Showing $\mathbb{Q}$ is homeomorphic to $\mathbb{Q}^2$

Question

Let $\mathbb{Q}$ be the set of rationals, as relative topology to $\mathbb{R}$.

How to show that $\mathbb{Q}$ is homeomorphic to $\mathbb{Q}^2$?

More general, if $T$ is a countable dense space(no isolated poiont) with every singleton closed, how to show that $T^2$ is homeomophic to $T$.

If $T$ is a countable dense space with every singleton closed then $T$ is homeomorphic to exactly one of the following: $[0,1]\cap\Bbb Q$; $(0,1]\cap\Bbb Q$; $(0,1)\cap\Bbb Q$. — Asaf Karagila, Apr 09 '13 at 09:39
@AsafKaragila: Aren't these three spaces homeomorphic to each other since they are metric countable with no isolated point ? — user10676, Apr 09 '13 at 13:25
@user10676: Hm, yeah, I suppose that I had in mind order isomorphism (or order-reversing). — Asaf Karagila, Apr 09 '13 at 13:29
@ Asaf Karagila . I found that the MO answer thru your link is very nice, especially its simplicity is surprising. — DanielWainfleet, Feb 06 '16 at 07:12

score 7 · Accepted Answer · edited Nov 15 '17 at 10:17

7

The easiest way is to prove that $\mathbb{Q}$ is the unique (up to homeomorphism) countable metric space without isolated points. Three proofs can be found here.

edited Nov 15 '17 at 10:17

M. Winter

30,828

answered Feb 06 '16 at 05:07

Stella Biderman

31,475

Showing $\mathbb{Q}$ is homeomorphic to $\mathbb{Q}^2$

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