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As the title says I was wondering (being vaguely inspired by a question from Hatcher asking about the fundamental group of the first space) whether $\Bbb R^2\setminus \Bbb Q^2$ and $\Bbb R^2\setminus \Bbb Q^2\cup \{(0,0)\}$ are homeomorphic.

My gut feeling is that they are, they have the same properties as far as connectedness, compactness and separation axioms are concerned, supporting this feeling, but I haven't been able to prove (or disprove) this fact.

Alessandro Codenotti
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  • I have the vague inkling that it will be easier to consider the one-point compactifications, i.e. $(\Bbb R^2 \cup {\infty}) \setminus \Bbb Q$. – Ben Grossmann Mar 30 '17 at 19:57
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    @ThePortakal $(t,\sqrt{2}t)$ with $t \in [0,1]$ is a path connecting $(0,0)$ to $(1,\sqrt{2})$ which doesn't leave the space in question. Remember that a point is in the space if at least one of its coordinates is irrational. – Ben Grossmann Mar 30 '17 at 20:00
  • Nice question. I just spent about an hour trying to construct a contradiction by extending the map to an automorphism over $\mathbb{R}^2$ and then subtracting a suitable set, using uncountability of the irrationals. However, either I forgot the contradiction, or I was hallucinating mathematically... I guess it was the latter :( – polynomial_donut Mar 30 '17 at 21:11
  • @polynomial_donut I would love to see a proof that you can extend it to an automorphisk of R^2. –  Mar 30 '17 at 21:21
  • @MikeMiller 'automorphism' as in 'automorphism in the category of topological spaces'? Might be nonsense, but here goes a short version, assuming a homeomorphism $f$ as in question: Enumerate the rationals. For the first, say $q$, pick a sequence $(s_n)_n$ in the codom with limit $q$ (when included in $\mathbb{R}^2$). Define $F(q)$ to be the limit of $(f(s_n))_n$. Should independent of the particular choice, because the spaces are T2. Define an extension of $\phi$ to $\mathbb{R}^2\setminus\mathbb{Q}\rightarrow\mathbb{R}\cup{(0,0),;F(q)}$. Proceed inductively to obtain a continuous map(?) – polynomial_donut Mar 30 '17 at 21:52
  • There's no reason to think that $f(s_n)$ should converge. –  Mar 30 '17 at 22:05
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    Take a look at theorem 7 in http://www.dwc.knaw.nl/DL/publications/PU00013058.pdf – Niels J. Diepeveen Mar 30 '17 at 22:23
  • @MikeMiller Shouldn't it work by continuity in both senses? – Daniel Robert-Nicoud Mar 30 '17 at 22:23
  • @DanielRobert-Nicoud Sure, if it extends continuously to all of R^2 (when we extend the codomain). I don't know why that should be possible. Seems like you could have a sequence going to (1/3,1/3) whose image is wildly divergent. –  Mar 30 '17 at 22:24
  • @MikeMiller We are given a topology coming from a metric on the domain and codomain. So if $(s_n)_n$ has a limit when included in $\mathbb{R}^2$, then this means it is a Cauchy sequence in the domain of $f$. Then, so is $(f(s_n))_n$ in the codomain, by 'epsilon-delta'. So, $(f(s_n)_n$ should indeed have a limit when included in $\mathbb{R}^2$. – polynomial_donut Mar 31 '17 at 00:27
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    @polynomial_donut You have no epsilon delta statement about the extension. You've made no assumption about its continuity! To be not clear, consider the map $S^1 \setminus {-1} \to (0,2\pi)$ given by taking a logarithm. This does not have an extension to a map $S^1 \to [0,2\pi]$. The only way you can extend continuous maps on metric spaces to their completions is by assuming (not necessarily convergent in your incomplete space) Cauchy sequences are sent to Cauchy sequences. This is usually false. –  Mar 31 '17 at 00:32
  • @MikeMiller Sure, I also haven't stated I had a bullet proof proof ;) Niels Diepeveen's comment above refers to a paper that seems to speak in favor of the possibility of such an extension, though, or maybe I completely misunderstood it at my first glance. Thanks for the example, by the way! – polynomial_donut Mar 31 '17 at 06:49
  • @polynomial_donut The paper gives an example of such a homeomorphism that does extend; it certainly doesn't imply all do. –  Mar 31 '17 at 07:14

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It is a theorem due to Brouwer (1913) that for any two dense countable subsets $A, B\subset R^n$, there is a homeomorphism $R^n\to R^n$ sending $A$ to $B$ bijectively. See also "General Topology" by Engelking (he has this as an exercise 4.5.2, with a detailed hint). If I remember it correctly, Hirsch in "Differential Topology" also has this as an exercise where instead of a homeomorphism he asks for a diffeomorphism. Lastly,

M. Morayne, Measure preserving analytic diffeomorphisms of countable dense sets in $C^n$ and $R^n$, Colloq. Math. 52 (1987), no. 1, 93–98.

proves that for any two countable dense subsets $A, B$ in $R^n$, $n\ge 2$, there exists an analytic volume-preserving diffeomorphism of $R^n\to R^n$ sending $A$ to $B$ bijectively.

So, the conclusion is that your spaces are homeomorphic.

Moishe Kohan
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