Topological Modules/Hahn–Banach theorems

Theorem (geometrical Hahn–Banach theorem):

Let $V$ be a real topological vector space, and let $U\subseteq V$ be open and convex so that $0\notin V$. Then there exists a hyperplane $W\leq V$ not intersecting $U$.

(On the condition of the axiom of choice.)

Proof: The set of all vector subspaces of $V$ that do not intersect is inductive and also nonempty (because of the zero subspace). Hence, by Zorn's lemma, pick a maximal vector subspace $W\leq V$ that does not intersect $U$. Claim that $W$ is a hyperplane. If not, $V/W$ has dimension $\geq 2$. Now the canonical map $p:V\to V/W$ is open, so that $U':=p(U)$ is an open, convex subset of $V/W$. We consider the cone

$C:=\bigcup _{\lambda >0}\lambda U'$

and note that it has a nonzero boundary point; for otherwise $C$ would be clopen in $V/W\setminus \{0\}$ which is path-connected (indeed by assumption $\operatorname {dim} V/W\geq 2$, so that for any two points $x,y\in V/W$ we find a 2-dimensional plane containing both, and by using a "corner point" $z$ when $x,y$ do lie on a line through the origin, we may connect them in $V/W\setminus \{0\}$, because a segment in a TVS yields a continuous path by continuity of addition and scalar multiplication), so that $C=V/W\setminus \{0\}$, which is impossible because for any $x\neq 0$ in $V/W$, we then have $x\in \mu U'$, $-x\in \lambda U'$ for $\mu ,\lambda >0$, so that $0\in U'$ by convexity, a contradiction. Hence, let $w\in \partial C$. Then the line $L$ generated by $w$ does not intersect $C$ and hence not $U'$, and $p^{-1}(L)$ is a larger subspace of $V$ that does not intersect $U$ than $W$ in contradiction to the maximality of the latter. $\Box$