Separable von Neumann algebra What is the simplest argument which shows that each infinite dimensional von Neumann algebra is not separable (in the norm topology)? It seems that this is a kind of folklore: at least I never saw the proof of this fact. 
 A: Suppose that we have an infinite-dimensional von Neumann algebra. In particular, this is a C*-algebra so it contains an infinite-dimensional abelian C*-algebra $A$. Consequently, $\overline{A}^{{\rm WOT}}$ is a non-separable abelian von Neumann algebra. No need to invoke the fact that it is isomorphic to $L_\infty(\mu)$ for some measure $\mu$, simply use the Borel functional calculus to build an uncountable, closed discrete subset in $\overline{A}^{{\rm WOT}}$ just like you would do in $L_\infty$.
The fact that every infinite-dimensional C*-algebra contains an infinite-dimensional abelian subalgebra is fairly elementary and can be found in Kadison-Ringrose.
A: I'm not sure if this would qualify as simple but the following argument proves the required result. There is a natural locally convex topology on each von Neumann algebra (the finest such that agrees with the weak operator topology on the unit ball) which has the property that the closed graph theorem holds for every linear mapping into a SEPARABLE Banach space.  Hence if the algebra were separable in the norm topology, then the latter  would coincide with the former  and this can only happen in the finite dimensional case.
A: This is proven for instance here, Corollary 1.3.17 p. 26.
I am far from being an expert, however it seems to me that the main ingredient in the proof is the fact that if a von Neumann algebra is infinite dimensional, then it contains
an infinite family of non-zero pairwise orthogonal projections (see Proposition 1.3.16). 
