NI

/NO, 6o 17

AXIOM OF CHOICE: EQUIVALENCES

AND SOME APPLICATIONS

THESIS

Presented to the Graduate Council of the

North Texas State University in Partial

Fulfillment of the Requirements

For the Degree of

MASTER OF ARTS

By

Denise T. Race, B.S.

Denton, Texas

August, 1983

Race, Denise T., Axiom of Choice: Equivalences and Some

Applications. Master of Arts (Mathematics), August, 1983,

60 pp., bibliography, 4 titles.

In this paper several equivalences of the axiom of choice

are examined. In particular, the axiom of choice, Zorn's lemma,

Tukey's lemma, the Hausdorff maximal principle, and the well-

ordering theorem are shown to be equivalent.

Cardinal and ordinal number theory is also studied. The

Schroder-Bernstein theorem is proven and used in establishing

order results for cardinal numbers. It is also demonstrated

that the first uncountable ordinal space is unique up to

order isomorphism.

We conclude by encountering several applications of the

axiom of choice. In particular, we show that every vector space

must have a Hamel basis and that any two Hamel bases for the

same space must have the same cardinality. We establish that

the Tychonoff product theorem implies the axiom of choice and

see the use of the axiom of choice in the proof of the Hahn-

Banach theorem.

TABLE OF CONTENTS

Chapter Page

I. INTRODUCTION AND EQUIVALENCES OFTHE AXIOM OF CHOICE I9.#. .9.9. . .. .. I

II. CARDINALS AND ORDINALS .. ,.. . .. . 10

III. APPLICATIONS INVOLVING THEAXIOM OF CHOICE.,..... .38

BIBLIOGRAPHY .4404 04 , #.......... 60

iii

;_

CHAPTER I

INTRODUCTION AND EQUIVALENCES

OF THE AXIOM OF CHOICE

The purpose of this thesis is to examine several equiv-

alences of the axiom of choice and to demonstrate their use

in basic results in various areas of mathematics. A general

knowledge of basic set theory, linear algebra, analysis, and

topology is assumed.

In the first chapter we establish several equivalences

of the axiom of choice. In particular, we show the axiom of

choice, Zorn's lemma, Tukey's lemma, the Hausdorff maximal

principle, and the well-ordering theorem are equivalent.

In the second chapter we study ordinal and cardinal

number theory. We begin by proving the Schroder-Bernstein

theorem. This result is used quite frequently in establishing

order results for cardinal numbers. In this context, we show

that any set of cardinal or ordinal numbers is linearly ordered.

We also demonstrate that the first uncountable ordinal space

is unique up to order isomorphism.

In the third chapter several applications of the axiom of

choice are encountered. In particular, we show that every

vector space must have a Hamel basis and that any two Hamel

bases for the same space must have the same cardinality. We

1

2

also demonstrate that the Tychonoff product theorem implies

the axiom of choice. Further, we investigate the existence

of {O,1}-valued finitely additive, non-countably additive

measures on the power class of the natural numbers in connec-

tion with certain transfinite constructions. We conclude

with a short discussion of connections between-the Hahn-Banach

theorem and the axiom of choice.

We use 4 and Y, to denote the natural and real numbers,

respectively.

Definition 1.1. Let {Ai}iEI be a family of sets. The

Cartesian product of this family, written X A. or It A., isiEI 1 iEI

the set of all functions f defined on I so that f(i)EA., iEI.

Sometimes f(i) will be denoted by f. and will be referred to

as the ith coordinate of f. Each such function f is referred

to as a choice function for the family {A.} .

Axiom of Choice. The Cartesian product of any non-empty

family of non-empty sets is non-empty; i.e., if {Ai}iEI is a

family of sets so that I 0 s and A. A for all iEI, then there

is at least one choice function for the family {A.}iEI.

Definition 1.2. Let each of A and I denote a set. Define

A' to be the set of all functions from I into A; i.e., AI = X A.,iEI

where A. = A for all iEI. If I = {1,2,...,n}, then we sometimes

write An instead of AI. Note that An can be written as a set

of n-tuples: An =f(xi:n :x.EA for all i}.i i=1 1

3

Definition 1.3. Let P denote a set. A partial ordering

on P is any relation < c PxP satisfying

(i) x < x;

(ii) if x<y and y < x, then x = y;

(iii) if x y and y <z, then x < z.

If < also satisfies

(iv) if x < y or y < x whenever x,yEP,

then < is called a linear ordering on P.

Remark 1.4. If P is a set of sets, we can define a relation

< by A < B if and only if Ac B. It follows easily from the

definition that this relation is a partial ordering and we refer

to this partial ordering by inclusion. In particular, if we

view functions as ordered pairs, f c g means D(f) c D(g) and

f(x) = g(x) for xED(f).

Definition 1.5. Let (P,<) be any partially ordered set

and let A c P. An element uEP is called an upper bound for A

if x < u for all xEA. An element mEP is called a maximal

element of P if m = x whenever xEP and m < x. Lower bound and

minimal element are defined analogously. A chain in P is a

linearly ordered subset of P. If < is a linear ordering on P so

that 0 # A c P implies that there exists an aEA so that a < x

for all xEA, then < is called a well-ordering of P.

Remark 1.6. By convention, every element of P is both

an upper and a lower bound for 0.

Definition 1.7. Let F be a family of sets ordered by

inclusion. Then Y is said to be a family of finite character

4

if for each set A we have AEY if and only if each finite subset

of A belongs to 7. Further, if f is a choice function such that

f : + {S(F):FEF} where S(F) is a non-empty family of strict

supersets of FEY, then a subfamily J of Y is said to be f-inductive

provided that

(i) 0E1;

(ii) f(A)E whenever AEf;

(iii) U{B:IBE2} Ef whenever 2 is a chain lying in J.

Remark 1.8. Of course we shall assume that the positive

integers are well-ordered; i.e., every non-empty subset of the

positive integers has a least member. An equivalent and useful

form of this assumption is the principle of finite induction

which may be stated as follows. Suppose that S is a subset of

the positive integers with the properties

(i) 1ES;

(ii) if nES, then.n+1ES.

Then S is the set of all positive integers.

We now show that a version of the induction principle is

valid for any well-ordered set.

Theorem 1.9. (Principle of Transfinite Induction) Let

(W,<) be a well-ordered set and let AcW satisfy the following

condition: aEA whenever I(a) = {xEW:x<a,xfa}cA. Then A=W,

Proof: If W = 0, then certainly W=A. Suppose then that

W 0 and let aEW such that a<x for all xEW. Hence I(a) = 0 C A

and it follows that aEA. Thus A 0. Suppose W\A 0 and let

bEW\A such that b<x for all xEW\A. Then b is the first element

5

of W\A so that I(b) c A, making bEA which contradicts our

assumption. Therefore A=W. Q.E.D.

The following rather technical lemma will be useful in

the proof of the main result in this chapter.

Lemma 1.10. Let Y be a family of finite character and

let . be a chain in F. Then U{B:BE2}EF.

Proof: Let A = U{B:BED'}. Let A' be a finite subset of A.

Denote A' by {a 1 ,a 2 ,..,an}. Now A'cA = U{B:BE }. Hence a1 EB1

for some B1iE. Likewise a 2 EB2 for some B2 E.. Now ,X is a chain

ordered by inclusion so that either B1cB2 or B2 cB1 , Without

loss of generality, assume B1cB 2 from which it follows that

{a1 ,a2 }cB2 . By induction, there exists some Bn E such that

A' = {a ,a2,...,a }cB . We have that B EC and A' is a finitea1,a2,..n -nn-

subset of Bn. Hence A'EF. Thus each finite subset of A is in

Y. Therefore, A = U{B:BE2}EF. Q.E.D.

We are now ready to establish the equivalences mentioned

above. We begin with formal statement of the propositions

which we shall study.

Tukey's Lemma. Every non-empty family of finite character

has a maximal element.

Hausdorff Maximal Principle. Every non-empty partially

ordered set contains a maximal linearly ordered subset.

Zorn's Lemma. Every non-empty partially ordered set in

which each chain has an upper bound must have a maximal element.

Well-ordering Theorem. Every set can be well-ordered,

6

Theorem 1.11. The following are equivalent:

(i) Axiom of Choice,

(ii) Tukey's Lemma,

(iii) Hausdorff Maximal Principle,

(iv) Zorn's Lemma,

(v) Well-ordering Theorem.

Proof: (i)-*(ii) Let the axiom of choice hold. Suppose

that Tukey's lemma fails. Let Y be a family of finite character

that has no maximal element. Then for all FEY, there exists a

GEF such that F G. For FEF, let S(F) denote all strict supersets

of F which belong to Y. Since Y has no maximal element, S(F) A g

Let f:Y + {S(F) :FEF} be a choice function for the family

{S(F) :FE}.

The first assertion is that Y itself is f-inductive. We

note that 0 is finite by definition. Let FEY. Then 0 is a

finite subset of F, and OEF. Also, if AEY, then f(A)ES(A)cF.

Let I be a chain in F. By lemma 1.10, U{B:BE.} EF. Thus F is

f-inductive, Let I = ff{I:I is f-inductive}. It follows easily

that I0 is the smallest f-inductive subfamily of Y.

Let H = {AEI 0 :if BEI and B A then f(B)c-A}. Note that 0EH,

For AEH, define GA = {CEI :CcA or Ac~f(A)cC}. Our second asser-

tion is that GA is f-inductive for any AEH. Let AEH. Clearly

0EGA. Let BEGA. Then BET0 so that f(B)EIo. Either BcA or

Acf (A)cB. If BcA, either B A or B=A. If B=A, then f(B)=f(A)

and Acf(A)cf(B); consequently, f(B)EGA. If B A, then f(B)cA

so that f(B)EGA. If Acf(A)cB, then Acf(A)cf(B) so that f(B)EGA.

. ..

7

Let ' be a chain lying in GA. Then U{B:BEX}EIo For each BE,

either BCA or Acf(A)CB. Either BcA for all BE2 or there, exists

some BE! such that Acf(A)cB. If BcA for all BE1, it follows that

U{B:BED'}cA and hence is in GA. If there exists some BED' such

that Acf (A) CB, then Acf,(A) cJ{B: B E.} and hence U{B: BEX,}EGA . Thus

GA is f-inductive. From the minimality of Io, it follows that

IocGA for all AEH. Consequently, Io=GA.

Our third assertion is that H is f-inductive. Now OEH.

Let AEH. Then AEI 0 and if BEIo such that B A, then f(B)cA,

Now f(A)EI0 Let CEIo such that C f(A) . Now CEIo=GA so that

CCA or Acf(A)cC. Thus CCA. If C=A, f(C)cf(A) . If CA, since

AEH, f(C)cAcf(A) . Thus f(C)cf(A) and hence f(A)EH. Let 2 be a

chain lying in H. Each BE. is in H so that BEIo. If DEI0 and

DB, then f(D)CB. Let CEIo such that CU{B:BE!}. Each BEH and

BEIo so that U{B:BE.}1EI0 . Also IJ=GB for all BE.. Since

CJ{B:BE}, there exists some B'E . such that B'fC. For each

BE., either CcB or Bcf(B)cC since BEH and Io=GB. From B't C, it

follows that CAB' . Thus since B'EH, f(C)cB' so that f(C)cU{B:BE'}.

Therefore, U{B:BED}EH and H is f-inductive.

