Theorem 8.8.5 (Open Mapping Theorem). Let $E, F$ be complete metric TVSs over $K \in \RC$, $T \in L(E; F)$ with $T(E)$ dense, then exactly one of the following holds:
$T(E)$ is meagre.
$T$ is open.
Proof. Suppose that $T(E)$ is not meagre. Let $r_{0} > 0$ and $r > 0$ such that $B_{E}(0, r) + B_{E}(0, r) \subset B_{E}(0, r_{0})$, then since $B_{E}(0, r)$ is absorbing,
By the Baire Category Theorem, there exists $N \in \natp$, $s > 0$, and $y \in nT(B_{E}(0, r))$ such that $B_{F}(y, s) \subset \overline{nT(B_E(0, r))}$. In which case,
By (TVS2), there exists $t > 0$ such that $n^{-1}B_{F}(0, s) \supset B_{F}(0, t)$, so $\overline{T(B_E(0, r_0))}\supset B_{F}(0, t)$.
Thus by Proposition 8.8.3, $B_{F}(0, t) \subset T(B_{E}(0, r)) \in \cn_{F}(0)$ for all $r > r_{0}$. As $r_{0} > 0$ is arbitrary, $T(U) \in \cn_{F}(0)$ for all $U \in \cn_{E}(0)$. Therefore $T$ is open by translation-invariance of the topology on $E$.$\square$