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1-1-1962
On the equivalence of the axiom of choice, Zorn's
Lemma, and the Well-ordering theorem
Grace Joy Traylor
Atlanta University
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Traylor, Grace Joy, "On the equivalence of the axiom of choice, Zorn's Lemma, and the Well-ordering theorem" (1962). ETD Collection
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ON THE EQUIVALENCE OF THE AXIOM OP CHOICE, ZORN'S LEMMA,
AM) THE 1ELL-ORDERING THEOREM
A THESIS
SUBMITTED TO THE FACULTY OP ATIANTA UNIVERSITY IN PARTIAL
FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF
MASTER OF SCIENCE
BY
GRACE JOY TRAYLGR
DEPARTMENT OF MATHEMATICS
ATLANTA UNIVERSITY
ATLANTA, GEORGIA
JANUARY 1962
PREFACE
In dealing with infinite sets there arises a logical question whether
it is possible to "order11 the elements of these sets.
Such a question has
led mathematicians, such as Russell, Godel, and Zermelo, to postulate the
"axiom of choice.1*
This axiom has other forms, which are more useful in
nature and in proofs of theorems; these forms are known as "Zorn's lemma1*
and the "well-ordering theorem.11
After some preliminary considerations,
these three statements will be shown to be equivalent.
The first chapter consists of definitions of an "order relation," a
"partially ordered" set, a "totally ordered" set, a "well-ordered" set and
"linear order," and a proof of transfinite induction.
The second chapter
consists of applications of the axiom of choice, Zorn's lemma and the wellordering theorem.
The third and final chapter oonsists of proofs showing
the equivalence of these three statements.
The writer is indebted to Dr. Lonnie Cross for his suggestion of this
subjeot and for his helpful criticisms.
I also wish to thank my typist,
Miss Geraldine Ellis, for doing such a wonderful job of typing.
ii
TABLE OF CONTENTS
Page
PREFACE
ii
GLOSSARY OF SYMBOLS
iii
Chapter
I.
II.
III.
17.
V.
ORDER RELATIONS
1
THE AXIOM OF CHOICE
6
ZORN'S LEMMA
8
THE WELL-ORDERING THEOREM
13
THE EQUIVALENCE OF THE AXIOM OF CHOICE,
LEMA, AND THE nlELL-ORDERING THEOREM
BIBLIOGRAPHY
ZORN'S
21
SO
iii
GLOSSARY OF SYMBOLS
SYMBOL
NAME OF SYMBOL
<=r
Set inclusion
<^~
Proper inclusion
£
is a member of
j.
Union
«
Intersection
0
Mull set
^
Such that
Implies that
There exists
If and only if
Therefore
Identical to
Totality of elements in S not in A
x does not precede y
^.
&.,n
^UJ
Not equal to
The first number in the brackets refers
to the number of tiie book in the biblio
graphical Hstj the second number if the
page reference
Quod erat demonstrandum (L., which was
to be demonstrated).
Iv
CHAPTER I
ORDER RELATIONS
Before we begin to solve the problem of "ordering" element of
an infinite set, we must have in mind a clear meaning of the term
"ordering.1*
The following explanation gives us a precise definition.
Letting S be a set and =^.be a binary relation defined on
the set S, we say that ^- is an order relation if the following
conditions are satisfied:
(1)
x-=& x for every x e S.
(2)
x :siy and y =£ x for x,y e S =S>x « y.
(3)
x^.y and y=&.z for
The above is to be read as follows:
In the set S, x^-y is to read as "x precedes y" and
is to be read as "y succeeds x."
If, however, the set S is
a set of real numbers, x,y £ S, then x=^y is to read as
"x less than or equal to y," and y-fcx is to be read as "y
greater than or equal to x."
These two relations are order
relations*
We have, however, three types of ordered sets:
totally-ordered, and well-ordered.
type*
partially-ordered,
We shall consider separately each
2
(1)
If it is possible to define an order relation in a
set S, then we say that the set is partially ordered
by the order relation.
(2)
If a partially ordered set has the additional property
that x^.y, x,ye S, y^x, then S is said to be
totally ordered by^. •
For example, let us consider the set
S = £Buick, Ford, Pontiac, Fiat, !G, Jaguar},
where x=<y, x,ye. S, if x does not occur after y in either the
United States made cars or the foreign made cars.
partially ordered, but not totally ordered.