Consequently, it follows that H=I0 . Now Ioc. which is

partially ordered, Let A,BEIo. Then A,BEH, Since Io=GA we

have BE GA so that BcA or Acf (A)cB. Thus Io is a chain. Let

M = U{I:IEI0 }. By lemma 1.10, MEF. Since I is a chain lying

in I which is f-inductive, U{I:IE 0 I} = MEI0. Thus f(M)EI0 and

f(M)CM which contradicts that f(M) is a strict superset of M.

Therefore Tukey's lemma holds.

8

(ii)+(iii) Let Tukey's lemma hold. Let S be a non-empty

partially ordered set. Let Y be the set of chains from S. Now

F is non-empty since S is non-empty. We assert that F is a

family of finite character. Let 4EF. Since any finite subset

of . represents a chain in S, it follows that any finite subset

of 4 is in . Suppose all finite subsets of a set A are in Y

and AJ. Then there exists no chain from S containing all

elements of A. It follows that there are two elements of A,

say a,b, which cannot be compared; i.e., neither a<b nor b<a.

Hence {a,b} does not form a chain and {a,b}4F which contradicts

our supposition. Thus F is a family of finite character.

By Tukey's lemma, Y has a maximal element, call it .

Then e must be a maximal chain in S and the Hausdorff maximal

principle holds,

(iii)+(iv) Let the Hausdorff maximal principle hold, and

suppose that S is a non-empty partially ordered set in which

each chain has an upper bound. Let $* be a maximal chain in S.

By hypothesis, 2* has an upper bound. Let kES be such a bound

and we have b<k for all bEJ*.

Suppose k is not a maximal element in S. We can choose

nES such that k<m and kim. Now mL * and b<m for all bE6* since

2* is linearly ordered. Consider .*U{m} which is linearly

ordered and hence a chain in S. Then 2* U{m} and we have

contradicted the maximality of 2*. Thus k is a maximal element

in S and Zorn's lemma holds.

.

9

(iv)+(v) Let Zorn's lemma hold. Let S be a non-empty

set and W the family of all well-ordered subsets of S. Now

W#O since for each xES, ({x},<{ })EW. Define a partial ordering <

on W by the following. If (U,<), (V,<V) EW, then (U,<U) <_ (V,<V)

provided either U=V and = <V or there is a yEV such that

U = {xEV:k y y} and < <V on U. The proof that < is a partial

ordering follows easily and hence will be omitted.

Let I be a, chain in W and M = U{U:B = (U,<)E2}. Clearly

M c S. Let 0 y N c M and BE2 such that B = (U,<) and tflN s.

Since U is well-ordered and UJf~N c U, we can find an element aE~flN

such that a <U b for all bEU~N. A messy but straightforward

approach proves that a is the least element of N. Then M is

well-ordered, (M,< )EW, and <- 4on each UEM. Thus

(U,<U) < (M,<) for each UEM; i.e., (M,<M) is an upper bound for 2.

By hypothesis, 2 has a maximal element, call it B* = (D,<D).

Suppose DDS and let yES\D. Let D' = DU{y}. Define <D, by

x <D, y for each xED and <D = <D, on D. We have D = {xED':x D'y}

so that (D,<D) < (D',KD,) and (D,<D) # (D',<D,) which contradicts

the maximality of B* = (D,<D). Then we have D=S, <S well-orders

S, and the well-ordering theorem holds.

(v)+(i) Let the well-ordering theorem hold. Let {A. Ii iEIbe a family of sets such that IAs and A.$0 for all iEI. By

hypothesis, we can choose a well-ordering of the set S = U Ai.iEI

Choose fiEA. such that f<a for each aEAi. Then there exists a

choice function for the family {Ai}iEI and the axiom of choice

holds. Q.E.D.

CHAPTER II

CARDINALS AND ORDINALS

In this chapter we establish some of the fundamental

properties of cardinal and ordinal numbers , We will begin

with a proof of the Schroder-Bernstein theorem. This theorem

will prove to be quite useful in some of the results which we

establish in this chapter.

Theorem 2.1. (Schrdder-Bernstein) Let each of A and B

be sets, f:A + B be a one-to-one function into B and g:B -+ A

be a one-to-one function into A. Then there is a one-to-one

function h:A + B onto B.

Proof: For each xEEA, define the sequence (cx) where

cXe + AUBUsin the following way: let cx = g~(x), c = f~1(cr),

x -1 x x -1 x.. c2n-1= g (c2n 2 ) and c2n= f (c2n-) for n>1 but terminating

the sequence if c = 0 for some i EA. This ensures that cxnEA2n-2

and c n-1EB so that succeeding terms are well-defined. Notice

that for nE!, either c= f 1 (cxn-) = or cnEA and either

cx - 1 xxc2n-1 g (c2n2) = 0 or c 2 nEB.Form the following sets:

A x)in dbf sA1 = {xEA: (c ) is terminated by c2n= 0 for some nb,

A2 = { xEA:.(cX) is terminated by cn=0 for some nEAG,nn_

A3 = {xEA: (cX) is infinite}C y A n

Clearly A 1 , A 2 , and A 3 partition A.

10

11

Likewise, for each yEB, form the sequence (dj) where

dy:.> AUBU(0}in the following way: let d1 = f-I(y), df = g 1 (d),

.. ,d 2n-1= f~ (d2n-2) and d2n= g~ (dn-1) for n>1 but terminating

the sequence if dY = 0 for some iEF. Notice that for nE,

either d2 = g~ (d2n-1 or d EB and either dn-1 -1 (d2n2)=0 or

d2n-1 EA. Form the following sets:

B1 = {yEB: (d) is terminated by dfn1= 0 for some nEt ,

B 2 = {ycB: (d ) is terminated by din= 0 for some nEll,

B3 = {yEB: (d ) is infinite}.

Clearly B1 , B2 , and B 3 partition B.

Define h:A + B by h (x) = g~1(x) when xEA1

f (x) when xEA2UA 3 .

Let h1 :B1 + A1 be given by h1 (y) = g(y), h2 :A2 + B2 be given

by h2 (x) = f(x), and h3 :A 3 + B3 be given by h3(x) = f(x).

Consider h1 :B1 + A1 defined by hj(y) = g(y) . By

hypothesis, h1 is one-to-one. Let xEA1 , Then (cX) is terminated

by czn= 0 for some nE!. Then c1 = g-1(x), c 2 = f 1(c 1) , ... ,

c n-1= g 1 (c2n-2), c2n = f (c2n-1) = 0. Consider y = c = g~1 (x)EB,

Then h1 (y) = g(y) = g(g- (x)) = x. Also consider the sequence

(dr). Note that d = fi 1(1 = f(cl) = cx, d = g (dl) =

g- (c2 ) = c3 ,...2dn-1 = zf (d2n-2)=f = 1~(2n-1) =c2n

Thus y EB1 and h1 is onto.

Consider h2 :A 2 -B 2 defined by h2 (x) = f(x). By hypothesis,

h2 is one-to-one. Let yEB 2 . Then (dj) is terminated by dfn= 0

for some nEf. Thus d = f-(y), d = g 1 (d*),..., dn-1= f (d n-2)and dn= g~ (dn-1) = 0. Let x = d = f- 1 (y) EA, and observe

12

that h2 (x) = f(x) = f(f-1(y)) = y. Then c1 = g~1(x --_g (dl) =

df, c2 = f-I(cl) = 2f(d2) = d3,'..., tn-1 g~ (c2n2) =

g (2n-1 n = 0. Thus x EA2 and h2 is onto.

Consider h3 :A3 -+ B3 defined by h3 (x) = f(x) . By hypothesis,

h3 is one-to-one. Let yEB3. Then (dn) is infinite; i .e.,

dl = f-I(y),d = g-1(da), ... ,d2d2 n= g(dzn-1)'

... for all nEX. Set x = dy = f-I(y)EA. Then h ~1 (yEA. hen 3 (x) = f (x)=f(f ~(y)) = y, and the sequence c1 = g(x) = g (di) = d2,

c2 = f-1(cl) = f~I(d2) = d3, ... ,cn- -1 x - 1dy2 =f (j( X = g (c2n-2) = g (d2n-1

dn, ... is infinite. Thus xEA3 and h3 is onto.

Since hj(y) = g(y) is one-to-one and onto, it follows that

1 -Ihi1 (x) = g (x) is one-to-one and onto. Therefore, h:A + B

is one-to-one and onto. Q.E.D.

Definition 2.2. With each set A we associate a symbol,

called the cardinal number of A, such that sets A and B have

the same symbol attached to them if and only if there is a

one-to-one function f from A onto B. We write card(A) to

denote the cardinal number of A. Specifically, we denote

card({1,2,...,n}) by n and card(o) by 0. Further, if each

of u and v is a cardinal number and each of U and V is a set

such that card(U) = u, card(V) = v, then we say that u<v if

and only if there exists a one-to-one function from U onto a

subset of V. We write u<v to mean that u<v and utv.

Remark 2.3. We use D(f) and R(f) to denote the domain

and range of a function f.

13

Theorem 2 4. Let u and v be cardinal numbers. Then

either u<v or v<u.

Proof: If either u=0 or v=0 then the conclusion is

clear. Suppose then that U and V are non-empty sets such

that card(U) = u and card(V) = v. Let F = {f:f:A -+ B is a

one-to-one function where AcV and BcU}. We assert that F

is a family of finite character. Now FAO since there exists

a one-to-one function f:{a} + {b} where aEV and bEU. Also

F is partially ordered by inclusion. If fEF, then any finite

subset of f is one-to-one and hence would be in F. Suppose f

is a function such that f:A + -U where AcV and each finite subset

of f belongs to F but f is not one-to-one. Then there exists

x,yED(f) such that xy and f(x)=f(y). Now {(x,f(x)),(y,f(y))}

is a finite subset of f and hence is in F. Thus f(x)Af(y)

which contradicts our supposition. Thus f is one-to-one and

hence in F, making F a family of finite character.

By Tukey's lemma, F has a maximal element, call it g. Then

g is a one-to-one function such that g:A -+ B and ACV, BcU.

Suppose D(g).V and let aEV\D(g). Either g(A)=U or g(A)gU. If

g(A)=U, then g-1:U -+ A is a one-to-one function and by definition

u<v. If g(A) U, we can choose bEU\g(A). Consider the function

h:AU[ta} -+ B(J{b} where h(x) = g(x) for xEA and h(a)=b. Clearly

h is one-to-one. Hence hEF, g<h, and we have contradicted the

maximality of g. Hence g:V + B so that v<u- Q.E.D,

Corollary 2.5,. The ordering < for cardinal numbers makes

any non-empty set of cardinal numbers linearly ordered.

14

Proof: Consider a non-empty set C of cardinal numbers.

Let UEC and U be a set such that card(U)=u. Now f(x)=x is a

one-to-one function from U onto U. Thus u<u. Let u,vEC and

U,V be sets such that card(U)=u and card(V)=v. Let u<v and

v<u, Then there is a one-to-one function f:U -+ V and a one-

to-one function g:V + U. By the Schr5der-Bernstein theorem,

there is a one-to-one function h:U -+ V onto V. Thus u=v,

Let u,v,wEC and U,V,W be sets such that card(U)=u, card(V)=v,

and card(W)=w. Let u<v and v<w. Then there are one-to-one

functions f :U -* V and g:V + W. Now gof:U + W is one-to-one

so that u<w. Finally, if u,vEC, then by theorem 2,4, u<v or

v<u. Thus C is linearly ordered by <, Q.E.D.Definition 2.6. A set S is said to be finite if S=0 or

card(S) = n = card({l,2,,,,,n}) for some nE#. If S is not finite,

it is said to be infinite. A set is said to be countable if it

is finite or can be put in a one-to-one correspondence with X.

