Hausdorff paradox

Let $S^2$ be the unit sphere in the Euclidean space ${ℝ}^3$. Thenit is possible to take “half” and “a third” of $S^2$ such thatboth of these parts are essentially congruent (we give a formalversion in a minute). This sounds paradoxical:wouldn’t that mean that half of the sphere’s area is equal to only athird? The “paradox” resolves itself if one takes into account thatone can choose non-measurable subsets of the sphere which ostensively are “half” and a “third” of it, using geometric congruence as means of comparison.

Let us now formally state the Theorem.

Theorem (Hausdorff paradox [H]).

There exists a disjoint of the unit sphere $S^{\mathrm{2}}$ in theEuclidean space ${\mathrm{R}}^{\mathrm{3}}$ into four subsets $A\mathrm{,}B\mathrm{,}C\mathrm{,}D$, such that thefollowing conditions are met:

1. Any two of the sets $A$, $B$, $C$ and $B\cup C$ are congruent.

2. $D$ is countable.

A crucial ingredient to the proof is the http://planetmath.org/node/310axiom of choice, so thesets $A$, $B$ and $C$ are not constructible. The theorem itself is acrucial ingredient to the proof of the so-called Banach-Tarskiparadox.