The Axiom of Choice and the Banach-Tarski Paradox · The Axiom of Choice and the Banach-Tarski Paradox Ben Babcock Department of Mathematical Sciences Lakehead University March 2011

The Axiom of Choice and the Banach-TarskiParadox

Ben Babcock

Department of Mathematical SciencesLakehead University

March 2011

Ben Babcock The Axiom of Choice and the Banach-Tarski Paradox

The Axiom of Choice

Axiom of Choice (Suppes 1960)For any set A there is a function f such that for any non-emptysubset B of A, f (B) ∈ B.

Choosing a shoe from each of an infinite set of pairs of shoes does not require theaxiom of choice: just choose the left shoe.

Choosing a sock from each of an infinite set of pairs of socks requires the axiom ofchoice.


What makes AC so special?

AC is consistent with ZF (Gödel 1935). Thus, if the axiomsof ZF do not result in any contradictions, ZFC is also free ofcontradictions.AC is independent of ZF (Gödel, Cohen 1963). Thus, ACcannot be proved true or false using the axioms of ZF.AC is non-constructive. It guarantees the existence of achoice function but provides no method for constructingsuch a function.AC can be used to prove counter-intuitive results, like theBanach-Tarski paradox.


Rigid Motions and Congruence in R3

Definition

A rigid motion is a mapping from R3 to R3 that is a translation,rotation, or combination of the two (but not a reflection).If a set Y is an image of a set X under some rigid motion, thenwe say X is congruent to Y .

DefinitionA partition of a set X is a collection of pairwise disjoint sets{Xi}ni=1, for some n ∈ N, such that X =

⋃ni=1{Xi}.


Equidecomposability

DefinitionTwo sets X ,Y are equidecomposable, denoted X ≈ Y , if andonly if there exist partitions {Xi}ni=1, {Yi}ni=1 such thatX =

⋃ni=1{Xi}, Y =

⋃ni=1{Yi}, and Xi is congruent to Yi for

each i = 1, . . . ,n.

The above two figures are equidecomposable.


Equidecomposability

DefinitionTwo sets X ,Y are equidecomposable, denoted X ≈ Y , if andonly if there exist partitions {Xi}ni=1, {Yi}ni=1 such thatX =

⋃ni=1{Xi}, Y =

⋃ni=1{Yi}, and Xi is congruent to Yi for

each i = 1, . . . ,n.

LemmaThe following hold:

1 ≈ is an equivalence relation.2 If X and Y are disjoint unions of X1,X2 and Y1,Y2,

respectively, and if Xi ≈ Yi for each i = 1,2, then X ≈ Y.3 If X1 ⊆ Y ⊆ X with X ≈ X1, then X ≈ Y.


The Banach-Tarski Paradox

Theorem (The Banach-Tarski Paradox, 1924)

Let U be a closed ball. Then there exists a partition {X ,Y} of Usuch that U ≈ X and U ≈ Y.

A “strong form” of the paradox that applies to any closedregion with non-empty interior.The decomposition uses only finitely many pieces.Holds in more than three dimensions.Holds in one and two dimensions only if countably manypieces are used.


The Secret Ingredient: Hausdorff’s Paradox

Theorem (Hausdorff’s Paradox, 1914)A sphere S can be decomposed into disjoint sets

S = A ∪ B ∪ C ∪Q

such that:1 the sets A,B,C are congruent to each other;2 the set B ∪ C is congruent to each of the sets A,B,C;3 Q is countable.


Take Two Rotations and Call Hausdorff in the Morning

Let ϕ be a rotation of 180◦ aboutsome axis aϕ.Let ψ be a rotation of 120◦ aboutsome axis aψ.Then ϕ2 = ψ3 = 1.Let G be the free product of thegroups {1, ϕ} and {1, ψ, ψ2}.Elements of G (words) look like this:

ϕψ±1 · · ·ϕψ±1

(since ψ2 = ψ−1).


A Partition of G

We construct a partition {G1,G2,G3} of G such that

G1 · ϕ = G2 ∪G3, G1 · ψ = G2, G1 · ψ2 = G3.

Let 1 ∈ G1, ϕ, ψ ∈ G2, ψ2 ∈ G3.Let α ∈ G. We assign αϕ and αψ to partitions based oncases.

1 α ends with ψ or ψ2

If α ∈ G1, put αϕ into G2.Otherwise, put αϕ into G1.

2 α ends with ϕIf α ∈ Gi , put αψ into Gi+1(mod 3) and αψ2 ∈ Gi+2(mod 3).


Fixed Points and Orbits

DefinitionFor a rotation α ∈ G, a point x ∈ S is fixed if x · α = x .

Let Q be the set of all fixed points under all rotations of G.Each rotation has two fixed points (one at either “pole”).Thus, Q is countable.

