a < p and x E M , , then i,(x) = x . In particular, i,(y,) (c) l f a < It then y, < 7,.

= y,.

Proof. (a) Since y c M,, i,(y) is the least ordinal in M,greater than or equal to y ; thus it suffices to show that y $ M, . If x E M,, then x = t[q . . ., q,,l where t is a Skolem term and the q's are either < y or elements of X,. For all such q, i ( q ) = q and hence i ( x ) = i ( t ( q l , . . . , q,,)) = t ( i ( q l ) ,. . . , i(q,,))= x. However, i ( Y ) # 7 and so Y $ M , * (b) Each x E M , is of the form t[q 1 , . . ., q,,l where the q's are either < y or in X,. If q < y, then clearly i,(q) = q. If q E X,, then because a < /3, we have IX, n q I = q and hence n,(q) = q ; in other words, i,(q) = q. Therefore i,(x) = x . (c) If u < j,then M a 2 M, and hence yu I y s . To see that y, # y,, note that because y , > y, we have i,(ya) > i , ( y ) = y , , while io(ya) = y, .

Lemma 30.17. If u < p, then there is an elementary embedding i,: L, -+ L, such that for every t; that is either < a or > we have i u a ( y e )= y c , and i,,(y,) = y,. Proof. Let M u , = S L K ( y au X,),and let i,, = nibS'where napis the transitive collapse of M,, . The mapping i,, is an elementary embedding of L. in L, . If q < y,, then clearly iUp(q)= q ; in particular ia,(ys) = ye if < a. If x E M , , 1, then x = t(ql, . . ., q,,) where the q's are either y or E X,, If q E X,, 1 , then 1 X, n q 1 = q and therefore ims(q)= q. Hence iua(x)= x for every x E M , , I , and in particular i+(ye) = yc if 5 > fl. Now we shall show that i,,(y,) = y , . Since M u , 2 M,, we have y , E Mu,; and since y . c Ma,, iu,(ya) is the least ordinal in M a , greater than or equal to y.; hence we have y . I iaa(ya) I y , . Thus it suffices to show that there is no ordinal 6 E Mu, such that y, I 6 < y s . Otherwise there is some 6 = t(< . . .,

...

where x = r(k), etc., to denote the thread r. Let us introduce an equivalence on threads: Two threads (xk+ xk+ I , . . .) and (y,, y l + ...) (where k I 1) are equioalent if x , = y,. Note that in each equivalence class of threads there exists one maximal thread; moreover, a thread (31.5) is maximal just in case there is no X E Ult'k- ') such that X k = i k - l , k ( X ) .

5.

364

MEASURABLE CARDINALS

Let M be the class of all maximal threads. We define a relation E over M as follows: If t and s are maximal threads, then

(31.6)

rE

s

for some n, r(n) E s(n)

iff

It should be clear that (31.6) holds just in case t ( n ) E s ( n ) for all n 2 k where k is the least element of dom(r) n dom(s). For each n c o,we define a mapping in, from U P ) into M as follows: For each x E Ult'"', consider the thread (x,,, x,+ . . .) where for each k 2 n, x, = ink(x).Let i,,Jx) be the maximal thread that extends (x,,, x,+ . . .). See Fig. 3 1.2.

Ulr'"'

FIGURE 31.2

We shall show that for each n, the mapping in,: U P ) + (M, E) is an elementary embedding. We prove this by induction on complexity of formulas, simultaneously for all n. Clearly, in, preserves both = and E: (31.7)

i,,(x) = i,,Jy)

iff

x =y

ino(x) E ino(y)

iff

x

E

Y

The induction step is easy to prove for propositional connectives. Thus consider a formula 32 q ( z , x , ...). Let n < w,let x, .. . E U P , and assume that (31.8)

M k 32 q ( z , i,,,(x), ...)

We want to show that (31.9)

Ult'") k 3 q ( z , x ,

. ..)

Let c be some maximal thread such that M C q(t, ...). If n E dom(t), then we simply let z = t(n), and have, by the induction hypothesis, Ult'")C q ( z , . . .). Otherwise, we consider some rn > n such that rn E dom(c). Since i,,(x) = imW(im(x)),we have U P ) C 3z q ( z , i,,,,(x)) and since i,,,,,: U l P + U l P ) is elementary, (31.9) follows. Thus (M, E) is a model of ZFC, elementarily equivalent t o each Ult(') (hence to V ) ; the mappings in, are elementary and commute with the i,,,,,.

