Hausdorff paradox

Let $S^2$ be the unit sphere in the Euclidean space ${ℝ}^3$. Then it is possible to take “half” and “a third” of $S^2$ such that both of these parts are essentially congruent (we give a formal version in a minute). This sounds paradoxical: wouldn’t that mean that half of the sphere’s area is equal to only a third? The “paradox” resolves itself if one takes into account that one 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 the Euclidean space ${\mathrm{R}}^{\mathrm{3}}$ into four subsets $A\mathrm{,}B\mathrm{,}C\mathrm{,}D$, such that the following 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 the sets $A$, $B$ and $C$ are not constructible. The theorem itself is a crucial ingredient to the proof of the so-called Banach-Tarski paradox.

Title	Hausdorff paradox
Canonical name	HausdorffParadox
Date of creation	2013-03-22 15:16:12
Last modified on	2013-03-22 15:16:12
Owner	GrafZahl (9234)
Last modified by	GrafZahl (9234)
Numerical id	9
Author	GrafZahl (9234)
Entry type	Theorem
Classification	msc 03E25
Classification	msc 51M04
Related topic	ChoiceFunction
Related topic	BanachTarskiParadox
Related topic	ProofofBanachTarskiParadox