# The Banach-Tarski Paradox

Katie Buchhorn

Supervised by Adam Sierakowski and David Robertson

### Abstract

In 1924, S. Banach and A. Tarski proved an astonishing, yet rather
counterintuitive paradox: given a solid ball in $\mathbb{R}^{3}$, it is
possible to partition it into finitely many pieces and reassemble them to form
two solid balls, each identical in size to the first.

This paper was written for the fulfilment of a _summer scholarship_ during an
undergraduate degree at the University of Wollongong. The aim of the paper is
to provide a comprehensive proof of the Banach-Tarski paradox thereby
introducing the notions of paradoxical and equidecomposable sets which are
phrased in terms of group actions. Once the reader has a firm grasp of these
ideas, the proof of the Banach-Tarski Paradox is fairly straight forward,
provided we have the Axiom of Choice at our disposal.

##  1 Paradoxical Decompositions

######  Definition 1.1.

A group $G$ is a set together with a binary operation

$\displaystyle G\times G\rightarrow$ $\displaystyle G$
$\displaystyle(g,h)\mapsto$ $\displaystyle gh$

satisfying the following axioms:

  * (i)

Closure: if $a$, $b\in G$, then $ab\in G$.

  * (ii)

Identity: There is an identity element $e\in G$ such that $ea=a=ae$, for every
$a\in G$.

  * (iii)

Inverse: There must exist an inverse for each group element,

i.e., for each $a\in G$, $\exists a^{-1}\in G$ such that $aa^{-1}=e=a^{-1}a$.

  * (iv)

Associativity: for all $a,b,c\in G$, $(ab)c=a(bc)$.

The operation with respect to which the group is formed is called the _group_
$operation$.

######  Definition 1.2.

Let $G$ be a group and let $X$ be a set. Then a (left) group action of $G$ on
$X$ is a function

$G\times X\rightarrow X,$ $(g,x)\mapsto g\cdot x$

that satisfies the following axioms:

  * (i)

Associativity: $(gh)\cdot x=g\cdot(h\cdot x)$ for all $g,h\in G$ and $x\in X$.

  * (ii)

Identity: $e\cdot a=a$ for all $x\in X$ (where $e$ denotes the identity
element in $G$).

######  Definition 1.3.

$G$ be a group acting on a set $X$ and suppose $E\subseteq X$. $E$ is _$G$
-paradoxical_ if for some positive integers $m,n$ there exists pairwise
disjoint subsets $A_{1},...,A_{n},B_{1},...,B_{m}$ of $E$ and
$g_{1},...,g_{n},h_{1},...,h_{m}\in G$ such that

$E=\bigcup_{i=1}^{n}g_{i}\cdot A_{i}\text{ and }E=\bigcup_{i=1}^{m}h_{i}\cdot
B_{i}$

A good way to interpret this is to picture $E$ containing two disjoint subsets
$\bigcup_{i=1}^{n}A_{i}$, $\bigcup_{i=1}^{m}B_{i}$ each of which can be broken
down into their namesakes $A_{i}$ or $B_{i}$ and then moved around
consecutively via elements of the group $G$ to cover all of $E$.

Note also that there is a stronger alternative; finding subsets
$A_{1},...,A_{n},B_{1},...,B_{m}$ of $E$ which in fact _partition_ $E$ (The
only extra condition being that these subsets initially union to give $E$).
This idea will be formally introduced later.

###  1.1 Free Groups

Free groups will be a recurring idea in the progression of the paper, so we
will formally introduce the concept. If $S$ is a set, the free group generated
by $S$ is the group of all reduced finite words with letters from
$\\{s,s^{-1}\mid s\in S\\}$. A word is called reduced if it contains no pairs
of adjacent letters $ss^{-1}$ or $s^{-1}s$. The group composition is
concatenation of words followed by reduction, that is, removing pairs of the
mentioned forms. The rank of a group is the size of the generating set.

######  Theorem 1.4.

A free group $\mathbb{F}_{2}$ of rank 2 is $\mathbb{F}_{2}$-paradoxical, where
$\mathbb{F}_{2}$ acts on itself by left multiplication.

###### Proof.

Let $\mathbb{F}_{2}$ be the free group with the generating set $\\{a,b\\}$ and
let $\rho\in\\{a,a^{-1},b,b^{-1}\\}$. Now let $\Psi(\rho)$ denote the set of
elements in $\mathbb{F}_{2}$ beginning with $\rho$.

Figure 1: The Cayley graph of the free group $\mathbb{F}_{2}$ on two
generators $a$ and $b$

It can be easily seen from Figure 1 that

$\mathbb{F}_{2}=\\{e\\}\bigcup\Psi(a)\bigcup\Psi(a^{-1})\bigcup\Psi(b)\bigcup\Psi(b^{-1})$
(1)

and that these subsets are pairwise disjoint. We are aiming to find group
elements of $\mathbb{F}_{2}$ that act on $\Psi(a)$ and $\Psi(a^{-1})$ to
rearrange the subsets in such a way that their union will give
$\mathbb{F}_{2}$, and similarly for $\Psi(b)$ and $\Psi(b^{-1})$.

Figure 2 The Cayley graph of the free group $\mathbb{F}_{2}$ and its subsets

For any element $h\in F\setminus\Psi(a)$, this implies that $h\notin\Psi(a)$.
So $h$ does not begin with the letter $a$. Then, if $a^{-1}$ acts on $h$ (on
the left), there will be no cancellation hence $a^{-1}\cdot h\in\Psi(a^{-1})$.

Now,

$h=(a\cdot a^{-1})\cdot h=a\cdot(a^{-1}\cdot h)\in a\cdot\Psi(a^{-1})$

So for any element in $\mathbb{F}_{2}\setminus\Psi(a)$, we can conclude that
the element must in fact be in $a\cdot\Psi(a^{-1})$. Similarly, any element in
$\mathbb{F}_{2}\setminus\Psi(b)$ implies that the element is in
$b\cdot\Psi(b^{-1})$. So,

$a\cdot\Psi(a^{-1})\cup\Psi(a)=\mathbb{F}_{2}\text{ and
}b\cdot\Psi(b^{-1})\cup\Psi(b)=\mathbb{F}_{2}$

Hence a free group $\mathbb{F}_{2}$ of rank 2 is $\mathbb{F}_{2}$-paradoxical.

∎

Notice that the following choice of subsets would have indeed _partitioned_
the free group $\mathbb{F}_{2}$:
$A_{1}=\Psi(a^{-1})\cup\\{e\\}\cup\\{a\\}\cup\\{a^{2}\\}\cup\\{a^{3}\\}...$
$A_{2}=\Psi(a)\setminus\\{e\\}\cup\\{a\\}\cup\\{a^{2}\\}\cup\\{a^{3}\\}...$
$B_{1}=\Psi(b^{-1})$
$B_{2}=\Psi(b)$

with $a\cdot A_{1}\cup A_{2}=\mathbb{F}_{2}$ and $b\cdot B_{1}\cup
B_{2}=\mathbb{F}_{2}$, offering an alternative proof of Theorem 1.2.

