# $M$ is compact, non-empty, perfect, and $M \cong M \times M$. Must $M$ be homeomorphic to the Cantor set, the Hilbert cube, or some combination?

Assume that $M$ is compact, non-empty, perfect, and homeomorphic to its Cartesian square, $M \cong M \times M$. Must $M$ be homeomorphic to the Cantor set, the Hilbert cube, or some combination of them?

An interesting triple-starred problem from Pugh's "Real Mathematical Analysis". This is not from an assignment or anything graded, I'm just curious as to what the right answer is and the route that one may take to get there.

• So from context I'm assuming that you want metric spaces only? – Henno Brandsma Apr 29 '13 at 6:39
• I'm guessing that Pugh intended as much, but if you know of any non-metric spaces that don't have to satisfy the homeomorphism claims, I'm definitely not one to discriminate. – WayMoreQuestionsThanAnswers Apr 29 '13 at 6:47
• dimension theory says that for compact metric spaces $M$ either $\dim(M \times M) = 2\dim(M)$ or $=2\dim(M)-1$, the latter are called exotic, and none exist of dimension less than $4$. This implies such $M$ must be zero-dimensional (and then it's a Cantor set), or infinite-dimensional. – Henno Brandsma Apr 29 '13 at 6:49
• larger Tychonov cubes are a candidate, e.g., if we allow non-metric compacta. – Henno Brandsma Apr 29 '13 at 6:51
• Dimension theory makes this problem rather trivial. Thanks for the insight! – WayMoreQuestionsThanAnswers Apr 29 '13 at 7:01

Take the countable product $$S=\prod_{i\in {\mathbb N}} S^1$$ of circles equipped with the product topology. Then $S$ is compact, metrizable, and perfect. However, $\pi_1(S)$ is nontrivial, hence, $S$ cannot be homeomorphic to (or even homotopy equivalent to) the Hilbert cube, since the later is contractible. Furthermore, $S$ is infinite dimensional, hence, it is not homeomorphic to the Cantor set. Lastly, $S\cong S\times S$. You can form more examples by taking, say, infinite products of spheres, etc.
However, I do not have any examples of contractible metrizable spaces $X$ satisfying $X^2\cong X$, besides the Hilbert cube.