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].
Technicalities
Theorem 1.
Equi-decomposability gives rise to an equivalence relation on the
subsets of Euclidean space.
Proof.
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.
∎
Theorem 2.
Given disjoint sets , such that and are equi-decomposable for , their unions and are equi-decomposable as well.
Proof.
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. ∎
Theorem 3.
Let , , be sets such that and are equi-decomposable and , then there exists such that and are equi-decomposable.
Proof.
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 .
∎
Theorem 4.
Let , and be sets such that and are equi-decomposable and . Then and are equi-decomposable.
Proof.
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. ∎
The proof
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 !).
References
- 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 |