##  2 Equidecomposibility

######  Definition 2.1.

Let $G$ be a group acting on a set $X$, and let $A,B\subseteq X$. We say that
$A$ and $B$ are _$G$ -congruent_ if there exists $g\in G$ such that $g\cdot
A=B$.

######  Definition 2.2.

Suppose $G$ acts on $X$ and $A,B\subseteq X$. $A$ and $B$ are _$G$
-equidecomposable_ (denoted $A\sim_{G}B$) if $A$ and $B$ can each be
partitioned into the same finite number of respectively $G$-congruent pieces.
Formally written, $A\sim_{G}B$ if $A=\bigcup_{i=1}^{n}A_{i}$ ,
$B=\bigcup_{i=1}^{n}B_{i}$ with

$A_{i}\cap A_{j}=\emptyset=B_{i}\cap B_{j}\text{ for }i<j\leq n,$

and there are $g_{1},...,g_{n}\in G$ such that, for each $i\leq n$,
$g_{i}\cdot A_{i}=B_{i}$.

In plain terms this definition means to say that if you can take one subset,
$A$, and break it apart into $n$ pieces, rearrange those $n$ pieces via the
group action of $G$ to form the $n$ pieces that union together to form another
subset $B$, then $A$ and $B$ are $G$-equidecomposable.

######  Lemma 2.3.

Suppose $G$ acts on $X$ and $E\subseteq X$. If the set $E$ has two disjoint
subsets $A$ and $B$ such that $A\sim_{G}E$ and $B\sim_{G}E$, then $E$ is
$G$-paradoxical.

Figure 3 An illustration of Lemma 2.3

###### Proof.

We know that, because $A\subseteq E$ and $E$ are $G$-equidecomposable, that
$A$ and $E$ can each be partitioned into the same number of respectively
$G$-congruent pieces, so

$A=\bigcup_{i=1}^{n}A_{i}\text{ and }E=\bigcup_{j=1}^{n}E_{j}\text{ for some
$A_{i}\subseteq A$ and $E_{j}\subseteq E$}$

$\text{and, for each $i\leq n$, }g_{i}\cdot A_{i}=E_{i}\text{ with }g_{i}\in
G$

Similarly, for $B\subseteq E$ and $E$,

$B=\bigcup_{k=1}^{m}B_{k}\text{ and }E=\bigcup_{l=1}^{m}E_{l}\text{ for some
$B_{k}\subseteq A$ and $E_{l}\subseteq E$}$

$\text{and, for each $i\leq m$, }h_{i}\cdot A_{i}=E_{i}\text{ with }h_{i}\in
G$

Therefore,

$E=\bigcup_{i=1}^{m}E_{i}=\bigcup_{i=1}^{m}h_{i}\cdot B_{i}\text{ and
}E=\bigcup_{i=1}^{n}E_{i}=\bigcup_{i=1}^{n}g_{i}\cdot A_{i}$

where each $A_{1},...,A_{n}$ and each $B_{1},...,B_{m}$ are pairwise disjoint,
and $A$ and $B$ are also disjoint, so we can conclude that subsets
$A_{1},...,A_{n},B_{1},...,B_{m}$ of $E$ are also pairwise disjoint. Hence $E$
is $G$-paradoxical. ∎

##  3 Banach-Schroder-Bernstein Theorem

This theorem is an interesting concept to visualise, and acts as a platform
for further study on equidecomposibility relationships. A detailed proof has
been included for the readers’ ease.

######  Lemma 3.1.

if $A\sim B$ then there is a bijection $g:A\rightarrow B$ such that $C\sim
g(C)$ whenever $C\subseteq A$

######  Lemma 3.2.

if $A_{1}\cap A_{2}=\emptyset=B_{1}\cap B_{2}$, and if $A_{1}\sim B_{1}$ and
$A_{2}\sim B_{2}$, then $A_{1}\cup A_{2}\sim B_{1}\cup B_{2}$.

######  Theorem 3.3 (Banach-Schroder-Bernstein Theorem).

Suppose $G$ acts on $X$ and $A,B\subseteq X$. If $A$ is G-equidecomposable to
a subset of $B$, and if $B$ is G-equidecomposable to a subset of $A$, then $A$
is G-equidecomposable to $B$.

###### Proof.

Denote $B_{1}\subseteq B_{2}$ and $A_{1}\subseteq A_{2}$. Define
$f:A\rightarrow B_{1}$ and $g:A_{1}\rightarrow B$ as bijections guaranteed by
Lemma 3.1 as $A\sim B_{1}$ and $B\sim A_{1}$.

Let $C_{0}=A\setminus A_{1}$ and define inductively $C_{n+1}=g^{-1}f(C_{n})$.
Now, let $C=\bigcup^{\infty}_{n=0}C_{n}$. We claim that

$g(A\setminus C)=B\setminus f(C)$ (2)

To see this, consider $x\in A_{1}$,

$\displaystyle g(x)\in B\setminus f(C)$ $\displaystyle\Leftrightarrow
g(x)\notin f(C)$ $\displaystyle\Leftrightarrow g(x)\notin
f(\bigcup^{\infty}_{n=0}C_{n})$ $\displaystyle\Leftrightarrow
g(x)\notin\bigcup^{\infty}_{n=0}f(C_{n})$ $\displaystyle\Leftrightarrow
g(x)\notin f(C_{n})\text{ for all }n\geq 0$ $\displaystyle\Leftrightarrow
g^{-1}g(x)\notin g^{-1}f(C_{n})\text{ for all }n\geq 0$
$\displaystyle\Leftrightarrow x\notin C_{n+1}\text{ for all }n\geq 0$
$\displaystyle\Leftrightarrow x\notin\bigcup^{\infty}_{n=0}C_{n+1}$
$\displaystyle\Leftrightarrow x\notin\bigcup^{\infty}_{n=1}C_{n}$
$\displaystyle\Leftrightarrow x\notin C\text{ by }\ast$
$\displaystyle\Leftrightarrow x\in A\setminus C$ $\displaystyle\Leftrightarrow
g(x)\in g(A\setminus C)$

$\ast$ provided $x\notin C_{0}\Leftrightarrow x\notin A\setminus A_{1}$. Which
is true because $x\in A_{1}$.

So our claim is true. We now wish to show that $g(A\setminus C)\sim A\setminus
C$. We can use Lemma 3.1 but first we need to convince ourselves that
$A\setminus C\subseteq A_{1}$.

$\displaystyle C=\bigcup^{\infty}_{n=0}C_{n}$ $\displaystyle\Rightarrow
C_{0}\subseteq C$ $\displaystyle\Leftrightarrow A\setminus C\subseteq
A\setminus C_{0}=A_{1}$

Indeed, $A\setminus C\sim g(A\setminus C)=B\setminus f(C)$* by equation (2).
We know $C\subseteq A$, so by Lemma 3.1 we have $C\sim g(C)$**.

The fact that $(A\setminus C)\cap C=\emptyset=(B\setminus f(C))\cap f(C)$ with
(*) and (**) allows us to utilise Lemma 3.2, so $(A\setminus C)\cup
C\sim(B\setminus f(C))\cup f(C)\Rightarrow A\sim B$.

