I feel that this is a very simple problem, but somehow I don't see the solution.
I want to show that if $A$ is a subalgebra of $B(H)$ containing $1$ then if $B\in SOTcl(A)$, for every n, $Lat(A^{(n)}) \subseteq Lat(B^{(n)})$. Where Lat is the set of the invariant subspaces of a set of operators, and $^{(n)}$ means the inflation of an operator or a subset of $B(H)$.
Ok I want to do an induction on the n.
First of all I want to prove the statement for $n=1$.
If $B\in SOTcl(A)$, then exists $B_n \in A$ such that $B_nx\rightarrow Bx\ \ \forall \ x \in X$.
If $B \in A$ surely $Lat(A) \subseteq Lat(B)$. My problem is the passage to the limit... My ability to go on is blocked!
Any help is appreciated.
I'm assuming you are considering the lattice to be the closed invariant subspaces. Otherwise I don't think the result is true.
Take a subspace $M\in\text{Lat}(A)$. Then $B_nM\subset M$. So, for each $x\in M$, the sequence $\{B_nx\}$ is in $M$ and it converges to $Bx$. As $M$ is closed, $Bx\in M$.
The $n$-amplification version runs exactly the same.