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

The elements:


 * $\psi = \begin{pmatrix} -\tfrac 1 2 & \tfrac{\sqrt 3} 2 & 0 \\ -\tfrac{\sqrt 3} 2 & -\tfrac 1 2 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}$


 * $\phi = \begin{pmatrix} -\cos \left({\vartheta}\right) & 0 & \sin \left({\vartheta}\right) \\ 0 & -1 & 0 \\ \sin \left({\vartheta}\right) & 0 & \cos \left({\vartheta}\right) \\ \end{pmatrix}$

form a basis for $R$ for some $\vartheta$.

We have:
 * $\psi^3 = \phi^2 = \mathbf I_3$

where $\mathbf I_3$ is the identity mapping in $\mathbb R^3$.

Therefore $\forall r \in R: r \ne \mathbf I_3, \psi, \phi, \psi^2$ there exists some $n \in \N$ and some set of numbers $m_k \in \left\{{1, 2}\right\}, 1 \le k \le n$ such that $r$ can written as one of the following:


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


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


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


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

Now we fix $\vartheta$ such that $\mathbf I_3$ cannot be written in any of the ways $\text{(a)}$, $\text{(b)}$, $\text{(c)}$, $\text{(d)}$.

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.

Therefore 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 $\mathbb S^2 - D$.

By Orbit is Equivalence Class, 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:


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

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 $\text{(a)}$ to $\text{(d)}$

Also, $A, B, C, B \cup C$ 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.

Whether you view this result as a veridical paradox or an antinomy depends your acceptance or otherwise of the Axiom of Choice.