TITLE: Are $\Bbb R^2\setminus \Bbb Q^2$ and $\Bbb R^2\setminus \Bbb Q^2\cup \{(0,0)\}$ homeomorphic?
QUESTION [24 upvotes]: 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.

REPLY [12 votes]: 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.