proof of BanachTarski paradox
We deal with some technicalities first, mainly concerning the properties of equidecomposability. We can then prove the paradox^{} in a clear and unencumbered line of argument: we show that, given two unit balls^{} $U$ and ${U}^{\prime}$ with arbitrary origin, $U$ and $U\cup {U}^{\prime}$ are equidecomposable, regardless whether $U$ and ${U}^{\prime}$ are disjoint or not. The original proof can be found in [BT].
Technicalities
Theorem 1.
Equidecomposability gives rise to an equivalence relation^{} on the subsets of Euclidean space.
Proof.
The only nonobvious part is transitivity. So let $A$, $B$, $C$ be sets such that $A$ and $B$ as well as $B$ and $C$ are equidecomposable. Then there exist disjoint decompositions ${A}_{1},\mathrm{\dots}{A}_{n}$ of $A$ and ${B}_{1},\mathrm{\dots}{B}_{n}$ of $B$ such that ${A}_{k}$ and ${B}_{k}$ are congruent^{} for $1\le k\le n$. Furthermore, there exist disjoint decompositions ${B}_{1}^{\prime},\mathrm{\dots},{B}_{m}^{\prime}$ of $B$ and ${C}_{1},\mathrm{\dots}{C}_{m}$ of $C$ such that ${B}_{l}^{\prime}$ and ${C}_{l}$ are congruent for $1\le l\le m$. Define
$${B}_{k,l}:={B}_{k}\cap {B}_{l}^{\prime}\text{for}1\le k\le n,1\le l\le m.$$ 
Now if ${A}_{k}$ and ${B}_{k}$ are congruent via some isometry^{} $\theta :{A}_{k}\to {B}_{k}$, we obtain a disjoint decomposition of ${A}_{k}$ by setting ${A}_{k,l}:={\theta}^{1}({B}_{k,l})$. Likewise, if ${B}_{l}^{\prime}$ and ${C}_{l}$ are congruent via some isometry ${\theta}^{\prime}:{B}_{l}^{\prime}\to {C}_{l}$, we obtain a disjoint decomposition of ${C}_{l}$ by setting ${C}_{k,l}:={\theta}^{\prime}({B}_{k,l})$. Clearly, the ${A}_{k,l}$ and ${C}_{k,l}$ define a disjoint decomposition of $A$ and $C$, respectively, into $nm$ parts. By transitivity of congruence^{}, ${A}_{k,l}$ and ${C}_{k,l}$ are congruent for $1\le k\le n$ and $1\le l\le m$. Therefore, $A$ and $C$ are equidecomposable. ∎
Theorem 2.
Given disjoint sets ${A}_{\mathrm{1}}\mathrm{,}\mathrm{,}\mathrm{\dots}{A}_{n}$, ${B}_{\mathrm{1}}\mathrm{,}\mathrm{\dots}\mathit{}{B}_{n}$ such that ${A}_{k}$ and ${B}_{k}$ are equidecomposable for $\mathrm{1}\mathrm{\le}k\mathrm{\le}n$, their unions $A\mathrm{=}\mathrm{\bigcup}_{k\mathrm{=}\mathrm{1}}^{n}{A}_{k}$ and $B\mathrm{=}\mathrm{\bigcup}_{k\mathrm{=}\mathrm{1}}^{n}{B}_{k}$ are equidecomposable as well.
Proof.
By definition, there exists for every $k$, $1\le k\le n$ an integer ${l}_{k}$ such that there are disjoint decompositions
$${A}_{k}=\bigcup _{i=1}^{{l}_{k}}{A}_{k,i},{B}_{k}=\bigcup _{i=1}^{{l}_{k}}{B}_{k,i}$$ 
such that ${A}_{k,i}$ and ${B}_{k,i}$ are congruent for $1\le i\le {l}_{k}$. Rewriting $A$ and $B$ in the form
$$A=\bigcup _{k=1}^{n}\bigcup _{i=1}^{{l}_{k}}{A}_{k,i},B=\bigcup _{k=1}^{n}\bigcup _{i=1}^{{l}_{k}}{B}_{k,i}$$ 
gives the result. ∎
Theorem 3.
Let $A$, $B$, $C$ be sets such that $A$ and $B$ are equidecomposable and $C\mathrm{\u228a}A$, then there exists $D\mathrm{\u228a}B$ such that $C$ and $D$ are equidecomposable.
Proof.
