# Convex hull of extremal points

Let $E$ be a locally convex Hausdorff space and $X$ a compact convex set. Is it true that $\overline{conv}(\overline{ex(X)})=\overline{conv}(ex(X))$? Here $conv(X)$ is the convex hull of the subset $X$ and $ex(X)$ is the extremal points set. I think that it's true, but I can't find a proof. I tried to first picking a convergent net $(x_\alpha)_{\alpha \in A}$ in the first set and then showing that there is another net $(y_\beta)_{\beta \in B}$ in the second set converging to the same limit, but I couldn't construct that second net.

Edit: I was read Lectures on Choquet's Theorem by R.Phelps, and I stumbled upon this question when he was proving that the Krein-Milman theorem is equivalent to an integral representation theorem. This question arises when you are trying to prove that this integral representation of the points implies Krein-Milman, so I can't use it to prove this!

Krein Milman says that $X=\overline{conv}(ex(X))$ since $X$ is compact thus closed, $\overline{ex(X)}\subset X$, you deduce that $conv(\overline{ex (X))}\subset conv(X)=\overline{conv}(e(X))=X$. Since $e(X)\subset\overline{e(X)}$, we deduce that $\overline{conv}(e(X))\subset\overline{conv}(\overline{e(X)})$ and henceforth $\overline{conv}(e(X))=\overline{conv}(\overline{e(X)})$.
Now I figure it out. I was looking in this site and found this Convex hull of extreme points . Just note that $ex(X) \subset conv(ex(X))$, the details are an exercise!
• So you don't need $X$ be convex and compact . – Red shoes Jul 3 '17 at 3:59