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