For an arbitrary (non discrete) metric space $M$, is $M$ never homeomorhpic to $M\times M$?
In my analysis class we learned that ${\mathbb{R}}$ is not homeomorhpic to $\mathbb{R}^{2}$ because if you remove the point (0) from each, ${\mathbb{R}}$ is unconnected while $\mathbb{R}^{2}$ stays connected. Does this hold in general for an arbitrary(non-discrete) metric space?
No, if $M$ is the Cantor set, then $M$ and $M \times M$ are homeomorphic.
The same holds if we take $M=\Bbb Q$ or $M=\Bbb P$ (the irrationals), in their usual topology inherited from $\Bbb R$.
A one-dimensional example is $M$ equal to the Erdős space, also homeomorphic to its square (and completely metrisable and separable too, as all previous examples).
Many infinite-dimensional examples exist, like $\ell^2$ and $\Bbb R^{\Bbb N}$ (in the product topology); though these examples are also topologically the same (homeomorphic).
A non-separable metric example is $\ell^\infty$, e.g.
