The Axiom of Choice - University of Southern Mississippi · 3.Other than that, the Axiom of Choice,...

Choice Functions Zorn’s Lemma Well-Ordering Theorem

The Axiom of Choice

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Choice

Why Do We Need More Axioms?

1. The final three axioms we discuss could be motivated bythe desire to “count past infinity” ...

2. ... and by the desire to have “standard sizes” for infinitesets.

3. Other than that, the Axiom of Choice, in its “Zorn’sLemma” incarnation is used every so often throughoutmathematics.

The Axiom of Choice

Why Do We Need More Axioms?1. The final three axioms we discuss could be motivated by

the desire to “count past infinity” ...

2. ... and by the desire to have “standard sizes” for infinitesets.

The Axiom of Choice

the desire to “count past infinity” ...2. ... and by the desire to have “standard sizes” for infinite

The Axiom of Choice

the desire to “count past infinity” ...2. ... and by the desire to have “standard sizes” for infinite

sets.3. Other than that, the Axiom of Choice, in its “Zorn’s

Lemma” incarnation is used every so often throughoutmathematics.

The Axiom of Choice

Axiom.

The Axiom of Choice. Let {Ai}i∈I be an indexed familyof sets. Then there is a function f : I→

⋃i∈I

Ai so that f (i) ∈ Ai

for all i ∈ I. The function is also called a choice function.

Definition. Let {Ai}i∈I be a family of sets. The product ∏i∈I

of the Ai is defined as the set of all functions f : I→⋃i∈I

Ai with

f (i) ∈ Ai for all i ∈ I.

The Axiom of Choice

Axiom. The Axiom of Choice.

Let {Ai}i∈I be an indexed familyof sets. Then there is a function f : I→

⋃i∈I

Ai with

The Axiom of Choice

Axiom. The Axiom of Choice. Let {Ai}i∈I be an indexed familyof sets.

Then there is a function f : I→⋃i∈I

Ai with

The Axiom of Choice

Axiom. The Axiom of Choice. Let {Ai}i∈I be an indexed familyof sets. Then there is a function f : I→

⋃i∈I

for all i ∈ I.

The function is also called a choice function.

Ai with

The Axiom of Choice

⋃i∈I

Ai with

The Axiom of Choice

⋃i∈I

Definition.

Let {Ai}i∈I be a family of sets. The product ∏i∈I

Ai with

The Axiom of Choice

⋃i∈I

Definition. Let {Ai}i∈I be a family of sets.

The product ∏i∈I

Ai with

The Axiom of Choice

⋃i∈I

Ai with

The Axiom of Choice

Theorem.

Intersection and union are completely distributive.Let {Ji}i∈I be a family of index sets and let {Cij}i∈I,j∈Ji be afamily of sets. Then the following hold.

1.⋂i∈I

⋃j∈Ji

Cij =⋃

f∈∏i∈I Ji

⋂i∈I

Cif (i).

2.⋃i∈I

⋂j∈Ji

Cij =⋂

f∈∏i∈I Ji

⋃i∈I

Cif (i).

The Axiom of Choice

Theorem. Intersection and union are completely distributive.

Let {Ji}i∈I be a family of index sets and let {Cij}i∈I,j∈Ji be afamily of sets. Then the following hold.

1.⋂i∈I

⋃j∈Ji

Cij =⋃

f∈∏i∈I Ji

⋂i∈I

Cif (i).

2.⋃i∈I

⋂j∈Ji

Cij =⋂

f∈∏i∈I Ji

⋃i∈I

Cif (i).

The Axiom of Choice

Theorem. Intersection and union are completely distributive.Let {Ji}i∈I be a family of index sets and let {Cij}i∈I,j∈Ji be afamily of sets.

Then the following hold.1.⋂i∈I

⋃j∈Ji

Cij =⋃

f∈∏i∈I Ji

⋂i∈I

Cif (i).

2.⋃i∈I

⋂j∈Ji

Cij =⋂

f∈∏i∈I Ji

⋃i∈I

Cif (i).

The Axiom of Choice

Theorem. Intersection and union are completely distributive.Let {Ji}i∈I be a family of index sets and let {Cij}i∈I,j∈Ji be afamily of sets. Then the following hold.

1.⋂i∈I

⋃j∈Ji

Cij =⋃

f∈∏i∈I Ji

⋂i∈I

Cif (i).

2.⋃i∈I

⋂j∈Ji

Cij =⋂

f∈∏i∈I Ji

⋃i∈I

Cif (i).

The Axiom of Choice

1.⋂i∈I

⋃j∈Ji

Cij =⋃

f∈∏i∈I Ji

⋂i∈I

Cif (i).

2.⋃i∈I

⋂j∈Ji

Cij =⋂

f∈∏i∈I Ji

⋃i∈I

Cif (i).

The Axiom of Choice

1.⋂i∈I

⋃j∈Ji

Cij =⋃

f∈∏i∈I Ji

⋂i∈I

Cif (i).

2.⋃i∈I

⋂j∈Ji

Cij =⋂

f∈∏i∈I Ji

⋃i∈I

Cif (i).

The Axiom of Choice

Proof (part 1).

Let x ∈⋂i∈I

⋃j∈Ji

Cij. Then for every i ∈ I there is a ji with x ∈ Ciji .

For each i ∈ I set g(i) := ji. Thenx ∈

⋂i∈I

Cig(i) ⊆⋃

f∈∏i∈I Ji

⋂i∈I

Cif (i). Conversely, let

x ∈⋃

f∈∏i∈I Ji

⋂i∈I

Cif (i). Then there is a choice function f ∈∏i∈I

that x ∈ Cif (i) for all i ∈ I. But then x ∈⋃j∈Ji

Cij for every i ∈ I.

Hence x ∈⋂i∈I

⋃j∈Ji

The Axiom of Choice

Proof (part 1).Let x ∈

⋂i∈I

⋃j∈Ji

Then for every i ∈ I there is a ji with x ∈ Ciji .

⋂i∈I

Cig(i) ⊆⋃

f∈∏i∈I Ji

⋂i∈I

x ∈⋃

f∈∏i∈I Ji

⋂i∈I

Hence x ∈⋂i∈I

⋃j∈Ji

The Axiom of Choice

⋂i∈I

⋃j∈Ji

⋂i∈I

Cig(i) ⊆⋃

f∈∏i∈I Ji

⋂i∈I

x ∈⋃

f∈∏i∈I Ji

⋂i∈I

Hence x ∈⋂i∈I

⋃j∈Ji

The Axiom of Choice

⋂i∈I

⋃j∈Ji

For each i ∈ I set g(i) := ji.

Thenx ∈

⋂i∈I

Cig(i) ⊆⋃

f∈∏i∈I Ji

⋂i∈I

x ∈⋃

f∈∏i∈I Ji

⋂i∈I

Hence x ∈⋂i∈I

⋃j∈Ji

The Axiom of Choice

⋂i∈I

⋃j∈Ji

For each i ∈ I set g(i) := ji. Thenx

∈⋂i∈I

Cig(i) ⊆⋃

f∈∏i∈I Ji

⋂i∈I

x ∈⋃

f∈∏i∈I Ji

⋂i∈I

Hence x ∈⋂i∈I

⋃j∈Ji

The Axiom of Choice

⋂i∈I

⋃j∈Ji

⋂i∈I

Cig(i)

⊆⋃

f∈∏i∈I Ji

⋂i∈I

x ∈⋃

f∈∏i∈I Ji

⋂i∈I

Hence x ∈⋂i∈I

⋃j∈Ji

The Axiom of Choice

⋂i∈I

⋃j∈Ji

⋂i∈I

Cig(i) ⊆⋃

f∈∏i∈I Ji

⋂i∈I

Cif (i).

Conversely, let

x ∈⋃

f∈∏i∈I Ji

⋂i∈I

Hence x ∈⋂i∈I

⋃j∈Ji

The Axiom of Choice

⋂i∈I

⋃j∈Ji

⋂i∈I

Cig(i) ⊆⋃

f∈∏i∈I Ji

⋂i∈I

x ∈⋃

f∈∏i∈I Ji

⋂i∈I

Cif (i).

Then there is a choice function f ∈∏i∈I

Hence x ∈⋂i∈I

⋃j∈Ji

The Axiom of Choice

⋂i∈I

⋃j∈Ji

⋂i∈I

Cig(i) ⊆⋃

f∈∏i∈I Ji

⋂i∈I

x ∈⋃

f∈∏i∈I Ji

⋂i∈I

that x ∈ Cif (i) for all i ∈ I.

But then x ∈⋃j∈Ji

Hence x ∈⋂i∈I

⋃j∈Ji

The Axiom of Choice

⋂i∈I

⋃j∈Ji

⋂i∈I

Cig(i) ⊆⋃

f∈∏i∈I Ji

⋂i∈I

x ∈⋃

f∈∏i∈I Ji

⋂i∈I

Hence x ∈⋂i∈I

⋃j∈Ji

The Axiom of Choice

⋂i∈I

⋃j∈Ji

⋂i∈I

Cig(i) ⊆⋃

f∈∏i∈I Ji

⋂i∈I

x ∈⋃

f∈∏i∈I Ji

⋂i∈I

Hence x ∈⋂i∈I

⋃j∈Ji

The Axiom of Choice

Proof (part 2).

Let x ∈⋃i∈I

⋂j∈Ji

Cij. Then there is an i0 ∈ I so that x ∈⋂

j∈Ji0

For every choice function f ∈∏i∈I

Ji we have

x ∈⋂

j∈Ji0

Ci0j ⊆ Ci0f (i0) ⊆⋃i∈I

Cif (i). Therefore x ∈⋂

f∈∏i∈I Ji

⋃i∈I

Cif (i).

For the reverse inclusion, let x 6∈⋃i∈I

⋂j∈Ji

Cij. Then for every i ∈ I

there is a ji ∈ Ji so that x 6∈ Ciji . Define g ∈∏i∈I

Ji by g(i) := ji.

Then x 6∈⋃i∈I

Cig(i) and hence x 6∈⋂

f∈∏i∈I Ji

⋃i∈I

Cif (i).

The Axiom of Choice

⋃i∈I

⋂j∈Ji

Then there is an i0 ∈ I so that x ∈⋂

j∈Ji0

Ji we have

x ∈⋂

j∈Ji0

f∈∏i∈I Ji

⋃i∈I

Cif (i).

⋂j∈Ji

Ji by g(i) := ji.

Then x 6∈⋃i∈I

f∈∏i∈I Ji

⋃i∈I

Cif (i).

The Axiom of Choice

⋃i∈I

⋂j∈Ji

j∈Ji0

Ji we have

x ∈⋂

j∈Ji0

f∈∏i∈I Ji

⋃i∈I

Cif (i).

⋂j∈Ji

Ji by g(i) := ji.

Then x 6∈⋃i∈I

f∈∏i∈I Ji

⋃i∈I

Cif (i).

The Axiom of Choice

⋃i∈I

⋂j∈Ji

j∈Ji0

Ji we have

∈⋂

j∈Ji0

f∈∏i∈I Ji

⋃i∈I

Cif (i).

⋂j∈Ji

Ji by g(i) := ji.

Then x 6∈⋃i∈I

f∈∏i∈I Ji

⋃i∈I

Cif (i).

The Axiom of Choice

⋃i∈I

⋂j∈Ji

j∈Ji0

Ji we have

x ∈⋂

j∈Ji0

⊆ Ci0f (i0) ⊆⋃i∈I

f∈∏i∈I Ji

⋃i∈I

Cif (i).

⋂j∈Ji

Ji by g(i) := ji.

Then x 6∈⋃i∈I

f∈∏i∈I Ji

⋃i∈I

Cif (i).

The Axiom of Choice

⋃i∈I

⋂j∈Ji

j∈Ji0

Ji we have

x ∈⋂

j∈Ji0

Ci0j ⊆ Ci0f (i0)

⊆⋃i∈I

f∈∏i∈I Ji

⋃i∈I

Cif (i).

⋂j∈Ji

Ji by g(i) := ji.

