Hahn-Banach Separation Theorem/Normed Vector Space/Real Case/Open Convex Set and Convex Set/Lemma

Lemma
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\R$.

Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $\struct {X, \norm \cdot}$.

Let $A \subseteq X$ be an open convex set.

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

Let $a_0 \in A$ and $b_0 \in B$.

Let:


 * $v_0 = b_0 - a_0$

Let:


 * $C = v_0 + A - B = \set {v_0 + a - b : a \in A, \, b \in B}$

Then:
 * $C$ is open, convex and contains $0$.

Proof
We first show that $C$ is open.

Note that we can write:


 * $\ds A - B = \bigcup_{b \in B} \set {a - b : a \in A} = \bigcup_{b \in B} \paren {A - b}$

Since $A$ is open, from Translation of Open Set in Normed Vector Space is Open, we have:


 * $A - b$ is open.

From Union of Open Sets of Normed Vector Space is Open, we therefore have:


 * $A - B$ is open.

From Translation of Open Set in Normed Vector Space is Open, we then obtain:


 * $C$ is open.

We now show that $C$ is convex.

Let $v, w \in C$ and $t \in \closedint 0 1$.

We can then write:


 * $v = v_0 + a - b$

and:


 * $w = v_0 + a' - b'$

for $a, a' \in A$ and $b, b' \in B$.

We then have:

Since $A$ and $B$ are convex, we have:


 * $t a + \paren {1 - t} a' \in A$

and:


 * $t b + \paren {1 - t} b' \in B$

so:


 * $t v + \paren {1 - t} w \in v_0 + A - B$

So $C$ is convex.

We finally show that $0 \in C$.

Since:


 * $-v_0 = a_0 - b_0 \in A - B$

We have:


 * $b_0 - a_0 + a_0 - b_0 = 0 \in C$