An Introduction to Independence for Analysts (London Mathematical Society Lecture Note Series)

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London Mathematical Society Lecture Note Series. 115

An Introduction to Independence for Analysts

H.G. DALES School of Mathematics, University of Leeds

W.H. WOODIN Department of Mathematics, California Institute of Technology

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V CONTENTS

Preface 1.

vii

Homomorphisms from algebras of continuous functions

2.

Partial orders, Boolean algebras, and ultraproducts

22

3.

Woodin's condition

43

4.

Independence in set theory

54

5.

Martin's Axiom

80

6.

Gaps in ordered sets

104

7.

Forcing

130

8.

Iterated Forcing

183

Bibliography

229

Index of notation

235

Index

237

vii

PREFACE

The purpose of this book is to explain what it means for a proposition to be independent of set theory, and to describe how independence results can be proved by the technique of forcing. We do this by presenting an application Our of forcing to a deep and interesting problem in analysis. application is, by current standards in set theory, fairly non-technical, and so it offers an excellent setting in which to exhibit to analysts these new techniques from set theory. Most analysts will have a certain acquaintance with They will know naive set theory up to the level of ordinals and cardinals. They will have heard logic and set theory.

that forcing is a powerful technique that enables one to

prove that certain propositions of set theory are independent of specified axioms, and, in particular, that Cohen developed the method of forcing in his proof that the Continuum Hypothesis theory,

(CH)

is independent of the basic axioms of set

They may also know of more recently formulated (MA), which can be used to axioms, such as Martin's Axiom ZFC.

establish independence results without the necessity of knowing any of the technicalities of forcing.

However, it is possible that analysts harbour two negative feelings about these matters. First, they may feel that, although logic and set theory are of interest in their own right, they have little to contribute concerning questions

which "really" arise in mathematical analysis, and so can be safely left to their disciples. But this is not true. For example, several natural questions about sets of real numbers cannot be resolved in the theory

ZFC.

For some of these

viii

questions the set-theoretic entanglements are quite subtle, in that these questions can be resolved by invoking the existence of large cardinal numbers. example.

is a continuous function, then for each Borel subset

f,g

:

Here is a specific

It is not difficult to show that, if

3t + 3t

B

f(B)

P.

of

Lebesgue measurable for each Borel set ZFC.

-. ]R

Now suppose that

Is f(]R\g(B))

are continuous functions.

cannot be decided in

f: 3t

is Lebesgue measurable

B?

This problem

However if there is a measurable

cardinal, then these sets are indeed all Lebesgue measurable. The main example that we present in this work did It concerns the automatic continuity of homomorphisms from a Banach algebra of continarise naturally in analysis.

uous functions into an arbitrary Banach algebra, and the formulation of the problem was such that a solution was expected (perhaps naively) in naive set theory.

But

eventually it was discovered that the existence of discontinuous homomorphisms from the algebras under consideration could not be decided in ZFC. Secondly, analysts may feel that the technicalities of forcing are too arcane to be readily accessible.

We seek

to challenge this view by presenting a reasonably complete account of forcing which is comprehensible to non-logicians: we hope that it will bring them to the point at which they can appreciate the application, which we shall give in detail, of these new ideas to the above automatic continuity problem. We should say, however, that it is not our intention to teach the practical use of forcing in general, but rather to explain quite explicitly how the method of forcing does yield independence and consistency results, and to exemplify this by the study of our chosen example.

Thus this book is directed towards non-logicians, and in particular to analysts.

We shall give an account of

the background in logic that we shall require, and we shall explain the key notions of proof, of consistency, and of independence. The approach to independence will be through the theory of models:

Godel's completeness theorem allows us

ix

to recast independence questions as problems of the construction of models with certain properties.

We shall give full proofs of the results about forcing and Martin's Axiom that we shall need. En route to our main theorem, we shall prove that CH is independent from

a result that we believe should be known to all

ZFC,

mathematicians.

We also hope that our account will be useful for students of set theory as a preliminary to other works.

Very

little knowledge of analysis is required to follow the details of our example, and the background that is required is given in Chapter 1.

We now describe the problem in analysis that we are Let X be a compact Hausdorff space, and let

to consider. C(X,C)

be the set of all continuous, complex-valued

functions on

X.

The set

C(X,C)

is a commutative algebra

with respect to the pointwise operations. IfIx = sup{If(x)l

Then

I.Ix

:

x E X}

is an algebra norm on

uniform norm - and

(f E C(X,C)).

C(X,C)

(C(X,C), I.IZ)

Set

- it is the The

is a Banach algebra.

question we ask is whether or not each algebra norm on C(X,C)

is necessarily equivalent to the uniform norm:

the topological structure of

C(X,C)

if so,

as a normed algebra

would be completely determined by its algebraic structure. This question was first discussed by Kaplansky in 1948, and it was eventually "resolved" independently by Dales and by Esterle in 1976:

if the Continuum Hypothesis holds, then,

for each infinite compact (Hausdorff) space algebra norm on

C(X,C)

X,

there is an

which is not equivalent to the

uniform norm.

The appeal to CH in this theorem was thought at the time (at least by those involved) to be an accident and a weakness of the given proofs.

However, also in 1976, it was

proved by Solovay, using a condition of Woodin, that, if there is a model of set theory, then there is a model of set

x

theory in which each algebra norm on each C(X,C) is equivalent to the uniform norm. Shortly afterwards a different, and easier, approach to the theorem was developed by Woodin;

the approach used some techniques of Kunen.

It

is this approach, which involves Martin's Axiom, that we present here.

These results are included in Woodin's thesis,

written at the University of California, Berkeley, and a presentation for experts in forcing is given in the recent volume [45] of Jech, but apart from this ours is the first account of the theorem to appear in print.

The original work

of Solovay has never been published.

A revised version of Woodin's thesis is to appear [72]; this memoir, which will also include several more complicated independence results, is written for logicians, and we can say with quiet confidence that it will be unintelligible to analysts. Chapter 1 contains only analysis:

following the

seminal work of Bade and Curtis, we shall analyze the

structure of an arbitrary homomorphism from C(X,C) into a Banach algebra. In Chapter 2, we shall discuss partially ordered sets, Boolean algebras, and ultraproducts, topics

which form the background to much of our later work, and in Chapter 3 we shall relate our question to one which is amenable to the techniques which are given to us in the theory of forcing:

if there is an algebra norm on any

C(X,C)

which is not equivalent to the uniform norm, then there is a free ultrafilter V on N and an isotonic map from a subset /V of the ultraproduct (R'J/V, satisfying certain rules: identifies

8

outside logic, one usually

with its underlying set

this in Chapters 1 - 6 of this book.

B,

and we shall do

However, in Chapters 7

and 8 it seems to be necessary for us to maintain the distinction between

8

and

B.

Here are some particular notations that we shall use.

(An index of notation is given on pages 235-236.) First, N = {1,2,3,...}, whereas Zz+ = {0,1,2,...} (so

that, again, we are following the analysts). allows the possibility that

S = T.

Second,

S c T

Third, in a phrase such

as

rad A = {a E A : e - ab E Inv A the bracket

"(b E A)"

means

"for all

(b E A) } b

in

A".

We give some references in the notes at the end of the chapters to the theorems proved. Although some results are new or have simplified proofs, it is of course not the

case that unattributed results are claimed as original.

The authors first discussed this work when the first author visited Caltech in 1984:

we are grateful to

Professor W. Luxemburg for arranging this visit.

We also

acknowledge the award of a Visiting Fellowship to the second author by the United Kingdom Science and Engineering Research Council that enabled him to visit Leeds in September, 1984.

The second author is an A.P. Sloan Fellow, and we are grateful that this fellowship provided financial support for us to meet at later times, both in the snows of Leeds and in the sunshine of Pacific Palisades, and for us to arrange the typing of this book. A number of people, both analysts and set theorists,

have read some chapters of a preliminary version of this work, and we are very grateful to them for their comments.

They

include William Bade, John Derrick, Peter Dixon, Peter We give special thanks to K.P. Hart and to the referee for reading

McClure, Neil Mowbray, John Truss and Tom Ransford.

the entire book in manuscript and for making many valuable suggestions: of course, the decisions on the approach, the (possibly tendentious) opinions, and the errors are all our responsibility.

We owe a special debt of thanks to our typist, Mrs. Joan Bunn. A large amount of information in a work of this type is carried by the details of the symbolism and the We have tried hard to eliminate errors in this area (presumably unsuccessfully): without the skill and accuracy of Mrs. Bunn's transcription, the process of correcting our fonts.

errors would surely have diverged.

H.G. Dales and W.H. Woodin, Leeds, April, 1987

1

1

HOMOMORPHISMS FROM ALGEBRAS OF CONTINUOUS FUNCTIONS

Throughout this book, we shall work in the naive set theory familiar to analysts:

the formalization of this

theory is the axiom system

which will be discussed in

ZFC,

Chapter 4.

We first summarize some elementary facts about Banach algebras that we shall use.

The account in the

standard text of Rudin (58, Chapters 10 and 11] covers essentially all that we shall require. All the algebras that we shall consider are linear

and associative, and their underlying field is either the An algebra is complex field C or the real field ]R. in this case, we often denote unital if it has an identity; e. The algebra formed by adjoining an identity to a non-unital algebra A is denoted by A#, and we take A# = A if A is unital. The set of invertible elements of a unital algebra A is denoted by Inv A. A character on a complex algebra A is a non-zero homomorphism from A onto C. The set of characters on A is the character space of A, written @A. Let A be a commutative algebra. An ideal I of A is modular if the quotient algebra A/I is unital. The (Jacobson) radical, written rad A, of A is defined to be the intersection of the maximal modular ideals of A; if A has no such ideals, then rad A = A, and in this case A is

the identity by

a radical algebra.

(So our convention is that a radical

algebra is necessarily commutative.) semisimple if

rad A = {O}.

It is standard that

The algebra A

is

2

rad A = {a E A : e - ab E Inv where

e

is the identity of

the fact that, if then

a E A

A#.

(b E A#) },

A#

Several times we shall use for some b E rad A,

ab = a

and

a = 0.

An element an = 0

of an algebra A

a

n E N.

for some

is nilpotent if for the set of

nil A

We write

nilpotent elements in A. If A is commutative, then is an ideal in A (it is the nilradical of A), and

nil A

nil A c rad A.

An ideal

P

in a commutative algebra A

is a

prime ideal if either a E P or b E P whenever a,b E A with ab E P. The algebra A is an integral domain if the zero ideal is a prime ideal in A. We shall occasionally use is an ideal in A (n E N), then there is

the standard algebraic fact that, if and if

a E A a prime ideal 1.1

is such that

with

P

I

an 4 I and

I c P

a 0 P.

DEFINITION

Let A be an algebra over a field

on A

is a map

p : A - Bt

(i)

p(a) > 0

(ii)

p(aa) =

(iii)

p(a + b) < p(a) + p(b)

(iv)

p(ab) < p(a)p(b)

A norm on A

p(a) # 0

The algebra A

A seminorm

(a E A);

(a E k, a E A);

Iaip(a)

is a seminorm

(v)

k.

such that:

p

(a,b E A);

(a,b E A).

such that

(a E A\{O}).

is seminormable (respectively, normable) if

there is a non-zero seminorm (respectively, a norm) on

A.

It would be more precise to say "algebra seminorm"

and "algebra norm", but this extra precision seems to be unnecessary for us.

A norm is usually denoted by

A Banach

3

Each normed

algebra is an algebra with a complete norm.

algebra A

is a dense, normed subalgebra of a Banach algebra,

A

the completion of

([8, 1.12]).

The starting point for the "automatic continuity" theory of Banach algebras is the following basic fact.

A be a Banach algebra, and let tinuous, and in fact 11+11

0 E

oD A.

Then

0

Let

is con-

4 1.

