the axiom of choice and some of its...
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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
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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
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Example
A tangent vector field on a circle S1
A tangent vector field on a sphere S2
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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.)
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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?
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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?
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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}.
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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}.
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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
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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.
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Equivalents
THE FOLLOWING ARE EQUIVALENT:
(1) The Axiom of Choice(2) Hausdorff Maximality Principle(3) Zorn’s Lemma(4) Zermelo’s Well-ordering Principle
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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.
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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
??��������
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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:
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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).
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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 .
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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).
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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 ).
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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.
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Thanks
Thank you very much, folks!
Jong Bum Lee The Axiom of Choice and some of its equivalents