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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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    O.K: H'about this: thereis a theorem by Serpinski, that any two countable metric spaces without isolated points are homeomorphic. This homeomorphism can be extended into a homeo of R with itself.2011-10-02

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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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    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.2013-07-20