# Is $\mathbb{Q}^2$ homeomorphic to $\mathbb{Q}^2\setminus \{0\}$?

I know that $\mathbb{R}^2$ and $\mathbb{R}^2\setminus\{(0,0)\}$ are not homeomorphic. (For examle $\pi_1(\mathbb{R}^2)=\{e\}$, but $\pi_1(\mathbb{R}^2\setminus\{(0,0)\})=\mathbb{Z}$).

But what can be said about $\mathbb{Q}^2$ and $\mathbb{Q}^2\setminus\{(0,0)\}$? Is there a homeomorphism?

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The rational numbers are the unique (up to isomorphism) metric space which is both countable and have no isolated points.

$\Bbb Q^2\setminus\{(0,0)\}$ is countable and without isolated points. Therefore the answer is yes. There is a homeomorphism.

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at.yorku.ca/p/a/c/a/25.pdf here are the proofs of the Sierpiński theorem you refer to. –  Damian Sobota Mar 7 '13 at 19:34
Unless I have misunderstood something, the same argument proves that there's a homeomorphism between $\mathbb Q$ and $\mathbb Q^2$. That seemed unlikely, but I did a Google search immediately produced mathoverflow.net/questions/26001/… . Thanks! –  MJD Mar 8 '13 at 20:34
@MJD: You're welcome, I guess. :-) –  Asaf Karagila Mar 8 '13 at 20:49