31. THE

MODEL

365

L[U]

Moreover, for each x E M there exists n < o and x, E UlP" such that x = ino(x,). We call (M, E) the direct limit of the system (31.4):

(31.10)

(M, E) = lim dir { Ult'"); i,,,,} n-#

I t is clear that the above construction works in general, for any system of models and elementary embeddings

{(Mn En);inrn}n.m < 3

(0

Moreover, there is no reason to restrict ourselves to an o-sequence of models. Let us assume that we have a system {(Mi Ea); i i , } a , , 7

< i.

of models and elementary embeddings i,,: M , M , where rl is a limit ordinal and the i,, commute. Then we define the direct limit --f

(31.11)

( M A ,E,) = lim dir { M a ;i,,} ,-a

analogously: Threads are now functions t with domain { a : a. Ia c A} and such that t , E M a for each a E dom(r) and t ( p ) = i,,(t(a)). Again, we obtain elementary embeddings i,,: M l + M, that commute with the i, and each x E M A is in the range of some i,, . There is no reason to expect that the direct limit of well-founded models is a well-founded model. However, as we shall see shortly, the direct limit of Ult(")is well-founded. Exercise 31.1. Let K be a measurable cardinal and j : V + M be the corresponding elementary embedding. Let M o = V , M I = M , and for each n < w, M , , I = j ( M , ) and i,,,, I = j ( M , . The direct limit of {Mn;in,,,} is not well founded. [iOlo(K), i l c u ( ~ ).,. . , ~,,JK), . . . is a descending sequence of ordinals in the model.]

Let us return to iterated ultrapowers. We started with a K-complete nonprincipal ultrafilter U over K and defined the models U P and elementary embeddings i,, . Let us denote by ( U P ) ,E'")) (or just Ult'")) the direct limit

(31.12)

( U P ) ,E'")) = lim dir {( Ult'"),E); i,,} n-w

Let K'") = iO&) and U(") = iou( U). Since Ult'") satisfies that U('") is a P)-complete nonprincipal ultrafilter over K("), we can construct, working inside the model U P ' ) , the ultrapower of Ult@')modulo U'")and the corresponding natural embedding j'"). [This is easy to understand if Ult'") is a transitive model. If ( U P ) ,E'")) is not well-founded, then the ultrapower and the mapping j(") are classes in the sense of the non-well-founded model. In general, note that a " class " C in a nonstandard model (M,E) is just a subclass of M ,namely {x E M : M b x belongs to C},and similarly, a " function " F in (M, E) is a function from M to M :{(x, y ) E M x M : M =! F(x) = y}.]

366

5.

MEASURABLE CARDINALS

Let us denote by ( U l t ( w + l )E'"')) , the ultrapower of (Ult'"), E'")) modulo U(,) and let i,,,+ be the corresponding natural embedding. For each n < o, let in+,,+ = i,,,, o in,,,. [Another way of looking at Ult'w+": Let E be the set algebra {extEI,,,(X) : U P ) k X E K ( ~ ) } and let U * be the ultrafilter { e x t ( X ): X E@') U(w)}.For each fsuch that Ult'") k f is a function on dW), let f * = { ( x , y ) : U P ) k f ( x ) = y } . Then Ult'"' is (isomorphic to) the ultrapower of Ult'") using U* and only the functions f *.] This procedure can be continued, and so we define the iterated ultrapower of V modulo U as follows: (3 1.13)

( U P ' , E ' O ' ) = ( V , E) (Ulf''+1), E ( " + 1 ) )= Ult"l.,(Ult'a',E'"')

(Ult'"', E'")

=

lim dir {( Ult'"),E'")); iua)

(A a limit)

"-11

where U(') = ion(U ) , for each a. We do not know yet that all the models Ult(') are well-founded; but we make a convention that if Ult'") is well-founded, then we identify it with its transitive collapse. The first lemma about iterated ultrapowers is the Factor Lemma below. If M is a transitive model of set theory and U is (in M ) a K-complete nonprincipal ultrafilter over K, we can construct, within M , the iterated ultrapowers. Let us denote by U l t t ' ( M ) the ath iterated ultrapower, constructed in M . (Thus Lilt'"' = Ult:.'( V)).