This set is
If we define ^- as
above and also say that every foreign made car precedes every
United States made car, then relative to this order relation, S
becomes totally ordered.
In connection with the above definition of an ordered relation ^,
we may define the following symbols:
(a)
xily means y=^x.
(b)
x <y means x^.y but x^ y.
(c)
x> y means y-< x,
(d)
x ?? y means x and y are comparable, that is, one
and only one of the three x< y, x-= y, x >- y,
holds true.
[1%YF$
From the definitions ofs£, >z»-< $ >■ » it is clear that if two
elemsnts are comparable, then one and only one of the three relations
y, x -=Ly, x y~y holds true*
3
Clearly every two elements of a partially ordered set are not
necessarily comparable, but every two elements of a totally ordered set
are necessarily comparable*
Let S be a set ordered by ^.(totally ordered or partially ordered)
and let A be any subset of S.
Relative to A we make the following
definitions.
(a)
If for every y €. A^ an a& A * az^y, then a is
the first element of A.
Similarly, if for every
y e A ^ a z fc A > z^c y, then z is the last element
of A.
(b)
If for a t A"* no elements ye A ^ y -< a, then a is
a minimal element.
Similarly, if for z fe A
no
elements ye Ad y > z then z is a maximal element
of A.
(c)
If ^ a b g S ^ b dLy for all y <s. A, then b is a
lower bound for A.
If 3 a g & S * it is the last
element of the set of lower bounds for A, then g is
the greatest lower bound (g.l.b.) for A; and we
write g ? g.l.b. A.
Similarly, if 3 a u 6 S
^
u ^.y for all y & A, then u is an upper bound for A*
If 3 a 1 e S ^ it is the first element of the set of
upper bounds for A, then 1 is the least upper bound
(l.u.b.) for Aj and we write lsl.u.b. A.
(3)
If a set S is a totally ordered relative to
and also
has the property that every non-null subset has a first
element, then S is said to be well-ordered.
h
An elementary property of well-ordered sets is that we may apply
the method of transfinite induction.
Before we can actually consider
transfinite induction, some terms must be defined.
(1)
A linear order L consists of a set, also designed by
L, and an order relation ^satisfying the following
conditions:
(2)
(a)
for every a 6 L, b & L, ^ a-=£ b, either a=£.b or
(b)
if a^.b, then it is not true that
(c)
if at£b and bt^-c, then a^c.
Two linearly ordered sets L and L' are said to be similar
if a one-to-one correspondence a* — f (a) exists between
them* if a^b, then f (a)^. f (b),
(3)
A set of similar linear ordered sets is called an order
(10
The order type of a well-ordered set is called an
ordinal number,
Now we are ready to prove the transfinite induction theorem.
Theorem.
If S is the set of all ordinal numbers less than a
given ordinal number, say$ , and A is a subset of S, and if
(a)
The first ordinal number 1 is in A,
(b)
For every ordinal number ^=c>, if all ordinal
numbers less than $ are in A then ^ is also in A,
then A~ S.
Proof.
Be
If S — A ^r$), then there is a first ordinal
S — A.
than
Q. E. D.
By (a), §^l 1.
But all ordinal numbers less
are in A so that, by (b),§€ A.". S— A=r<£>,
CHAPTER II
THE AXIOM OF CHOICE
In order to understand the axiom of choice, we mast know what is
meant by a choice function.
non-null set.
For instance, we take S and let it be a
We letCPbe a mapping (function) of S onto S which
associates with every subset A of S an element a -=CP(A) of S.
assume, also, that Op (A) e A*
for S.
We
, therefore, is a choice function
A choice function, then, is a method of selecting one element
from every non-null subset of a given set.
The axiom of choice states that every set has a choice function.
As an illustration of the use of the axiom of choice, let us consider
the following theorem:
Theorem.
If a set is infinite, then it has a subset
equivalent to the set of positive integers, that is, the
subset is denuraerable.
Proof.
Let S be an infinite set and let CP be a choice
function for S as proposed by the axiom of choice.
LetCp
be a unique mapping on the set of positive integers, that
is, CP assigns 1 to the element xt selected from S, then
2 to the element 2g selected from S- £x^ } then 3 to the
element X3 selected from S - \x^, X£J ) and so forth.
