Hausdorff paradox
Let ${S}^{2}$ be the unit sphere^{} in the Euclidean space ${\mathbb{R}}^{3}$. 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 nonmeasurable 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 socalled BanachTarski paradox^{}.
References
 H F. Hausdorff, Bemerkung über den Inhalt von Punktmengen, Math. Ann. 75, 428–433, (1915), http://dzsrv1.sub.unigoettingen.de/sub/digbib/loader?did=D28919http://dzsrv1.sub.unigoettingen.de/sub/digbib/loader?did=D28919 (in German).
Title  Hausdorff paradox^{} 

Canonical name  HausdorffParadox 
Date of creation  20130322 15:16:12 
Last modified on  20130322 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 