# separable weakly dense subalgebra of a von Neumann algebra

This is an exercise of "A short course on spectral theory" by William Arveson.

For any von Neumann algebra $\mathcal{M}$ on a Hilbert space $H$, I need to show there exists a unital $C^\ast$-subalgebra $\mathcal{A}$ such that $\mathcal{A}$ is separable (has countable norm-dense subset) and is weakly dense in $\mathcal{M}$.

I really have no idea how to do this. Any suggestion is welcome.

What you are missing from the statement in the book is that $H$ is separable.
Consider first $B(H)$. Then $K(H)$ is weakly dense in $B(H)$, and it is not hard to show that $K(H)$ is separable.
Now consider a von Neumann algebra $M\subset B(H)$. You have an inclusion of unit balls $M_1\subset B(H)_1$. The previous exercise in the book proves that, since $H$ is separable, the weak operator topology is metrizable on bounded sets. On metric spaces, subsets of separable sets are separable. So $M_1$ is weakly separable. Finally, $M=\overline{\bigcup_n nM_1}$, so $M$ is separable.