Convex open neighborhood of compact convex subset I'm stuck on what ought to be a straightforward topology problem. Say $X$ is a compact convex subset of a locally convex space (everything in sight is assumed Hausdorff). Say $Y\subseteq X$ is a compact convex subset, and $U\supseteq Y$ is relatively open in $X$. I want to find a relatively open $V\supseteq Y$ with $V\subseteq U$ and $V$ convex.
 A: It seems the following. 
Your are right, but you need some more experience dealing with such a stuff. The proof is natural. For each point $y\in Y$ choose an open convex neighborhood $U_y$ of the zero such that $(y+U_y+U_y)\cap X\subset U$. The family $\{y+U_y : y\in Y\}$ is an open cover of a compact set $Y$. Therefore there exists a finite subset $F$ of the set $Y$ such that $Y\subset \bigcup\{y+U_y : y\in F\}$. Put $U_0=\bigcap\{U_y : y\in F\}$.  The set $U_0$ is open as an intersection of a finite family of open sets and it is convex as an intersection of a family of convex sets. The set $Y+U_0$ is convex as a sum of convex  sets and it is open as a union of a family $\{y+U_0: y\in Y\}$ of open sets. The set $V=(Y+U_0)\cap X$ is relatively open in $X$ and it is convex as an intersection of two convex sets. Clearly, that $V\supset Y$. It rests to show $V\subset U$. Let $x\in V\subset Y+U_0$ be an arbitrary point. Therefore there exist a point $z\in Y$ such that $x\in z+U_0$. Also there exists a point $y\in F$ such that $z\in y+U_y$. Then  $x\in y+U_y+U_0\subset  y+U_y+U_y$. Since $x\in X,$ we see that $x\in U$.
