Equivalence of the Axiom of Choice, the Well-ordering Theorem ...

Equivalence of the Axiom of Choice, the
Well-ordering Theorem, and Zorn’s Lemma
H. Krieger, Mathematics 331, Harvey Mudd College
1 Definitions
Fall, 2003
Definition 1 A set S is partially ordered (by the relation ≤) ⇐⇒
1. x ≤ x
2. x ≤ y and y ≤ x =⇒ x = y
3. x ≤ y and y ≤ z =⇒ x ≤ z
Definition 2 Let S be partially ordered and let B ⊂ S.
1. x is an upper bound for B ⇐⇒ x ∈ S and y ≤ x for all y ∈ B.
2. x is the least upper bound (lub or sup) for B ⇐⇒ x is an upper bound
for B and whenever z is an upper bound for B then x ≤ z.
3. x is the last element (greatest element) of B ⇐⇒ x ∈ B and x is an upper
bound for B ⇐⇒ x ∈ B and x is the least upper bound for B.
4. x is a maximal element of B ⇐⇒ x ∈ B and whenever y ∈ B, y ≥ x,
then y = x.
There are corresponding definitions for lower bound, greatest lower bound (glb
or inf), first element (smallest element), and minimal element.
Definition 3 A set S is linearly (totally) ordered (by the relation ≤) ⇐⇒ S is
partially ordered (by the relation ≤) and x ≤ y or y ≤ x for all x ∈ S, y ∈ S
⇐⇒
1. x ≤ y or y ≤ x
2. x ≤ y and y ≤ x =⇒ x = y
3. x ≤ y and y ≤ z =⇒ x ≤ z
A linearly ordered set is also called a chain.
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Definition 4 A set S is well ordered (by the relation ≤)⇐⇒ S is partially
ordered (by the relation ≤) and every non-empty subset of S has a first element
⇐⇒ S is linearly ordered (by the relation ≤) and every non-empty subset of S
has a first element.
Definition 5 A set S is inductively ordered (by the relation ≤) ⇐⇒ S is partially
ordered (by the relation ≤) and every linearly ordered subset (chain) in S
has a least upper bound.
2 Theorem Statements
W.O.T. Every set can be well-ordered.
A.C.1. If S is a non-empty collection of disjoint non-empty sets S, then there
is a set R which has as its elements exactly one element x from each set
S in S.
A.C.2. If S is a non-empty collection of non-empty sets A, then there is a
function ϕ : S → ∪{A : A ∈ S} such that ϕ(A) ∈ A for all A ∈ S.
A.C.3. If I is a non-empty set and, for each i ∈ I, Si is a non-empty set, then
the cartesian product
Si = {f : I → ∪{Si : i ∈ I} such that f(i) ∈ Si for all i ∈ I}
i∈I
is non-empty.
Z.L.1. If S is a partially ordered set such that each chain in S has an upper
bound, then S has a maximal element.
Z.L.2. If S is inductively ordered, then S has a maximal element.
3 Equivalence of these Statements
A.C.3. =⇒ A.C.2. Since the cartesian product {A : A ∈ S} is non-empty,
there is a function ϕ : S → ∪{A : A ∈ S} such that ϕ(A) ∈ A for all
A ∈ S.
A.C.2. =⇒ A.C.1. Since there is a function ϕ : S → ∪{S : S ∈ S} such that
ϕ(S) ∈ S for all S ∈ S and the sets in in S are disjoint, the range R of ϕ
is a set which has for its elements exactly one element from each set S in
S.
A.C.1. =⇒ A.C.3. For each i ∈ I, the set Ti = {(i, x) : x ∈ Si} is non-empty.
Thus the collection S = {Ti : i ∈ I} is a non-empty collection of disjoint
non-empty sets. Hence there is a set f which has as its elements exactly
one element (i, f(i)) from each set Ti. Then f : I → {Si : i ∈ I} with
f(i) ∈ Si for all i ∈ I.
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Z.L.1. =⇒ Z.L.2. If S is inductively ordered, then every chain in S has an
upper bound. Thus S has a maximal element.
Z.L.2. =⇒ Z.L.1. Suppose S is partially ordered and every chain in S has an
upper bound. Let T = {C : C is a chain in S}. Since T ⊂ 2 S and 2 S
is partially ordered by inclusion, so is T . If C0 is a maximal element of
T , then there is an upper bound x0 for C0. Then x0 ∈ C0 and x0 is a
maximal element of S. Hence our problem is reduced to showing that T
has a maximal element. But if C is a chain in T and C0 = ∪{C : C ∈ C},
then C0 is the least upper bound of C. Hence T is inductively ordered
which means T does have a maximal element.
W.O.T. =⇒ A.C.2. Let S be a non-empty collection of non-empty sets A.
Let S = ∪{A : A ∈ S} and, then, well-order S. Hence each A is then a
non-empty subset of the well-ordered set S. If ϕ(A) is the first element of
A, then ϕ : S → ∪{A : A ∈ S} such that ϕ(A) ∈ A for all A ∈ S.
Z.L.2. =⇒ W.O.T. Let S be a set. For each subset A of S which can be
well-ordered, let WA be the collection of well-orderings of A and then let
W = {(A, ≤A) : A ⊂ S, ≤A∈ WA}.
