The Axiom of Choice and some of its equivalentsmaths.sogang.ac.kr/.../Set/Axiom_of_Choice_slide.pdf · Partially ordered sets A relation 4on a set A (i.e., 4ˆA A) is called apartial

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.


Example

A tangent vector field on a circle S1

A tangent vector field on a sphere S2


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.)


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?


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?


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}.


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}.


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}.


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.


Equivalents

THE FOLLOWING ARE EQUIVALENT:

(1) The Axiom of Choice(2) Hausdorff Maximality Principle(3) Zorn’s Lemma(4) Zermelo’s Well-ordering Principle


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.


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

??��



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:



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).



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 .



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).



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 ).



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.


Thanks

Thank you very much, folks!


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The Axiom of Choice and some of its equivalentsmaths.sogang.ac.kr/.../Set/Axiom_of_Choice_slide.pdf · Partially ordered sets A relation 4on a set A (i.e., 4ˆA A) is called apartial