In most people’s minds, or at least in the minds of those with a mathematical bent, every set of points in (a fancy way of writing three-dimensional Euclidean space) has a volume. Some sets, such as any set consisting of a finite number of points, has zero volume, whereas other sets, such as the set of all points less than one unit away from the origin, have finite but non-zero volume, and still other sets, such as itself, have infinite volume. But does every set necessarily have a volume? It turns out the answer is no, even in one-dimensional Euclidean space; I showed this in a previous post. Instead, I’ll show you the Banach-Tarski paradox, which states that it is possible, via translations and rotations, to dissect a three-dimensional sphere (technically a 3-ball) and rearrange it into two spheres of the same radius as the original.

For simplicity, we’ll start by only looking at the surface of the sphere, . Let refer to rotation by (that is, pi degrees) about the -axis and let refer to rotation by about the -axis. The exact degree of the rotation isn’t important, only that it should be impossible via any combination of and to go back to the origin. Let be the set of all rotations that you can get by combining , , and their inverses. Then and form what’s known as the ”free group on two generators“, which is composed of all strings using , , , and as symbols, with the provision that and cannot appear together for . The multiplication for this group is just writing the strings together, subject to the rule that and cancel, as do and . So, for example, . The name arises from the fact that there are two ‘fundamental’ symbols, and , and it is ‘free’, since they don’t commute (that is, ).

Now, one very weird fact about the free group on two generators (written ) is that you can break it into four pieces, rearrange those four pieces, and then reassemble them into two copies of it. To elaborate, let denote the set of strings in that start with , and similarly for the other three symbols; let denote the empty string. Then obviously

. But we also have (if we include in ) The reason for this is that if the string doesn’t start with , then it’s the same string as and so is in ; otherwise, it’s in . We also have, for the same reason,

How is that relevant? Well, partitions into orbits, where an orbit is a collection of points such that each element of moves points from the orbit into the orbit, and points not in the orbit into another point not in the orbit. For each orbit, we can pick a point inside it; let the set of all these points be . Then we can turn the decomposition of into a decomposition of ; using to indicate the set of all rotations in that start with , we have

These four sets together make up the sphere (except for itself; we’ll get back to that), but if we pick two of them and rotate one of those two, we also get back the sphere. Therefore, we can break down the sphere into four pieces and recompose them into two spheres. If we then draw lines joining the sphere to the origin, we can turn this into a decomposition of two solid spheres.

Now, there are two omissions in this proof: first off, I ignored the origin. It turns out that you can cut the sphere without the origin up, rearrange it, and then put it back together so that you have the origin. Second, I ignored the axes of the rotations in ; they are the fixed points of these rotations, and so might cause problems; it turns out that if we call those points ; we can again cut up and rearrange it into .

So what does this mean? If we want to have any reasonable definition of volume, then we want translations and rotations to leave it the same. Then, we must have that some of these sets don’t have a well-defined volume; much like how you can construct a subset of the real line that doesn’t have one. We break the sphere down into a number of sets, some of which don’t have a defined volume; then we move them around and reassemble them. There’s no contradiction here, since two sets that don’t have a well-defined volume (or non-measurable sets, as they’re known) might have a union that has a volume. Of course, this is physically meaningless, since you can’t actually perform the required divisions; atoms are not infinitely small, after all.