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The following questions occurred to me while browsing this site and looking at Exercise 20 here.

Question 1. Let $n>1$. Does there exist a countable dense subset $A\subset\mathbb{R}^n$ for which the complement $\mathbb{R}^n\setminus A$ is disconnected?

EDIT. I deleted an erroneous paragraph. Let me add two more questions, the first one being Gerald Edgar's comment here, the second one correcting the erroneous paragraph.

Question 2. Let $n>1$. Is it true that for any subset $A\subset\mathbb{R}^n$ of Hausdorff dimension less than $n-1$ the complement $\mathbb{R}^n\setminus A$ is connected?

It seems that Joel's argument answers this in the affirmative as well.

Question 3. Let $n>1$. Are there two countable dense subsets $A,B\subset\mathbb{R}^n$ whose complements are not homeomorphic?

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I don't think exercise 20 implies that the complements are not homeomorphic. – Joseph Van Name Jan 3 '13 at 3:58
@Joseph: You are right, this is all connected to my very silly initial thought that $\mathbb{R}^n\setminus\mathbb{Q}^n$ is equal to $(\mathbb{R}\setminus\mathbb{Q})^n$ which is homeomorphic to $\mathbb{N}^\mathbb{N}$ and is totally disconnected. So I cut another erroneous line, not affecting the questions themselves. – GH from MO Jan 3 '13 at 4:06
Question 3 is very nice. – Joel David Hamkins Jan 3 '13 at 4:10
For question 3 it seems like one could adapt the proof that all countable dense total orders are isomorphic to this case. For instance, one could rotate $\mathbb{R}^{n}$ so that the sets $A$ and $B$ are both totally ordered in all $n$-variables. i.e., so that the projections onto the $i$-th coordinate is injective. By density, it seems like there would be bijection from $A$ to $B$ that is order preserving in each variable that extends to a homeomorphism from $\mathbb{R}^{n}$ to itself which is order preserving in all variables. i.e. You map cubes to cubes. – Joseph Van Name Jan 3 '13 at 5:01

The answer to Question 3 is negative; this is an immediate consequence of the following classical theorem:

Theorem. For all $n\ge 1$, if $A$ and $B$ are countable dense subsets of $\Bbb{R}^n$, then there is a homomeomorphism $f: \Bbb{R}^n\rightarrow \Bbb{R}^n$ such that $f(A)=B$.

Historical note: In the above theorem, the $n=1$ case is due to Cantor; later and independently the general case was established by Fréchet [Les dimensions d’un ensemble abstrait, Math. Ann. 68 (1910), 145–168] and Brouwer [Some remarks on the coherence type $\eta$, Proc. Akad. Amsterdam 15 (1913), 1256–1263].

The above theorem also holds for the Hilbert cube sitting in for $\Bbb{R}^n$, as shown by M.K. Fort in his paper Homogeneity of infinite products of manifolds with boundary, Pacific J. Math. 12 (1962), 879–884.

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Thank you very much, this is very interesting and useful! – GH from MO Jan 3 '13 at 7:01
Ali, would it be possible for you to sketch the proof? I imagine a back-and-forth construction combined with extra convergence information, successively finer promises about the map, to make sure that things work out in the limit. For example, at any stage, one has mapped finitely many points correctly from $A$ to $B$, and also made promises about how, say, successively smaller circles surrounding those points but in the common complement will transfer. – Joel David Hamkins Jan 3 '13 at 13:30
@Joel & GH: Since I am "on the road", I will just point out that the proof, as sketched by Brouwer in the 9-page paper below (in Theorem 7, p.1260), only takes a paragraph or so. Note that he defines $C_m$ on p.1259, where he writes " an arbitrary countable set of points, lying everywhere dense in $R_n$ can construct a cartesian system of coordinates $C_m$ with the property that no $R_{n-1}$ parallel to a coordinate space contains more than one point of the set" (note that Brouwer uses $R_{n}$ for our $\Bbb{R}^{n}$). – Ali Enayat Jan 4 '13 at 1:41
I sketched this proof already in the comments. – Joseph Van Name Jan 4 '13 at 16:16
@Joseph: yes, your outline does the job, and is a good summary of the proof given in Brouwer's paper. – Ali Enayat Jan 5 '13 at 18:24

Every countable subset $A\subset \mathbb{R}^n$, with $n\gt 1$, and indeed, every subset of size less than the continuum, has a path-connected complement. This is because for any two points in the complement, there is a foliation of continuum many paths joining them, and so most of these paths lie entirely in the complement of $A$.

This observation also appears to answer the exercise in your link.

The argument in your final paragraph appears to conflate $\mathbb{R}^n-B^n$ with $(\mathbb{R}-B)^n$, but these are not generally the same and they cannot be equal when $B$ is countable.

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You need $n>1$ for this. – Pablo Zadunaisky Jan 3 '13 at 3:14
Thanks, I have edited. – Joel David Hamkins Jan 3 '13 at 3:15
It is also important to have in mind that the fact pointed out by Joel was first noted by Georg Cantor, who noted that the complement of $\Bbb{Q}^2$ in $\Bbb {R}^2$ is path connected; Cantor was impressed with this fact, and thought it revealed an important feature of physical space (I learned about this discovery of Cantor from J. Dauben's Biography of Cantor). – Ali Enayat Jan 3 '13 at 3:28
Sorry, I got confused about $B^n$ and this is why your argument did not occur to me (it is in fact quite similar to my response to…). I fixed that and added two more questions. – GH from MO Jan 3 '13 at 3:48
Hey, don't worry about it! I've said some things too, but I'm glad to say that this comment field is simply too short to provide links to specific instances... – Joel David Hamkins Jan 3 '13 at 20:42

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