Then x 6∈⋃i∈I

f∈∏i∈I Ji

⋃i∈I

Cif (i).

The Axiom of Choice

⋃i∈I

⋂j∈Ji

j∈Ji0

Ji we have

x ∈⋂

j∈Ji0

Cif (i).

Therefore x ∈⋂

f∈∏i∈I Ji

⋃i∈I

Cif (i).

⋂j∈Ji

Ji by g(i) := ji.

Then x 6∈⋃i∈I

f∈∏i∈I Ji

⋃i∈I

Cif (i).

The Axiom of Choice

⋃i∈I

⋂j∈Ji

j∈Ji0

Ji we have

x ∈⋂

j∈Ji0

f∈∏i∈I Ji

⋃i∈I

Cif (i).

⋂j∈Ji

Ji by g(i) := ji.

Then x 6∈⋃i∈I

f∈∏i∈I Ji

⋃i∈I

Cif (i).

The Axiom of Choice

⋃i∈I

⋂j∈Ji

j∈Ji0

Ji we have

x ∈⋂

j∈Ji0

f∈∏i∈I Ji

⋃i∈I

Cif (i).

⋂j∈Ji

Then for every i ∈ I

Ji by g(i) := ji.

Then x 6∈⋃i∈I

f∈∏i∈I Ji

⋃i∈I

Cif (i).

The Axiom of Choice

⋃i∈I

⋂j∈Ji

j∈Ji0

Ji we have

x ∈⋂

j∈Ji0

f∈∏i∈I Ji

⋃i∈I

Cif (i).

⋂j∈Ji

there is a ji ∈ Ji so that x 6∈ Ciji .

Define g ∈∏i∈I

Ji by g(i) := ji.

Then x 6∈⋃i∈I

f∈∏i∈I Ji

⋃i∈I

Cif (i).

The Axiom of Choice

⋃i∈I

⋂j∈Ji

j∈Ji0

Ji we have

x ∈⋂

j∈Ji0

f∈∏i∈I Ji

⋃i∈I

Cif (i).

⋂j∈Ji

Ji by g(i) := ji.

Then x 6∈⋃i∈I

f∈∏i∈I Ji

⋃i∈I

Cif (i).

The Axiom of Choice

⋃i∈I

⋂j∈Ji

j∈Ji0

Ji we have

x ∈⋂

j∈Ji0

f∈∏i∈I Ji

⋃i∈I

Cif (i).

⋂j∈Ji

Ji by g(i) := ji.

Then x 6∈⋃i∈I

Cig(i)

and hence x 6∈⋂

f∈∏i∈I Ji

⋃i∈I

Cif (i).

The Axiom of Choice

⋃i∈I

⋂j∈Ji

j∈Ji0

Ji we have

x ∈⋂

j∈Ji0

f∈∏i∈I Ji

⋃i∈I

Cif (i).

⋂j∈Ji

Ji by g(i) := ji.

Then x 6∈⋃i∈I

f∈∏i∈I Ji

⋃i∈I

Cif (i).

The Axiom of Choice

⋃i∈I

⋂j∈Ji

j∈Ji0

Ji we have

x ∈⋂

j∈Ji0

f∈∏i∈I Ji

⋃i∈I

Cif (i).

⋂j∈Ji

Ji by g(i) := ji.

Then x 6∈⋃i∈I

f∈∏i∈I Ji

⋃i∈I

Cif (i).

The Axiom of Choice

Definition.

Let X be an ordered set. A totally ordered subset Cof X is also called a chain. An element m ∈ X so that for allx ∈ X we have that m≤ x implies m = x is called a maximalelement of X.

Lemma. Let X be a set, and let Z ⊆P(X) be a nonempty set ofsubsets of X, ordered by set containment ⊆ and with thefollowing properties.

1. For every set C ∈ Z we have that every subset of C is anelement of Z.

2. For every chain (with respect to set containment) C ⊆ Z,the union

⋃C of C is an element of Z.

Then Z has a maximal element (with respect to setcontainment).

The Axiom of Choice

Definition. Let X be an ordered set.

A totally ordered subset Cof X is also called a chain. An element m ∈ X so that for allx ∈ X we have that m≤ x implies m = x is called a maximalelement of X.

The Axiom of Choice

Definition. Let X be an ordered set. A totally ordered subset Cof X is also called a chain.

An element m ∈ X so that for allx ∈ X we have that m≤ x implies m = x is called a maximalelement of X.

The Axiom of Choice

Definition. Let X be an ordered set. A totally ordered subset Cof X is also called a chain. An element m ∈ X so that for allx ∈ X we have that m≤ x implies m = x is called a maximalelement of X.

The Axiom of Choice

Lemma.

Let X be a set, and let Z ⊆P(X) be a nonempty set ofsubsets of X, ordered by set containment ⊆ and with thefollowing properties.

The Axiom of Choice

Proof (initial setup).

Consider the indexed family{i}i∈P(X)\{ /0}. The union of this family is X. By the Axiom ofChoice, there is a choice function f : P(X)\{ /0}→ X so thatf (A) ∈ A holds for all A ∈P(X)\{ /0}.For each C ∈ Z, define the set EC :=

{x ∈ X \C : C∪{x} ∈ Z

}and let

g(C) :={

f (EC)}

; if EC 6= /0,C; if EC = /0.

If M ∈ Z satisfies g(M) = M, then there is no element x ∈ X \Mso that M∪{x} ∈ Z, which means that M is maximal in Z. Theproof will be done once we find an M ∈ Z with g(M) = M.

The Axiom of Choice

Proof (initial setup). Consider the indexed family{i}i∈P(X)\{ /0}.

The union of this family is X. By the Axiom ofChoice, there is a choice function f : P(X)\{ /0}→ X so thatf (A) ∈ A holds for all A ∈P(X)\{ /0}.For each C ∈ Z, define the set EC :=

{x ∈ X \C : C∪{x} ∈ Z

}and let

g(C) :={

f (EC)}

; if EC 6= /0,C; if EC = /0.

The Axiom of Choice

Proof (initial setup). Consider the indexed family{i}i∈P(X)\{ /0}. The union of this family is X.

By the Axiom ofChoice, there is a choice function f : P(X)\{ /0}→ X so thatf (A) ∈ A holds for all A ∈P(X)\{ /0}.For each C ∈ Z, define the set EC :=

{x ∈ X \C : C∪{x} ∈ Z

}and let

g(C) :={

f (EC)}

; if EC 6= /0,C; if EC = /0.

The Axiom of Choice

Proof (initial setup). Consider the indexed family{i}i∈P(X)\{ /0}. The union of this family is X. By the Axiom ofChoice, there is a choice function f : P(X)\{ /0}→ X

so thatf (A) ∈ A holds for all A ∈P(X)\{ /0}.For each C ∈ Z, define the set EC :=

{x ∈ X \C : C∪{x} ∈ Z

}and let

g(C) :={

f (EC)}

; if EC 6= /0,C; if EC = /0.

The Axiom of Choice

Proof (initial setup). Consider the indexed family{i}i∈P(X)\{ /0}. The union of this family is X. By the Axiom ofChoice, there is a choice function f : P(X)\{ /0}→ X so thatf (A) ∈ A holds for all A ∈P(X)\{ /0}.

For each C ∈ Z, define the set EC :={

x ∈ X \C : C∪{x} ∈ Z}

and let

g(C) :={

f (EC)}

; if EC 6= /0,C; if EC = /0.

The Axiom of Choice

Proof (initial setup). Consider the indexed family{i}i∈P(X)\{ /0}. The union of this family is X. By the Axiom ofChoice, there is a choice function f : P(X)\{ /0}→ X so thatf (A) ∈ A holds for all A ∈P(X)\{ /0}.For each C ∈ Z, define the set EC :=

{x ∈ X \C : C∪{x} ∈ Z

and let

g(C) :={

f (EC)}

; if EC 6= /0,C; if EC = /0.

The Axiom of Choice

{x ∈ X \C : C∪{x} ∈ Z

}and let

g(C) :={

f (EC)}

; if EC 6= /0,C; if EC = /0.

The Axiom of Choice

{x ∈ X \C : C∪{x} ∈ Z

}and let

g(C) :={

f (EC)}

; if EC 6= /0,C; if EC = /0.

If M ∈ Z satisfies g(M) = M, then there is no element x ∈ X \Mso that M∪{x} ∈ Z

, which means that M is maximal in Z. Theproof will be done once we find an M ∈ Z with g(M) = M.

The Axiom of Choice

{x ∈ X \C : C∪{x} ∈ Z

}and let

g(C) :={

f (EC)}

; if EC 6= /0,C; if EC = /0.

If M ∈ Z satisfies g(M) = M, then there is no element x ∈ X \Mso that M∪{x} ∈ Z, which means that M is maximal in Z.

Theproof will be done once we find an M ∈ Z with g(M) = M.

The Axiom of Choice

{x ∈ X \C : C∪{x} ∈ Z

}and let

g(C) :={

f (EC)}

; if EC 6= /0,C; if EC = /0.

The Axiom of Choice

Proof (towers).

A subset T ⊆ Z will be called a tower iff1. /0 ∈ T , and2. If C ∈ T , then g(C) ∈ T , and3. If C ⊆ T is a chain in T , then

⋃C ∈ T .

The set Z contains at least one tower, because Z itself is a tower.Moreover, the intersection of any set of towers is a tower, too.Let T0 be the intersection of all towers that are contained in Z.Then T0 is not empty, because /0 ∈ T0.Call an element C ∈ T0 comparable iff for all A ∈ T0 we haveA⊆ C or C ⊆ A. First note that, clearly, /0 is a comparable set.

The Axiom of Choice

Proof (towers). A subset T ⊆ Z will be called a tower iff

1. /0 ∈ T , and2. If C ∈ T , then g(C) ∈ T , and3. If C ⊆ T is a chain in T , then

⋃C ∈ T .

The Axiom of Choice

Proof (towers). A subset T ⊆ Z will be called a tower iff1. /0 ∈ T , and

2. If C ∈ T , then g(C) ∈ T , and3. If C ⊆ T is a chain in T , then

⋃C ∈ T .

The Axiom of Choice

Proof (towers). A subset T ⊆ Z will be called a tower iff1. /0 ∈ T , and2. If C ∈ T , then g(C) ∈ T , and

3. If C ⊆ T is a chain in T , then⋃

C ∈ T .The set Z contains at least one tower, because Z itself is a tower.Moreover, the intersection of any set of towers is a tower, too.Let T0 be the intersection of all towers that are contained in Z.Then T0 is not empty, because /0 ∈ T0.Call an element C ∈ T0 comparable iff for all A ∈ T0 we haveA⊆ C or C ⊆ A. First note that, clearly, /0 is a comparable set.

The Axiom of Choice

Proof (towers). A subset T ⊆ Z will be called a tower iff1. /0 ∈ T , and2. If C ∈ T , then g(C) ∈ T , and3. If C ⊆ T is a chain in T , then

⋃C ∈ T .

The Axiom of Choice

⋃C ∈ T .

The set Z contains at least one tower, because Z itself is a tower.

Moreover, the intersection of any set of towers is a tower, too.Let T0 be the intersection of all towers that are contained in Z.Then T0 is not empty, because /0 ∈ T0.Call an element C ∈ T0 comparable iff for all A ∈ T0 we haveA⊆ C or C ⊆ A. First note that, clearly, /0 is a comparable set.

The Axiom of Choice

⋃C ∈ T .

The set Z contains at least one tower, because Z itself is a tower.Moreover, the intersection of any set of towers is a tower, too.

Let T0 be the intersection of all towers that are contained in Z.Then T0 is not empty, because /0 ∈ T0.Call an element C ∈ T0 comparable iff for all A ∈ T0 we haveA⊆ C or C ⊆ A. First note that, clearly, /0 is a comparable set.

The Axiom of Choice

⋃C ∈ T .

The set Z contains at least one tower, because Z itself is a tower.Moreover, the intersection of any set of towers is a tower, too.Let T0 be the intersection of all towers that are contained in Z.