The character space

is a locally compact space with

to

respect to the relative weak *-topology from the dual space of

(We adopt throughout the convention that a locally

A.

compact space is Hausdorff.)

unital Banach algebra, then

A

Let 4P

A

# 0,

A

If

is a commutative,

is compact and non-empty.

to

be a commutative Banach algebra.

then the map

f+ ker 0

4

onto the set of maximal modular ideals of

rad A = (1{ker 0

:

If

is a bijection from

4

A

and so

A,

0 E IA} .

(1)

If 4A = 0, then A is a radical algebra. Since each character is continuous on A, each maximal modular ideal is

closed, and so

is a closed ideal in

rad A

Let A be an algebra, and let the spectrum of

a,

A.

a E A.

Then

a(a),

is the set

a(a) = {z E C

ze - a 0 Inv

:

A#}.

Let A be a Banach algebra. ([58, 10.13])

subset of

C

asserts that for each

r(a) = sup{I zI

Then

r(a)

a(a)

a E A.

:

A fundamental theorem is a non-empty, compact

Set

z E a(a)}

is the spectral radius of

radius formula asserts that

(a E A).

a,

and the spectral

4

r(a) = lim IlanIil" = inf IIanIIl/n

(a E A).

n+=

If

A

is commutative, then

a(a) = {q(a)

:

0 E D #},

and

A

Thus a a E rad A if and only if r(a) = 0 ([58, 11.9]). commutative Banach algebra is a radical algebra if and only so

IIanII1/n - O

if

as

n - -

for each

a E A.

We now describe in more detail the specific Banach algebras that we are studying. Let

be a topological space.

X

(respectively,

C(X))

Then

C(X,C)

denotes the set of continuous,

complex-valued (respectively, real-valued) functions on X. (In making this definition, we implicitly assume that X is non-empty.) Then C(X,C) and C(X) are, respectively, complex and real algebras with respect to the pointwise algebraic operations on X. The constant function 1 is the identity of these algebras.

Let function

on

f

be a subset of

S

S,

set

IfIS = sup{If(x)I

Then

Let

x E S}.

S.

be a compact (Hausdorff) space.

X

set

a

x

(f)

(f E C(X,C)).

= f(x)

homeomorphism of X onto Oc(x,C), identify these two spaces. Let z

and set

f E C(X,C),

is the complex conjugate of

E C(X,C).

A subalgebra A

of

For

Then it is a

standard result ([58, 11.13(a)]) that the map

where

Then

is a commutative, unital Banach algebra.

(C(X,C),I.Ix)

x E X,

:

is the uniform norm on

I.IS

For each bounded

X.

x r e

x

is a

and we shall henceforth f(x)

= f x

z E C.

C(X,C)

(x E X),

Then

is self-adjoint if

We shall use the standard fact ([58, 11.18]) (f E A). E A that each closed, self-adjoint subalgebra of C(X,C) which for a contains 1 is isometrically isomorphic to C(Y,C) certain compact space Y. Let f E C(X,C). Then Z(f) = f-1'({O}) is the

zero-set of

f.

A subset

F

of

X

is a zero-set if

5

F = Z(f)

f E C(X,C).

for some

Let

be an ideal in

I

the closed subset of

h(I) _ (1{Z(f) Let

The hull of

is

I

f E U.

:

be a closed set in

F

C(X,C).

X:

Then we throughout set

X.

I(F) = {f E C(X,C)

:

Z(f) z F}

J(F) = {f E C(X,C)

:

Z(f)

and

is a neighbourhood

of F}. It is easily seen that

and

I(F)

the maximal and minimal ideals in that

Let

is closed, and that

I(F)

C(X,C)

respectively.

Jx and We see that

whose hull is

is dense in

J(F)

Then we write

x E X.

I({x}),

are, respectively,

J(F)

Mx

for

{M

: x E X}

x

F,

I(F).

and

J({x})

is the

C(X,C).

family of maximal ideals in

We use the same notations I(F), J(F), Jx and Mx C(X) when we are considering

for the analogous ideals in real-valued functions.

We write

(respectively,

L'(C)

k7)

for the set

of bounded, complex-valued (respectively, real-valued) sequences on

(alN Then

and we set

IN,

= sup( IanI

:

n E N}

(a = (an) E 1-(C)) .

is a complex, commutative, unital Banach Temporarily, we denote its character space by @.

(i°'(C),1.I3N)

algebra.

For n E N, a E RF(C),

set

set En(a) = an a(4) = (a)

(a = (an) E £°°(C)), (0 E @).

are standard and are easily proved;

and, for

The following results

they follow from the

Gelfand-Naimark theorem ([58, 11.18]), for example, and are given in greater generality in [35, 8.3]. (i)

The map

n F+ En, N - $,

is a homeomorphism

6

with dense range.

The map

(ii)

a N &, km(C)

is an

-r

isometric isomorphism.

Thus the compact space

@

define

to be

SIN

has the properties of the Stone-

BN

Cech compactification

N.

of

In our approach, we

different, topological

t:

characterizations are given in [36] and (69], for example, and we shall give a characterization in terms of ultrawith a filters in Chapter 2. Henceforth, we identify N

BN

and

then

MP

subset of p E BIN ,

Let

k (C)

with C(BN ,C).

is a maximal ideal in be a subset of

a

characteristic function of

the closure of

XQ (p) = 1} = a,

and let

IN,

Clearly

Q.

in

a

For example, if

k M. be the

XQ

{p E MN: BN . Thus,

if {a,T}

is a partition of N , then {o,r} is a partition of BIN into clopen sets: we shall use this fact several times. It implies that a n T = Q n T for a,T c N , and that, for a c N ,p E o} is a base of neighbourhoods of p E BIN, {a p in BIN consisting of clopen sets. The existence of this base shows that BN is a totally disconnected space. We note that, if p E BN , then JP is a prime ideal in k (C). For take f,g E k (C) with fg E J J. Then :

p E {n

:

{n

:

and so either or

p E {n

:

(fg) (n) = 0}

f(n) = 0}

p E {n

g(n) = O}

U {n

f(n) = 0}

:

,

:

,

g(n) = O}

,

in which case

in which case

g E Jp.

f E Jp,

Thus

JP

is prime.

We write

c0(C)

(respectively,

for the set

c0)

of complex-valued (respectively, real-valued) sequences which converge to zero.

Then co(C) is an ideal in k (Q), our identification equates it with the ideal I(BN \N)

and in

C(BN C) . coo (C) and coo for the ideals in respectively, consisting of sequences

We write co(C)

and

such that

co,

an = 0

for all but finitely many

n.

(a n)

7

The question that we shall study in this book is the following.

QUESTION X

Let

algebra

be a compact space.

C(X,C)

Is each norm on the

necessarily equivalent to the uniform norm

I,IX?

The question was raised by Kaplansky in lectures

around 1948, and it was Kaplansky who gave the first partial result ([48)). 1.2

THEOREM Let

on

C(X,C).

be a compact space, and let

X

Then

be a norm

11.11

(f E C(X,C)).

IfIX 'IIfII

Proof

Let A be the completion of the normed algebra Then A is a commutative, unital Banach

(C(X,C), 11.11).

algebra.

The map 0 .. 4IC(X,C), 0A uous injection, and so we may regard space of x E X\@A. f(@A

Then there exist By (1),

and

f

such that

f E rad A,

g

C(X,(r),

A

P

# X,

A

in

g(x) = 1,

and so

is a contin-

as a closed sub-

Suppose, if possible, that

= g(4A) = {O},

fg = g.

Thus

X.

4D

C(X,C)

and take

such that

and such that

g = 0,

a contradiction.

0A = X.

Let

i f (x) I < I I f I I ,

and let

f E C(X,C),

and so I

I X< I I f II ,

x E X.

Since

x E 4

as required.

At

It follows immediately from 1.2 (via the open mapping theorem) that each complete norm on

C(X,C)

is equiv-

alent to the uniform norm, and so there is an inequivalent norm on C(X,C) if and only if there is an incomplete norm. Let be an incomplete norm on C(X,C), and

let A be the completion of bedding

C(X,C) -. A

(C(X,C), II.II).

Then the em-

is a discontinuous monomorphism.

other hand, suppose that

6

On the

is a discontinuous homomorphism

8

from

into a Banach algebra.

C(X,C)

= max{If IX, IIe(f)II}

IIf1I Then

11.11

Set (f E C(X,C)).

is an incomplete norm on

C(X,C).

Thus our

question is equivalent to the question of the existence of a discontinuous homomorphism from C(X,C) into a Banach algebra, and it will usually be convenient to formulate it in this way.

The next advance in the study of norms on

C(X,C)

after the work of Kaplansky was the seminal paper of Bade and Curtis of 1960 ([2]).

In this paper, the structure of an

arbitrary homomorphism from

into a Banach algebra

C(X,C)

was analysed, and the key notions of singularity set and radical homomorphism (see below) were introduced. Later, in 1976, Johnson ([47]) extended Bade and Curtis's theorem, and used the extension to show that, if there is an incomplete norm on

for any compact space X, then there exists a radical Banach algebra R, and a non-zero

C(X,C)

P E $N\N, homomorphism

9

from

c0(C)

into

R

such that

ker 9

J

.

It is this reduction theorem that we shall need in Chapter 3.

Our proof is slightly more direct, and gives a little more information, than that obtained by following the original route.

The main technical device in the proof of Bade and Curtis's theorem is the following main boundedness theorem, which originates in [2, Theorem 2.1]. 1.3

THEOREM

A and B be Banach algebras, and let 6 be a homomorphism from A into B. Suppose that there are sequences (a (b and in A such that a b = 0 n n m n (m,n E IN, m $ n). Then there exists a constant C such Let

)

)

that 11 6 (anbn) I I

< C II an 11 II bn II

(n E 10 .

9

Proof

(n E Ilan II = Ilbn II = 1 Suppose, if possible, that the result is false.

We can suppose that Then for each

E N x N,

(i,j)

such that the map

we may choose

n(i,j) E N

is injective and such

I+ n(i,j)

(i,j)

that

Ile(uijvij) II > 4i+j where

and

uij = an(i,j)

(i,j E N)

vij = bn(i,j).

Set

m

fi so that each

=

v

E

R=1

(i E N) ,

it /2R

Choose

fi E A.

IIe(fi) II < 2j (i)

j(i) E N

so that

(i E N)

and set

g=

k k=1

so that

g E A.

uk,j(k)/2

Now, for

i E IN,

k+R

cc

=

1l RL1 k=1

u

i,j(i)vi,j(i) /2i+j(i)

and so

Ile(gfi)II

> 4i+j(i)/2i+j(i) = 2i+j(i).

Ila(gfi) II


2i

IIa(g) II Ila(f1) II < 2j(i) Ile(g) II , for all

Thus the result holds.

i E N , I

N) .

a contradiction.

10

Let

We introduce some further notation. We write

compact space.

N

neighbourhoods of a point set of

X,

those functions of

C(X,T)

a closed ideal in

C(X,T).

1.4

Let

of

is an open sub-

U

If

X.

consists of

KU

so that

which vanish off

U,

KU

and

is

K0 = {0}.

For example,

be a homomorphism from

8

A point

if, for each

x E X

U E Nx, BlKU

C(X,C)

into a

is a singularity point for is discontinuous. The set of

singularity points is the singularity set for 1.5

be a

DEFINITION

Banach algebra. 8

x

x

KU = I(X\U),

then

X

for the family of open

8.

DEFINITION

A radical homomorphism at morphism from a maximal ideal

M

is a non-zero homo-

x

of

x

C(X,T)

into the

radical of a commutative Banach algebra. Let

and let

x E X,

: M

8

into a commutative Banach algebra

be a homomorphism We claim that 0 is a - A

X

A.

radical homomorphism if and only if 8lJx = 0. For suppose that 8(M ) c rad A, and take f E J Then there exists x x g E Mx with fg = f, and 0(f)8(g) = 8(f), so that .

0(f) = 0.

Thus

BjJx = 0,

and take

elJx = 0. E

4D

Conversely, suppose that Then 4o8 E 0M U {O}, A.

and

x

4oe)l.x = 0.

Since

8(Mx) c ker 0.

Thus

homomorphism.