Lemma 2.7. Every infinite set has a countably infinite

subset.

Proof: Let A be an infinite well-ordered set. Let Al = {a1 }

where a1 is the first element of A. Then card(A1 ) = 14 Suppose

for kE that there is a set AkOA such that card(Ak) = k. Let

ak+1 be the first element of A\Ak. Let Ak+l1= AkU{ak+1} so that

AK+1CA and card(Ak+1) = k+1. Thus for each nE, there is an

AncA such that card(A ) = n,n- n

Let (A11 n=l be such a sequence of sets. Define B1 = A1 = {a1 }.

Let B 2 = {a 2} where a2 EA2 \A1 . (Since card(A 1 ) = 1 < card(A2 ) = 2,

15

we can choose such an a 2 EA2 \A1 so that B 2 A0.) Let B3 = {a3}

where a3 = A3 \(B1 UB2 ). Since card(A3 ) = 3 > card(B1 UB2 ) = 2,

n-iA3 \(BiUB2 ) . Let Bn = {anI where anEAn \U B.. Each B. has

i=1n-1

cardinality 1 so that n = card(An) > card( U B.) = n-i. Thusi=i

n-1 n-iAn\ U Bi g 0. Choose anEAn \U B. and put Bn = {an}. Therefore

i=1 i=1 n

for each nE there is a finite sequence (B .)n=of singleton

sets so that

(*) B.1B. = 0 if ij and B.c:A., i=1,2,...,n.:i j 1- 1

Frequently the finite induction principle is abused in

similar constructions at precisely this point. That is, the

assertion is made that by induction there is an infinite

sequence (B ) i of singleton sets so that BinOB. = 0 if ifj

and Bic:Ai for each i. Clearly the lemma follows from this

assertion, We conclude this argument by using the Hausdorff

maximal principle to verify this assertion.

Let C be the collection of all such finite sequences

satisfying (*). Partially order C by containment, and let M

be a maximal linearly ordered subset of C. Of course, every such

finite sequence may be identified with a one-to-one function

defined on an initial segment of f and having range in A. Let

the identification be made, and set g = U{f:fEM}. Then g must

be one-to-one. Further D(g) = I, for otherwise we use our

construction above to extend g and contradict the maximality of

M. Finally, put B. = {g(i)}, i=i,2,..,, and we have U B.

a countably infinite subset of A. Q.E.D.

. ;. .

16

Remark 2.8. The cardinal number of I is denoted by N

(aleph naught).

Corollary 2.9. If a is any infinite cardinal number,

then N.< a.

Proof: Let S be a set such that card(S) = a. By lemma 2.7,

S contains a countably infinite subset B. Enumerate B and call

it (bn)n= Define f:f + S by f(n) = bn. Clearly f is one-to-one

so that N< a. Q.E.D.

The following theorem states an alternate definition of finite

and infinite without making mention of .

Theorem 2,10. Let S be a set. Then

(i) S is finite if and only if each non-empty family

of subsets of S has a minimal element;

(ii) S is infinite if and only if S can be put into a

one-to-one correspondence with a proper subset of itself.

Proof: (i) Let S be finite. If S=0, the only non-empty

family of subsets of S is {0} which clearly contains a minimal

element. Assume SAO and let n = card(S). Let F be a non-empty

family of subsets of S. Let m be the smallest natural number

for which there is a subset of S in F with cardinality m. Choose

AEF such that card(A) = m. Now if BEF\{A}, card(B) > m. Thus

there does not exist a BEF\{A} such that B-A. Hence A is a

minimal element of F.

Suppose each non-empty family of subsets of S has a minimal

element and that S is infinite. By lemma 2.7, S has a countably

infinite subset. Enumerate such a subset and call it (an)n 1 .

17

Consider the family F of S given by F = {(bn)0i: (bn)0 is

an infinite subsequence of (an) n=i} Clearly (an)n=16F so

that FAQ. By hypothesis, F has a minimal element, say (b )0 ,n n=l100 00Consider the subsequence (c )= = (b )= which is in F.n n=1 n n=2

Then (cn) n=1 (bn)n=1 and we contradict the minimality of

(bn) n=i, Thus S must be finite.

(ii) Let S be infinite and < a well-ordering of S. Select

a countably infinite subset of S, say (s n n=1. Define f:S +S\{s1}by sj1 fr s=1s 10

Sf(s) = n n=1 and f(x) = x for XES\(snn=1.

Hence S can be put in a one-to-one correspondence with a proper

subset of itself,

Suppose S can be put into a one-to-one correspondence with

a proper subset of itself and that S is finite. Let card(S) = n.

Then any proper subset of S has cardinality less than n. Hence

S cannot be put into a one-to-one correspondence with any proper

subset of itself. Thus S must be infinite. Q.E.D.

Theorem 2,11. Any subset of a countable set is countable,

Proof: Let A be a countable set. If A is finite, any

subset of A is finite and hence countable. Assume A can be put

into a one-to-one correspondence with /, Any finite subset of

A is countable so that we can restrict our attention to an

infinite subset of A. Let B be such a set. Let f:#-- A be a

bijection. Then f- :B + / is a one-to-one function. Thus

card(B) < card(.) = . But card(B) is infinite and by corollary 2;9,

N< card(B). Thus we have card(B) = t and B is countable. Q.E.D.

18

Remark 2.12. It follows directly from theorem 2.11 that

every infinite subset of f can be put into a one-to-one corre-

spondence with I.

As we shall see later, the following result is an intro-

duction to the idea of multiplying cardinal numbers.

Theorem 2.13. # x sis countable.

Proof: Define g:Ax#÷ % by g(i,j) = 2133. Clearly g is

well-defined. Also R(g) = {2'33 :i,jE/c. Suppose (i,j),

(m,n)EJx J such that (i, j) (m,n) and g (i, j) = 2133 = 2 m3 n = g (m,n).

Then 21-m = 3n-j so that i-m=0 and n-j=0. Thus we have (i,j) =

(m,n) which contradicts our supposition. Therefore g is one-to-

one, Now we have g onto R(g) so that card(fx4) = card(R(g)).

By theorem 2,11, R(g)cf is countable which gives us that hf

is countable, Q.E.D.

Theorem 2.14. If A is non-empty and countable, then there

exists a function f from / onto A.

Proof: Either A is finite or A can be put into a one-to-one

correspondence with J. If A is finite, there exists a one-to-one

function from A onto a subset of . Let f be such a function.

Let M = f(A) c .. Consider f-I:M -+ A which is clearly one-to-one

and onto. Let aEA and define g:-- A by g(x) ={fl1(x) xEM

a x EJ\M.

Since g(M) = f (M) = A, we have a function g from # onto A. Q.E.D.Theorem 2.15. If A and B are non-empty sets and f maps A

onto B, then card(B) < card(A).

Proof: Let Ab = fx:f(x) = b} for each bEB. Now Ab00 for

each bEB since f is onto B. Let g be a choice function defined

19

on B so that g(b)EAb. Then g is a bijection from B to a

subset of A so that card(B) < card(A). Q.E.D.

Definition 2.16. Let a, be cardinal numbers and A, B

be disjoint sets such that card(A) = a and card(B) = . We

define a+ = card(AUB), oa3 = card(AxB), and aR = card(AB).

The following theorem contains a list of properties of

these arithmetic operations on cardinal numbers. Each conclusion

can be proven by defining appropriate functions. We omit the

proof.

Theorem 2.17. Let u,v, and w be any three cardinal numbers.

Then

(i) u + (v + w) = (u + v) + w;

(ii) u + v = v + u;

(iii) u(v + w) = uv + uw;

(iv) u(vw) = (uv)w;

(v) uv = vu;

(vi) uVuW =uV ;

(vii) uWvV = (uv) W;

(viii) (uV) = uV;

(ix) u<v implies u + w < v + w;

(x) u<v implies uw < vw;

(xi) u<v implies uW < vW;

.. . u v(xii) u<v implies w < w

Remark 2.18. If A is a set, we use P(A) to denote the

power class of A.

20

Theorem 2.19. If A is any set, then card(A) < card(P(A)).

Proof: We will begin by showing that card(2A) = card(P(A)).

Define a function f:P(A) + 2A by f(B) = g where g:A + fO,1} is

defined by g(x) = 1 XEB

fO XEA\B.

Clearly f is well-defined. Let B,CEP(A) such that BAC. Either

\Cg or C\B ! , Without loss of generality, assume B\CA0 and

choose bEB\C. If f(B) = g and f(C) = h, then we have that

g(b) = 1 and h (b) = 0. Thus f (B) f(C) , and f is one-to-one,

Let gE2A, and consider the set B = {x:g(x) = 1} c P(A) . Then

f(B) = g so that f is onto. Thus card(2A) = card(P(A)).

We assert that card(A) < card(2A), Define g:A + 2A by

g(a) = fa where fa:A + {0,1} is defined by a = 1 x = a

o xEA\{a}

Clearly g is one-to-one so that card(A) < card(2A), Suppose

that g:A +* 2 is one-to-one and onto. Let g(a) = f for all

aEA. Define h:A -+ {0,1} by h(x) =(1 f (x) = 0

x0 fi(x) = 1.

Then h f = g(x) since h(x) f (x) for each xEA, Thus

hE2 A and hig(A), and we have contradicted our supposition.

Therefore, card(A) < card(2A) so that card(A) < card(P(A)).Remark 2,20, We denote the cardinality of 1R by c.

Theorem 2,21, card([0,11) = card((0,1)) = card([0,1)) =

Proof: Clearly card(L0,1)) < card([0,1J). Define

f: [0,1 + [0,1) by f(x) = x/2 for xEt0,1. If f(x) = f(y) we

Q.E.D.

WA.F- I MW I I

c,

21

have x/2 = y/2 or x = y, Thus f is one-to-one so that card(L0,11)

< card({0,1)), Therefore, card([0,1J) = card(LO,1)).

Clearly card((0,1)) < card([0,1)). Define f:L0,1) -4 (0,1)

by f(x) = (x + 1)/2 for xE[0,1). If f(x) = f(y), then (x + 1)/2

= (y + 1)/2 so that x = y. We have f one-to-one and hence

card([0,1)) < card((0,1)). Therefore, card((0,1)) = card([0,1)).

Consider the interval [a,b). Define f:[0,1) } [a,b) by

f(x) = (b - a)x + a. Clearly f is one-to-one and onto so that

card([0,1)) = card([a,b)),

Now card((0,1)) < card(IR). Also card(Ln-1,n)) =

card([1/n+1,1/n+2)) for nEJ. For each n, choose a bijection

fn: [n-1,n) + tl/n+1,1/n+2). Consider f:L0,o) + (0,1/2J defined

by f = UfnnEf. Since D(fn)fD(fm) = 0 = R(fn)~R(fm) for nm, it

follows that f is one-to-one. Also since (0,1/2] = U(1/n+2,1/n+1],

nEr, f is onto. Thus card([0,o)) = card((0,1/2J). Similarly, it

can be shown that card((-«o,0)) = card((1/2,1)), Therefore,

card((0,1)) = card((0,1/23) + card((1/2,1)) = card([0,o)) +

card((-o,0)) = card(IR) = c. Q.E.D.

In the next two theorems we demonstrate some peculiar

properties of cardinal arithmetic.

