Hausdorff Paradox/Proof 2

Lemma $2$
Let $Q$ be the set of all fixed points of $\Bbb S^2$ of all rotations $\alpha \in G$.

Each such $\alpha$ has $2$ fixed points.

Hence $Q$ is countable.

The set $\Bbb S^2 \setminus Q$ is a disjoint union of all orbits $P_x$ of $G$:
 * $P_x = \set {x \alpha: \alpha \in G}$

By the, there exists a set $M$ which contains exactly $1$ element in each $P_x$ where $x \in \Bbb S^2 \setminus Q$.

Let:

It then follows from that:
 * $A$, $B$ and $C$ are disjoint
 * $A$, $B$ and $C$ are congruent to each other
 * $B \cup C$ is congruent to each of $A$, $B$ and $C$

and moreover:
 * $S = A \cup B \cup C \cup Q$