The Axiom of Choiceand some of its equivalents
Jong Bum Lee
Sogang University, Seoul, KOREA
June, 2017
Jong Bum Lee The Axiom of Choice and some of its equivalents
Axiom of Choice
For any nonempty set S whose elements are nonempty sets,
there exists a function
f : S −→⋃
A∈SA
such that f (A) ∈ A for all A ∈ S.
Such a function f is called a choice function.
Jong Bum Lee The Axiom of Choice and some of its equivalents
Example
A tangent vector field on a circle S1
A tangent vector field on a sphere S2
Jong Bum Lee The Axiom of Choice and some of its equivalents
Partially ordered sets
A relation 4 on a set A (i.e., 4 ⊂ A× A)is called a partial order relation if 4 is
(i) reflexive(ii) anti-symmetric (i.e., a 4 b and b 4 a⇒ a = b)(iii) transitive
A set together with a partial order relationis called a partially ordered set (in short, a poset.)
Jong Bum Lee The Axiom of Choice and some of its equivalents
Totally ordered sets
A total order relation 4 on a set Ais a partial order relation on A such that
∀ a,b ∈ A, a 4 b or b 4 a.
A set together with a total order relationis called a totally ordered set (in short, a toset.)
ExampleFor any set X , the ordinary inclusion ⊂ is a partial order relationon P(X ). Why?
(i) reflexive(ii) anti-symmetric(iii) transitiveIs the partial order relation ⊂ a total order relation?When?
Jong Bum Lee The Axiom of Choice and some of its equivalents
More examples
ExampleThe ordinary relation ≤ on R is a total order relation.
ExampleDefine a relation 4 on C by
a + bi 4 x + y i ⇔ a ≤ x , b ≤ y .
Show that 4 on C is a partial order relation.(i) reflexive(ii) anti-symmetric(iii) transitive
Is this a total order relation?
Jong Bum Lee The Axiom of Choice and some of its equivalents
Sub-posets and Chains
DefinitionLet (A,4) be a poset and B ⊂ A.Define
4B := 4⋂
(B × B).
Then 4B is a partial order relation on B (Why?).The poset (B,4B ) is called a sub(-po)set of (A,4).The subset B is called a chainif the partial order relation 4B is a total order relation.
Example
Let A = {1,2,3}.
Jong Bum Lee The Axiom of Choice and some of its equivalents
Maximal elements and minimal elements
DefinitionLet (A,4) be a poset.An element e ∈ A is called a maximal elementif e 4 a⇒ e = a.
Example (Maximal elements and minimal elements)
Let A = {1,2,3}.
Jong Bum Lee The Axiom of Choice and some of its equivalents
Hausdorff Maximality Principle
Theorem (Hausdorff Maximality Principle)
Let (A,4) be a poset.Let T be the set of all chains of (A,4).Then the poset (T ,⊂) has a maximal element.
Example (all chains)
Let A = {1,2,3}.
Jong Bum Lee The Axiom of Choice and some of its equivalents
Zorn’s Lemma and Zermelo’s Well-ordering Principle
Theorem (Zorn’s Lemma)
Let (A,4) be a poset in which every chain has an upper bound.Then A has a maximal element.
Theorem (Zermelo’s Well-ordering Principle)Every set can be well-ordered,that is, there exists a total order relationsuch that every nonempty subset has a minimal element.
RemarkThe ordinary order relation ≤ on R is a total order relation,but not a well-ordered relation (Why?).
Zermelo’s Well-ordering Principle⇒ the existence of a well-ordered relation 4 on R.
Jong Bum Lee The Axiom of Choice and some of its equivalents
Equivalents
THE FOLLOWING ARE EQUIVALENT:
(1) The Axiom of Choice(2) Hausdorff Maximality Principle(3) Zorn’s Lemma(4) Zermelo’s Well-ordering Principle
Jong Bum Lee The Axiom of Choice and some of its equivalents
Application of Zorn’s Lemma
TheoremLet A and B be two sets. Then we have either
cardA ≤ cardB
orcardB ≤ cardA.
CorollaryThe cardinal order relation ≤is a total order relation on the cardinal numbers.
