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Let $S$ be a countable dense subset of $\mathbb R$. Must there exist a homeomorphism $f: \mathbb R \rightarrow \mathbb R$ such that $f(S) = \mathbb Q$? More weakly, must $S$ be homeomorphic to $\mathbb Q$?

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This is certainly an ambiguously phrased question. It could mean "Is there any countable dense subset of $\mathbb{R}$ that is ambiently homeomorphic to $\mathbb{Q}$?", or it could mean "Is it the case that any countable dense subset of $\mathbb{R}$ is ambiently homeomorphic to $\mathbb{Q}$?". If the latter, merely saying "every" rather than "any" would eliminate the ambiguity, and if the former, then "some" would accomplish the purpose. –  Michael Hardy Oct 1 '11 at 4:18
I think this should do it too: Let D be a dense countable subset of $\mathbb R$ since the only connected subsets of the real line are the intervals and all intervals are countable, then D contains no intervals, so that D is totally-disconnected, with components $d_n$ , where the $d_i$'s are the points. But the same is true for $\mathbb Q$, i.e., its components are the points $q_i$ . Then any bijection between $\mathbb Q$ and D is trivially a homeomorphism between the connected components of the respective sets, and I think this implies it is a homeomorphism between D and $\mathbb Q$. –  MAK Oct 1 '11 at 6:28
@MAK, it is not true that a bijection between totally disconnected spaces in automatically an homeo. –  Mariano Suárez-Alvarez Oct 1 '11 at 6:31
@Mariano: but isn't a map that restricts to a homeomorphism on each component, itself a homeomorphism? –  MAK Oct 1 '11 at 6:39
@MAK, consider the map from Q to itself which fixes every point except that it interchanges 0 and 1: it is obviously an homeo when restricted to each connected component... –  Mariano Suárez-Alvarez Oct 1 '11 at 6:44

1 Answer 1

up vote 12 down vote accepted

Two countable totally ordered, densely ordered sets without endpoints are isomorphic---this is a theorem of Cantor (Gesammelte Ahbandlungen, chp. 9, page 303 ff. Springer, 1932) Thus two countable dense subsets of $\mathbb R$ are homeomorphic, since their topology is induced by their orders.

Now, if $A$, $B\subset\mathbb R$ are countable dense subsets, fix an order isomrphism $f:A\to B$ and extend it by continuity. What you get is an homeomorphism $\mathbb R\to\mathbb R$.

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Okay, thanks, so that answers the second part. But must there exist an ambient homeomorphism? –  user15464 Oct 1 '11 at 3:25
@user15464: What do you mean by "ambient homeomorphism"? There exists an order preserving isomorphism between $S$ and $\mathbb Q$, which is an order-topology homeomorphism; and this can be extended uniquely to an automorphism of $\mathbb R$. –  Asaf Karagila Oct 1 '11 at 6:31
Note that there's a little work to do in the phrase "extend it by continuity"; continuous functions on a dense subset don't in general extend continuously to the whole space. Some kind of uniformity is needed. This step is where you need to use the fact that $B$ is dense. –  Nate Eldredge Jul 20 '13 at 19:15

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