Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
3  
@AsafKaragila all countable metrisable spaces without isolated points are homeomorphic, so the spaces mentioned too. –  Henno Brandsma Mar 11 '12 at 22:19
2  
@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

1 Answer 1

up vote 2 down vote accepted

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 $0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.