Jong Bum Lee The Axiom of Choice and some of its equivalents
Proof of Theorem, p.1
If A or B is the empty set, then there is noting to prove.Consequently, we shall assume that both A and B arenonempty sets.
It suffices to show that either ∃A� B or ∃B � A.
Consider
X = {(Aα, fα) | Aα ⊂ A, fα : Aα� B}
and define a relation . on X as follows:
(Aα, fα) . (Aβ, fβ)⇔ Aα ⊂ Aβ, fα ⊂ fβ⇔ the following diagram is commutative
Aβ
fβ // B
Aα
∪OO
fα
??��������
Jong Bum Lee The Axiom of Choice and some of its equivalents
Proof of Theorem, p.2
Then (X ,.) is a poset:(i) . is reflexive(ii) . is anti-symmetric(iii) . is transitive
Next, we will show that the poset (X ,.) satisfies the conditionfor Zorn’s Lemma, that is, every chain of (X ,.) has an upperbound.
Let T = {(Aγ , fγ) | γ ∈ Γ} be a chain of X . We will construct anupper bound of T as follows:
Jong Bum Lee The Axiom of Choice and some of its equivalents
Proof of Theorem, p.3
PutA1 =
⋃γ∈Γ
Aγ , f1 =⋃γ∈Γ
fγ .
Then f1 : A1 → B is given by
f1(x) = fγ(x) if x ∈ Aγ and (Aγ , fγ) ∈ T .
Need to check first that f1 is indeed a function, that is, f1 is afunction by the “Pasting Lemma”.If x ∈ Aδ and (Aδ, fδ) ∈ T , then since T is a chain, we haveeither (Aγ , fγ) . (Aδ, fδ) or (Aδ, fδ) . (Aγ , fγ). In either case,fγ(x) = fδ(x).
Jong Bum Lee The Axiom of Choice and some of its equivalents
Proof of Theorem, p.4
WithA1 =
⋃γ∈Γ
Aγ , f1 =⋃γ∈Γ
fγ ,
we will show that f1 is injective.
Assume that f1(x) = f1(y) for some x , y ∈ A1.Then ∃(Aγ , fγ), (Aδ, fδ) ∈ T such that x ∈ Aγ and y ∈ Aδ.Since T is a chain, either (Aγ , fγ) . (Aδ, fδ) or (Aδ, fδ) . (Aγ , fγ).We may assume that (Aγ , fγ) . (Aδ, fδ).Then f1(x) = fγ(x) = fδ(x) as fγ ⊂ fδ, and f1(y) = fδ(y). Sincef1(x) = f1(y), we have fδ(x) = fδ(y). This implies that x = y asfδ is injective.
We have proven that f1 is injective.Consequently, (A1, f1) ∈ X and is an upper bound of T .
Jong Bum Lee The Axiom of Choice and some of its equivalents
Proof of Theorem, p.5
By Zorn’s Lemma, the poset (X ,.) has a maximal element,say (A, f ).If A = A then since (A, f ) ∈ X , f : A = A→ B is injective.
If A 6= A then we will show that f : A→ B is bijective, which
shows that B f−1→ A ⊂ A is injective.
Since f is already injective, it remains to show that f issurjective.Assume that f is not surjective. Choose y0 ∈ B − f (A).From the assumption that A 6= A, we can choose x0 ∈ A− (A).
Jong Bum Lee The Axiom of Choice and some of its equivalents
Proof of Theorem, p.6
Now we construct a function
f : A⋃{x0} → B
by
f (x) =
{f (x) if x ∈ Ay0 if x = x0
It is clear that f is injective. Hence
(A⋃{x0}, f ) ∈ X
and(A, f ) . (A
⋃{x0}, f ).
Jong Bum Lee The Axiom of Choice and some of its equivalents
Proof of Theorem, p.7
Because (A, f ) is a maximal element of X , we must have
(A, f ) = (A⋃{x0}, f ).
This is a contradiction.
Consequently, f is surjective and so bijective.
Jong Bum Lee The Axiom of Choice and some of its equivalents
Thanks
Thank you very much, folks!
Jong Bum Lee The Axiom of Choice and some of its equivalents