拓撲模/哈恩-巴拿赫定理

定理（幾何哈恩-巴拿赫定理）:

設 $V$ 為實拓撲向量空間，且 $U\subseteq V$ 為開凸集，使得 $0\notin V$。那麼存在一個超平面 $W\leq V$ 不與 $U$ 相交。

（在選擇公理的條件下。）

證明： $V$ 中所有不與 $U$ 相交的向量子空間的集合是歸納的，並且非空的（因為零子空間）。因此，由佐恩引理，選取一個最大向量子空間 $W\leq V$，它不與 $U$ 相交。斷言 $W$ 是一個超平面。如果不是，則 $V/W$ 的維度 $\geq 2$。現在，正則對映 $p:V\to V/W$ 是開對映，因此 $U':=p(U)$ 是 $V/W$ 中的開凸集。我們考慮錐

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

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. $\Box$