The axiom of choice - uni-bonn.de · What is choice and (why) do we need it? Some equivalences of...

The axiom of choiceHow (not) to choose in�nitely many socks

Regula Krapf

University of Bonn

April 27, 2016

Regula Krapf (University of Bonn) The axiom of choice April 27, 2016 1 / 26

Content

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

2 Permutation models

3 The second Fraenkel model

4 Homework


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

Motivation

The axiom of choice is necessary to select a set from an in�nite number of

socks but not an in�nite number of shoes.

Bertrand Russell



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 (Xi )i∈I is a family of non-empty sets, then there is a family (yi )i∈I suchthat yi ∈ Xi for every i ∈ I .



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 (Xi )i∈I is a family of non-empty sets, then there is a family (yi )i∈I suchthat yi ∈ Xi for every i ∈ I .



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



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.

Proof.

Left as an excercise.



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.

Proof.

Left as an excercise.



But the axiom of choice implies weird things...

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



But the axiom of choice implies weird things...

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



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<ℵ0ℵ0 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



König's lemma

Theorem (Dénes König, 1927)

Every �nitely branching tree that contains in�nitely many vertices has an

in�nte branch.

It was shown by Pincus in 1972 that König's lemma is equivalent to the

axiom AC<ℵ0ℵ0 .



König's lemma

Theorem (Dénes König, 1927)

Every �nitely branching tree that contains in�nitely many vertices has an

in�nte branch.

It was shown by Pincus in 1972 that König's lemma is equivalent to the

axiom AC<ℵ0ℵ0 .



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 in�nite set X , then there is an

in�nite subset Y of X which is monochromatic.

It holds that ACℵ0 ⇒ RPP⇒ KL.























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.



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.


Permutation models

Set theory with atoms (ZFA)

The language of ZFA is given by {∈,A}. The axioms are given by the

axioms of ZF with a modi�ed version of the axiom of the empty set andthe 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))].


Permutation models

Models of set theory with atoms

For a set S we de�ne a hierarchy by

P0(S) = S

Pα+1(S) = P(Pα(S))

Pα(S) =⋃β<α

Pβ(S), β limit

P∞(S) =⋃

α∈OrdPα(S).

Then V = P∞(A) is a model of ZFA, V̂ = P∞(∅) is a model of ZF.


Permutation models

Models of set theory with atoms

For a set S we de�ne a hierarchy by

P0(S) = S

Pα+1(S) = P(Pα(S))

Pα(S) =⋃β<α

Pβ(S), β limit

P∞(S) =⋃

α∈OrdPα(S).

Then V = P∞(A) is a model of ZFA, V̂ = P∞(∅) is a model of ZF.


Permutation models

Normal �lters of permutation groups

Let G be a group of permutations of A.

De�nition

Let F be a set of subgroups of G . Then F is said to be a normal �lter onG , 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 .


Permutation models


De�nition

For a subset E ⊆ A we de�ne

�xG (E ) = {π ∈ G | πa = a for all a ∈ E}.

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

H ∈ F ⇐⇒ ∃E ⊆ A �nite such that �xG (E ) ⊆ H

is a normal �lter on G , denoted F�n.

From now on, let F = F�n.


Permutation models


De�nition

For a subset E ⊆ A we de�ne

�xG (E ) = {π ∈ G | πa = a for all a ∈ E}.

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

H ∈ F ⇐⇒ ∃E ⊆ A �nite such that �xG (E ) ⊆ H

is a normal �lter on G , denoted F�n.

From now on, let F = F�n.


Permutation models

By trans�nite induction we de�ne for every set x and for every π ∈ G the

set πx by

πx =

∅ if x = ∅,πx if x ∈ A,

{πy | y ∈ x} otherwise.

For every set x we de�ne

symG (x) = {π ∈ G | πx = x}.

De�nition

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

symmetric, if symG (x) ∈ F .hereditarily symmetric, if x is symmetric and every element of x is

hereditarily symmetric.


Permutation models


set πx by

πx =

∅ if x = ∅,πx if x ∈ A,



symG (x) = {π ∈ G | πx = x}.

De�nition





Permutation models


set πx by

πx =

∅ if x = ∅,πx if x ∈ A,



symG (x) = {π ∈ G | πx = x}.

De�nition





Permutation models

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


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


Permutation models

Symmetric sets

Lemma

A set x is symmetric if and only if there is a �nite set E ⊆ A such that

�xG (E ) ⊆ symG (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 .


Permutation models

Permutation models

De�nition

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 and V̂ ⊆ VF .

Theorem (Jech-Sochor)

Permutation models can always be embedded into models of ZF.


