Let $Q$ follow subspace topology from $R$ Then I think it is true that $Q^n$ and $Q^m$ (with product topology) are not homeomorphic.I also think it will be possible to define "rational" homotopy groups by considering $[0,1]\cap Q$ etc. But I can't find any reference regarding topology of product spaces of rationals in general . So I was wondering if someone can give a sketch of a proof and give some references.
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2$\begingroup$ They're homeomorphic. $\endgroup$– ApolloCommented Oct 4, 2012 at 13:50
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1$\begingroup$ can you explain how? $\endgroup$– MarcusCommented Oct 4, 2012 at 13:55
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2$\begingroup$ They're all countable metric spaces with no isolated points. See at.yorku.ca/p/a/c/a/25.htm $\endgroup$– ApolloCommented Oct 4, 2012 at 14:12
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$\begingroup$ The result mentioned by Apollo above also shows that for any nonempty countable metric $X$, $X \times \mathbb Q$ is homeomorphic to $\mathbb Q$ - This is one of the results I try to keep in mind to remind myself that my intuition regarding topological spaces is not to be trusted! $\endgroup$– Julien MellerayCommented Oct 4, 2012 at 15:24
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Edit: The following is a second attempt to repair problems in earlier proposed solutions, as pointed out by Gerald Edgar in comments. Hopefully this time I've gotten it right this time.
It's classical (by a back-and-forth argument; see for instance A Shorter Model Theory by Hodges) that any two countable dense unbounded linear orders without endpoints are isomorphic. For example the linear order $\mathbb{Q}$ is isomorphic to $L = \mathbb{Z}_{(2), (5)} \cap (0, 1)$, referring here to the integers localized at the primes 2 and 5 (i.e., rationals whose decimal expansions don't have an infinite tail of 9's or 0's -- this is for technical reasons). $L$ and $\mathbb{Q}$ are homeomorphic under their order topologies.
Let $f: L \times L \to \mathbb{Q} \cap (0, 1)$ be the map that takes a pair of elements $\alpha, \beta \in L$ and forms a rational number by interleaving their decimal expansions, with the decimal expansion of $\alpha$ appearing in odd places and that of $\beta$ in the even places. Let $I$ be the image of $f$. The map $f: L \times L \to I$ is a homeomorphism, and $I$ is again a countable dense linear order without endpoints, hence homeomorphic to $\mathbb{Q}$ again. Then the evident composite
$$\mathbb{Q} \times \mathbb{Q} \cong L \times L \stackrel{f}{\to} I \cong \mathbb{Q}$$
is a homeomorphism $\mathbb{Q} \times \mathbb{Q} \cong \mathbb{Q}$.
answered Oct 4, 2012 at 14:20
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2$\begingroup$ Am I confused? $f : \mathbb{Q} \times \mathbb{Q} \to \mathbb{Q}(\sqrt{2})$ sending $(r,s)$ to $r+s\sqrt{2}$ is a homeomorphism? Surely you can have $r_n \to \infty$ and $s_n \to -\infty$ in such a way that $r_n+s_n\sqrt{2} \to 0$? $\endgroup$ Commented Oct 4, 2012 at 14:51
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$\begingroup$ Sorry! You are right. I will edit and try to fix this thing. $\endgroup$ Commented Oct 4, 2012 at 15:13
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$\begingroup$ I have substantially edited my answer. Apologies again. If there is still an error, then please (nemesiso) unaccept this answer and I will delete it. I really just wanted something semi-constructive. $\endgroup$ Commented Oct 4, 2012 at 15:33
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$\begingroup$ Isn't the product topology on QxQ the same as the order topology on QxQ determined by the lexicographic order? This, together with back-and-forth, would do the trick. $\endgroup$ Commented Oct 4, 2012 at 15:46
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1$\begingroup$ "Forbidding tails of zeros" ... a slight problem. No pair of numbers interleaves to u = 0.40404040404040... despite its being eventually periodic and not having a tail of zeros. $\endgroup$ Commented Oct 4, 2012 at 19:31