Theorem 1 (Hahn-Banach)

— Let A be an open convex non empty set of the topological vector space E and let M be a non-empty linear variety which does not meet A. Then there exists a closed hyperplane H which contains M and does not meet A.

By translation the problem can be reduced to the case $0 \in A$, so that $A$ is absorbent. Let $p$ be the gauge of $A$ (II, p. 20) so that $A$ is the set of points $x \in E$ such that $p(x) < 1$. On the other hand, let $V$ be the vector subspace of $E$ generated by $M$; thus $M$ is a hyperplane in $V$ that does not contain $0$, and hence there is a unique linear form $f$ on $V$ such that $M$ is the set of points $y \in V$ for which $f(y) = 1$. The hypothesis $M \cap A = \varnothing$ implies therefore that for all $y \in V$ for which $f(y) = 1$, we have $p(y) \geqslant 1$; as $f$ and $p$ are positively homogeneous we have $f(y) \leqslant p(y)$ for all $y \in V$ such that $f(y) > 0$; finally as $p(y) \geqslant 0$ for all $y \in V$, we see that $f(y) \leqslant p(y)$ for all $y \in V$. By the analytical form of the Hahn-Banach theorem (II, p. 22, th. 1) there exists a linear form $h$ on $E$ which extends $f$ and is such that, for all $x \in E$, $h(x) \leqslant p(x)$. Let $H$ be the hyperplane in $E$ with the equation $h(x) = 1$. Clearly $H \cap V = M$ and $H \cap A = \varnothing$. On the other hand the complement of $H$ in $E$ contains the open non-empty set $A$, therefore $H$ is closed in $E$ (I, p. 11, corollary).

Q.E.D.