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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:

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

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4 Answers 4

up vote 46 down vote accepted

Yes, Sierpinski proved that every countable metric space without isolated points is homeomorphic to the rationals: http://at.yorku.ca/p/a/c/a/25.htm .

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 $p$) :-)

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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 $([a_0,a_2,...], [a_1,a_3,...])$.

Mind that this is a homeomorphism, but not an isometry (cf comment on Tom's answer).

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.

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Why the construction does not work for $\mathbb R$ instead of $\mathbb Q$? –  Wadim Zudilin May 26 '10 at 12:40
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Each positive rational has two different continued fraction expansions. 5+1/2 = [5,2] maps to ([5],[2]) = (5,2). But 5+1/2 = [5,1,1] also, which maps to ([5,1],[1]) = (6,1). –  Gerald Edgar May 26 '10 at 13:20
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@Xandi: On the contrary, it works with infinite continued fractions to show that the space of irrationals is homeomorphic to its square. –  Gerald Edgar May 26 '10 at 13:22
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Sry but i was wondering why this is - surjective: only $a_0$ might be negative; hence you never get $a_1$ negative and hence the map is not surjective. - continuous : Note that $a_n:=[1;n]=1+1/n$ converges to $1$, but $([1],[n])=(1,n)$ diverges. I don't know how to repair this. Still it would be nice to have an explicit homeomorphism. –  HenrikRüping May 26 '10 at 14:22
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The irrationals have a nice characterisation as well (the rationals are the unique countable metric space without isolated points): the irrationals are the unique 0-dimensional [base of clopen sets] separable metric space that is nowhere locally compact [no non-empty open set has compact closure]. It is a way of showing that the irrationals are homeomorphic to N^N and hence to any finite or countable power of itself. –  Henno Brandsma May 26 '10 at 17:55

As Gerald Edgar points out in his comment on Xandi Tuni's answer, the continued fraction trick works for the irrationals, not for the rationals. But this turns out to be enough:

1) An irrational number has a unique continued fraction expansion. Therefore the irrationals are isomorphic to an infinite direct product of ${\mathbb Z}$ with itself. Therefore (writing ${\mathbb I}$ for the irrationals) we have ${\mathbb I}={\mathbb I}^2$.

2) Therefore ${\mathbb I}^2$ imbeds as a dense subset of the reals. Now map ${\mathbb Q}\times {\mathbb Q}$ to ${\mathbb I}\times {\mathbb I}$ by (say) adding $\sqrt{2}$ to each component. This imbeds ${\mathbb Q}\times {\mathbb Q}$ as a countable dense subset of the reals.

3) Now use the fact that every countable dense subset of the reals is homeomorphic to ${\mathbb Q}$.

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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 $\mathbb{R}^2\to\mathbb{R}$?

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Your argument only shows that there is no isometry. –  Xandi Tuni May 26 '10 at 12:35
    
You're quite right. I thought it sounded too good to be true... –  Tom Smith May 26 '10 at 12:36
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This will give a really good exercise for a topology course! –  Xandi Tuni May 26 '10 at 12:48
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It also shows there is no homeomorphism uniformly continuous in both directions. –  Gerald Edgar May 26 '10 at 13:24

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