Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
@YaakovBaruch and Ben, I think this idea can be made to work. Ben, you've essentially reduced the problem of partitioning $[0,1]^{n+1}$ into copies of $(0,1)^n$ to the problem of partitioning $S^{n-1}$ into copies of $(0,1)^{n-2}$. Now, partitioning $S^{n-1}$ into a closed hemisphere and an open hemisphere, we get a copy of $(0,1)^{n-1}$ and a copy of $[0,1]^{n-1}$. The former is easily partitioned into copies of $(0,1)^{n-2}$, and the latter can be partitioned into copies of $(0,1)^{n-2}$ by induction.
@TimothyChow: I meant a full rotation of $\pi$ radians (or more) about the vertical line in the same plane as the square, the axis of symmetry for the M-shaped part of the partition. I'll edit the question to make this more clear.
Unfortunately, I still don't think this works. In (5), it looks like you've reduced things down to partitioning $[0,1]^{n-2}$ (not $[0,1]^{n-1}$) into copies of $(0,1)^{n-2}$. And this is impossible.
I'm confused about (4). One of your two leftover bits is homeomorphic to $(0,1)^n$, but it seems to me like the other is homeomorphic to $S^{n-1} \times (0,1)^2$, not $S^1 \times (0,1)^n$. Maybe I'm missing something? (It's definitely a neat way of doing $[0,1]^3$, which I hadn't noticed before.)
@AlessandroDellaCorte: I mean that each piece is homeomorphic to $(0,1)^2$, and therefore can be partitioned into copies of $(0,1)$. (For example, $\{\{x\} \times (0,1) :\, x \in (0,1)\}$ is such a partition.) Perhaps it's even simpler just to say: by taking the horizontal slices of each of the four leftover pieces, we obtain a partition into copies of $(0,1)$.
. . . If it is not available, he can just randomly select any other chocolate. If I haven't made any mistakes, this strategy works with probability $1$. I have no idea whether it's possible to turn this idea into a deterministic winning strategy, which is presumably what you're looking for.
Here is a non-deterministic winning strategy for the Glutton, when the set of chocolates has size $\aleph_1$. Just like in my answer, let $\{c_\alpha :\, \alpha < \omega_1\}$ enumerate the chocolate types. On even-numbered turns, the Glutton plays the chocolate with the largest index. On odd-numbered turns, the Glutton first selects an index $\alpha$ of an already-played chocolate, completely randomly. Then he looks at an enumeration $\{\beta_m :\, m < \omega\}$ of the ordinals below $\alpha$, and plays the chocolate with index $\beta_m$ with probability $1/2^m$, if it is available to play....
@JoelDavidHamkins: It seems the inductive argument doesn't have any difficulties at cardinals of cofinality $\omega$, I think because the chocolatier only presents finitely many chocolates at once. If she were to present countably many instead, I think $\square$ could come into play. :) I will think about getting rid of the eating-history, although the induction arguments seems to really need it when the set of chocolates is large.
@JoelDavidHamkins: Yes, I was just thinking about that. Furthermore, I think you can use $\square$-like principles to push the induction past $\aleph_\omega$, and show that, consistently, the Glutton can win with sets of any size.
