12

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 $\mathbb Q$?

user15464
  • 11,682
  • 1
    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 01 '11 at 04:18
  • 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 01 '11 at 06:28
  • @MAK, it is not true that a bijection between totally disconnected spaces in automatically an homeo. – Mariano Suárez-Álvarez Oct 01 '11 at 06:31
  • @Mariano: but isn't a map that restricts to a homeomorphism on each component, itself a homeomorphism? – MAK Oct 01 '11 at 06:39
  • @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-Álvarez Oct 01 '11 at 06:44
  • @Mariano, I see, so we also need to have an order-isomorphism. What if we had two disconnected, but not totally-disconnected subspaces X,Y, with non-trivial components. Is the same then the case, or would then a map f: X-->Y that restricts to a homeomorphism h| X_i-->Y_i ,between the components of X,Y be aglobal homeo. from X to Y? – MAK Oct 01 '11 at 06:49
  • @MAK: Take $X=\bigcup{[2^{-(2n+1)},2^{-2n}]:n\in\omega}\cup\bigcup{[2+2^{-(2n+1)},2+2^{-2n}]:n\in\omega}\cup{0,2}$ and let $h:X\to X$ interchange $0$ and $2$ and be the identity on $X\setminus{0,2}$. (This is really the same idea that Mariano used.) – Brian M. Scott Oct 01 '11 at 07:13
  • @Brian: Thanks; but {0,2} is totally disconnected. I will drop this, if it is distracting from the OP, but I think the result is true if there are no trivial --singleton--component, because for trivial components, the order matters. – MAK Oct 01 '11 at 07:22
  • Never mind; this is too long and getting too far off-topic; thanks for both your comments, though. – MAK Oct 01 '11 at 07:37
  • @MAK: The example is exactly what you required: the space is disconnected but not totally disconnected, and $h$ is a homeomorphism on each component but not a homeomorphism on $X$. If you prefer, replace $0$ by $[-1/2,0]$ and $2$ by $[3/2,2]$ and let $h$ interchange these intervals in the obvious way ($x\leftrightarrow x+2$) and be the identity on the rest of $X$. Now all components of $X$ are non-trivial, $h$ is a homeomorphism on each, and $h$ is not a homeomorphism of $X$ to itself. – Brian M. Scott Oct 01 '11 at 07:38
  • 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. – MAK Oct 02 '11 at 09:39

1 Answers1

14

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 $\mathbb R\to\mathbb R$.

Mariano Suárez-Álvarez
  • 135,076
  • Okay, thanks, so that answers the second part. But must there exist an ambient homeomorphism? – user15464 Oct 01 '11 at 03:25
  • 4
    @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 01 '11 at 06:31
  • 1
    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