Then T0 is not empty, because /0 ∈ T0.Call an element C ∈ T0 comparable iff for all A ∈ T0 we haveA⊆ C or C ⊆ A. First note that, clearly, /0 is a comparable set.

The Axiom of Choice

⋃C ∈ T .

The set Z contains at least one tower, because Z itself is a tower.Moreover, the intersection of any set of towers is a tower, too.Let T0 be the intersection of all towers that are contained in Z.Then T0 is not empty, because /0 ∈ T0.

Call an element C ∈ T0 comparable iff for all A ∈ T0 we haveA⊆ C or C ⊆ A. First note that, clearly, /0 is a comparable set.

The Axiom of Choice

⋃C ∈ T .

The set Z contains at least one tower, because Z itself is a tower.Moreover, the intersection of any set of towers is a tower, too.Let T0 be the intersection of all towers that are contained in Z.Then T0 is not empty, because /0 ∈ T0.Call an element C ∈ T0 comparable iff for all A ∈ T0 we haveA⊆ C or C ⊆ A.

First note that, clearly, /0 is a comparable set.

The Axiom of Choice

⋃C ∈ T .

The Axiom of Choice

Proof (C comparable implies g(C) comparable).

Let C ∈ T0be a fixed comparable set. Consider the setU :=

{A ∈ T0 : A⊆ C or g(C)⊆ A

}. We will prove that U is a

tower, which implies that U = T0, which implies that g(C) iscomparable. Clearly, /0 ∈ U. Now let A ∈ U. Because C iscomparable, we have A = C or A⊂ C or C ⊂ A. In case A = C,we have g(A) = g(C)⊇ g(C), which means g(A) ∈ U. In caseA⊂ C, because C is comparable, g(A)⊆ C or C ⊂ g(A). Strictcontainment C ⊂ g(A) would mean (by A⊂ C) that C has atleast one more element than A and g(A) has at least one moreelement than C, which is impossible. Thus, in case A⊂ C wemust have g(A)⊆ C, which means g(A) ∈ U. In the last case,C ⊂ A, we note that A 6⊆ C. Thus, by definition of U,g(C)⊆ A⊆ g(A) and g(A) ∈ U.

The Axiom of Choice

Proof (C comparable implies g(C) comparable). Let C ∈ T0be a fixed comparable set.

Consider the setU :=

{A ∈ T0 : A⊆ C or g(C)⊆ A

The Axiom of Choice

Proof (C comparable implies g(C) comparable). Let C ∈ T0be a fixed comparable set. Consider the setU :=

{A ∈ T0 : A⊆ C or g(C)⊆ A

We will prove that U is atower, which implies that U = T0, which implies that g(C) iscomparable. Clearly, /0 ∈ U. Now let A ∈ U. Because C iscomparable, we have A = C or A⊂ C or C ⊂ A. In case A = C,we have g(A) = g(C)⊇ g(C), which means g(A) ∈ U. In caseA⊂ C, because C is comparable, g(A)⊆ C or C ⊂ g(A). Strictcontainment C ⊂ g(A) would mean (by A⊂ C) that C has atleast one more element than A and g(A) has at least one moreelement than C, which is impossible. Thus, in case A⊂ C wemust have g(A)⊆ C, which means g(A) ∈ U. In the last case,C ⊂ A, we note that A 6⊆ C. Thus, by definition of U,g(C)⊆ A⊆ g(A) and g(A) ∈ U.

The Axiom of Choice

{A ∈ T0 : A⊆ C or g(C)⊆ A

, which implies that U = T0, which implies that g(C) iscomparable. Clearly, /0 ∈ U. Now let A ∈ U. Because C iscomparable, we have A = C or A⊂ C or C ⊂ A. In case A = C,we have g(A) = g(C)⊇ g(C), which means g(A) ∈ U. In caseA⊂ C, because C is comparable, g(A)⊆ C or C ⊂ g(A). Strictcontainment C ⊂ g(A) would mean (by A⊂ C) that C has atleast one more element than A and g(A) has at least one moreelement than C, which is impossible. Thus, in case A⊂ C wemust have g(A)⊆ C, which means g(A) ∈ U. In the last case,C ⊂ A, we note that A 6⊆ C. Thus, by definition of U,g(C)⊆ A⊆ g(A) and g(A) ∈ U.

The Axiom of Choice

{A ∈ T0 : A⊆ C or g(C)⊆ A

tower, which implies that U = T0

, which implies that g(C) iscomparable. Clearly, /0 ∈ U. Now let A ∈ U. Because C iscomparable, we have A = C or A⊂ C or C ⊂ A. In case A = C,we have g(A) = g(C)⊇ g(C), which means g(A) ∈ U. In caseA⊂ C, because C is comparable, g(A)⊆ C or C ⊂ g(A). Strictcontainment C ⊂ g(A) would mean (by A⊂ C) that C has atleast one more element than A and g(A) has at least one moreelement than C, which is impossible. Thus, in case A⊂ C wemust have g(A)⊆ C, which means g(A) ∈ U. In the last case,C ⊂ A, we note that A 6⊆ C. Thus, by definition of U,g(C)⊆ A⊆ g(A) and g(A) ∈ U.

The Axiom of Choice

{A ∈ T0 : A⊆ C or g(C)⊆ A

tower, which implies that U = T0, which implies that g(C) iscomparable.

Clearly, /0 ∈ U. Now let A ∈ U. Because C iscomparable, we have A = C or A⊂ C or C ⊂ A. In case A = C,we have g(A) = g(C)⊇ g(C), which means g(A) ∈ U. In caseA⊂ C, because C is comparable, g(A)⊆ C or C ⊂ g(A). Strictcontainment C ⊂ g(A) would mean (by A⊂ C) that C has atleast one more element than A and g(A) has at least one moreelement than C, which is impossible. Thus, in case A⊂ C wemust have g(A)⊆ C, which means g(A) ∈ U. In the last case,C ⊂ A, we note that A 6⊆ C. Thus, by definition of U,g(C)⊆ A⊆ g(A) and g(A) ∈ U.

The Axiom of Choice

{A ∈ T0 : A⊆ C or g(C)⊆ A

tower, which implies that U = T0, which implies that g(C) iscomparable. Clearly, /0 ∈ U.

Now let A ∈ U. Because C iscomparable, we have A = C or A⊂ C or C ⊂ A. In case A = C,we have g(A) = g(C)⊇ g(C), which means g(A) ∈ U. In caseA⊂ C, because C is comparable, g(A)⊆ C or C ⊂ g(A). Strictcontainment C ⊂ g(A) would mean (by A⊂ C) that C has atleast one more element than A and g(A) has at least one moreelement than C, which is impossible. Thus, in case A⊂ C wemust have g(A)⊆ C, which means g(A) ∈ U. In the last case,C ⊂ A, we note that A 6⊆ C. Thus, by definition of U,g(C)⊆ A⊆ g(A) and g(A) ∈ U.

The Axiom of Choice

{A ∈ T0 : A⊆ C or g(C)⊆ A

tower, which implies that U = T0, which implies that g(C) iscomparable. Clearly, /0 ∈ U. Now let A ∈ U.

Because C iscomparable, we have A = C or A⊂ C or C ⊂ A. In case A = C,we have g(A) = g(C)⊇ g(C), which means g(A) ∈ U. In caseA⊂ C, because C is comparable, g(A)⊆ C or C ⊂ g(A). Strictcontainment C ⊂ g(A) would mean (by A⊂ C) that C has atleast one more element than A and g(A) has at least one moreelement than C, which is impossible. Thus, in case A⊂ C wemust have g(A)⊆ C, which means g(A) ∈ U. In the last case,C ⊂ A, we note that A 6⊆ C. Thus, by definition of U,g(C)⊆ A⊆ g(A) and g(A) ∈ U.

The Axiom of Choice

{A ∈ T0 : A⊆ C or g(C)⊆ A

tower, which implies that U = T0, which implies that g(C) iscomparable. Clearly, /0 ∈ U. Now let A ∈ U. Because C iscomparable, we have A = C or A⊂ C or C ⊂ A.

In case A = C,we have g(A) = g(C)⊇ g(C), which means g(A) ∈ U. In caseA⊂ C, because C is comparable, g(A)⊆ C or C ⊂ g(A). Strictcontainment C ⊂ g(A) would mean (by A⊂ C) that C has atleast one more element than A and g(A) has at least one moreelement than C, which is impossible. Thus, in case A⊂ C wemust have g(A)⊆ C, which means g(A) ∈ U. In the last case,C ⊂ A, we note that A 6⊆ C. Thus, by definition of U,g(C)⊆ A⊆ g(A) and g(A) ∈ U.

The Axiom of Choice

{A ∈ T0 : A⊆ C or g(C)⊆ A

tower, which implies that U = T0, which implies that g(C) iscomparable. Clearly, /0 ∈ U. Now let A ∈ U. Because C iscomparable, we have A = C or A⊂ C or C ⊂ A. In case A = C,we have g(A) = g(C)⊇ g(C), which means g(A) ∈ U.

In caseA⊂ C, because C is comparable, g(A)⊆ C or C ⊂ g(A). Strictcontainment C ⊂ g(A) would mean (by A⊂ C) that C has atleast one more element than A and g(A) has at least one moreelement than C, which is impossible. Thus, in case A⊂ C wemust have g(A)⊆ C, which means g(A) ∈ U. In the last case,C ⊂ A, we note that A 6⊆ C. Thus, by definition of U,g(C)⊆ A⊆ g(A) and g(A) ∈ U.

The Axiom of Choice

{A ∈ T0 : A⊆ C or g(C)⊆ A

tower, which implies that U = T0, which implies that g(C) iscomparable. Clearly, /0 ∈ U. Now let A ∈ U. Because C iscomparable, we have A = C or A⊂ C or C ⊂ A. In case A = C,we have g(A) = g(C)⊇ g(C), which means g(A) ∈ U. In caseA⊂ C, because C is comparable, g(A)⊆ C or C ⊂ g(A).

Strictcontainment C ⊂ g(A) would mean (by A⊂ C) that C has atleast one more element than A and g(A) has at least one moreelement than C, which is impossible. Thus, in case A⊂ C wemust have g(A)⊆ C, which means g(A) ∈ U. In the last case,C ⊂ A, we note that A 6⊆ C. Thus, by definition of U,g(C)⊆ A⊆ g(A) and g(A) ∈ U.

The Axiom of Choice

{A ∈ T0 : A⊆ C or g(C)⊆ A

tower, which implies that U = T0, which implies that g(C) iscomparable. Clearly, /0 ∈ U. Now let A ∈ U. Because C iscomparable, we have A = C or A⊂ C or C ⊂ A. In case A = C,we have g(A) = g(C)⊇ g(C), which means g(A) ∈ U. In caseA⊂ C, because C is comparable, g(A)⊆ C or C ⊂ g(A). Strictcontainment C ⊂ g(A) would mean (by A⊂ C) that C has atleast one more element than A

and g(A) has at least one moreelement than C, which is impossible. Thus, in case A⊂ C wemust have g(A)⊆ C, which means g(A) ∈ U. In the last case,C ⊂ A, we note that A 6⊆ C. Thus, by definition of U,g(C)⊆ A⊆ g(A) and g(A) ∈ U.

The Axiom of Choice

{A ∈ T0 : A⊆ C or g(C)⊆ A

tower, which implies that U = T0, which implies that g(C) iscomparable. Clearly, /0 ∈ U. Now let A ∈ U. Because C iscomparable, we have A = C or A⊂ C or C ⊂ A. In case A = C,we have g(A) = g(C)⊇ g(C), which means g(A) ∈ U. In caseA⊂ C, because C is comparable, g(A)⊆ C or C ⊂ g(A). Strictcontainment C ⊂ g(A) would mean (by A⊂ C) that C has atleast one more element than A and g(A) has at least one moreelement than C

, which is impossible. Thus, in case A⊂ C wemust have g(A)⊆ C, which means g(A) ∈ U. In the last case,C ⊂ A, we note that A 6⊆ C. Thus, by definition of U,g(C)⊆ A⊆ g(A) and g(A) ∈ U.