Jx

Mx, 08 = 0,

is dense in

8(Mx) c rad A,

and

8

and so

is a radical

The claim is established.

It is further clear that there is a radical homomorphism at x if and only if M /J is seminormable. x

x

We shall several times use the elementary fact that, if F is an infinite subset of a regular space X, then there exist x E F and U E N such for n E N n

that

u

m

n u

n

= 0

Then there exist

and

U U V = X.

so we can choose

n

(m

U E N

n).

x

x

For take

and V E N y

n

x,y E F with x # y. with x 0 V, Y E U,

Either u fl F or V a F is infinite, and x1 E F

and an open set

W1

such that

11

is infinite and

W1 fl F

W

set

such that

n

Set

xn )t Wn.

U

For

x1 fE W1.

successively choose xn E W1 fl

...

w n ... n w n 1_ = X\W1 and set

Un = (Wl n ...

n = 2,3,...,

and an open

fl Wn-1 A F fl F

fl Wn-1)\Wn

is infinite and

(n = 2,3,...).

There is one final remark before we prove the Let A be a Banach algebra, let

theorem of Bade and Curtis. 6

be a homomorphism and let I be an ideal of of the form KU or KU n J. Suppose that there is

C(X,C) - A

:

C(X,C)

(g E I). Take a constant k such that < kIgIX2 II6(g2)II f E I. Then f = f1 - f2 + i(f3 - f4), where fj E I and f(X) c [O,IfIX] (j = 1,...,4). Further, there exist gj E I with gj2 = fj, and so

II6(fj) II < kIf .IX Thus

116(f)II

< 4k I f I I.

then, for each n E N, > nlgn1X2.

II 6(gn2)II

6II is discontinuous, there exists g E I with

Hence, if

Similarly, we see that, if

g E I

then there exists

(j = 1,...,4).

such that 6(g 2)

611 # 0,

0.

We can now give the theorem of Bade and Curtis ([2, Theorem 4.3]). 1.6

THEOREM

Let X be a compact space, let homomorphism from F

C(X,C)

into a Banach algebra

be the singularity set of 6. Then: (i) F is empty if and only if F

(ii)

Now suppose that

be a

6

6

A,

and let

is continuous;

is finite. 0

is discontinuous and that

F = (xI'...,xn}.

(iii) There are a continuous homomorphism C(X,C)

that and

viIMx

A

and linear maps v1,...,vn : C(X,C) -' A such is a radical homomorphism at xi (i = 1,...,n)

0 = µ + vl + ... + vn

12

Proof

versely, suppose that exists

F = 0

Clearly

(i)

is continuous.

6

Then, for each

F = 0.

such that

Ux E Nx

if

Con-

there

x E X,

X

is continuous, and so

6IKU X

V such that 6IKv is continuous Using a partition of unity subordinate to

has a finite open cover

V E

for each V,

V.

we see easily that 6 is continuous. Suppose, if possible, that (ii)

Then there are an infinite sequence

is infinite.

F

(xn) c F

Un E Nx

and

n

Um n Un = 0

such that I

I

6(fn`) I

(n E N) .

n Ifn Ix`

>

I

(m # n).

such that

fn E KU

Choose

Since fm fn = 0

n

(m # n)

Thus

this contradicts the main boundedness theorem.

F

is

finite.

Let

(iii)

Ux E Nx

such that

x E F. 6I(KU

We claim that there exists

n Jx)

is continuous.

For suppose

x

that this is not the case. sequences n E N ,

Then we may inductively choose (fn) c Jx

and

(V n) c Nx

such that, for each and

fn E Kv

(i = 1,...,n),

fi(Vn+1) = {O} Ix 2. II 6(f n2) II > n Ifn Since f mf n = 0 contradicts the main boundedness theorem. n

,

this again

(m # n),

Thus the claim

holds.

It is immediate from the claim that

continuous, say Fix

Wi E Nx

6IJ(F)

is

(f E J(F)).

< klflx

110(f) II

such that

(i = 1,...,n) i

Wi n wj = 0

(i # j),

and fix

ei E Kw

such that i

Ietlx = 1

ei = 1

and

be the set of functions f - f(xi)1 E Jx

on some neighbourhood of Then

s algebra of J(P),

C(X,(C).

Let

such that

f E C(X,(C)

(i = 1,...,n).

x

is a dense sub-

B

n

For

f E B,

f(xi)ei

f -

belongs to

and so

110(f) II

< kIf -

i=1

f(xi)eilx +

< [(n + 1)k +

Ii

f(xi)6(ei) I

I

=1

II 6 (ei) II ]if lx.

B

13

Thus

is continuous on

a

Let C(X,C),

B.

be the continuous extension of

g

Then

v = e - M.

and let

continuous homomorphism, and Take

C(X,C) -. A

:

is a

vjJ(F) = 0.

and

f E I(F)

LL

to

e1B

g E J(F).

Then

e(f)LL(g) = e(fg) = M(fg) = M(f)LL(g),

and so v(f)LL(g) = (e(f) - M(f))M(g) = 0.

Since

LL

J F

is continuous and (f,g E I(F)).

v(f)LL(g) = 0

we have

= I(F),

f,g E I(F),

Thus, for

v(fg) = e(f)e(g) - M(f)M(g) (LL(f)

_

+ V(f)HM(g) + v(g)) - LL(f)LL(g)

= v(f)v(g), and so

is a homomorphism.

vII(F)

Set

vi,...,vn

Then

i = 1,...,n). (f E C(X,C), vi(f) = v(fei) C(X,C) + A are linear maps. Since :

n

v = vi + ... + vn.

ei E J(F),

1 -

Let

i E {1,...,n}

1=1

and let

Then

f,g E MX .

fei, gei E I(F),

and so

i

v(fei) v(gei) = v(fgei2) . But

e

2

- ei E J(F),

V (f)vi(g) = vi(fg), i

If

f E JX ,

then

and so and

Vi

v(fge2) = v(fgei). Thus : MX A is a homomorphism.

fei E J(F),

i

and so

vi(f) = 0.

Thus

i

vilMx

is a radical homomorphism.

I

i

1.7

COROLLARY

Let

X

be a compact space.

Then the following are

equivalent: (a)

there is a norm on

C(X,C)

which is not

14

equivalent to the uniform norm;

there is a discontinuous homomorphism from

(b)

into a Banach algebra;

C(X,C)

there is a radical homomorphism from a maximal

(c)

ideal of

C(X,C);

there exists

(d)

seminormable.

x E X

such that M /J x

x

is

I

THEOREM

1.8

Assume that there is a Let X be a compact space. discontinuous homomorphism from C(X,C) into a Banach algebra. Then there exists p E RN\ N, a radical Banach algebra R, and a non-zero homomorphism 0 from c0(C) into R such that ker 6 m J. fl co(C). Proof Let

0

be a discontinuous homomorphism from A. Then there exists

into a Banach algebra

C(X,C)

(f n ) c C(X,(C)

such that

if

=1

and

containing

if n, fn

Then

n E 3N).

:

isometrically isomorphic to space

Since

Y.

clearly

with

Mx -*,R Y.

Let

such that

Y

is

is metrizable, and

there is a singularity point

x E Y,

a

and a non-zero homomorphism vIJx = 0. Certainly x is not an isolated

radical Banach algebra :

B

is discontinuous.

0IB

By 1.6,

point of

and so

C(X,C),

for a certain compact

C(Y,C)

is separable,

B

is a closed,

B

separable, self-adjoint subalgebra of

v

Let

110(f )II

nIX n be the smallest closed, unital subalgebra of C(X,C)

B

Set

R,

Xo = Y\{x}.

be a sequence of open subsets of

(Un)

Um fl Un = 0

S = in E N

and let

(m # n),

:

X0

# 0}.

V IKU

n

For each

n E S,

there exists

gn E KU

such that n

v(gn2)

VIJx = 0,

0.

Since

KU

fl Jx

is dense in

we may suppose that

11v(gn2)11

KU

,

and since

> nIgnI Y. 2

Since

15

gmgn = 0

be a metric defining the topology on

d

(x ) c X n

< d

n+1

is finite.

S

Let

and take 46

it follows from the main boundedness

(m # n),

theorem that

such that

o

where

(n E N) ,

n

6

+ 0

n

Set

= d(x,xn).

d

Y,

and

It

Vn = {y E Xo

< d(y,x) < 26 }

:

(n E N).

26

The sets V Vm n v

are non-empty and open, and {a : m E N } (m # n). Let

n

= 0

n

into infinite subsets, and let Then

Um = U{Vn : n E Qm}.

sets of

X

such that

o

u

is a sequence of open sub-

(Um) f1 U

m

n

and so

(m # n),

= 0

for all but finitely many

= 0

vIKU

be a countable

m

partition of N

m.

By passing to a

m

subsequence of

V = V{V

n

we may suppose that

(xn),

where

vIKV = 0,

n E N }.

:

For n F

n

set

14,

= {y E X o

26

:

n+1

< d(y,x)

ld 2n

Suppose that V U {x}

F = 0 for all but finitely many n. Then n is a neighbourhood of x in Y, and so, if

f E Mx,

there exists

f = fl + f2, diction.

whence

For n

{1}, n

d(y,x) < 6n+i'

with

f2 E Kv

and

v(f) = v(fI) + v(f2) = 0,

Thus we may suppose that

F = U Fn U {x}, e (F

fl E Jx

a closed set in n E IN,

e (y) n

n

= 0

# 0

a contra-

Let

(n E 24).

Y.

en E Jx

take

and with

F

if

with

IenIY = 1,

d(y,x)

> 6

n

with

or

Set CO

ry(a)

= n11anen

(a = (an) E co(C)) .

For each

y E Xo, there exists W E NY with en (W) = 0 for all but two values of n, and so ry(a) is continuous on X0. Clearly, ty(a) E Mx, and voip c0(C) + R is a linear :

map.

For each

en2 - en E Kv

y E Xo\V, and

en(y) E {0,1}, and so v(en2 - en) = 0 (n E N). Also

16

(m # n),

emen = 0

and so

is a homomorphism.

vo

p(coo(C)) c ix c ker v, coo (C) = 0. Choose g E MX with v(g) 0, and let so that (Bn) E co. Take an > 0 (n E N), an = Ig IF n such that a = (a n) E c o and (6n /a n) E C o t set f = g/ a n Since

on

and set

F , n

f(x)

= 0.

Then

is continuous on the

f

in Y, and so f extends to a function in MX. Since *(a) f = g on F, (a) v(f) = v(g), and so (voi) (a) # 0. Thus v°V : c0(0) -+ R is a non-zero homomorphism with (voiy) Ic 00 (C) = 0.

closed set

F

We now regard

c

0

(C)

as an ideal in

C(BN ,C).

Set

E = {p E 61N

Since

vo p # 0,

E c $N \N . If sequence U

in

(U

fl Un = 0

Since

:

(v-,) I (KU n co(C) ) # 0

E # 0. Since E

(m O n)

and

there would exist a

(vole) Icoo(C) = 0,

such that

(n E N).

n

the argument of the third paragraph Thus

(vop) I (J (E) n co(C) ) = O.

Let

e = 1

gN

n c0(Q) ) # 0

(voip) I(KU

leads to a contradiction. such that

(vol,) Icoo(C) = 0,

were infinite,

of non-empty, open subsets of

)

n

(U E Np)

E

is finite.

p E E,

on a neighbourhood of

Clearly,

and take e E C (SN,C) p

and a= 0

neighbourhood of E\{p}. Set 0(f) = (vop)(fe) Then a is the required homomorphism.

on a

(f E c(0)).

Bade and Curtis left open the question of the existence of discontinuous homomorphisms from the algebras C(X,C)

into Banach algebras.

