I am in the middle of working through the details of the proof of the Krein-Milman theorem in Rudin's Functional Analysis (Theorem 3.23), and I am stuck on one detail. I will state the theorem and write down the relevant parts of the proof.
The Krein-Milman Theorem states: Suppose $X$ is a topological vector space on which $X^*$ separates points. If $K$ is a nonempty compact convex set in $X$, then $K$ is the closed convex hull of the set of its extreme points.
Proof (partial). Let $\mathscr{P}$ be the collection of all compact extreme sets of $K$. Since $K\in \mathscr{P}$, $\mathscr{P}\not =\emptyset$. We shall use the following property of $\mathscr{P}$:
(a) The intersection $S$ of any nonempty subcollection of $\mathscr{P}$ is a member of $\mathscr{P}$, unless $S=\emptyset$.
Choose some $S\in \mathscr{P}$. Let $\mathscr{P}'$ be the collection of all members of $\mathscr{P}$ that are subsets of $S$. Since $S\in \mathscr{P}'$, $\mathscr{P}'$ is not empty. Partially order $\mathscr{P}'$ by set inclusion, let $\Omega$ be a maximal totally ordered subcollection of $\mathscr{P}'$, and let $M$ be the intersection of all members of $\Omega$. Since $\Omega$ is a collection of compact sets with finite intersection property, $M\not =\emptyset$. By (a), $M\in \mathscr{P}'$.
Here's my question: Why does the maximality of $\Omega$ imply that no proper subset of $M$ belongs to $\mathscr{P}$, as claimed in the next line of the proof?
$\text{ }$
Here is what I started on. Suppose $A\subset M$. By means of contradiction, suppose $A\in \mathscr{P}$. Since $M\subset S$, $A\subset S$ and $A\in \mathscr{P}'$. Then $\Omega \subset \Omega \cup \{A\}$, which would contradict the maximality of $\Omega$ unless $A\in \Omega$. But I cannot prove that $A\not \in \Omega$. In fact, it must be that $A\in \Omega$ since $\Omega \subset \Omega \cup \{A\}$ and $\Omega$ is maximal. Any ideas or hints would be greatly appreciated.
If $A \subset \Omega$, then the intersection of all members of $\Omega$ would be contained in $A$, and so $M \subseteq A$.