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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 $\mathbb{Q}$ ?

Ricardo Andrade
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HenrikRüping
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4 Answers4

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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 $p$) :-)

Robin Chapman
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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 ${\mathbb Q}$.

Steven Landsburg
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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 general Theorem in point set topology stating that all countable topological spaces "of the same kind as $\mathbb Q$" are homeomorphic.

Glorfindel
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Xandi Tuni
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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
  • My vague memory is Sierpinskis theorem cited by Robin. – Xandi Tuni May 26 '10 at 12:42
  • @Wadim: The map still exists over the reals, but it is not a homeomorphism (The continued fractions will be infinite, and that kills continuity). – Xandi Tuni May 26 '10 at 12:47
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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
  • So, how to struggle against the double representation of rationals? (There should be a trick: what about taking the even length only? BTW, (6,1) is also represented by [6,1]...) – Wadim Zudilin May 26 '10 at 13:54
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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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    You are right, there should be an additional trick to make the argument work... I'm out of wisdom for the moment. – Xandi Tuni May 26 '10 at 14:23
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    There is no way to make this work for rationals. (That's a strong statement.) It is fine for irrationals, however. – Gerald Edgar May 26 '10 at 15:38
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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
  • @HenrikRüping maybe constraining the sign to match or not match the parity is the trick to make it work? – it's a hire car baby Apr 09 '22 at 02:37
  • @GeraldEdgar how definite is your statement this can't work for rationals? I wasn't sure if that was meant as "I'd bet my bottom dollar there's no trick to make it work" or "I know a complicated reason that proves it can't work which is too long to state here". – it's a hire car baby Apr 09 '22 at 10:54
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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 $\mathbb{R}^2\to\mathbb{R}$?

Tom Smith
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