proof of Hausdorff paradox

We start by outlining the general ideas, followed by the strict proof.

General approach

The unit sphereMathworldPlanetmath S2 in the Euclidean space 3 is finite in the sense that it is and has a finite volume: 43π.

On the other hand, S2 is an infiniteMathworldPlanetmath point set. As everyone who has ever visited Hilbert’s Hotel knows, it is possible to split an infinite set, such as the natural numbersMathworldPlanetmath, in pieces and all pieces are, in a sense, equal to the original set, for example if

3:={nn is divisible by 3},

then naïvely 3 has a third of the size of , and half of the size of 3. There exists, however, a bijectionMathworldPlanetmath from to 3, so 3 has, again naïvely, both half and a third of the size of .

To do the same thing with S2, bijections won’t suffice, however: we need isometriesMathworldPlanetmath to establish congruenceMathworldPlanetmathPlanetmathPlanetmathPlanetmath. We can almost reduce this problem to the case of bijections in the following way. The group R of rotationsMathworldPlanetmath (rotations are isometries) in 3 acts on S2 (say, from the right). Given an countably infiniteMathworldPlanetmath subgroupMathworldPlanetmathPlanetmath G of R which acts freely (or almost so) on S2, take a set of representatives M of the action of G on S2 (this requires the of choiceMathworldPlanetmath). Then a disjoint decomposition G1,Gn of G into countable sets acting on M yields a disjoint decomposition of S2. Doing similarMathworldPlanetmathPlanetmathPlanetmath juggling with G1,,Gn as we did with , 3 and 3 above should yield the result, if all the G1,Gn are related by fixed isometries, for example if G1=G2φ for some isometry φ, and similar relationsMathworldPlanetmathPlanetmath for the other pieces. This last restrictionPlanetmathPlanetmathPlanetmath will make the proof possible only in three or more dimensionsPlanetmathPlanetmathPlanetmath, and indeed, Banach and Tarski later showed in [BT] that analogous theorems on the line or in the plane do not hold.

The proof

Let G be the subgroup of rotations of 3 generated by a half rotation φ and a one third rotation ψ around different axes. We fix an orthonormal basis for the rest of the proof, so we may identify φ and ψ with their rotation matricesMathworldPlanetmath acting from the right on the row vectorsMathworldPlanetmath of 3, such that without restriction


where λ=cos2π3=-12, μ=sin2π3=123 and ϑ is arbitrary for now.

Since φ2=1 and ψ3=1, there exists for each element g from G other than the unity, ϕ, ψ or ψ2 a positive integer n and numbers mk{1,2}, 1kn, such that g can be written in one of the following ways:

  • (α)


  • (β)


  • (γ)


  • (δ)

    ψm1(k=2nφψmk) (n2).

To have completePlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath control over the structureMathworldPlanetmath of G, we fix ϑ in such a way, that g can be written uniquely in one of the ways (α)–(δ), with fixed n and mk. In other words: we fix ϑ such that the unity 1 cannot be written in one of the ways (α)–(δ).

To see how to do that, let us see to where the vector (0,0,1) is transported by a transformationMathworldPlanetmath of type (α). It is easily verified that

φψ=(-λcosϑ-μcosϑsinϑμ-λ0λsinϑμsinϑcosϑ) and φψ2=(λcosϑ-μcosϑsinϑμλ0-λsinϑμsinϑcosϑ),

so that (0,0,1)φψ=(λsinϑ,μsinϑ,cosϑ) and (0,0,1)φψ2=(-λsinϑ,μsinϑ,cosϑ). More generally, let n be a positive integer, p1,p2 polynomials of degree n-1 and p3 a polynomial of degree n, then


By inductionMathworldPlanetmath, it follows that (0,0,1) is transported by a transformation of type (α) to a vector whose third componentMathworldPlanetmathPlanetmathPlanetmath is a nonconstant polynomial p in cosϑ. If we restrict ϑ such that 0ϑπ, there are only finitely many values for ϑ such that p(cosϑ)=1. Given all possible combinationsMathworldPlanetmathPlanetmath of n and mk, 1kn, there are in total only countably many problematic values for ϑ, so we can easily fix hereby an unproblematic one. Now that the case (α) has been dealt with, so have been the others automatically. For assume one could write a 1 of type (γ), one could convert it to type (δ) by 1=φ1φ, and from type δ to type (α) or type (β) by 1=ψ3-m11ψm1, and lastly from type (β) to type (α) by 1=φ1φ, completing the contradictionMathworldPlanetmathPlanetmath.

Now that we know the structure of G, how does it act on S2? Certainly not freely, since every rotation other than the unity has its axis as fixed pointPlanetmathPlanetmathPlanetmath set, which in case of the sphere makes two fixed points per rotation. The productPlanetmathPlanetmath of two rotations is again a rotation. Furthermore G is finitely generatedMathworldPlanetmath and so is countableMathworldPlanetmath. So there are only countably many points of S2 where the action of G fails to be free. Denote the set of these points by D, so that G acts freely on E:=S2D. The action creates a partitionMathworldPlanetmathPlanetmath on E into orbits. The allows us to choose a set M, such that M meets any orbit in precisely one element. We have then the disjoint unionMathworldPlanetmath


where Mg is the image of M under the action of the group element g. We now define three sets A, B, C to be the smallest sets satisfying:

  • M1=MA;

  • if Mg is a subset of A, B or C, then Mgφ is a subset of B, A or A, respectively;

  • if Mg is a subset of A, B or C, then Mgψ is a subset of B, C or A, respectively;

  • if Mg is a subset of A, B or C, then Mgψ2 is a subset of C, A or B, respectively.

The sets A, B and C are well-defined because we ensured the uniqueness of the representations (α)–(δ) above.

The sets A, B, C and BC are all congruentMathworldPlanetmath by virtue of


Since S2=ABCD, a disjoint union, and D is countable, the theorem is proven.


  • BT St. Banach, A. Tarski, Sur la décompositionMathworldPlanetmathPlanetmath des ensembles de points en parties respectivement congruentes, Fund. math. 6, 244–277, (1924).
  • H F. Hausdorff, Bemerkung über den Inhalt von Punktmengen, Math. Ann. 75, 428–433, (1915).
Title proof of Hausdorff paradox
Canonical name ProofOfHausdorffParadox
Date of creation 2013-03-22 15:16:18
Last modified on 2013-03-22 15:16:18
Owner GrafZahl (9234)
Last modified by GrafZahl (9234)
Numerical id 6
Author GrafZahl (9234)
Entry type Proof
Classification msc 03E25
Classification msc 51M04
Classification msc 20F05
Related topic ChoiceFunction
Related topic BanachTarskiParadox