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Hausdorff paradox
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(Theorem)
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Let $S^2$ be the unit sphere in the Euclidean space $\mbb{R}^3$ . 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 1 (Hausdorff paradox [ H]) There exists a disjoint decomposition of the unit sphere $S^2$ in the Euclidean space $\mbb{R}^3$ into four subsets $A,B,C,D$ , such that the following conditions are met:
- Any two of the sets $A$ , $B$ , $C$ and $B\cup C$ are congruent.
- $D$ is countable.
A crucial ingredient to the proof is the axiom 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.
- 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).
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Cross-references: Banach-Tarski paradox, proof, countable, disjoint, theorem, subsets, area, sphere's, paradoxical, congruent, Euclidean space, unit sphere
There are 3 references to this entry.
This is version 6 of Hausdorff paradox, born on 2005-05-15, modified 2006-12-24.
Object id is 7057, canonical name is HausdorffParadox.
Accessed 4228 times total.
Classification:
| AMS MSC: | 51M04 (Geometry :: Real and complex geometry :: Elementary problems in Euclidean geometries) | | | 03E25 (Mathematical logic and foundations :: Set theory :: Axiom of choice and related propositions) |
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Pending Errata and Addenda
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