# Separability of von Neumann algebra

In a lecture note, from where I am studying theory of von Neumann algebras, the author has commented that the following are equivalent. Let $$A$$ be a von Neumann algebra.

1. $$A$$ is SOT separable.

2. $$A$$ is WOT separable.

3. $$A$$ is $$\sigma$$-WOT separable.

4. The predual is a separable Banach space.

Clearly, 1. implies 2. But I do not know how to prove the other implications. Can someone help me to prove this in the most elementary way?

It doesn't look like a research level question, but I wasn't able to track down a simple statement of the (easy) equivalences (1) $$\Leftrightarrow$$ (2) $$\Leftrightarrow$$ (3) in the standard references, so I think it is worthwhile to give an answer.
Lemma. Let $$X_0$$ be a countable subset of a complex topological vector space $$E$$. Then the set of all finite linear combinations of elements of $$X_0$$ with complex rational (in $$\mathbb{Q} + i\mathbb{Q}$$) coefficients is countable and its closure is a linear subspace of $$E$$. (Exercise)
The equivalence of (1), (2), and (3) now follows from the fact that all three topologies have the same continuous linear functionals (a standard fact). E.g., for (2) $$\Rightarrow$$ (1), if $$X_0$$ is a countable WOT dense subset of $$A$$ then its complex rational span $$X$$ is still countable, and the SOT closure of $$X$$ is a SOT closed subspace such that any (WOT-continuous, hence SOT-continuous) linear functional that vanishes on it must be zero. Hence $$X$$ is SOT dense in $$A$$.
For (4) $$\Rightarrow$$ (3), use the fact that the unit ball of the dual of any separable Banach space is weak* compact and weak* metrizable (exercise). Thus if the predual of $$A$$ is separable then its unit ball is weak* (= $$\sigma$$-WOT) separable and hence $$A$$ is weak* separable.
However, (3) does not imply (4). For instance, the von Neumann algebra $$l^\infty[0,1]$$ is weak* separable ($$C[0,1]$$ is weak* dense) but its predual, $$l^1[0,1]$$, is not separable.