Theorem: Slices form Neighborhood Basis of Extreme Point of Compact Convex Subset of Hausdorff Locally Convex Space

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \PP}$ be a Hausdorff locally convex space over $\GF$ with its standard topology.
Let $X^\ast$ be the topological dual space of $X$.
Let $K$ be a compact convex subset of $X$.
Let $x_0$ be an extreme point of $K$.
Let $V$ be a open neighborhood of $x_0$ in $K$. 
Let:
:$V_{f, \alpha} = \set {x \in K : \map \Re {\map f x} < \alpha}$
Then there exists $f \in X^\ast$ and $\alpha \in \R$ such that:
:$x_0 \in V_{f, \alpha} \subseteq V$

Proof:
From Intersections of Slices form Neighborhood Basis for Standard Topology of Compact Convex Subset of Hausdorff Locally Convex Space, there exists $f_1, \ldots, f_n \in X^\ast$ and $\alpha_1, \ldots, \alpha_n \in \R$ such that:
:$\ds x_0 \in \bigcap_{i \mathop = 1}^n \set {x \in K : \map \Re {\map {f_i} x} < \alpha_i} \subseteq V$
Let:
:$K_i = \set {x \in K : \map \Re {\map {f_i} x} \ge \alpha_i}$ for $i \in \set {1, \ldots, n}$.
For each $i \in \set {1, \ldots, n}$, $\map \Re {f_i}$ is continuous from Real Part of Continuous Function is Continuous.
Hence $K_i$ is closed in $K$ for each $i \in \set {1, \ldots, n}$.
From Closed Subspace of Compact Space is Compact, $K_i$ is compact.
Since:
:$\map \Re {\map {f_i} {x_0} } < \alpha_i$ for each $i \in \set {1, \ldots, n}$
we have $x_0 \not \in K_i$ for each $i \in \set {1, \ldots, n}$.
Hence we have:
:$\ds x_0 \not \in \bigcup_{i \mathop = 1}^n K_i$
By Extreme Point of Convex Set not contained in Convex Hull of Subset not containing Point, we have:
:$\ds x_0 \not \in \map {\operatorname {conv} } {\bigcup_{i \mathop = 1}^n K_i}$
For $1 \le i \le n$, let $\pi_i^\R : \R^n \to \R$ be the projection onto the $i$th coordinate. 
Equip $K^n$ with the product topology.
For $1 \le i \le n$, let $\pi_i^K : K^n \to K$ be the projection onto the $i$th coordinate.
From Projection from Product Topology is Continuous: General Result, we have that $\pi_i^\R$ and $\pi_i^K$ are continuous for each $i$.
Equip $\R^n \times K^n$ with the product topology.
Let $\pi_1 : \R^n \times K^n \to \R^n$ and $\pi_2 : \R^n \times K^n \to K^n$ be the projection.
Define $\phi : \R^n \times K^n \to X$ by:
:$\ds \map \phi {t, x} = \sum_{i \mathop = 1}^n \map {\pi_i^\R} t \map {\pi_i^K} x = \sum_{i \mathop = 1}^n \map {\pi_i^\R} {\map {\pi_1} {t, x} } \map {\pi_i^K} {\map {\pi_2} {t, x} }$
for each $\tuple {t, x} \in \R^n \times K^n$. 
From Product of Continuous Functions on Topological Ring is Continuous and Composite of Continuous Mappings is Continuous, $\tuple {t, x} \mapsto \map {\pi_i^\R} {\map {\pi_1} {t, x} } \map {\pi_i^K} {\map {\pi_2} {t, x} }$ is continuous for each $1 \le i \le n$.
From Sum of Continuous Functions on Topological Ring is Continuous, we then have that $\phi$ is continuous.
Define also the function $\psi : \R^n \to \R$ by:
:$\ds \map \psi x = \sum_{i \mathop = 1}^n \map {\pi_i^\R} x$ 
for each $x \in \R^n$.
