Is $F(H)$ dense in $B(H)$?

Question

Let $F(H)$ be the set of finite rank operators on Hilbert sapce $H$,$K(H)$ is the set of compact operators on $H$,$B(H)$ is the set of bounded linear operators on $H$.I know the fact :$F(H)$ is dense in $K(H)$.

Is $K(H)$ dense in $B(H)$?If it is ,then $F(H)$ is dense in $B(H)$

Answer 1

If you use the norm topology, $K(H)$ is not dense in $B(H)$ (easy proof: no compact $T$ is invertible, so $\|I-T\|\geq1$).

If you use the sot or wot topologies, then $F(H)$ is dense in $B(H)$, as a consequence of the Double Commutant Theorem.

Answer 2

$K(H)$ is not dense in $B(H)$. In fact $K(H)$ is closed in $B(H)$ and $K(H)=B(H)$ iff $H$ is finite dimensional.