Slightly adjusted Krein-Milman.

src/fa/lc/compact.tex

@@ -24,7 +24,7 @@
 
 \begin{definition}[Extreme Point]
 \label{definition:extreme-point}
-    Let $E$ be a vector space over $\real$, $K \subset E$, and $x \in K$, then $x$ is \textbf{extremal} if there exists no $y, z \in K$ such that $x \in (y, z) \subset K$. 
+    Let $E$ be a vector space over $\real$, $K \subset E$, and $x \in K$, then $x$ is \textbf{extremal} if there exists no $y, z \in K \setminus \bracsn{x}$ such that $x \in (y, z) \subset K$. 
 \end{definition}
 
 \begin{definition}[Extreme Subset]
@@ -32,6 +32,13 @@
     Let $E$ be a vector space over $\real$, $K \subset E$ be convex, and $A \subset K$, then $A$ is \textbf{extreme set} if for any $x \in A$ and $y, z \in K$ such that $x \in (y, z)$, $y, z \in A$ as well.  
 \end{definition}
 
+\begin{lemma}
+\label{lemma:extreme-subset-extreme-point}
+    Let $E$ be a vector space over $\real$, $K \subset E$ be convex, $A \subset K$ be extreme, and $x \in A$ be an extreme point of $A$, then $x$ is an extreme point of $K$. 
+\end{lemma}
+\begin{proof}
+    Let $y, z \in K$ such that $x \in (y, z)$. Since $A$ is extreme, $y, z \in A$. As $x$ is extremal in $A$, $x = y = z$, so $x$ is extremal in $K$. 
+\end{proof}
 
 \begin{lemma}
 \label{lemma:extremal-face}
@@ -135,11 +142,3 @@
 
     Now, let $A'$ be the closed convex hull of $\seq{x_n}$ with respect to $E_B$. Since the inclusion $E_B \to E$ is continuous, $A'$ is a compact convex set in $E$ by \autoref{proposition:compact-extensions}. As such, $A' = A$ by \autoref{proposition:closure-of-image} and \autoref{proposition:compact-closed}. Therefore $A$ is a compact subset of $E_B$. 
 \end{proof}
-
-\begin{lemma}
-\label{lemma:auxiliary-weak-dense}
-    Let $E$ be a separated locally convex space over $K \in \RC$, $A \subset E$ be compact, convex, and circled, and $
-\end{lemma}
-
-
-