Since
the range of the choice of a choice function is always the
6
7
subset of a given set, we see that C-P is an onto mapping*
Let us suppose ~l
a finite set N
S — N"=-
Then Cf establishes that
S H
N.
This is contrary to the hypothesis that S is infinite.
Therefore, for every
S — N % $•
from this ire conclude that the range of Cfi is equivalent to
the set of positive integers, but -foe range of C-? is a
subset of S, which means that the theorem is established*
Q. E. D.
CHAPTER in
ZORN'S
LEMMA
Zorn's lemma is a maximal principle named for Zorn.
Unfortunately,
the history of it and some closely related maximal principles is very
tangled.
However, it is certain that Zorn was essentially anticipated
by Felix Hausdorff.
Zorn's lemma can be stated in two forms*
The first form says that
"every partially ordered set S contains a maximal (with respect to9)
totally ordered subset."
The second form says that "if S is a partially
ordered set such that every totally ordered subset has an upper bound in
S, then S has at least one maximal element,"
equivalent.
These two forms are
Let us now prove this equivalence and make an application
of this lemma.
Consider the following theorem and its proof:
Theorem.
Proof.
The two forms of Zorn's lemma are equivalent.
The first form of Zorn's lemma ==>the second form of
Zorn's lemma.
Let the first form of Zorn's lemma hold true*
Let S be partially ordered by -=1 and assume that every
totally ordered subset of S has an upper bound in S.
By the
first form of Zorn's lemma S has a maximal subset, which we
shall denote by A*
S.
By hypothesis A has an upper bound u in
We see that for all m in A m^.u.
Suppose that u Jfe. A.
We claim that u€ A.
Let \ti\ denote a set consisting of a
8
9
single element u, and A+- A tt ^ti]f. This set A is a totally
ordered subset in S which properly contains A.
This is a
contradiction, for A is a maximal, and therefore it cannot
be contained in any other set.
The tipper bound u & S.
If there existed a u4" in S a
u+>-u, then A**= AU&*9( would be a totally ordered subset
in S which contains A.
This contradicts the maximal
character of A*
Conversely, the second form of Zorn's lemma =f> the
first form.
Let the second form hold true.
partially ordered by ^ •
Let C be the collection of all
totally ordered subsets of S.
by-S. .
Let S be
Now C is partially ordered
Let C6 be a totally ordered subset in C relative
to £= , and let us consider the set A« •£. S,
Ao"=- U A, where A €
Co.
We claim that A* is totally ordered relative to^- .
If
x & A» and y 6- Aft then x e A and y e A+for some A and A*1 in
C .
But C
is totally ordered by .£. •
Hence A and A are
comparable and either x and y are in A or A+.
For the sake of argument let x and y be in A+.
A* £ C A is totally ordered relative to ^. •
and y are comparable and hence A
to =£u and A* € C.
Because
Therefore, x
is totally ordered relative
Thus every totally ordered subset C of
(relative to& ) has an upper bound Ao in C.
By the second
form of Zorn's lemma, C has at least one maximal element
relative to£= .
Q. E. D.
10
Before we can make an application of Zorn's lemma we must clear
the way by considering certain definitions*
(1)
A group 6 is a collection of objects on which a binary
operation # is defined.
This operation is subject to
the following laws:
(a)
Closure*
If a and b are in G, then a * b is in G.
(b)
Associativity*
If a,b and c are in G, then
(a # b) # c= a # (b # c).
(c)
Identity*
There exists a unique element e in G
(called the identity element) such that for all
a's in G, a •& e = e -* a = a.
(d)
Inverse.
For every a in G, there exists a unique
element af in G called the inverse of a such that
a * a' c a' # a — e.
(2)
An additive group G is a group that indicates that the
binary operation is addition.
(3)
A group G is an abelian group if for all a,b in G,
a # b~b •» a.
(U)
A ring R is an additive abelian group with the additional
properties:
(a)
The group E is closed with respect to a second
binary operation designated multiplication.
(b)
Multiplication is associative, that is,
(ab)c ~ a(bc
for all elements a,b,c in R.
n
(c)
Multiplication is distributive with respect to
addition on both the left and right side, that is,
a(b-v c)=
ab -v ac
(b * c)a •= ba + ca
for all elements a,b,c in R.