The Axiom of Choice

{A ∈ T0 : A⊆ C or g(C)⊆ A

tower, which implies that U = T0, which implies that g(C) iscomparable. Clearly, /0 ∈ U. Now let A ∈ U. Because C iscomparable, we have A = C or A⊂ C or C ⊂ A. In case A = C,we have g(A) = g(C)⊇ g(C), which means g(A) ∈ U. In caseA⊂ C, because C is comparable, g(A)⊆ C or C ⊂ g(A). Strictcontainment C ⊂ g(A) would mean (by A⊂ C) that C has atleast one more element than A and g(A) has at least one moreelement than C, which is impossible.

Thus, in case A⊂ C wemust have g(A)⊆ C, which means g(A) ∈ U. In the last case,C ⊂ A, we note that A 6⊆ C. Thus, by definition of U,g(C)⊆ A⊆ g(A) and g(A) ∈ U.

The Axiom of Choice

{A ∈ T0 : A⊆ C or g(C)⊆ A

tower, which implies that U = T0, which implies that g(C) iscomparable. Clearly, /0 ∈ U. Now let A ∈ U. Because C iscomparable, we have A = C or A⊂ C or C ⊂ A. In case A = C,we have g(A) = g(C)⊇ g(C), which means g(A) ∈ U. In caseA⊂ C, because C is comparable, g(A)⊆ C or C ⊂ g(A). Strictcontainment C ⊂ g(A) would mean (by A⊂ C) that C has atleast one more element than A and g(A) has at least one moreelement than C, which is impossible. Thus, in case A⊂ C wemust have g(A)⊆ C

, which means g(A) ∈ U. In the last case,C ⊂ A, we note that A 6⊆ C. Thus, by definition of U,g(C)⊆ A⊆ g(A) and g(A) ∈ U.

The Axiom of Choice

{A ∈ T0 : A⊆ C or g(C)⊆ A

tower, which implies that U = T0, which implies that g(C) iscomparable. Clearly, /0 ∈ U. Now let A ∈ U. Because C iscomparable, we have A = C or A⊂ C or C ⊂ A. In case A = C,we have g(A) = g(C)⊇ g(C), which means g(A) ∈ U. In caseA⊂ C, because C is comparable, g(A)⊆ C or C ⊂ g(A). Strictcontainment C ⊂ g(A) would mean (by A⊂ C) that C has atleast one more element than A and g(A) has at least one moreelement than C, which is impossible. Thus, in case A⊂ C wemust have g(A)⊆ C, which means g(A) ∈ U.

In the last case,C ⊂ A, we note that A 6⊆ C. Thus, by definition of U,g(C)⊆ A⊆ g(A) and g(A) ∈ U.

The Axiom of Choice

{A ∈ T0 : A⊆ C or g(C)⊆ A

tower, which implies that U = T0, which implies that g(C) iscomparable. Clearly, /0 ∈ U. Now let A ∈ U. Because C iscomparable, we have A = C or A⊂ C or C ⊂ A. In case A = C,we have g(A) = g(C)⊇ g(C), which means g(A) ∈ U. In caseA⊂ C, because C is comparable, g(A)⊆ C or C ⊂ g(A). Strictcontainment C ⊂ g(A) would mean (by A⊂ C) that C has atleast one more element than A and g(A) has at least one moreelement than C, which is impossible. Thus, in case A⊂ C wemust have g(A)⊆ C, which means g(A) ∈ U. In the last case,C ⊂ A, we note that A 6⊆ C.

Thus, by definition of U,g(C)⊆ A⊆ g(A) and g(A) ∈ U.

The Axiom of Choice

{A ∈ T0 : A⊆ C or g(C)⊆ A

The Axiom of Choice

Proof (C comparable implies g(C) comparable, concl.).

Finally, let A ⊆ U be a chain. If C ⊇ A for all A ∈A , thenC ⊇

⋃A and

⋃A ∈ U. Otherwise, there is an A ∈A so that

C ⊂ A. But then A 6⊆ C, which implies g(C)⊆ A⊆⋃

hence⋃

A ∈ U. Thus U ⊆ T0 is a tower. By definition of T0,T0 ⊆ U and hence U = T0. Thus for all A ∈ T0 we haveA⊆ C ⊆ g(C) or g(C)⊆ A. So if C ∈ T0 is comparable, theng(C) is comparable, too.

The Axiom of Choice

Proof (C comparable implies g(C) comparable, concl.).Finally, let A ⊆ U be a chain.

If C ⊇ A for all A ∈A , thenC ⊇

⋃A and

hence⋃

The Axiom of Choice

Proof (C comparable implies g(C) comparable, concl.).Finally, let A ⊆ U be a chain. If C ⊇ A for all A ∈A , thenC ⊇

⋃A and

⋃A ∈ U.

Otherwise, there is an A ∈A so that

hence⋃

The Axiom of Choice

⋃A and

C ⊂ A.

But then A 6⊆ C, which implies g(C)⊆ A⊆⋃

hence⋃

The Axiom of Choice

⋃A and

C ⊂ A. But then A 6⊆ C

, which implies g(C)⊆ A⊆⋃

hence⋃

The Axiom of Choice

⋃A and

hence⋃

The Axiom of Choice

⋃A and

hence⋃

A ∈ U.

Thus U ⊆ T0 is a tower. By definition of T0,T0 ⊆ U and hence U = T0. Thus for all A ∈ T0 we haveA⊆ C ⊆ g(C) or g(C)⊆ A. So if C ∈ T0 is comparable, theng(C) is comparable, too.

The Axiom of Choice

⋃A and

hence⋃

A ∈ U. Thus U ⊆ T0 is a tower.

By definition of T0,T0 ⊆ U and hence U = T0. Thus for all A ∈ T0 we haveA⊆ C ⊆ g(C) or g(C)⊆ A. So if C ∈ T0 is comparable, theng(C) is comparable, too.

The Axiom of Choice

⋃A and

hence⋃

A ∈ U. Thus U ⊆ T0 is a tower. By definition of T0,T0 ⊆ U

and hence U = T0. Thus for all A ∈ T0 we haveA⊆ C ⊆ g(C) or g(C)⊆ A. So if C ∈ T0 is comparable, theng(C) is comparable, too.

The Axiom of Choice

⋃A and

hence⋃

A ∈ U. Thus U ⊆ T0 is a tower. By definition of T0,T0 ⊆ U and hence U = T0.

Thus for all A ∈ T0 we haveA⊆ C ⊆ g(C) or g(C)⊆ A. So if C ∈ T0 is comparable, theng(C) is comparable, too.

The Axiom of Choice

⋃A and

hence⋃

A ∈ U. Thus U ⊆ T0 is a tower. By definition of T0,T0 ⊆ U and hence U = T0. Thus for all A ∈ T0 we haveA⊆ C ⊆ g(C) or g(C)⊆ A.

So if C ∈ T0 is comparable, theng(C) is comparable, too.

The Axiom of Choice

⋃A and

hence⋃

The Axiom of Choice

Proof (existence of maximal elements).

Let C ⊆ T0 be a chainof comparable elements and let A ∈ T0. If there is a C ∈ C withA⊆ C, then A⊆ C ⊆

⋃C . Otherwise for all C ∈ C we have

C ⊆ A, which means⋃

C ⊆ A. Consequently, if C ⊆ T0 is a

chain of comparable elements, then the union⋃

C iscomparable.Thus the set of comparable elements in T0 is a tower. BecauseT0 is the intersection of all towers, every element of T0 iscomparable. By definition of comparable elements, T0 is achain. Because T0 is a tower, we have

⋃T0 ∈ T0. Hence

)∈ T0, which means that g

(⋃T0

)⊆⋃

T0. By

definition of g, g(⋃

)⊇⋃

T0, so g(⋃

Hence M :=⋃

T0 is the desired maximal element.

The Axiom of Choice

Proof (existence of maximal elements). Let C ⊆ T0 be a chainof comparable elements and let A ∈ T0.

If there is a C ∈ C withA⊆ C, then A⊆ C ⊆

⋃T0 ∈ T0. Hence

(⋃T0

)⊆⋃

T0. By

)⊇⋃

T0, so g(⋃

Hence M :=⋃

The Axiom of Choice

Proof (existence of maximal elements). Let C ⊆ T0 be a chainof comparable elements and let A ∈ T0. If there is a C ∈ C withA⊆ C, then A⊆ C ⊆

⋃C .

Otherwise for all C ∈ C we have

⋃T0 ∈ T0. Hence

(⋃T0

)⊆⋃

T0. By

)⊇⋃

T0, so g(⋃

Hence M :=⋃

The Axiom of Choice

C ⊆ A.

Consequently, if C ⊆ T0 is a

⋃T0 ∈ T0. Hence

(⋃T0

)⊆⋃

T0. By

)⊇⋃

T0, so g(⋃

Hence M :=⋃

The Axiom of Choice

C iscomparable.

Thus the set of comparable elements in T0 is a tower. BecauseT0 is the intersection of all towers, every element of T0 iscomparable. By definition of comparable elements, T0 is achain. Because T0 is a tower, we have

⋃T0 ∈ T0. Hence

(⋃T0

)⊆⋃

T0. By

)⊇⋃

T0, so g(⋃

Hence M :=⋃

The Axiom of Choice

C iscomparable.Thus the set of comparable elements in T0 is a tower.

BecauseT0 is the intersection of all towers, every element of T0 iscomparable. By definition of comparable elements, T0 is achain. Because T0 is a tower, we have

⋃T0 ∈ T0. Hence

(⋃T0

)⊆⋃

T0. By

)⊇⋃

T0, so g(⋃

Hence M :=⋃

The Axiom of Choice

C iscomparable.Thus the set of comparable elements in T0 is a tower. BecauseT0 is the intersection of all towers, every element of T0 iscomparable.

By definition of comparable elements, T0 is achain. Because T0 is a tower, we have

⋃T0 ∈ T0. Hence

(⋃T0

)⊆⋃

T0. By

)⊇⋃

T0, so g(⋃

Hence M :=⋃

The Axiom of Choice

C iscomparable.Thus the set of comparable elements in T0 is a tower. BecauseT0 is the intersection of all towers, every element of T0 iscomparable. By definition of comparable elements, T0 is achain.

Because T0 is a tower, we have⋃

T0 ∈ T0. Hence

(⋃T0

)⊆⋃

T0. By

)⊇⋃

T0, so g(⋃

Hence M :=⋃

The Axiom of Choice

⋃T0 ∈ T0.

(⋃T0

)⊆⋃

T0. By

)⊇⋃

T0, so g(⋃

Hence M :=⋃

The Axiom of Choice

⋃T0 ∈ T0. Hence

)∈ T0

, which means that g(⋃

)⊆⋃

T0. By

)⊇⋃

T0, so g(⋃

Hence M :=⋃

The Axiom of Choice

⋃T0 ∈ T0. Hence

(⋃T0

)⊆⋃

)⊇⋃

T0, so g(⋃

Hence M :=⋃

The Axiom of Choice

⋃T0 ∈ T0. Hence

(⋃T0

)⊆⋃

T0. By

)⊇⋃

, so g(⋃

Hence M :=⋃

The Axiom of Choice

⋃T0 ∈ T0. Hence

(⋃T0

)⊆⋃

T0. By

)⊇⋃

T0, so g(⋃

Hence M :=⋃

The Axiom of Choice

⋃T0 ∈ T0. Hence

(⋃T0

)⊆⋃

T0. By

)⊇⋃

T0, so g(⋃

Hence M :=⋃

The Axiom of Choice

⋃T0 ∈ T0. Hence

(⋃T0

)⊆⋃

T0. By

)⊇⋃

T0, so g(⋃

Hence M :=⋃

The Axiom of Choice

Theorem.

Zorn’s Lemma. Let X be a nonempty ordered set sothat every chain in X has an upper bound. Then X has amaximal element.

