Are the rationals homeomorphic to any power of the rationals? 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 $\mathbb{Q}$ ?
 A: 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 $p$) :-)
A: 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 ${\mathbb Q}$.
A: 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.
A: 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  $\mathbb{R}^2\to\mathbb{R}$? 
