Connectedness of the complement of small subsets (extended question) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T00:50:09Z http://mathoverflow.net/feeds/question/117923 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/117923/connectedness-of-the-complement-of-small-subsets-extended-question Connectedness of the complement of small subsets (extended question) GH 2013-01-03T03:08:14Z 2013-01-12T18:29:48Z <p>The following questions occurred to me while browsing this site and looking at Exercise 20 <a href="http://www.math.unt.edu/~sjackson/6010f11/polish.pdf" rel="nofollow">here</a>. </p> <p><strong>Question 1.</strong> 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?</p> <p><strong>EDIT.</strong> I deleted an erroneous paragraph. Let me add two more questions, the first one being Gerald Edgar's comment <a href="http://mathoverflow.net/questions/117840/injective-with-finite-discontinuities-mapping-from-mathbb-rn-to-0-1/117844" rel="nofollow">here</a>, the second one correcting the erroneous paragraph.</p> <p><strong>Question 2.</strong> 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?</p> <p>It seems that Joel's argument answers this in the affirmative as well.</p> <p><strong>Question 3.</strong> Let $n>1$. Are there two countable dense subsets $A,B\subset\mathbb{R}^n$ whose complements are not homeomorphic?</p> http://mathoverflow.net/questions/117923/connectedness-of-the-complement-of-small-subsets-extended-question/117924#117924 Answer by Joel David Hamkins for Connectedness of the complement of small subsets (extended question) Joel David Hamkins 2013-01-03T03:12:29Z 2013-01-03T03:24:37Z <p>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 $A$. </p> <p>This observation also appears to answer the exercise in your link.</p> <p>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. </p> http://mathoverflow.net/questions/117923/connectedness-of-the-complement-of-small-subsets-extended-question/117929#117929 Answer by Ali Enayat for Connectedness of the complement of small subsets (extended question) Ali Enayat 2013-01-03T06:39:37Z 2013-01-12T18:29:48Z <p>The answer to Question 3 is negative; this is an immediate consequence of the following classical theorem:</p> <blockquote> <blockquote> <p><strong>Theorem.</strong> 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 $f(A)=B$.</p> </blockquote> </blockquote> <p><strong>Historical note:</strong> In the above theorem, the $n=1$ case is due to Cantor; later and independently the general case was established by Fréchet [<em>Les dimensions d’un ensemble abstrait</em>, Math. Ann. 68 (1910), 145–168] and Brouwer [<em>Some remarks on the coherence type</em> $\eta$, Proc. Akad. Amsterdam 15 (1913), 1256–1263]. </p> <p>The above theorem also holds for the Hilbert cube sitting in for $\Bbb{R}^n$, as shown by M.K. Fort in his paper <em>Homogeneity of infinite products of manifolds with boundary</em>, Pacific J. Math. 12 (1962), 879–884.</p>