Lemma 31.4 (Factor Lemma). Let us assume that Ulr'"' is well-founded. Thenfor each fi, the iterated ultrapower Ultf,$(Ult('))taken in Ult'") is isomorphic to the iterated ultrapower Ult''+ 0). Moreover, there is for each /?an isomorphism e r ) such that iffor all 5 and q, iyi denotes the elementary embedding of Ult$/,l(Ult'"))into Ultg),,(Ult'')), then the following diagram (Fig. 3 1.3) commutes :

FIGURE 31.3

Proof: Although the lemma sounds complicated, it is actually quite simple, and also easy to prove. The proof is by induction on p. If /I = 0, then the 0th iterated ultrapower in U l P ) is Ult'");and we let eg) be the identity mapping. If Ult',fLI and Ult:"+P'are isomorphic and e;) is the isomorphism, then UltjJI: and

367

31. THE MODEL LEU]

U f t g + f l +l ) are ultrapowers of Uftf,&and Ult',"+8), respectively, modulo itJ(Va)) and i,,, +8( U ) , respectively; and since io.a+fl(U ) = eF)(i$J(Uca))),the isomorphism e f ) induces an isomorphism efi between Ultg,: and U l t g + B + l ) . If I is a limit ordinal, then Ult$!, is the direct limit (in Ult'")) of {UlrVJ,;if,!}B,y ) ) K('). To show that the sequence is continuous, let A be a limit ordinal; we want to show that K(') = lim a+* I&'). If y < K('), then y = i,@) for some a < 1 and 6 < K('). Hence y = 6 and so y < K").

Lemma 31.8. Let D be a normal measure on K , and let for each a, Ult(') be the ath iterated ultrapower mod D, K(') = i o a ( K ) , and D(')= ioa(D).Let 1 be an infinite limit ordinal. Then for each X (31.18)

X E D(')

iff

E

Ult('), X

c K(*),

(hr < l ) X 3 {dY) : a I y < A}

Proof. Since for no X can both X and its complement contain a final segment of the sequence ( K ( ~ :) y < A), it suffices to show that if X E D('),then there is a such that K ( ~ E) X for all y 2 a.

369

31. THE MODEL L [ U ]

There exists a < A such that X = iaA(Y) for some Y E Da).Let me show that X for all y, a I y < 1. Let y 2 a and let Z = &,( Y). Then Z E DtY)and since D(Y) is a normal measure on K ( ~ in ) U / T ( ~we ) , have K ( , ) E i,,,, l(Z). However, i,.,, l(Z)G i,, l,d(iy,y+ l(Z))= X and hence K ( , ) E X . ~ ( 7E )

Representation of Iterated Ultrapowers We shall now give an alternative description of each of the models Ult") by means of a single ultrapower of the universe modulo an ultrafilter on a certain Boolean algebra of subsets of ' K . This will enable us to obtain more precise information about the embeddings ion: V + Ult(') and also to generalize the iteration to the case when U is an M-ultrafilter for some model M. We shall deal first with the finite iterations. Let U be a K-complete nonprincipal ultrafilter over K. Let us use the symbol V*a for "for almost all a < K " : (31.19)

If X

V*a q ( a )

{ a < K : cp(a)}E U

iff

< K , let

E " K and a

~ ( ~ ) = { ( a ~ , . . . , a , - ~ ) :, .( .a.,, O aL ~,-~)EX}

(31.20)

We define ultrafilters U , over ,K, by induction on n:

u, = u

(31.21)

X

E

U,+l

iff

V*a X t u )E

U,

Each U , is a nonprincipal K-complete ultrafilter over "Z E U , . It is easy to see that for all X C "K,

x E U,

iff

and if Z E U , then

'K,

V*a, V*a1 ... V*anP1( a o , ..., a,- ,) E x

Although the present definition varies slightly from definition (29.10), it is practically the same because for each n, U , concentrates on increasing n-sequences: { ( a o , .. ., a,-

(31.22)

I) E "K:

a. < ... < a,-

1}

(because Vao(Va, > a ) ... (Van- > a,-z)[ao < ... < a,-

E

U,

I]).

Lemma 31.9. For every n,

mu"( V ) = Ult'"'( V ) and jum= ion.

[Here j u mis the natural embedding j : V + Ulfu,(V).] Proof: By induction on n. The case n = 1 is trivial. Let us assume that the lemma is true for n and let us consider Ult,,, ,. Letfbe a function on ' K . For each t = (ao, . .., a,E "K, let As be the function on K defined by A,(() = f ( a o , ..., a,- (), and let F be a function on "K such that F ( t ) =f(?,for all "

,,

+

370

5.

MEASURABLE CARDINALS

In UlrUn= Ult'"), the function F represents a function on j u . ( K ) = ti("): Let .f= [Flu". This way we assign to each function .f on ""K a function . f U~ P on ti("). Conversely, if h E UIt(")is a function on d"),there is f on 'K such that h =$ There exists F on "K such thatf= [F],, and that for each I E "K, F ( r ) is a at a,, . function on K ; thus we let f ( a o , .. ., a,) be the value of F ( a o , . . ., a,We shall show that the correspondence [ f l u . +I w [flu,., is an isomorphism U P ) ) .We have between Ult,,, I and UlP" = Ulf,,,")( I E "ti.

