# A Banach-Tarski game

This is partially inspired by the question https://math.stackexchange.com/questions/1383397/cutting-a-banach-tarski-cake, which I find intriguing if unclearly written.

A paradoxical family of subsets is a finite family $(Z_i)_{i<n}$ of bounded subsets of $\mathbb{R}^3$ such that they have images $f_{0, i}(Z_i)=Z_{0, i}$ and $f_{1, i}(Z_i)=Z_{1, i}$, where each $f_{j, i}$ is a rigid motion, with the property that $\bigcup Z_{0, i}$ and $\bigcup Z_{1, i}$ are both measurable, but have different measures. In the standard Banach-Tarski paradox, these two unions are the closed ball of radius 1 and two closed balls of radius 1.

Consider the following game between two players Alice and Bob. On turn $n$ (for $n\in\omega$), Alice partitions the unit sphere into two pieces, $X_{n,0}$ and $X_{n, 1}$; Bob then picks one of these two sets to be his $Y_n$. Alice wins if the family $(Y_n)_{n\in\omega}$ contains a paradoxical subfamily, and Bob wins otherwise.

(Note that if we switch the win condition, so that Alice is trying to prevent Bob from building a paradoxical set, then the game is trivial: Alice simply "plays nicely" by making sure all $X_i^n$s are, say, Borel.)

My question is:

Who wins the game?

Note that there's no reason for either player to have a winning strategy: since this is a game whose non-triviality relies crucially on the axiom of choice, it would be very reasonable for it to be undetermined in ZFC. Still, I don't see how to show this.

There are of course many variations on this general theme; I'm also interested in information on similar games, although this is the game I care about most right now.

• I edited to avoid confusion between subsets and families of subsets. But I still don't understand: in the 3-space all subsets of the 2-sphere have measure 0. So is everything on the 2-sphere? Or does Alice play with subsets of the 3-ball instead of sphere? – YCor Aug 15 '15 at 7:48
• Don't you want to say that $bigcup Z_{0,i}$ and $\bigcup Z_{1,i}$ are disjoint unions? – bof Aug 15 '15 at 12:20
• It seems that one of the players must have a winning strategy. Either Alice has a forced win in finitely many moves, or Bob can win by consistently choosing an option that doesn't give her such a possibility. – Johan Wästlund Aug 15 '15 at 20:51
• The original Hausdorff paradox provides a set which is (except for a null set) at the same time a "half" and a "third" of a sphere (or ball). If we allow the pieces to intersect in a null set, then Alice should win in at most six moves (six such pieces can be assembled to either 2 or 3 spheres/balls). – Johan Wästlund Aug 16 '15 at 13:39