Hello!


*I recently started (it's purely self-education) reading a **"Mathematical programming and optimizations"** 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:*


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.


Assume we have some set $S \subset \mathbb{R\_n}$ and for this set: $S = \overline{S}$ *(set closure equals the set itself)*.


*Now, there exists only one projection of arbitrary point $y$ which doesn't belong to the set* $S :$


$\forall y \in \mathbb{R\_n}, \space y \notin S: \space \exists ! \space p = \pi\_S(y) $


This should mean that $S$ is a ***convex*** set.


Could someone please point me if I'm wrong *(or right, but with some limitations for this statement)* and help me proving it if I'm right.


*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 :)*


Thank you.



Convex sets and projections
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.