" +

[flu,,1

=

[slu.+ I

V*ao ... V*a,,-, V*5

iff

f ( a o , . . . , a n - 1, 5 ) = g(ao, ..., a,,-', 5 ) iff

v * a 1and y' < a for all y < a. By Lemma 31.12, if a E K, then ion@) = a and i0@) = p for all p E K greater than a. Hence if a, fl E K, then &,(a) = p and i,,(y) = y for all y E K that are greater than p or less than a. Now if D is a normal measure on K, then because K < 1,we have i,,(D) = D for all a, p E K. Thus each i,, (a, p E K), restricted to YD], is an elementary embedding of YO] in YO] such that i,,(a) = p andi,,(y) for all y E K below a or above p. Using these embeddings i,, (as in the proof of Lemma 30.19), one shows that the elements of K are indiscernibles for the model 401. Since some elements of K are regular cardinals, and some are limit cardinals, it follows that all elements of K are inaccessible cardinals in 4 0 1 . In the above argument, it was not necessary that K be a measurable cardinal, only that K be measurable in YO]. Thus we have proved :

Lemma 31.19. Let (31.40)

K

be a measurable cardinal, and assume that.

For some y < K , there is D G P ( y ) such that YO] k D is a normal measure on y.

Then there are arbitrarily large successor cardinals that are inaccessible in 4 0 1 .

We have proved in Lemma 31.18 that if U is a nonprincipal K-complete ultrafilter over K, then j&) < io&), where ioo is the natural embedding of LED] in Ult',")(YD]). We can now prove a stronger statement:

Lemma 31.20. If there is a K-complete nonprincipal ultrafilter U over K such that 2 i0&), then (31.40) holds.

j&)

Proof. Let us work in the model M = Ult,(V). The cardinal&) is measurable . F be the collection of all while ioo(tc) has cofinality w, and so i0&) < j ( ~ )Let

32.

381

LARGE CARDINALS BELOW A MEASURABLE CARDINAL

subsets X of i0&) such that X 2 { i O n ( K ) : n 2 no} for some n o . Using Lemma 31.8, we proceed as in the proof of Lemma 31.14 to show that

YF]1 F n q F 3 is a normal measure on ioO(.) Thus (31.40) holds in M for j ( K ) . Sincej is an elementary embedding, (31.40) holds in V for K. Corollary. ff K is a measurable cardinal and 2" > K + , then (31.40) holds. Consequently, it is impossible to prove consistency of 2" > K + relative t o ZFC + '' K is a measurable cardinal." Proof. On the one hand, I i0&) 1 = ( K + ) " ~ I IK + ; on the other hand, if LI is any K-complete nonprincipal ultrafilter over K , we have jU(K) > 2" > K + . W

We shall return to the problem of 2" > K + at a measurable cardinal in Section 36. 32. LARGE CARDINALS BELOW A MEASURABLE CARDINAL

Weakly Compact Cardinals In Section 29 we defined weakly compact cardinals as the uncountable cardinals satisfying the partition property K -,( K) '. We have shown that weakly compact cardinals are inaccessible, and in fact, exactly those inaccessible cardinals that have the tree property, that is, every tree of height K whose levels have size less than K has a branch of length K. We have shown that every measurable cardinal is weakly compact and moreover if D is a normal measure on K, then the set of all weakly compact cardinals below K is in D. We shall now investigate weakly compact cardinals in some detail. First we give a characterization of weakly compact cardinals which explains the name "weakly compact." This aspect of weakly compact cardinals has, as many other large cardinal properties, motivation in model theory. We shall consider infinitary languages which are generalizations of the ordinary first order language. Let K be an infinite cardinal number. The language YK*,, consists of: (a) K variables; (b) various relation, function, and constant symbols; (c) logical connectives and the infinitary connectives a < K (infinite disjunction and conjunction); (d) quantifiers 30, Vv.

v5 K and K is the least ordinal moved by i[It can be proved that the assumption that K is inaccessible in L is redundant and follows from the other assumptions.] K

Proof: Let u < K ' . We may assume that ( L a ,E) is a model of ZF- (by taking a larger a' that has this property). Let ( X , : 5 < K ) be an enumeration of P(K) n L * For each t : y {O, l}, y < K, let -+

x f = n Y= 5 ye > ... is impossible, we have yo < y1 < ... < yc < .... We shall reach a contradiction by showing that G is homogeneous for g. For each k, consider the formula


) 6. W Before I start the proof of the theorem, let us go back to iterated ultrapowers. Let D be a normal measure on K, and let ion denote, for each a, the elementary embedding io,: V + Ult'");let K(') = i O , ( K ) and D@)= i@). First recall (31.18): If I is a limit ordinal, then X E U/?(')belongs to D(')just in case X 2 {K(,): a I y < I ) for some o! < I . Let (33.8)

