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It is well known that $\mathbb{R}$ is homeomorphic to every interval $(a,b) \subset \mathbb{R}$. What can we say about $\mathbb{Q}$? Is it homeomorphic to some of its subintervals?

Crostul
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  • Isn't it clear that the map $x\mapsto \frac{x}{x+1}$ that maps $(0,\infty)$ to $(0,1)$ works for $\Bbb Q$ just as it does for $\Bbb R$? – MJD Jun 26 '14 at 17:35
  • You can get all of $\mathbb Q$ from @MJD's example by using $x\mapsto \frac{x}{1+|x|}$, which maps onto the range $(-1,1)$ in $\mathbb Q$. – Thomas Andrews Jun 26 '14 at 17:38
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    Ok, this example works for intervals of the form $(a,b)$ with $a,b \in \mathbb{Q}$. What about other intervals (with irrational sup and/or inf)? – Crostul Jun 26 '14 at 17:45
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    @Crostul for those cases, I think it may suffice to show that $\mathbb{Q}$ is homeomorphic to $\mathbb{Q} + x$ and $\mathbb{Q} \cdot x$ for all $x \in \mathbb{R}$. – Ben Grossmann Jun 26 '14 at 17:52
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    $\Bbb Q$ is also homeomorphic to any of its non-empty open subsets – mercio Jun 26 '14 at 18:18

1 Answers1

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Yes, and there are three ways that one can show that:

  1. You can first prove that a countable linear order which is both dense and without endpoints is order isomorphic to the rational numbers. Therefore homeomorphic in the order topology to the rational numbers.

  2. You can also prove Sierpinski's theorem that every countable metric space which has no isolated points is homeomorphic to the rational numbers. (See proof here, for example).'

  3. You can find an explicit bijection. For bounded intervals you can use Thomas Andrew's suggestion from the comments, and for unbounded intervals MJD's suggestion reduces the case for a bounded interval again.

Asaf Karagila
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