This is probably the first question regarding the theorem in the title...
If $K$ is a compact set in a locally convex space $X$, and if $\bar{co}(K)$ is also compact, then every extreme point of $\bar{co}(K)$ lies in $K$.
I struggle to understand the very first 3 lines of the proof.
Assume that some extreme point $p$ of $\bar{co}(K)$ is not in $K$. Then there is a convex balanced neighborhood $V$ of $0$ in $X$ such that $$ (p + \overline{V}) \cap K = \emptyset $$
Can anyone explain why can we find such a $V$? My guess is that it's due to compacteness of both $\overline{co}(K)$ and $K$ plus the fact that $K \subset \overline{co}(K)$.
Compactness is not required. $K^{c}$ is an open set containing $p$ and this gives existence of such a set $V$. This has been proved in earlier sections of Rudin's book$.