K

= {v

: v is a strong limit cardinal, v > 2", and cf v > K }

By Lemma 31.12, if v E K , then K(') = v, and io,(v) = v for all a < v. Thus if yo < y1 < ... < yn < . . * are elements of the class K , and if I = limn-,x y n , then K(') = I , and X E Ulr@)belongs to D(') just in case X 2 {y,,: no I n} for some no. If A is a set of ordinals of order type w, A = { Y , , ) ; = ~ ,we define a filter F ( A ) over A = sup A as follows (33.9)

X

E

F(A)

3noVn 2 no(?, E X )

iff

The above discussion leads us to this: If A c K has order type w , and if I = sup A, then for every X E U P (33.10)

X

E

D(')

iff

X

E

F(A)

In other words, (33.1 1 )

D(') = F ( A ) n Ulf(')

Hence F ( A ) n U l P ) E Ult"); and so by Lemma 15.4, L [ F ( A ) ]= L[D('"].Thus F ( A ) n L [ F ( A ) ]= D") n L[D'')],and we have (33.12) L [ F ( A ) ] k F ( A ) n L [ F ( A ) ] is a normal measure on 1. The only assumption needed to derive (33.12) is that K is measurable and A is a subset of the class K . We shall now use Lemma 33.2 and a similar construction to obtain a model with two measurable cardinals.

33.

403

COMPACT CARDINALS

Suppose that A = (y,},“,o is as above, and that A’ = {yb},“=o is another subset of K of order type w, such that 1 ~>’ ~1 = sup A ; let 1’= sup A’. Let F = F ( A ) and F’ = F ( A ’ ) . Our intention is to choose A and A’ such that the model L[F, F’] has two measurable cardinals, namely I and A’, and that F n L[F, F’] and F’ n L[F, F‘] are normal measures on 1 and A’, respectively. The argument leading to (33.12) can again be used to show that F n L[F, F‘] is a normal measure on 1 in L[F, F‘]. This is because we have again

D(L= ) F n lJ1$A) moreover, i o i ( y i )

=

y: for each n, and hence ioi(A’)= A’ and we have ioA(F‘)= F‘ n U I P )

(33.1 3)

Therefore

L[F,F’] = L[D‘”’,F‘]= L[D‘”, ioA(F’)] and

F n L[F, F‘] =

(33.14)

n L[D‘”’, ioA(F‘)]

which gives (33.15)

L[F, F’] k F n L[F, F’] is a normal measure on 1.

In order to find A, A’ so that F‘ also gives a normal measure in L[F, F‘], let us make the following observation: Let us think for a moment that A c K and A’ c K . Then i o i , ( A ) = A and D‘”)= F’ n Ult(”), and the same argument as above shows that (33.16)

L[F, F‘] k F‘ n L[F, F‘] is a normal measure on 1’.

We shall use this observation below. Let us define the following classes of cardinals (compare with (30.34)): KO = K

(33.17)

K 3 1 + , = { v ~ K ZI K: , n v ( = v )

K,

=

0K ,

(y a limit)

l K. Now we let yn = the least element of K ,

(33.18)

A = {y,,};=o,

1 = lim y, n-m

and let A‘

= {Y;},”=~

be a subset of K , ,

6.

404

OTHER LARGE CARDINALS

Let us consider the model L[A, A'], and let for each n I w (33.19) M, = the Skolem hull of K , in = the

class of all x

L[A, A']

4 A , A']

E L [ A , A']

x = t[v,,

***,

such that

vk, y o ,

***,

&, A, A'] K,.

**.,

yk,

where r is a Skolem term and vlr ..., vk

E

(Let us not worry about the problem whether (33.19) is expressible in the language of set theory; it can be shown that it is, similarly as in the case of ordinal definable sets. Alternatively, we can consider the model L,[A, A'] where 8 is some large enough cardinal in K,, ,.) Each M , is an elementary submodel of L[A, A ' ] ; let IT,be the transitive collapse of M , ; then nn[Mn]= L[n,(A), 7rn(A')]and j , = IT;' is an elementary embedding (33.20)

j n : L[n,(A), xn(A')] + L[A, A']

