# $\mathcal{K}(\mathcal{H})$ is separable?

I am reading Arveson's Notes On Extensions of $C^*$-algebras, in proving the Corollary to Thm2, he seems to assume that $\mathcal{K}(\mathcal{H})$, the space of compact operators on a separable Hilbert space, is separable as a topological space, i.e., it contains a countable dense set.

This is a little bit surprising to me. If we just take $\mathcal{H}=\ell^2$ and $F$ defined by:$$F(\sum\alpha_n e_n)=(\sum\alpha_n f_n)e_1,$$
where $\{e_n\}$ is the canonical basis for $\ell^2$. That is, $F$ maps everything to the first coordinate, $F e_n=f_n e_1$. Since $(f_n)$ can be be arbitrary sequences in $\ell^{\infty}$, which is a non-separable space.

My argument seems to imply that even the space of rank-one operators is not separable.

Where did I make a mistake? Thanks!

-
How can $f_n$ be an arbitrary sequence from $\ell^\infty$? Certainly $f_n = 1$ for all $n$ is not allowed. It must be a sequence from $\ell^2$... On the right hand side you write a functional on $\ell^2$ times $e_1$, after all. As you should know, the finite rank operators are dense in $\mathcal{K(H)}$ and it's not too hard to find a dense sequence there. – t.b. Jul 11 '12 at 18:36

The space of rank-one operators is separable, being a continuous image of $\mathcal H\oplus \mathcal H$ under the map $(u,v)\mapsto T_{u,v}$ where $T_{u,v}x=\langle x,u\rangle v$. Hence the space of finite-rank operators is separable (they are finite sums of rank-1 operators), and its closure in $B(\mathcal H)$ is $\mathcal K(\mathcal H)$.