Is the strong operator topology metrizable? 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 $x\in X$. 
 A: A quick proof using the open mapping theorem: It follows easily from the uniform boundedness principle that $(B(X),SOT)$ is sequentially complete. If it were metrizable it would thus be a Fréchet space and the open mapping theorem implies that the continuous identity $(B(X),\|\cdot\|_{op})\to (B(X),SOT)$ would be open, i.e., $SOT$ coincides with the operator norm topology which is not true if $X$ is infinite dimensional.
EDIT. You do not need a sledgehammer to show that $SOT$ differs from the operator norm topology: Fixing a non-zero element $y\in X$ we have an embedding $X^*\to B(X)$, $f\mapsto f\otimes y$ where $f\otimes y$ maps $x$ to $f(x)y$. The induced topologies on $X^*$ are the weak$^*$ topology and the dual norm topology. They differ because every continuous semi-norm of the weak$^*$ topology has a huge kernel.
A: 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. 
A: No, it is not sequential (hence non-metrisable) unless $X$ is finite-dimensional. Otherwise, let $(z_n)_{n=1}^\infty$ be a linearly independent sequence that is dense in $X$. For each $k$ we may consider the subspace $Z_k$ of $B(X)$ comprising operators mapping ${\rm span}\{z_1, \ldots, z_k\}$ to itself that are 0 on some fixed complement of ${\rm span}\{z_1, \ldots, z_k\}$ (so that $\dim Z_k = k^2$). We may then choose a finite $\tfrac{1}{k}$-net $T_{k,j}$ of the sphere $\{T\in Z_k\colon \|T\|=k\}$. Let $S$ be the union of all the nets picked above. We claim that 0 is in the SOT-closure of $S$. 
Indeed, suppose that $U$ is a SOT-open neighbourhood of 0. Let $x_1, \ldots, x_n\in X$ and let $\varepsilon > 0$ be such that
$$\{T\in B(X)\colon \max_{i\leqslant n} \|Tx_i\| < 2\varepsilon \}\subseteq U.$$
Take $k$ with $1/k <\varepsilon$. When $k$ is large enough, by the density of $(z_n)_{n=1}^\infty$ there must be $T_k\in Z_k$ with $\|T_k\|=k$ such that $\|T_kx_i\|<\varepsilon$ for all $i$. Pick $j$ so that $\|T_{k,j} - T_k\|\leqslant\tfrac{1}{k}$. Thus,
$$\|T_{k,j} x_i\| =  \|T_{k,j} x_i - T_kx_i + T_k x_i\|\leqslant \tfrac{1}{k}+\varepsilon<2\varepsilon, $$
that is, $T_{k,j}\in U$.
Consequently, the set $S$ is not SOT-closed however it is SOT-sequentially closed.
