Extreme Set in Compact Convex Set contains Extreme Point/Lemma

Lemma

Let $X$ be a vector space over $\R$.

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

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

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

Define an ordering $\preceq$ on $P$ by $A \preceq B$ if and only if $B \subseteq A$.

Then:

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

Proof

Let $C$ be a non-empty chain in $\struct {P, \preceq}$.

Let:

$\ds D = \bigcap_{E \in C} E$

We first show that:

$D \ne \O$

Aiming for a contradiction, suppose suppose that $D = \O$.

Then, from de Morgan's laws:

$\ds K = \bigcup_{E \in C} \paren {K \setminus E}$

Since each $E \in C$ is closed, we have that:

$K \setminus E$ is open in $K$.

So:

$\set {K \setminus E : E \in C}$ is an open cover for $K$.

Since $K$ is compact, we have that:

there exists a finite subcover $\set {K \setminus E_i : i \in \set {1, 2, \ldots, n} }$ of $\set {K \setminus E : E \in C}$ for $K$.

That is:

$\ds K = \bigcup_{i \mathop = 1}^n \paren {K \setminus E_i}$

From de Morgan's laws, we then have:

$\ds \O = \bigcap_{i \mathop = 1}^n E_i$

However, from Finite Non-Empty Subset of Ordered Set has Maximal and Minimal Elements: Corollary, there exists $i, j$ such that:

$E_i \preceq E_k \preceq E_j$

for each $k \in \set {1, 2, \ldots, n}$.

Then:

$\ds E_j \subseteq E_k$

for each $k \in \set {1, 2, \ldots, n}$.

So, from Set Intersection Preserves Subsets, we have:

$\ds E_j \subseteq \bigcap_{i \mathop = 1}^n E_i$

Since $E_j$ is non-empty, we have:

$\ds \bigcap_{i \mathop = 1}^n E_i \ne \O$

a contradiction.

So, we have:

$D \ne \O$

We now show that $D$ is an extreme set of $K$, and that it is an upper bound for $C$.

From Intersection of Closed Sets is Closed in Normed Vector Space, $D$ is closed.

Let $x, y \in K$ and $t \in \openint 0 1$ be such that $t x + \paren {1 - t} y \in D$.

Then, from the definition of set intersection, we have:

$t x + \paren {1 - t} y \in E$

for each $E \in C$.

Since $E$ is an extreme set, we have $x, y \in E$ for each $E \in C$.

So, $x, y \in D$.

So $D$ is an extreme set of $K$

From Intersection is Subset, we then have:

$\ds D = \bigcap_{E \in C} E \subseteq F$

for each $F \in C$.

So, we have:

$F \preceq D$

for each $F \in C$.

So $D$ is an upper bound for $C$.

Since $C$ was an arbitrary non-empty chain in $\struct {P, \preceq}$, we have:

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

$\blacksquare$