Banach?Tarski paradox

First stated by Stefan Banach and Alfred Tarski in 1924, the Banach?Tarski paradox or Hausdorff?Banach?Tarski paradox is the famous "doubling the ball" paradox,

A sketch of the proof

Essentially, the paradoxical decomposition of the ball is achieved in four steps:

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

• Find a paradoxical decomposition of the free group in two generators.
• Find a group of rotations in 3-d space isomorphic to the free group in two generators.
• Use the paradoxical decomposition of that group and the axiom of choice to produce a paradoxical decomposition of the hollow unit sphere.
• Extend this decomposition of the sphere to a decomposition of the solid unit ball.

We now discuss each of these steps in more detail.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Step 1. The free group with two generators a and b consists of all finite strings that can be formed from the four symbols a, a-1, b and b-1 such that no a appears directly next to an a-1 and no b appears directly next to a b-1. Two such strings can be concatenated and converted into a string of this type by repeatedly replacing the "forbidden" substrings with the empty string. For instance: abab-1a-1 concatenated with abab-1a yields abab-1a-1abab-1a, which gets reduced to abaab-1a. One can check that the set of those strings with this operation forms a group with neutral element the empty string e. We will call this group F_2.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The group F_2 can be "paradoxically decomposed" as follows: let S(a) be the set of all strings that start with a and define S(a-1), S(b) and S(b-1) similarly. Clearly,

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

:F_2={e}cup S(a)cup S(a^{-1})cup S(b)cup S(b^{-1})

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

but also

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

:F_2=aS(a^{-1})cup S(a), and

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

:F_2=bS(b^{-1})cup S(b).

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

(The notation a S(a-1) means take all the strings in S(a-1) and concatenate them on the left with a.) Make sure that you understand this last line, because it is at the core of the proof. Now look at this: we cut our group F_2 into four pieces (Forget about e for now, it doesn't pose a problem.), then "shifted" some of them by multiplying with a or b, then "reassembled" two of them to make F_2 and reassembled the other two to make another copy of F_2. That's exactly what we want to do to the ball.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Step 2. In order to find a group of rotations of 3D space that behaves just like (or "isomorphic to") the group F_2, we take two orthogonal axes and let A be a rotation of arccos(1/3) about the first and B be a rotation of arccos(1/3) about the second. (This step cannot be performed in two dimensions.) It is somewhat messy but not too difficult to show that these two rotations behave just like the elements a and b in our group F_2. We'll skip it. The new group of rotations generated by A and B will be called H. Of course, we now also have a paradoxical decomposition of H.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Step 3. The unit sphere S2 is partitioned into orbits by the action of our group H: two points belong to the same orbit if and only if there's a rotation in H which moves the first point into the second. We can use the axiom of choice to pick exactly one point from every orbit; collect these points into a set M. Now (almost) every point in S2 can be reached in exactly one way by applying the proper rotation from H to the proper element from M, and because of this, the paradoxical decomposition of H then yields a paradoxical decomposition of S2.

Related Topics:
Orbit - Action - Axiom of choice

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Step 4. Finally, connect every point on S2 with a ray to the origin; the paradoxical decomposition of S2 then yields a paradoxical decomposition of the solid unit ball. (The center of the ball needs a bit more care, but we omit this part in the sketch.)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

NB. This sketch glosses over some details. One has to be careful about the set of points on the sphere which happen to lie on an axis of rotation of some matrix in H. On the one hand, there are countably many such points so they "do not matter", and on the other hand it is possible to patch up even those points. The same applies to the center of the ball.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~