24

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?

Oiale
  • 1,563

1 Answers1

20

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.

Asaf Karagila
  • 405,794
  • 11
    http://at.yorku.ca/p/a/c/a/25.pdf here are the proofs of the Sierpiński theorem you refer to. – Damian Sobota Mar 07 '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 http://mathoverflow.net/questions/26001/are-the-rationals-homeomorphic-to-any-power-of-the-rationals/26009#26009 . Thanks! – MJD Mar 08 '13 at 20:34
  • @MJD: You're welcome, I guess. :-) – Asaf Karagila Mar 08 '13 at 20:49