I have just read this question Is a von Neumann algebra just a C*-algebra which is generated by its projections? and am wondering about Robert Israel's answer when he says that a subalgebra of $C(X)$ is not a Von Neumann algebra. Some abstract definition must be being used, but he also says weak closure (by which I assume he means weak operator closure) and that only exists in $B(H)$. When someone says that a C* algebra $A$ is a von Neumann algebra, is he saying that $A$ is $*$-isomorphic (i.e. a bijective map preserving the 3 algebraic operations, is an isometry, and preserves $*$) to a (concrete) von Neumann algebra? ($*$-subalgebra of $B(H)$ equal to its bicommutant) That doesn't seem like what people would mean since such an isomorphism would say nothing about how $A$ lies within $B(H)$ topologically. On the other hand, $A$ isn't known to lie in some external space so one can't hope for a big isomorphism that happens to take $A$ to a von Neumann algebra, and the bigger space containing $A$ to $B(H)$. I am similarly confused about what "weak topology" means.
I have also heard of a predual characterization of Von Neumann algebras, although I've never heard the equivalence of this notion and mine formally stated nor proved. If that's what's going on here, could someone please direct me to a sufficiently precise description of what everything means?