Theorem 2.22. Let a be an infinite cardinal number and

e a finite cardinal. Then

(i) a + 3 = a;

(ii) a + a = a;

(iii) if ip is any cardinal such that < < a, then a = a + p.

Proof: Let A and B be disjoint sets such that card(A) = a

and card(B) = .

22

(i) Now A contains a countably infinite subset; call suchCOa set C. Let (ci ) be an enumeration of C. Represent B by

cx =b.EB

{b1 ,b2'''.,b}. Define f:AUB + A by f(x)=c x = CkEC

x xEA\C.

Clearly f is one-to-one and onto so that a + = card(AUB) =

card(A) = a.

(ii) Let $ be the set of one-to-one functions f with

D(f)cA and R(f) = D(f)x{0,1}. Choose a countably infinite

subset C of A. Then Cx{0,1} is countable. Thus card(C) = N =

card(Cx{0,1}). Hence there exists a bijection f:C + Cx{O,l},

and Y#0. Partially order . by inclusion and choose a maximal

chain 6 from 97. Let F = U{f:fE&} . Now D(F) = UfD(f) : fE&} .

Let x,yED(F) such that xfy. There is an fEt with x,yED(f) .

Thus F(x) = f(x) A f(y) = F(y), and F is one-to-one. If yER(F),

then there is some fc& so that yER(f) = D(f)x{0,1}. Since

D(f)x{0,1} c: D(F)x{0,1}, it follows that yED(F)x{O,1}. Conversely,

suppose that (a,b) ED(F)x{0,1}. Then there is some fE& so that

aED(f). But then (a,b)ED(f) x {0,1} = R(f) C R(F). Therefore

R(F) = D(F)x{0,1}, and FE&.

Suppose A A D(F). First, we will consider the case when

A\D(F) is finite. Since D(F) is infinite, by the first part

of this theorem, we have card(D(F)) = card(D(F)) + card(A\D(F)) =

card(A) and card(D(F)x{0,1}) = card(D(F)x{0,1}) +

card((A\D(F))x{0,1}) = card(Ax{0,1}). Thus card(A) = card(D(F)) =

card(D(F)x{0,1}) = card(Ax{0,1}). Then we have a = a + a in

23

this case. Now suppose A\D(F) is infinite. Let C be a countably

infinite subset of A\D(F). Let g:C - Cx{0,1} be a bijection,

and define G : (D (F)UC) + (D (F)UC)xf{ 0 , 1} by G (x) = F(x) xE D (F)

lg(x) XE C.

Clearly G is one-to-one and onto so that G EF. Also F<G which

contradicts the maximality of 6. Thus D(F) = A so that

F:A + Ax{0,1} is one-to-one and onto. Therefore we have

a = card(A) = card(Ax{0,1}) = a + a.

(iii) Clearly a < a + i. Let A1 ,A 2 ,C be pairwise disjoint

sets such that card(A1 ) = a = card(A2 ) and card(C) = J. Since

a + a = a, we can choose a bijection f:A1 UA2 + A1 . Since 4, < a,

we can choose a one-to-one function g:C + A1 . Now f restricted

to g (C)UA 2 is a one-to-one mapping into A1 so that card (g (C)UA2) <

card(A1). Since card(C) = card(g(C)), we have card(CUA2) ~

card(C) + card(A 2 ) = card(g(C)) + card(A2 ) = card(g(C)UA 2 ) I

card(A 1 ), Therefore ip + a < a and we have a + = a. Q.E.D.

Theorem 2.23. If a is any infinite cardinal number, then2

a = a.

Proof: Let A be a set such that card(A) = a. Let t be

the set of one-to-one functions f with D(f) c. A and R(f) = D(f)xD(f).

Choose a countably infinite subset C of A. Then CxC is countable

and card(C) = card(CxC). Thus there exists a bijection

f0:C -+ CxC, and FAt. Partially order F by inclusion and choose

a maximal chain & from F which contains f . Let F = U{f:fE }.

Let x,yED(F) = U{D(f):f&} such that x/y. We can choose f¬&

with x,y(D(f). Thus F(x) = f(x) t f(y) = F(y) so that F is

24

one-to-one onto its range. A straightforward argument shows

that R(F) = D(F)xD(F). Thus F is onto D(F)xD(F) and FEY.

Consequently, card(D(F)) = card(D(F)xD(F))

Suppose D(F) A A. First, we will consider the case when

card(A\D(F)) < card(D(F)). By theorem 2.22, card(D(F)) =

card(A\D(F)) + card(D(F)) = card(A) so that card(A) = card(AxA)

2i e., a = a . Now suppose card(A\D(F)) > card(D(F)). Let

G c A\D(F) with card(G) = card(D(F)). Since card(D(F)) =

card(D(F)xD(F)), it follows that card(GxD(F)) = card(GxG) =

card(D(F)xD(F)) = card(D(F)) = card(G). Thus by theorem 2.22,

card(G) = card(D(F)xG) + card(GxD(F)) + -card(GxG) =

card((D(F)xG)U(GxD(F))U(GxG)). Let g be a bijection where

g: G + (D(F) xG)U (GxD (F))U (GxG). Define h: (D(F)UG) + (D (F)UG) x (D(F)UG)

by h(x) = F(x) xED(F)

g(x) xEG.

Clearly h is one-to-one and onto so that hEY. Also F<h,

contradicting the maximality of . Thus D(F) = A, and card(AxA) =

card(D(F)xD(F)) = card(D(F)) = card(A); i.e., a2 = a. Q.E.D.

We state our next lemma without proof. The reader may

consult Hewitt and Stromberg (1, p. 46) for an outline of a proof.

Lemma 2.24. Every xE[0,1J has a binary representation.

Our next result is related quite closely to theorem 2.19.

Specifically, we show that card(P(.f)) = card(IR),

Theorem 2,25, 2 = c.

Proof: Let gE{0,1} . Consider S = {n:g(n) = 1} c ,

Then 0 < g) < = 1/2 so thatvg(n L E [0,1). DefinenES 3 nE43 nE.4 3

25

f:{0,1}+[0,1) by f(g) = .nN. We can easily view C g(n)nE./ 3n~ n Esl3

in its ternary representation. Let g,hE{0,1} and gh. Let

n be the first positive integer such that g(n) A h(n). Without

loss of generality, we can assume g(n) = 1 and h(n) = 0. Then

f (g) n-1f1g (k) + 1 + + g(k)k=1 3 3n k>n 3

and

f(]) = n -h(k) + 0 + v h(k)k=1 3 k>n 3

n 1(k) + 0 + h(k)k=1 3 k>n 3

n-iS g(k) + 0 +

k=1 3 k>n 3

n-1< K k + 0 +1

k=1 3 3n

n71 (k) + 1 -_ + g(k)

k=1 3 3n k>n 3

=f(g).

Thus f is one-to-one and card({0,1} ) < card([0,1)),

Let yE[0,1) and y(n) where y(n) E {0,1} for each nE6n A 2n

be a binary representation for y. Suppose y(n) = 0 only for

finitely many n. Choose N such that for all n>N, y(n) = 1

where y(N-1) = 0. Thus y = .xxx...xxO111... = .xxx...xx1000..

so that each y E[0,1) has a binary representation where y(n) = 0

for infinitely many n. Note that if y(N-1) = 1 for all N>2,

26

then y = 111 .. = 1 ( [0,1). Define f:[0,1) +{0,1} by

f (x) =_xt where x(n) is the sequence from {9,11 chosenncS 2n

so that x(n) = 0 for infinitely many n when two representations

exist. Now f is one-to-one and x(n) is unique fqr each x.

Thus card([0,1)) < card({0,1} ) and we have c = yard([0,1)) =

card({0,1}) 2 Q.E.D.

In the remainder of the chapter, we turn ouy attention to

ordered sets and bijections which preserve the older structure.

Definition 2.26. Let A and B be linearly ordered sets.

An order isomorphism from A onto B is a one-to-or e function

from A onto B such that x<y in A implies f(x) < f(y) in B

Write A~B if such an order isomorphism exists,

Theorem 2.27, The relation "~" is reflexive, symmetric,

and transitive.

Proof: Let A,B,C be linearly ordered sets. Clearly

f:A + A defined by f(a) = a is an order isomorphism making A=A.

Let A~B and f:A + B be an order isomorphism from A onto B. Now

-1f :B + A is one-to-one and onto. Let z,w EB such that z<w.

Then z = f(x) <_ f(y) = w where x,yEA. Suppose y<x. Then

w = f(y) < f(x) = z which is a contradiction, Thus f (z) - x <

-I -1y = f (w), and f is an order isomorphism from B onto A.

Consequently, B~-A,

Let A=B, B=C, and f:A + B, g:B + C be order isomorphisms

from A onto B and B onto C, respectively. Now h = gof:A + C

is one-to-one and onto. Let x,y EA such that x<y in A. Then

27

f(x) < f(y) and h(x) = g(f(x)) < g(f(y)) = h(y) so that h is

an order isomorphism from A onto C; hence A~C. Q.E.D.

Definition 228. Associated with each linearly ordered

set we have a symbol called the order tte of A such that two

sets A and B have the same order type if and only if A~B. We

say that A and B are order isomorphic or have the same order

type. Write ord(A) to denote the order type of A. Further,

if A is well-ordered, we call ord(A) an ordinal number.

Example 2_.29. Let {l,2 ,..,n}, f, and Q, the set of

rational numbers, have their usual orderings. We write

ord( 0 ) = 0, ord({1,2,...,n}) = n, ord(P) = o, and ord(') = p.

Then 0, n, and w are ordinal numbers but r is not since C is

not well-ordered.

Definition 2.30. Let A be a linearly ordered set and

xEA. We call A = {y EA:y < x} the initial segment determined

by x. If a and 5 are ordinal numbers and A and B are well-

ordered sets such that a = ord(A) and = ord(B), write a<

to denote that there exists x EB such that A~BV. Write a<$ to

denote that either a= or a<S.

Remark 2.31. By definition we also refer to a linearly

ordered set itself as a segment.

Our next result demonstrates an important mapping property

of order isomorphisms on well-ordered sets.

Theorem 2.32. If A is a well-ordered set and f is an order

isomorphism from A into A, then x _< f(x) for all xE.A.

Proof: Suppose there exists an xEA such that f(x) < x.

Let y be the first such element of A. Then f(y) < y and,

28

consequently, y cannot be the first element of A. If zEA such

that z<y, then z <_ f(z) . Hence f (y) <_ f(f (y)) since f (y) EA.

Since f is order-preserving, f(f(y)) < f(y), which is a contra-

diction. Thus x < f(x) for all xEA. Q.E.D.

Next we shall begin our discussion of the class of successor-

preserving maps. In theorems 2.34 and 2.40, we show that successor-

preserving maps are closely related to order isomorphisms.

Definition 2.33. Let X and Y be well-ordered sets. A

function f:X + Y is called successor-preserving if for each

xEX, the element f(x) is the first element of Y not in

f({z:z < x}).

Theorem 2.34. Let A and B be well-ordered sets and f:A + B

be an order isomorphism from A onto B. Then f is successor-

preserving.

Proof: Let xEA and zEA ={zEA:z < x}. Then f(z) < f(x).

Let y be the first element of B\f(A ). Since f is onto, there

exists an a EA such that f(a) = y. Also y/f(Ax) so that x<a

and f (x) <_ f(a) = y. Now x/A, so that f(x)E B\f (A ) . Hencex xf(x) = y since y is the first element of B\f(Ax). Q.E.D.

