TITLE: $M$ 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?
QUESTION [7 upvotes]: 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.

REPLY [3 votes]: I am not quite sure what "some combination" means (I assume, it means the disjoint union), but here is an example:
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.