Banach-Tarski Paradox

The Banach-Tarski Paradox is the famous "doubling the ball" paradox, which claims 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, using only rotation and translation, reassemble the pieces into two balls the same size as the original. Or short: 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-space) with non-empty interior are equi-decomposable.

In other words, a marble can be cut up into finitely many pieces and reassembled into a planet. Or you could disassemble a telephone and reassemble it to make a water lily.

It is interesting to note that this theorem depends on three dimensions; while intuitively the two-dimensional case seems to be easier, it is in fact not true that all bounded subsets of the plane with non-empty interior are equi-decomposable.

Note that in the decomposition, the pieces won't be measurable?, and so they will 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.

In 1924, [Stefan Banach]? and [Alfred Tarski]? described this paradox, building on earlier work by Felix Hausdorff who managed to "chop up" the unit interval into countably many pieces which (by translation only) can be reassembled into the interval of length 2. Usually, the purpose of these paradoxical decompositions is to show that not all sets are measurable.

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. Because its proof prominently uses the axiom of choice, opponents of the axiom of choice present this counter-intuitive conclusion as an argument against adoption of that axiom.

A sketch of the proof follows. Essentially, the paradoxical decomposition of the ball is achieved in four steps:

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

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 an 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 ε. We will call this group G.

The group G 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,

G = {ε} ∪ S(a) ∪ S(a^-1) ∪ S(b) ∪ S(b^-1)

but also

G = a S(a^-1) ∪ S(a)   and   G =b S(b^-1) ∪ S(b)

(The notation a S(a^-1) means: take all the strings in S(a^-1) and concatenate them from 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 G into four pieces (forget about ε for now, it doesn't pose a problem), then "rotated" some of them by multiplying with a or b, then "reassembled" two of them to make G and reassembled the two others to make another copy of G. That's exactly what we want to do to the ball. This finishes step 1.

In order to find a group of rotations of 3-d space that behaves just like (or "isomorphic to") the group G, 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 G. 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 number 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 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. (This sketch glosses over some details.)

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 (minus the origin, but that can be dealt with easily).

---

References:

• [The Banach-Tarski Paradox by Francis E. Su] (PDF)
• S. Wagon, "The Banach-Tarski Paradox", Cambridge University Press, 1986