In the next theorem we show that if A and B are order

isomorphic, then there is a unique order isomorphism between

A and B.

Theorem 2.35. Let A and B be well-ordered sets. Then

(i) A is order isomorphic to no initial segment of A;

(ii) A A implies x = y;X y

(iii) if A~B, then there exists a unique order isomorphism

from A onto B.

29

Proof: (i) Suppose there exists an xEA such that A A.X

Let f:A -* Ax be an order isomorphism from A onto A . By theorem

2.32, x < f (x) . Since f (A) = Ax, it follows that f(x) EAx so

that f(x) < x. Thus we have contradicted our supposition, and

(i) follows.

(ii) Let Ax ~ A and f:A, + Ay be an order isomorphism from

Ax onto A . Suppose y<x. Since f(A ) = Ay, it follows that

f(y)EA , and again we contradict theorem 2.32. Therefore y>x.

If we consider an order isomorphism from A onto Ax, we concludeyx

that x>y; thus x=y.

(iii) Let A~B and f:A + B be an order isomorphism mapping

A onto B. By theorem 2.34, f is successor-preserving. Suppose

g:A + B is an order isomorphism from A onto B different from f.

Hence g is successor-preserving also. Let xEA be the first

element such that f(x) A g(x). If S = {zEX:z < x}, we have

f (S) = g (S) . Now f (x) is the first element of B\f(S) = B\g (S) .

Consequently f(x) = g(x), and we have contradicted our supposition

above. Thus f is unique. Q.E.D.

It follows easily from the proof of theorem 2.35 that there

is at most one successor-preserving map of A into B.

The following lemmas can be proven easily by the definition

of well-ordering.

Lemma 2.36. Any subset of a well-ordered set is well-ordered.

Lemma 2.37. If < is a partial ordering on X with the property

that every non-empty subset of X has a least element, then < is

a linear ordering and, consequently, a well-ordering.

- - -- - ---IW-l-l----.--l---,---,40414 wam WN RIWAIWAN 0-0 -11 1

30

Lemma 2.38. The union of segments is a segment.

Theorem 2,39. The range of a successor-preserving map

is a segment.

Proof: Let X and Y be well-ordered sets and f:X + Y be

successor-preserving. If f(X) = Y, then f(X) is a segment.

We only need to consider when f(X) $ Y. Let z be the first

element of Y\f(X), Suppose there exists a tEf(X) such that

t>z. Since z f(X), t>z. Choose xEX such that f (x) = t. Now

f(x) is the first element of Y not in f(fa:a < x}). Also

z(Y\f({a:a < x}) and z < t = f(x) which contradicts how f(x)

was chosen. Suppose there exists tEY such that t<z and tgf(X).

Then tEY\f(X). Now z<t since z is the first element of Y\f(X),

contradicting our supposition. Thus if tcf(X), t<z and if t<z,

tEf(X). Also since z<y for all yEY\f(X), it follows that

f(X) = {ycY:y < z}, and f(X) is a segment. Q.E.D.

The following theorem contains the converse of theorem 2.34.

Theorem 2.40. Let X and Y be well-ordered sets and f:X + Y

be successor-preserving. Then f is an order isomorphism from

X into Y.

Proof: Let x,yEX where xfy. Without loss of generality,

assume x<y. Now f(y) is the first element of Y not in f({z:z < y})

and f(x)Ef({z:z < y}). Thus f(x) A f(y) and f is one-to-one.

Again, let x<y. Then f(x) is the first element of Y not in

f(f z: z < x}). Now ytz for all z<x so that f (y) A f(z) since

f is one-to-one, Thus f(yff({z:z < x}) and hence f(y)EY\f(fz:z < x}

Then f (x) <_ f (y) by the definition of f. Since xAy, f (x) < f (y),

31

and it follows that f is order-preserving. The argument in

theorem 2.27 shows that f~ is an order isomorphism from f(X)

onto X. Q.F4D,

Lemma 2.41. If f is a successor-preserving map of X into

Y, then the restriction of f to a segment is also successor-

preserving.

The proof of this lemma is obvious.

The next result is the counterpart to theorem 2.35.

Theorem 2.42. If X and Y are well-ordered sets, then

there is a successor-preserving map from one of them onto a

segment of the other.

Proof: Let 6 be the collection of all segments of X on

which there is a successor-preserving map into Y. Let a1 and

a2 be the first and second elements of X. Then {a1} = {zEX:z < a2}

is a segment of X. Define f:{a1 } -+ Y by f(al) = b1 where b1 is

the first element of Y. Then {a1 }ES and SA0. Let S = U{A:AES}.

By lemma 2.38, S is a segment. Let U,VE& and f,g be corresponding

successor-preserving functions. Without loss of generality,

assume UcV. Now g restricted to U is also successor-preserving.

Since successor-preserving maps are unique, then g restricted to

U is f, making g an extension of f. Consider Y = {f:f is a

successor-preserving map defined for some segment of X and having

range in Y}.. Now J is linearly ordered by inclusion. Let

F = U{f: fE} and we have D(F) = S.

Let xED(F) Then xED(f) for some fEY. Now f(x) is the

first element of Y not in f({z:z < x}) so that F is successor-

preserving from D(F) into Y. Suppose D(F) X and let z be

32

the first element of X\D(F) . Either F(S) = F({y:y < z}) = Y

or F(S) c Y. Suppose F(S) Y. Define G by G(x) = F(x) if

xED(F) and G(z) to be the first element of Y not in F(S). Thus

G is a successor-preserving map defined on the segment SU{z}

and G which is a contradiction. Thus either D(F) = S = X

or F(S) = Y.

If S = X, then F is a successor-preserving map from X onto

a segment of Y, Now suppose that F(S) = Y, SAX, and consider

F :Y + S. By theorem 2.40 we know that F:S + Y is an order

isomorphism, and by theorem 2.27 we know that F~:Y + S is an

order isomorphism. Then from theorem 2.34 we conclude that

F- :Y + S is successor-preserving. Q.E.D.

Corollary 2.43. Any set of ordinal numbers is linearly

ordered.

Proof: Let & be a collection of ordinal numbers. Let

a,3,4,E& and AB,C be well-ordered sets such that ord(A) = a,

ord(B) = , and ord(C) = k.

Clearly c=a. Suppose a<I3, f3<o, and c3, Choose ycB, XEA

such that A~B and B~A, Let f,g be the unique order isomorphisms

from A onto B and B onto A , respectively. Consider gIB which

yis an order isomorphism from By onto an initial segment of Ax.

Thus gof is an order isomorphism from A onto an initial segment

of AX, which contradicts theorem 2.35. Thus a=B. Suppose a<_

and 3<1 . Then gof is an order isomorphism from A onto an

initial segment of C. Thus a<' .

33

By theorem 2.42, there exists a successor-preserving map

from one of A or B onto a segment of the other. This map is

an order isomorphism by theorem 2.40. Thus either a<3 or $<a

and we have shown ' is linearly ordered. Q.E.D.

Let a be an ordinal number and let P. denote the set of

ordinals less than a.

Theorem 2.44. The set Pa is well-ordered and ord(Pa) = a.

Proof: From theorem 2.43 it follows that Pa is linearly

ordered. Consequently all we must show to see that P. is well-

ordered is that every non-empty subset of Pa has a first

element. Let A be a well-ordered set such that ord(A) = a. If

PEPa, 3<a and we can choose a well-ordered set B such that

ord(B) = and a unique xEA such that B~A . We note that the

choice of x is independent of the choice of B. For example,

if B and B' are well-ordered sets such that ord(B) = $ = ord(B'),

then B~Ax and B'~A for some x,yEA. We also have B~B' so that

B'~A ~=A and x=y.x y

Let S be a non-empty subset of P. and T = {xEA:there exists

a ES and a well-ordered set B such that ord(B) = R and B~A}.

Let y be the first element of T, i the ordinal number in S, and

C a well-ordered set such that CAy and ord(C) = ord(Ay) = 4).

Now Ay c Ax for all XET. Thus there exists an order isomorphism

from A into A, for all xET and hence we have an order isomorphism

from C into each set B corresponding to each xET. Thus g for

all $ ES. Therefore, P is well-ordered.

34

Define f:P + A by f(3) = x where B is a well-ordered set

such that ord(B) = and xEA such that B=A . The fact that

f is well-defined follows from the independence of x with

respect to B established above. Let $3,jE(P and BC be well-

ordered sets such that ord(B) = and ord(C) = ip. Suppose

3+p. Without loss of generality, assume 8<i. Choose x,yEA

such that B~A, and C~Ay. Let g,h be order isomorphisms from

Ax onto B and C onto Ay respectively. If Ayc Ax, then goh

is an order isomorphism from C onto an initial segment of B.

Thus t<3 which contradicts our supposition. Thus A c A andx y

we have x<y or f() < f(4). Hence f is one-to-one and order-

preserving. Let x(A. Consequently, we have A~A, and hence

ord(A ) < ord(A) = ao. Thus ord(Ax)EP and f is onto. Therefore,

f is an order isomorphism from P onto A. It follows that Pj=A

and ord(P ) = ord(A) = a. Q.E.D.

Our next result establishes a connection between cardinal

numbers and initial segments of ordinals.

Theorem 2.45, Let a be a cardinal number. Then there

exists an ordinal a such that card(P ) = a.

Proof: Let A be a set such that card(A) = a. Well-order

A and let a = ord(A) . Then ord(P ) = a = ord(A) and 4e have

P ~A. Thus there exists an order isomorphism f from Te onto

A. Thus f is a bijection from P to A and we have cald(P ) =

card(A) = a. Q.E.D.

Theorem 2.46. There is an uncountable set X which is well-

ordered by a relation < in such a way that

....................

35

(i) there is a last element c in X;

(ii) if xEX and x Q, then the set {yEX:y < x} is countable.

Proof: Let Y be an uncountable set and < a well-ordering

of Y. If Y has no last element,, choose z Y. Now consider

the uncountable set YU{z} and extend the ordering < by letting

y<z for all yEY. Let A {y:{x(Y:x < y} is uncountable} . Now

Af0 since zEA, Let S be the first element of A and let

X = {xEY:x < ,}{S} . Then X satisfies the theorem. Q.E.D.

Definition 247. In the set X, defined in theorem 2,46,

the last element c is called the first uncountable ordinal and

X is called the set of ordinals less than or equal to the first

uncountable ordinal. The elements x<G are called countable

ordinals. If {y:y < x} is finite, then x is a finite ordinal.

If co is the first nonfinite ordinal, then {x:x < c0} is the set

of finite ordinals and is equivalent, as an ordered set, to .f.

Theorem 2.48. Let Y be the set of ordinals less than the

first uncountable ordinal; i.e., Y = {xEX:x < 0}. Then every

countable subset E of Y has an upper bound in Y and hence a

least upper bound.

Proof: Let E be a countable subset of Y. Suppose E has

no upper bound. Then for all yEY, there exists an aEE such

that y< a. For each aEE, let Ya be the countable set {fxEX:x < a}

Let B = U Ya which is countable. If bEY, there exists some aEEaEE

such that b<a. Thus bEYa and bEB. Thus Yc=B which contradicts

that B is countable. Thus E has an upper bound. Consider the

36

set M = {bEY:x _ b for all xEE}. Since Y is well-ordered, M

has a least element; hence M has a least upper bound. Q.E.D.

Our next result shows that the ordinal space constructed

in theorem 2.46 is unique up to order isomorphism.,

Theorem 2 49. The well-ordered set X defined in theorem 2.46

is unique up to isomorphism.

