Existence of an extreme point of a compact convex set The Krein-Milman theorem shows that a compact convex set in a Hausdorff locally convex topological vector space is the convex hull of its extreme points.
It seems this implies that a compact convex set in such a space must have an extreme point.
I am interested in whether there is a very simple elementary argument that shows that a compact convex set must have an extreme point.
I have such an argument, but since it uses compactness of the unit ball, it is not so good if the space is infinite dimensional.
In point of fact, I am using this in R^n, but if there is a way to put it that can generalize to infinite dimensions then that would seem preferable for the students.
 A: The main question is what generality you want. As soon as you have just one strictly convex function in the space, any point where it attains its maximum on $K$ is an extreme point. If we were talking about separable normed spaces, the construction of such function would be trivial: $F(x)=\sum_j 2^{-j}(1+\|x_j\|)^{-1}\|x-x_j\|$ where $x_j$ is any countable dense set would work just fine. In a strictly convex normed space $F(x)=\|x\|$ would be an even simpler example. The problem is that you want it in an abstract locally convex topological linear space, so some form of AC seems, indeed, inavoidable.
A: In fact, the standard proof of Krein–Milman first proves the existence of an extreme point. Note (or recall) that a face $F$ of a convex set $K$ is defined by the requirement that, if $tx+(1-t)y\in F$ for $0<t<1$ and $x,y\in K$, then $x,y\in F$. Hence an extreme point is just a singleton face. The existence of an extreme point is shown by using Zorn's lemma (or Hausdorff maximality lemma) to show the existence of a minimal, nonempty, closed face. Use Hahn–Banach to show the minimal face is a singleton: If a continuous linear functional separates two points in $F$, then the set where it achieves its maximum in $F$ is a smaller face. (And to get started, note that $K$ is a face in itself, so faces do indeed exist.)
A: The beef of the Krein-Milman theorem is the fact that each face of your compact convex set K has an extreme point; the statement about the (closed) convex hull then follows from a swift application of Hahn-Banach. Now notice that a face of K is itself compact and convex. So the difficulty of proving the Krein-Milman theorem is pretty much equal to the difficulty of proving that every compact convex set has an extreme point.
A: What is easy is going from "there is at least one extreme point" to "closed convex hull of extreme points".  So what you are asking for is essentially the proof of the whole thing.
A: See Prof. Greg Hjorth's Measure theory notes 2010. It contains a full proof.
A: 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)
A: just suppose that point x is the point in the compact set that has the largest norm (Euclidean norm, L2). Let's now prove that it is actually a extreme point because otherwise we can choose x = 1/2 ( y + z) where y, z are points not identical to x. But because x has the largest norm, one can prove that under condition y and z has norm less or equal than x, but x is the middle point of the segment [y, z], it leads to the conclusion that x = y = z, which contradicts!