Proof. Let Z be the set of all chains in X, ordered by inclusion.If C ∈ Z, then every subset of C is in Z, too. Moreover, theunion of every chain in Z is again an element of Z. Hence Z hasa maximal element M with respect to inclusion. This set M hasan upper bound m in X, and M∪{m} is a chain in X, that is,M∪{m} ∈ Z. But M is maximal in Z, so m ∈M. Now let x ∈ Xsatisfy x≥m. Then M∪{x} ∈ Z. So x ∈M and then m≥ x, thatis, m = x. Therefore, m is maximal in X.

The Axiom of Choice

Theorem. Zorn’s Lemma.

Let X be a nonempty ordered set sothat every chain in X has an upper bound. Then X has amaximal element.

The Axiom of Choice

Theorem. Zorn’s Lemma. Let X be a nonempty ordered set sothat every chain in X has an upper bound.

Then X has amaximal element.

The Axiom of Choice

Theorem. Zorn’s Lemma. Let X be a nonempty ordered set sothat every chain in X has an upper bound. Then X has amaximal element.

The Axiom of Choice

Proof.

Let Z be the set of all chains in X, ordered by inclusion.If C ∈ Z, then every subset of C is in Z, too. Moreover, theunion of every chain in Z is again an element of Z. Hence Z hasa maximal element M with respect to inclusion. This set M hasan upper bound m in X, and M∪{m} is a chain in X, that is,M∪{m} ∈ Z. But M is maximal in Z, so m ∈M. Now let x ∈ Xsatisfy x≥m. Then M∪{x} ∈ Z. So x ∈M and then m≥ x, thatis, m = x. Therefore, m is maximal in X.

The Axiom of Choice

Proof. Let Z be the set of all chains in X, ordered by inclusion.

If C ∈ Z, then every subset of C is in Z, too. Moreover, theunion of every chain in Z is again an element of Z. Hence Z hasa maximal element M with respect to inclusion. This set M hasan upper bound m in X, and M∪{m} is a chain in X, that is,M∪{m} ∈ Z. But M is maximal in Z, so m ∈M. Now let x ∈ Xsatisfy x≥m. Then M∪{x} ∈ Z. So x ∈M and then m≥ x, thatis, m = x. Therefore, m is maximal in X.

The Axiom of Choice

Proof. Let Z be the set of all chains in X, ordered by inclusion.If C ∈ Z, then every subset of C is in Z, too.

Moreover, theunion of every chain in Z is again an element of Z. Hence Z hasa maximal element M with respect to inclusion. This set M hasan upper bound m in X, and M∪{m} is a chain in X, that is,M∪{m} ∈ Z. But M is maximal in Z, so m ∈M. Now let x ∈ Xsatisfy x≥m. Then M∪{x} ∈ Z. So x ∈M and then m≥ x, thatis, m = x. Therefore, m is maximal in X.

The Axiom of Choice

Proof. Let Z be the set of all chains in X, ordered by inclusion.If C ∈ Z, then every subset of C is in Z, too. Moreover, theunion of every chain in Z is again an element of Z.

Hence Z hasa maximal element M with respect to inclusion. This set M hasan upper bound m in X, and M∪{m} is a chain in X, that is,M∪{m} ∈ Z. But M is maximal in Z, so m ∈M. Now let x ∈ Xsatisfy x≥m. Then M∪{x} ∈ Z. So x ∈M and then m≥ x, thatis, m = x. Therefore, m is maximal in X.

The Axiom of Choice

Proof. Let Z be the set of all chains in X, ordered by inclusion.If C ∈ Z, then every subset of C is in Z, too. Moreover, theunion of every chain in Z is again an element of Z. Hence Z hasa maximal element M

with respect to inclusion. This set M hasan upper bound m in X, and M∪{m} is a chain in X, that is,M∪{m} ∈ Z. But M is maximal in Z, so m ∈M. Now let x ∈ Xsatisfy x≥m. Then M∪{x} ∈ Z. So x ∈M and then m≥ x, thatis, m = x. Therefore, m is maximal in X.

The Axiom of Choice

Proof. Let Z be the set of all chains in X, ordered by inclusion.If C ∈ Z, then every subset of C is in Z, too. Moreover, theunion of every chain in Z is again an element of Z. Hence Z hasa maximal element M with respect to inclusion.

This set M hasan upper bound m in X, and M∪{m} is a chain in X, that is,M∪{m} ∈ Z. But M is maximal in Z, so m ∈M. Now let x ∈ Xsatisfy x≥m. Then M∪{x} ∈ Z. So x ∈M and then m≥ x, thatis, m = x. Therefore, m is maximal in X.

The Axiom of Choice

Proof. Let Z be the set of all chains in X, ordered by inclusion.If C ∈ Z, then every subset of C is in Z, too. Moreover, theunion of every chain in Z is again an element of Z. Hence Z hasa maximal element M with respect to inclusion. This set M hasan upper bound m in X

, and M∪{m} is a chain in X, that is,M∪{m} ∈ Z. But M is maximal in Z, so m ∈M. Now let x ∈ Xsatisfy x≥m. Then M∪{x} ∈ Z. So x ∈M and then m≥ x, thatis, m = x. Therefore, m is maximal in X.

The Axiom of Choice

Proof. Let Z be the set of all chains in X, ordered by inclusion.If C ∈ Z, then every subset of C is in Z, too. Moreover, theunion of every chain in Z is again an element of Z. Hence Z hasa maximal element M with respect to inclusion. This set M hasan upper bound m in X, and M∪{m} is a chain in X

, that is,M∪{m} ∈ Z. But M is maximal in Z, so m ∈M. Now let x ∈ Xsatisfy x≥m. Then M∪{x} ∈ Z. So x ∈M and then m≥ x, thatis, m = x. Therefore, m is maximal in X.

The Axiom of Choice

Proof. Let Z be the set of all chains in X, ordered by inclusion.If C ∈ Z, then every subset of C is in Z, too. Moreover, theunion of every chain in Z is again an element of Z. Hence Z hasa maximal element M with respect to inclusion. This set M hasan upper bound m in X, and M∪{m} is a chain in X, that is,M∪{m} ∈ Z.

But M is maximal in Z, so m ∈M. Now let x ∈ Xsatisfy x≥m. Then M∪{x} ∈ Z. So x ∈M and then m≥ x, thatis, m = x. Therefore, m is maximal in X.

The Axiom of Choice

Proof. Let Z be the set of all chains in X, ordered by inclusion.If C ∈ Z, then every subset of C is in Z, too. Moreover, theunion of every chain in Z is again an element of Z. Hence Z hasa maximal element M with respect to inclusion. This set M hasan upper bound m in X, and M∪{m} is a chain in X, that is,M∪{m} ∈ Z. But M is maximal in Z, so m ∈M.

Now let x ∈ Xsatisfy x≥m. Then M∪{x} ∈ Z. So x ∈M and then m≥ x, thatis, m = x. Therefore, m is maximal in X.

The Axiom of Choice

Proof. Let Z be the set of all chains in X, ordered by inclusion.If C ∈ Z, then every subset of C is in Z, too. Moreover, theunion of every chain in Z is again an element of Z. Hence Z hasa maximal element M with respect to inclusion. This set M hasan upper bound m in X, and M∪{m} is a chain in X, that is,M∪{m} ∈ Z. But M is maximal in Z, so m ∈M. Now let x ∈ Xsatisfy x≥m.

Then M∪{x} ∈ Z. So x ∈M and then m≥ x, thatis, m = x. Therefore, m is maximal in X.

The Axiom of Choice

Proof. Let Z be the set of all chains in X, ordered by inclusion.If C ∈ Z, then every subset of C is in Z, too. Moreover, theunion of every chain in Z is again an element of Z. Hence Z hasa maximal element M with respect to inclusion. This set M hasan upper bound m in X, and M∪{m} is a chain in X, that is,M∪{m} ∈ Z. But M is maximal in Z, so m ∈M. Now let x ∈ Xsatisfy x≥m. Then M∪{x} ∈ Z.

So x ∈M and then m≥ x, thatis, m = x. Therefore, m is maximal in X.

The Axiom of Choice

Proof. Let Z be the set of all chains in X, ordered by inclusion.If C ∈ Z, then every subset of C is in Z, too. Moreover, theunion of every chain in Z is again an element of Z. Hence Z hasa maximal element M with respect to inclusion. This set M hasan upper bound m in X, and M∪{m} is a chain in X, that is,M∪{m} ∈ Z. But M is maximal in Z, so m ∈M. Now let x ∈ Xsatisfy x≥m. Then M∪{x} ∈ Z. So x ∈M and then m≥ x, thatis, m = x.

Therefore, m is maximal in X.

The Axiom of Choice

Theorem.

Let A be an infinite set. Then(A×{0}

)∪(A×{1}

)= A×{0,1} is equivalent to A.

Proof. Let F be the set of all bijective functionsf : X×{0,1}→ X, where X ⊆ A. F 6= /0, because it contains allthe bijective functions f : X×{0,1}→ X, where X ⊆ A iscountable. F is ordered by set inclusion. Moreover for anychain C ⊆F we can form the union u :=

⋃C , and it will be a

bijective function u : Xu×{0,1}→ Xu for some subset Xu ⊆ A(good exercise). Now u is an upper bound for C in F . Thus thehypotheses of Zorn’s Lemma are satisfied.Let h : X×{0,1}→ X be a maximal element of F . Supposefor a contradiction that A\X contains a countably infinite set C.Let b : C×{0,1}→ C be a bijective function. Then t : h∪b is abijective function between (X∪C)×{0,1} and X∪C, that is,t ∈F , contradiction.

The Axiom of Choice

Theorem. Let A be an infinite set.

Then(A×{0}

)∪(A×{1}

The Axiom of Choice

Theorem. Let A be an infinite set. Then(A×{0}

)∪(A×{1}

= A×{0,1} is equivalent to A.

The Axiom of Choice

)∪(A×{1}

)= A×{0,1}

is equivalent to A.

The Axiom of Choice

)∪(A×{1}

The Axiom of Choice

)∪(A×{1}

Proof.

Let F be the set of all bijective functionsf : X×{0,1}→ X, where X ⊆ A. F 6= /0, because it contains allthe bijective functions f : X×{0,1}→ X, where X ⊆ A iscountable. F is ordered by set inclusion. Moreover for anychain C ⊆F we can form the union u :=

The Axiom of Choice

)∪(A×{1}

Proof. Let F be the set of all bijective functionsf : X×{0,1}→ X, where X ⊆ A.

F 6= /0, because it contains allthe bijective functions f : X×{0,1}→ X, where X ⊆ A iscountable. F is ordered by set inclusion. Moreover for anychain C ⊆F we can form the union u :=

The Axiom of Choice

)∪(A×{1}

Proof. Let F be the set of all bijective functionsf : X×{0,1}→ X, where X ⊆ A. F 6= /0, because it contains allthe bijective functions f : X×{0,1}→ X, where X ⊆ A iscountable.

F is ordered by set inclusion. Moreover for anychain C ⊆F we can form the union u :=

The Axiom of Choice

)∪(A×{1}

Proof. Let F be the set of all bijective functionsf : X×{0,1}→ X, where X ⊆ A. F 6= /0, because it contains allthe bijective functions f : X×{0,1}→ X, where X ⊆ A iscountable. F is ordered by set inclusion.

Moreover for anychain C ⊆F we can form the union u :=

The Axiom of Choice

)∪(A×{1}

, and it will be abijective function u : Xu×{0,1}→ Xu for some subset Xu ⊆ A(good exercise). Now u is an upper bound for C in F . Thus thehypotheses of Zorn’s Lemma are satisfied.Let h : X×{0,1}→ X be a maximal element of F . Supposefor a contradiction that A\X contains a countably infinite set C.Let b : C×{0,1}→ C be a bijective function. Then t : h∪b is abijective function between (X∪C)×{0,1} and X∪C, that is,t ∈F , contradiction.

