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:

$$ \text{a}:

r = \prod_{k=1}^n \phi \psi^{m_k} $$

$$

\text{b}:

r = \psi^{m_1} \left( \prod_{k=2}^n \phi \psi^{m_k} \right)  \phi $$

$$

\text{c}:

r = \left( \prod_{k=1}^n \phi \psi^{m_k} \right) \phi

$$

$$ \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

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

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.