Hausdorff paradox

Let S2 be the unit sphereMathworldPlanetmath in the Euclidean space 3. Then it is possible to take “half” and “a third” of S2 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 (Hausdorff paradox [H]).

There exists a disjoint of the unit sphere S2 in the Euclidean space R3 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 BC are congruent.

  2. 2.

    D is countableMathworldPlanetmath.

A crucial ingredient to the proof is the of choiceMathworldPlanetmath, 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 paradoxMathworldPlanetmath.


  • H F. Hausdorff, Bemerkung über den Inhalt von Punktmengen, Math. Ann. 75, 428–433, (1915), (in German).
Title Hausdorff paradoxMathworldPlanetmath
Canonical name HausdorffParadox
Date of creation 2013-03-22 15:16:12
Last modified on 2013-03-22 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