The Axiom of Choice

)∪(A×{1}

bijective function u : Xu×{0,1}→ Xu for some subset Xu ⊆ A

(good exercise). Now u is an upper bound for C in F . Thus thehypotheses of Zorn’s Lemma are satisfied.Let h : X×{0,1}→ X be a maximal element of F . Supposefor a contradiction that A\X contains a countably infinite set C.Let b : C×{0,1}→ C be a bijective function. Then t : h∪b is abijective function between (X∪C)×{0,1} and X∪C, that is,t ∈F , contradiction.

The Axiom of Choice

)∪(A×{1}

bijective function u : Xu×{0,1}→ Xu for some subset Xu ⊆ A(good exercise).

Now u is an upper bound for C in F . Thus thehypotheses of Zorn’s Lemma are satisfied.Let h : X×{0,1}→ X be a maximal element of F . Supposefor a contradiction that A\X contains a countably infinite set C.Let b : C×{0,1}→ C be a bijective function. Then t : h∪b is abijective function between (X∪C)×{0,1} and X∪C, that is,t ∈F , contradiction.

The Axiom of Choice

)∪(A×{1}

bijective function u : Xu×{0,1}→ Xu for some subset Xu ⊆ A(good exercise). Now u is an upper bound for C in F .

Thus thehypotheses of Zorn’s Lemma are satisfied.Let h : X×{0,1}→ X be a maximal element of F . Supposefor a contradiction that A\X contains a countably infinite set C.Let b : C×{0,1}→ C be a bijective function. Then t : h∪b is abijective function between (X∪C)×{0,1} and X∪C, that is,t ∈F , contradiction.

The Axiom of Choice

)∪(A×{1}

bijective function u : Xu×{0,1}→ Xu for some subset Xu ⊆ A(good exercise). Now u is an upper bound for C in F . Thus thehypotheses of Zorn’s Lemma are satisfied.

Let h : X×{0,1}→ X be a maximal element of F . Supposefor a contradiction that A\X contains a countably infinite set C.Let b : C×{0,1}→ C be a bijective function. Then t : h∪b is abijective function between (X∪C)×{0,1} and X∪C, that is,t ∈F , contradiction.

The Axiom of Choice

)∪(A×{1}

bijective function u : Xu×{0,1}→ Xu for some subset Xu ⊆ A(good exercise). Now u is an upper bound for C in F . Thus thehypotheses of Zorn’s Lemma are satisfied.Let h : X×{0,1}→ X be a maximal element of F .

Supposefor a contradiction that A\X contains a countably infinite set C.Let b : C×{0,1}→ C be a bijective function. Then t : h∪b is abijective function between (X∪C)×{0,1} and X∪C, that is,t ∈F , contradiction.

The Axiom of Choice

)∪(A×{1}

bijective function u : Xu×{0,1}→ Xu for some subset Xu ⊆ A(good exercise). Now u is an upper bound for C in F . Thus thehypotheses of Zorn’s Lemma are satisfied.Let h : X×{0,1}→ X be a maximal element of F . Supposefor a contradiction that A\X contains a countably infinite set C.

Let b : C×{0,1}→ C be a bijective function. Then t : h∪b is abijective function between (X∪C)×{0,1} and X∪C, that is,t ∈F , contradiction.

The Axiom of Choice

)∪(A×{1}

bijective function u : Xu×{0,1}→ Xu for some subset Xu ⊆ A(good exercise). Now u is an upper bound for C in F . Thus thehypotheses of Zorn’s Lemma are satisfied.Let h : X×{0,1}→ X be a maximal element of F . Supposefor a contradiction that A\X contains a countably infinite set C.Let b : C×{0,1}→ C be a bijective function.

Then t : h∪b is abijective function between (X∪C)×{0,1} and X∪C, that is,t ∈F , contradiction.

The Axiom of Choice

)∪(A×{1}

bijective function u : Xu×{0,1}→ Xu for some subset Xu ⊆ A(good exercise). Now u is an upper bound for C in F . Thus thehypotheses of Zorn’s Lemma are satisfied.Let h : X×{0,1}→ X be a maximal element of F . Supposefor a contradiction that A\X contains a countably infinite set C.Let b : C×{0,1}→ C be a bijective function. Then t : h∪b is abijective function between (X∪C)×{0,1} and X∪C

, that is,t ∈F , contradiction.

The Axiom of Choice

)∪(A×{1}

bijective function u : Xu×{0,1}→ Xu for some subset Xu ⊆ A(good exercise). Now u is an upper bound for C in F . Thus thehypotheses of Zorn’s Lemma are satisfied.Let h : X×{0,1}→ X be a maximal element of F . Supposefor a contradiction that A\X contains a countably infinite set C.Let b : C×{0,1}→ C be a bijective function. Then t : h∪b is abijective function between (X∪C)×{0,1} and X∪C, that is,t ∈F

, contradiction.

The Axiom of Choice

)∪(A×{1}

The Axiom of Choice

Proof (cont.).

Therefore A\X cannot be infinite. If A\X = /0,then the function h is the desired bijection between A×{0,1}and A. Finally consider the case that A\X 6= /0. By the above,A\X is finite. Let C ⊆ X be a countably infinite subset of X.Let R⊆ C be an |A\X|-element subset of C. Then C \R is stillcountably infinite. Let p : h−1[C]→ C \R be a bijectivefunction and let q : (A\X)×{0,1}→ A\X∪R be a bijectivefunction. Then t :=

(h\h|h−1[C]

)∪p∪q is the desired bijective

function with domain A×{0,1} and range A.

The Axiom of Choice

Proof (cont.). Therefore A\X cannot be infinite.

If A\X = /0,then the function h is the desired bijection between A×{0,1}and A. Finally consider the case that A\X 6= /0. By the above,A\X is finite. Let C ⊆ X be a countably infinite subset of X.Let R⊆ C be an |A\X|-element subset of C. Then C \R is stillcountably infinite. Let p : h−1[C]→ C \R be a bijectivefunction and let q : (A\X)×{0,1}→ A\X∪R be a bijectivefunction. Then t :=

(h\h|h−1[C]

The Axiom of Choice

Proof (cont.). Therefore A\X cannot be infinite. If A\X = /0

,then the function h is the desired bijection between A×{0,1}and A. Finally consider the case that A\X 6= /0. By the above,A\X is finite. Let C ⊆ X be a countably infinite subset of X.Let R⊆ C be an |A\X|-element subset of C. Then C \R is stillcountably infinite. Let p : h−1[C]→ C \R be a bijectivefunction and let q : (A\X)×{0,1}→ A\X∪R be a bijectivefunction. Then t :=

(h\h|h−1[C]

The Axiom of Choice

Proof (cont.). Therefore A\X cannot be infinite. If A\X = /0,then the function h is the desired bijection between A×{0,1}and A.

Finally consider the case that A\X 6= /0. By the above,A\X is finite. Let C ⊆ X be a countably infinite subset of X.Let R⊆ C be an |A\X|-element subset of C. Then C \R is stillcountably infinite. Let p : h−1[C]→ C \R be a bijectivefunction and let q : (A\X)×{0,1}→ A\X∪R be a bijectivefunction. Then t :=

(h\h|h−1[C]

The Axiom of Choice

Proof (cont.). Therefore A\X cannot be infinite. If A\X = /0,then the function h is the desired bijection between A×{0,1}and A. Finally consider the case that A\X 6= /0.

By the above,A\X is finite. Let C ⊆ X be a countably infinite subset of X.Let R⊆ C be an |A\X|-element subset of C. Then C \R is stillcountably infinite. Let p : h−1[C]→ C \R be a bijectivefunction and let q : (A\X)×{0,1}→ A\X∪R be a bijectivefunction. Then t :=

(h\h|h−1[C]

The Axiom of Choice

Proof (cont.). Therefore A\X cannot be infinite. If A\X = /0,then the function h is the desired bijection between A×{0,1}and A. Finally consider the case that A\X 6= /0. By the above,A\X is finite.

Let C ⊆ X be a countably infinite subset of X.Let R⊆ C be an |A\X|-element subset of C. Then C \R is stillcountably infinite. Let p : h−1[C]→ C \R be a bijectivefunction and let q : (A\X)×{0,1}→ A\X∪R be a bijectivefunction. Then t :=

(h\h|h−1[C]

The Axiom of Choice

Proof (cont.). Therefore A\X cannot be infinite. If A\X = /0,then the function h is the desired bijection between A×{0,1}and A. Finally consider the case that A\X 6= /0. By the above,A\X is finite. Let C ⊆ X be a countably infinite subset of X.

Let R⊆ C be an |A\X|-element subset of C. Then C \R is stillcountably infinite. Let p : h−1[C]→ C \R be a bijectivefunction and let q : (A\X)×{0,1}→ A\X∪R be a bijectivefunction. Then t :=

(h\h|h−1[C]

The Axiom of Choice

Proof (cont.). Therefore A\X cannot be infinite. If A\X = /0,then the function h is the desired bijection between A×{0,1}and A. Finally consider the case that A\X 6= /0. By the above,A\X is finite. Let C ⊆ X be a countably infinite subset of X.Let R⊆ C be an |A\X|-element subset of C.

Then C \R is stillcountably infinite. Let p : h−1[C]→ C \R be a bijectivefunction and let q : (A\X)×{0,1}→ A\X∪R be a bijectivefunction. Then t :=

(h\h|h−1[C]

The Axiom of Choice

Proof (cont.). Therefore A\X cannot be infinite. If A\X = /0,then the function h is the desired bijection between A×{0,1}and A. Finally consider the case that A\X 6= /0. By the above,A\X is finite. Let C ⊆ X be a countably infinite subset of X.Let R⊆ C be an |A\X|-element subset of C. Then C \R is stillcountably infinite.

Let p : h−1[C]→ C \R be a bijectivefunction and let q : (A\X)×{0,1}→ A\X∪R be a bijectivefunction. Then t :=

(h\h|h−1[C]

The Axiom of Choice

Proof (cont.). Therefore A\X cannot be infinite. If A\X = /0,then the function h is the desired bijection between A×{0,1}and A. Finally consider the case that A\X 6= /0. By the above,A\X is finite. Let C ⊆ X be a countably infinite subset of X.Let R⊆ C be an |A\X|-element subset of C. Then C \R is stillcountably infinite. Let p : h−1[C]→ C \R be a bijectivefunction

and let q : (A\X)×{0,1}→ A\X∪R be a bijectivefunction. Then t :=

(h\h|h−1[C]

The Axiom of Choice

Proof (cont.). Therefore A\X cannot be infinite. If A\X = /0,then the function h is the desired bijection between A×{0,1}and A. Finally consider the case that A\X 6= /0. By the above,A\X is finite. Let C ⊆ X be a countably infinite subset of X.Let R⊆ C be an |A\X|-element subset of C. Then C \R is stillcountably infinite. Let p : h−1[C]→ C \R be a bijectivefunction and let q : (A\X)×{0,1}→ A\X∪R be a bijectivefunction.

Then t :=(

h\h|h−1[C]

The Axiom of Choice

Proof (cont.). Therefore A\X cannot be infinite. If A\X = /0,then the function h is the desired bijection between A×{0,1}and A. Finally consider the case that A\X 6= /0. By the above,A\X is finite. Let C ⊆ X be a countably infinite subset of X.Let R⊆ C be an |A\X|-element subset of C. Then C \R is stillcountably infinite. Let p : h−1[C]→ C \R be a bijectivefunction and let q : (A\X)×{0,1}→ A\X∪R be a bijectivefunction. Then t :=

(h\h|h−1[C]

The Axiom of Choice

Proof (cont.). Therefore A\X cannot be infinite. If A\X = /0,then the function h is the desired bijection between A×{0,1}and A. Finally consider the case that A\X 6= /0. By the above,A\X is finite. Let C ⊆ X be a countably infinite subset of X.Let R⊆ C be an |A\X|-element subset of C. Then C \R is stillcountably infinite. Let p : h−1[C]→ C \R be a bijectivefunction and let q : (A\X)×{0,1}→ A\X∪R be a bijectivefunction. Then t :=

(h\h|h−1[C]

The Axiom of Choice

Definition.

Let S be a set and let ≤⊆ S×S be an order relation.Then ≤ is called a well-order (relation) iff it is a total orderand every nonempty subset of S has a smallest element withrespect to ≤.