Define a partial ordering on W by letting (A, ≤A) ≤ (B, ≤B)
⇐⇒ A ⊂ B, ≤A=≤B |A, and a ∈ A, b ∈ B − A =⇒ a ≤B b.
Now if C is a chain in W , C = {(C, ≤C)}, let C0 = ∪{C : (C, ≤C) ∈ C}
and let ≤C0= ∪{≤C: (C, ≤C) ∈ C}, i.e. if x and y are in C0, then
x ≤C0 y ⇐⇒ x ≤C y for some C such that (C, ≤C) ∈ C. Then ≤C0 is a
well-ordering of C0 and (C0, ≤C0) is the least upper bound of C. Hence,
W is inductively ordered which means that W has a maximal element.
Let (A, ≤A) be a maximal element of W . Then A = S, for if S − A = φ,
we consider B = A ∪ {x}, where x ∈ S − A, and define ≤B by y ≤B x
for all y ∈ A, ≤B |A =≤A. However, in this case (A, ≤A) ≤ (B, ≤B) and
(A, ≤A) = (B, ≤B). Hence, (S, ≤S) ∈ W for some well-ordering ≤S, i.e.
S can be well-ordered.
Lemma: Let S be inductively ordered and let f : S → S such that f(x) ≥ x
for every x ∈ S. Then there is an x0 ∈ S such that f(x0) = x0. In fact, if
a ∈ S, there is an x0 ≥ a such that f(x0) = x0.
Proof: Let a ∈ S and let Ba be the collection of all subsets B of S such that
i) a ∈ B,
ii) f[B] ⊂ B,
iii) if C is a chain in B, then lub C ∈ B.
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Note that Ba is not empty, since {x : x ≥ a} ∈ Ba. Let A = ∩{B : B ∈
Ba}. Then it is easy to see that A ∈ Ba. Our objective is to show that A
is a chain, for then if x0 = lub A, it follows that x0 ∈ A =⇒ f(x0) ≤ x0 ≤
f(x0), i.e. f(x0) = x0. The intuitive idea is that A consists only of a and
its successive images under f. Hence, we let
P = {p ∈ A : y ∈ A and y < p =⇒ f(y) ≤ p}.
Note that P ⊂ A but we would like to have equality. As a start, we show
that if p ∈ P and z ∈ A, then either z ≤ p or z ≥ f(p) (i.e. elements of P
and A are comparable and there are no elements of A between p and f(p)
if p ∈ P ). To see this, let p ∈ P and let
Then,
i) a ≤ p =⇒ a ∈ B(p).
B(p) = {z ∈ A : either z ≤ p or z ≥ f(p)}.
ii) If z ∈ A and z < p, then f(z) ≤ p; if z ∈ A and z = p, then
f(z) = f(p) ≥ f(p); if z ∈ A and z ≥ f(p), then f(z) ≥ z ≥ f(p).
Hence, z ∈ B(p) =⇒ f(z) ∈ B(p), i.e., f[B(p)] ⊂ B(p).
iii) Let C be a chain in B(p) and let c0 = lub C. Then c0 ∈ A. If c ≤ p
for all c ∈ C, then c0 ≤ p =⇒ c0 ∈ B(p). Otherwise, there is some
c ∈ C such that c ≥ f(p) =⇒ c0 ≥ c ≥ f(p) =⇒ c0 ∈ B(p). Thus
c0 = lub C ∈ B(p). Consequently, B(p) ∈ Ba and, since B(p) ⊂ A,
we have B(p) = A.
Now we can show that P = A.
i) a ∈ P since there is no y ∈ A with y < a.
ii) Let p ∈ P . Suppose that y ∈ A and y < f(p). Note that f(p) ∈ A.
Then we must have y ≤ p. If y < p then f(y) ≤ p ≤ f(p) and if
y = p then f(y) = f(p) ≤ f(p). Hence f(p) ∈ P , i.e. f[P ] ⊂ P .
iii) Let C be a chain in P . Then c0 = lub C ∈ A. Suppose that y ∈ A
and y < c0. If y ∈ P , then c0 ≥ f(y). Otherwise, for each c ∈ C,
either y < c or y ≥ f(c) ≥ c. But we can’t have y ≥ c for all c ∈ C,
since c0 = lub C. Hence, for some c ∈ C, y < c =⇒ f(y) ≤ c ≤ c0.
Therefore, c0 ∈ P .
Consequently, P ∈ Ba and, since P ⊂ A, we have P = A. Finally, if x ∈ A
and y ∈ A, then either x ≤ y or x ≥ f(y) ≥ y and hence A is a chain.
A.C.2. =⇒ Z.L.2. let S be inductively ordered. Let S be the collection of all
non-empty subsets of S and let ϕ : S → ∪{A : A ∈ S} be a mapping with
ϕ(A) ∈ A for all A ∈ S. Define f : S → S by letting f(x) = x if x is a
maximal element and letting f(x) = ϕ({y : y > x}) if x is not a maximal
element. Then f(x) ≥ x for all x ∈ S. Hence by the lemma, there is an
x0 ∈ S such that f(x0) = x0. Then, x0 must be a maximal element of S.
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