∎

######  Theorem 3.4.

$\mathbb{N}\cup\\{0\\}$ and $\mathbb{Z}$ are G-equidecomposable, where G is
the group of all bijections on $\mathbb{Q}$.

###### Proof.

Let $N_{1}$ be the set of even numbers (including zero), and $N_{2}$ be the
set of odd numbers. So

$N_{1}\cup
N_{2}=\\{0,2,4,6,8....\\}\cup\\{1,3,5,7,9....\\}=\mathbb{N}\cup\\{0\\}$

Let $Z_{1}$ be the set of positive integers (including zero) and $Z_{2}$ be
the set of odd integers. So

$Z_{1}\cup Z_{2}=\\{0,1,2,3,4....\\}\cup\\{-1,-2,-3,-4....\\}=\mathbb{Z}$

and $N_{1}\cap N_{2}=\emptyset=Z_{1}\cap Z_{2}$.

Define the bijection $g_{1}:N_{1}\rightarrow Z_{1}$ as

$\displaystyle g_{1}x$ $\displaystyle=\frac{x}{2}$ $\displaystyle\therefore
g_{1}\cdot N_{1}$ $\displaystyle=Z_{1},\text{ where }g_{1}\in G$

Define the bijection $g_{2}:N_{2}\rightarrow Z_{2}$ as

$\displaystyle g_{2}x$ $\displaystyle=-\frac{x+1}{2}$ $\displaystyle\therefore
g_{2}\cdot N_{2}$ $\displaystyle=Z_{2},\text{ where }g_{2}\in G$

So $\mathbb{N}\cup\\{0\\}$ and $\mathbb{Z}$ each can be partitioned into two
respectively $G$-congruent pieces, hence
$(\mathbb{N}\cup\\{0\\})\sim\mathbb{Z}$ ∎

######  Theorem 3.5.

$S^{1}$ is $SO_{2}$-equidecomposable to $S^{1}\setminus\\{1\\}$, where
$SO_{2}$ = group of rotations in $\mathbb{R}^{2}$, and
$S^{1}=\\{(x_{1},x_{2})\in\mathbb{R}^{2}:x_{1}^{2}+x_{2}^{2}=1\\}$ the circle
radius 1 centred about the origin.

###### Proof.

Let $\theta$ be a counterclockwise rotation by, say, 1 radians around the
origin. Since $2\pi$ is irrational we see that $\theta^{n}\cdot 1$ never comes
back to coincide with $1$ for all $n\in\mathbb{N}.$

Let

$\displaystyle A_{1}$ $\displaystyle=\\{1,\theta\cdot 1,\theta^{2}\cdot
1,\theta^{3}\cdot 1,....\\}\text{ and }A_{2}=S^{1}\setminus A_{1}$

so that $A_{1}\cup A_{2}=S^{1}$ and $A_{1}\cap A_{2}=\emptyset$

Let

$\displaystyle B_{1}$ $\displaystyle=\theta\cdot A_{1}=\\{\theta\cdot
1,\theta^{2}\cdot 1,\theta^{3}\cdot 1,....\\}$ $\displaystyle\text{ and
}B_{2}$ $\displaystyle=1\cdot A_{2}=S^{1}\setminus A_{1},\text{where 1 is the
identity element in }SO_{2}$

so that $B_{1}\cup B_{2}=S^{1}\setminus\\{1\\}$ and $B_{1}\cap
B_{2}=\emptyset$

∎

This theorem illustrates how you can take a circle, partition it into no more
than two pieces, rearrange one piece via the group of rotations to give you
two pieces of that original circle only now with a point missing. The choice
of rotation was paramount here, ensuring that no multiple of $\theta$ will be
the identity rotation and hence fill the hole in the circle at 1.

######  Theorem 3.6.

$B^{3}$ is $SO_{3}$-equidecomposable with $B^{3}\setminus\\{0\\}$, where
$SO_{3}$ is the group of rotations in $\mathbb{R}^{3}$

This theorem describes how a solid ball can be divided into pieces and
rearranged to give the same solid ball minus a single point.

###### Proof.

Let $C^{1}$ be a circle passing through the origin contained in $B^{3}$. Let
$A_{0}=B^{3}\setminus C^{1}$ and $A=C^{1}$ so that $A_{0}$, $A$ partition
$B^{3}$. Let $B_{0}=A_{0}=B^{3}\setminus C^{1}$ and $B=C^{1}\setminus\\{0\\}$
so $B_{0}$, $B$ partition $B^{3}\setminus\\{0\\}$.

Theorem 3.5 gives us that $A$ and $B$ are $SO_{2}$-equidecomposable. Hence $A$
and $B$ can be partitioned into a finite number, say $n$, of
$SO_{2}$-congruent pieces. i.e,

$\displaystyle A$ $\displaystyle=\bigcup^{n}_{i=1}A_{i}\text{ and
}B=\bigcup^{n}_{i=1}B_{i}$ $\displaystyle B_{j}$ $\displaystyle=h_{j}\cdot
A_{j}$ $\displaystyle\text{for }0<j\leq n,h_{j}\in SO_{2}$

Since $SO_{2}\subseteq SO_{3}$, $h_{j}\in SO_{2}\Rightarrow h_{j}\in SO_{3}$
for $0<j\leq n$. $B_{0}=1\cdot A_{0}$, where 1 is the identity element in
$SO_{3}$. Therefore,

$\displaystyle
B^{3}=A_{0}\cup(\bigcup^{n}_{i=1}A_{i})=\bigcup^{n}_{i=0}A_{i}\text{ and
}B^{3}\setminus\\{0\\}$
$\displaystyle=B_{0}\cup(\bigcup^{n}_{i=1}B_{i})=\bigcup^{n}_{i=0}B_{i}$

and

$\displaystyle B_{j}$ $\displaystyle=h_{j}\cdot A_{j}$ $\displaystyle\text{for
}0\leq j\leq n,h_{j}\in SO_{3}$

∎

It can now be seen that a solid ball in $\mathbb{R}^{3}$ can be partitioned
into two pieces; one a ring passing through the origin, and the other being
the remainder of the ball. The ring, as seen before, is equidecomposable to
the ring minus a point. So using that phenomenon on a ring, as a piece of the
ball, yields the same result in a higher dimension.

##  4 The importance of independence

A set $S$ of elements in a group are called _independent_ if no non-trivial,
reduced words using letters from $S$ and their inverses is the identity.
Hence, a pair of independent elements will generate a free subgroup or rank 2.

######  Theorem 4.1.

There are two independent rotations about the axis through the origin in
$\mathbb{R}^{3}$. Hence, $SO_{3}$ contains a free subgroup of rank 2.

###### Proof.

Let $\sigma$ be a counter clockwise rotation about the z-axis through an angle
of $\theta=cos^{-1}\left(\frac{3}{5}\right)$. Let $\tau$ be a counter
clockwise rotation about the x-axis through the same angle $\theta$.

