convexity of a relatively open subset of a compact set

Question

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 $\hat{C}=C\cap \Omega$?

Answer 1

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 $\hat{C}=C\cap\Omega'$.