# 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 $G_{1},\ldots G_{n}$ of $G$ into countable sets acting on $M$ yields a disjoint decomposition of $S^{2}$. Doing similar    juggling with $G_{1},\ldots,G_{n}$ as we did with $\mathbb{N}$, $\mathbb{N}_{3}$ and $\mathbb{N}\setminus\mathbb{N}_{3}$ above should yield the result, if all the $G_{1},\ldots G_{n}$ are related by fixed isometries, for example if $G_{1}=G_{2}\varphi$ for some isometry $\varphi$, 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 $\varphi$ and a one third rotation $\psi$ around different axes. We fix an orthonormal basis for the rest of the proof, so we may identify $\varphi$ and $\psi$ with their rotation matrices  acting from the right on the row vectors  of $\mathbb{R}^{3}$, such that without restriction

 $\psi=\left(\begin{array}[]{rrr}\lambda&\mu&0\\ -\mu&\lambda&0\\ 0&0&1\end{array}\right),\qquad\varphi=\left(\begin{array}[]{rrr}-\cos\vartheta% &0&\sin\vartheta\\ 0&-1&0\\ \sin\vartheta&0&\cos\vartheta\end{array}\right),$

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

Since $\varphi^{2}=1$ and $\psi^{3}=1$, there exists for each element $g$ from $G$ other than the unity, $\phi$, $\psi$ or $\psi^{2}$ a positive integer $n$ and numbers $m_{k}\in\{1,2\}$, $1\leq k\leq n$, such that $g$ can be written in one of the following ways:

• ($\alpha$)

$\left(\prod\limits_{k=1}^{n}\varphi\psi^{m_{k}}\right)$,

• ($\beta$)

$\psi^{m_{1}}\left(\prod\limits_{k=2}^{n}\varphi\psi^{m_{k}}\right)\varphi$,

• ($\gamma$)

$\left(\prod\limits_{k=1}^{n}\varphi\psi^{m_{k}}\right)\varphi$,

• ($\delta$)

$\psi^{m_{1}}\left(\prod\limits_{k=2}^{n}\varphi\psi^{m_{k}}\right)$ ($n\geq 2$).

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

 $\varphi\psi=\left(\begin{array}[]{rrr}-\lambda\cos\vartheta&-\mu\cos\vartheta&% \sin\vartheta\\ \mu&-\lambda&0\\ \lambda\sin\vartheta&\mu\sin\vartheta&\cos\vartheta\end{array}\right)\text{ and }\varphi\psi^{2}=\left(\begin{array}[]{rrr}\lambda\cos\vartheta&-\mu\cos% \vartheta&\sin\vartheta\\ \mu&\lambda&0\\ -\lambda\sin\vartheta&\mu\sin\vartheta&\cos\vartheta\end{array}\right),$

so that $(0,0,1)\varphi\psi=(\lambda\sin\vartheta,\mu\sin\vartheta,\cos\vartheta)$ and $(0,0,1)\varphi\psi^{2}=(-\lambda\sin\vartheta,\mu\sin\vartheta,\cos\vartheta)$. More generally, let $n$ be a positive integer, $p_{1},p_{2}$ polynomials of degree $n-1$ and $p_{3}$ a polynomial of degree $n$, then

 $\displaystyle(p_{1}(\cos\vartheta)\sin\vartheta,p_{2}(\cos\vartheta)\sin% \vartheta,p_{3}(\cos\vartheta))\varphi\psi=\\ \displaystyle((\lambda p_{3}(\cos\vartheta)+\mu p_{2}(\cos\vartheta)-\lambda% \cos\vartheta p_{1}(\cos\vartheta))\sin\vartheta,(\mu p_{3}(\cos\vartheta)-% \lambda p_{2}(\cos\vartheta)-\mu\cos\vartheta p_{1}(\cos\vartheta))\sin% \vartheta,(1-\cos^{2}\vartheta)p_{1}(\cos\vartheta)+p_{3}(\cos\vartheta)),$
 $\displaystyle(p_{1}(\cos\vartheta)\sin\vartheta,p_{2}(\cos\vartheta)\sin% \vartheta,p_{3}(\cos\vartheta))\varphi\psi^{2}=\\ \displaystyle((-\lambda p_{3}(\cos\vartheta)+\mu p_{2}(\cos\vartheta)+\lambda% \cos\vartheta p_{1}(\cos\vartheta))\sin\vartheta,(\mu p_{3}(\cos\vartheta)+% \lambda p_{2}(\cos\vartheta)-\mu\cos\vartheta p_{1}(\cos\vartheta))\sin% \vartheta,(1-\cos^{2}\vartheta)p_{1}(\cos\vartheta)+p_{3}(\cos\vartheta)).$

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 $\cos\vartheta$. If we restrict $\vartheta$ such that $0\leq\vartheta\leq\pi$, there are only finitely many values for $\vartheta$ such that $p(\cos\vartheta)=1$. Given all possible combinations   of $n$ and $m_{k}$, $1\leq k\leq n$, 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 $1=\varphi 1\varphi$, 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 $1=\varphi 1\varphi$, 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  $E=\bigcup\limits_{g\in G}Mg,$

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 $Mg\varphi$ 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 well-defined 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

 $A\psi=B,\qquad A\psi^{2}=C\quad\text{and}\quad A\varphi=B\cup C.$

Since $S^{2}=A\cup B\cup C\cup D$, a disjoint union, and $D$ is countable, the theorem is proven.

## References

Title proof of Hausdorff paradox ProofOfHausdorffParadox 2013-03-22 15:16:18 2013-03-22 15:16:18 GrafZahl (9234) GrafZahl (9234) 6 GrafZahl (9234) Proof msc 03E25 msc 51M04 msc 20F05 ChoiceFunction BanachTarskiParadox