(5)
/?} 5U7
A ring R that has a tinit element is a ring with the
element e such that ea •*- ae ■= a is true*
(6)
If N is an additive subgroup of a ring H such that both
ra and ar are in N for all elements a in N and all ring
elements r in R, then N is an ideal.
(7)
^2$737
An ideal N is maximal if it is not contained properly in
any other ideal.
Now let us consider the following theorem from the theory of rings*
Theorem,
let R be a ring with a unity element.
Then every
ideal properly contained in R is contained in a maximal ideal.
Proof.
Let R be a ring which contains properly N and let C
be the collection of all ideals containing N and unequal to R*
Assume C is partially ordered byS: and by Zorn's lemma it
contains a maximal totally ordered subset Co*
*"
We see that Q
where Q e Co.
is an ideal since C
We claim that Q* is a maximal ideal.
is totally ordered by S •
Assert first Q ^ R.
Suppose that the unity element 1 were in Q .
in some Qt c ,
Let
But if le.§ , the ideal
Then 1 would be
would become the
whole ring E (if an ideal contains the unity element, it then
becomes the whole ring), which is a contradiction.
We assert
32
second that there exists no ideal Q
Suppose such an ideal did exist.
such that QCQ
c R*
The set of ideals ^Co > Q
would be totally ordered—which contradicts the maximal
character of Co•
Hence Q + is a maximal ideal and therefore
every ideal properly contained in R is contained in a
maximal ideal.
Q. E. D.
J
CHAPTER IV
THE 1ELL-CRDERING THEOREM
Since Zermelo was the first to give a rigorous proof of the welLordering theorem, we often refer to the theorem as "Zermelo's theorem.n
Its proof is presented in two forms.
Because of the difficulty in
following the steps of the second form of the proof, the first form
only will be reproduced here.
This proof reveals to us only the
existence of a well-ordered set; it does not demonstrate to us how to
obtain a well-ordered set.
For example, no one has exhibited to us a
well-ordering of the set of real numbers*
Let us consider the following:
Definition.
A system of subsets of S is called a tower if
it has the following properties:
(a)
It contains the null set<t> •
(b)
If it contains a class of subsets of S, then it
contains their union.
(c)
If it contains a proper subset A C_S, then it
contains the union of A and the element CP (A),
where Cjp (A) is a mapping of S into S — A.
Well-ordering Theorem.
Eroof.
^j6j*7
Every set can be well-ordered.
Let S be an infinite set.
Let Q) (A) be a mapping
(function) of the elements a e S —A into every proper subset
13
A of S, including the null set.
We shall show that the function Cp(A) establishes an
onto mapping between S and a class C of subsets of S which
is shown to be a well-ordered set with respect to d •
Let C be the intersection of all towers.
We claim
that C itself is a tower.
(a)
C contains the null set, since it is the inter
sections of classes each of which contains the
null set.
(b)
If A* £ C for every uA, then O A4 & C, -i e A.
For A*,JU A belongs to every tower, so that
VJ A«i, -l £ A, belongs to every tower, whence
(c)
If A 6 C, then AU^.CP(a)"|j € C.
every tower, so that A U$.Q(A)1[
For A belongs to
belongs to every
tower; hence A L>9j(A) J e C.
Therefore, C is a tower.
Since C is the intersection of all
toners, it follows that any subset of 0 which is a tower must
be identical with C.
By A 6
C comparable we mean that for every Y e
either I C A or A C Y.
C
We show that if A is comparable
then for every Y e C either Y c A or AU$,C0(A)^c Y.
show this we show that the sets of Y
property form a tower*
(a) <$ C. A.
C which have this
To
(b)
Given X^fc A.
If Yjl C A for every^fe A, then
UYjCA,-lfeA.
If Y^D AU §#A) 1[
for someJL6 A,
then U YjtD AO5.Q(A) \ ,-le A.
(c) If Y D AO$G?(A)]f , then YO^Q(Y)^ A U^>(A) j .
If Y •=. A, then YOjCfcCY)"} H A O^A) J . If Y is a
proper subset of A, then YU^Cp(Y) J C. A.
For if
YU^CD(Y)}if not a subset of A, it contains some
element not in A, but Y is a subset of A, so that
this element does not belong to Y.
Since Y is a
proper subset of A there is an element in A not in
Y which belongs to YU^CP(Y)^, since A is conparable and so is a subset of Yb$_C0(Y) J •
Hence
at least two elements not belonging
to Y.