Eventually, such homomorphisms

were constructed for each infinite, compact space by Dales ([16]) and by Esterle ([26], [24], (27]) in 1976. However, both constructions required the assumption of the Continuum Hypothesis (CH), and are thus theorems of the system ZFC + CH. Neither Dales nor Esterle discussed whether or not CH was required. We now state without proof the results of Dales and of Esterle. We denote the cardinality of a set S by ISI;

17

a review of cardinal numbers will be given in Chapter 5. Theorems whose proof is given in the theory ZFC + CH

are labelled

(see Chapter 4)

which hold in

THEOREM

1.9

Let

X

(CH)

be a compact space, and let

maximal, prime ideal in Then the algebra

but theorems

"CH",

are not labelled.

ZFC

C(X,C)

X

be a non-

IC(X,C)/Pl = 2 0.

such that

is normable.

C(X,(Z)/P

Now, if

P

is infinite, then

C(X,C)

does contain K

non-maximal, prime ideals

such that

P

Further, it is easily seen that, if then the hull of

C(X,C),

exists

x E X

such that

IC(X,C)/Pl = 2 0.

is any prime ideal in

P

is a singleton, and so there

P

since J= M,

J c P c M ; x x

is closed if and only if

P

P

X x is equal to the maximal ideal

Thus it follows from 1.9 that, if the Continuum

Mx.

Hypothesis holds, then the four properties (a) - (d) of Corollary 1.7 are true for each infinite, compact space

X.

In particular, we have the following result.

THEOREM

1.10

Let

X

(CH)

be an infinite compact space.

exists a discontinuous homomorphism from Banach algebra, and there is a norm on

C(X,C)

Then there into a

which is not

C(X,C)

equivalent to the uniform norm.

Let

X

be a compact space.

X

If

is separable,

x

then

IC(X,Q)l = 2 0.

Mx # Jx

x

(so that

Suppose that is not a

Definition 2.22), and take

x

c P 1.11

f E M x\J x.

and so there is a prime ideal

(n E N), J

x E X

P-point of

and

f (E P.

THEOREM Let

X

Then P

is such that

X - see fn t Jx in

C(X,C)

Hence we have two further results. (CH)

be a separable, infinite compact space.

with

18

Each non-maximal, prime ideal of

(i)

C(X,C)

is

the kernel of a discontinuous homomorphism into a Banach algebra.

1.12

and if

x E X

If

(ii)

M /J

quotient algebra

x

THEOREM

M

x

# J , x

then the

is seminormable.

x

(CH)

(C)/j

p E aN\N. Then the algebras co(C)/(Jp n co(C)) are normable. Let

P

and

We have stated the above four theorems as results of ZFC + CH. In fact, they all follow from a more general result on the normability of integral domains that can be proved in

This more general result will be discussed

ZFC.

at the end of Chapter 6.

It is an immediate consequence of Theorem 1.8 that,

if there is a discontinuous homomorphism from any compact space

morphism from

X,

c (C). 0

C(X,C) for then there is a discontinuous homoIt is natural to enquire whether or

not all infinite compact spaces are equivalent for our problem. The first result in this direction is the following theorem. 1.13

THEOREM

Assume that there is a discontinuous homomorphism from

t"(C)

Then there is a discon-

into a Banach algebra.

tinuous homomorphism from

C(X,C)

each infinite compact space

into a Banach algebra for

X.

Proof Since

discrete subspace

X

is infinite, it contains an infinite, {xn}.

(f) (n) = f (xn) Then

For

f E C(X,C),

set

(n E N) .

T(f) E £°(C), and T : C(X,C) + £ (C) homomorphism with range B, say. For each

is a continuous a E B

and each

19

there exists f E C(X,(C) with TM = a and IfIX < IalN + E, and so B is a closed subalgebra of

E > 0,

k`°(C). Since there is a discontinuous homomorphism from

p E BN\N and a non-zero homomorphism into a Banach algebra A such that elm # 0

there exists

2°°(C),

from k°(C) and 611 = 0. 8

P

Take such that for

0
proof from T. It is now clear that, if T I- 0 and T }- s,

then

T F-

0 A

is a

: use the logical axiom

'0 - (* - (4' A j,)) ' and modus ponens twice. It is also true that T + 0 p if and only if T f- (4 » *). 4.3

DEFINITION

Let

be a theory in Y. inconsistent if there is a sentence T }-

otherwise,

4 A (-10);

sentence

Then

T

T

4,

T

such that

is consistent.

is inconsistent, then T T has little interest;

If

T

0,

and so

is

4,

for each we prefer

consistent theories. Note that T I- ¢ if and only if T + (10) T is inconsistent. Equivalently, if and 4,

only if

T + (10)

We write

is consistent.

60

Con T if the theory

is consistent.

T

A sentence

relatively consistent with a theory Con T i.e., either

T

T

is

0

if

Con(T + 0.

implies

is inconsistent or

T + 0

is consistent.

(Note that every sentence is relatively consistent with an We shall remark shortly that it

inconsistent theory.)

cannot be proved that ZFC is consistent, and so it is natural to consider whether or not a given sentence 0 is relatively consistent with

ZFC.

We now come to the notions of model and satisfaction, which will be central to much of our later work. 4.4

DEFINITION

A model N is a pair (M,E), where M is a nonempty set and E is a subset of M x M. The set M is the underlying set of N. The model N is countable if M is countable.

Throughout, we shall adopt the convention that a model is denoted by an upper-case fraktur letter and that the underlying set is denoted by the same roman letter. The subset

binary relation on

E M :

of

M x M can be regarded as a

aEb

if and only if

(a,b) E E.

Formally, our definition of a model has no connection with anything preceding the definition.

However,

it is formulated with a connection to the language

V

in

mind (and more general definitions of model would be appropriate if we were considering more general languages). "model" is sometimes termed an literature.

".5,-structure"

in the

Our

61

4.5

DEFINITION Let

m

McN

and if

and

N,

be models.

N = (N,EN)

and

9 = (M,EM)

is a submodel of

is an extension of

N

Then if

2,

EN n (M x m) = EM

For example,

(3t,< )

is a model, and

is a

(@,

is

logically derivable if and only if it is true in every model. More generally, if T is a theory, then if each model of T is also a model of

T 4>.

if and only This is the

F-

4>

importance of Godel's completeness theorem : it identifies

64

the notions of proof and of truth.

Our approach to the proof of the relative consistency of a sentence from a model of

ZFC

with

a model of

ZFC

will be to construct It is because we

ZFC + 0.

are willing to apply Godel's completeness theorem that we can step from this to the assertion that 0 is relatively consistent with

ZFC.

Models of quite a number of theories are studied, and we shall give a few rather elementary examples.

But the examples that we are seriously interested in here are "models of set theory itself".

More precisely, we shall eventually

be concerned with models of the theory

ZFC

this is

:

Zermelo-Fraenkel set theory with the Axiom of Choice.

(There

are other set theories that could equally well be considered, such as

GB,

Godel-Bernays set theory, but

ZFC

seems to be

the most popular.)

We first list axioms for

ZFC.

We give a formal

version of each axiom, and an informal interpretation:

it is not expected that the reader will always easily verify, and

it is not always true, that the formal version is a precise formalization of the interpretation. There are a number of variants on this list that are current;

we do not wish to

discuss the possible variations.

Our choice is made for the sake of later technical convenience. The point is that the

axioms are an attempt to capture basic intuitions about sets, and a judgement on the success of this attempt must be subjective.

AXIOM 0

Set existence.

3x (x = x) 'There is a set'.

AXIOM 1

Extensionality.

YxYy((x n y) 06 Yz((z E x)

40

(z E y)))

65

'Two sets are equal if and only if they have the same elements'.

AXIOM 2 Let

Comprehension.

4,(xl,.... xn)

be a formula in

Y.

Vx2...lxnVxn+l3xn+2Vx1((xl E xn+2) a ((x1 E xn+1) n 'If

{B E A : P(B)

A

is a set and

P

4,)

is a "property", then

is true} is a set.'

In fact, Axiom 2 is an infinite scheme of axioms,

one axiom corresponding to each formula 4i(xl,...,xn). The language V and the formally stated axiom make precise our intuitive notion of "property".

It follows from Comprehension that there is a set with no elements, and, by Extensionality, that this set is unique. It is denoted by 0. AXIOM 3

Pairing.

Vxdy3z((x E z) A (y E Z)) 'If

with A E C

A

and

A and

and

B

are sets, then there is a set

C

B E C.'

It follows from Pairing and Comprehension that, if B are sets, then so is {A,B}. We introduce the abbreviation

Vt(t E z AXIOM 4

z c y

for

t E Y) . Union.

Vx3yVz(z E x " z c y) 'If

A

is a set, then there is a set containing

the members of the members of

A.'

66

It follows from Union and Comprehension that, if A is a set, then so is U{B : B E A}; we often denote this latter set by UA. AXIOM 5

Power set.

Vx3ydz (z a x + z E Y) 'If

A

is a set, then there is a set whose

elements include all the subsets of AXIOM 6 Let

A.'

Replacement.

4(x1,x2)

be a formula in

..

Vx33x4Vx1((x1 E x3 A 3x2(4)) + 3x2(x2 E x4 A 4)) The essence of Replacement is the following. function

F

whenever

P(A,B)

is defined by a property

P

if

(A

F(A) = B

holds.)

'The image of a set under a function defined by a property is itself a set.'

Our formulation of Replacement is a slight variant of this, but each formulation can be proved from the other by using Axioms 0 - 5, and the Axiom of Regularity given below. Axiom 6 is also an infinite scheme of axioms. AXIOM 7

Regularity (or Foundation).

Vx3yVz ((z i x) v (y E X A (z E y - z i x)) ) 'Each non-empty set A contains an element B such that there are no elements belonging to both B and

A.'

It follows from Regularity and Comprehension that

67

A E A is prohibited. This fact is used in the theory of ordinals, to be given in Chapter 5. Infinity

AXIOM 8

3xdy3z((z E X) A (y E x

y E Z))

A

'There is a non-empty set ment of

form:

W

A

such that each ele-

is an element of an element of

A.'

This axiom is often stated in a slightly different W such that the empty set belongs to

'There is a set

and such that

element of

A U (Al belongs to W whenever A is an This form can be deduced from our version by

W.'

using the other axioms.

The purpose of the axiom is to

the set W establish the existence of an infinite set: contains 0, {0}, {O,{Oi}}, {O,{{},{(,(O}}},..., and hence "really" contains AXIOM 9

Vx3y3z (x (y

2Z+.

Axiom of Choice

AC

0 v (z E X A Z

p)

v

z A y f1 z

XAZEXAy

0)

(Vt3uVv(t i x v (u E t A U E Y A (v i t v v IE y v v where

'x = 0'

abbreviates

W

abbreviates

'Vx(x i y A x

'Vy(y E x)'

and

v

u))))) ,

'y n z 2 0'

W.

'For each non-empty set

A

of pairwise disjoint,

non-empty sets, there is a set B such that the intersection of B with each member of A contains exactly one element.' Assuming Axioms 1 - 8, AC is equivalent to 'For each non-empty set A, there is a function f : A + UA such that f(B) E B for each B E A with B 1 0.' Such a

68

is a choice function for A. It is manifest that the formal version of

function

f

AC

is

cumbersome and non-enlightening. 4.9

DEFINITION

The theory consisting of the set of sentences which are proved by Axioms 0 - 8 is

ZF,

and the theory

consisting of the set of sentences which are proved by Axioms 0 - 9

ZFC.

is

Thus

ZFC

is an infinite theory in the language

So.

We write "AI I Comprehension" if a model &I satisfies each sentence of the infinite axiom scheme of Comprehension, etc.

Our point of view is that models (and, in partZFC) are just algebraic objects. They

icular, models of

have the same ontological status as partially ordered sets, for example. Necessarily a model of the whole of ZFC will be complicated, but there are familiar algebraic objects which are models of some subtheories of the theory give some examples. 4.10

ZFC:

we

EXAMPLES

(i) ft

Let

It =

(]R, i a

then

for each

1 < a

n= Then

i=o Ql

("a

is

is the element such that no element

M=

a"). (2Z+,

where, for m,n E a+

E) ,

mth. digit in the binary expansion of

if the

i.e., if

Then

a = 1,

If

is an element of

b

(N, )

...pk k,

= p1

number, and write of the formula

pi r

for

4>

0.

where

i

r

ir(4>);

is the

prime

ith

i

is the Godel number

4>

For example,

x1)) = 25385372113138176198234294. (The coding map

it

Certainly

is not very efficient!).

it

is an injection.