Example. N is well-ordered.

Theorem. Well-ordering Theorem. Every set can bewell-ordered. That is, for every set S, there is a well-orderrelation ≤⊆ S×S.

The Axiom of Choice

Definition. Let S be a set and let ≤⊆ S×S be an order relation.

Then ≤ is called a well-order (relation) iff it is a total orderand every nonempty subset of S has a smallest element withrespect to ≤.

The Axiom of Choice

Definition. Let S be a set and let ≤⊆ S×S be an order relation.Then ≤ is called a well-order (relation) iff it is a total order

and every nonempty subset of S has a smallest element withrespect to ≤.

The Axiom of Choice

Definition. Let S be a set and let ≤⊆ S×S be an order relation.Then ≤ is called a well-order (relation) iff it is a total orderand every nonempty subset of S has a smallest element withrespect to ≤.

The Axiom of Choice

Example.

N is well-ordered.

The Axiom of Choice

Theorem.

Well-ordering Theorem. Every set can bewell-ordered. That is, for every set S, there is a well-orderrelation ≤⊆ S×S.

The Axiom of Choice

Theorem. Well-ordering Theorem.

Every set can bewell-ordered. That is, for every set S, there is a well-orderrelation ≤⊆ S×S.

The Axiom of Choice

Theorem. Well-ordering Theorem. Every set can bewell-ordered.

That is, for every set S, there is a well-orderrelation ≤⊆ S×S.

The Axiom of Choice

Proof.

Let X be the set of all well-order relations ≤⊆ D×D,where D is a subset of S. Then X 6= /0. For any two well-orderrelations ≤1⊆ D1×D1 and ≤2⊆ D2×D2 in X define ≤1v≤2iff D1 ⊆ D2, every d2 ∈ D2 \D1 is a strict ≤2-upper bound ofD1, and ≤2 |D1×D1 =≤1.Then v is an order relation on X (good exercise). Let C ⊆ X bea chain and let ≤:=

⋃C .

≤ is an order relation: Reflexivity is trivial. Let D be thedomain of the relation ≤. For antisymmetry, let x,y ∈ D be sothat x≤ y and y≤ x. Then there is a ≤′∈ C with domain D′ sothat x,y ∈ D′. Hence x≤′ y and y≤′ x, which implies x = y.Transitivity is proved similarly.

The Axiom of Choice

Proof. Let X be the set of all well-order relations ≤⊆ D×D,where D is a subset of S.

Then X 6= /0. For any two well-orderrelations ≤1⊆ D1×D1 and ≤2⊆ D2×D2 in X define ≤1v≤2iff D1 ⊆ D2, every d2 ∈ D2 \D1 is a strict ≤2-upper bound ofD1, and ≤2 |D1×D1 =≤1.Then v is an order relation on X (good exercise). Let C ⊆ X bea chain and let ≤:=

⋃C .

The Axiom of Choice

Proof. Let X be the set of all well-order relations ≤⊆ D×D,where D is a subset of S. Then X 6= /0.

For any two well-orderrelations ≤1⊆ D1×D1 and ≤2⊆ D2×D2 in X define ≤1v≤2iff D1 ⊆ D2, every d2 ∈ D2 \D1 is a strict ≤2-upper bound ofD1, and ≤2 |D1×D1 =≤1.Then v is an order relation on X (good exercise). Let C ⊆ X bea chain and let ≤:=

⋃C .

The Axiom of Choice

Proof. Let X be the set of all well-order relations ≤⊆ D×D,where D is a subset of S. Then X 6= /0. For any two well-orderrelations ≤1⊆ D1×D1 and ≤2⊆ D2×D2 in X define ≤1v≤2iff

D1 ⊆ D2, every d2 ∈ D2 \D1 is a strict ≤2-upper bound ofD1, and ≤2 |D1×D1 =≤1.Then v is an order relation on X (good exercise). Let C ⊆ X bea chain and let ≤:=

⋃C .

The Axiom of Choice

Proof. Let X be the set of all well-order relations ≤⊆ D×D,where D is a subset of S. Then X 6= /0. For any two well-orderrelations ≤1⊆ D1×D1 and ≤2⊆ D2×D2 in X define ≤1v≤2iff D1 ⊆ D2

, every d2 ∈ D2 \D1 is a strict ≤2-upper bound ofD1, and ≤2 |D1×D1 =≤1.Then v is an order relation on X (good exercise). Let C ⊆ X bea chain and let ≤:=

⋃C .

The Axiom of Choice

Proof. Let X be the set of all well-order relations ≤⊆ D×D,where D is a subset of S. Then X 6= /0. For any two well-orderrelations ≤1⊆ D1×D1 and ≤2⊆ D2×D2 in X define ≤1v≤2iff D1 ⊆ D2, every d2 ∈ D2 \D1 is a strict ≤2-upper bound ofD1

, and ≤2 |D1×D1 =≤1.Then v is an order relation on X (good exercise). Let C ⊆ X bea chain and let ≤:=

⋃C .

The Axiom of Choice

Proof. Let X be the set of all well-order relations ≤⊆ D×D,where D is a subset of S. Then X 6= /0. For any two well-orderrelations ≤1⊆ D1×D1 and ≤2⊆ D2×D2 in X define ≤1v≤2iff D1 ⊆ D2, every d2 ∈ D2 \D1 is a strict ≤2-upper bound ofD1, and ≤2 |D1×D1 =≤1.

Then v is an order relation on X (good exercise). Let C ⊆ X bea chain and let ≤:=

⋃C .

The Axiom of Choice

Proof. Let X be the set of all well-order relations ≤⊆ D×D,where D is a subset of S. Then X 6= /0. For any two well-orderrelations ≤1⊆ D1×D1 and ≤2⊆ D2×D2 in X define ≤1v≤2iff D1 ⊆ D2, every d2 ∈ D2 \D1 is a strict ≤2-upper bound ofD1, and ≤2 |D1×D1 =≤1.Then v is an order relation on X

(good exercise). Let C ⊆ X bea chain and let ≤:=

⋃C .

The Axiom of Choice

Proof. Let X be the set of all well-order relations ≤⊆ D×D,where D is a subset of S. Then X 6= /0. For any two well-orderrelations ≤1⊆ D1×D1 and ≤2⊆ D2×D2 in X define ≤1v≤2iff D1 ⊆ D2, every d2 ∈ D2 \D1 is a strict ≤2-upper bound ofD1, and ≤2 |D1×D1 =≤1.Then v is an order relation on X (good exercise).

Let C ⊆ X bea chain and let ≤:=

⋃C .

The Axiom of Choice

Proof. Let X be the set of all well-order relations ≤⊆ D×D,where D is a subset of S. Then X 6= /0. For any two well-orderrelations ≤1⊆ D1×D1 and ≤2⊆ D2×D2 in X define ≤1v≤2iff D1 ⊆ D2, every d2 ∈ D2 \D1 is a strict ≤2-upper bound ofD1, and ≤2 |D1×D1 =≤1.Then v is an order relation on X (good exercise). Let C ⊆ X bea chain and let ≤:=

⋃C .

The Axiom of Choice

⋃C .

≤ is an order relation:

Reflexivity is trivial. Let D be thedomain of the relation ≤. For antisymmetry, let x,y ∈ D be sothat x≤ y and y≤ x. Then there is a ≤′∈ C with domain D′ sothat x,y ∈ D′. Hence x≤′ y and y≤′ x, which implies x = y.Transitivity is proved similarly.

The Axiom of Choice

⋃C .

≤ is an order relation: Reflexivity is trivial.

Let D be thedomain of the relation ≤. For antisymmetry, let x,y ∈ D be sothat x≤ y and y≤ x. Then there is a ≤′∈ C with domain D′ sothat x,y ∈ D′. Hence x≤′ y and y≤′ x, which implies x = y.Transitivity is proved similarly.

The Axiom of Choice

⋃C .

≤ is an order relation: Reflexivity is trivial. Let D be thedomain of the relation ≤.

For antisymmetry, let x,y ∈ D be sothat x≤ y and y≤ x. Then there is a ≤′∈ C with domain D′ sothat x,y ∈ D′. Hence x≤′ y and y≤′ x, which implies x = y.Transitivity is proved similarly.

The Axiom of Choice

⋃C .

≤ is an order relation: Reflexivity is trivial. Let D be thedomain of the relation ≤. For antisymmetry, let x,y ∈ D be sothat x≤ y and y≤ x.

Then there is a ≤′∈ C with domain D′ sothat x,y ∈ D′. Hence x≤′ y and y≤′ x, which implies x = y.Transitivity is proved similarly.

The Axiom of Choice

⋃C .

≤ is an order relation: Reflexivity is trivial. Let D be thedomain of the relation ≤. For antisymmetry, let x,y ∈ D be sothat x≤ y and y≤ x. Then there is a ≤′∈ C with domain D′ sothat x,y ∈ D′.

Hence x≤′ y and y≤′ x, which implies x = y.Transitivity is proved similarly.

The Axiom of Choice

⋃C .

≤ is an order relation: Reflexivity is trivial. Let D be thedomain of the relation ≤. For antisymmetry, let x,y ∈ D be sothat x≤ y and y≤ x. Then there is a ≤′∈ C with domain D′ sothat x,y ∈ D′. Hence x≤′ y and y≤′ x, which implies x = y.

Transitivity is proved similarly.

The Axiom of Choice

⋃C .

The Axiom of Choice

Proof (cont.).

Now let ≤′∈ C and let D′ be the domain of ≤′.Clearly, D′ ⊆ D. Let d ∈ D\D′ and let d′ ∈ D′. There is a≤′′∈ C with domain D′′ so that ≤′v≤′′ and d ∈ D′′ \D′. Butthen d ≥′′ d′, which means d > d′. Hence d is a strict ≤-upperbound of D′. Finally, because ≤=

⋃C , ≤ |D′×D′ =≤′. This

does not establish that ≤ is an upper bound of C , because westill do not know if ≤∈ X.For ≤∈ X, let A⊆ D be a nonempty subset of D. Then there is a≤′∈ C with domain D′ so that A∩D′ 6= /0. Because ≤′ is awell-order, A∩D′ has a ≤′-smallest element a. Because≤ |D′×D′ =≤′, a is the ≤-smallest element of A∩D′. Becauseall elements of D\D′ are ≤-strict upper bounds of D′, a is the≤-smallest element of A. Therefore (simple exercise, maybe toosimple) ≤ is a well-order. Hence it is a v-upper bound of C .

The Axiom of Choice

Proof (cont.). Now let ≤′∈ C and let D′ be the domain of ≤′.

Clearly, D′ ⊆ D. Let d ∈ D\D′ and let d′ ∈ D′. There is a≤′′∈ C with domain D′′ so that ≤′v≤′′ and d ∈ D′′ \D′. Butthen d ≥′′ d′, which means d > d′. Hence d is a strict ≤-upperbound of D′. Finally, because ≤=

⋃C , ≤ |D′×D′ =≤′. This

The Axiom of Choice

Proof (cont.). Now let ≤′∈ C and let D′ be the domain of ≤′.Clearly, D′ ⊆ D.

Let d ∈ D\D′ and let d′ ∈ D′. There is a≤′′∈ C with domain D′′ so that ≤′v≤′′ and d ∈ D′′ \D′. Butthen d ≥′′ d′, which means d > d′. Hence d is a strict ≤-upperbound of D′. Finally, because ≤=

⋃C , ≤ |D′×D′ =≤′. This

The Axiom of Choice

Proof (cont.). Now let ≤′∈ C and let D′ be the domain of ≤′.Clearly, D′ ⊆ D. Let d ∈ D\D′ and let d′ ∈ D′.

There is a≤′′∈ C with domain D′′ so that ≤′v≤′′ and d ∈ D′′ \D′. Butthen d ≥′′ d′, which means d > d′. Hence d is a strict ≤-upperbound of D′. Finally, because ≤=

⋃C , ≤ |D′×D′ =≤′. This

The Axiom of Choice

Proof (cont.). Now let ≤′∈ C and let D′ be the domain of ≤′.Clearly, D′ ⊆ D. Let d ∈ D\D′ and let d′ ∈ D′. There is a≤′′∈ C with domain D′′ so that ≤′v≤′′ and d ∈ D′′ \D′.

