Question

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 $\mathbb{Q}$ ?

Answer 1

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 $p$) :-)

Answer 2

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 $\mathbb{R}^2\to\mathbb{R}$?

Answer 3

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 ceneral Theorem in point set topology stating that all coutable topological spaces "of the same kind as $\mathbb Q$" are homeomorphic.