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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.

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  • $\begingroup$ I have the vague inkling that it will be easier to consider the one-point compactifications, i.e. $(\Bbb R^2 \cup \{\infty\}) \setminus \Bbb Q$. $\endgroup$
    – Ben Grossmann
    Mar 30, 2017 at 19:57
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    $\begingroup$ @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. $\endgroup$
    – Ben Grossmann
    Mar 30, 2017 at 20:00
  • $\begingroup$ 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 :( $\endgroup$
    – polynomial_donut
    Mar 30, 2017 at 21:11
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    $\begingroup$ Take a look at theorem 7 in dwc.knaw.nl/DL/publications/PU00013058.pdf $\endgroup$
    – Niels J. Diepeveen
    Mar 30, 2017 at 22:23
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    $\begingroup$ @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. $\endgroup$
    – user98602
    Mar 31, 2017 at 0:32

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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.

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