We have

$\sigma=\left(\begin{array}[]{ccc}\frac{3}{5}&\frac{-4}{5}&0\\\
\frac{4}{5}&\frac{3}{5}&0\\\ 0&0&1\end{array}\right)$

and

$\tau=\left(\begin{array}[]{ccc}1&0&0\\\ 0&\frac{3}{5}&\frac{-4}{5}\\\
0&\frac{4}{5}&\frac{3}{5}\end{array}\right)$

We wish to show that no non-trivial reduced word in
$\\{\sigma,\sigma^{-1},\tau,\tau^{-1}\\}$ is the identity homeomorphism. We
claim that conjugation by $\sigma$ does not effect whether or not a word is
the identity. To see this, let $y$ be some non-trivial, reduced word in
$\\{\sigma,\sigma^{-1},\tau,\tau^{-1}\\}$ and let $\gamma=\sigma^{-1}\cdot
y\cdot\sigma$. There are two cases;

  * (i)

$y$ contains either $\tau$ or $\tau^{-1}$. Then $\gamma$ must be non-trivial
when reduced.

  * (ii)

$y$ does not contain either $\tau$ or $\tau^{-1}$. Then, since $y$ was assumed
to be reduced, $y$ can only be either a sequence of $\sigma$ or a sequence of
$\sigma^{-1}$ of length $>$ 0\. No complete reduction can occur in $\gamma$
(leaving $\gamma$ non-trivial when reduced), hence our claim is true.

In order to prove the theorem, we need only to consider words ending (on the
right) with $\sigma^{\pm 1}$ (words ending in $\sigma^{-1}$ after conjugation
by $\sigma$ for the case when $y=...\sigma^{-1}\cdot\sigma^{-1}$, or
$y=\sigma^{-1}$). We now claim that for any such word $\omega$ the point
$\omega(1,0,0)$ will take the form $(a,b,c)/5^{n}$, where n is the integer
length of $\omega$; $a,b$ and $c$ are all integers with $b$ not divisible by
5. Given that zero is indeed divisible by 5, this claim implies that
$\omega(1,0,0)\not=(1,0,0)$, so $\omega$ is not the identity homeomorphism.

Let us prove this claim by induction;
For $n=1$, we have $\omega=\sigma^{\pm 1}$ and so

$\displaystyle\omega(1,0,0)$ $\displaystyle=\sigma^{\pm 1}(1,0,0)$
$\displaystyle=\frac{1}{5}\left(\begin{array}[]{ccc}3&\mp 4&0\\\ \pm 4&3&0\\\
0&0&5\end{array}\right)\left(\begin{array}[]{c}1\\\ 0\\\ 0\end{array}\right)$
$\displaystyle=(3,\pm 4,0)/5$

which is of the required form.

Fix $k\in\mathbb{N}\setminus\\{1\\}$. Assume claim is true for $n=1,2,..,k-1$.
Let $\omega_{n}$ be any non-trivial, reduced word ending in $\sigma^{\pm 1}$.
By assumption, $\omega_{n}(1,0,0)=(a_{n},b_{n},c_{n})/5^{n}$, where
$a_{n},b_{n},c_{n}\in\mathbb{Z}$ and $b_{n}$ not divisible by 5.

Write $\omega_{k}=\phi\omega_{k-1}$ for any reduced word $\omega_{k}$, where
$\mid\omega_{k}\mid=k$ and
$\phi\in\\{\sigma,\sigma^{-1},\tau,\tau^{-1}\\}$.There are four cases:

  * Case 1:

$\omega_{k}=\sigma\omega_{k-1}$

$\displaystyle\omega_{k}(1,0,0)$ $\displaystyle=\sigma\omega_{k-1}(1,0,0)$
$\displaystyle=\sigma(a_{k-1},b_{k-1},c_{k-1})/5^{k-1}$
$\displaystyle=\frac{1}{5^{k}}\left(\begin{array}[]{c}{3a_{k-1}-4b_{k-1}}\\\
{4a_{k-1}+3b_{k-1}}\\\ {5c_{k-1}}\end{array}\right)$

We can see that $a_{k}=3a_{k-1}-4b_{k-1}$ , $b_{k}=4a_{k-1}+3b_{k-1}$ ,
$c_{k}=5c_{k-1}$ are all integers. We must check now to see if

$b_{k}=4a_{k-1}+3b_{k-1}$ (3)

is divisible by 5. To do this, we need to look now at 3 more cases;
$\omega_{k}=\sigma\tau^{\pm 1}\omega_{k-2}$ ,
$\omega_{k}=\sigma\sigma\omega_{k-2}$, where $\omega_{k-2}$ is possibly the
empty word.

For the first two cases, we already know that $b_{k-1}$ is not divisible by 5,
so what we really want to check is if $a_{k-1}$ is divisible by 5, for then
the addition of the two terms will yield a number not divisible by 5. We have

$\displaystyle\omega_{k}(1,0,0)$ $\displaystyle=\sigma\tau^{\pm
1}\omega_{k-2}(1,0,0)$
$\displaystyle=\sigma\frac{1}{5^{k-2}}\times\frac{1}{5}\left(\begin{array}[]{ccc}5&0&0\\\
0&3&\mp 4\\\ 0&\pm 4&3\end{array}\right)\left(\begin{array}[]{c}{a_{k-2}}\\\
{b_{k-2}}\\\ {c_{k-2}}\end{array}\right)$
$\displaystyle=\sigma\frac{1}{5^{k-1}}\left(\begin{array}[]{c}{5a_{k-2}}\\\
{3b_{k-2}\mp 4c_{k-2}}\\\ {4b_{k-2}\pm 3c_{k-2}}\end{array}\right)$

Since $a_{k-1}=5a_{k-2}$, $a_{k-1}$ is divisible by 5 when
$\omega_{k}=\sigma\tau^{\pm 1}\omega_{k-1}$.

For the second case, we check to see if $b_{k}$ is divisible by 5 based on the
assumption that both $b_{k-1}$ and $b_{k-2}$ are not divisible by 5. We have

$\displaystyle\omega_{k}(1,0,0)$
$\displaystyle=\sigma\sigma\omega_{k-2}(1,0,0)$
$\displaystyle=\sigma\frac{1}{5^{k-2}}\times\frac{1}{5}\left(\begin{array}[]{ccc}3&-4&0\\\
4&3&0\\\ 0&0&5\end{array}\right)\left(\begin{array}[]{c}{a_{k-2}}\\\
{b_{k-2}}\\\ {c_{k-2}}\end{array}\right)$
$\displaystyle=\sigma\frac{1}{5^{k-1}}\left(\begin{array}[]{c}{3a_{k-2}-4b_{k-2}}\\\
{4a_{k-2}+3b_{k-2}}\\\ {5c_{k-2}}\end{array}\right)$

So now,

$\displaystyle a_{k-1}$ $\displaystyle=3a_{k-2}-4b_{k-2}$ (4) $\displaystyle
b_{k-1}$ $\displaystyle=4a_{k-2}+3b_{k-2}\Rightarrow
4a_{k-2}=b_{k-1}-3b_{k-2}$ (5)

Subbing equations (3) and (4) into (2) gives

$\displaystyle b_{k}$ $\displaystyle=4(3a_{k-2}-4b_{k-2})+3b_{k-1}$
$\displaystyle=3(b_{k-1}-3b_{k-2})-16b_{k-2}+3b_{k-1}$
$\displaystyle=6b_{k-1}-25b_{k-2}$

Hence $b_{k}$ is not divisible by 5 for Case 1.

