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 ${ℝ}^3$ isfinite 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 hasever visited Hilbert’s Hotel knows, it is possible to split aninfinite set, such as the natural numbers, in pieces and all piecesare, 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 $\mathbb{N}$, andhalf 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 ofthe size of $\mathbb{N}$.

To do the same thing with $S^2$, bijections won’t suffice, however: weneed isometries to establish congruence. We can almost reducethis problem to the case of bijections in the following way. The group$R$ of rotations (rotations are isometries) in ${ℝ}^3$ 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/1517orbitrepresentatives $M$ of the action of $G$ on $S^2$ (this requires thehttp://planetmath.org/node/310axiom of choice). Then a disjoint decomposition$G_1,\mathrm{\dots }{G}_n$ of $G$ intocountable 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 shouldyield the result, if all the $G_1,\mathrm{\dots }{G}_n$ are related by fixedisometries, for example if $G_1=G_2\phi$ for some isometry$\phi$, and similar relations for the other pieces. This lastrestriction will make the proof possible only in three or moredimensions, 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 ${ℝ}^3$ generated by ahalf rotation $\phi$ and a one third rotation $\psi$ arounddifferent axes. We fix an orthonormal basis for the restof the proof, so we may identify $\phi$ and $\psi$ with theirrotation matrices acting from the right on the row vectors of${ℝ}^3$, such that without restriction

where $\lambda =\cos \frac{2\pi }{3}=-\frac{1}{2}$,$\mu =\sin \frac{2\pi }{3}=\frac{1}{2}\sqrt{3}$ and $\vartheta$ isarbitrary for now.

Since ${\phi }^2=1$ and ${\psi }^3=1$, there existsfor each element $g$ from $G$ other than the unity, $\varphi$, $\psi$ or ${\psi }^2$ apositive 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:

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

• ${\psi }^{m_1}\left(\prod _{k=2}^n\phi {\psi }^{m_k}\right)\phi$,

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

• ${\psi }^{m_1}\left(\prod _{k=2}^n\phi {\psi }^{m_k}\right)$($n\ge 2$).

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 theways ($\alpha$)–($\delta$), with fixed $n$ and $m_k$. In other words: wefix $\vartheta$ such that the unity $1$ cannot be written inone of the ways ($\alpha$)–($\delta$).

To see how to do that, let us see to where the vector $(0,0,1)$ istransported by a transformation of type ($\alpha$). It is easilyverified that

so that$(0,0,1)\phi \psi =(\lambda \sin \vartheta ,\mu \sin \vartheta ,\cos \vartheta )$and$(0,0,1)\phi {\psi }^2=(-\lambda \sin \vartheta ,\mu \sin \vartheta ,\cos \vartheta )$.More generally, let $n$ be a positive integer, $p_1,p_2$ polynomialsof degree $n-1$ and $p_3$ a polynomial of degree $n$, then

$$\begin{array}{c}(p_1(\cos \vartheta )\sin \vartheta ,p_2(\cos \vartheta )\sin \vartheta ,p_3(\cos \vartheta ))\phi \psi =\hfill \\ \hfill ((\lambda p_3(\cos \vartheta )+\mu p_2(\cos \vartheta )-\lambda \cos \vartheta p_1(\cos \vartheta ))\sin \vartheta ,(\mu p_3(\cos \vartheta )-\lambda p_2(\cos \vartheta )-\mu \cos \vartheta p_1(\cos \vartheta ))\sin \vartheta ,(1-{\cos }^2\vartheta )p_1(\cos \vartheta )+p_3(\cos \vartheta )),\end{array}$$

$$\begin{array}{c}(p_1(\cos \vartheta )\sin \vartheta ,p_2(\cos \vartheta )\sin \vartheta ,p_3(\cos \vartheta ))\phi {\psi }^2=\hfill \\ \hfill ((-\lambda p_3(\cos \vartheta )+\mu p_2(\cos \vartheta )+\lambda \cos \vartheta p_1(\cos \vartheta ))\sin \vartheta ,(\mu p_3(\cos \vartheta )+\lambda p_2(\cos \vartheta )-\mu \cos \vartheta p_1(\cos \vartheta ))\sin \vartheta ,(1-{\cos }^2\vartheta )p_1(\cos \vartheta )+p_3(\cos \vartheta )).\end{array}$$

By induction, it follows that $(0,0,1)$ is transported by atransformation of type ($\alpha$) to a vector whose third component isa nonconstant polynomial $p$ in $\cos \vartheta$. If we restrict $\vartheta$ such that$0\le \vartheta \le \pi$, there are only finitely many values for$\vartheta$ such that $p(\cos \vartheta )=1$. Given all possiblecombinations of $n$ and $m_k$, $1\le k\le n$, there are in totalonly countably many problematic values for $\vartheta$, so we caneasily fix hereby an unproblematic one. Now that the case ($\alpha$)has been dealt with, so have been the others automatically. For assumeone 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 $1={\psi }^{3-m_1}1{\psi }^{m_1}$, andlastly 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 hasits axis as fixed point set, which in case of the sphere makes twofixed points per rotation. The product of two rotations is again arotation. Furthermore $G$ is finitely generated and so iscountable. So there are only countably many points of $S^2$ where theaction 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 apartition on $E$ into orbits. The allows us to choosea set $M$, such that $M$ meets any orbit in precisely one element. Wehave 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 asubset of $B$, $A$ or $A$, respectively;

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

• if $Mg$ is a subset of $A$, $B$ or $C$, then $Mg{\psi }^2$ is asubset of $C$, $A$ or $B$, respectively.

The sets $A$, $B$ and $C$ are well-defined because we ensured theuniqueness 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 $D$ iscountable, the theorem is proven.