Lemma 33.4. For each n < w, n,,(y,,)< (2")' Proof. By induction on n. First let n = 0. Let a < yo be in M,. Then a = t(v,, . ..,vk, A, A ' ) for some Skolem term t and some vtr . . .,vk E K O .Let ioa be the natural embedding into Ultg) for some U over K . Since y o is the least element of K O ,we have a < v for all v E K and hence ioa(v ) = v for all v E K o . Hence also iou(A)= A and iou(A')= A' and it follows that iou(a)= a. Now ioa(a)= a is possible only if a < K. Hence each a < yo in M, is less than K and therefore ITo(Y0) 5 K < (2")' and let us show that IT,+ ,(yn+ I ) < NOWlet us assume that IT,,(~,,) (2")'. By Lemma 33.2 there exists U such that j U ( K ) > n,(y,). We shall show that n,(a) <jU(.) for all a < yn+, in M,,,; since IT,, l(a) I n,(a) it follows that IT,+ 1(Yn+ 1) = SuP{nn+,(a): a < Y n + 1 and a E Mn+ 1) s j u ( K ) < (2")'. First notice that it follows from the definition of K,+' in (33.17) that n,(v) = v for all v E K,, Note also that y,,, E K,+ for all rn 2 n 1, and A' c K,, Let a < Y , , + ~ be in M , + l . Then (in L[A, A']),

,

,.

a=t(yO,

Ynr

v1, ..., vki A, A ' )

where t is some Skolem term and v l , . . ., vk E K,, nn(a) = f(nn(rO), *

-

- 9

+

nn(yn),

vlr

Hence (in L[n,(A),nn(A')]) vk, nn(A)? A')

Now we argue inside the model U f t , ( V ) (which contains both zn(A)and A'): Consider the ath iterated ultrapower (modulo some measure on j u ( K ) ) . Since n,(yo),,.., n,(y,) are all less than j U ( K ) , we have ioa(nn(yi))= zn(yi)for all i = 0, . .., n. We also have iod(v) = v for each v E K,, (because a < v for each

33.

405

COMPACT C A R D I N A L S

,

,

v E K , , and K , , c K ) . It follows that ioz(a,,(a))= n,,(a). Now (because a,,(a) I a ) this is only possible if x,(a) < j r , ( ~ ) . W

We can now complete the proof of Theorem 80 for 6 = 2. Let us consider the model M,, the Skolem hull in 4 A , A'] of K , . Let a, be the transitive collapse of M, and B = n,(A). Since A' c K,, we have n,(A') = A', and j , = a; is an elementary embedding

,,

j,:

YE, A'] + 4 A , A']

By Lemma 33.4, n,(y,) I n,,(yn)< (2")' for all n, and hence .,(A) < (2")'. Let U be a ti-complete ultrafilter over ti such that j&) > a,@) (by Lemma 33.2). In U l t , , B is a subset of j , ( t i ) and A' is a subset of the class K . Thus we can apply (33.16) and get Ult, 1 ( 4 F ( B ) ,F(A')] =! F ( A ' ) n 4 F ( B ) , F(A')] is a normal measure on A')

Hence L[B, A'] k

(L[F(B),F ( A ' ) ] 1 F ( A ' ) n L[F(B),F ( A ' ) ] is a normal measure on A') and applying j,, we get

44 A']

1

( L [ F ( A ) ,F ( A ' ) ] 1 F ( A ' ) n L [ F ( A ) ,F ( A ' ) ] is a normal measure on A') Therefore F' n q F , F'] is (in q F , F ' ] ) a normal measure on A'. This completes the proof of the theorem for 6 = 2. The proof of the general case is completely analogous for 6 < K . Instead of A, A', one uses a sequence ( A , : < 6 ) , where $, the nth element of A,, is defined as the least element of K , , , , , . One uses the models M,,.,,, to prove an analog of Lemma 33.4: ~,,,,+,,(y!,~)) < (2")'. The assumption 8 < ti is important because when one uses the ultrapower U l t , , one needs to know that ( A S: 5 < 6 ) E Ult, . Once the theorem is proved for all 6 < ti, then the general version is derived as follows: If 8 is arbitrary, let us consider U P ( V ) for some a > 8. Then the model Ult'")(V )has a strongly compact cardinal greater than 8 and therefore has a transitive model with 8 measurable cardinals. W


K is a singular cardinal, then 2"' A < A implies Acf A = A'. (Consequently, i f A > ti is a singular strong limit cardinal, then 2l = A'.) We shall prove the theorem in a sequence of lemmas. An ultrafilter over a cardinal A is uniform if every set in the ultrafilter has size A.

406

6.

