Takesaki theorem 2.6 clarification

Consider the following theorem from Takesaki's book "Theory of operator algebra I":

Question is simple: in (iv) should the term "closed" be understood as "closed in $\mathscr{M}$"? Or as closed in $\mathscr{L}(\mathfrak{H})$? I suspect it is the former, as it allows us to apply the Krein-Smulian theorem.

It's irrelevant, due to $\mathscr M$ being closed in all those topologies.

So, if $K$ is closed in $B(H)$, this means that $B(H)\setminus K$ is open. Then $\mathscr M\setminus K=\mathscr M\cap(B(H)\setminus K)$ is relatively open in $\mathscr M$.

Conversely, if $K$ is closed in $\mathscr M$, then $\mathscr M\setminus K$ is relatively open. That is, $\mathscr M\setminus K=\mathscr M\cap V$ for some open $V$. Then $$ B(H)\setminus K=(B(H)\setminus\mathscr M)\cup(\mathscr M\setminus K) =(B(H)\setminus\mathscr M)\cup(\mathscr M\cap V)=(B(H)\setminus\mathscr M)\cup V $$ is open.