proof of Hausdorff paradox
We start by outlining the general ideas, followed by the strict proof.
General approach
The unit sphere^{} ${S}^{2}$ in the Euclidean space ${\mathbb{R}}^{3}$ is finite in the sense that it is http://planetmath.org/node/4826bounded and has a finite volume: $\frac{4}{3}\pi $.
On the other hand, ${S}^{2}$ is an infinite^{} point set. As everyone who has ever visited Hilbert’s Hotel knows, it is possible to split an infinite set, such as the natural numbers^{}, in pieces and all pieces are, in a sense, equal to the original set, for example if
$${\mathbb{N}}_{3}:=\{n\in \mathbb{N}\mid n\text{is divisible by}3\},$$ 
then naïvely ${\mathbb{N}}_{3}$ has a third of the size of $\mathbb{N}$, and half of the size of $\mathbb{N}\setminus {\mathbb{N}}_{3}$. There exists, however, a bijection^{} from $\mathbb{N}$ to $\mathbb{N}\setminus {\mathbb{N}}_{3}$, so ${\mathbb{N}}_{3}$ has, again naïvely, both half and a third of the size of $\mathbb{N}$.
To do the same thing with ${S}^{2}$, bijections won’t suffice, however: we need isometries^{} to establish congruence^{}. We can almost reduce this problem to the case of bijections in the following way. The group $R$ of rotations^{} (rotations are isometries) in ${\mathbb{R}}^{3}$ acts on ${S}^{2}$ (say, from the right). Given an countably infinite^{} subgroup^{} $G$ of $R$ which acts freely (or almost so) on ${S}^{2}$, take a set of http://planetmath.org/node/1517orbit representatives $M$ of the action of $G$ on ${S}^{2}$ (this requires the http://planetmath.org/node/310axiom of choice^{}). Then a disjoint decomposition ${G}_{1},\mathrm{\dots}{G}_{n}$ of $G$ into countable sets acting on $M$ yields a disjoint decomposition of ${S}^{2}$. Doing similar^{} juggling with ${G}_{1},\mathrm{\dots},{G}_{n}$ as we did with $\mathbb{N}$, ${\mathbb{N}}_{3}$ and $\mathbb{N}\setminus {\mathbb{N}}_{3}$ above should yield the result, if all the ${G}_{1},\mathrm{\dots}{G}_{n}$ are related by fixed isometries, for example if ${G}_{1}={G}_{2}\phi $ for some isometry $\phi $, and similar relations^{} for the other pieces. This last restriction^{} will make the proof possible only in three or more dimensions^{}, and indeed, Banach and Tarski later showed in [BT] that analogous theorems on the line or in the plane do not hold.
The proof
Let $G$ be the subgroup of rotations of ${\mathbb{R}}^{3}$ generated by a half rotation $\phi $ and a one third rotation $\psi $ around different axes. We fix an orthonormal basis for the rest of the proof, so we may identify $\phi $ and $\psi $ with their rotation matrices^{} acting from the right on the row vectors^{} of ${\mathbb{R}}^{3}$, such that without restriction
$$\psi =\left(\begin{array}{ccc}\hfill \lambda & \hfill \mu & \hfill 0\\ \hfill \mu & \hfill \lambda & \hfill 0\\ \hfill 0& \hfill 0& \hfill 1\end{array}\right),\phi =\left(\begin{array}{ccc}\hfill \mathrm{cos}\vartheta & \hfill 0& \hfill \mathrm{sin}\vartheta \\ \hfill 0& \hfill 1& \hfill 0\\ \hfill \mathrm{sin}\vartheta & \hfill 0& \hfill \mathrm{cos}\vartheta \end{array}\right),$$ 
where $\lambda =\mathrm{cos}\frac{2\pi}{3}=\frac{1}{2}$, $\mu =\mathrm{sin}\frac{2\pi}{3}=\frac{1}{2}\sqrt{3}$ and $\vartheta $ is arbitrary for now.
Since ${\phi}^{2}=1$ and ${\psi}^{3}=1$, there exists for each element $g$ from $G$ other than the unity, $\varphi $, $\psi $ or ${\psi}^{2}$ a positive integer $n$ and numbers ${m}_{k}\in \{1,2\}$, $1\le k\le n$, such that $g$ can be written in one of the following ways:

($\alpha $)
$\left(\prod _{k=1}^{n}\phi {\psi}^{{m}_{k}}\right)$,

($\beta $)
${\psi}^{{m}_{1}}\left(\prod _{k=2}^{n}\phi {\psi}^{{m}_{k}}\right)\phi $,

($\gamma $)
$\left(\prod _{k=1}^{n}\phi {\psi}^{{m}_{k}}\right)\phi $,

($\delta $)
${\psi}^{{m}_{1}}\left(\prod _{k=2}^{n}\phi {\psi}^{{m}_{k}}\right)$ ($n\ge 2$).
To have complete^{} control over the structure^{} of $G$, we fix $\vartheta $ in such a way, that $g$ can be written uniquely in one of the ways ($\alpha $)–($\delta $), with fixed $n$ and ${m}_{k}$. In other words: we fix $\vartheta $ such that the unity $1$ cannot be written in one of the ways ($\alpha $)–($\delta $).
To see how to do that, let us see to where the vector $(0,0,1)$ is transported by a transformation^{} of type ($\alpha $). It is easily verified that
$$\phi \psi =\left(\begin{array}{ccc}\hfill \lambda \mathrm{cos}\vartheta & \hfill \mu \mathrm{cos}\vartheta & \hfill \mathrm{sin}\vartheta \\ \hfill \mu & \hfill \lambda & \hfill 0\\ \hfill \lambda \mathrm{sin}\vartheta & \hfill \mu \mathrm{sin}\vartheta & \hfill \mathrm{cos}\vartheta \end{array}\right)\text{and}\phi {\psi}^{2}=\left(\begin{array}{ccc}\hfill \lambda \mathrm{cos}\vartheta & \hfill \mu \mathrm{cos}\vartheta & \hfill \mathrm{sin}\vartheta \\ \hfill \mu & \hfill \lambda & \hfill 0\\ \hfill \lambda \mathrm{sin}\vartheta & \hfill \mu \mathrm{sin}\vartheta & \hfill \mathrm{cos}\vartheta \end{array}\right),$$ 
so that $(0,0,1)\phi \psi =(\lambda \mathrm{sin}\vartheta ,\mu \mathrm{sin}\vartheta ,\mathrm{cos}\vartheta )$ and $(0,0,1)\phi {\psi}^{2}=(\lambda \mathrm{sin}\vartheta ,\mu \mathrm{sin}\vartheta ,\mathrm{cos}\vartheta )$. More generally, let $n$ be a positive integer, ${p}_{1},{p}_{2}$ polynomials of degree $n1$ and ${p}_{3}$ a polynomial of degree $n$, then
$$\begin{array}{c}({p}_{1}(\mathrm{cos}\vartheta )\mathrm{sin}\vartheta ,{p}_{2}(\mathrm{cos}\vartheta )\mathrm{sin}\vartheta ,{p}_{3}(\mathrm{cos}\vartheta ))\phi \psi =\hfill \\ \hfill ((\lambda {p}_{3}(\mathrm{cos}\vartheta )+\mu {p}_{2}(\mathrm{cos}\vartheta )\lambda \mathrm{cos}\vartheta {p}_{1}(\mathrm{cos}\vartheta ))\mathrm{sin}\vartheta ,(\mu {p}_{3}(\mathrm{cos}\vartheta )\lambda {p}_{2}(\mathrm{cos}\vartheta )\mu \mathrm{cos}\vartheta {p}_{1}(\mathrm{cos}\vartheta ))\mathrm{sin}\vartheta ,(1{\mathrm{cos}}^{2}\vartheta ){p}_{1}(\mathrm{cos}\vartheta )+{p}_{3}(\mathrm{cos}\vartheta )),\end{array}$$ 
$$\begin{array}{c}({p}_{1}(\mathrm{cos}\vartheta )\mathrm{sin}\vartheta ,{p}_{2}(\mathrm{cos}\vartheta )\mathrm{sin}\vartheta ,{p}_{3}(\mathrm{cos}\vartheta ))\phi {\psi}^{2}=\hfill \\ \hfill ((\lambda {p}_{3}(\mathrm{cos}\vartheta )+\mu {p}_{2}(\mathrm{cos}\vartheta )+\lambda \mathrm{cos}\vartheta {p}_{1}(\mathrm{cos}\vartheta ))\mathrm{sin}\vartheta ,(\mu {p}_{3}(\mathrm{cos}\vartheta )+\lambda {p}_{2}(\mathrm{cos}\vartheta )\mu \mathrm{cos}\vartheta {p}_{1}(\mathrm{cos}\vartheta ))\mathrm{sin}\vartheta ,(1{\mathrm{cos}}^{2}\vartheta ){p}_{1}(\mathrm{cos}\vartheta )+{p}_{3}(\mathrm{cos}\vartheta )).\end{array}$$ 
By induction^{}, it follows that $(0,0,1)$ is transported by a transformation of type ($\alpha $) to a vector whose third component^{} is a nonconstant polynomial $p$ in $\mathrm{cos}\vartheta $. If we restrict $\vartheta $ such that $0\le \vartheta \le \pi $, there are only finitely many values for $\vartheta $ such that $p(\mathrm{cos}\vartheta )=1$. Given all possible combinations^{} of $n$ and ${m}_{k}$, $1\le k\le n$, there are in total only countably many problematic values for $\vartheta $, so we can easily fix hereby an unproblematic one. Now that the case ($\alpha $) has been dealt with, so have been the others automatically. For assume one could write a $1$ of type ($\gamma $), one could convert it to type ($\delta $) by $1=\phi 1\phi $, and from type $\delta $ to type ($\alpha $) or type ($\beta $) by $1={\psi}^{3{m}_{1}}1{\psi}^{{m}_{1}}$, and lastly from type ($\beta $) to type ($\alpha $) by $1=\phi 1\phi $, completing the contradiction^{}.
Now that we know the structure of $G$, how does it act on ${S}^{2}$? Certainly not freely, since every rotation other than the unity has its axis as fixed point^{} set, which in case of the sphere makes two fixed points per rotation. The product^{} of two rotations is again a rotation. Furthermore $G$ is finitely generated^{} and so is countable^{}. So there are only countably many points of ${S}^{2}$ where the action of $G$ fails to be free. Denote the set of these points by $D$, so that $G$ acts freely on $E:={S}^{2}\setminus D$. The action creates a partition^{} on $E$ into orbits. The allows us to choose a set $M$, such that $M$ meets any orbit in precisely one element. We have then the disjoint union^{}
$$E=\bigcup _{g\in G}Mg,$$ 
where $Mg$ is the image of $M$ under the action of the group element $g$. We now define three sets $A$, $B$, $C$ to be the smallest sets satisfying:

•
$M1=M\subseteq A$;

•
if $Mg$ is a subset of $A$, $B$ or $C$, then $Mg\phi $ is a subset of $B$, $A$ or $A$, respectively;

•
if $Mg$ is a subset of $A$, $B$ or $C$, then $Mg\psi $ is a subset of $B$, $C$ or $A$, respectively;

•
if $Mg$ is a subset of $A$, $B$ or $C$, then $Mg{\psi}^{2}$ is a subset of $C$, $A$ or $B$, respectively.
The sets $A$, $B$ and $C$ are welldefined because we ensured the uniqueness of the representations ($\alpha $)–($\delta $) above.
The sets $A$, $B$, $C$ and $B\cup C$ are all congruent^{} by virtue of
$$A\psi =B,A{\psi}^{2}=C\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}A\phi =B\cup C.$$ 
Since ${S}^{2}=A\cup B\cup C\cup D$, a disjoint union, and $D$ is countable, the theorem is proven.
References
 BT St. Banach, A. Tarski, Sur la décomposition^{} des ensembles de points en parties respectivement congruentes, Fund. math. 6, 244–277, (1924).
 H F. Hausdorff, Bemerkung über den Inhalt von Punktmengen, Math. Ann. 75, 428–433, (1915).
Title  proof of Hausdorff paradox 

Canonical name  ProofOfHausdorffParadox 
Date of creation  20130322 15:16:18 
Last modified on  20130322 15:16:18 
Owner  GrafZahl (9234) 
Last modified by  GrafZahl (9234) 
Numerical id  6 
Author  GrafZahl (9234) 
Entry type  Proof 
Classification  msc 03E25 
Classification  msc 51M04 
Classification  msc 20F05 
Related topic  ChoiceFunction 
Related topic  BanachTarskiParadox 