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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 $K \subset \overline{co}(K)$.

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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$.

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  • $\begingroup$ The subtlety I'm not getting is the closure of $V$ not $V$ itself. $\endgroup$ Feb 20 '19 at 10:34
  • $\begingroup$ $V$ is an open neighborhood of $0$. It is fact that every open neighborhood of $0$ contains the closure of another open neighborhood of $0$. This is what is being used here. $\endgroup$ Feb 20 '19 at 10:37

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