A question about unbounded connected subsets of the plane. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T06:18:39Z http://mathoverflow.net/feeds/question/36587 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/36587/a-question-about-unbounded-connected-subsets-of-the-plane A question about unbounded connected subsets of the plane. Garabed Gulbenkian 2010-08-24T19:43:51Z 2010-08-27T20:18:38Z <p>A number of clever examples have been given of unbounded connected subsets of the Euclidean plane containing no <em>infinite</em> bounded subsets that are connected. None of those that I have seen are completely metrizable. Does anybody know if such can exist or if their existence can be ruled out by some theorem? I know that no such completely metrizable sets can exist if they are the graphs of functions of the form y=f(x) in the Cartesian plane. But does this prohibition extend to all unbounded connected planar sets?</p> http://mathoverflow.net/questions/36587/a-question-about-unbounded-connected-subsets-of-the-plane/36818#36818 Answer by Pietro Majer for A question about unbounded connected subsets of the plane. Pietro Majer 2010-08-26T22:21:32Z 2010-08-27T20:18:38Z <p>The following gives a partial answer: no such unbounded connected set may exist with the further assumption that it is <em>closed</em>. Actually, the argument generalizes for any locally compact metric space. I'm not completely sure that a much simpler or even trivial proof may exist, though. </p> <p>Let $\Gamma$ be a closed unbounded connected subset of the plane. Let $x\in\Gamma$ and let $B:=B(x,r)$ be an open ball around $x$. I claim that the connected component of $x$ in $\Gamma\cap \bar{B}$ meets $\partial B$, which shows that $\Gamma$ does contain non-trivial bounded connected subsets. </p> <p>For any $\epsilon>0$, consider the $\epsilon$-neighborhood of $\Gamma,$ that is $\Gamma_\epsilon:=\cup_{y\in\Gamma}B(y,\epsilon).$ It is an open unbounded connected subset of the plane. Let $U_\epsilon$ be the connected component of $x$ in $\Gamma_\epsilon\cap B$. Since the latter is locally connected, $U_\epsilon$ is both an open and closed subset of it in the relative topology. It is therefore an open subset of $\Gamma_\epsilon$; however it is not closed in it, because $\Gamma_\epsilon$ is connected. Therefore $\bar U_\epsilon$ is a closed connected set that meets $\partial B,$ and of course contains $x$. Since the set of all connected closed subsets of a compact metric space is compact in the Hausdorff distance, taking a limit as $\epsilon\to0$ we get a bounded connected subset of $\Gamma$ connecting $x$ with $\partial B$ (this also passes to the limit).</p> <p><strong>Rmk</strong> One could state the above in terms of the one-point compactification of $\Gamma$, and more generally for compact connected metric spaces. The trick of approximating a metric space with a locally connected metric space is made possible via the Kuratowski embedding (one defines $X_\epsilon$ as an $\epsilon$ nbd of $X$ in the embedding).</p> <p>PS: Of course the same affirmative conclusion holds, even more directely, if $\Gamma$ is assumed to be open, which is another case included in the original assumption of completely metrizable. </p>