  * Case 2:

$\omega_{k}=\sigma^{-1}\omega_{k-1}$

$\displaystyle\omega_{k}(1,0,0)$
$\displaystyle=\sigma^{-1}\omega_{k-1}(1,0,0)$
$\displaystyle=\frac{1}{5^{k}}\left(\begin{array}[]{c}{3a_{k-1}+4b_{k-1}}\\\
{-4a_{k-1}+3b_{k-1}}\\\ {5c_{k-1}}\end{array}\right)$

Again, we see that $a_{k}=3a_{k-1}+4b_{k-1}$ , $b_{k}=-4a_{k-1}+3b_{k-1}$ ,
$c_{k}=5c_{k-1}$ are all integers. We must check now to see if

$b_{k}=-4a_{k-1}+3b_{k-1}$ (6)

is divisible by 5. So we take the cases $\omega_{k}=\sigma^{-1}\tau^{\pm
1}\omega_{k-2}$ and $\omega_{k}=\sigma^{-1}\sigma^{-1}\omega_{k-2}$. The same
working from Case 1 applies here in that $a_{k-1}=5a_{k-2}$, so $a_{k-1}$ is
divisible by 5 when $\omega_{k}=\sigma^{-1}\tau^{\pm 1}\omega_{k-1}$. Now we
have

$\displaystyle\omega_{k}(1,0,0)$
$\displaystyle=\sigma^{-1}\sigma^{-1}\omega_{k-2}(1,0,0)$
$\displaystyle=\sigma^{-1}\frac{1}{5^{k-1}}\left(\begin{array}[]{c}{3a_{k-2}+4b_{k-2}}\\\
{-4a_{k-2}+3b_{k-2}}\\\ {5c_{k-2}}\end{array}\right)$

So now,

$\displaystyle a_{k-1}$ $\displaystyle=3a_{k-2}+4b_{k-2}$ (7) $\displaystyle
b_{k-1}$ $\displaystyle=-4a_{k-2}+3b_{k-2}\Rightarrow
4a_{k-2}=3b_{k-2}-b_{k-1}$ (8)

Subbing equations (6) and (7) into (5) gives

$\displaystyle b_{k}$ $\displaystyle=-4(3a_{k-2}+4b_{k-2})+3b_{k-1}$
$\displaystyle=-3(3b_{k-2}-b_{k-1})-16b_{k-2}+3b_{k-1}$
$\displaystyle=6b_{k-1}-25b_{k-2}$

Hence $b_{k}$ is not divisible by 5 for Case 2.

  * Case 3:

$\omega_{k}=\tau\omega_{k-1}$

  * Case 4:

$\omega_{k}=\tau^{-1}\omega_{k-1}$

A similar proof can be used to show that $b_{k}$ is not divisible by 5 for
Case 3 and Case 4, and that $a_{k}$, $b_{k}$, $c_{k}$ are all integer values.

∎

######  Definition 4.2.

A set $S$ is _countable_ if it has the same cardinality as a subset of
$\mathbb{N}$. i.e, if $\mid S\mid=\mid N\mid$, $N\subseteq\mathbb{N}$

######  Lemma 4.3.

Let $D=\\{x\in S^{2}:\text{ x is fixed by some element in
}F_{2}\setminus\\{e\\}\\}$. Then $D$ is countable and $S^{2}$ is
$SO_{3}$-equidecomposable with $S^{2}\setminus D$.

###### Proof.

Each non-identity rotation in $F_{2}$ fixes precisely two points on $S^{2}$,
namely the intersection of the axis of rotation with the sphere. $D$ is the
collection of such points.

Let $D_{n}$ denote the points on $S^{2}$ fixed by words in
$F_{2}\setminus\\{e\\}$ of length $n\in\mathbb{N}$.

$\displaystyle\mid D_{1}\mid=$ $\displaystyle 4$ $\displaystyle\mid
D_{2}\mid=$ $\displaystyle 4\times 3$ $\displaystyle\mid D_{3}\mid=$
$\displaystyle 4\times 3^{2}$ $\displaystyle\vdots$ $\displaystyle\mid
D_{n}\mid=$ $\displaystyle 4\times 3^{n-1}$

Then $D=\bigcup^{\infty}_{n=1}D_{n}$ which is a countable union of finite
sets, hence $D$ is countable.

We claim that $S^{2}$ is not countable. Define a bijection $f$ as

$\displaystyle f:S^{1}$ $\displaystyle\rightarrow[0,1)$
$\displaystyle(cos\theta,sin\theta)$ $\displaystyle\mapsto\frac{\theta}{2\pi}$

So $\mid S^{1}\mid=\mid[0,1)\mid\geq\mid\mathbb{N}\mid$, given that the
interval [0,1) is uncountable. Hence $S^{1}$ is uncountable. Then, since
$S^{1}\subseteq S^{2}$, $S^{2}$ is uncountable and we have proved our claim.

It follows, since $D$ is countable, and $S^{2}$ is uncountable, that
$S^{2}\setminus D$ is non-empty and uncountable. Hence there exists a pair of
antipodal points on $S^{2}$ which aren’t elements of $D$. Let $\ell$ be the
axis through the centre and these points. We wish to find a rotation $\mu$
about the axis $\ell$ such that $D$, $\mu D$, $\mu^{2}D$ … are disjoint.

Let $A=\\{\rho:\rho\text{ is a rotation about }\ell\text{ and
}\rho^{m}D\bigcap_{m\in\mathbb{N}}D\not=\emptyset\\}$. Now, $\rho^{m}D\cap
D\not=\emptyset\Rightarrow\exists d_{1},d_{2}\in D:\rho^{m}d_{1}=d_{2}$. Since
$d_{1}$ and $d_{2}$ are part of a countable set, then $A$ is countable. Choose
a rotation $\mu$ about $\ell$ that is not in $A$. So

$\displaystyle\mu^{m-n}D\cap
D=\emptyset\Rightarrow\mu^{m}D\cap\mu^{n}D=\emptyset\text{ for all
}m,n\in\mathbb{N}$

as required.

Let $A_{1}=D\cup\rho D\cup\rho^{2}D\cup...$ and $A_{2}=S^{2}\setminus A_{1}$
so that $A_{1}\cap A_{2}=\emptyset$ and $A_{1}\cup A_{2}=S_{2}$. Let
$B_{1}=\rho A_{1}=\rho D\cup\rho^{2}D\cup...$ and $B_{2}=A_{2}$ so that
$B_{1}\cap B_{2}=\emptyset$ and $B_{1}\cup B_{2}=S_{2}\setminus D$. We have
partitioned $S_{2}$ and $S_{2}\setminus D$ into two $SO_{3}$-congruent pieces,
hence $S_{2}\sim S_{2}\setminus D$.

∎

##  5 The Banach-Tarski Paradox

As mentioned in the Abstract, the Banach-Tarski Paradox illustrates how one
can cut a solid ball in $\mathbb{R}^{3}$ into finitely many pieces, where each
piece can only be rotated and translated to form two balls identical to the
original.

