拓扑模/哈恩-巴拿赫定理

定理（几何哈恩-巴拿赫定理）:

设 $V$ 为实拓扑向量空间，且 $U\subseteq V$ 为开凸集，使得 $0\notin V$。那么存在一个超平面 $W\leq V$ 不与 $U$ 相交。

（在选择公理的条件下。）

证明： $V$ 中所有不与 $U$ 相交的向量子空间的集合是归纳的，并且非空的（因为零子空间）。因此，由佐恩引理，选取一个最大向量子空间 $W\leq V$，它不与 $U$ 相交。断言 $W$ 是一个超平面。如果不是，则 $V/W$ 的维度 $\geq 2$。现在，正则映射 $p:V\to V/W$ 是开映射，因此 $U':=p(U)$ 是 $V/W$ 中的开凸集。我们考虑锥

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

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. $\Box$