Extreme Subsets and Extreme Points of a Set in a LCTVS

# Extreme Subsets and Extreme Points of a Set in a LCTVS

 Definition: Let $X$ be a locally convex topological vector space and let $K$ be a nonempty convex subset of $X$. A subset $E$ of $K$ is an Extreme Subset of $K$ if it is nonempty, closed, convex, and that whenever $x \in E$ is a convex combination of $u, v \in K$ then $u, v \in E$.

The last condition can be restated as follows: If $x \in E$ is such that there exists $u, v \in K$ with $x = \lambda u + (1 - \lambda) v$ where $\lambda \in [0, 1]$ then $u, v \in E$.

 Definition: Let $X$ be a locally convex topological vector space and let $K$ be a nonempty convex subset of $X$. A point $x \in K$ is an Extreme Point of $K$ if $\{ x \}$ is an extreme subset of $K$.

If $K$ is a nonempty convex subset of $X$ then we can visualize extreme points of $K$ as follows. $x \in K$ is an extreme point of $K$ provided $x$ is not contained in the interior of any line segment containing $x$ that is fully contained in $K$. Proposition 1: Let $X$ be a locally convex topological vector space and let $K$ be a nonempty convex subset of $X$. Let $\{ E_i : i \in I \}$ be an arbitrary collection of extreme subsets of $K$. If $\displaystyle{\bigcap_{i \in I} E_i \neq \emptyset}$ then $\displaystyle{\bigcap_{i \in I} E_i}$ is an extreme subset of $K$.
• Proof: By assumption we have that $\displaystyle{\bigcap_{i \in I} E_i \neq \emptyset}$. Since each $E_i$ is an extreme subset of $K$ we have by definition each $E_i$ is closed. So an arbitrary intersection of closed sets is closed, i.e., $\displaystyle{\bigcap_{i \in I} E_i}$ is closed. Furthermore, since each $E_i$ is a convex subset of $K$ we have that an arbitrary intersection of convex sets is convex, i.e., $\displaystyle{\bigcap_{i \in I} E_i}$ is convex.
• So let $\displaystyle{x \in \bigcap_{i \in I} E_i}$ and suppose that $x = \lambda u + (1 - \lambda) v$ for some $u, v \in K$ and $\lambda \in [0, 1]$. Since $x \in E_i$ for each $i \in I$ and since $E_i$ is an extreme subset of $K$ we have that $u, v \in E_i$. So $u, v \in \bigcap_{i \in I} E_i$.
• Thus $\displaystyle{\bigcap_{i \in I} E_i}$ is an extreme subset of $K$. $\blacksquare$
 Proposition 2: Let $X$ be a locally convex topological vector space. If $A$ is an extreme subset of $B$ and $B$ is an extreme subset of $K$ then $A$ is an extreme subset of $K$.
• Proof: Since $A$ is an extreme subset of $B$ we have that $A$ is a nonempty, closed, convex subset of $B$ and hence a nonempty, closed, convex subset of $K$.
• Let $x \in A$ and suppose that $x = \lambda u + (1 - \lambda) v$ for some $\lambda \in [0, 1]$ and $u, v \in K$. Since $x \in A$ and $A \subseteq B$ we have that $x \in B$. Since $B$ is an extreme subset of $K$ we have that $u, v \in B$. So $x \in A$ with $x = \lambda u + (1 - \lambda) v$ where $\lambda \in [0, 1]$ and $u, v \in B$. But since $A$ is an extreme subset of $B$ we have that $u, v \in A$. Thus $A$ is an extreme subset of $K$. $\blacksquare$