OTHER LARGE CARDINALS

Lemma 33.5. ff K is a compact cardinal and I > K is a regular cardinal, then there exists a K-complete uniform ultrajilter D over I with the property that almost all (mod D )ordinals a < I have cojinality less than K . ProoJ: Let U be a fine measure on P,(I). Since U is fine, every a < I belongs to almost all (mod V ) P E P K ( I ) Let . us consider the ultrapower Ult,( V ) and letf be the least ordinal function in U l t , greater than all the constant functions c , , y

K is a regular cardinal, then there exists a K-complete nonprincipal ultrajilter D over I , and a collection { M , : a < I ) such that

(33.24) (a) I M , I < K for all a < I , (b) for euery y

-= I , y belongs to M,for almost all a (mod D)

[An ultrafilter D that has a family { M , : a < I } with property (33.24) is called (K, I)-regular.] Proof: Let D be the ultrafilter over I from Lemma 33.5. For almost all a (mod D), there exists A, c_ a of size less than K and cofinal in a. If cf a 2 K, let A, = 0. Let A be set of ordinals represented in Ult,(V) by the function ( A , : a < I ) . The set A is cofinal in the ordinal represented by the diagonal

33.

407

COMPACT CARDINALS

function d ; and since [dl > for all 7 < A, it follows that for each q < A there is q' > q such that A n {< :,jD(q)I < j D ( q ' ) } is nonempty. We construct a sequence ( q y : y < A) of ordinals < A as follows: let qo = 0 and q, = lima*y qa if y is a limit; let q, + be some ordinal such that there exists E A such that j&,) I t <j&+ l). In other words, if we denote I , the interval (5 : q, I5 < q y + then for every y , the interval I , has nonempty intersection with almost every A,. Thus if we let j,()l)


K is an arbitrary cardinal, then we have, by Lemma 33.7, A < K

5

(A+)-

= A+

In particular, we have AN" I 1 ' for every A > K. By Theorem 23 (or rather by Lemma 8.3), this implies that the singular cardinals hypothesis holds for every A > K. Supercompact Cardinals We proved in Lemma 33.1 that a compact cardinal K is characterized by the property that every PK(A)has a fine measure. If we require the fine measure to satisfy a normality condition, then we obtain a stronger notion-a supercornpact cardinal. Ultrapowers modulo normal measures on PK(A)induce elementary embeddings that can be used to derive strong consequences of supercompact cardinals. For instance, Theorems 79 and 80 become almost trivial if existence of a strongly compact cardinal is replaced by existence of a

6.

408

OTHER LARGE CARDINALS

supercompact cardinal. It is consistent to assume that a compact cardinal is not supercompact, or that every compact cardinal is supercompact, but it is not known whether supercompact cardinals are consistent relative to compact cardinals. A fine measure U on P , ( A ) is normal if wheneverf P , ( A ) + A is such that f ( P ) E P for almost all P , then f is constant on a set in U.A cardinal K is supercompact if for every A such that I A 1 2 K, there exists a normal measure on P,(A). Exercise 33.3. If

K

is measurable, then there is a normal measure on P r ( ~ ) .

Normality of filters over P , ( A ) is a generalization of normality of filters over K . Many results on normality generalize from K to P,(A) (but not the existence of a normal measure on a measurable cardinal). For instance, let us call C G PK(A)closed unbounded if (a) for every P in P,(A) there is Q E C such that P c_ Q, and (b) if D s C is a nonempty directed (that is, V P 1 , P , E D 3Q E D such that Q 2 P I u P 2 )set of size < K , then D E C. [It is not difficult to show that (b) can be replaced by: closed under unions of chains of length < K . ] The closed unboundedfilter over P , ( A ) is the collection of all subsets of P,(A) that have a closed unbounded subset. Stationary sets are defined as sets that intersect every closed unbounded set.

u

Exercise 33.4.

(a) Each = {Q : P E Q} is closed unbounded. (b) The closed unbounded filter is K-complete. (c) Iff(P) E P on a stationary set, thenfis constant on a stationary set. [For (c), prove that the closed unbounded filter is closed under the diagonal intersections A a G C, A = { P : P E C, for all a E P}.]

Let A 2 K be a cardinal and let us consider the ultrapower Ult,( V ) by a normal measure U over PK(A);let j = j , be the corresponding elementary embedding. Clearly, a set X G PK(A)belongs to U just in case [ d ] E j ( X ) , where d, the “diagonal function” is the function d ( P ) = P .

Lemma 33.8. If U is a normal measure on PK(A),then [ d ] = { j ( y ): y < A} and hencefor eoery X 5 PK(A), (33.26)

X

E

U

iff

{ j ( y ) :y < A} E j ( X )

Proof. On the one hand, if y < A, then y E P for almost all P and hence j ( y ) E [ d ] . On the other hand, if [ f ] E [ d ] , then f ( P ) E P for almost all P and by normality, there is y < 1 such that [f]= j ( y ) . H Let G = { j ( y ): y < A}. It follows from (33.26)that iffand g are functions on PK(A),then

I![

=

[sl

iff

( d ) ( G )= ( j g ) ( G )

33.

