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

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    $\begingroup$ 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. $\endgroup$ – Hans Engler Apr 10 at 21:56
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    $\begingroup$ @HansEngler Are you sure this works? I don't find it convincing. $\endgroup$ – Kavi Rama Murthy Apr 10 at 23:24
  • $\begingroup$ @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 $1$) $\endgroup$ – Giulio Marchi Apr 11 at 11:31
  • $\begingroup$ On Hilbert spaces, the Hilbert projection theorem might prove to be useful for this problem $\endgroup$ – Berni Waterman Apr 11 at 14:37
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    $\begingroup$ @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. $\endgroup$ – Paul Frost Apr 15 at 22:45
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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 $\hat{C}=C\cap\Omega'$.

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