By Sum of Continuous Functions on Topological Ring is Continuous, $\psi$ is continuous. 
Set:
:$\ds C = \set {x \in \R^n : \map {\pi_i^\R} x \ge 0 \text { for each } i \text { and } \sum_{i \mathop = 1}^n \map {\pi_i^\R} x = 1}$
We can rewrite:
:$\ds C = \psi^{-1} \sqbrk {\set 1} \cap \bigcap_{i \mathop = 1}^n \paren {\pi_i^\R}^{-1} \sqbrk {\hointr 0 \infty}$
Since $\psi$ is continuous and $\set 1$ is closed, $\psi^{-1} \sqbrk {\set 1}$ is closed.
Since each $\pi_i^\R$ is continuous and $\hointr 0 \infty$ is closed, $\paren {\pi_i^\R}^{-1} \sqbrk {\hointr 0 \infty}$ is closed.
Hence $C$ is closed as the intersection of closed sets.
Finally, for $x \in C$ we have:
:$\ds 0 \le \map {\pi_i^\R} x \le \sum_{i \mathop = 1}^n \map {\pi_i^\R} x = 1$ 
Hence $x \in \closedint 0 1^n$.
From the Heine-Borel Theorem and Tychonoff's Theorem, we have that $\closedint 0 1^n$ is compact.
From Closed Subspace of Compact Space is Compact, $C$ is compact. 
We now have:
:$\ds \psi \sqbrk {C \times \prod_{i \mathop = 1}^n K_i} = \set {\sum_{i \mathop = 1}^n t_i x_i : t_i \ge 0, \, \sum_{i \mathop = 1}^n t_i = 1, \, x_i \in K_i}$
From Convex Hull of Finite Union of Convex Sets, we therefore have:
:$\ds \psi \sqbrk {C \times \prod_{i \mathop = 1}^n K_i} = \map {\operatorname {conv} } {\bigcup_{i \mathop = 1}^n K_i}$
From Tychonoff's Theorem, we have that $\ds C \times \prod_{i \mathop = 1}^n K_i$ is compact.
Since $\psi$ is continuous:
:$\ds \map {\operatorname {conv} } {\bigcup_{i \mathop = 1}^n K_i}$ is compact
from Continuous Image of Compact Space is Compact.
In particular, from Compact Subspace of Hausdorff Space is Closed:
:$\ds \map {\operatorname {conv} } {\bigcup_{i \mathop = 1}^n K_i}$ is closed. 
From Hahn-Banach Separation Theorem: Hausdorff Locally Convex Space: Compact Convex Set and Closed Convex Set (Complex Case), there exists $f \in X^\ast$ and $\alpha \in \R$ such that:
:$\ds \map \Re {\map f {x_0} } < \alpha < \inf_{x \in \map {\operatorname {conv} } {\bigcup_{i \mathop = 1}^n K_i} } \map \Re {\map f x}$
In particular, we have:
:$\ds x_0 \in \set {x \in K : \map \Re {\map f {x_0} } < \alpha} \subseteq K \setminus \map {\operatorname {conv} } {\bigcup_{i \mathop = 1}^n K_i}$
From Set Complement inverts Subsets, we have:
:$\ds K \setminus \map {\operatorname {conv} } {\bigcup_{i \mathop = 1}^n K_i} \subseteq K \setminus \bigcup_{i \mathop = 1}^n K_i$
Recall that:
:$\ds \bigcap_{i \mathop = 1}^n \set {x \in K : \map \Re {\map {f_i} x} < \alpha_i} \subseteq V$
That is:
:$\ds K \setminus \bigcup_{i \mathop = 1}^n K_i \subseteq V$ 
Hence we have:
:$\ds \set {x \in K : \map \Re {\map f x} < \alpha} \subseteq V$
as desired.
{{qed}}
Category:Locally Convex Spaces
Category:Extreme Points of Convex Sets
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