Hahn-Banach Separation Theorem/Hausdorff Locally Convex Space/Real Case/Open Convex Set and Convex Set

Theorem
Let $\struct {X, \PP}$ be a Hausdorff locally convex space over $\R$ equipped with its standard topology.

Let $X^\ast$ be the topological dual space of $\struct {X, \PP}$. Let $A \subseteq X$ be an open convex set.

Let $B \subseteq X$ be a convex set disjoint from $A$.

Then there exists $f \in X^\ast$ and $c \in \R$ such that:


 * $A \subseteq \set {x \in X : \map f x < c}$

and:


 * $B \subseteq \set {x \in X : \map f x \ge c}$

That is:


 * there exists $f \in X^\ast$ and $c \in \R$ such that $\map f a < c \le \map f b$ for each $a \in A$ and $b \in B$.

Proof
Fix $a_0 \in A$ and $b_0 \in B$.

From Sum of Set and Open Set in Topological Vector Space is Open, we have:


 * $A - B$ is open.

From Translation of Open Set in Topological Vector Space is Open, we have:


 * $A - B + \paren {b_0 - a_0}$ is open.

Note that since $A \cap B = \O$, we have $0 \not \in A - B$.

Hence $b_0 - a_0 \not \in A - B + \paren {b_0 - a_0}$.

From Sum of Convex Sets in Vector Space is Convex: Corollary, we have that $A - B$ is convex.

From Translation of Convex Set in Vector Space is Convex, we have that $A - B + \paren {b_0 - a_0}$ is convex.

Applying Open Convex Set in Hausdorff Locally Convex Space is Separated from Points outside Set by Continuous Linear Functional, there exists $f \in X^\ast$ such that:


 * $\map f v < \map f {b_0 - a_0}$

for all $v \in A - B + \paren {b_0 - a_0}$.

That is:


 * $\map f {a - b + b_0 - a_0} < \map f {b_0 - a_0}$

for each $a \in A$ and $b \in B$.

Since $f$ is linear, we have:


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

for each $a \in A$ and $b \in B$.

Then we have:


 * $\map f {a_0} < \map f b$

for all $b \in B$.

Hence:


 * $\ds \inf_{b \in B} \map f b > -\infty$

Let:


 * $\ds c = \inf_{b \in B} \map f b$

Then, we have:


 * $\map f a \le c \le \map f b$

It remains to show that $\map f a < c$ for each $a \in A$.

Let $a \in A$.

Since $A$ is open, there exists an open neighborhood $V_x$ of ${\mathbf 0}_X$ such that $x + V_x$ is an open neighborhood of $x$ in $A$.

Now, since we have:


 * $\map f {a_0} < \map f b$

for each $b \in B$, we in particular have that $f$ is not constant ${\mathbf 0}_X$.

Hence there exists $v \in X$ such that:


 * $\map f v \ne 0$

Replacing $v$ by $-v$ if necessary, suppose that:


 * $\map f v > 0$

From Multiple of Vector in Topological Vector Space Converges, we have:


 * $\ds \frac v n \to {\mathbf 0}_X$ as $n \to \infty$

So for some $N \in \N$, we have:


 * $\ds \frac v N \in V_x$

by the definition of a convergent sequence.

Set:


 * $\ds v' = \frac v N$

Then, we have $\map f {v'} > 0$ and $x + v' \in A$.

We therefore have:

In particular, $\map f a < c$ for each $a \in A$.

Hence we have found $f \in X^\ast$ such that:


 * $\map f a < c \le \map f b$ for each $a \in A$ and $b \in B$.