TITLE: Is the strong operator topology metrizable?
QUESTION [10 upvotes]: Let $X$ be a separable Banach space. Is the strong operator topology  metrizable on $B(X)$, the space of all bounded operators on $X$?
SOT-$\lim T_i=0~$ if and only if $~\lim \|T_ix\|=0$ for every $x\in X$.

REPLY [2 votes]: Not sure whether this is equivalent to former answers. For maps between metric spaces, continuous is equivalent to sequentially continuous (see this wiki page). Consider the composition of operators in $B(X)$, it is a map from $B(X) \times B(X) \to B(X)$. It can be shown that this map is sequentially continuous (by playing with the definitions), but it is not continuous (See a counterexample here). So $B(X)$ is not metrizable.