# Homeomorphism between $\mathbb{Q}$ and $\mathbb{Q}(>0)$, and $\mathbb{Q}(\ge 0)$

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. –  Asaf Karagila Mar 11 '12 at 21:30
@ 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't –  you Mar 11 '12 at 22:01
Possibly related question: Given two linear orders and their induced topologies, is an order-isomorphism also a homeomorphism? –  you Mar 11 '12 at 22:08
@AsafKaragila all countable metrisable spaces without isolated points are homeomorphic, so the spaces mentioned too. –  Henno Brandsma Mar 11 '12 at 22:19
@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... :\ –  Asaf Karagila Mar 11 '12 at 22:20

The following function is an 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 an homeomorphism between $\mathbb{Q}_{\geq 0}$ and $\mathbb{Q}$, I will rather construct an 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 an homemomorphism 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 an homemomorphism from $]\frac{\sqrt{2}}{2n},\frac{\sqrt{2}}{2n-1}[ \cap \mathbb{Q}$ to $]-\frac{\sqrt{2}}{n+1},-\frac{\sqrt{2}}{n}[ \cap \mathbb{Q}$.

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 $0$.

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