I asked myself, which spaces have the property that $X^2$ is homeomorphic to $X$. I started to look at some examples like $\mathbb{N}^2 \cong \mathbb{N}$, $\mathbb{R}^2\ncong \mathbb{R}, C^2\cong C$ (for the cantor set $C$). And then I got stuck, when I considered the rationals. So the question is:
Is $\mathbb{Q}^2$ homeomorphic to $\mathbb{Q}$ ?
Yes, Sierpinski proved that every countable metric space without isolated points is homeomorphic to the rationals: http://at.yorku.ca/p/a/c/a/25.htm .
An amusing consequence of Sierpinski's theorem is that $\mathbb{Q}$ is homeomorphic to $\mathbb{Q}$. Of course here one $\mathbb{Q}$ has the order topology, and the other has the $p$-adic topology (for your favourite prime $p$) :-)
As Gerald Edgar points out in his comment on Xandi Tuni's answer, the continued fraction trick works for the irrationals, not for the rationals. But this turns out to be enough:
1) An irrational number has a unique continued fraction expansion. Therefore the irrationals are isomorphic to an infinite direct product of ${\mathbb Z}$ with itself. Therefore (writing ${\mathbb I}$ for the irrationals) we have ${\mathbb I}={\mathbb I}^2$.
2) Therefore ${\mathbb I}^2$ imbeds as a dense subset of the reals. Now map ${\mathbb Q}\times {\mathbb Q}$ to ${\mathbb I}\times {\mathbb I}$ by (say) adding $\sqrt{2}$ to each component. This imbeds ${\mathbb Q}\times {\mathbb Q}$ as a countable dense subset of the reals.
3) Now use the fact that every countable dense subset of the reals is homeomorphic to ${\mathbb Q}$.
Yes, they are homeomorphic. To construct a homeomorphism from $\mathbb Q$ to $\mathbb Q^2$, one can proceed roughly as follows: express $q\in \mathbb Q$ as a continued fraction $[a_0, a_1,a_2,...]$ (of finite length) and associate with it the pair $([a_0,a_2,...], [a_1,a_3,...])$.
Mind that this is a homeomorphism, but not an isometry (cf comment on Tom's answer).
I vaguely remember that there is a general Theorem in point set topology stating that all countable topological spaces "of the same kind as $\mathbb Q$" are homeomorphic.
I don't think so: the completion of $\mathbb{Q}^2$ is $\mathbb{R}^2$, so that a homeomorphism $\mathbb{Q}^2\to\mathbb{Q}$ would give a homeomorphism $\mathbb{R}^2\to\mathbb{R}$?