Let ${A}_{1},\mathrm{\dots},{A}_{n}$ and ${B}_{1},\mathrm{\dots}{B}_{n}$ be disjoint decompositions of $A$ and $B$, respectively, such that ${A}_{k}$ and ${B}_{k}$ are congruent via an isometry ${\theta}_{k}:{A}_{k}\to {B}_{k}$ for all $1\le k\le n$. Let $\theta :A\to B$ a map such that $\theta (x)={\theta}_{k}(x)$ if $x\in {A}_{k}$. Since the ${A}_{k}$ are disjoint, $\theta $ is welldefined everywhere. Furthermore, $\theta $ is obviously bijective^{}. Now set ${C}_{k}:=C\cap {A}_{k}$ and define ${D}_{k}:={\theta}_{k}({C}_{k})$, so that ${C}_{k}$ and ${D}_{k}$ are congruent for $1\le k\le n$, so the disjoint union^{} $D:={D}_{1}\cup \mathrm{\cdots}\cup {D}_{n}$ and $C$ are equidecomposable. By construction, $\theta (C)=D$. Since $C$ is a proper subset^{} of $A$ and $\theta $ is bijective, $D$ is a proper subset of $B$. ∎
Theorem 4.
Let $A$, $B$ and $C$ be sets such that $A$ and $C$ are equidecomposable and $A\mathrm{\subseteq}B\mathrm{\subseteq}C$. Then $B$ and $C$ are equidecomposable.
Proof.
Let ${A}_{1},\mathrm{\dots},{A}_{n}$ and ${C}_{1},\mathrm{\dots}{C}_{n}$ disjoint decompositions of $A$ and $C$, respectively, such that ${A}_{k}$ and ${C}_{k}$ are congruent via an isometry ${\theta}_{k}$ for $1\le k\le n$. Like in the proof of theorem 3, let $\theta :A\to C$ be the welldefined, bijective map such that $\theta (x)={\theta}_{k}(x)$ if $x\in {A}_{k}$. Now, for every $b\in B$, let $\mathcal{C}(b)$ be the intersection^{} of all sets $X\subseteq B$ satisfying

•
$b\in X$,

•
for all $x\in X$, the preimage^{} ${\theta}^{1}(x)$ lies in $X$,

•
for all $x\in X\cap A$, the image $\theta (x)$ lies in $X$.
Let ${b}_{1},{b}_{2}\in B$ such that $\mathcal{C}({b}_{1})$ and $\mathcal{C}({b}_{2})$ are not disjoint. Then there is a $b\in \mathcal{C}({b}_{1})\cap \mathcal{C}({b}_{2})$ such that $b={\theta}^{r}({b}_{1})={\theta}^{s}({b}_{2})$ for suitable integers $s$ and $r$. Given ${b}^{\prime}\in \mathcal{C}({b}_{1})$, we have ${b}^{\prime}={\theta}^{t}({b}_{1})={\theta}^{t+sr}({b}_{2})$ for a suitable integer $t$, that is ${b}^{\prime}\in \mathcal{C}({b}_{2})$, so that $\mathcal{C}({b}_{1})\subseteq \mathcal{C}({b}_{2})$. The reverse inclusion follows likewise, and we see that for arbitrary ${b}_{1},{b}_{2}\in B$ either $\mathcal{C}({b}_{1})=\mathcal{C}({b}_{2})$ or $\mathcal{C}({b}_{1})$ and $\mathcal{C}({b}_{2})$ are disjoint. Now set
$$D:=\{b\in B\mid \mathcal{C}(b)\subseteq A\},$$ 
then obviously $D\subseteq A$. If for $b\in B$, $\mathcal{C}(b)$ consists of the sequence^{} of elements $\mathrm{\dots},{\theta}^{1}(b),b,\theta (b),\mathrm{\dots}$ which is infinite^{} in both directions, then $b\in D$. If the sequence is infinite in only one direction, but the final element lies in $A$, then $b\in D$ as well. Let $E:=\theta (D)$ and $F:=B\setminus D$, then clearly $E\cup F\subseteq C$.
Now let $c\in C$. If $c\notin E$, then ${\theta}^{1}(c)\notin D$, so $\mathcal{C}({\theta}^{1}(c))$ consists of a sequence $\mathrm{\dots},{\theta}^{2}(c),{\theta}^{1}(c),\mathrm{\dots}$ which is infinite in only one direction and the final element does not lie in $A$. Now ${\theta}^{1}(c)\in \mathcal{C}({\theta}^{1}(c))$, but since ${\theta}^{1}(c)$ does lie in $A$, it is not the final element. Therefore the subsequent element $c$ lies in $\mathcal{C}({\theta}^{1}(c))$, in particular $c\in B$ and $\mathcal{C}(c)=\mathcal{C}({\theta}^{1}(c))\u2288A$, so $c\in B\setminus D=F$. It follows that $C=E\cup F$, and furthermore $E$ and $F$ are disjoint.