Butthen d ≥′′ d′, which means d > d′. Hence d is a strict ≤-upperbound of D′. Finally, because ≤=

⋃C , ≤ |D′×D′ =≤′. This

The Axiom of Choice

Proof (cont.). Now let ≤′∈ C and let D′ be the domain of ≤′.Clearly, D′ ⊆ D. Let d ∈ D\D′ and let d′ ∈ D′. There is a≤′′∈ C with domain D′′ so that ≤′v≤′′ and d ∈ D′′ \D′. Butthen d ≥′′ d′, which means d > d′.

Hence d is a strict ≤-upperbound of D′. Finally, because ≤=

⋃C , ≤ |D′×D′ =≤′. This

The Axiom of Choice

Proof (cont.). Now let ≤′∈ C and let D′ be the domain of ≤′.Clearly, D′ ⊆ D. Let d ∈ D\D′ and let d′ ∈ D′. There is a≤′′∈ C with domain D′′ so that ≤′v≤′′ and d ∈ D′′ \D′. Butthen d ≥′′ d′, which means d > d′. Hence d is a strict ≤-upperbound of D′.

Finally, because ≤=⋃

C , ≤ |D′×D′ =≤′. Thisdoes not establish that ≤ is an upper bound of C , because westill do not know if ≤∈ X.For ≤∈ X, let A⊆ D be a nonempty subset of D. Then there is a≤′∈ C with domain D′ so that A∩D′ 6= /0. Because ≤′ is awell-order, A∩D′ has a ≤′-smallest element a. Because≤ |D′×D′ =≤′, a is the ≤-smallest element of A∩D′. Becauseall elements of D\D′ are ≤-strict upper bounds of D′, a is the≤-smallest element of A. Therefore (simple exercise, maybe toosimple) ≤ is a well-order. Hence it is a v-upper bound of C .

The Axiom of Choice

Proof (cont.). Now let ≤′∈ C and let D′ be the domain of ≤′.Clearly, D′ ⊆ D. Let d ∈ D\D′ and let d′ ∈ D′. There is a≤′′∈ C with domain D′′ so that ≤′v≤′′ and d ∈ D′′ \D′. Butthen d ≥′′ d′, which means d > d′. Hence d is a strict ≤-upperbound of D′. Finally, because ≤=

⋃C , ≤ |D′×D′ =≤′.

Thisdoes not establish that ≤ is an upper bound of C , because westill do not know if ≤∈ X.For ≤∈ X, let A⊆ D be a nonempty subset of D. Then there is a≤′∈ C with domain D′ so that A∩D′ 6= /0. Because ≤′ is awell-order, A∩D′ has a ≤′-smallest element a. Because≤ |D′×D′ =≤′, a is the ≤-smallest element of A∩D′. Becauseall elements of D\D′ are ≤-strict upper bounds of D′, a is the≤-smallest element of A. Therefore (simple exercise, maybe toosimple) ≤ is a well-order. Hence it is a v-upper bound of C .

The Axiom of Choice

⋃C , ≤ |D′×D′ =≤′. This

does not establish that ≤ is an upper bound of C , because westill do not know if ≤∈ X.

For ≤∈ X, let A⊆ D be a nonempty subset of D. Then there is a≤′∈ C with domain D′ so that A∩D′ 6= /0. Because ≤′ is awell-order, A∩D′ has a ≤′-smallest element a. Because≤ |D′×D′ =≤′, a is the ≤-smallest element of A∩D′. Becauseall elements of D\D′ are ≤-strict upper bounds of D′, a is the≤-smallest element of A. Therefore (simple exercise, maybe toosimple) ≤ is a well-order. Hence it is a v-upper bound of C .

The Axiom of Choice

⋃C , ≤ |D′×D′ =≤′. This

does not establish that ≤ is an upper bound of C , because westill do not know if ≤∈ X.For ≤∈ X, let A⊆ D be a nonempty subset of D.

Then there is a≤′∈ C with domain D′ so that A∩D′ 6= /0. Because ≤′ is awell-order, A∩D′ has a ≤′-smallest element a. Because≤ |D′×D′ =≤′, a is the ≤-smallest element of A∩D′. Becauseall elements of D\D′ are ≤-strict upper bounds of D′, a is the≤-smallest element of A. Therefore (simple exercise, maybe toosimple) ≤ is a well-order. Hence it is a v-upper bound of C .

The Axiom of Choice

⋃C , ≤ |D′×D′ =≤′. This

does not establish that ≤ is an upper bound of C , because westill do not know if ≤∈ X.For ≤∈ X, let A⊆ D be a nonempty subset of D. Then there is a≤′∈ C with domain D′ so that A∩D′ 6= /0.

Because ≤′ is awell-order, A∩D′ has a ≤′-smallest element a. Because≤ |D′×D′ =≤′, a is the ≤-smallest element of A∩D′. Becauseall elements of D\D′ are ≤-strict upper bounds of D′, a is the≤-smallest element of A. Therefore (simple exercise, maybe toosimple) ≤ is a well-order. Hence it is a v-upper bound of C .

The Axiom of Choice

⋃C , ≤ |D′×D′ =≤′. This

does not establish that ≤ is an upper bound of C , because westill do not know if ≤∈ X.For ≤∈ X, let A⊆ D be a nonempty subset of D. Then there is a≤′∈ C with domain D′ so that A∩D′ 6= /0. Because ≤′ is awell-order, A∩D′ has a ≤′-smallest element a.

Because≤ |D′×D′ =≤′, a is the ≤-smallest element of A∩D′. Becauseall elements of D\D′ are ≤-strict upper bounds of D′, a is the≤-smallest element of A. Therefore (simple exercise, maybe toosimple) ≤ is a well-order. Hence it is a v-upper bound of C .

The Axiom of Choice

⋃C , ≤ |D′×D′ =≤′. This

does not establish that ≤ is an upper bound of C , because westill do not know if ≤∈ X.For ≤∈ X, let A⊆ D be a nonempty subset of D. Then there is a≤′∈ C with domain D′ so that A∩D′ 6= /0. Because ≤′ is awell-order, A∩D′ has a ≤′-smallest element a. Because≤ |D′×D′ =≤′, a is the ≤-smallest element of A∩D′.

Becauseall elements of D\D′ are ≤-strict upper bounds of D′, a is the≤-smallest element of A. Therefore (simple exercise, maybe toosimple) ≤ is a well-order. Hence it is a v-upper bound of C .

The Axiom of Choice

⋃C , ≤ |D′×D′ =≤′. This

does not establish that ≤ is an upper bound of C , because westill do not know if ≤∈ X.For ≤∈ X, let A⊆ D be a nonempty subset of D. Then there is a≤′∈ C with domain D′ so that A∩D′ 6= /0. Because ≤′ is awell-order, A∩D′ has a ≤′-smallest element a. Because≤ |D′×D′ =≤′, a is the ≤-smallest element of A∩D′. Becauseall elements of D\D′ are ≤-strict upper bounds of D′, a is the≤-smallest element of A.

Therefore (simple exercise, maybe toosimple) ≤ is a well-order. Hence it is a v-upper bound of C .

The Axiom of Choice

⋃C , ≤ |D′×D′ =≤′. This

does not establish that ≤ is an upper bound of C , because westill do not know if ≤∈ X.For ≤∈ X, let A⊆ D be a nonempty subset of D. Then there is a≤′∈ C with domain D′ so that A∩D′ 6= /0. Because ≤′ is awell-order, A∩D′ has a ≤′-smallest element a. Because≤ |D′×D′ =≤′, a is the ≤-smallest element of A∩D′. Becauseall elements of D\D′ are ≤-strict upper bounds of D′, a is the≤-smallest element of A. Therefore (simple exercise, maybe toosimple) ≤ is a well-order.

Hence it is a v-upper bound of C .

The Axiom of Choice

⋃C , ≤ |D′×D′ =≤′. This

The Axiom of Choice

Proof (concl.).

By Zorn’s Lemma, X has a v-maximal element≤. Then ≤ is a well-order with domain D. Suppose for acontradiction that D 6= S and let s ∈ D\S. Define ≤′ to be anorder relation on D∪{s} so that ≤′ |D×D =≤ and so that s is astrict ≤′-upper bound of D. Then ≤′∈ X is a strict v-upperbound of ≤, contradicting the maximality of ≤. Hence ≤ mustbe a well-order for S.

The Axiom of Choice

Proof (concl.). By Zorn’s Lemma, X has a v-maximal element≤.

Then ≤ is a well-order with domain D. Suppose for acontradiction that D 6= S and let s ∈ D\S. Define ≤′ to be anorder relation on D∪{s} so that ≤′ |D×D =≤ and so that s is astrict ≤′-upper bound of D. Then ≤′∈ X is a strict v-upperbound of ≤, contradicting the maximality of ≤. Hence ≤ mustbe a well-order for S.

The Axiom of Choice

Proof (concl.). By Zorn’s Lemma, X has a v-maximal element≤. Then ≤ is a well-order with domain D.

Suppose for acontradiction that D 6= S and let s ∈ D\S. Define ≤′ to be anorder relation on D∪{s} so that ≤′ |D×D =≤ and so that s is astrict ≤′-upper bound of D. Then ≤′∈ X is a strict v-upperbound of ≤, contradicting the maximality of ≤. Hence ≤ mustbe a well-order for S.

The Axiom of Choice

Proof (concl.). By Zorn’s Lemma, X has a v-maximal element≤. Then ≤ is a well-order with domain D. Suppose for acontradiction that D 6= S and let s ∈ D\S.

Define ≤′ to be anorder relation on D∪{s} so that ≤′ |D×D =≤ and so that s is astrict ≤′-upper bound of D. Then ≤′∈ X is a strict v-upperbound of ≤, contradicting the maximality of ≤. Hence ≤ mustbe a well-order for S.

The Axiom of Choice

Proof (concl.). By Zorn’s Lemma, X has a v-maximal element≤. Then ≤ is a well-order with domain D. Suppose for acontradiction that D 6= S and let s ∈ D\S. Define ≤′ to be anorder relation on D∪{s} so that ≤′ |D×D =≤ and so that s is astrict ≤′-upper bound of D.

Then ≤′∈ X is a strict v-upperbound of ≤, contradicting the maximality of ≤. Hence ≤ mustbe a well-order for S.

The Axiom of Choice

Proof (concl.). By Zorn’s Lemma, X has a v-maximal element≤. Then ≤ is a well-order with domain D. Suppose for acontradiction that D 6= S and let s ∈ D\S. Define ≤′ to be anorder relation on D∪{s} so that ≤′ |D×D =≤ and so that s is astrict ≤′-upper bound of D. Then ≤′∈ X is a strict v-upperbound of ≤, contradicting the maximality of ≤.

Hence ≤ mustbe a well-order for S.

The Axiom of Choice

Proof (concl.). By Zorn’s Lemma, X has a v-maximal element≤. Then ≤ is a well-order with domain D. Suppose for acontradiction that D 6= S and let s ∈ D\S. Define ≤′ to be anorder relation on D∪{s} so that ≤′ |D×D =≤ and so that s is astrict ≤′-upper bound of D. Then ≤′∈ X is a strict v-upperbound of ≤, contradicting the maximality of ≤. Hence ≤ mustbe a well-order for S.

The Axiom of Choice

Proof (concl.). By Zorn’s Lemma, X has a v-maximal element≤. Then ≤ is a well-order with domain D. Suppose for acontradiction that D 6= S and let s ∈ D\S. Define ≤′ to be anorder relation on D∪{s} so that ≤′ |D×D =≤ and so that s is astrict ≤′-upper bound of D. Then ≤′∈ X is a strict v-upperbound of ≤, contradicting the maximality of ≤. Hence ≤ mustbe a well-order for S.

The Axiom of Choice

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