proof of Banach-Tarski paradox

We deal with some technicalities first, mainly concerning the properties of equi-decomposability. We can then prove the paradoxMathworldPlanetmath in a clear and unencumbered line of argument: we show that, given two unit ballsMathworldPlanetmath U and U with arbitrary origin, U and UU are equi-decomposable, regardless whether U and U are disjoint or not. The original proof can be found in [BT].


Theorem 1.

Equi-decomposability gives rise to an equivalence relationMathworldPlanetmath on the subsets of Euclidean space.


The only non-obvious part is transitivity. So let A, B, C be sets such that A and B as well as B and C are equi-decomposable. Then there exist disjoint decompositions A1,An of A and B1,Bn of B such that Ak and Bk are congruentMathworldPlanetmathPlanetmath for 1kn. Furthermore, there exist disjoint decompositions B1,,Bm of B and C1,Cm of C such that Bl and Cl are congruent for 1lm. Define

Bk,l:=BkBl for 1kn,1lm.

Now if Ak and Bk are congruent via some isometryMathworldPlanetmath θ:AkBk, we obtain a disjoint decomposition of Ak by setting Ak,l:=θ-1(Bk,l). Likewise, if Bl and Cl are congruent via some isometry θ:BlCl, we obtain a disjoint decomposition of Cl by setting Ck,l:=θ(Bk,l). Clearly, the Ak,l and Ck,l define a disjoint decomposition of A and C, respectively, into nm parts. By transitivity of congruenceMathworldPlanetmathPlanetmath, Ak,l and Ck,l are congruent for 1kn and 1lm. Therefore, A and C are equi-decomposable. ∎

Theorem 2.

Given disjoint sets A1,,An, B1,Bn such that Ak and Bk are equi-decomposable for 1kn, their unions A=k=1nAk and B=k=1nBk are equi-decomposable as well.


By definition, there exists for every k, 1kn an integer lk such that there are disjoint decompositions


such that Ak,i and Bk,i are congruent for 1ilk. Rewriting A and B in the form


gives the result. ∎

Theorem 3.

Let A, B, C be sets such that A and B are equi-decomposable and CA, then there exists DB such that C and D are equi-decomposable.


Let A1,,An and B1,Bn be disjoint decompositions of A and B, respectively, such that Ak and Bk are congruent via an isometry θk:AkBk for all 1kn. Let θ:AB a map such that θ(x)=θk(x) if xAk. Since the Ak are disjoint, θ is well-defined everywhere. Furthermore, θ is obviously bijectiveMathworldPlanetmathPlanetmath. Now set Ck:=CAk and define Dk:=θk(Ck), so that Ck and Dk are congruent for 1kn, so the disjoint unionMathworldPlanetmath D:=D1Dn and C are equi-decomposable. By construction, θ(C)=D. Since C is a proper subsetMathworldPlanetmathPlanetmath of A and θ is bijective, D is a proper subset of B. ∎

Theorem 4.

Let A, B and C be sets such that A and C are equi-decomposable and ABC. Then B and C are equi-decomposable.


Let A1,,An and C1,Cn disjoint decompositions of A and C, respectively, such that Ak and Ck are congruent via an isometry θk for 1kn. Like in the proof of theorem 3, let θ:AC be the well-defined, bijective map such that θ(x)=θk(x) if xAk. Now, for every bB, let 𝒞(b) be the intersectionMathworldPlanetmath of all sets XB satisfying

  • bX,

  • for all xX, the preimageMathworldPlanetmath θ-1(x) lies in X,

  • for all xXA, the image θ(x) lies in X.

Let b1,b2B such that 𝒞(b1) and 𝒞(b2) are not disjoint. Then there is a b𝒞(b1)𝒞(b2) such that b=θr(b1)=θs(b2) for suitable integers s and r. Given b𝒞(b1), we have b=θt(b1)=θt+s-r(b2) for a suitable integer t, that is b𝒞(b2), so that 𝒞(b1)𝒞(b2). The reverse inclusion follows likewise, and we see that for arbitrary b1,b2B either 𝒞(b1)=𝒞(b2) or 𝒞(b1) and 𝒞(b2) are disjoint. Now set


then obviously DA. If for bB, 𝒞(b) consists of the sequenceMathworldPlanetmath of elements ,θ-1(b),b,θ(b), which is infiniteMathworldPlanetmathPlanetmath in both directions, then bD. If the sequence is infinite in only one direction, but the final element lies in A, then bD as well. Let E:=θ(D) and F:=BD, then clearly EFC.

Now let cC. If cE, then θ-1(c)D, so 𝒞(θ-1(c)) consists of a sequence ,θ-2(c),θ-1(c), which is infinite in only one direction and the final element does not lie in A. Now θ-1(c)𝒞(θ-1(c)), but since θ-1(c) does lie in A, it is not the final element. Therefore the subsequent element c lies in 𝒞(θ-1(c)), in particular cB and 𝒞(c)=𝒞(θ-1(c))A, so cBD=F. It follows that C=EF, and furthermore E and F are disjoint.

It now follows similarly as in the preceding proofs that D and E=θ(D) are equi-decomposable. By theorem 2, B=DF and C=EF are equi-decomposable. ∎

The proof

We may assume that the unit ball U is centered at the origin, that is U=𝔹3, while the other unit ball U has an arbitrary origin. Let S be the surface of U, that is, the unit sphereMathworldPlanetmath. By the Hausdorff paradoxMathworldPlanetmath, there exists a disjoint decomposition


such that B, C, D and CD are congruent, and E is countableMathworldPlanetmath. For r>0 and A3, let rA be the set of all vectors of A multiplied by r. Set


These sets give a disjoint decomposition of the unit ball with the origin deleted, and obviously B, C, D and CD are congruent (but E is of course not countable). Set


where 0 is the origin. B and CD are trivially equi-decomposable. Since C and B as well as D and C are congruent, CD and BC are equi-decomposable. Finally, B and CD as well as C and B are congruent, so BC and BCD are equi-decomposable. In total, B and BCD are equi-decomposable by theorem 1, so A1 and U are equi-decomposable by theorem 2. Similarly, we conclude that C, D and BCD are equi-decomposable.

Since E is only countable but there are uncountably many rotations of S, there exists a rotation θ such that θ(E)BCD, so F:=θ(E) is a proper subset of BCD. Since C and BCD are equi-decomposable, there exists by theorem 3 a proper subset GC such that G and F (and thus E) are equi-decomposable. Finally, let pCG an arbitrary point and set


a disjoint union by construction. Since D and BCD, G and E as well as {0} and {p} are equi-decomposable, A2 and U are equi-decomposable by theorem 2. A1 and A2 are disjoint (but A1A2U!).

Now U and U are congruent, so U and A2 are equi-decomposable by theorem 1. If U and U are disjoint, set H:=A2. Otherwise we may use theorem 3 and choose HA2 such that H and UU are equi-decomposable. By theorem 2, A1H and U(UU)=UU are equi-decomposable. Now we have


so by theorem 4, U and UU are equi-decomposable.


  • BT St. Banach, A. Tarski, Sur la décomposition des ensembles de points en parties respectivement congruentes, Fund. math. 6, 244–277, (1924).
Title proof of Banach-Tarski paradox
Canonical name ProofOfBanachTarskiParadox
Date of creation 2013-03-22 15:19:03
Last modified on 2013-03-22 15:19:03
Owner GrafZahl (9234)
Last modified by GrafZahl (9234)
Numerical id 7
Author GrafZahl (9234)
Entry type Proof
Classification msc 51M25
Classification msc 03E25
Related topic HausdorffParadox