Here is Greg's proof of the Krein-Milman theorem (only the existence part).

**Proposition 1:** Let $A\subseteq V$ be a non-empty compact convex set of an hausdorff locally convex semi-topological vector space (over some field which contains the reals topologically). Then $A$ has an extreme point.

Specifically we shall show that every non-empty convex open-in-$A$ proper subset $U^{-}$, has an extension to a convex open-in-$A$ proper subset $U^{+}\supseteq U^{-}$, for which $A\setminus U^{+}=\{e\}$ some extreme point $e$.

**Proof (Prop 1):** If $A$ is a singleton, we are done. So suppose otherwise. First notice we can separate two points by a convex open set, $U$, yielding $U^{-}=U\cap A$ a convex proper subset open in $A$. So fix such a set $U^{-}$.

Now let $\mathbb{P}:=\{W:U^{-}\subseteq W\mbox{ convex proper subset open in }A\}$. We claim that the p.o. $(\mathbb{P},\subseteq)$ has the Zorn property. Clearly the union of any chain of convex subsets open in $A$, is again a convex subset open in $A$. Suppose \it{per contra} that a union of a chain $\mathcal{C}\subseteq\mathbb{P}$ is equal to $A$. Then since $A$ is compact, there must be a finite sub(chain) $\mathcal{C}'\subseteq\mathcal{C}$ for which $\operatorname{max}(\mathcal{C}')=\bigcup\mathcal{C}'=A$, which contracts the properties of members of $\mathbb{P}$. Thus $\mathbb{P}$ is non-empty and has the Zorn property. So fix $U^{+}$ open such that $U^{+}\cap A\in\mathbb{P}$ maximal.

**Sub-claim 1:** $U^{+}\cap A\subset\operatorname{cl}_{A}U^{+}$ strictly.

**Proof (sub-claim 1):** Let $x\in U^{+}\cap A$ and $y\in A\setminus U^{+}$. Consider the map (here we need the underlying field to contain the reals topologically)

$$\begin{array}[t]{lrclll}
&f &: &\mathbb{F} &\to &V\\
&&: &s &\mapsto &s~x+(1-s)~y\\
\end{array}$$

which is continuous and affine. Thus the set

$$S:=\{s\in[0,1]:s\mapsto s~x+(1-s)~y\in U^{+}\cap A\} =[0,1]\cap f^{-1}U^{+}\cap f^{-1}A)$$

is a convex. The set $f^{-1}A$ is closed convex and contains $0,1$, thus contains $[0,1]$. So $S=[0,1]\cap f^{-1}U^{+}$ which is convex open in $[0,1]$. Since $0\notin S$ and $1\in S$, then $S=(a,1]$ some $a\in[0,1)$. Clearly $f(a)=\lim_{s\searrow a}f(s)$ which is a limit of vectors from $U^{+}\cap A$, and thus lies in $\operatorname{cl}U^{+}$. It also clearly lies in $A$, since $f^{-1}A\supseteq[0,1]$. We also have $f(a)\notin U^{+}$. Thus $\operatorname{cl}_{A}U^{+}\setminus U^{+}=A\cap\operatorname{cl}U^{+}\setminus U^{+} \supseteq\{f(a)\}$. Thus the containment is strict. QED (sub-claim 1)

**Sub-claim 2:** If $W\subseteq A$ is convex, then $U^{+}\cup W$ is convex.

**Proof (sub-claim 2):** Fix $x\in U^{+}\cap A,t\in(0,1)$. Consider the map

$$\begin{array}[t]{lrclll}
&T &: &V &\to &V\\
&&: &y &\mapsto &t~x+(1-t)~y\\
\end{array}$$

This is continuous and affine. By the proposition below, we also see that $T(\operatorname{cl}(U^{+}))\subseteq U^{+}$. So $A\cap T^{-1}U^{+}$ is convex, open in $A$, and contains $A\cap\operatorname{cl}(U^{+})$ which strictly contains $U^{+}\cap A$ by Claim 1. Thus by maximality of $U^{+}$ in $\mathbb{P}$, $A\cap T^{-1}U^{+}\overset{\mbox{must}}{=}A$. In particular, $T^{-1}U^{+}\supseteq A$, and so $TW\subseteq TA\subseteq U^{+}$. Utilising such maps as $T$, we see that a convex-linear combination of any pair of elements from $U^{+}\cup W$ is contained in $U^{+}\cup W$. QED (sub-claim 2)

**Sub-claim 3:** $A\setminus U^{+}$ is a singleton.

**Proof (sub-claim 3):** Else, let $x_{1},x_{2}\in A\setminus U^{+}$ distinct. Let $W$ be a convex open set separating $x_{1}$ from $x_{2}$. Then by Claim 3, $U^{+}\cap A\cup W\cap A$ is convex open in $A$. Since $x_{1}\in W\setminus U^{+}$, this convex set is strictly large than $U^{+}\cap A$. By maximality of $U^{+}$ in $\mathbb{P}$, we have that $U^{+}\cap A\cup W\cap A=A\ni x_{2}$. But this contradicts the fact that $x_{2}\notin U^{+}\cup W$. QED (sub-claim 3)

At last, we claim that the point $e\in A\setminus U^{+}$ is extreme in A. Consider $x,y\in A$ and $t\in(0,1)$ for which $tx+(1-t)y=e$. Case by case, we see that if $x,y\in U^{+}$ then $e=tx+ty\in U^{+}$. If $x\in A\setminus U^{+}$, $y=\frac{e-tx}{1-t}=\frac{e-te}{1-t}=e=x$. If $y\in A\setminus U^{+}$, then $x=e=y$ similarly. Thus $e$ is extreme. QED(Prop 1)

**Proposition 2:** Let $A\subseteq V$ be a convex subset of a semi-topological space, and $x,y$ be vectors lying respectively in the closure and interior of $A$. Then for any $t\in(0,1)$, we have $tx+(1-t)y\in A$.

**Proof:** Let $U$ be an open neighbourhood of $y$, contained in $A$. Observe that $x-t^{-1}(1-t)(U-y)$ is an open neighbourhood of $x$, and thus meets $A$, say at $x'$. Thus $x\in x'+t^{-1}(1-t)(U-y)$ implying $tx+(1-t)y\in tx'+(1-t)U\subseteq tA+(1-t)A\subseteq A$, since $A$ is convex. QED(Prop 2)

2more comments