The question which sentences of ZFC

a polynomial

PZFC

are theorems of

.zf

It is known that there is

now has a simple formulation.

in finitely many variables, say in

variables, with coefficients in

Z;,

such that, for each r

sentence

4>,

range of

PZFC

4>

E Thm(ZFC)

on

Nk.

k

if and only if

i

0

is in the

The existence of this polynomial

arises from Matijacevic's solution to Hilbert's tenth problem : it is not possible to find an algorithm for testing

an arbitrary polynomial equation

P(X1,.... Xn) = 0

with

coefficients in Z; for the possession of a solution in integers. One can take k to be nine, and PzFC could be written down explicitly if one had sufficient fortitude. r Thus, ZFC is consistent if and only if 3x1(1(x1 = x1)) is not in the range of PZFC' and it is tempting to attempt But Godel's second incompleteness theorem shows that the attempt would be futile. to prove that this is the case.

The question of the independence of sentences also has a simple formulation

:

a sentence

0

is independent of

78

i

f if and only if neither

ZFC

of

r

nor

4

7

'14

is in the range

PZFC'

In fact, this number-theoretic approach has not been successful in proving new independence results, and success does not seem to be imminent.

The second approach is algebraic, and involves the building of models.

We have already given an example of this approach when we showed in Example 4.15 that the theory of dense total orders without endpoints is complete. In this example, we used Godel's completeness theorem.

The point of this approach is that ZFC + ¢ is has a model, and one consistent if and only if ZFC + tries to build a model of

ZFC + 4,

given a model of

ZFC.

Godel was the first to be successful in this approach when he proved (in 1938) that Con ZFC implies Con(ZFC + CH). In fact, we have the following theorem. 4.19

THEOREM Assume that

submodel

Dl

of

N

ZFC

has a model

such that

K.

Then there is a

2 = ZFC + CH.

that, if ZFC has a then there is an extension Al of K such that

(We shall prove in 7.38

model 2

N,

ZFC + CH.)

In essence, Godel discovered an effective method for building new models of ZFC a given model.

by constructing submodels of

What remained undiscovered for 25 years was

an effective method for building extensions of a given model.

The technique of forcing was discovered by Cohen in 1963 forcing does give a method for building extensions, and it is the only such non-trivial method known today. Cohen's method :

allows one to extend a model by adding new elements in a controlled manner: the sentences which are true in the extended model are exactly those that are forced to be true. Forcing will be explained in Chapter 7.

79

4.20

NOTES

We have, in the formal definition of taken the alphabet to consist of a set of positive integers. This is merely to avoid some metamathematical considerations, and is not important.

For a general background in mathematical logic, see For model theory, see Bell and

[23] and [62], for example.

Slomson ([7]), and Chang and Keisler ([10]), for example. Godel's completeness theorem 4.8 is [3, 4.9],

The theorem applies to more r in the form that a theory is consistent if and only if it has a model. Lists of the axioms of ZFC are given in [13], [44] and [49]. For a discussion of the Axiom of Choice, There is an including many other formulations, see [43]. [10, 1.3.21], and [23, §2.5].

general languages than

interesting historical account of the origins of, and the controversies surrounding, this axiom in [53]. The Lindenbaum algebra is discussed in [7, Chapter 3, §4] and [44, §17].

For a discussion of Godel's incompleteness theorems, see Smorynski [66] and the standard texts. The formulation of

NDH

is due to Solovay.

An account of Matijacevic's work and of later results is given by Davis in [19]; We make one final remark.

see also [20].

As we stated on page 64,

we shall use Godel's completeness theorem to step from an assertion about the existence of models to the fact that a sentence is relatively consistent with

ZFC.

In fact,

because of the technical details of the method by which our models are constructed, it is possible in the cases which we consider to avoid the use of this theorem.

80

5

MARTIN'S AXIOM

In this chapter, we shall introduce Martin's Axiom (MA).

first, a key step in our

We have three motives:

NDH, Theorem 6.25, is given in

proof of the independence of the theory

second, the techniques used here

ZFC + MA + iCH;

in the applications of

MA

are similar to those to be used

and third, we believe that later in the study of forcing; some of the applications of MA to analysis that are given will be of general interest.

The axiom MA

was first studied (as "Axiom A") by

Martin and Solovay in [52].

The purpose of Martin and

Solovay was to settle questions in otherwise be resolved.

ZFC + iCH

that could not

In this aim, they were eminently

turned out to be an interesting and powerful axiom that does resolve many questions in ZFC + iCH, and successful:

MA

indeed it is only rather recently that some combinatorial statements that are independent of discovered.

MA + iCH

have been

At the present time, the study of some general-

izations of

MA is playing an important role in set theory. We shall prove in Chapter 8 that the theory

ZFC + MA + iCH

is relatively consistent with

Before dealing with

MA,

ZFC.

we shall first review the

elementary theory of ordinals and cardinals that we require, and we shall formally state the Continuum Hypothesis.

The

proofs that we omit of "easily checked" results can all be found in many texts, for example in [49, Chapter 11. We shall then introduce

MA

in the form of a

version of the Baire category theorem for compact spaces, and we shall show that it is equivalent to MA', the original

81

Axiom A. analysis:

Finally, we shall give some applications of MA in we hope these applications, all of which are well

known, will give the reader some feeling how the axiom is used.

However, it should be stressed that none of these

applications will be used in the proofs of our main theorems, and so they could be omitted by a single-minded reader. 5.1

DEFINITION Let

(P, 2 0.

DEFINITION

Martin's Axiom Let

x

(MA)

is the sentence:

be a compact topological space

satisfying the countable chain condition, and let

U

be a family of dense, open sets in

X

93

with in

< 2 °

JUI

is dense

Then n{U : U E U)

.

X.

implies

CH

Certainly,

We shall see in due

MA.

is consistent, then so is the theory ZFC + MA + -1CH, and so MA does not imply CH. First, however, we give a sentence which is equivalent to MA in ZFC. It is less natural, but easier to use, than the actual statement. course that, if

ZFC

DEFINITION

5.15

Let

A subset

(i)

each

with

a,b E Q

ccc

of

P

is an antichain if, for and b are incompatible. P

P

is countable;

be a subset of

V

V-generic if

is

is then

(P, 2°. 8

Thus whether or not in

=

X

(MA)

be a complete,

ccc

be a family of dense, open sets in

metric space, and let X with

K JUI

< 2 °.

Then

fl{U

:

U E U}

is dense in

X.

103

Proof V

Let

be the Stone-Cech compactification of

OX

as before, we write S

of

Since

X.

for the closure in

S

X

ccc,

is

suffices to show that

set in

V

with V fl x = V.

OX

and, for each

S = 0Vn) n X. X,

say

n,

X

y = x,

5.30

x E X,

$X.

V be an open

let

let

in

and let

X,

x =flWn.

is complete,

x EflWn,

Take

x E V.

with S

be the

Vn

Set

is a singleton in Then y x.

Suppose, if possible, that

there is a neighbourhood V of But eventually Vn a V, and so So

X,

1/n

We claim that choose Vn E Vn

Since

S = {y}.

in

For n E 1q,

family of open balls of radius Wn = U{V : V E Vn}.

and so it

GS-subset of

is a

X

For each open set

of a subset

OX

ccc,

is

$X

X;

and

y

in

X

such that

Vn - V c V,

X =flWn,

x 9 V.

a contradiction.

giving the result.

I

NOTES

The theory of ordinals and cardinals that we have Proposition 5.9 is a form of

given is quite standard.

Konig's theorem ([49,I.10.40)).

Most of the results about MA that we have given were first proved by Martin and Solovay [52], and are included in [44] and [49].

prehensive account of

MA.

The book [34] of Fremlin is a comIt includes our applications, and

many more, often in more general form, and it has a detailed account of the historical sources. Two articles about MA are (57] and [63];

a more recent account is [70].

Generalizations of

MA

to larger classes of

partially ordered sets have been studied. A certain maximal generalization is called MM ('Martin's Maximum'). See (32] and [45, Chapter 7]. It is known that, assuming the consistency of certain large cardinal axioms,

MM + ICH

is also K

consistent.

Curiously,

MM + iCH

Another variant of

implies that

2 ° =

2.

MA is PFA (the "Proper Forcing Axiom"): see [45] and [61]. In fact, MM implies PFA and PFA implies NDH: even the definitions of MM and of PFA are beyond the scope of this book.

104

GAPS IN ORDERED SETS

6

Our main story about the existence of discontinuous homomorphisms from the algebras C(X,C) into Banach algebras was left in Chapter 3. In the language of Chapter 4, the consistency of the axiom

with

NDH

by the construction of a model of of

will be established

ZFC

given a model

ZFC + NDH

We take a major step towards this goal in the

ZFC.

present chapter by proving in Theorem 6.25 that, with MA + -iCH,

NDH

follows from the fact that there is no

(NN, a).

Thus is

q(W)

a < w,

For

and

(a a

(a > T),

x < x,

xT = 0

So

za=0

a1 set.

so

(R, J u = w]J A

First, we want the

Secondly, we want

to be related to the value v(u) for u E dom v. I u E v] A first attempt at the definition of E U E v] is to set

I u E v I] = v (u)

for u E dom v

(and

[ u E v I] = 0 for

u !E dom v).

But this definition fails to satisfy our

desiderata.

For example, it could happen that

but that

v(w) > 0

for some

in this case, we must have

w E dom. u

with

J u E v I] > v(w).

v(u) = 0, I u = w I]

= 1:

We are

essentially forced by these considerations to our form of (2); a similar justification applies to (3).

hold for

We next show that the analogues of 7.1(i) - (v) do u,v,w E VB. This appears to prove that (VB,E,tiB

140

is a

$-valued model, where

E8(u,v) = jf u E vI] and but this is not the case because

ti8(u,v) _ I u = v] ,

V0

is a class and not a set.

PROPOSITION

7.8

Let

u,v E V0.

Iu (ii) Iu

Then:

UT = 1;

(i)

vI _ Iv ° uI].

Proof

If p(u) = 0, then Tu uI] = 1 by 7.7(i). (i) Now suppose that p (u) > 0 and that [Iv = v I] = 1 for all v E V8 with p(v) < p(u). For each t E dom u,

It E

uI]

= V{u(s) A It = sI]

:

s E dom u}

> u(t) A It = tI] = u(t) because

I t = t I] = 1, and so

u(t)' v It E u] > u(t)' Thus

T u c u I] = 1,

and so

V u(t) = 1.

E u = u I] = 1.

The induction

continues. This is immediate from the definition in

(ii)

7.7(v). 7.9

t

PROPOSITION Let

u,v,w E V8.

(i)

Iu = VI A

(ii)

Iu

(iii) Iu

Then:

TV = wI] < Jfu = wI];

VI A (IV vl] A [[w

w]1 < Ju VI < 1w

WI;

u]].

Proof

The proof is by induction on the position of (p(u),p(v),p(w))

in

(Ord3, Iv ° uD

and

[Lw c v]] > Iv = wI],

condition (i) follows.

Condition (ii) is immediate if Now suppose that

IV E W I] = 0.

t E dom w.

p(w)

w = 0,

for then

and take

> 0,

By the inductive hypothesis,

lu

Tv -t]] 0,

(6)

j w E v I] = O.

t E dom v.

Now

Then

TV c u] 5 v(t)' v It E u]], and so

Iv C u]] A (v(t) A [Lw = t]])

t]] A ItEu]] u (t)

and so

u(t)' v It E uI] > u(t)' v (u(t) u

a

Ifu c uaI] > a, I u a c uI] = 1,

Then

v

For

u,v E VB,

set

equivalent if

u tiv.

if

and

If u = uaI] % a. I

VO:

u

= 1.

and

v

are

We write

supp u = {t E dom u

(t E supp u).

and

if l[u = V1

u ' v

is an equivalence relation on

u ti v

A a) > a,

(t) ' v It E U I > (u(t) A a)' v u(t) = 1.

