Two Spheres In One

I suppose in recent posts involving the Axiom of Choice, I promised that I would give a proof of the Banach-Tarski paradox here. Not that anybody actually cares to read it, but the point of this blog is for me to feel smart while I rant about math, so I will do that.

First of all, we need to do a bit of group theory, because really all of the important math is just group theory. First, consider rotations of the three dimensional unit ball. These rotations form a non-commuting group, as you should know. Specifically, let A be a rotation of the sphere around the x-axis by 1 radian, and let B be a rotation of the sphere around the y-axis by 1 radian. A and B do not commute, and Aⁿ and Bⁿ are never equal to the identity (since there are 2π radians in the full rotation and 2kπ cannot ever be rational for rational k). It is also important to know that no string of A's and B's and their inverses (for example AB³A^-2BA²) will ever be equal to the identity, unless the inverses cancel exactly (so A^-2BB^-1AB^-1BA is the identity trivially, but ABA^-1B^-1 will not be).

Alright, now let us consider S to be the set of all strings involving A, A^-1, B, B^-1 with no term right next to its own inverse (so if there were to be any of those, cancel them). The empty string is also in S, to be the identity element. The strings in S can be arbitrarily long, though none of them are actually infinitely long. S clearly forms a group (the free group on two elements, to be specific).

Next, let us decompose S into 5 sets. S(A) will be elements of S staring with A, S(B) will be elements of S starting with B, S(A^-1) and S(B^-1) defined similarly, and S(e) has the last element, the empty string. Clearly S is the union of these 5 disjoint sets. Next, let us denote multiplication of an element x by a set P as the set you get when you multiply x by each element in P (so A times {A^-1, BA, A, B^-1} = {e, ABA, A², AB^-1} with e as the empty string).

Now, one can see that A times S(A^-1) gives you every element of S, except the ones that start with A (as S(A^-1) never has an element that starts A^-1A, but it will have plenty that start A^-1B , for example). Even the empty string is in A S(A^-1). So we can say that S is the union of the disjoint sets S(A) and A S(A^-1). Similarly, S is the union of the disjoint sets S(B) and B S(B^-1).

Something a bit weird has happened here, since now we see S = S(A) ∪ S(B) ∪ S(A^-1) ∪ S(B^-1) ∪ S(e) = S(A) ∪ A S(A^-1) = S(B) ∪ B S(B^-1). Technically, this isn't a problem, as all of those sets (except S(e)) are infinitely large anyway, so there isn't a problem with S being one-to-one with some of its own subsets. Anyway, enough group theory, back to the sphere.

Now, consider a point on the surface of the sphere. Consider all the places you can reach from that point by using elements of S. There is an element that is A away, an element that is B away, an element that is ABA^-3B²A away, and so on. These elements cannot cover the entire sphere (there are only countably many elements of S, and uncountably many places on the sphere to reach). We will consider an equivalence class on the sphere (as we always do when we are about to use the axiom of choice), two points on (or in) the sphere will be equivalent if you can move one into the other just by using rotations in S. Now use the Axiom of Choice to select one element from each equivalence class and put them into a set V. So for each point on the unit ball (the solid sphere, that is), there is exactly one element of V that can reach that point with exactly one of the rotations in S, and no two elements of V can reach eachother using rotations in S.

We will break the unit ball U up into 5 pieces. Places in U that can be reached from an element of V and a rotation in S(A) will be called U(A). Places in B that can be reached from an element of V and a rotation in S(A^-1) will be called U(A^-1). Similarly we define U(B), and U(B^-1). U(e) is just V itself. So U is the union of the 5 disjoint sets U(A), U(B), U(A^-1), U(B^-1), and U(e).

Now is where the "magic" happens. Since we know that A^-1S(A) is S(A) ∪ S(B) ∪ S(B^-1), then A^-1 applied to U(A) gives us U(A) ∪ U(B) ∪ U(B^-1) ∪ U(e). Similarly, B^-1 applied to U(B) gives us U(B) ∪ U(A) ∪ U(A^-1) ∪ U(e). Thus, U(B) and U(B^-1) can construct the entire unit ball, and U(A) and U(A^-1) can also give us the entire unit ball.

I suppose I have dropped off U(e), as well as the issue of the fixed points under rotations A and B (specifically, the x-axis and the y-axis). I'm not going to handle those properly, but the main point has already been made anyway.

It may seem very strange that you can rotate U(A) and get U(A) unioned with other sets, but this actually can happen without the Axiom of Choice. Consider the unit circle on the 2-plane. Starting with the point (1,0) on the x-axis, consider the set of points that you hit by rotating that point counter-clockwise 1 radian at a time. So you get a countably infinite set of points, as that list never crosses itself, but does not cover the whole circle (it never hits the point at -1 radian, or the point at π radians, for example). Now rotate that set clockwise by 1 radian. The point at 1 radian moves to the point at 0, the point at 6 radians moves to the point at 5 radians, and the initial point at zero radians moves to the point at -1 radians. This new set covers all the points in the old set, and has one extra point also, very strange (actually, this is how you can handle the set U(e), you basically "rotate it out of existence" by combining it with U(A) or something). Cardinality and measure are not an issue here, as the set is countable with measure zero both before and after the rotation.

Anyway, thats really it for the proof. You can also extend the proof to take any finite sized object, cut it into finitely many pieces, and reassemble them into any other finite object (the standard example is to arrange a pea into the sun). Note that this does require three dimensions or more, as you need two non-commuting rotations.