# convexity of a relatively open subset of a compact set

I'm struggling with the following problem: it seems to be true but I'm not able to prove it! Let $$C$$ be a compact convex subset of a locally convex metric vector space and $$\hat{C}$$ be a relatively open subset of $$C$$, i.e. there exists an open set $$\Omega$$ such that $$\hat{C}=C\cap\Omega$$. Clearly if $$\Omega$$ is convex then $$\hat{C}$$ is also convex; is the converse true? I mean, if I assume that $$\hat{C}$$ is convex, can I suppose the existence of an open and convex set $$\Omega$$ such that $$\hat{C}=C\cap \Omega$$?

• Hello and welcome to math.stackexchange. Let $x \in \hat C$, then there exists an open ball $B$ about $x$ with maximal radius such that $B \cap C \subseteq \hat C$. Now take the union of all these balls and show that it is convex. – Hans Engler Apr 10 at 21:56
• @HansEngler Are you sure this works? I don't find it convincing. – Kavi Rama Murthy Apr 10 at 23:24
• @HansEngler Thank you for your fast reply. Unfortunately take the square $C=[-2,2]\times[0,4]$ and $\hat{C}=\{(x,y)\in (-2,2)\times[0,2):x<y\}$ (that is open in $C$). Then the union of all the proposed balls is the open set $\hat{C}\cup B((-1,0);1)$ which is not convex (where $B((-1,0);1)$ is the open ball with center at $(-1,0)$ and radius $1$) – Giulio Marchi Apr 11 at 11:31
• On Hilbert spaces, the Hilbert projection theorem might prove to be useful for this problem – Berni Waterman Apr 11 at 14:37
• @GiulioMarchi You should write an official answer to your own question and accept it (math.stackexchange.com/help/accepted-answer). This will clear the question from the unanswered queue. – Paul Frost Apr 15 at 22:45

## 1 Answer

I wrote that it seemed to be true but, after a week, I have found a counterexample! Take the square $$C=[-2,2]\times[-2,2]$$ and the open set $$\Omega=A\cup B$$ where $$A=(-2,2)\times(-2,2)$$ and $$B=(-1,1)\times\mathbb{R}$$. Then $$\hat{C}=A\cup[(-1,1)\times\{\pm 2\}]$$ is convex and open in $$C$$ but clearly there is no open and convex set $$\Omega'$$ such that $$\hat{C}=C\cap\Omega'$$.