Proof: Let X and Y be uncountable sets satisfying the

properties in theorem 2.46. In particular, let a be the last

element of X and Q be the last element of Y. By theorem 2.42,

there is a successor-preserving map from one of X or Y onto a

segment of the other. Without loss of generality, assume f is

a successor-preserving map such that f:X -- Y. Since f(X) is a

segment, either f(X) = Y, f(X) = {yEY:y < z} for some zEY\{ ,

or f(X) = {yEY:y < 2 }. Suppose f(X) = {yEY:y < z} for zEY\{y}1.

Then f(X) is countable. Since f is one-to-one and X is uncountable,

it follows that f(X) must be uncountable which is a contradiction.

Suppose f(X) ={yEY:y < S }. Let kEY\{2y} such that f(Q ) = k.

Now f(X) = f({xEX:x <_ 0}) ={yEY:y < k} ={yEY:y < k}U{k} which

is countable since k < Sy, and we have again contradicted the

fact that f(X) must be uncountable. Thus it must be true that

f(X) = Y. Then f is a one-to-one, successor-preserving map

from X onto Y, and from theorem 2.40 it follows that X and Y

are order isomorphic. Q.E.D.

m ...

CHAPTER BIBLIOGRAPHY

1. Hewitt, E. and Stromberg, K .~, Real and Abstract Analysis,Graduate Texts in Mathematics, New York7 pringerVerlag, 1975.

37

CHAPTER III

APPLICATIONS INVOLVING THE

AXIOM OF CHOICE

In this chapter we will consider applications from various

areas of mathematics which involve the use of the equivalences

of the axiom of choice. We will begin with applications from

linear algebra.

Definition 3.1. Let X be a vector space over IR. A subset

S of X is said to be linearly independent if for each finite

subset {x1,x2 ,...,xn} of distinct elements of A and every

sequence (a,c2,.'''.,an) of members of IR, the equality

n

ai ax= 0 implies al = a2 = . = an = 0.

Remark 32. By definition, 0 is independent.

Definition 3.3. A non-empty independent set B so that

B E c X implies that E is not linearly independent is called

a Hamel basis for X (over IR) . That is, a Hamel basis is a

maximal non-empty linearly independent set.

Theorem 3.4. Every vector space X with at least two

elements contains a Hamel basis.

Proof: Define L = {ScX:S is linearly independent}. Suppose

that xEX\{0}, ax = 0, and aER\{0}. Then x = 0 which contradicts

our supposition. Thus a = 0 and {x} is linearly independent; i.e.,

LAO. Partially order L by inclusion. By the Hausdorff maximal

38

39

principle, L contains a maximal chain, call it 2. Set

M = U{B:BE2}. Let {x1 ,x 2 ,,,,Xn} c M where xi # x. for all i/j

nand a 1x. = 0 where aiE1R for each i = ,2 ,...,n NowxEBi-1 l 1

for some B E.' and x2EB2 for some B2 Either B c B or B2cBS 22 - l"

Without loss of generality, assume B1 c B2 and we have {x1,x2 } c B2 .

By induction we can find BnE. such that {x1 ,x 2 , ...,x}cB

Since Bn is linearly independent, it follows that ac = =n1 2 . .an = 0 so that M is linearly independent,

Suppose there exists a set E c X where M E and B is

linearly independent. Then EEL and M = U{B:BE2} cE, making

,RU{E} a chain in L which contradicts the maximality of 2. Thus

M is a maximal linearly independent set. Q.E.D.

The following alternate description of a Hamel basis will

prove useful. The proofs of theorems 3,5 and 3.6 are straight-

forward and will be omitted.

Theorem 3.5 Let X be a vector space with at least two

elements. Then 0 B c X is a Hamel basis if and only if every

element of X can be written uniquely as a finite linear combi-

nation of elements from B.

Theorem 3.6. If A is a linearly independent subset of a

vector space X with the property that each element of X is a

finite linear combination of elements from A, then each

representation is unique,

Theorem 3.7. Let X be a vector space over 1R. Let A be a

non-empty linearly independent subset of X and let StX such that

,

40

each element of X is a finite linear combination of elements

from S. Suppose AcS. Then X has a Hamel basis B such that

A c B cS.

Proof: If S is linearly independent, it follows from

theorem 3.5 and 3.6 that S itself is a Hamel basis. Assume

S is not linearly independent. Define L = {DCS:D is linearly

independent and ACD}. Now LS0 since AEL. Partially order L

by inclusion and by the Hausdorff maximal principle, we can

choose a maximal chain & in L. Set B = U{C:CE } and we have

that B is linearly independent, AcB, and BCS so that BEL. Now

B is a maximal linearly independent set in S. Suppose xES\B

cannot be written as a finite linear combination of elements

of B. Then BL){x} is linearly independent contradicting the

maximality of B in S. Thus every element in S can be written

as a linear combination of elements of B. Hence every element

of X can be written as a finite linear combination of B. Thus

B is a maximal linearly independent set in X; i.e., B is a

basis in X. Q.E.D.

The dimension of a vector space X is usually defined to be

the cardinality of a Hamel basis for X. The following theorem

shows that this notion of dimension is well-defined.

Theorem 3.8. Let X be a vector space over IR containing at

least two elements. Then any two Hamel bases have the same

cardinality.

Proof: Let A and B be bases of X. Let C be the class of

all functions f satisfying the following:

41

(i) f is one-to-one,

(ii) D(f) c A,

(iii) R(f) c B,

(iv) (B\(R(f))) U D(f) is independent.

Let g be the identity function on ATB. It follows easily that

g0 EC. If A=B, then we are finished. Also since A and B are

bases, A ( B and B 1-A and we can assume A\BA0 and B\A/0. Let

a1 EA\B. Then a /0 and can be written as a finite linear combi-

nnation of elements of B; i.e., a1 = a ib. where a.0 for

1=1

i = 1,2,,.,n, Furthermore, at least one of the bV must be in

nB\A. Let b1 be such an element and we have bi = (a - i12a b /

so that every element of X can be written as a linear combination

of (B\{b1 }) U {al}. Also (B\{b1 }) U {ay} is linearly independent

since B\{b1 } is linearly independent and is not a basis. Define

f: (AQB)U{a 1 1 + (AQB)U{b1 } by f(x) = go(x) A

bl x=a1,

Since B\((AOfB)U{b 1 })U((Aql)U{al}) = B\{b1 }U{al} is linearly

independent, we have fEC.

Partially order by containment the set of all functions

in C which extend g0 . We have observed that this set is non-

empty. Let L be a maximal linearly ordered subset. Set

fo = U{f:f EL}. Let x,y ED(f0 ) such that x#y. Since L is ordered

by inclusion, we can find f1 EL such that x,yED(f1 ) , Hence

f0 (-x) = f 1 (x) f1 (y) = f0 (y), making f0 one-to-one. Clearly

42

D(f0 ) = U{D(f) :fEL} C A and R(fc) = U{R(f) :fEL} C B. Let

{a1 ,,,,an. ,b 1 ,,.,bm} be a finite subset of (B\R(f0)) U D(f ),0 0

where {a,,,.an} c D(fo) and {b 1 , ,.,,,b} C B\R(f0 ). Choose

fEL such that {aI,,..,an,b1 ,,,,,bm} C B\R(f)) U D(f) which is

independent. Hence any finite subset of (B\R(f0)) U D(f ) islinearly independent. Thus no element of (B\R(fo)) U D(f0) can

be written as a finite linear combination of elements of

(B\R(fo)) U D(f ). Therefore, (B\R(fo)) U D(f0) is independent

so that f0 EC.

Suppose R(f0) B and let b0EB\R(fo). (Therefore, boNA.)

Now bo cannot be written as a finite linear combination of

elements of B\(R(fo)U{bo})UD(f ) since (B\R(fo)) U D(fo) is

an independent set, Since A is a basis, b0 can be written as

a finite linear combination of elements of A so that there

exists an a0 EA\D(f 0 ) which must be a term in the expansion for

bo and which cannot be written as a finite linear combination

of elements of B\(R(f )U{bo})UD(:f ). Hence B\(R(f )U{bo})U

(D(f0 )U{a0 }) is independent. Define f*:D(f0 )U{a } + R(f0)U{b }

by f*(x) = fo(x) xED(f0)

b x=ao

Clearly f*EC and fo<f* which contradicts the maximality of L.

Thus R(fo) = B. Now f0 is a one-to-one function and f0 :D(fo) - B.

Then f :B + A is a one-to-one function from B into A.

Similarly, consider the class of functions f such that

(i) f is one-to-one;

(ii) D(f) c B;

43

(iii) R(f) C A;

(iv) (A\R(f)) U D(f) is independent.

In an analogous fashion, we can find a one-to-one function

from A into B. Thus by the Schr6der-Bernstein theorem, there

exists a one-to-one function from A onto B; i.e., card(A) =

card(B). Q.E.D.

Remark 3.9. We assume that the reader has a basic knowl-

edge of general topology. We refer the reader to Royden (3,

Chapters 7-9) for other definitions and well-known theorems

which may be needed in the following discussion. The standard

proof of the following theorem makes strong use of the axiom

of choice. We refer the reader to Royden (3, p. 166) for the

details of this argument.

Theorem 3.10. (Tychonoff) The Cartesian product of

compact topological spaces is compact.

The reader should recall that a space X is compact if and

only if each family of closed subsets of X with the finite

intersection property has non-empty intersection.

Theorem 3.11. The Tychonoff product theorem implies the

axiom of choice.

Proof: For each aEA, let Xa be non-empty. To each Xa'

adjoin the single point c, letting Ya = Xa U {c}. Assign a

topology for Ya by letting the open sets be {{c},s,Xa ,Ya

Clearly Ya is compact with this topology.

For each aEA, let Za be the subset of X Yb consisting ofbEA

all points whose a-th coordinate lies in Xa; i.e., all points

44

whose a-th coordinate is not {c)}. Since Xa is closed in Ya'

Za is closed in X Yb. Also, if B is a finite subset of A,bEA

the intersection fl Za is non-empty since each Xa # 0 and byaEB

the finite axiom of choice, we can choose x EX for aEB andaa

set xa = c for aEA\B. Hence {Za :aEA} is a family of closed

subsets of X Yb with the property that the intersection ofbEA

any finite subfamily is non-empty. By the hypothesis, since

each Ya is compact, X Yb is compact. Thus the intersectionbEA

a Za is non-empty. Consequently, l Za C X Xb, and theaAaEA a-bEA

product is non-empty. Q.E.D.,

We remark that this argument is taken from Kelley (1).

Next we investigate certain topological properties of

the ordinal space when it is endowed with its natural order

topology.

Let X be a well-ordered uncountable set which has a last

element 0, such that every predecessor of Q has at most countably

many predecessors. We saw in theorem 2.46 that such a set

exists. For xEX, let P be the set of all predecessors of x

and Sx be the set of all successors of x. We will call a subset

of X open if it is a Px, a Sy, a PXEy, or a union of such sets.

The following theorems refer to the set X and the topology

defined on X. The space X is called the ordinal space.

We remark that if xEX and x has no immediate predecessor,

then x is referred to as a limit ordinal.

--- - --------

45

Theorem 3.12. X is a compact Hausdorff space.