This is impossible.
The property of comparable sets just obtained allows us to
show that the sets A & C which are comparable form a tower,
so that all members of G are comparable.
(a) 9^
(b)
is comparable.
Suppose Ajl, J-€A, is comparable.
C.
If
Ajl C Y for all J & A, then U Aj. C Y,<* e A.
If
Aj. 3 Y for some d. e A then
Hence,
(c)
o
U
Let Y €
A j. D Y, «i fc A.
A^,oifeA, is comparable.
Suppose A 6. C is comparable.
Then for every Y e
either Y C A or Ab^QCA) ^C Y, so that A U
is comparable.
C
16
We now know that for every A e C, B e C, either A C B or
BC A.
Using HC" as the order relation, the members of
C form a linear order.
We show that this linear order forms a well-ordered
sets
Let Aj^,^€ A, be any class of sets in C.
A= A Aj,«ie A.
B=UB^ c
Let
Consider all B^ 6 C,^BiB»CA.
A, ^6 B.
If
Then
B-=Aj.for someJife A, then B is
the first member of this set.
If B is not one of the A^ ,
then B U l^CB)^ is one and must be the first.
there is a ^fe A > A^cAa for everyJ-fc A.
In any case,
Hence C, with
« c» as an order relation, is a well-ordered set.
Finally we show that there is a one-to-one correspondence
between C and S.
correspondence.
The function a —CP(A) gives the desired
For if A G C, B <b C, and A is a proper subset
of B, then CP (A) 6 B, so that C?(A) ^=C/>(B).
then C? (A) — Cp(B).
a "=Cp(A).
Hence if A = B,
Next, for every a <s S there is an A 6 C*
For let A be the union of all sets in C which do
not contain a.
If a=rCp(A), then a« A(J WCA) ] is a
contradiction.
Hence a =<-P(A)•
This one-to-one correspond
ence produces a well-ordering of S.
Q. E. D.
Let us consider the following problem as an application of the
well-ordering theorem*
Problem.
Find a function f(x) such that
(1)
f(x*y)-^f(x) + f(y).
The function ex obviously has this property, for every
constant cj and if only continuous functions are sought, there
17
are no other solutions of this problem.
For, put f(l)"= c.
then from (1) we get
f (2) =r f (1+1) ■= f (1) + f (1) = c + c r 2c,
f (3) - £(2+ 1) ^ f (2) + f (1) -=2c V c = 3c,
and in general, for every natural number n,
(2)
Further,
f(n)^= nc.
f(n*O) = f(n) ■* f(0), and hence f(0) zr 0,
Conse
quently
f fr-n) 4 f (n) s f(0) - 0,
so that (2) holds for every integer n.
Moreover, analogous
considerations for every integer q"=£ 0 shows that
f(n)r= f(q-$)
q .
and hence
so that
f(r)- re
for every rational r.
For a continuous function, however,
f (x) •=. ex.
For discontinuous function this functional equation (1) has
solutions other than those just found.
To prove this fact,
we employ the well-ordering theorem.
We first obtain a Hamel basis for the real numbers.
By this we
mean a set H of real numbers is called a Hamel basis if it has the
following properties:
(a)
For any finite number of basis elements b,, ba,
..., bm, and arbitrary rational numbers r( , r^,
18
..., r^
(b)
not all zero, we never have
For every number z "=£ 0 there are among the basis
elements a finite number of basis numbers b| ,
..., bm, and, corresponding to these, rational
numbers r, ,
..., r^
z rr
all different from zero,
a Tk,
bv •
First we have to show that such a basis exists.
We do this by-
building up a basis as a well-ordered set by means of transfinite
induction.
The starting-point is the well-ordering
ft:
of the set of all real numbers.
xas
H.
Let x, e
H if x, i 0.
Otherwise, let
For a given**. , suppose for every ^ ^-<Athat the decision has
been made whether or not x$ £ H.
Then xj e H if and only if x is not a
linear combination with rational coefficients of any finite number of
those x$ , ^^-"J, which are in H.
The set His a basis.
For suppose that an equation of the sort
described under (a) held for some finite number of basis numbers.
Then
one of these finitely many basis numbers would be the last to have a
nonzero coefficient, and could then not be admitted into H, contrary to
our assumption.
Condition (a) is fulfilled.
is seen as follows:
not.
z=b.
