4
$\begingroup$

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?

$\endgroup$
4
  • 2
    $\begingroup$ I don't think exercise 20 implies that the complements are not homeomorphic. $\endgroup$ Jan 3, 2013 at 3:58
  • $\begingroup$ @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. $\endgroup$
    – GH from MO
    Jan 3, 2013 at 4:06
  • 1
    $\begingroup$ Question 3 is very nice. $\endgroup$ Jan 3, 2013 at 4:10
  • 2
    $\begingroup$ 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. $\endgroup$ Jan 3, 2013 at 5:01

2 Answers 2

9
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Thank you very much, this is very interesting and useful! $\endgroup$
    – GH from MO
    Jan 3, 2013 at 7:01
  • $\begingroup$ 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. $\endgroup$ Jan 3, 2013 at 13:30
  • 1
    $\begingroup$ @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 "...to 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 $\Bbb{R}^{n}$). dwc.knaw.nl/DL/publications/PU00013058.pdf $\endgroup$
    – Ali Enayat
    Jan 4, 2013 at 1:41
  • 1
    $\begingroup$ I sketched this proof already in the comments. $\endgroup$ Jan 4, 2013 at 16:16
  • 1
    $\begingroup$ @Joseph: yes, your outline does the job, and is a good summary of the proof given in Brouwer's paper. $\endgroup$
    – Ali Enayat
    Jan 5, 2013 at 18:24
6
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ You need $n>1$ for this. $\endgroup$ Jan 3, 2013 at 3:14
  • 1
    $\begingroup$ 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). $\endgroup$
    – Ali Enayat
    Jan 3, 2013 at 3:28
  • $\begingroup$ 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 mathoverflow.net/questions/117840/…). I fixed that and added two more questions. $\endgroup$
    – GH from MO
    Jan 3, 2013 at 3:48
  • $\begingroup$ I should have said: $\mathbb{R}^n-B^n$ and $(\mathbb{R}-B)^n$ are never the same for $n\gt 1$ unless $B$ is empty or everything. $\endgroup$ Jan 3, 2013 at 13:54
  • 1
    $\begingroup$ 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... $\endgroup$ Jan 3, 2013 at 20:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.