The axiom of choice How (not) to choose in nitely many socks Regula Krapf

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The axiom of choice

How (not) to choose innitely many socks

Regula Krapf

University of Bonn

April 27, 2016

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Content

1 What is choice and (why) do we need it?

2 Permutation models

3 The second Fraenkel model

4 Homework

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Motivation

The axiom of choice is necessary to select a set from an innite number of socks but not an innite number of shoes.

Bertrand Russell

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The axiom of choice

The axiom of choice AC states the following:

∀x[∅∈/ x → ∃f :x →[

x(∀y ∈x(f(y)∈y))].

Or, in a less cryptic way,

If (X_i)_i∈I is a family of non-empty sets, then there is a family (y_i)_i∈I such that y_i ∈X_i for every i ∈I .

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The axiom of choice

The axiom of choice AC states the following:

∀x[∅∈/ x → ∃f :x →[

x(∀y ∈x(f(y)∈y))].

Or, in a less cryptic way,

If (X_i)_i∈I is a family of non-empty sets, then there is a family (y_i)_i∈I such that y_i ∈X_i for every i ∈I .

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The axiom of choice

The Axiom of Choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn's lemma?

Jerry Bona

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Some equivalences of the axiom of choice

Theorem

The following statements are equivalent.

1 The axiom of choice.

2 The well-ordering principle.

3 Zorn's lemma.

4 Every vector space has a basis.

5 Every non-trivial unital ring has a maximal ideal.

6 Tychono's theorem.

7 Every connected graph has a spanning tree.

8 Every surjective map has a right inverse.

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Some equivalences of the axiom of choice

Theorem

The following statements are equivalent.

1 The axiom of choice.

2 The well-ordering principle.

3 Zorn's lemma.

4 Every vector space has a basis.

5 Every non-trivial unital ring has a maximal ideal.

6 Tychono's theorem.

7 Every connected graph has a spanning tree.

8 Every surjective map has a right inverse.

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But the axiom of choice implies weird things...

... so maybe it would be nice to have a weaker choice principle?

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But the axiom of choice implies weird things...

... so maybe it would be nice to have a weaker choice principle?

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Weaker forms of choice

The following choice principles are strictly weaker then AC.

ACℵ₀ Every countable family of non-empty sets has a choice function (Axiom of countable choice).

AC^<ℵ_ℵ₀⁰ Every countable family of non-empty nite sets has a choice

function.

ACⁿℵ₀ Every family of countably many n-element sets has a choice function.

KL König's lemma

RPP Ramsey's partition principle

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König's lemma

Theorem (Dénes König, 1927)

Every nitely branching tree that contains innitely many vertices has an innte branch.

It was shown by Pincus in 1972 that König's lemma is equivalent to the axiom AC^<ℵ_ℵ₀⁰.

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König's lemma

Theorem (Dénes König, 1927)

Every nitely branching tree that contains innitely many vertices has an innte branch.

It was shown by Pincus in 1972 that König's lemma is equivalent to the axiom AC^<ℵ_ℵ₀⁰.

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Ramsey's partition principle

A 2-coloring of a set X is a map f : [X]² → {0,1}. A subset Y of X is said to be monochromatic (or homogeneous) with resprect to f , if f [Y]² is constant.

Theorem (Ramsey's partition principle)

If f : [X]→ {0,1} is a 2-coloring of an innite set X , then there is an innite subset Y of X which is monochromatic.

It holds that ACℵ₀ ⇒RPP⇒KL.

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Ramsey's partition principle

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Ramsey's partition principle

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Stronger forms of choice

In class theory, one often uses the axiom of global choice:

There is a class function F :V\ {∅} →V such that F(x)∈x for every set x ∈V\ ∅.

This is equivalent to postulating that there is a global well-order of the set-theoretic universe V.

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Stronger forms of choice

In class theory, one often uses the axiom of global choice:

There is a class function F :V\ {∅} →V such that F(x)∈x for every set x ∈V\ ∅.

This is equivalent to postulating that there is a global well-order of the set-theoretic universe V.

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Set theory with atoms (ZFA)

The language of ZFA is given by {∈,A}. The axioms are given by the axioms of ZF with a modied version of the axiom of the empty set and the extensionality axiom

∃x[x ∈/ A∧ ∀y(y ∈/ x)]

∀x,y[(x,y ∈/A)→ ∀z(z ∈x ↔z ∈y)→x =y] and the additional axiom of atoms given by

∀x[x ∈A↔(x 6=∅ ∧ ¬∃y(y ∈x))].

