Proposition 1

— Let $A$ be an open non-empty convex set and let $B$ be a non-empty convex set in a real topological vector space $E$; if $A$ does not meet $B$ then there exists a closed hyperplane that separates $A$ from $B$.

For the set $C = A - B$ is open, convex (II, p. 9, prop. 7) and non-empty, also $0 \notin C$. By theorem 1 of II, p. 36, there exists a continuous linear form $f \neq 0$ on $E$ such that $f(z) > 0$ in $C$. Then, for all $x \in A$, and $y \in B$, we have $f(x) > f(y)$. Write $\alpha = \inf_{x \in A} f(x)$; $\alpha$ is finite and we have $f(x) \geq \alpha$ for all $x \in A$ and $f(y) \leq \alpha$ for all $y \in B$; the closed hyperplane $H$ with the equation $f(z) = \alpha$ separates $A$ from $B$.