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