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Countable Proof Set Euclidean Vector Space Disjoint Sphere Congruence Area Paradox Banach Tarski Paradox

1	Euclidean Vector Space
1	Proof
1	Disjoint
1	Countable
1	Set
2	Sphere
2	Area
2	Congruence
3	Paradox
9	Banach Tarski Paradox

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Hausdorff Paradox

Let be the unit sphere in the Euclidean space $\mathbb{R}^3$ . Then it is possible to take “half” and “a third” of 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 1 (Hausdorff paradox [H]) There exists a disjoint decomposition of the unit sphere in the Euclidean space $\mathbb{R}^3$ into four subsets , such that the following conditions are met:

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

A crucial ingredient to the proof is the axiom of choice, so the sets , and are not constructible. The theorem itself is a crucial ingredient to the proof of the so-called Banach-Tarski paradox.

Bibliography

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=D28919 (in German).