Proof: Let & be an open cover of X consisting only of sets

of the form Px, Sy, or P PS<, Let C1 E such that n EC1 , Since

{f} is not open, C1 must contain an element of X preceding csince 0 is a limit ordinal and has no predecessor. Then C1 = Sx

1

for some x1 EX. (Or, C1 = X and we are finished.) Choose C2 E6

such that x1 EC2 , If x1 is a limit ordinal, C2 must contain an

element preceding x1 . If not, x1 has a predecessor. Let x2 be

the first element of C2 if it is different from x1 ; otherwise,

let x2 be the predecessor of- x1 . Choose C3 E& such that x2EC3

and let x3 be chosen in a manner analogous to the manner in

which x2 was chosen. Continue this process, forming a sequence.

Let a denote the first element of X. If a is in U C., thennEl I

there exists an nE!' such that aECn and {fC1} i covers X. Other-

wise, U C. is a subset of X which has no first element since

(C. 1_is a decreasing sequence. This contradicts the fact

that X is well-ordered and it follows that ' must have a finite

subcover. Since the situation when & is a general open cover

can easily be reduced to the case just discussed, it follows

that X is compact.

Let x,yEX and assume x<y. Let t be the first element of

T = {zEX:x < z}. Then x<t<y. Since xEPt, yES , andPP S = 0,

X is Hausdorff., QE.D.

The reader should recall that a subset of a topological

space is said to be a-compact if it is the union of a countable

number of compact sets.

,.--

46

Theorem 3.13. The complement of the point Q is an open

set which is not a-compact.

Proof: Clearly X\{c} = PO is open. Suppose P is o-compact

and let (Ai) gbe a sequence of compact sets such that P = U Ai EY

kkLet & be an open cover of Pg . Choose {Ck} c- g which covers Akand from this collection, choose a finite subcover, {C }n

covering Ak. Hence there must exist a countable subcover ofPQ = iA. from 6. Consider the open cover ' = {P :xEX\{Q}} .

Since PO is uncountable, 6 has no countable subcover which is

a contradiction. Hence PQ is not a-compact. Q.E.D.Next we investigate a curious property of the space of

real-valued continuous functions defined on X. We denote this

space by C(X).

Theorem 3.14. If fEC(X), then there is an xEX\{f} such

that f is constant on Sy.

Proof: We assert that for each nEr, there exists an

xnEX\{Q} such that if xy>xn, then j f(x) - f(y)f < 1/n. Wewill prove this assertion by contradiction. Let NE! such thatfor all xEX\{,} there exists a yEX\{Q} such that y>x and

ff(x) - f(y)J > 1/N. Let x1 be the first element of X. Choosex2 EX\{Q} such that x2>x1 and If(x1) - f(x 2 )J1 > 1/N and x3 EX\fQ}

such that x 3 >x 2 and jf(x 2 ) - f (x 3 )J > 1/N. Continue this

process, thereby obtaining the sequence (x)=. t r 4nn=1 By theorem 2.48,we know that the lub(x )CO e

n n=1 exists; call it x. Let U be aneighborhood of f(x), say U = {y:Jy - f(x)j < 1/3N}. Since f

47

is continuous, there exists an open set V c X containing x such

that f(V) c U. Choose xkxk+lE) n=1 such that xkxk+1EV.(Recall that x = lub (x) )n=1) Then If(xk) - f(xk+1) -

If (xk+l) - f(x) I + If(x) - f(xk) < 1/3N + 1/3N < 1/N,

and we have a contradiction. Thus we have for each nE4 that

there exists an xnEX\{Q} such that Ijf(x) - f(y) < 1/n whenever

x,y>xn'

Now choose x1EX\{Q} such that if x,y>x1 , then Jf(x) - f(y)j < 1,

Choose x2EX\{Q} such that x2 >x1 and if x,y>x 2 , then j f(x) - f(y)I <

1/2. Continue this process, choosing x EX\{} such that x >xn n n-iand for all xYy>xn, If(x) - f(y)j < 1/n. Let x = lub(x )Ci

If a,b>x, then If(b) - f(a) I < 1/n for all n; hence f(b) = f(a).

Suppose x = P, Now {yEX:y < xn} is countable for each n. Thus

if x = Q, then U n{yX:y < x} = {yEX:y < } is countable.nEs1

Consequently x Q, and f is constant on S. Q.E.D.Theorem 3.15. The intersection of every countable collection

{Kn n=1 of uncountable compact subsets of X is uncountable.

Proof: Choose a sequence from U K in the following way.nEJ n

Let x11EK \{[} . Choose xlnEKn\{} such that xln>1(1). Let

y= lub ((xn n=1). Now y1 Q. Choose x2 1 EK 1 \{Q} such that

x21 > y1 and x 2 nEKn\{c} such that x2n > X2(n-1) . Let y2 =00

lub((x 2 n)n=1) . Continue this process, choosing xmiEKi\{c}

such that xml > 7m-1 at the m-th step; hence we form a countable

set ((xmn) n=1m=1 which intersects each Kn in infinitely manypoints. Now from this set ((xmn) n=1m=1, choose an increasing

48

sequence in the following manner. Choose x1 EK1 ; then x2 EK1

such that x2 :> x1 and x3 EK2 such that x3 > x2 then x4 EK1 ,X5EK2 , x6 EK3 , each xn >Xn-1. Continue this process, as

illustrated in the following diagram, forming an increasing

sequence (xn)i0 intersecting each Kn in infinitely many points.

K1: x1 x2 x4 x7 - -

K2' x3 x 5 x 8

K3 x6 9 '

K4: x10

Let S = { (b )n=: (b) = is increasing and (b~)PK. isLeS={bn=1 n n=1ln 1

infinite for all i}. The preceding construction shows that S

is non-empty. Define the relation < on S by (bn)n n1< (cn)nif and only if bn = cn for each n or there exists a ckE(c )rknn n=1such that ck > bi for all biE (bn )n The proof that (S,<) isa partially ordered set is straightforward and will be omitted,

Let & be a maximal chain in S. Suppose A = U{C:CE'} is

countable and let z = lub(A) / s. Form a sequence ((xm) )0 imn n=1 m=1

from nU Kn as done previously with x1 1 > z. Thus an increasing

sequence from ((xmn)n=1)m=1, say (xn)n=1 has the property that( CO < ( CO n0n(n0 n=1 n i=1 for all (cn )n=1E, contradicting the maximality

of &, Thus A is uncountable which tells us that there must be

an uncountable number of elements in . Also if (x )CO1 (y)00 En n=1' Ynn=1

49

where (xnn=1 ( nn=1 then x = lub (xn)n=1 lub (yn n=1 = Y.Thus T = {t:t = lub(xn n=1 for some (xn)n=1 E }is uncountable.

If t = lub(x )m_1, then xm + t, Therefore, it follows that

(for each n) the subsequence of (x) which lies in K mustmm=1wi enconverge to t. But each Kn is compact; hence Kn is closed.

Thus tEKn for each n so that tE (E) Kn. Hence Tc cf Kn, makingnEIS nEs n

n K uncountable. Q.E.D.nEfn

Let M be the collection of all ECX such that either EU{}

or EcU{Q} contains an uncountable compact set. In the first

case, define ?,(E) = 1; in the second case, define X (E) = 0 .

The next several theorems deal with this M and A.

Theorem 3.16. M is a c--algebra containing all Borel sets

in X.

Proof: Since X is uncountable and compact, OEM. If E1EM,

it follows easily that EEM. Let E 1 ,E 2 EM. Either E1 U{Q} or

EU{c} and E2L{Q} or EU{Q} contains an uncountable compact set.

Without loss of generality, assume E1U{M} and E 2U{c} contain

an uncountable compact set. Let C1 , C2 be such sets. By

theorem 3.15, C1f C2 is uncountable. Since X is Hausdorff and

C1 , C2 are compact, it follows that C1 , C2 are closed, making

C1(~C2 closed. Since X is compact, C5QC2 is compact. Hence

CnC2 c (fE1u{})f(E 2u{Q}) = (E1 q 2 )U{Q} so that EE 2 EM. Now

if E 1 ,E2EM, EB EM and E EEM so that (EfEl)c = E 1 UE 2 EM.

Also E1 E = E 1 \E 2 EM. Let (E )i be a sequence from M. Certainly

if E {0} contains an uncountable compact set, then ( U E.)U{}iEr

. : ...; .. .. , ,,., . .> .; :.p, .:1,-.lE rc,.'.s' cr'," 1 wf " . ,, ,. _-. :ifae r wa _, . . ,.. s sr .ca 9 a1 " N ... r. a ,,,., ,..., a9 9fa e rSYSMBi riLu .. ,

50

will contain such a set. Hence U E-EM, and it follows thatiEf

M is a c-algebra

Next we consider PX for xEX\{Q}. Note that Pc is closed,

compact, and contains an uncountable set; therefore, PREM.

Since PQU{Q} = X, we have P EM. Consider Sx for xEX. Now

S = Px+1 so that S EM and S EM. Thus M contains all Borel

sets in X. Q.E.D.

Definition 3.17. Let X be a compact Hausdqrff space,

E be the Borel subsets of X, and p: + JR be a cQuntably additive

real-valued measure. We say p is regular provided

-p(A) = inf{p(0) :A c 0, 0 an open set}

= sup{p (F):F c A, F a compact set};

i.e., if >O0, there exists a compact set F and an open set 0

such that F c A c 0 and p(0\F) < E.

Theorem 3.18. The set function x defined on M is a countably

additive measure which is not regular and f(Q) = f f dX for everyx

fEC(X).

Proof: Let EEM and suppose EU{Q} and ECU{Q} each contain

an uncountable compact set, C1 and C2 . Now CfC 2 is uncountable

and C{ C2 c { } which is a contradiction. Thus only one of Eu{2}

and ECU{Q} contains an uncountable compact set, and x is well-

defined,

Let (E )i' be a sequence of pairwise disjoint sets from M.

For each iEj, either E U{c} or E U{2} contains an uncountable

compact set. Without loss of generality, assume E1 U{h} contains

51

an uncountable compact set and E U{} does not. Now EU{Q} C

E U{Q} for i>1 so that EU{Q} must contain an uncountable compact

set since E U{} does not. Since M is a o-algebra, U E.EM.

i~f'Also E1U{M} c UE UQ} must contain an uncountable compact

CO

set so that ( U B.) = 1. But X x(E.) = X(E1) + X(E.) =iE4 i=i i>2

CO

1 + 0 = 1. Therefore x( U E) = x(Ei) so that A is countablyiEy i=1

additive in this case. From theorem 3,15 we see that X( U E.) = 0

if x(E.) = 0 for each i. Therefore, x is countably additive.

Since P. is open, {2} is compact. Also {Q}cU{c} = X

contains an uncountable compact set; thus x({c}) = 0. Any open

set containing {S} contains a set of the form Sx for some xEX\{c}

and A(S ) = 1 for all x. Thus x ({c}) inf{x (0) : {c} c 0, 0 an

open set} = 1. Therefore, A is not regular.

Let fEC(X) and choose x Q such that f is constant on Sx.

Let k = f(x+1) = f(Q) . Now A(Px+1) = 0 and x(S ) = 1 so that

f f dx = f f dx + f f da = f f dx = f k d = f f() dx =X Px+1 Sx Sx Sx Sx

f(c). Q.E.D,

The next set of theorems deal with filters. We will show

that every filter is contained in an untrafilter. Then we will

use the existence of an ultrafilter to find an example of a

bounded finitely additive real-valued measure defined on P(I)

which fails to be countably additive. Finally, we will show

that the existence of such a measure implies the existence of

-:

52

an ultrafilter. We remark that it is known that the ultrafilter

theorem for Boolean algebras is strictly weaker than the axiom

of choice (2)

Definition 319. A filter F on a set X is a collection of

non-empty subsets of X such that

(i) if h1, kEF, then bf~kEF;

(ii) if hEF and h c; k, then kEF,

If F1 and F2 are filters, then F1 is finer than F2 if and only

if F2 c F1 . If F is a filter and G c. F such that if hEF, there

exists a gEG such that g cc h, then G is a filter base for F.

