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I want to ask about the homeomorphism between $\mathbb{Q}$, $\mathbb{Q}_{>0}$: the rationals greater than $0$, and $\mathbb{Q}_{\geqslant 0}$: the rationals $\geqslant 0$?

For $\mathbb{Q}$ and $\mathbb{Q}_{>0}$, I can use the back and forth map to have an order-isomorphism, and as a result, a homeomorphism. Is there a direct formula for the map?

It's very surprising that $\mathbb{Q}_{\geqslant 0}$ is homeomorphic to $\mathbb{Q}_{>0}$ and $\mathbb{Q}$, I don't know how to show they are homeomorphic.

Could anyone please help me with the homeomorphisms between them?

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    How can they be homeomorphic (or order-isomorphic)? $\mathbb Q_{\geq0}$ has an end point but $\mathbb Q_{>0}$ doesn't.2012-03-11
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    @ Asaf: good point, I deleted the erroneous comment. What IS completely clear is, like you said, it is impossible for $\mathbb{Q}$ and $\mathbb{Q}_{\geq 0}$ to be order isomorphic, since one satisfied "$\exists x\ \forall y\ :\ y\geq x$" and the other doesn't2012-03-11
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    Possibly related question: Given two linear orders and their induced topologies, is an order-isomorphism also a homeomorphism?2012-03-11
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    @you yes, but not reversely: a homeomorphism need not preserve (or reverse) order.2012-03-11
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    @AsafKaragila all countable metrisable spaces without isolated points are homeomorphic, so the spaces mentioned too.2012-03-11
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    @Henno: Hmmm. What you say is true, and what's worse is that not even a year ago I knew this fact very well. Screw this memory... :\2012-03-11
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    I wrote an answer, then deleted it; now I'm editing and will re-post.2012-03-11
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    My first thought on homeomorphisms between $\mathbb{Q}\cap(0,1)$ and $\mathbb{Q}\cap[0,1)$ is that some of the mappings used in constructive proofs of the Cantor--Schroeder--Bernstein theorem might serve. Still ironing out the details.....2012-03-11
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    I feel like a proof should exploit the fact that $\mathbb{Q}$ is totally disconnected. Like you could write is as a topological sum $\mathbb{Q}=(-\infty,0)\dot{\cup}[0,\infty)$ (as subsets of $\mathbb{Q}$ of course), and then give homeomorphisms $(-\infty,0)\rightarrow (\sqrt{2},\infty)$ and $[0,\infty)\rightarrow [0,\sqrt{2})$2012-03-11
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    @you : I look at your topological sum and think: open neighborhoods of $0$ intersect _both_ terms in the sum.2012-03-11
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    http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2009;task=show_msg;msg=35572012-03-11

1 Answers 1

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The following function is a homeomorphism between $\mathbb{Q}$ and $\mathbb{Q}_{>0}$ : $$f(x) = \begin{cases} \frac{-1}{x}, & \text{if } x<-1 \\ x+2, & \text{if } x \geq -1 \end{cases}.$$

For a homeomorphism between $\mathbb{Q}_{\geq 0}$ and $\mathbb{Q}$, I will rather construct a homeomorphism between $[0,\sqrt{2}[ \cap \mathbb{Q}$ and $]-\sqrt{2},\sqrt{2}[ \cap \mathbb{Q}$. Let $f : [0,1[ \cap \mathbb{Q} \rightarrow \mathbb{Q}$ defined by :

  • $f(0) =0$.
  • For all $n \in \mathbb{N_{\geq 1}}$, $f$ is a homeomorphism from $]\frac{\sqrt{2}}{2n+1},\frac{\sqrt{2}}{2n}[ \cap \mathbb{Q}$ to $]\frac{\sqrt{2}}{n+1},\frac{\sqrt{2}}{n}[ \cap \mathbb{Q}$.
  • For all $n \in \mathbb{N_{\geq 1}}$, $f$ is a homeomorphism from $]\frac{\sqrt{2}}{2n},\frac{\sqrt{2}}{2n-1}[ \cap \mathbb{Q}$ to $]-\frac{\sqrt{2}}{n},-\frac{\sqrt{2}}{n+1}[ \cap \mathbb{Q}$.

Then :

  • The function $f$ is one to one from $[0,\sqrt{2}[ \cap \mathbb{Q}$ to $]-\sqrt{2},\sqrt{2}[ \cap \mathbb{Q}$.

  • It is continuous on $]0,\sqrt{2}[ \cap \mathbb{Q}$ because it is continuous on the open covering $(]\frac{\sqrt{2}}{k+1},\frac{\sqrt{2}}{k}[)_{k \geq 1}$.

  • It is continuous at $0$ because $f([0,\frac{\sqrt{2}}{2k}[) \subset (]-\frac{\sqrt{2}}{k},\frac{\sqrt{2}}{k}[)$.

  • For the same reasons $f^{-1}$ is continuous on $(]-\sqrt{2}[ \cup ]\sqrt{2}[) \cap \mathbb{Q}$ and at $0$.