That (b) too is fulfilled
A given number z^O is either a basis number b or
In the first case, (b) is fulfilled trivially by the equation
m the second case, an equation of the kind described in (a) z
and the basis numbers preceding it in R.
On account of property (a)
already established for the basis numbers, z has a nonzero coefficient
19
in this equation.
The equation can therefore be solved for z, and if
we leave out the term with zero coefficients, assertion (b) is obtained.
The representation which exists, according to (b), for every
number z ■£ 0, is uniquely determined by a.
For if there were two
distinct representations, subtraction of the two equations would result
in a contradiction of (a).
Now we define the function f(x) first only for the basis numbers,
and this is done in an arbitrary manner.
x
Further, we let f(0)— 0.
If
0 is an arbitrary real number, it can be represented in terms of a
finite number of basis numbers with rational coefficients, in the form
Here we shall also admit terms whose coefficients are zero; the
uniqueness of the representation is then obviously still valid.
Now
we put
f(x)- irt ftt^).
Then, if
y-^s^ b*.
is a representation of y in terms of basis elements,
is such a representation of x + y, and, as a matter of fact, due to the
uniqueness, precisely the representation.
-f(x)
Consequently,
+ f(y).
The function f is thus a solution of the functional equation.
It is a
discontinuous solution if we put, say, f(b,) ~ 0, f(b^) =: 1 for two
basis elements b,, b.;j for if f were a continuous solution, the argument
20
presented at the beginning shows that we should have
f(b, ):
f 0^) = b(:
b^ which is not the case.
CHAPTER V
THE EQUIVALENCE OF THE AXIOM OF CHOICE, ZORN'S TEWk
AND THE WELL-CRDERING THEOREM
We know that if we prove the equivalence of the axiom of choice
and Zorn's lemma and the equivalence of the axiom of choice and the
well-ordering theorem, then the equivalence of the wen- ordering
theorem and Zorn's lemma automatically follows.
axiom of choice and Zorn's lemma.
and a lemma.
Let us consider the
We must begin with a definition
This information is as follows:
Definition.
relative to
Suppose S is a non-null partially ordered set
and suppose that every totally ordered
subset of S has a least upper bound in S.
Suppose further
that there exists a function ^ which maps S into S
^~
x=S&.ci(x) for an xe S, and finally, suppose there exist no
elements x and y •*
Let ok
x <-< y -< ^.(x).
be any fixed element of S.
A subset L of S with the
properties that:
(a) <k€
L.
(b) X £ L =S>o|(x) €
(c)
l.
If C s L is a totally ordered subset of S, then
l.u.b. C e L.
21
22
will be called an M«t -chain" or simply a chain if the
element is understood.
If I, and Lxare chains, then so is L,f\ Ljj»
the intersection of all chains, i.e., K =• HL.
Let K be
Then K is
the smallest chain, that is, if L is any chain, then K £= L.
The element 4. is the first element of K.
Proof.
Let M be the set of all elements x in S^«i^ x.
Clearly M is an «•* -chain.
Therefore, K -£s M, and this
proves the assertion.
Lemma.
The set K introduced above is well-ordered with
respect to^- and,
<^ (l.u.b. K) rz l.u.b. K.
Proof.
(A)
We shall prove the statement:
If be K is comparable with every x e K, then so is <\ (b)»
Let L consist of all elements x in K ^ x and \ (b) are
comparable.
We shall show that L is a chain.
Clearly L &. K.
The element must be in L since 4 is comparable with every
element in K («< is the first element of K).
Now we shall prove that x £ L=5^ (x) 6. L.
arbitrary element in L.
Let x be an
Then either
^ (b) * x
or
Now since x e K, we know that x^.<\(x).
then o^ (b) :^L c^ (x) and thus
«^ (x) e L.
Hence if <k (b)=d x,
23
Now we show, even the case x-<<^(b), that <^
(x) €.
L.
Since b is comparable with every x 6. K and <^ (x) e K, we
know that b £ <j (x).
Hence either*^ (x) z£ b or b-<<^(x).
Now suppose b<<^(x)•
Then either xVb or x-<b since
b is comparable with every x in K.
If x-<b, then
x -< b-<«^(x),
which contradicts the assertion about <^. •
Thus xXb, that
is, either x -= b or x > b.
If x =b, then «^(x) =«^ (b) and since «^. (b) e
have -^ (x) £ L.
