convexity - Existence of an extreme point of a compact convex set

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

Proposition 1: Let $Asubseteq V$ 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 $Asetminus U^{+}={e}$ 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 $U^{-}=Ucap A$ a convex proper subset open in $A$. So fix such a set $U^{-}$.

Now let $mathbb{P}:={W:U^{-}subseteq Wmbox{ convex proper subset open in }A}$. 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 $mathcal{C}subseteqmathbb{P}$ is equal to $A$. Then since $A$ is compact, there must be a finite sub(chain) $mathcal{C}'subseteqmathcal{C}$ for which $operatorname{max}(mathcal{C}')=bigcupmathcal{C}'=A$, 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 $U^{+}cap Ainmathbb{P}$ maximal.

Sub-claim 1: $U^{+}cap Asubsetoperatorname{cl}_{A}U^{+}$ strictly.

Proof (sub-claim 1): Let $xin U^{+}cap A$ and $yin Asetminus U^{+}$. 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

$$S:={sin[0,1]:smapsto s~x+(1-s)~yin U^{+}cap A} =[0,1]cap f^{-1}U^{+}cap f^{-1}A)$$

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 $0notin S$ and $1in S$, then $S=(a,1]$ some $ain[0,1)$. Clearly $f(a)=lim_{ssearrow a}f(s)$ which is a limit of vectors from $U^{+}cap A$, and thus lies in $operatorname{cl}U^{+}$. It also clearly lies in $A$, since $f^{-1}Asupseteq[0,1]$. We also have $f(a)notin U^{+}$. Thus $operatorname{cl}_{A}U^{+}setminus U^{+}=Acapoperatorname{cl}U^{+}setminus U^{+} supseteq{f(a)}$. Thus the containment is strict. QED (sub-claim 1)

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

Proof (sub-claim 2): Fix $xin U^{+}cap A,tin(0,1)$. 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 $Acap T^{-1}U^{+}$ is convex, open in $A$, and contains $Acapoperatorname{cl}(U^{+})$ which strictly contains $U^{+}cap A$ by Claim 1. Thus by maximality of $U^{+}$ in $mathbb{P}$, $Acap T^{-1}U^{+}overset{mbox{must}}{=}A$. In particular, $T^{-1}U^{+}supseteq A$, and so $TWsubseteq TAsubseteq U^{+}$. 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: $Asetminus U^{+}$ is a singleton.

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

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

Proposition 2: Let $Asubseteq V$ 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 $tin(0,1)$, we have $tx+(1-t)yin A$.

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 $xin x'+t^{-1}(1-t)(U-y)$ implying $tx+(1-t)yin tx'+(1-t)Usubseteq tA+(1-t)Asubseteq A$, since $A$ is convex. QED(Prop 2)