Is $B(H)$ sot separable

Question

To prove that the unit ball of $B(H)$ is separable in strong operatior topology using the fact that $K(H)$ is separable and also is sot- dense in $B(H)$.

I think we can conclude that $B(H)$ is also sot- separable. In fact if the sequence $\{A_n\}$ is dense in $K(H)$ then $B(H) = sot -\lim \{A_n\}$.

Is it correct?

Answer 1

Your reasoning is correct. Note that it is implicitly assumed that $H$ itself is separable.

A more self-contained proof is to take an orthonormal basis $(e_i)$ and define, for a given $A\in B(H)$, $$ A_n = P_n AP_n,\quad P_n(x) = \sum_{i=1}^n (x,e_i)e_i $$ For any fixed $x$, $P_n x \to x$ in the norm, hence $A_nx\to Ax$ in the norm. And the operators $A_n$ are just linear combinations of rank-one operators, which form a separable space as long as $H$ is separable.

Answer 2

It seems this argument works too: Let us note that $B(H)=\bigcup_{1}^{\infty} B(H)_{||\cdot||\leq n}$. Since closed balls $B(H)_{||\cdot||\leq n}$ are all SOT-separable then $B(H)$ is SOT-separable as well.