It is the theorem from a dead-tree math-textbook where it is used as a step in a Banach-Tarski Paradox proving.
The textbook is not in English. Here's a translation of the theorem:
An orbit O [1] can be decomposed into 4 sets: A, B, C, D. Using rotation these sets can be combined into 2 orbits:
A ∪ aB = O; C ∪ bD = O
Proof:
A = H(a)x; B = H(a')x; C = H(b)x; D = H(b')x
The theorem statement follows from the fact that the free group H can be decomposed into 4 parts:
H(a),H(a'),H(b),H(b')
and doubled by rotations:
H = aH(a') ∪ H(a); H = bH(b') ∪ H(b)
∎
__ [1]: The term `orbit` is used in the same sense as in [2]. `H` is a free group similar to the one from the step 3 in [2], and `a`, `b` are rotations defined similar to Step 2 in [2] i.e., they are generators of H:
I can't follow either of the proofs yet, but one question does pose itself: given a decomposition of the sphere, don't you get a decomposition of the ball, just by projecting towards the origin?
Of course what happens at the origin itself is not clear. Perhaps there's no way of getting around that difficulty. If you could deal with that, however, a proof on the sphere would be equivalent to a proof on the ball.
So would I. I think it's wrong; I'm pretty sure that the Hausdorff paradox, which is kinda-sorta the equivalent on the sphere, requires some Choice. But maybe I'm confused.
On the other hand, you don't need any Choice for the Sierpinski-Mazuriewicz paradox, which says that there's a subset A of the plane, which can be written as the disjoint union of two "smaller" subsets B and C, but where A, B, and C are all congruent. (But A is countable, so this isn't nearly as startling as Banach-Tarski.) Specifically, let T be translation by (1,0), and let R be rotation about the origin by 1 radian; then let A be the set of points you can get from (0,0) by applying some sequence of T and R, let B consist of the points where the last operation in the sequence was T rather than R, and let C be everything else. Then B = T(A) and C = R(A). (The key point here is that it turns out that the sequence of Ts and Rs is unique, so that B is well-defined.)