Thus

Set

and

:

u(t) # O}.

supp u = supp v and u(t) = v(t) Then % is an equivalence relation on

V$,

u tiv if u tiv.

Recall from 5.15(1) that a subset A of a Boolean A c B\{O} and if a A b= 0 a is an antichain if whenever a,b E A with a # b. The following result is algebra

sometimes called the mixing lemma. PROPOSITION

7.11

algebra

Let A be an antichain in the complete Boolean and let f : A + V8 be a function. Then there

g,

exists u E VB with If u = f(a) I] >a

(a E A).

Proof Set t E dom u,

dom u = V{dom f(a)

:

a E A},

and, for

set

u(t) = V{f(a) (t) A a : a E A with t E dom f(a)}.

144

Take

Since

a E A.

A

is an antichain, we have

u( t)

A a =

f (a) (t) A a if t E dom f (a) , and u(t) A a = 0 if t ,E dom f(a). Now ua % f(a)a, and so I ua = f(a)a I] = 1. By 7.10, [u = uaI] >a and f f(a) = f(a)a] 3 a, and so,

by 7.9(i),

1u = f(a) ] 3 a.

be a formula of ., and let at ul,...,un E V8. We define the truth of (ul,.... un) in V8 to be an element of B, to be again denoted by Let

(xl,.... xn)

11

as if

were a

(V8,E8,'8)

Formally, the

8-valued model.

rules are exactly those set out on pages 132/133. is required with Rule 3, however.

Some care The problem is that we

must be sure that each supremum of the form exists in

B:

proper class.

V{...

:

u E V8}

we are apparently taking a supremum over a To show the suprema do exist, it is sufficient

to show that

{a E B

:

a = D[ul....,ui-l,u,ui+ll...,un] 1J for some

is actually a set. formula

u E V8},

We shall note on page 156 that, for each there is a formula **(xi,.... xm+2)

p(xl,.... xm),

such that, for vl,...,vm E V8, I*[vl,...,vm] I] = a in

B

is true in V, and so the **[vl,...,vm,a,B] result follows from the Axiom of Comprehension because B is

if and only if a set.

(The formula * does not assert that

D[vl,.... vm]I] exists, but it specifies the conditions on a E B that ensure that a = I [v1,...,vm11. The construction of

**

depends on both

i4

and

in,

and involves

working through the finitely many subformulae of

*.)

Since this is a book about forcing, we should at least mention the standard notation used in the subject.

Let

145

B

be a complete Boolean algebra, and let

b E B.

Then one

writes

b IF '4[u1,...,un]' for b < D(ul,...,un] I] in

V0

However, we shall not use this notation.

On page 65, we introduced the abbreviation x c y for Vt(t E x r t E y). We now see that this is consistent with the notation in Definition 7.7. We must show that, for each u,v E V$,

Iu c VI] = A{Iw E uI]' v Iw E vI] This is immediate if

u = 0.

:

Now suppose that

w E V$}. (7) p(u) > O.

By

f[u c VI] A Irw E u] < jw E vJ], and so ju a vI] j w E u l ' v j w E v j for w E V$ . Also, for t E dom u,

(6),

u(t) 6 I t E uI,

and so

Iu C VI > A{It E uJ]' v It E v]

:

t E dom u}.

Thus (7) follows. and so u = 0 I] has two 0 E V$ possible interpretations: the first involves the equality of

Observe that

u and 0 as terms, the second is where O(x) = 14(u] I] Vz(z j x). Fortunately, both interpretations give the same

value.

Part (i) of the following result for a formula 4(x)

is

(u = v V

4 (u) )

4 (v) I]

= 1.

Part (ii) is the fullness lemma for to the Axiom of Choice,

V$ : its proof appeals

and indeed the result can be shown to

be equivalent to the Axiom of Choice.

Let

4(x1,...,xn) be a fixed formula of .'.

146

THEOREM

7.12

Let

be a complete Boolean algebra, and let

B

u11...,un E VB. For each

(i)

u E V

u ° ui11 A Em[u1,...,un] I S ¢[u

i-1,u,u

1

u E V0

There exists

(ii)

I

such that

[u,...,u1 1] = D[ul,...,ui-1,u,ui+l,...,unI I .

Proof

Let

(i)

D

be the set of subformulae

such that the result holds for By 7.9,

4D

of

p

V.

contains all the atomic formulae.

It

is an immediate consequence of the definition of truth in VB and X that contains V v X whenever t contains V contains and p v X is a subformula of 0, and that (D

4)

3xfor each variable 3x.*

is a subformula of Suppose that

of

say

y, = V,(x1)

whenever

x

0

contains

j

and

c.

and that -p is a subformula p E for notational convenience. Set 4D

a = FLu = u111, b = EVi[u11 1, and c= FLiy(u] 1]. Then a A b < a A C. Also, a = I u1 ° u 1] , and so a A c < a A b. Thus a A b = a A c, and so a A b' = a A c'. This implies

that

-j tp E

4D.

It follows that all subformulae V of and in particular the result holds for c

to

(ii)

suppose that

Again for the sake of notational sanity, we 0 _ (xl).

Let S = {I [u] 11 S

is a subset of

minimum ordinal Then

f

:

belong itself.

S - Ord

so that, as before,

For each b E S, let f(b) be the with b = F [u] ]1 for some u E V is a function defined by a property, and

B. a

u E VB },

147

so, by the Axiom of Replacement, the image of a

with

13X 4 j = V{ JL

Lu] 1J

So there is an ordinal

f(b) 4 a :

is a set.

S

We have

(b E S).

u E Va}.

Let

I = {b E B : b
0.

Set

such that, for

150

dom v = U{dom t and, for

set

s E dom v,

For

:

t E dom u}, v(s) = 1.

t E dom u, It c v I] = 1 because

I s E V I > V (S) A I s = s I] = 1 for s E dom t.

Take

Then

w E V$.

tI] A It cVI

tD < Jw

u(t) A fw

< FIw c vI, and so

I w E u I] < j w c v I] ,

as required.

Power set

5

Vx 3y dz (z c x * z E Y) u E V$,

Given

each w E

V8,

we require

We can suppose that and, for

define

set

t E dom v,

f:

dom u -+ B

p(u)

Set

> 0.

Take

v(t) = 1.

dom v = Bdom u,

w E V8,

and

by

f(t) = It e wI] For t E dom f, f(t)' v It Also, f E dom v, and so For

such that, for

v E VB

T W c U1 < T W E V I] .

(t E dom u).

E wI] = 1,

f c wI] =1.

and so

f E V I] ) v (f) A ff f

f I] = 1.

t E V0,

It E wI] A It E uI] p (u) , then (ii) If a > P (u) , then (i)

fFa E u I] = 0. cEav

..=

u I]

= 0.

Proof

We prove (i) and (ii) simultaneously by induction on the position of

(a,P(u)) in (Ord2 P(u),

j[a = t I] =o, 6 = P (u) .

Then

then

P(t) < a

whence

tLa E

(t E dom u),

and so

u j =O. If a> p (u) , set

166

Ia

= u l < . v c .1 = A{ U

E U] : w E a}

I a E U I] = 0. The induction continues. 7.22

LEMMA Let

and suppose that

u E V8,

I ' u is an ordinal' II # 0. Then there exists

a E Ord

with

., v I u = al]

0.

Proof V

For

a E Ord,

L 'a

is an ordinal']] = 1

by

and so

7.20(ii),

Eu E al v E. = a]] v Es E ul] 1'6v

3 I 'u

= I'u is an ordinal'] Take

a E Ord with

a E uI] = 0, and so j u

a< a

with

7.23

is an ordinal'

is an ordinal']] A # 0.

> p (u) . l)

0.

By 7.21,

u Il

By (9), there exists

I u 2 a I] # O.

DEFINITION Set

Then

Ordv = {a Ordv

:

I

a E Ord}.

is a class which is contained in

V$,

and 7.22 shows that it satisfies the condition specified on in fact, page 162:

Du is an ordinal' l= V{ j u

v

a I]

:

a E Ord}

=V{{u2al] : a a' I] = 1.

(v)

Pw is the least infinite ordinal']] = 1.

168

Proof

It is easy to check Examples (i) - (iii). were false, there would exist

8

with

Ia E s11 n [ES E (a+1)v]]

0,

Example (v) follows from

and, by 7.18, this is impossible. (iv).

If (iv)

I

However, other basic properties of sets may not be preserved in the passage from

V

to

These include the

V$.

power set operation (in general,

E['P(u) = (P(u))v']l # 1) There is a general theorem

and the cardinality of a set.

that asserts that properties specified by

"1 -formulae"

are

we shall return to this point in a more general

preserved:

context in Chapter S.

It follows from 7.24(iii) and (v) that, if

is a

a

v

countable ordinal, then I'a is countable' = 1. However, v if a is an uncountable ordinal, it is not always true that

T'va

is uncountable']]

0:

it is particularly important for

More generally, if

us to deal with this possibility. u,v E V,

then

_

Jul

I'iul

lvJ

implies that

lvi

does not imply that

=

JvJ']] = 1,

but Jul

7.25

DEFINITION Let

VS

if

f'lul # ivJ'I] = 1.

K

be a cardinal.

Then

K

is preserved in

is a cardinal' ]] = 1.

J[' K

We have seen that is preserved, then

R0

is preserved in

V8.

If

ktl

v

'H1

is the least uncountable cardinal']] = 1,

and we write this as

I'itl = !tl'I] = 1.

(To recap, in this

169 v

expression

is the image under the embedding v of the V, whereas K1 is the first

K1

first uncountable ordinal in

uncountable ordinal in the universe

is pre-

If

VD.)

2

served, then all we can say in general is that

'!t2

v

or h2 = K1'

= h2

= 1, V

but, if

and

h1

are preserved, then

R2

I'K2 =

14 21 I]

= 1.

To study the preservation of cardinals, we consider terms for subsets of where

Let

fv and, for

:

set

and for functions from

u E V.

For

tHI

V

u

to

v,

of = {s

:

we set

V E V$,

u -+B,

we define

f: u -+ B,

dom

Then

u

u,v E V.

of E V8

s E u},

by

Vf(s) = f(s).

TV

= 1. For VvE V$ E v f I] = v*f (s) and T V f c u I] v*I] g = f v and = T V c u I] , = v . Then T V = 4 v

and so

v ti v*

if

T V c u I]

= 1.

Thus, up to equivalence, v

we have obtained a canonical set of terms for subsets of

u.

We essentially repeat this construction in the next theorem, obtaining a canonical set of terms for bijections. 7.26

THEOREM

Let u,v E V.

Then

[' Iu I

and only if there is a function

(i)

Of (r, s)

:

(ii)

V{f (r, s)

:

sEv}=b rEu}=b

(iii) f(r,s1) A f(r,s2) = 0 (iv) f(r1,s) A f(r2,s) = 0

=

in V$ such that:

jv j' I] 3 b

f: u x v -+ B

(r E u) ; (s E v) ;

(r E U, sl

(s E v, rl

in v); r2 in u). s2

if

170

Proof

',u,

Suppose that

lemma, there exists

=

v,']j

By the fullness

b.

with

w E VS

[['w is a bijection from u to v' ]J > b. Define

by setting

f: u x v + B

f(r,s) = ['s is the value of Then

f

(iii) holds because

11w

is an injection' I]

(i)

[['w

dom w = {(r,s)v V

Clearly

:

f: u x v + B

w((r,s)v) = f(r,s).

and it is easily checked that

is a bijection from u

vJ' I

=

satisfies

v

IV

I'1uI

and (iv)

Z b.

r E u, s E v},

11'W c u x v'I] = 1, [['w

Thus

is a function' I] 3 b,

Conversely, suppose that - (iv). Define w E V8 by

I] A b.

1 'ran w = v' ]] ) b,

(ii) holds because

holds because

r'

(i) holds because

satisfies conditions (i) - (iv):

'dom w = u' ]] > b,

at

w

b,

to

v' I]

as required.

