Convex sets and projections - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T02:14:07Z http://mathoverflow.net/feeds/question/41606 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/41606/convex-sets-and-projections Convex sets and projections MasterOfOrion 2010-10-09T15:57:31Z 2010-10-09T23:59:58Z <p>Hello!</p> <p><em>I recently started (it's purely self-education) reading a <strong>"Mathematical programming and optimizations"</strong> book, did a vast part of the exercises related to the theoretical part and at one moment I got the following question about convex sets:</em></p> <p>I'm almost sure this statement is correct, but unfortunately, couldn't find something similiar on the internet and I tried to prove it, but I couldn't.</p> <p>Assume we have some set $S \subset \mathbb{R_n}$ and for this set: $S = \overline{S}$ <em>(set closure equals the set itself)</em>.</p> <p><em>Now, there exists only one projection of arbitrary point $y$ which doesn't belong to the set</em> $S :$</p> <p>$\forall y \in \mathbb{R_n}, \space y \notin S: \space \exists ! \space p = \pi_S(y)$</p> <p>This should mean that $S$ is a <strong><em>convex</em></strong> set.</p> <p>Could someone please point me if I'm wrong <em>(or right, but with some limitations for this statement)</em> and help me proving it if I'm right.</p> <p><em>I also understand that this question be a too "basic" to post here, but I've just started educating myself in this sphere and hope that sometimes I'll get smart enough to ask really bright questions :)</em></p> <p>Thank you.</p> http://mathoverflow.net/questions/41606/convex-sets-and-projections/41608#41608 Answer by Robin Chapman for Convex sets and projections Robin Chapman 2010-10-09T16:13:34Z 2010-10-09T16:13:34Z <p>I presume what you want to prove is the following. Let $S$ be a nonempty closed subset of $\mathbb{R}^n$. Then if there is a point $y\in\mathbb{R}^n$ and there are at least two points $p$ and $q$ in $S$ with Euclidean distance $d$ from $y$ (where $d$ is the distance of $y$ from $S$), then $S$ is not convex. To see this, note that the midpoint $r$ of the line segment $pq$ is closer to $y$ than $p$ of $q$ is, and so cannot lie in $S$. Hence $S$ isn't convex.</p>