Hausdorff Paradox

Theorem
There is a disjoint decomposition of the sphere $\mathbb{S}^2 \ $ into four sets $A, B, C, D \ $ such that $A, B, C, B \cup C \ $ are all congruent and $D \ $ is countable.

Proof
Let $R \subset \mathbb{SO}(3) \ $ be the group generated by the $\pi \ $ and $\tfrac{2\pi}{3} \ $ rotations around different axes.

The elements

$\psi = \begin{pmatrix} -\tfrac{1}{2} & \tfrac{\sqrt{3}}{2} & 0 \\ -\tfrac{\sqrt{3}}{2} & -\tfrac{1}{2} & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}, \phi = \begin{pmatrix} -\text{cos}(\vartheta) & 0 & \text{sin}(\vartheta) \\ 0 & -1 & 0 \\ \text{sin}(\vartheta) & 0 & \text{cos}(\vartheta) \\ \end{pmatrix} \ $

form a basis for $R \ $ for some $\vartheta \ $. We have $\psi^3 = \phi^2 = \mathbf{I}_3 \ $, so $\forall r \in R | r \neq \mathbf{I}_3, \psi, \phi, \psi^2 \ $ some natural number $n \ $ and some set of numbers $m_k \in \left\{{1,2}\right\}, 1 \leq k \leq n \ $ such that $r \ $ can written as one of the following:


 * $\displaystyle \text{a}: r = \prod_{k=1}^n \phi \psi^{m_k}$


 * $\displaystyle \text{b}: r = \psi^{m_1} \left({\prod_{k=2}^n \phi \psi^{m_k}}\right) \phi$


 * $\displaystyle \text{c}: r = \left({\prod_{k=1}^n \phi \psi^{m_k}}\right) \phi$


 * $\displaystyle \text{d}: r = \psi^{m_1} \left({\prod_{k=2}^n \phi \psi^{m_k}}\right)$

Now we fix $\vartheta \ $ such that $\mathbf I_3 \ $ cannot be written in any of the ways A, B, C, D.

The action of $R \ $ on $\mathbb{S}^2 \ $ will leave two points unchanged for each element of $R \ $ (the intersection of the axis of rotation and the sphere, to be exact).

Since $R \ $ is finitely generated, it is a countable group, and so the set of points of $\mathbb{S}^2 \ $ which are unchanged by at least one element of $R \ $ is also countable.

We call this set $D \subset \mathbb{S}^2 \ $, so that $R \ $ acts freely on $\mathbb{S}^2 - D \ $. This partitions $\mathbb{S}^2 - D \ $ into orbits. By the Axiom of Choice, there is a set $X \ $ containing one element of each orbit.

For any $r \in R \ $, let $X_r \ $ be the action of $r \ $ on $X \ $.

We have:


 * $\displaystyle \mathbb{S}^2 - D = \bigcup_{r \in R} X_r \ $

Define the sets $A, B, C \ $ to be the smallest sets satisfying


 * $X \subseteq A \ $


 * $\text{If } X_r \subset A, B, C, \text{ then } X_{r\phi} \subset B, A, C, \text{respectively.} \ $


 * $\text{If } X_r \subset A, B, C, \text{ then } X_{r\psi} \subset B, C, A, \text{respectively.} \ $


 * $\text{If } X_r \subset A, B, C, \text{ then } X_{r\phi^2} \subset C, A, B, \text{respectively.} \ $

These sets are defined due to the uniqueness of the properties a-d, and $A, B, C, B \cup C \ $ are congruent since they are rotations of each other, namely:


 * $A_{\psi} = B, B_{\psi^2} = C, A_{\phi} = B \cup C \ $.

Hence we have constructed the sets $A, B, C, D \ $ of the theorem.