= b.

I

We can now show that cardinals are preserved in V8 provided that 8 satisfies the key countable chain condition introduced in Chapter 4.

antichain in

Recall that

8

is

ccc

if each

is countable.

B

THEOREM Let 8 be a complete ccc each cardinal is preserved in V8. 7.27

Boolean algebra.

Then

Proof Let

with

a < K.

K

be a cardinal, and let

Set

b = J'JaJ = JKJ'I].

a

If

be an ordinal K = H0,

then

171

by 7.24(v).

b = 0

Now suppose that K > $°. Let f: a X K - B be For a < a, set fa(S) _ the function specified in 7.26.

f(a,s)

and set

(8 < K), Sa = {B

By (iii) in 7.26,

f(a,8) # O},

:

fa :

is an injection and

Sa + B\{O}

is an antichain in

fa(Sa)

S = US C'.

Since

B.

B

ccc,

is

ISI < max{a,K°} < K. Take countable, and so Then b < V{f(a,8) a < a} = 0. that a 0 S.

Thus, in each case 'K

in

Let

IKI']] = 0,

and so

is preserved

K

be any complete Boolean algebra.

8

is preserved in

is a cardinal with

K

The above

K > 1BI,

then

V8.

THEOREM

7.28

Let K =

such

V$.

proof shows that, if K

c['Ial =

This shows that

is a cardinal']] = 1.

is

S01

S < K

K IBI °.

be a complete Boolean algebra, and set

8

Then K V 12 ° < K' ]] = 1 in V8.

Proof Set

A = K+,

the successor cardinal to

K.

We

V

first claim that

if'2 0 > x'I] = 0

the case, there exists

't

b E B\{O}

and

is an injection from

By the fullness lemma, for each va E V8

in

V8.

t E VB v A

a < A,

into

If this is not such that P(3N)

there exists

such that v

'the value of

t

at

a

is

va']] > b,

]] > b.

172

and then

m'Va C

fa

if fa = f8,

Inv

nF

:

then

(a < X).

3b

I]

N'

and so a =

b,

vSI]

I va

a < X,

set

N + B.

E vaI],

[ ' t is an injection' I] 3 b.

For

Thus

H

B

a I fa,

:

because

A + ON ,

x

is an injection, and so

of the hypothesis that

a contradiction A < IB NI _ JBI 0, K K < JBI °. Hence the claim holds.

The result is immediate from the claim. It can be shown that, if

is a complete Boolean

8

h

algebra and

is infinite, then

JBI

° =

JBI

It follows from 7.26 that, if is preserved in

CBI.

then

K % CBI,

K

and it then follows easily from 7.28

V8,

that, if

1'2x°

= (280)v' I] = 1 in JBI = 280, then We shall not need this remark, but we shall need the

VB.

following corollary of the theorem.

COROLLARY

7.29

Let

B

(i)

If

(CH)

be a complete Boolean algebra.

(ii) If

K

< K2,

then

1 ' 2 ° < K2' ]] = 1 in VB .

JBI < K1,

then

[[CH] = 1

IB I

in

VB.

Proof

and so

(i)

Since

K

K

CH

holds,

K2 ° = K2

by 5.11, K

JBI 0 < K2 ° = K2.

By the theorem,

v

1-2 ° < K2'I1 =1.

V

Certainly

I'K2 < K2'I] = 1, (ii)

Again using

12 0 < Kl' I] = 1. [['2 ° < K1' I] = 1,

and so the result follows. CH,

Certainly

i.e.,

JBI ° = Kl,

and so

J ' K1 < K1' II = 1,

[[CHI] = 1.

and so

B

We can now construct the complete Boolean algebra

173

B

such that

in V.

E[ CH I = O

We must show that

I' I P(NV) I > ttl' -fl = 1

in

V8

The following theorem gives a condition which reduces the problem to a combinatorial question about Boolean algebras.

be a complete Boolean algebra, let be the f,g : N - B be functions, and let v ,v corresponding terms, (see page 169), so that dom of = dom v g = {n : n E N }, vf(n) = f(n), and ) = g (n) Let

g

v(n

.

Then

Vf c vg I]

_ A{vf(n)' v In E vg]] = A{f(n) ' v g(n)

:

:

n E N}

n E N} ,

and so

I Vf # Vg1] = V{f(n) c g(n) where

a A b = (a' n b)

:

n E N} ,

(a,b E B).

v (a n b')

(11) This shows

that condition (12), below, arises naturally. 7.30

THEOREM Let

B

be a complete

Suppose that there is a set

Boolean algebra.

ccc

S c BN

such that

IS1 = K2

and such that

V{f(n) A g(n) Then

I CH 1]

= 0 in

:

n E N} = 1

(f,g E S, f # g) .

(12)

VB.

Proof

Suppose, if possible, that

By 7.28,

I CH I]

1 ' x 1 = K 1' I] = 1,

'there is a surjection from

ttl

= b # 0

in

VB.

and so onto

P(N V)'I] = b.

174

w E VB with

By the fullness lemma, there exists Rv

is a surjection from

11w

= b.

V

onto

1

I'vf E P(Nv)'II = IVf cIN VI] = 1 for f E S, there exists of < wl such that Since

T'vf is the value of Since and

# 0.

I]

= K21 there is a subset T of S with IS a < wl such that bf # 0, where, for f E T,

ITI = ti2

I

bf

=

I 'v f is the value of

Take

Uv f

vaf'

at

w

with

f,g E T

= v g I] = 0, and so

is an antichain in

B

b

f

f

g.

as required.

b = 0,

.

By (11) and (12),

n b = 0.

Thus

g

of cardinality

contradiction of the hypothesis that

va'

at

w

f

ccc,

is

B

{b

:

f E T}

But this is a

K2'

and so

I

Our choice of a Boolean algebra

B

to satisfy the

conditions of Theorem 7.30 was implicitly given in the original arguments of Cohen. 7.31

DEFINITION

Let 2N

be the Cantor set

{O,1)3N,

and set

Xc = (2N ) w2. The set

Xc

is a compact Hausdorff space with

respect to the product topology.

Xc

is homeo-

morphic to

2 2, the Cantor cube of weight w2. as the set of functions F: w x N - {0,1}; a < w2, set Fa(n) = F(a,n).

We regard

Xc

for

Of course,

w

The regular-open algebra

space

X

there that

of a topological was introduced in Example 2.7, and it was noted R(X)

R(X)

is a complete Boolean algebra.

175

PROPOSITION

7.32

The Boolean algebra

R(X

is

d

ccc.

Proof

Let 2 N.

T

Clearly

T

Let

P

be the standard base for the topology of is countable. be the collection of maps whose domain is w2 and whose range is contained in T.

a finite subset of For t E P, set

Ut = {F E XC Then

:

Fa E t(a)

is a clopen set in

Ut

base for the topology of

s(a) = t(a)

XC.

(a E dom t)}.

and

XC,

{Ut

:

Note that, if

then

(a E dom s fl dom t),

Suppose, if possible, that

t E P}

us nut

R(XC)

is a

s,t E P

and if

# W.

is not

ccc.

Then there exists an uncountable subset A of P such that Ug n Ut = 0 whenever s,t E A with s # t. Let F = {dom t Then, t E Al. Suppose that F is uncountable. :

by the

A-system lemma, Proposition 6.1, there is a finite

subset

T

w2 and an uncountable subfamily G dom s n dom t = T whenever dom s and

of

such that

are distinct elements of F. subset A* of A such that and

t

F

and

u. n Ut # 0 Thus R(XC) 7.33

A*.

s

This conclusion also

is countable.

Since s,t E A*

F

dom t

Thus there is an uncountable dom s n dom t = T whenever

are distinct elements of

holds if

of

T

is countable, we may suppose that, for s(a) = t(a). But this implies that

a E T,

a contradiction of the hypothesis.

(s,t E A*), is

ccc.

I

THEOREM There is a complete Boolean algebra

J CHI] = 0 in

8

such that

V$ .

Proof Set $

= R(XC),

so that

0

is a complete

ccc

176

we shall apply Theorem 7.30.

Boolean algebra.

a < w2,

For

define fa : N - P(X

fa(n) = {F E Xc Clearly each

f

a

(n)

F(a,n) = 1}

is clopen in

by setting

(n E N) . and so

XC,

fa(n) E B.

Set S = {fa : a < w2}. Since fa # fs if a # g, Take a,$ < w2 with a 8. Then

U{fa(n) o fa(n)

ISI = K2

n E N}

= {F E XC : F(a,n) # F(5,n) a set which is clearly dense in

for some n E N} ,

It follows from the

X C.

first of equations (3) of Chapter 2 (page 29) that

V{fa(n) n f6 (n) and so

S

n E N} = 1,

satisfies the conditions in 7.30.

follows.

The result

a

We have now achieved the following goal. 7.34

THEOREM Assume that there is a model

there is a model, extending

Ut,

of

Al

of

ZFC.

ZFC + -CH.

Then

I

We noted in Theorem 4.19 the result of Godel that, if there is a model pt

a

which is a model of

of

ZFC,

ZFC + CH.

then there is a submodel of

We shall prove, in Theorem

7.38, that, if there is a model of ZFC then there is a model of ZFC + CH; we shall also prove a related result, Theorem 7.39, that will be required in Chapter 8. 7.35

DEFINITION Let

injective maps ran f c P(K),

K

be a cardinal.

Then

f such that dom f E K+ the power set of K. For

Q(K)

is the set of

and such that f,g E Q(K),

set

177

f 6 g

if

dom g c dom f

Certainly By 2.10,

f1dom g= g.

and

is a partially ordered set.

(Q(K),b

and

I ' b* = O in u' I] Since

Eb* E v I] = 1,

and then

b*

Set

subalgebra of

b' .

we may, by (b), suppose that b* E C*,

is uniquely specified by these properties. B* = {b* : b E B}. C*;

isomorphism, and so

the map 8*

Then

b F* b*,

H*

fl + 8*,

is a Boolean is a Boolean

is complete.

We first claim that

C*

is complete.

be a subset of C*, so that Is E v I] = 1 (8 E S) and [ES# cvI] = 1. By the fullness lemma 7.12(11), there exists c E V$ such that IL'c = VSO in u' I] = 1, Let

S

192

1. E S#]] > S#(s) = 1, that

and that

d E C*

'c

u

and so s< c s o d

c < d.

so

is complete.

C*

(b E T),

Thus

c = VS

This establishes that

in

C*,

and

is a complete sub-

8*

C*.

Take c = V{b*

Now suppose

C*.

Then

s E S } = 1,

:

Secondly, we claim that algebra of

s E S,

(s E S#1' ]]

d

u

= A l l ' s

' b*
b.

I'c = 1

in

u'I] = 1,

c' ]] A I ' b * = 1

and so

c = 1.

in u' I]

This is

sufficient to establish the second claim.

We now choose a complete Boolean algebra

C conas a complete subalgebra such that there is a Boolean isomorphism ir: C - C* with ir(b) = b* (b E B). We finally claim that

taining

8

!'u

is isomorphic to

Take

c1,c2 E C

and

/tiGI I]

b E B.

= 1.

Then

b A c1 = b A c2

if and only if b* A 1T(c1) = b* A 7r(c2) in C*, 8.6(iv), if and only if b 4 J[c11G = (c21G]

I'b* A I(c1) = b* A 7r(c2) Since

j['b* = 1 in u' I] ) b,

and so, by

in u' ]] = 1.

(3) holds if and only if

(3)

193

= rt(c2) I] ) b.

7r(c1

Thus

[c2] GI] _ f 7r(c1)

rL [c1IG

Set S = {[c] G

f

c E C} and

:

S - D; f is well defined, for if C1 = c2

by 8.6(iii).

(4),

satisfies (2), and so

f

7r(c2) I]. (c)G 1+ 7T (c) ,

(c 11 G = (c2] G, then f is a homomorphism.

By 8.6(i),

Cv/tiG

where

f is the term corresponding to claim holds, and (i) is proved.