409

COMPACT CARDINALS

and

[fl E [sl

iff

( X ) ( G )E ( M ( G )

Consequently,

(33.27) [f 1 = ( X ) ( G ) for every function f on P I @ ) . For each P E P K ( l ) ,let us denote K , = P ~ K

1,

= the

order type of P

Note that if we let G = ( j ( y ) : y < A}, then the order type of G is 1and hence by (33.27), 1 is represented in the ultrapower by the function P - A p . Also, since 1, < K for all P, we have j ( K ) > 1.By K-completeness of U , we havej(y) = y for all y < K ; and since K is moved byj, it follows that G n j ( K ) = K and therefore K is represented by the function Pt-+ K ~ . This gives the following characterization of supercompact cardinals:

Lemma 33.9. Let 1 2 K . A normal measure on P K ( 2 exists ) ifand only ifthere i s an elementary embedding j : V M such that -+

(33.28)

(i) j ( y ) = y for all y < K ; (ii) j ( ~>)A; (iii) ' M G M; i.e., every sequence ( a , : a < 1 ) of elements of M is a member of M .

Proof: (a) Let U be a normal measure on P K ( l )We . let M = Ult,( V) and let j be the natural embeddingj: V + Ult. We have already proved (i) and (ii). To prove (iii), it suffices to show that whenever ( a , : a < 1)is such that a , E M for all a < A, then the set {a, : a < 1) belongs to M. Let h , a < 1, be functions representing elements of M: [fa] E M. We consider the function f on P J 1 ) defined as follows: f ( P ) = { j J P ) : a E P } ; I claim that [f ] = {aa: a < A}. On the one hand, if a < 1,then a E P for almost all P and hence [fa] E [f]. On the other hand, if [g] E [f], then for almost all P, g ( P ) = f , ( P ) for some a E P . By normality, there exists y < 1such that g ( P ) = f y ( P ) for almost all P , and hence [g] = a ? . (b) Let j : V -+ M be an elementary embedding that satisfies (i), (ii), and (iii). By (iii), the set { j ( y ) : y < A} belongs to M and so the following defines an ultrafilter over P K ( l ) :

(33.29)

XEU

iff

( j ( y ) : y < 1)E j ( X )

A standard argument shows that U is a K-complete ultrafilter. U is a fine measure because for every a E 1,{ P : a E P } is in U . Finally, U is normal: If

6.

410

OTHER LARGE CARDINALS

f ( P ) E P for almost all P, then ( j f ) ( G )E G where G = { j ( y ) : y < I ) . Hence ( j f ) ( G )= j ( y ) for some y < I , and s o f ( P ) = y for almost all P . Exercise 33.5. I f U is a normal measure on PK(2).then every closed unbounded subset of PK(2)is in U . [If C G PI(2) is closed unbounded, then D = ( j ( P ): P E Cl is a directed subset of . j ( C )and ID1 = 2'" < j ( K ) . Hence D E j ( C ) , and since D = { j ( y ) : y < 2 ) . we have

cE

u

U.]

u

We have seen several examples how large cardinals restrict the behavior of the continuum function (e.g., if K is measurable and 2" > K + , then 2' > a+ for cofinally many a < K ) . This is more so for supercompact cardinals: Exercise 33.6. If 12 K and if U is a normal measure on PK(2),then the ultraproduct Ult,( (V,, , E): P E P,(2)) is isomorphic to (V,, E).

As a typical application, if K is supercompact and if there is a regular cardinal a such that 2" = a + , then there is a regular cardinal a < K such that 2' = a + . The following two results illustrate the technique of elementary embeddings.

Lemma 33.10. If K is supercompact, then there exists a normal measure D on K such that almost every a < K (mod D ) is measurable. Proof. Let I = 2" and let j : V + M be an elementary embedding with properties (33.28). Let D be defined by D = { X : K E j ( X ) } ,and let j , : V + UltD be the corresponding elementary embedding. Let k : Ult, + M be the elementary embedding defined in Lemma 28.6: k([flD)

= (jf)(K)

Note that k ( ~ =) K . Now, P ( K )c M and every subset of M of size iis in M ; hence every U G P ( K )is in M and it follows that in M , K is a measurable cardinal. Since k is elementary and k ( ~=) K , we have UltD k K is a measurable cardinal, and the lemma follows. Corollary. If

K

is supercompact, then K is the Kth measurable cardinal.

Incidentally, it is consistent that there exists a compact cardinal with no measurable cardinals below it. Lemma 33.11. If K is supercompact, then for every W normal measure D on K such that W E ultD(v): Proof. Assume on the contrary that 3W

G

(33.30) For every normal measure U on

G

P ( K ) there exists a

P(K)@(K,W ) ,where