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 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.

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

1. 1.

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

2. 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.

## References

• H F. Hausdorff, Bemerkung über den Inhalt von Punktmengen, Math. Ann. 75, 428–433, (1915), http://dz-srv1.sub.uni-goettingen.de/sub/digbib/loader?did=D28919http://dz-srv1.sub.uni-goettingen.de/sub/digbib/loader?did=D28919 (in German).
Title Hausdorff paradox HausdorffParadox 2013-03-22 15:16:12 2013-03-22 15:16:12 GrafZahl (9234) GrafZahl (9234) 9 GrafZahl (9234) Theorem msc 03E25 msc 51M04 ChoiceFunction BanachTarskiParadox ProofofBanachTarskiParadox