User raj - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T04:41:47Z http://mathoverflow.net/feeds/user/9293 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/15654/extreme-point-compact-convex-set/39058#39058 Answer by Raj for Extreme point compact convex set. Raj 2010-09-17T05:07:20Z 2010-09-17T05:07:20Z <p>Here is Greg's proof of the Krein-Milman theorem (only the existence part). My apologies about the TeX: i don't know what does or doesn't work on this site, so I've excluded the beginâ€¦end environments, but placed headings to guide the reader.</p> <p>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.</p> <p>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 $e$.</p> <p>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 $U^{-}$.</p> <pre><code>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. \textbf{Sub-claim 1:} $U^{+}\cap A\subset\operatorname{cl}_{A}U^{+}$ strictly.\\ \textbf{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{math}\begin{array}[t]{lrclll} &amp;f &amp;: &amp;\mathbb{F} &amp;\to &amp;V\\ &amp;&amp;: &amp;s &amp;\mapsto &amp;s~x+(1-s)~y\\ \end{array}\end{math}\\ 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) \textbf{Sub-claim 2:} If $W\subseteq A$ is convex, then $U^{+}\cup W$ is convex.\\ \textbf{Proof (sub-claim 2):} Fix $x\in U^{+}\cap A,t\in(0,1)$. Consider the map\\ \begin{math}\begin{array}[t]{lrclll} &amp;T &amp;: &amp;V &amp;\to &amp;V\\ &amp;&amp;: &amp;y &amp;\mapsto &amp;t~x+(1-t)~y\\ \end{array}\end{math}\\ 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 $\mathbb{P}$, $A\cap T^{-1}U^{+}\overset{\mbox{\tiny must}}{=}A$. 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) \textbf{Sub-claim 3:} $A\setminus U^{+}$ is a singleton.\\ \textbf{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. </code></pre> <p>QED(Prop 1)</p> <p>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 $tx+(1-t)y\in A$.</p> <p>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)</p>