We write

(ii)

C

C*

for is

By

onto U11=1,

['f is an isomorphism from

We prove that

(4)

Thus the final

f.

2 0.

ccc

and has cardinality

t:

this is sufficient. Let S C C*\{O}

be an antichain in

S

and

s A t = 0

E'S#\{O} Since

j 'U

is

with s# t.

s,t E S

is an antichain in

ccc'I] = 1,

'S#\{O}

for

so that

C*,

Then

u']] = 1.

it follows that

is countable']] = 1.

By the fullness lemma, there exists

t E VB

with

Nv

['t

is a surjection from

Take

n E N.

Then there exists

I 'the value of t at For

sot E S

It n

with

onto

n

to E V8

with

is tn' I] = 1.

s # to I's A t = 0

= s I A Itn

S#\{O}'I] = 1.

t j < It n

in

u']] = 1,

=01

and so

194

I 't n

But Since

many

B

n

t :

Then a # 0, and so is and so for some n E N,

# O

]

n E N},

whence

Itn = sI]

Sn = {s E S

set

n E IN,

a C S.

is

S = U{S

For

S.

Take

# 0).

is

in

a

Thus

E S#\ {O }' I] = 1, and so It n = 81 A It n = t I] = O. is ccc, It n sI] = 0 for all but countably 0 I]

is a countable set.

S

# 1.

Thus

C*

ccc.

By hypothesis, li ' Iv I < t' I] = 1. Since IB I = t, it follows from 7.28 that li ' c < ' I] = 1. Thus, by the fullness lemma, there exists t E VB with ['t and, for each

is a surjection from

fs

s E C*,

takes the value

= 1,

a' I]

= 1.

define

v

:

v' I]

is the least ordinal at which t

For

onto

is E VB with

there exists

s E C*,

['t

C

c - B.

a F 11 a = is I] ,

If a # 6, then IVa = is I] n Isv= is I] = 0, and so, since B is ccc, {a : fs(a) # O} is countable. Since Tv

18 E v I] = 1, if

s,t E C*

that

s = t.

V{

8 = t8 I]

with

fs = ft,

Thus the map

a < c l = 1. This implies that,

:

then a

t

f

8

s

'

t,

and hence, by (a),

is an injection.

It follows that IC*I < 1{f E Bt : N

=

Since

1{a

:

f(a) # Oil ' H°}1

K

IBI °=t°=t.

IC*I = t, as required. = IBI, This completes the proof of the theorem.

IC*I % IB*I

195

We now fix a complete Boolean algebra complete Boolean algebra subalgebra, and a term

T' W=

C/ ti G'

(so that, by 8.7,

containing

C

8,

a

as a complete

8

such that

w E V8

10 = 1 is a complete Boolean algebra'

[['w

The next step is to define a class

V(w)

=1).

by

setting

Vow) = {t E V8

:

I't E Vw' ]JB = 1}.

(To be precise, ['t E VW' IJ 8 algebra and

belongs to

x

t, w) ]J B,

=

is the formula which formalizes

'y

where o (x, y)

is a complete Boolean

VY'.)

Our proof of the iteration theorem involves a map e

from VC

to V. We shall give a quite explicit

definition of

texts in logic avoid, perhaps wisely, the

e;

detailed verifications that we give, for they infer the existence of define

from general considerations.

e

eIVa by induction on the ordinal First set

In fact, we

a.

e(Q) = 0.

ejVa

Now suppose that

t E Va+1 \ V. Define

by

ft

dom ft = dom t,

has been defined, and take

and,

for s E dom ft, (5)

ft(s) = V{1[e(s) Then

ft

is a function from

1s E For

e(r) ]J$ A [[ r E tJ]C

tIC < ft(s)

61,82 E dom t,

:

to

C.

(a E dom

t).

dom t

r E dom t}. Certainly

(6)

we have

lEe(s1) = e(52) IJ8 < I (ft(81)1G = [ft(s2)1GI8:

(7)

196

to see this, it is sufficient by 8.6(ii) to check that

e(sl)

8

e(s2) ]

n ft(sl) =

Ie(s1) ° e(a2) ]

A ft(s2 (The reason

and this is immediate from the definition (5). for defining fail.)

ft

is that the analogue of (7) for

may

Now set e(t) = {{{e(s)}#, {e(s),(ft(s))G}#}#

(We are determined to be quite explicit!)

defined

t

e(t)

:

s E dom t}#

Thus we have

so that

D'e(t)

is a function with domain

{e(s) and so that, for each

:

s E dom t}#' I]$

= 1,

(8)

s E dom t,

['the value of e(t) at e(s) is [f t WIG' It follows that

I 'e(t)

The inductive definition of

term in 8.9

and so

E Vw'I]8 = 1, e

(9) e(t)

E V.

continues.

In summary, for each term t in VC, e(t) is a VS which is also a term for a "term in Vw". LEMMA

e(a) = e(t) ]]B s = tI]C (s,t E VC). (ii) For a E 1/C, fs(p) _ [p sIC (p E dom s). (i)

Proof

The proof is by induction on the position of in (Ord2, 0 because 1[ e(s) is vacuous if t = 0.

if and only if p(s),p(t) > 0

Now suppose that for each p E dom. s.

p,q E V$ Then

with

s # 0,

and (ii)

and that (i) holds

(p(p),p(q)) < (P(8),P(t)).

Take

fs(p) = y{[e(p) = e(r) 18 A Ir E sI]C : r E dom s} a and

{q = 010 > a' .

V

['p(q) < g in V'' 10 = 1, and so, by the inductive C VC such that hypothesis, there exists p E VC $ a $ B+1 a, I 'e (p) ti q in Vw' ]] = 1. We have I e (p) 2 r I] Then

and so

[[ ' [e(p)

E

8Jw A E[e(p) j e(t)

By definition,

Dw

dom t = Va,

follows from (9) and the fact that

> O' I]8 >. a.

and so

(11)

p E don t.

It

that

t(p) = ft(p)

If'Ife(p) E e(t):n' > (t(p)]G'IS = 1. Thus

If' Ie(p) E e(t) IIw 3 Ie(p) E aI W' 118 = 1. But this is a contradiction of (11). The proof of the lemma is complete.

4)(x1,...,x) be a formula.

Let 8.12

i

THEOREM Let

8

be a complete Boolean algebra, let

be a

t

complete Boolean algebra containing $ as a complete subv algebra and let w E V8 be such that I 'w = CA,G'I] = 1. Then, for

t1,.... tn E VC

and

c E C,

[Egt1,...,tnI IIc = c if and only if I ' I0[e(tI),...,e(tn)] IIw = [c]G' ] 0 = 1. Proof Let

0

be the set of subformulae

V

of

0

such

202

that the result holds for ,. By 8.10, contains all the atomic formulae in $. 0 By 8.6(i), contains i v X whenever 0 contains i and 0 and

X

is a subformula of

v X

whenever

0

contains

and

contains

0

c.

Similarly,

mp

is a subformula.

Suppose that contains p(x1) and that 3x1p is a subformula. Set c = 1 3x l* I] E, and take d E C with I' 13X J W = [d]al I] 8 = 1. By the fullness lemma, there exists I t E VC with f * [ t] 11 C = c. Since I' 14 (e(t) II w

I o = 1, it follows that c < d. By the fullness

[c]

lemma again,

'there exists t E Vw with J4 [tl

[d] G' J] s = 1,

fl u ' =

and so, by a third application of the fullness lemma, there exists

s E

V8 with

I' j * [ s] j W = [d] Go I S = I ' s E Vw-18 = 1. Now

s E

Ow),

u E VE with I i [ u] IC = d, 3xly, E

and so it follows from 8.11 that there exists

in

E[ ' e (u) ti s

and so d < c.

Vw' II

8

Thus

Since * E c = d, and hence

= 1.

,

(P.

It follows that all subformulae of

and in particular the result holds for completes the proof of the theorem.

belong to

¢

@,

This

itself.

I

We need one last lemma before we can complete our analysis of

V(w)

in terms of

Vf.

u E V8 with

Let

I l u is a complete Boolean algebra' II

a

= 1.

vt

For

t E Vol

we have

E V8.

But inside

v

"t

relative to

u".

V8 we can form

In an attempt to avoid confusion, we

denote this latter term by defined up to equivalence in

(t)u

V, 8

Of course,

(t)v U

is only

but this will not matter

203

because

8.13

will only appear in expressions of the form

(t)v

LEMMA

Let t E V$ .

J ' e (t) ', (t) v in

Then

Vw' ]]

= 1.

Proof

Let s E dom t. The main point of the proof is that the following equation holds:

e(a) E e(t)I]w = To see this, set

(s)' E

1. (12)

b = Is E t] Is.

By 8.10(1),

[[ ' fre(e) E e(t) I]w = [b]G' is = 1, and so

[E' ff e(s) E e(t) jw = 1' I]8 = b,

Ie(a) E e(t) Iw = O' II$ = b'. Also, we have

If 'If (a) I (a)w

1' ]J$ = s E tJj = b, (t) ;I]w = 0' 1$ = 1s

tIB = b'o,

and so (12) follows.

We now prove the lemma by induction on result is trivial if

p(t) = 0.

Suppose that

that the result holds for each Take

a E dom t.

p(t).

The

p(t) > 0

and

with

p(8) < p(t). s E V$ By the inductive hypothesis and

(12),

lL'tLe(8) E e(t) I W = Ife(s) and so, taking account of (12), we have

(t)vIW' $ = 1,

204

e(t) c (t)w ]

= 1' II$ = 1.

Also by the inductive hypothesis and (12),

(s)w E (t),Iw = I(s)" E e(t) I W' 113 we have

It c (dom t)#I]g = 1,

and so, since

I' I (t)w c e(t) 1w = 1' I$ = 1. The result for

follows, and the induction continues.

t

Let

4(x1,...,xn)

be a formula.

We need an observation: complete Boolean algebras, if isomorphism, and if

n

it(I4(a1,...,a] I81) 8.14

let

if

81

it: 81 -r 82

a ,...,a 1

I

E V,

=

and

82

are

is a Boolean

then

TI 10[a1...'an] 1182

(Iteration theorem)

THEOREM (i)

Let

8

t1,...,t

and

u

be a complete Boolean algebra, and be terms in VS such that

is a complete Boolean algebra' D S

EL'u

[(t11u,..., (tn)u] Th = 1' 11, = 1. Then there is a complete Boolean algebra C containing a complete subalgebra such that EL 4[tl...,tn] IC = 1. Suppose, further, that

(ii)

8

ccc

is

and has

ti

cardinality

2 °

and that K

EL'u

is

ccc

and has cardinality

Then we can suppose that Ro 2

.

C

is

ccc

8

< 2 °'I]8 = 1.

and has cardinality

as

205

Proof

By 8.8, there is a complete Boolean algebra containing 8 as a complete subalgebra such that is isomorphic to

'u

t

C/tiG']]$ = 1.

V

Choose I 'u

with

w E VB

11'w = E/,v G']]B = 1.

is isomorphic to

$ = 1,

w'

Since

the above observation

shows that

IjW = 1']]$ = 1.

]]u = l' ]]B = 1, and so, by

By 8.13, Theorem 8.12,

t[4[tl,.... tn] ]]f = 1. This follows from 8.8(11).

(ii)

1

Our strategy for building the complete Boolean algebra A for which I MA ]]A = 1 is to express A as a union of complete subalgebras which are themselves built by successively eliminating potential counter-examples.

To

carry out this strategy we shall require the following result. 8.15

PROPOSITION Let

be an ordinal and let

F

sequence of complete

ccc

be a

Boolean algebras such that:

whenever

BS

whenever

Ba

a < E.

is a

ccc

Boolean algebra, and

B

a

B

denotes the completion of

I]$ = 1, Q(t).

Proof Since ccc.

is

s

By 5.16,

I's

is a gap, Theorem 6.12 proves that 8

is a

Q(t)

Boolean algebra, and so

ccc

is an (8 1,K1)-pregap in (NN,