TITLE: Are the rationals minus a point homeomorphic to the rationals?
QUESTION [38 upvotes]: A while ago I was dreaming up point-set topology exam questions, and this one came to mind:
Is $\mathbb Q\setminus \{0\}$ homeomorphic to $\mathbb Q$? (Where both sets have the subspace topology induced from the standard topology on $\mathbb R$.)
However, I couldn't figure this out at the time, and I'm curious to see whether anyone has a nice argument. I'm not even willing to take a guess as to whether they are or aren't homeomorphic.

REPLY [20 votes]: The ordered approach is fine. A classical theorem by Sierpinski says that all countable metric spaces without isolated points are homeomorphic. $\mathbf{Q}$ is such a space.
It also implies $\mathbf{Q} \setminus \{0\}$ is homeomorphic to $\mathbf{Q}$ and $\mathbf{Q} \times \mathbf{Q}$ e.g., or any finite product for that matter. 
A proof is at the topology atlas, topology explained