In proving this, most of the work will be in proving the phenomenon for a
sphere in $\mathbb{R}^{3}$ where the radius of the sphere is irrelevant. So we
will endeavour to partition a sphere into two pieces, each of which can be
rearranged to form the original sphere. Once we have this we can form a ball
(minus the origin) by the infinite union of spheres.

######  Lemma 5.1.

Let $x\in S^{2}\setminus D$ and $f,g\in\mathbb{F}_{2}$. Then

$f\cdot x=g\cdot x\Rightarrow f=g$

###### Proof.

$\displaystyle D$ $\displaystyle=\\{x\in S^{2}:x\text{ is fixed by some point
in }\mathbb{F}_{2}\setminus\\{e\\}\\}$ $\displaystyle S^{2}\setminus D$
$\displaystyle=\\{x\in S^{2}:x\text{ is fixed only by }\\{e\\}\\}$

So when $f\cdot x=g\cdot x\Rightarrow g^{-1}f\cdot x=x$, $x$ is fixed by
$g^{-1}f$. Hence $g^{-1}f=e\Rightarrow g=f$. ∎

######  Lemma 5.2.

Let $x\in S^{2}$ and $f\in\mathbb{F}_{2}$. Then

$x\in D\Leftrightarrow f\cdot x\in D$

###### Proof.

($\Leftarrow$)

$\displaystyle f\cdot x\in D$ $\displaystyle\Rightarrow\exists
g\in\mathbb{F}_{2}\setminus\\{e\\}\text{ with }gf\cdot x=f\cdot x$
$\displaystyle\Rightarrow f^{-1}gf\cdot x=x$

In order to show $x\in D$, since $x$ is fixed by $f^{-1}gf$ we are only
required to show that $f^{-1}gf\in\mathbb{F}_{2}\setminus\\{e\\}$. If $f=e$,
then $f^{-1}gf=g\in\mathbb{F}_{2}\setminus\\{e\\}$. If $f\not=e$, then, since
$g$ is a non-trivial word in the free group $\mathbb{F}_{2}$, conjugation by
$f$ does not affect whether or not a word is the identity (conjugation result
discussed in Theorem 4.1); $f^{-1}gf\not=e$. So $x\in D$.
($\Rightarrow$)

$\displaystyle x\in D$ $\displaystyle\Rightarrow\exists
h\in\mathbb{F}_{2}\setminus\\{e\\}\text{ with }h\cdot x=x$
$\displaystyle\Rightarrow(fhf^{-1})f\cdot x=f(h\cdot x)=f\cdot x$

Similarly, if $f=e$, then $fhf^{-1}=h\in\mathbb{F}_{2}\setminus\\{e\\}$. If
$f\not=e$, then $f^{-1}hf\not=e$. So $f\cdot x\in D$ ∎

######  Definition 5.3.

An _$\mathbb{F}_{2}$ \- orbit_ is a set of the form

$F_{x}:=\\{f\cdot x:f\in\mathbb{F}_{2}\\}\text{ for some }x\in S^{2}\setminus
D$

Denote a family of $\mathbb{F}_{2}$-orbits by $(F_{i})_{i\in I}$ for some
interval $I$.

######  Lemma 5.4.

Two $\mathbb{F}_{2}$-orbits are either disjoint or equal

###### Proof.

Assume two $\mathbb{F}_{2}$-orbits say, $F_{x}$ with $x\in S^{2}\setminus D$
and $F_{y}$ with $y\in S^{2}\setminus D$, are not disjoint. So $F_{x}\cap
F_{y}\not=\emptyset$. Fix

$w\in F_{x}\cap F_{y}\Rightarrow w=g\cdot x=h\cdot y\text{ for some
}g,h\in\mathbb{F}_{2}$ (9)

We wish to show that $F_{x}=F_{y}$.

$\displaystyle\text{Fix }z\in F_{x}\Rightarrow$ $\displaystyle z=f\cdot
x\text{ for some }f\in\mathbb{F}_{2}$ $\displaystyle\Rightarrow$
$\displaystyle f^{-1}\cdot z=x$ subbing into (9) gives $\displaystyle
g(f^{-1}\cdot z)=h\cdot y$ $\displaystyle\Rightarrow$ $\displaystyle
z=(gf^{-1})^{-1}h\cdot y\in F_{y}$

Therefore, $F_{x}\subseteq F_{y}$.

$\displaystyle\text{Fix }z\in F_{y}\Rightarrow$ $\displaystyle z=f\cdot
y\text{ for some }f\in\mathbb{F}_{2}$ $\displaystyle\Rightarrow$
$\displaystyle f^{-1}\cdot z=y$ subbing into (9) gives $\displaystyle
h(f^{-1}\cdot z)=g\cdot x$ $\displaystyle\Rightarrow$ $\displaystyle
z=(hf^{-1})^{-1}g\cdot x\in F_{x}$

Therefore, $F_{y}\subseteq F_{x}$ and $F_{y}=F_{x}$.

The other direction of the proof was to show that two disjoint
$\mathbb{F}_{2}$-orbits are not equal, which is trivial. ∎

The implication of this Lemma is that if two $\mathbb{F}_{2}$\- orbits have a
single point in common, then the orbits are equivalent.

######  Lemma 5.5.

$\mathbb{F}_{2}$-orbits in $S^{2}\setminus D$ union to give $S^{2}\setminus D$

###### Proof.

$F_{x}=\\{f\cdot x:f\in\mathbb{F}_{2}\\}$ for some $x\in S^{2}\setminus D$. So
$f\cdot x\in S^{2}\setminus D$ by Lemma 5.2. Since $x$ is arbitrary,
$\bigcup_{i\in I}F_{i}\subseteq S^{2}\setminus D$

$\displaystyle\text{Now fix }x\in S^{2}\setminus D$ $\displaystyle\Rightarrow
F_{x}=\\{f\cdot x:f\in\mathbb{F}_{2}\\}\text{ is the $\mathbb{F}_{2}$-orbit}$
$\displaystyle\Rightarrow F_{x}=F_{i}\text{ for some }i\in I\text{ by Lemma
5.4}$ $\displaystyle\Rightarrow x\in F_{x}\text{ because }e\in\mathbb{F}_{2}$

So $x\in F_{x}\subseteq\bigcup_{i\in I}F_{i}$. Hence, $\bigcup_{i\in
I}F_{i}=S^{2}\setminus D$. ∎

###  5.1 Revisiting $\mathbb{F}_{2}$; Free group of rank 2

######  Lemma 5.6.

There exists subsets $P_{1},P_{2}\subseteq\mathbb{F}_{2}$ where
$P_{1}\sim_{\mathbb{F}_{2}}\mathbb{F}_{2}$ and
$P_{2}\sim_{\mathbb{F}_{2}}\mathbb{F}_{2}$, and $P_{1},P_{2}$ partition
$\mathbb{F}_{2}$.

We can divide $\mathbb{F}_{2}$ into two sets, each of which has pieces that
can be acted upon in such a way so that each set forms $\mathbb{F}_{2}$ again.
We began looking at free groups of rank 2 in Theorem 1.4, although a similar
concept, this lemma is a much stronger version. Let us adopt the notation
defined previously.

