Why is any von Neumann algebra SOT-closed?

Question

I am trying to figure out why any von Neumann is SOT closed, which is stated in Conway. Do we need to use the Double Commutant Theorem on this implication aswell?

Is there a simpler explanation?

Answer 1

A von Neumann algebra is often defined as a follows:

Let $A$ be a $*$-subalgebra of $\mathcal{B(H)}$ acting non-degenerately on $\mathcal{H}$. Then $A$ is a von Neumann algebra if $A=A''$.

The bicommutant theorem says that

Let $A$ be a $*$-subalgebra of $\mathcal{B(H)}$ acting non-degenerately on $\mathcal{H}$. Then the following are equivalent:

1. $A=A''$,
   2. $A$ is weakly closed,
   3. $A$ is strongly closed.

The implication from $2$ to $3$ is trivial. Showing $1$ to $2$ is an easy exercise.

Could you specify the definition you use and which step is causing troubles?