L, we must
If x >b we have b-< x-oj(b), which is the
same contradiction as the one before regarding<^ •
Now we must consider the caseo^(x)=£-b.
since we have b=£ % (b) and therefore,
This is simple,
°± (x) ^. «^(b)»
Since «^, (x) and *\(b), are comparable, <^,(x) is in L.
Hence we see that if x fe L, then «^.(x) & L.
We have, there
fore, proved the two requirements that L is a chain.
To
prove the third, let C£ Lbe any totally ordered subset of
L and let
Jf-i
Now if"^
since V
l.u.b. C.
an x in C ^ x ^Gi(b), then we must have *^<^(b),
is an upper bound.
But if every x in C has the
property that x-< %(b), then Y -^ (b) is a bound since «^ (b)
is a bound.
Since V
we have * fc L.
is comparable with<=^ (b) in either case,
This completes the proof of the statement
2U
that L is a chain.
Thus K&- L.
But L£K by definition.
Hence, L = K, which proves the statement (a).
We shall next show that:
(B)
K is totally ordered.
Let M be the set of all elements in K which are comparable
with every element of K.
In particular oi e, M.
If b Q M, then «^ (b) e M by statement (A) above,
let C&M be totally
l.u.b. C.
ordered subset of M and let* be
Now if y £ C, then y is comparable with every x
in K, since y € 11.
Consider any fixed x in K.
C ^ x=^.y, then, of course, x ^Jf.
If not, then y^. x
for all y 6 C and x is an upper bound for 0.
In either case* & M.
Hence K Ss. M.
If"^ a y in
ThenV^x.
Thus we have shown that M is a chain.
But M ■£* K by construction.
Therefore, K ~ M
and statement (A) is proved.
Finally we shall shows
(C)
K is well-ordered.
Let N& K be any non-null subset of K.
IfeieN, then
is the first element of N and our statement (G) is true*
Suppose^- & N.
Let Q be the set of all elements x in K
x =i u for all u s.
N.
Clearly ck 6 Q»
Let P be any subset of Q.
p -< l.u.b. P.
Then B
Then P is totally ordered.
We shall show that p e. Q.
a u in N 3 p >> u.
Let
For suppose p £
Q.
But every u in N is an upper bound
for Q, so that
l.u.b. Qi. u -<* p.
25
But p £. Q.
Hence l.tub. P*< p, Taihich is a contradiction.
Therefore, p c Q.
Thus the set Q satisfies conditions (a)
and (c) for a chain.
chain.
If condition (c) is true, then Q is a
Suppose for the moment that this is not true.
there must be an x in Q ^ *J (x) 4
Q.
Then
(This is the only
possibility since conditions (a) and (c) of an <H -chain are
true).
Let
q "= l.u.b. Q.
If x is not equal to q, we assert that °± (x) €:
<^
Q.
For if
(x) &. Q, then there would exist a u in N* u-S^(x).
But x £ Q.
Hence x ^. u.
Equality may be rejected since
x— u *==?x e W, and since x e Q we infer x— q.
Hence
x-< u-<<^(x), which is again a contradiction.
We have now q <= Q and % (q) e Q.
Since q e Q, we must
have q ^.u for all u <=. N and since <J (q) fe Q, we must have
°^ (q) y u for some u e. M.
u in N.
Mow if q 4 N, then q < u for every
Then
for some u in N.
This is a contradiction*
Hence q e
N and
q is the first element of N.
Thus K is well-ordered and we have proved proposition
(C).
This also proves the first part of the lemma.
Now we must show that <^ (l.u.b. K)= l.u.b. K.
l.u.b. K.
that k€. K.
Let K
Since K is a chain, and totally ordered, we know
(Conditions (a) and (c) of an-I
-chain).
26
Further, <Y(k) 6- K by condition (b) of axi^L -chain and
k :^(k) by definition of the «^ function.
which completes our proof.
Hence k = <^(k),
Q. E. D. ^?;17U-lY67"
We are now ready to prove the equivalence of these two theorems*
Let us re-state each*
Axiom of Choices
Every set has a choice function.
Zorn's Lemma (first form):
Every totally ordered set contains a
maximal totally ordered (relative to-S) subset*
Theorem*
Proof.
The axiom of choice =£> Zorn's lemma (first form).
