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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, which states that by using the axiom of choice it is possible to take a solid ball in 3-dimensional space, cut it up into finitely many pieces and, moving them using only rotations and translations, reassemble the pieces into two balls of the same radius as the original. Banach and Tarski intended for this proof to demonstrate that the axiom of choice was incorrect, but the nature of the proof is such that most mathematicians take it to mean that the axiom of choice merely results in bizarre and unintuitive consequences.

Formally: Let A and B be two subsets of Euclidean space. We call them equi-decomposable if they can be represented as finite unions of disjoint subsets $A=\cup_{i=1}^n A_i$ and $B=\cup_{i=1}^n B_i$ such that, for any i, the subset $A i$ is congruent to $B i$ . Then, the paradox can be reformulated as follows:

The ball is equi-decomposable with two copies of itself.

For the ball, five pieces are sufficient to do this; it cannot be done with fewer than five. There is an even stronger version of the paradox:

Any two bounded subsets of 3-dimensional Euclidean space with non-empty interior are equi-decomposable.

In other words, a shape of marble can be cut up into finitely many pieces and reassembled into a planet, or a telephone could be cut up and reassembled as a water lily. These transformations are not possible with real objects made of a finite number atoms and bounded volumes, but it is possible with their geometric shapes. The Banach-Tarski paradox is made somewhat less bizarre by pointing out that there is always a function that can map one-to-one the points in one shape to another. For example, two spheres can be transformed bijectively to a similarly infinite subset of itself (such as one sphere). However, such transformations in general are non-isometric or involve an uncountably infinite number of "pieces". The surprising consequence of the Banach-Tarski paradox is that it can be done with only rotation and translation (isometric mapping) of a finite number of pieces.

What makes the paradox possible is that the pieces are infinitely convoluted. Technically, they are not measurable, and so they do not have "reasonable" boundaries nor a "volume" in the ordinary sense. It is impossible to carry out such a disassembly physically because disassembly "with a knife" can create only measurable sets. This pure existence statement in mathematics points out that there are many more sets than just the measurable sets familiar to most people.

The paradox also holds in all dimensions higher than three. It does not hold for subsets of the Euclidean plane. (The statement above does not apply to a two-dimensional subset of three-dimensional space, since such a subset would have empty interior.) Still, there are some paradoxical decompositions in the plane: a disc can be cut into finitely many pieces and reassembled to form a solid square of equal area; see Tarski's circle-squaring problem.

The paradox shows that it is impossible to define "volume" on all bounded subsets of Euclidean space such that equi-decomposable sets will have equal volume.

The proof is based on the earlier work of Felix Hausdorff, who found a closely related paradox 10 years earlier; in fact the Banach-Tarski paradox is a simple corollary of the technique developed by Hausdorff.

Logicians most often use the term "paradox" for a statement in logic which creates problems because it causes contradictions, such as the Liar paradox or Russell's paradox. The Banach-Tarski paradox is not a paradox in this sense but rather a proven theorem; it is a paradox only in the sense of being counter-intuitive. Because its proof prominently uses the axiom of choice, this counter-intuitive conclusion has been presented as an argument against adoption of that axiom.

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 3-d 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.

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.

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