Can a von Neumann algebra $A$ have an infinite sequence $A_0 \subset A_1 \subset A_2 \subset ...$ of sub-vN-algebras such that every other sub-vN-algebra $B \subseteq A$ satisfies, for some $n \geq 0$ and finite $F \subseteq A$, either (1) $B \subseteq (A_n \cup F)''$, or (2) $(A_n \cup B \cup F)'' = A$?
Since the double-commutant $X''$ is the smallest vN algebra that includes $X$ and its adjoints, this question could be phrased: Can the $A_n$'s determine a maximal directed set of $A$'s proper sub-vN-algebras, "modulo finite sets"?
[I ask this question because, as a student in general topology, I've been looking in vain for something analogous in other spaces (measure spaces, Hilbert spaces, etc.). Loosely speaking, I've found you can always diagonalize out some elements of your $A_n$'s to make a new subspace $B$ that violates (appropriate analogs of) both (1) and (2) above, for all $n$ and $F$. I believe this procedure works with abelian vN algebras too. But I'm informed that not all vN algebras might be vulnerable to it, because their "weaker" topologies might make any diagonalized subalgebra have a "big enough" closure to satisfy (2) for some $n$ and $F$. Sounds plausible, but having skimmed through Kadison & Ringrose's text, it seems like it'd be hard work to make this precise, so ... is anyone aware of relevant results, positive or negative? Or even hints or hunches? I would be so, so, grateful!]
I wouldn't expect a straightforward answer to your question.
Suppose that $A$ is finitely generated, i.e. $A=F_0''$ for some finite $F_0$. Then, for any sequence $\{A_n\}$, for any $B\subseteq A$, for any $n$, you can take $F=F_0$ and then $B\subseteq A=F_0''=(A_n\cup F_0)''$.
The problem is that it is not known if every von Neumann algebra is finitely generated; it is known that all but type II$_1$ direct summands are singly generated. A counterexample to your question would provide an example of a von Neumann algebra $A$ that is not finitely generated; no such is example is known to this day.