A filter is generated by taking supersets of all elements in

its filter base. An ultrafilter is a filter F with the property

that if G is finer than F, then G = F. A filter F is fixed if

and only if f{f:fEF} / 0 and free if and only if off:fEF} = 0.

Lemma 3.20. If F is an ultrafilter on X and k c X, then

kEF or kcEF

Proof: If kF, then no subset a of k is in F. Suppose

f(ik 0 for all fEF. Define a filter base for G by BG =

{f(:fEF} . If fEF, then since fJ1k EBG cC G and fk c f, it follows

that fEG. Thus F c G. But F is an ultrafilter, making F = G.

Since fqlkEBG c: G = F we have kEF which is a contradiction. Thus

there exists an fEF such that flk = 0 which implies f c kc and

kcEF. Q.E.D,

Corollary 3.21. If F is a filter and for each k c X

either kEF or kc EF, then F is an ultrafilter.

53

Lemma 3.22. Let F be an ultrafilter. If 4~1b = 0 and

albEF, then either aEF or bEF but not both.

Proof: Suppose a,bEF. Then aflb = OEF which contradicts

the definition of filter. Suppose neither a nor b is in F.

By the previous lemma, ac and bc are in F. Thus acbc = (aUb)c

must be in F. This gives us that a.UbF which again is a

contradiction. Thus exactly one of a or b is in F. Q.E.D.

Theorem 3.23. Every filter is contained in an ultrafilter.

Proof: Let F be a filter on X and define S = {G:G is finer

than F}. We observe that SAO since FES. Partially order S by

G1 < G2 if and only if G c G2 . Let C be a chain from S. Let

H = U{G:GEC}. Let h,kEH and we can find GEC such that h,kEG,

putting hFrEG c H. Clearly if hEH and h c k, then kEH. Hence

H is a filter and an upper bound for C. By Zorn's lemma, C

contains a maximal element which is an ultrafilter containing

F. Q.E.D.

Theorem 3.24. Let P(,f) be the power class of %. There

exists a bounded finitely additive measure p:P(Q(p + which

fails to be countably additive.

Proof: Let F be the cofinite subsets of J; i.e., F =

{(n,oo):nEJ,}. Clearly F is a filter. Let G be an ultrafilter

containing F. Define pi:P(4) + {0,1} by pA(a) = fi aEG

0 otherwise.

Clearly p is bounded. Let a,bEP(f) such that alb = 0. First,

we consider when aUbEG. By lemma 3.22, exactly one of a or b

is in G. Thus y(aub) = 1 = y(a) + p(b). Now consider when

54

aUb4G. Then neither a nor b is in G. Hence p (aUb) = 0 =

S(.a) + p(b) so that p is finitely additive.

Next, consider U {n} =f EG. Certainly p( U {n}) = 1.n EfnEsf

Now {n}, G for nEJ since (n+1,o>) EG. Thus ya({n}) = 0 for all

nE4. Therefore, p( U {n}) u ({n}) which implies that VnEr n=1

is not countably additive. Q.ED.

Theorem 3,.25. Let u:P(X) + {0,1} be a finitely and

non-countably additive measure defined on the power class of a

set X. The existence of such a measure implies the existence

of an ultrafilter.

Proof: Now pi(a) = 1 for some aEP(X); otherwise, p = 0 and

p would be countably additive. Then U = {aEP(X):yp(a) = 1} is

non-empty. We assert that U is a free ultrafilter. Let aEU

and a c0 b. Then pj(b) = -p(a) + pi(b\a) > 1 since pi(a) = 1. Thus

p (b) = 1, and b is in U. Let a,b EU so that Up(a) = 1 = y (b) .

Now 1 = pi(a) = u (G\b) + up(afb) = p(b\a) + u (af'b) = p (b) = 1.

Thus y(a\b) = yp(b\a) . Suppose i(a\b) = y (b\a) = 1. Now

1 = p(afi) =p(a\b) + u(b\a) + up(alb) > 2 which contradicts

the definition of p . Thus p (a\b) = p(b\a) = 0 and it follows

that p (aFlb) = 1; i.e., afbEU. Clearly XEU. Let c be an arbitrary

subset of X. Suppose pi(c) = 0 = u(cc) . Now 1 = P(X) = ii(c) +

(cc) = 0 which contradicts our supposition. Thus either cEUCor c 'EU, and U is an ultrafilter.

Next we demonstrate that U is a free ultrafilter. Since

p is not countably additive, we may (and shall) choose a disjoint

55

00

sequence {c}C such that (U c.) yp(c.) . Suppose1E Y=1

y( u ci) = 0. Then 2 y(c.) > I so that we can choose c.IEJ 1=1 1 -1

such that }(c ) = 1. Hence ciEU and it follows that U c.EU.iE

Consequently ,p( U c.) = 1, and we have a contradiction. Thus1 E1

00 00

ii(U c.) = 1. Then either p (c.) = 0 or u y (c.) > 1.E 11i=1 1

00

Suppose p P(ci) > 1. Choose c1, c2 such that V (c1) = 1 = p(c

Then c1 ,c 2 EU so that 0 = c1F)c2 EU, contradicting the definition

of a filter. Thus we must have u ( U c.) = 1 and j(c.) = 0.iEr1 i=1

Now .Uc EU. Also p(ci) = 0 for each i so that cc EU since U

is an ultrafilter. Now f{a:aEU} c ((l cl)F(U c.)iEY iE. r

(U ci)c cpci) = 0. Thus f{a:aEU} = 0, and U is a free

ultrafilter. Q.E.D.

We conclude our applications of the axiom of choice with

a proof of the Hahn-Banach theorem.

Definition 3.26. Suppose that X is a real linear space.

A function f:X -+ IR is called a linear functional if f(ax + by) =

af(x) + bf(y) for every x,yEX and a,bEJR,

Remark 3.27. We use span(x) to denote the subspace

generated by the element x in X.

Theorem 3,28. (Hahn-Banach) Let p be a real-valued

function on X satisfying

56

(i) p(x + y) < p(x) + p(y) for all x,yEX;

(ii) p(ax) = ap(x) for all a>o.

Suppose S is a subspace of X and f:S + is a linear functional

so that f(s) < p(s) for all sES. Then there exists a linear

functional F:X + so that F(x) < p(x) for all xEX and FlS = f.

Proof: Suppose S X and let uEX\S. We will begin by

showing that f can be extended to S + span(u) and still main-

tain domination by p, Now S + span(u) is a subspace. Let

wES + span(u) and suppose w = x + au = y + bu where x,yES and

a,buER Then x - y = (b - a)uES. If b = a, then x = y so that

w has a unique representation, If b - a o 0, then uES which

contradicts our supposition. Thus for each wES + span(u),

there exists a unique xES and aEIR such that w = x + au.

If g were a linear extension of f to S + span(u), we would

have g(x + au) = g(x) + ag(u) = f(x) + ag(u). Thus we would

only need to determine g(u) to define g. Now if g were domi-

nated by p, then g(x + u) = g(x) + g(u) = f(x) + g(u) < p(x + u), or

g(u) < p(x + u) - f(x). Also g(y - u) = f(y) - g(u) < p(y - u),

or f(y) - p(y - u) < g(u), Hence g(u) would satisfy the following

inequality': f(y) - p(y - u) < g(u) < p(x + u) - f(x) for all

x,yES.

Suppose sup{f(y) - p(y - u)} > inf{p(x + u) - f(x)}. ThenyES xES

there exists an x,yES such that f(y) - p(y - u) > p(x + u) - f(x),

Thus we have f(y) + f(x) = f(y + x) > p(x + u) + p(y - u) > p(x + y)

which contradicts the hypothesis. Thus it follows that

.,,

57

sup{f(y) - p(y - u)} < inf{p(x + u) - f(x)} and we can chooseYES xES

a real number k such that sup{f(y) - p(y - u)} < k <yES

inf{p (x + u) - f (x)}, Thus we define g (u) = k and g (x + au) =xES

g(x) + ak = f (x) + ak.

Clearly g15 = f and g is a linear functional. Thus it

only remains to show that g is dominated by p. Let wES + span(u),

w = x + au, so that g(x + au) = f(x) + ak, Let a>O and suppose

g(x + au) > p(x + au). Then g(x + au) = f(x) + ak > p(x + au)

and ak > p (x + au) - f(x) so that k > (p (x + au) - f (x))/a =

p (x/a + u) - f (x/a) . Now k < inf{p (x + u) - f (x)} <

~x ES

p(x/a + u) - f(x/a) and we have g(x + au) < p (x + au) for a>O.

If a=0, w = x + auES and g(w) = f(x) < p(x) = p(w). Let a<O

and suppose g (x + au) > p (x + au) . Then g (x + au) = f (x) + ak >

p(x + au) and ak > p(x + au) - f(x). Thus k < (p(x + au) - f(x))/a =

-p(-x/a - u) + f(-x/a) = f(-x/a) - p(-x/a - u) <

sup{f(y) - p(y - u)} < k. Thus g(x + au) <_ p(x + au) for a<O.yES

Therefore, g is a linear functional extending f to S + span(u)

which is dominated by p.

Let L be the set of all linear functionals g which are

extensions of f to a subspace of X such that g(x) < p(x) for all

xED(g). We have shown L 0. Partially order L by inclusion

and let C be a maximal chain in L. Set F = U{g:gEC} and we

have D(F) = U{D(g):gEC} which is a subspace of X since each D(g)

58

is a subspace and the subspaces are ordered by containment.

Also F is a linear functional for if x,yED(F), then x,yED(g)

for some gEC where g is a linear functional. Also if xED(F),

xED(g) for some gEC so that F(x) = g(x) < p(x).

Now if D(F) t X, we can find uEX\D(F) and extend F to

D(F) + span(u) by the above argument which would contradict

the maximality of C. Thus we must have F defined on X, Q.E.D,

CHAPTER BIBLIOGRAPHY

1. Kelley, J. L., "The Tychonoff Product Theorem Implies theAxiom of Choice," Fundamenta Mathematica, 37 (1950),75-76.

2. Mathias, A. R. D., "A Survey of Recent Results in Set Theory,Proceedings of the 1967 American Mathematical SocietySummer Institute on Set Theory, Providence, American

Mathematical Society,T96T

3. Royden, H. L,, Real Analysis, New York, MacMillan PublishingCompany, Inc., 1968.

59

BIBLIOGRAPHY

Books

Hewitt, E. and Stromberg, K., Real and Abstract Analysis,Graduate Texts in Mathematics, New Yo, Springer-Verlag, 1975.

Mathias, A. R. D., "A Survey of Recent Results in Set Theory,'Proceedings of the 1967 American Mathematical SocietySummer Institute on Set Theory, Providence,7AmericanMathematical Society~19~6T

Royden, H. L., Real Analysis, New York, MacMillan PublishingCompany, Inc., 168.

Articles

Kelley, J. L., "The Tychonoff Product Theorem Implies theAxiom of Choice," Fundamenta Mathematica, 37 (1950),75-76.

60