###### Proof.

$\displaystyle A_{1}$
$\displaystyle=\Psi(a^{-1})\cup\\{e\\}\cup\\{a\\}\cup\\{a^{2}\\}\cup\\{a^{3}\\}...$
$\displaystyle A_{2}$
$\displaystyle=\Psi(a)\setminus\\{e\\}\cup\\{a\\}\cup\\{a^{2}\\}\cup\\{a^{3}\\}...$
$\displaystyle B_{1}$ $\displaystyle=\Psi(b^{-1})$ $\displaystyle B_{2}$
$\displaystyle=\Psi(b)$

Let $P_{1}=A_{1}\cup A_{2}$, $P_{2}=B_{1}\cup B_{2}$. $A_{1}$ and $A_{2}$ are
disjoint so clearly $A_{1}$, $A_{2}$ partition $P_{2}$. As previously
mentioned in Theorem 1.4, $(a\cdot A_{1})\cup A_{2}=\mathbb{F}_{2}$. Hence
$P_{1}\sim_{\mathbb{F}_{2}}\mathbb{F}_{2}$. Similarly $B_{1}$, $B_{2}$
partition $P_{2}$ and $b\cdot B_{1}\cup B_{2}=\mathbb{F}_{2}$. Hence
$P_{2}\sim_{\mathbb{F}_{2}}\mathbb{F}_{2}$. Also,

$\displaystyle(A_{1}\cup A_{2})\cup(B_{1}\cup B_{2})=\mathbb{F}_{2}$
$\displaystyle\Rightarrow P_{1}\cup P_{2}=\mathbb{F}_{2}$
$\displaystyle\text{and }(A_{1}\cap A_{2})\cap(B_{1}\cap B_{2})=\emptyset$
$\displaystyle\Rightarrow P_{1}\cup P_{2}=\emptyset$

So $P_{1},P_{2}$ partition $\mathbb{F}_{2}$. ∎

###  5.2 Tying it all together

######  Theorem 5.7 (Banach-Tarski Paradox with spheres).

There exists subsets $S_{1},S_{2}\subseteq S^{2}$ that partition $S^{2}$ and
$S_{1}\sim_{SO_{3}}S^{2}$, $S_{2}\sim_{SO_{3}}S^{2}$

We wish to find a set $M\subseteq S^{2}$ through which we act on to translate
the equidecomposibility properties of $\mathbb{F}_{2}$ over to $S^{2}$. By the
Axiom of Choice, $\exists M\subseteq S^{2}\setminus D\text{ such that }\forall
i\in I$, $|M\cap F_{i}|=1$. Plainly put, $M$ is manifested by taking one
element from each $F$-orbit, $F_{i}$.

######  Lemma 5.8.

For $P_{1},P_{2}\subseteq\mathbb{F}_{2}$ and $M\subseteq S^{2}\setminus D$,
define $P_{1}M=\\{f\cdot m:f\in P_{1},m\in M\\}$, $P_{2}M=\\{f\cdot m:f\in
P_{2},m\in M\\}$. Then, $P_{1}M$, $P_{2}M\subseteq S^{2}\setminus D$ and

$\displaystyle P_{1}M\cup P_{2}M$ $\displaystyle=(P_{1}\cup P_{2})M$
$\displaystyle P_{1}M\cap P_{2}M$ $\displaystyle=(P_{1}\cap P_{2})M$

###### Proof.

$\displaystyle f\in P_{1}M\cup P_{2}M\Leftrightarrow$ $\displaystyle
f\in\\{\\{f\cdot m:f\in P_{1},m\in M\\}\cup\\{f\cdot m:f\in P_{2},m\in
M\\}\\}$ $\displaystyle\Leftrightarrow$ $\displaystyle f\in\\{f\cdot m:f\in
P_{1}\text{ or }f\in P_{2},m\in M\\}$ $\displaystyle\Leftrightarrow$
$\displaystyle f\in\\{f\cdot m:f\in P_{1}\cup P_{2},m\in M\\}$
$\displaystyle\Leftrightarrow$ $\displaystyle f\in(P_{1}\cup P_{2})M$

Similar working shows that $f\in P_{1}M\cap P_{2}M\Leftrightarrow
f\in(P_{1}\cap P_{2})M$.

Lemma 5.2 gives for $m\in M\subseteq(S^{2}\setminus D)\subseteq S^{2}$ and
$f\in P_{1}\subseteq\mathbb{F}_{2}$, that $m\in S^{2}\setminus
D\Leftrightarrow f\cdot m\in S^{2}\setminus D$. Hence, $P_{1}M=\\{f\cdot
m:f\in P_{1},m\in M\\}\subseteq S^{2}\setminus D$. Similarly, $P_{2}M\subseteq
S^{2}\setminus D$. ∎

###### Proof of the Banach-Tarski paradox with spheres.

It will be shown that, because $P_{1}$ and $P_{2}$ union to give
$\mathbb{F}_{2}$, then $P_{1}M$ and $P_{2}M$ union to give $\mathbb{F}_{2}M$.

$\displaystyle P_{1}M\cup P_{2}M$ $\displaystyle=(P_{1}\cup P_{2})M$ (Lemma
5.7) $\displaystyle=\mathbb{F}_{2}M$ $\displaystyle=\\{f\cdot
m:f\in\mathbb{F}_{2},m\in M\subseteq(S^{2}\setminus D)\\}$
$\displaystyle=\text{union of all $F$-orbits in $S^{2}\setminus D$}$
$\displaystyle=S^{2}\setminus D$ (Lemma 5.5)

Also, since $P_{1}$ and $P_{2}$ are disjoint in $\mathbb{F}_{2}$, then
$P_{1}M$ and $P_{2}M$ are disjoint in $S^{2}$

$\displaystyle P_{1}M\cap P_{2}M$ $\displaystyle=(P_{1}\cap P_{2})M$ (Lemma
5.7) $\displaystyle=\emptyset M$ $\displaystyle=\emptyset$

######  Lemma 5.9.

$P_{1}M\sim_{SO_{3}}\mathbb{F}_{2}M$ and $P_{2}M\sim_{SO_{3}}\mathbb{F}_{2}M$

###### Proof.

Using $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$ from the proof of Lemma 5.6, we know
that $A_{1}\cup A_{2}=P_{1}$, $A_{1}\cap A_{2}=\emptyset$ and $a\cdot
A_{1}\cup A_{2}=\mathbb{F}_{2}$. So
$A_{1}M\cup A_{2}M=(A_{1}\cup A_{2})M=P_{1}M$
$A_{1}M\cap A_{2}M=(A_{1}\cap A_{2})M=\emptyset$
and, $a\cdot A_{1}M\cup A_{2}M=(a\cdot A_{1}\cup A_{2})M=\mathbb{F}_{2}M$.
Hence $P_{1}M\sim_{SO_{3}}\mathbb{F}_{2}M$.

