Let $R \subset S$ be distinct von Neumann algebras having a separating vector in the separable Hilbert space $H$ on which they act. In what cases (if any) does there exist a topology $\tau$ on $S$ such that:
(1) every $\tau$-open set meets $R$ (has nonempty intersection with $R$);
(2) $\tau$'s subspace topology on $R$ is $WOT(R)$ (the weak operator topology on $R$ -- see below);
(3) $\tau \not \subseteq WOT(S)$?
I'm going to replace the vague notes I originally posted towards a solution to this question with a simple attempt to clarify it. The weak operator topology (WOT) on the set $B(H)$ of all bounded operators on $H$ is the one whose basic open sets have form
$ \{ T \in B(H) : | \langle (T - T_0) x_i, y_i \rangle | < \epsilon, i = 1, ..., n \}, $
for some $T_0 \in B(H)$, $\epsilon > 0$, and vectors $x_i, y_i$ in $H$. By $WOT(R)$ I mean $WOT$'s subspace topology on $R$ -- that is,
$WOT(R) = \{ O \cap R : O \in WOT \},$
and similarly for $WOT(S)$.
The subspace topology of $WOT(S)$ on $R$ is $WOT(R)$, so $WOT(S)$ meets condition (2) above. However I believe that some nonempty open sets of $WOT(S)$ fail to meet $R$, so neither $WOT(S)$ nor any finer topology satisfies condition (1) above. I also believe there exists a weaker topology than $WOT(S)$ on $S$ that does satisfy (1) and (2). But I don't have any intuition about whether any $\tau$ can jointly satisfy (1), (2), and (3).