Separability of unit ball in strong operator topology Let $X$ be a separable Banach space.  Let $B_1(X)$ be the set of all bounded linear operators $X \to X$ with operator norm $\leq 1$.  Does $B_1(X)$ have to be separable in the strong operator topology?
 A: It turns out the answer is yes.  Let $(x_k)_{k \geq 1}$ be a countable dense subset of $X$ and consider the set $A = \{ (Tx_k)_{k \geq 1} : T \in B_1(X) \} \subseteq X^{\mathbb{N}}$.  Since $X$ is separable and metrizable, $X^{\mathbb{N}}$ is also separable and metrizable, and therefore $A$ is also separable.  Let $(T_n)_{n \geq 1}$ be an enumeration of the operators corresponding to a countable dense subset of $A$.
The set $\{T_n\}_{n \geq 1}$ is dense in $B_1(X)$ for the strong operator topology.  To see this, fix $T \in B_1(X)$ and consider the sequence $(Tx_k)_{k \geq 1} \in A$.  By construction there is a subsequence $(T_{n_j})_{j \geq 1}$ such that $(T_{n_j} x_k)_{k \geq 1} \to (Tx_k)_{k \geq 1}$ in $A$ as $j \to \infty$.  In particular this implies that $T_{n_j} x_k \to T x_k$ as $j \to \infty$ for each fixed $k$.  Finally because $\{x_k\}_{k \geq 1}$ is dense in $X$ this implies that $T_{n_j} \to T$ in the strong operator topology.
A: Let me add that it is not separable in the uniform operator topology. The example is in Section 1.2 in 1.
Stochastic Equations in Infinite Dimensions , DA PRATO, ZABCZYK