DefinitionFor each x ∈ S \Q, the orbit of x , Px , is the image of x underevery rotation in G:

Px = {x · α | α ∈ G}.


Invoking the Axiom of Choice

We can show that the orbits Px for all x ∈ S \Q form apartition of S \Q.Let M be the set containing exactly one point from eachorbit.Each orbit contains uncountably many points. Thecollection of these orbits is more like a collection of pairs ofsocks than pairs of shoes.So, to guarantee that M exists, we need the axiom ofchoice.


The Secret Ingredient: Hausdorff’s Paradox

Theorem (Hausdorff’s Paradox, 1914)A sphere S can be decomposed into disjoint sets

S = A ∪ B ∪ C ∪Q

such that:1 the sets A,B,C are congruent to each other;2 the set B ∪ C is congruent to each of the sets A,B,C;3 Q is countable.

We already have Q, the set of all fixed points of rotations inG. Now we just need A, B, and C.


A Decomposition of the Sphere

Let A = M ·G1,B = M ·G2,C = M ·G3.We’ve used M to partition S \Q according to our partitionof G.Recall

G1 · ϕ = G2 ∪G3, G1 · ψ = G2, G1 · ψ2 = G3,

soA · ϕ = B ∪ C, A · ψ = B, A · ψ2 = C.

Hence, A,B,C, and B ∪C are congruent to each other, and

S = A ∪ B ∪ C ∪Q.


Decomposing the Closed Ball

Theorem (The Banach-Tarski Paradox)

Let U be a closed ball. Then there exists a partition {X ,Y} of Usuch that U ≈ X and U ≈ Y.

Let X ⊂ S. Let X denote the set of all x ∈ U, other thanthe centre, such that the projection of x onto the surface isin X .Then from S = A ∪ B ∪ C ∪Q we obtain

U = A ∪ B ∪ C ∪ Q ∪ {c},

where c is the centre.


Decomposing the Closed Ball

S = A ∪ B ∪ C ∪Q, U = A ∪ B ∪ C ∪ Q ∪ {c}

From the congruence of A,B,C and B ∪ C, we have

A ≈ B ≈ C ≈ B ∪ C.

Let X = A ∪Q ∪ {c} and Y = U \ X .

Lemma(2)If X and Y are disjoint unions of X1,X2 and Y1,Y2, respectively,and if Xi ≈ Yi for each i = 1,2, then X ≈ Y .

So A ≈ A ∪ B ∪ C, and thus

X ≈ U.


Showing Y ≈ U

S = A ∪ B ∪ C ∪Q, U = A ∪ B ∪ C ∪ Q ∪ {c}

X = A ∪Q ∪ {c}, Y = U \ X

We have partitioned the ball U into X and Y and shownU ≈ X . Now we have to show U ≈ Y .Start by taking a rotation τ /∈ G such that Q ∩ (Q · τ) = ∅No fixed point of any rotation in G is fixed under τ , so

Q · τ ⊆ A ∪ B ∪ C.


Showing Y ≈ U

S = A ∪ B ∪ C ∪Q, U = A ∪ B ∪ C ∪ Q ∪ {c}

X = A ∪Q ∪ {c}, Y = U \ X

No fixed point of any rotation in G is fixed under τ , so

Q · τ ⊆ A ∪ B ∪ C,

Also, C ≈ A ∪ B ∪ C.Then there exists T ⊆ C such that T is congruent to Q · τand thus to Q. Hence, T ≈ Q.


Showing Y ≈ U

S = A ∪ B ∪ C ∪Q, U = A ∪ B ∪ C ∪ Q ∪ {c}

X = A ∪Q ∪ {c}, Y = U \ X

Take any p ∈ C \ T . Then:

A ≈ B, Q ≈ T , {c} ≈ {p},

soA ∪Q ∪ {c} ≈ B ∪ T ∪ {p}.


Showing Y ≈ U

S = A ∪ B ∪ C ∪Q, U = A ∪ B ∪ C ∪ Q ∪ {c}

X = A ∪Q ∪ {c}, Y = U \ X

Lemma(3)If X1 ⊆ Y ⊆ X with X ≈ X1, then X ≈ Y .

We have X ≈ B ∪ T ∪ {p} ⊆ Y ⊂ U, so

Y ≈ U.


Is the Axiom of Choice worth the trouble?

Before you choose to reject AC, remember that it isequivalent to:

the Well-Ordering Theoremthe Numeration Theoremthe Law of TrichotomyZorn’s Lemma

all of which are important and useful theorems with wideapplications.

xkcd #804 demonstrates how the Banach-Tarski paradox applies to pumpkin carving.


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The Axiom of Choice and the Banach-Tarski Paradox · The Axiom of Choice and the Banach-Tarski Paradox Ben Babcock Department of Mathematical Sciences Lakehead University March 2011