Permutation models

Permutation models

De�nition








Permutation models

Permutation models

De�nition








The second Fraenkel model


We now construct a speci�c permutation model with atoms given by

A =⋃n∈ω

Pn,

where Pn = {an, bn} consists of two elements for all n ∈ ω and

Pn ∩ Pm = ∅ for n 6= m.

Let G be the group of all permutations π of A such that πPn = Pn for all

n ∈ ω, and let F = F�n. We call the corresponding permutation model the

second Fraenkel model VF2 .




We now construct a speci�c permutation model with atoms given by

A =⋃n∈ω

Pn,

where Pn = {an, bn} consists of two elements for all n ∈ ω and

Pn ∩ Pm = ∅ for n 6= m.

Let G be the group of all permutations π of A such that πPn = Pn for all

n ∈ ω, and let F = F�n. We call the corresponding permutation model the

second Fraenkel model VF2 .




Lemma

The following statements hold in VF2 .

1 For all n ∈ ω, Pn ∈ VF2 .

2 {Pn | n ∈ ω} ∈ VF2 .

Proof.

1 By de�nition we have πPn = Pn for all π ∈ G , so Pn is symmetric. SincesymG (an), symG (bn) ∈ F , Pn is also hereditarily symmetric.

2 For every π ∈ G we have

π{Pn | n ∈ ω} = {πPn | n ∈ ω} = {Pn | n ∈ ω},

so {Pn | n ∈ ω} is symmetric and by (1) also hereditarily symmetric.




Lemma



2 {Pn | n ∈ ω} ∈ VF2 .

Proof.



π{Pn | n ∈ ω} = {πPn | n ∈ ω} = {Pn | n ∈ ω},





Lemma



2 {Pn | n ∈ ω} ∈ VF2 .

Proof.



π{Pn | n ∈ ω} = {πPn | n ∈ ω} = {Pn | n ∈ ω},




A failure of AC2ℵ0

Theorem

In VF2 the axiom AC2ℵ0 fails.

Proof.

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

contradiction that f : ω →⋃

n∈ω Pn is a choice function, i.e. f (n) ∈ Pn for

all n ∈ ω. Let Ef be a support of f . WLOG we may assume that E is of

the form E = {a0, b0, . . . , ak , bk} for some k ∈ ω. Let π ∈ �xG (E ) withπak+1 = bk+1. Since π ∈ �xG (E ) ⊆ symG (f ) we have πf = f and hence

π〈k + 1, f (k + 1)〉 = 〈π(k + 1), πf (k + 1)〉 = 〈k + 1, πf (k + 1)〉 ∈ f ,

so f (k + 1) = πf (k + 1), a contradiction.




Theorem


Proof.






π〈k + 1, f (k + 1)〉 = 〈π(k + 1), πf (k + 1)〉 = 〈k + 1, πf (k + 1)〉 ∈ f ,





Theorem


Proof.






π〈k + 1, f (k + 1)〉 = 〈π(k + 1), πf (k + 1)〉 = 〈k + 1, πf (k + 1)〉 ∈ f ,





Theorem


Proof.






π〈k + 1, f (k + 1)〉 = 〈π(k + 1), πf (k + 1)〉 = 〈k + 1, πf (k + 1)〉 ∈ f ,




A failure of König's lemma

Theorem

In VF2 there is an in�nite binary tree which does not have an in�nite

branch. In particular, in VF2 König's lemma fails.

Proof.

For every n ∈ ω consider

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

The vertices are given by V =⋃

n∈ω Vn. Furthermore, we de�ne s ≺ t i� there isn ∈ ω such that s ∈ Vn, t ∈ Vn+1 and t extends s. Then T = 〈V ,≺〉 is an in�nitebinary tree with the property that if s ∈ V , then s ∪ {〈n, an〉}, s ∪ {〈n, bn〉} ∈ V .

But every in�nite branch through T would give a choice function on {Pn | n ∈ ω}contradicting our previous theorem.




Theorem



Proof.


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







Theorem



Proof.


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







Theorem



Proof.


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






A failure of the partition principle

Theorem

In VF2 there is a 2-coloring of [A]2 such that no in�nite subset of A is

homogeneous.

Proof.

Consider f : [A]2 → {0, 1} given by

f ({a, b}) =

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

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

Let E be a support of B and let k ∈ ω be such that E ∩ Pk = ∅ andB ∩ Pk 6= ∅. Then there is π ∈ �xG (E ) such that πak = bk . But then

π /∈ symG (B).




Theorem


homogeneous.

Proof.


f ({a, b}) =




π /∈ symG (B).




Theorem


homogeneous.

Proof.


f ({a, b}) =




π /∈ symG (B).



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.























Homework

Homework

In�nitely 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.

Is there a possibility for (almost) all dwarves to be freed?


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The axiom of choice - uni-bonn.de · What is choice and (why) do we need it? Some equivalences of...

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