Show that weak operator topology is the weakest locally convex topology on $B(H)$ such that every $\phi\in F(H)$ is continuous. (F(H) means finite rank operators on $H$).
To show it , let $\tau$ be the collection of all unions of finite intersections of set $\phi^{-1}(V)$, where $\phi\in F(H)$ and $V$ is an open subset of $B(H)$. By this definition $\tau$ is the smallest topology that every $\phi\in F(H)$ is continuous.
$\tau$ is the smallest topology so every $\tau $- open set is also wot-open.
Conversely, for $x_0\in B(H)$, $\phi_1,...,\phi_n\in F(H)$ and $\epsilon> 0$, $N=N(x_0,\phi_1,...,\phi_n,\epsilon)=\{x\in B(H) ; ~~ \phi_i(x-x_0)<\epsilon\}$ is an open basis element of weak operator topology. I show that $N$ is $\tau-$ open. $$N= \cap_{i=1}^n \{x\in B(H) ; ~~|\phi_i(x-x_0)|<\epsilon\} $$ $$=\cap \phi_i^{-1}((\phi(x_0), \epsilon))~~~~~~~~~~~~~~~~~~~~~~~~~~~$$ which shows $N$ is $\tau-$ open.
Is my process correct? Thanks.