An interesting fact popped out of a paper I'm writing: if the invariant subspace problem for Hilbert space operators has a positive solution, then every $A \in B(\mathcal{H})$ can be made "upper triangular" in the sense that there is a maximal chain of closed subspaces of $\mathcal{H}$, each of which is invariant for $A$.


The proof is quite easy, almost trivial, yet I had never heard of it before. (It isn't mentioned in the answers to [this question](https://mathoverflow.net/questions/48908/is-the-invariant-subspace-problem-interesting), for example.) It was a surprise to notice that the ISP has such a strong consequence for the structure of arbitrary operators because one thinks of is as merely the first, most basic question on that topic.


Surely this is known?



Is this consequence of the invariant subspace problem known?
Under the assumption that the invariant subspace problem has a positive solution, this is an immediate consequence of Lemma 7.1.11 (The Triangularization Lemma) in Radjavi and Rosenthal's book Simultaneous Triangularization: 


Say that a property $P$ of families of operators is inherited by quotients if whenever $N \subseteq M$ are invariant subspaces for a family of operators with $P$, then the family of induced operators on the quotient $M/N$ also has $P$.


Then if $P$ is a property of families of operators that is inherited by quotients, and if every family of operators satisfying $P$ (and acting on a space of dimension greater than 1) has a non-trivial closed invariant subspace, then every family satisfying $P$ is triangularizable in the sense that there is a maximal chain of closed subspaces, each of which is invariant for the family. 