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Models of set theory with atoms

For a set S we dene a hierarchy by P⁰(S) =S

P^α+¹(S) =P(P^α(S)) P^α(S) = [

β<α

P^β(S), β limit

P^∞(S) = [

α∈Ord

P^α(S).

Then V ^∞ A is a model of ZFA,Vˆ ^∞ is a model of ZF.

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Models of set theory with atoms

For a set S we dene a hierarchy by P⁰(S) =S

P^α+¹(S) =P(P^α(S)) P^α(S) = [

β<α

P^β(S), β limit

P^∞(S) = [

α∈Ord

P^α(S).

Then V ^∞ A is a model of ZFA,Vˆ ^∞ is a model of ZF.

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Normal lters of permutation groups

Let G be a group of permutations of A.

Denition

LetF be a set of subgroups of G. ThenF is said to be a normal lter on G, if for all subgroups H,K of G the following statements hold.

(A) G ∈ F

(B) H ∈ F and H ⊆K ⇒K ∈ F (C) H,K ∈ F ⇒H∩K ∈ F

(D) π∈G and H∈ F ⇒πHπ⁻¹ ∈ F (E) for all a∈A, {π ∈G |πa=a} ∈ F.

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Normal lters of permutation groups

Denition

For a subset E ⊆A we dene

x_G(E) ={π ∈G |πa=a for all a∈E}.

Then the lter F generated by {x_G(E)|E ⊆A nite}, i.e.

H ∈ F ⇐⇒ ∃E ⊆A nite such that x_G(E)⊆H is a normal lter on G, denoted F_n.

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Normal lters of permutation groups

Denition

For a subset E ⊆A we dene

x_G(E) ={π ∈G |πa=a for all a∈E}.

Then the lter F generated by {x_G(E)|E ⊆A nite}, i.e.

H ∈ F ⇐⇒ ∃E ⊆A nite such that x_G(E)⊆H is a normal lter on G, denoted F_n.

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By transnite induction we dene for every set x and for everyπ ∈G the set πx by

πx =







∅ if x=∅,

πx if x∈A,

{πy |y ∈x} otherwise.

For every set x we dene

sym_G(x) ={π ∈G |πx =x}.

Denition

Let F be a normal lter on G. A set x is said to be

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πx =







∅ if x=∅,

πx if x∈A,

Denition

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πx =







∅ if x=∅,

πx if x∈A,

Denition

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Symmetric sets

Lemma

The following statements hold.

1 Every atom a∈A is symmetric.

2 A set x is hereditarily symmetric if and only if for every π∈G,πx is hereditarily symmetric.

3 For every set x ∈V and for everyˆ π∈G,πx =x.

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Symmetric sets

Lemma

A set x is symmetric if and only if there is a nite set E ⊆A such that x_G(E)⊆sym_G(x). Such a set E is said to be a support of x.

Note that if E is a support of x then every nite set F with E ⊆F ⊆A is also a support of x.

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Permutation models

Denition

Let F be a normal lter on a group G of permutations of A. Then the class VF of all hereditarily symmetric sets in V=P^∞(A) is a model of ZFA called a permutation model.

We have A∈VF andVˆ ⊆VF. Theorem (Jech-Sochor)

Permutation models can always be embedded into models of ZF.

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Permutation models

Denition

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Permutation models

Denition

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The second Fraenkel model

We now construct a specic permutation model with atoms given by A= [

n∈ω

Pn,

where P_n={a_n,b_n} consists of two elements for all n∈ω and Pn∩Pm=∅for n6=m.

Let G be the group of all permutations π of A such that πP_n=P_n for all n∈ω, and letF =F_n. We call the corresponding permutation model the second Fraenkel model V_F₂.

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The second Fraenkel model

We now construct a specic permutation model with atoms given by A= [

n∈ω

Pn,

where P_n={a_n,b_n} consists of two elements for all n∈ω and Pn∩Pm=∅for n6=m.

Let G be the group of all permutations π of A such that πP_n=P_n for all n∈ω, and letF =F_n. We call the corresponding permutation model the second Fraenkel model V_F₂.

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The second Fraenkel model

Lemma

The following statements hold in V_F₂.