It now follows similarly as in the preceding proofs that $D$ and $E=\theta (D)$ are equidecomposable. By theorem 2, $B=D\cup F$ and $C=E\cup F$ are equidecomposable. ∎
The proof
We may assume that the unit ball $U$ is centered at the origin, that is $U={\mathbb{B}}_{3}$, while the other unit ball ${U}^{\prime}$ has an arbitrary origin. Let $S$ be the surface of $U$, that is, the unit sphere^{}. By the Hausdorff paradox^{}, there exists a disjoint decomposition
$$S={B}^{\prime}\cup {C}^{\prime}\cup {D}^{\prime}\cup {E}^{\prime}$$ 
such that ${B}^{\prime}$, ${C}^{\prime}$, ${D}^{\prime}$ and ${C}^{\prime}\cup {D}^{\prime}$ are congruent, and ${E}^{\prime}$ is countable^{}. For $r>0$ and $A\subseteq {\mathbb{R}}^{3}$, let $rA$ be the set of all vectors of $A$ multiplied by $r$. Set
$$ 
These sets give a disjoint decomposition of the unit ball with the origin deleted, and obviously $B$, $C$, $D$ and $C\cup D$ are congruent (but $E$ is of course not countable). Set
$${A}_{1}:=B\cup E\cup \{0\}$$ 
where $0$ is the origin. $B$ and $C\cup D$ are trivially equidecomposable. Since $C$ and $B$ as well as $D$ and $C$ are congruent, $C\cup D$ and $B\cup C$ are equidecomposable. Finally, $B$ and $C\cup D$ as well as $C$ and $B$ are congruent, so $B\cup C$ and $B\cup C\cup D$ are equidecomposable. In total, $B$ and $B\cup C\cup D$ are equidecomposable by theorem 1, so ${A}_{1}$ and $U$ are equidecomposable by theorem 2. Similarly, we conclude that $C$, $D$ and $B\cup C\cup D$ are equidecomposable.
Since ${E}^{\prime}$ is only countable but there are uncountably many rotations of $S$, there exists a rotation $\theta $ such that $\theta ({E}^{\prime})\u228a{B}^{\prime}\cup {C}^{\prime}\cup {D}^{\prime}$, so $F:=\theta (E)$ is a proper subset of $B\cup C\cup D$. Since $C$ and $B\cup C\cup D$ are equidecomposable, there exists by theorem 3 a proper subset $G\u228aC$ such that $G$ and $F$ (and thus $E$) are equidecomposable. Finally, let $p\in C\setminus G$ an arbitrary point and set
$${A}_{2}:=D\cup G\cup \{p\},$$ 
a disjoint union by construction. Since $D$ and $B\cup C\cup D$, $G$ and $E$ as well as $\{0\}$ and $\{p\}$ are equidecomposable, ${A}_{2}$ and $U$ are equidecomposable by theorem 2. ${A}_{1}$ and ${A}_{2}$ are disjoint (but ${A}_{1}\cup {A}_{2}\ne U$!).
Now $U$ and ${U}^{\prime}$ are congruent, so ${U}^{\prime}$ and ${A}_{2}$ are equidecomposable by theorem 1. If $U$ and ${U}^{\prime}$ are disjoint, set $H:={A}_{2}$. Otherwise we may use theorem 3 and choose $H\u228a{A}_{2}$ such that $H$ and ${U}^{\prime}\setminus U$ are equidecomposable. By theorem 2, ${A}_{1}\cup H$ and $U\cup ({U}^{\prime}\setminus U)=U\cup {U}^{\prime}$ are equidecomposable. Now we have
$${A}_{1}\cup H\subseteq U\subseteq U\cup {U}^{\prime},$$ 
so by theorem 4, $U$ and $U\cup {U}^{\prime}$ are equidecomposable.
References
 BT St. Banach, A. Tarski, Sur la décomposition des ensembles de points en parties respectivement congruentes, Fund. math. 6, 244–277, (1924).
Title  proof of BanachTarski paradox 

Canonical name  ProofOfBanachTarskiParadox 
Date of creation  20130322 15:19:03 
Last modified on  20130322 15:19:03 
Owner  GrafZahl (9234) 
Last modified by  GrafZahl (9234) 
Numerical id  7 
Author  GrafZahl (9234) 
Entry type  Proof 
Classification  msc 51M25 
Classification  msc 03E25 
Related topic  HausdorffParadox 