Hausdorff paradox

Let $S^{2}$ be the unit sphere in the Euclidean space $\\mathbb{R}^{3}$ . 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 $S^{2}$ in theEuclidean space $\\mathbb{R}^{3}$ into four subsets $A, B, C, D$ , 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.

References

H F. Hausdorff, Bemerkung über den Inhalt vonPunktmengen, 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).

单词	HausdorffParadox
释义	Hausdorff paradox Let $S^{2}$ be the unit sphere in the Euclidean space $\\mathbb{R}^{3}$ . 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 $S^{2}$ in theEuclidean space $\\mathbb{R}^{3}$ into four subsets $A, B, C, D$ , 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. References H F. Hausdorff, Bemerkung über den Inhalt vonPunktmengen, 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).
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