I want to prove that if $N$ is normal, then $W^\ast(N)=\{N\}''$, where $W^\ast(N)$ is the Von Neumann algebra generated by $N$ and $\{N\}''$ is the bicommutant of $N$.
for the inclusion $W^\ast(N) \subseteq\{N\}''$: is simple to see that $\{N\}''$ is an abelian $*$-algebra WOT closed containing 1 and $N$, i.e. a Von Neumann algebra.
I'm blocked for the second inclusion. It's likely I have to use the bicommutant theorem. Can you give me a hand please?
By the spectral theorem, $C^*(I, N)$ is a commutative C*-algebra. We have $W^*(N) = W^*(C^*(N,I))$. By the bicommutant theorem,
$$W^*(C^*(N,I)) = C^*(N,I)^{\prime \prime} = \{N\}^{\prime \prime}.$$