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Is it possible to embed $\mathbb{Q}$ in $(0,1)$ (both with usual topology) ?

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$\mathbb{R}$ is homeomorphic to $(0, 1)$. –  Qiaochu Yuan Aug 9 '11 at 14:45
    
@Theo Buehler: ah, didn't know that. Can you please post your reply as an answer? –  user10 Aug 9 '11 at 14:54
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2 Answers 2

up vote 8 down vote accepted

On user10's request I'm posting an earlier comment of mine as an answer—even if it is more than overkill. Of course simply using the fact that $(0,1)$ and $\mathbb{R}$ are homeomorphic is the way to go. You can do this as suggested in Ross's answer.


Sierpiński proved in 1920 (apparently he proved it already in 1915 according to a footnote in the original article) that $\mathbb{Q}$ is up to homeomorphism the only countable metrizable space without isolated points.

In particular $\mathbb{Q}$ is homeomorphic to $\mathbb{Q}^n$, $\mathbb{Q} \cap (0,1)$ or indeed to any countable dense set in a perfect metric space.

The original paper is freely available here:

Wacław Sierpiński, Sur une propriété topologique des ensembles dénombrables denses en soi, Fund. Math. 1 (1920), 11-16 (pdf here).

As ccc points out a simpler proof along these lines would be to appeal to a result of Cantor's (I quote a comment to this answer):

[...] if you feel like using slightly less overkill, you can handle this particular problem with Cantor's theorem on the uniqueness of dense linear orders without endpoints, which I suppose can be viewed as a sort of "one dimensional" precursor to the quoted theorem of Sierpiński's.


A write-up of Sierpiński's theorem and a few related things is e.g. contained in (jstor-link, needs university subscription):

Carl Eberhart, Some Remarks on the Irrational and Rational Numbers, The American Mathematical Monthly, 84 (1) (Jan., 1977), pp. 32-35, MR644644.

Bruno recommends the article by Abhijit Dasgupta, Countable metric spaces without isolated points from Topology Explained, June 2005, published by the Topology Atlas which is an excellent source for everything related to topology.

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thank you a lot, you have been very helpful! love this site. –  user10 Aug 9 '11 at 15:28
    
I am in awe of your ability to dig up references so quickly. Anyway, if you feel like using slightly less overkill, you can handle this particular problem with Cantor's theorem on the uniqueness of dense linear orders without endpoints, which I suppose can be viewed as a sort of "one dimensional" precursor to the quoted theorem of Sierpiński's. –  user83827 Aug 9 '11 at 15:29
    
@ccc: Thanks for that pointer, that's certainly easier and a nice way of looking at it. Finding stuff is not so difficult if one knows what one is looking for (I sort of remembered that it was Sierpiński or Kuratowski -- a Polish mathematician for sure)... –  t.b. Aug 9 '11 at 15:43
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Aha! I seemed to recall reading about this interesting thereom a couple of years ago. Of course, it was in Henno Brandma's collection of notes on topology. This one in particular is at.yorku.ca/p/a/c/a/25.htm . –  Bruno Stonek Aug 10 '11 at 1:41
    
@Bruno: Thanks a lot! I added the reference to the answer. –  t.b. Aug 10 '11 at 1:51
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Following Qiaochu Yuan's comment, you can take $x \in \mathbb{Q} \to \frac{1}{2}+\frac{\arctan(x)}{\pi}$ for an explicit solution

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Even $\frac12\left(1+\frac{x}{\sqrt{1+x^2}}\right)$, $\frac12(1+\tanh\,x)$, $\frac1{1+\exp(-x)}$ or any other "sigmoidal curves" ought to work... –  J. M. Aug 10 '11 at 7:07
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