proof of Banach-Tarski paradox
We deal with some technicalities first, mainly concerning the properties of equi-decomposability. We can then prove the paradox in a clear and unencumbered line of argument: we show that, given two unit balls and with arbitrary origin, and are equi-decomposable, regardless whether and are disjoint or not. The original proof can be found in [BT].
The only non-obvious part is transitivity. So let , , be sets such that and as well as and are equi-decomposable. Then there exist disjoint decompositions of and of such that and are congruent for . Furthermore, there exist disjoint decompositions of and of such that and are congruent for . Define
Now if and are congruent via some isometry , we obtain a disjoint decomposition of by setting . Likewise, if and are congruent via some isometry , we obtain a disjoint decomposition of by setting . Clearly, the and define a disjoint decomposition of and , respectively, into parts. By transitivity of congruence, and are congruent for and . Therefore, and are equi-decomposable. ∎
Given disjoint sets , such that and are equi-decomposable for , their unions and are equi-decomposable as well.
By definition, there exists for every , an integer such that there are disjoint decompositions
such that and are congruent for . Rewriting and in the form
gives the result. ∎
Let , , be sets such that and are equi-decomposable and , then there exists such that and are equi-decomposable.
Let and be disjoint decompositions of and , respectively, such that and are congruent via an isometry for all . Let a map such that if . Since the are disjoint, is well-defined everywhere. Furthermore, is obviously bijective. Now set and define , so that and are congruent for , so the disjoint union and are equi-decomposable. By construction, . Since is a proper subset of and is bijective, is a proper subset of . ∎
Let , and be sets such that and are equi-decomposable and . Then and are equi-decomposable.
Let and disjoint decompositions of and , respectively, such that and are congruent via an isometry for . Like in the proof of theorem 3, let be the well-defined, bijective map such that if . Now, for every , let be the intersection of all sets satisfying
for all , the preimage lies in ,
for all , the image lies in .
Let such that and are not disjoint. Then there is a such that for suitable integers and . Given , we have for a suitable integer , that is , so that . The reverse inclusion follows likewise, and we see that for arbitrary either or and are disjoint. Now set
then obviously . If for , consists of the sequence of elements which is infinite in both directions, then . If the sequence is infinite in only one direction, but the final element lies in , then as well. Let and , then clearly .
Now let . If , then , so consists of a sequence which is infinite in only one direction and the final element does not lie in . Now , but since does lie in , it is not the final element. Therefore the subsequent element lies in , in particular and , so . It follows that , and furthermore and are disjoint.
It now follows similarly as in the preceding proofs that and are equi-decomposable. By theorem 2, and are equi-decomposable. ∎
We may assume that the unit ball is centered at the origin, that is , while the other unit ball has an arbitrary origin. Let be the surface of , that is, the unit sphere. By the Hausdorff paradox, there exists a disjoint decomposition
such that , , and are congruent, and is countable. For and , let be the set of all vectors of multiplied by . Set
These sets give a disjoint decomposition of the unit ball with the origin deleted, and obviously , , and are congruent (but is of course not countable). Set
where is the origin. and are trivially equi-decomposable. Since and as well as and are congruent, and are equi-decomposable. Finally, and as well as and are congruent, so and are equi-decomposable. In total, and are equi-decomposable by theorem 1, so and are equi-decomposable by theorem 2. Similarly, we conclude that , and are equi-decomposable.
Since is only countable but there are uncountably many rotations of , there exists a rotation such that , so is a proper subset of . Since and are equi-decomposable, there exists by theorem 3 a proper subset such that and (and thus ) are equi-decomposable. Finally, let an arbitrary point and set
a disjoint union by construction. Since and , and as well as and are equi-decomposable, and are equi-decomposable by theorem 2. and are disjoint (but !).
- 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|
|Date of creation||2013-03-22 15:19:03|
|Last modified on||2013-03-22 15:19:03|
|Last modified by||GrafZahl (9234)|