most recent 30 from http://mathoverflow.net 2013-05-19T12:08:35Z http://mathoverflow.net/feeds/question/26001 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/26001/are-the-rationals-homeomorphic-to-any-power-of-the-rationals HenrikRüping 2010-05-26T12:22:55Z 2010-05-26T16:51:51Z

<p>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:</p> <p>Is $\mathbb{Q}^2$ homeomorphic to $\mathbb{Q}$ ?</p>

http://mathoverflow.net/questions/26001/are-the-rationals-homeomorphic-to-any-power-of-the-rationals/26004#26004 Tom Smith 2010-05-26T12:32:21Z 2010-05-26T12:32:21Z

<p>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 $\mathbb{R}^2\to\mathbb{R}$? </p>

http://mathoverflow.net/questions/26001/are-the-rationals-homeomorphic-to-any-power-of-the-rationals/26006#26006 Xandi Tuni 2010-05-26T12:35:05Z 2010-05-26T12:35:05Z

<p>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 $([a_0,a_2,...], [a_1,a_3,...])$.</p> <p>Mind that this is a homeomorphism, but not an isometry (cf comment on Tom's answer).</p> <p>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.</p>

http://mathoverflow.net/questions/26001/are-the-rationals-homeomorphic-to-any-power-of-the-rationals/26009#26009 Robin Chapman 2010-05-26T12:37:35Z 2010-05-26T16:51:51Z

<p>Yes, Sierpinski proved that every countable metric space without isolated points is homeomorphic to the rationals: <a href="http://at.yorku.ca/p/a/c/a/25.htm" rel="nofollow">http://at.yorku.ca/p/a/c/a/25.htm</a> .</p> <p>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 $p$) :-)</p>