We also know that $B_{1}\cup B_{2}=P_{2}$, $B_{1}\cap B_{2}=\emptyset$ and
$b\cdot B_{1}\cup B_{2}=\mathbb{F}_{2}$. So
$B_{1}M\cup B_{2}M=(B_{1}\cup B_{2})M=P_{2}M$
$B_{1}M\cap B_{2}M=(B_{1}\cap B_{2})M=\emptyset$
and similarly, $b\cdot B_{1}M\cup B_{2}M=(b\cdot B_{1}\cup
B_{2})M=\mathbb{F}_{2}M$. Hence $P_{2}M\sim_{SO_{3}}\mathbb{F}_{2}M$
∎

We have used the fact that $P_{1}$ and $P_{2}$ are
$\mathbb{F}_{2}$-equidecomposable to $\mathbb{F}_{2}$ to show that $P_{1}M$
and $P_{2}M$ are both $SO_{3}$-equidecomposable to
$\mathbb{F}_{2}M=S^{2}\setminus D$.

_Proof of the Banach-Tarski paradox with spheres continued:_ Define
$S_{1}:=P_{1}M\cup D$ and $S_{2}:=P_{2}M$. We need to show that $S_{1},S_{2}$
partition $S^{2}$.

$\displaystyle S_{1}\cup S_{2}=$ $\displaystyle(P_{1}M\cup D)\cup P_{2}M$
$\displaystyle=$ $\displaystyle P_{1}M\cup P_{2}M\cup D$ $\displaystyle=$
$\displaystyle\mathbb{F}_{2}M\cup D$ $\displaystyle=$ $\displaystyle
S^{2}\setminus D\cup D$ $\displaystyle=$ $\displaystyle S^{2}$

$\displaystyle S_{1}\cap S_{2}=$ $\displaystyle(P_{1}M\cup D)\cap P_{2}M$
$\displaystyle=$ $\displaystyle(P_{1}M\cap P_{2}M)\cup(D\cap P_{2}M)$ (Lemma
5.8) $\displaystyle=$ $\displaystyle\emptyset$

Now, $S_{1}=P_{1}M\cup D$, and we know from Lemma 5.9 that
$P_{1}M\sim\mathbb{F}_{2}M$ and $D\sim D$ is trivial to see. So Lemma 3.2
gives $(P_{1}M\cup D)\sim(\mathbb{F}_{2}M\cup D)$. We have

$\displaystyle S_{1}\sim$ $\displaystyle\mathbb{F}_{2}M\cup D$
$\displaystyle=$ $\displaystyle S^{2}\setminus D\cup D$ $\displaystyle=$
$\displaystyle S^{2}$

We also have

$\displaystyle S_{2}=$ $\displaystyle P_{2}M$ $\displaystyle\sim$
$\displaystyle\mathbb{F}_{2}M$ (Lemma 5.9) $\displaystyle=$ $\displaystyle
S^{2}\setminus D$ $\displaystyle\sim$ $\displaystyle S^{2}$ (Lemma 4.3)

∎

######  Theorem 5.10 (Banach-Tarski paradox).

There exists subsets $\mathcal{B}_{1},\mathcal{B}_{2}\subseteq B^{3}$ that
partition $B^{3}$ and $\mathcal{B}_{1}\sim_{SO_{3}}B^{3}$,
$\mathcal{B}_{2}\sim_{SO_{3}}B^{3}$

###### Proof of the Banach-Tarski paradox.

There are 5 pieces of the sphere; $A_{1}M$, $A_{2}M$, $B_{1}M$, $B_{2}M$, $D$.
Remembering that these pieces are grouped accordingly so $S_{1}=A_{1}M\cup
A_{2}M\cup D$, $S_{2}=B_{1}M\cup B_{2}M$. Fix $0<r\leq 1$. Define
$rS_{i}=\\{rx|x\in S_{i}\\}$ for $i=1,2$.

Define $\mathcal{B}_{1}=\bigcup_{0<r\leq 1}rS_{1}$ and
$\mathcal{B}_{2}=\bigcup_{0<r\leq 1}rS_{2}$ so that
$\mathcal{B}_{1},\mathcal{B}_{2}\subseteq B^{3}$. Since $S_{1}$ and $S_{2}$
are disjoint, then by construction $\mathcal{B}_{1}$ and $\mathcal{B}_{2}$ are
disjoint. To see that $\mathcal{B}_{1}$ and $\mathcal{B}_{2}$ partition
$B^{3}\setminus\\{0\\}$ note that

$\displaystyle\mathcal{B}_{1}\cup\mathcal{B}_{2}$
$\displaystyle=\bigcup_{0<r\leq 1}r(S_{1}\cup S_{2})$
$\displaystyle=\bigcup_{0<r\leq 1}rS^{2}$
$\displaystyle=B^{3}\setminus\\{0\\}$

The three pieces of the ball that partition $\mathcal{B}_{1}$ are
$(\bigcup_{0<r\leq 1}rA_{1}M)$, $(\bigcup_{0<r\leq 1}rA_{2}M)$ and
$(\bigcup_{0<r\leq 1}rD)$. They can be rearranged as follows;
$(\bigcup_{0<r\leq 1}r(a\cdot A_{1})M)$, $(\bigcup_{0<r\leq 1}rA_{2}M)$ and
$(\bigcup_{0<r\leq 1}rD)$. It can be seen that these rearranged pieces
partition $B^{3}\setminus\\{0\\}$. Hence
$\mathcal{B}_{1}\sim_{SO_{3}}B^{3}\setminus\\{0\\}$. Recall from Theorem 3.6
that $B^{3}\setminus\\{0\\}\sim_{SO_{3}}B^{3}$, so
$\mathcal{B}_{1}\sim_{SO_{3}}B^{3}$.

Two pieces of the ball, $(\bigcup_{0<r\leq 1}rB_{1}M)$ and $(\bigcup_{0<r\leq
1}rB_{2}M)$ partition $\mathcal{B}_{2}$. When the pieces are rearranged to
form $(\bigcup_{0<r\leq 1}r(b\cdot B_{1})M)$ and $(\bigcup_{0<r\leq
1}rB_{2}M)$, then these pieces partition $B^{3}\setminus\\{0\\}$. Hence
$\mathcal{B}_{1}\sim_{SO_{3}}B^{3}\setminus\\{0\\}\sim_{SO_{3}}B^{3}$. ∎

## Consequences of the Banach-Tarski Paradox

The corresponding paradox for free groups, while somewhat surprising, is
certainly not something that anyone would argue with. When this paradox is
applied to 3- dimensional space it does go against our intuition, but very
often our intuition is flawed.

By allowing the use of the Axiom of Choice, we can obtain sets which are so
complicated that they cannot be assigned a measure consistent with the desired
properties. This is precisely what happens in the Banach-Tarski paradox; most
of the sets we have constructed are _nonmeasurable_ ; not Lebesgue measurable.
So the Banach-Tarski Paradox doesn’t double the volume of the ball because the
pieces in the decomposition cannot be assigned a volume - such sets would be
far more intricate than the atomic structure of matter, and as such, could
never be realised in real life. Such beauty exists only in the realm of
mathematics.

## References

[1] Stan Wagon. The Banach-Tarski Paradox. Cambridge University Press, 1985.