Let S be a set partially ordered with respect to
and let Cf be a choice function for S.
Let us consider C,
the set of all totally ordered subsets of S.
For each
A £■ C, let F(^ be the set of all elements x inS> AU tx\
is still totally ordered set.
one A&S, the set F^
We shall show that for at least
is the null set.
Let y be a mapping
of C into G defined by the equations
<^ (A) O
A
<\ (A)
if
Ffc+<J>
if
F^ -Ct>
We see that we have fulfilled the conditions of our lemma with
C playing the role of S relative to-Q ,
e^ playing the role
of ov and any A in G will play the role of oL .
Now from the
second part of the lemma we know "^ an element A
in C
For this element A we have F^+ -<fj so that A is a maximal
totally ordered subset of S.
27
Conversely, Zorn's lemma => the axiom of choice.
let S be a non-null set.
Again
Now in certain subsets A of S some
choice functions can be defined, for example, the finite
subsets*
Consider the set E of all pairs (A,Cp), where A is a subset
of S and Of is a choice function for A*
We define the following order relation for E:
(A,Q) =^ (U,T)
is to mean:
A S. U, and for every subset X ^. U
we have «f~ (X) —Cp(A P\ X).
From our definition of
we know that =^. is an order relation.
ordered set.
Thus E is a partially
Now, by Zorn's lemma (first form), let C be a
maximal totally ordered (relative to iL ) subset of E.
Consider
A*^- <-> A,
(A,Co) e C,
and let (A,, Q? ) and (Aa, C£ ) be any two elements of C.
The elements (A,,O> ) and (Aa,Cp^) are comparable, since C
is totally ordered.
For the sake of argument let
(A,,Cp, ) -k (A^CfJ.
Let us consider now any X Q- it ♦
A,. r\ TL*4>, then A^rs X^^and
cp^k^n x) ^^(a, r\ (Aa r\ xJJ^q^a, r\ x).
With the above equation in mind we give the following
definition:
Let X be any non-null subset of A •
(A,Q> ) t C * kf\ X *<b.
define <&*~ by
Choose any
Clearly such A»s exist.
Then we
If
28
q
=cp (ah x).
The functionQ? is a choice function for A+.
X) =cPP,(Al r\ X) show that Q>* (X) is independent
of the choice of A.
.'.
(A~,Q> ) <s E.
Furthermore,
(A, q>) ^ (A*", c/ ) for all (A, Q) & C. Hence A \J %AW,Q+ )}
is also totally ordered.
Since C is a maximal, (A+,Q*) must
belong to C and (A^jQ ) most be the unique maximal element of C.
We show now that A^= S.
be an element of S — A .
Suppose that A* - S and let J.
Let us consider the set
A^- A* Vj 5.x}
Let us define a choice function
for A
For every non-null subset S of A
as follows:
let
^ O+(av n x), if x =+ <L
^
, if X -S.
Clearly CP^ is a choice function for A.
since A+ -< A^»
But
Thus C U ^(A^, C^) would be totally ordered
which contradicts the maximal character of C.
Thus A =
S*
Q. E. D.
In Chapter IV, in order to prove the wel!L- ordering theorem, we
made use of the axiom of choice.
the well-ordering theorem.
We also showed that the axiom of choice
Now we have only to show that the well-
ordering theorem j=r>the axiom of choice.
Theorem.
Proof.
If a set is well-ordered, then it has a choice function.
If A £
S is a non-null subset of S, then 1 has a first
element, a, since S is well-ordered.
Let C% (A) — a.
The
29
mapping C-P is a choice function for S.
Q. E. D.
You can see, then that we have proved the following:
The axiom
of choice, Zorn's lenana, and the well-ordering theorem are equivalent.
Q. £. D.
B3BU0GRAPHI
1.
Casper Gof flnan, Real Functions, New York, Rinehart and Company,
1953.
2.
Kenneth S. Miller, Elements of Modern Abstract Algebra, New York,
Harper and Brothers Publishers, 1958.
3.
Paul R. Halmos, Waive Set Theory, Mew York, D. Van Nostrand,
I960.
U.
E. Kamke:
£.
Patrick Suppes, Axiomatic Set Theory, New York, D. Van Nostrand
Translated by Frederick Bagendhl, Theory of Sets,
New York, Dover Publication, Inc., I960.
Company, Inc., I960.
30