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:

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

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.

From Sum of Convex Sets in Vector Space is Convex: Corollary, we have:

$A - B$ is convex.

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

$v_0 + A - B = C$ is convex.

We finally show that $0 \in C$.

Since:

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

We have:

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

$\blacksquare$