1 For all n∈ω, P_n∈V_F₂.

2 {P_n|n∈ω} ∈V_F₂.

Proof.

1 By denition we haveπPn=Pn for allπ∈G, so Pn is symmetric. Since sym_G(an),sym_G(bn)∈ F, Pnis also hereditarily symmetric.

2 For everyπ∈G we have

π{P_n|n∈ω}={πP_n|n∈ω}={P_n|n∈ω},

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The second Fraenkel model

Lemma

2 {P_n|n∈ω} ∈V_F₂.

Proof.

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The second Fraenkel model

Lemma

2 {P_n|n∈ω} ∈V_F₂.

Proof.

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A failure of AC

²ℵ₀

Theorem

In V_F₂ the axiom AC²ℵ₀ fails.

Proof.

We show that there is no choice function on {Pn|n ∈ω}. Suppose for a contradiction that f :ω→S

n∈ωP_n is a choice function, i.e. f(n)∈P_n for all n∈ω. Let E_f be a support of f . WLOG we may assume that E is of the form E ={a0,b0, . . . ,a_k,b_k}for some k ∈ω. Letπ∈xG(E) with πa_k+1 =b_k+1. Since π ∈x_G(E)⊆sym_G(f) we haveπf =f and hence

πhk+1,f(k+1)i=hπ(k+1), πf(k+1)i=hk+1, πf(k+1)i ∈f,

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A failure of AC

²ℵ₀

Theorem

Proof.

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A failure of AC

²ℵ₀

Theorem

Proof.

n∈ωP_n is a choice function, i.e. f(n)∈P_n for all n∈ω. Let E_f be a support of f . WLOG we may assume that E is of the form E ={a0,b0, . . . ,a_k,b_k}for some k ∈ω. Letπ∈xG(E) with πa_k+1 =b_k+1. Sinceπ ∈x_G(E)⊆sym_G(f) we haveπf =f and hence

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A failure of AC

²ℵ₀

Theorem

Proof.

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A failure of König's lemma

Theorem

In V_F₂ there is an innite binary tree which does not have an innite branch. In particular, in V_F₂ König's lemma fails.

Proof.

For every n∈ω consider

V_n={s:{0, . . . ,n−1} →A| ∀i<n[s(i)∈P_i]}.

The vertices are given by V =S

n∈ωV_n. Furthermore, we dene s ≺t i there is n∈ωsuch that s ∈V_n,t ∈V_n+1and t extends s. Then T =hV,≺iis an innite binary tree with the property that if s ∈V , then s∪ {hn,a_ni},s∪ {hn,b_ni} ∈V .

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A failure of König's lemma

Theorem

Proof.

V_n={s:{0, . . . ,n−1} →A| ∀i<n[s(i)∈P_i]}.

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A failure of König's lemma

Theorem

Proof.

V_n={s:{0, . . . ,n−1} →A| ∀i<n[s(i)∈P_i]}.

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A failure of König's lemma

Theorem

Proof.

V_n={s:{0, . . . ,n−1} →A| ∀i<n[s(i)∈P_i]}.

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A failure of the partition principle

Theorem

In V_F₂ there is a 2-coloring of[A]² such that no innite subset of A is homogeneous.

Proof.

Consider f : [A]² → {0,1}given by

f({a,b}) =

(1 if{a,b}=P_n for some n∈ω, 0 otherwise.

Suppose that B ⊆A is an innite homogeneous set. Clearly, f [B]² ≡0.

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A failure of the partition principle

Theorem

Proof.

f({a,b}) =

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A failure of the partition principle

Theorem

Proof.

f({a,b}) =

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Further statements that are consistent in the absence of AC

All sets of reals are Lebesgue measurable.

There is a countable union of countable sets which is not countable.

The reals cannot be well-ordered.

The Baire category theorem fails.

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Further statements that are consistent in the absence of AC

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Further statements that are consistent in the absence of AC

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Further statements that are consistent in the absence of AC

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Homework

Innitely many dwarves are standing in a straight line. Every dwarf wears a hat of color either red or blue and sees the color of the hats of all the dwarves standing in front of him. There is explicitly a rst dwarf, who has to start guessing the color of his hat and then the guessing proceeds with the next one in the line.

If a dwarf guessed correctly, it is freed; if he guessed wrong, it is fried.

Every dwarf can hear the voice of all other dwarves without a problem.

Everybody is only allowed to speak out either the color red or blue, but no further information.