Extreme Set in Compact Convex Set contains Extreme Point

Theorem
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\R$.

Let $K$ be a non-empty compact convex subset of $X$.

Let $E$ be an extreme set of $K$.

Then $E$ contains an extreme point of $K$.

Proof
Let $P$ be the set of extreme sets in $K$ that are contained in $E$.

Since $E \in P$, $P$ is certainly non-empty.

Define a relation $\preceq$ on $P$ by $A \preceq B$ $B \subseteq A$.

From Subset Relation is Ordering and Dual Ordering is Ordering, we have:


 * $\struct {P, \preceq}$ is an ordered set.

We show that every non-empty chain in $\struct {P, \preceq}$ has an upper bound.

We will then invoke Zorn's Lemma.

Lemma
From Zorn's Lemma, we have that $P$ has a maximal element.

Let $M$ be a maximal element of $P$.

Then $M$ is an extreme set of $K$ such that:


 * whenever $M_\ast \in P$ has $M \preceq M_\ast$ we have $M_\ast = M$

and:


 * $M \subseteq E$

That is:


 * whenever a extreme set $M_\ast \subseteq E$ of $K$ has $M_\ast \subseteq M$, we have $M_\ast = M$.

So $M$ does not properly contain any other extreme set of $K$.

With view to apply Point in Convex Set is Extreme Point iff Singleton is Extreme Set, we show that $M$ is a singleton.

suppose that $a, b \in M$ with $a \ne b$.

Then from Normed Dual Space Separates Points, there exists an $f \in X^\ast$ with:


 * $\map f a \ne \map f b$

where $X^\ast$ is the normed dual of $X$.

So either:


 * $\map f a > \map f b$

or:


 * $\map f a < \map f b$

suppose that $\map f a < \map f b$.

Otherwise, swap $a$ and $b$.

Define:


 * $\ds M^f = \set {x \in M : \map f x = \max_{y \in M} \map f y}$

From Preimage of Maximum of Bounded Linear Functional on Extreme Set in Convex Compact Set is Extreme Set, we have that:


 * $M^f$ is an extreme set in $K$.

We also have:


 * $M^f \subseteq M \subseteq E$

However, from the definition of maximum, we have:


 * $\ds \map f b \le \max_{y \in M} \map f y$

So we have:


 * $\ds \map f a < \max_{y \in M} \map f y$

So:


 * $a \not \in M^f$

So:


 * $M^f$ is a proper subset of $M$

and:


 * $M^f$ is an extreme set of $K$ contained in $E$.

This contradicts the maximality of $M$.

So $M$ is a singleton, and:


 * $M = \set x$

for some $x \in X$.

Since $M \subseteq E$, in particular we have $x \in E$.

From Point in Convex Set is Extreme Point iff Singleton is Extreme Set, we have:


 * $x$ is an extreme point in $K$ that is contained in $E$.