Let $H$ be a (separable) infinite dimensional Hilbert space, and $B(H)$ the space of bounded operators on $H$. Is $B(H)$ separable in the operator norm topology? What about in the strong and weak operator topologies? In the latter cases (which are not metrizable, correct?), what about second countability?
I recall hearing that the answer to the first question was no, but I cannot see why.
This is an answer only to the first question.
Since $H$ is separable it has countable orthonormal basis $\{e_n:n\in\mathbb{N}\}$. Now for each $\lambda\in\ell_\infty$ consider diagonal operator $T_\lambda:H\to H$ well defined by equalities $T_\lambda(e_n)=\lambda_n e_n$. It is easy to check that $$ T_{\alpha' \lambda'+\alpha''\lambda''}=\alpha'T_{\lambda'}+\alpha''T_{\lambda''} $$ $$ \Vert T_\lambda\Vert=\Vert \lambda\Vert_\infty. $$ for all $\alpha',\alpha''\in\mathbb{C}$ and $\lambda',\lambda''\in\ell_\infty$. Thus we have an isometric inclusion $$ i:\ell_\infty\to\mathcal{B}(H):\lambda\to T_\lambda, $$ i.e. we can consider $\ell_\infty$ as subspace of $\mathcal{B}(H)$. Since $\ell_\infty$ is not separable then $\mathcal{B}(H)$ can't be separable too.
Hint for your second question: In the strong operator topology, the finite rank operators are dense in $B(H)$.
For the first question you may consider a Hilbert basis for $H$ and consider the map $\Phi:l^{\infty}\to B(H)$ such that $\Phi((a_i)_i)=T; \ Te_i=a_ie_i$. It is an isometry.
