The invariant subspace problem (ISP) for Hilbert spaces asks whether every bounded linear operator $A$ on $l^2$ (with complex scalars) must have a closed invariant subspace other than $\{0\}$ and $l^2$. A subspace $E$ is invariant for $A$ if $A(E) \subseteq E$.

Some time ago I noticed a reformulation that has a set-theoretic flavor. Let $P$ be the set of all linear operators $A$ from a finite-dimensional subspace of $l^2$ into $l^2$ such that (1) $\|A\| < 1$ and (2) if $E$ is a nonzero subspace of the domain of $A$ then $A(E) \not\subseteq E$. Order $P$ by reverse inclusion and for any unit vectors $v,w \in l^2$ define $$D_{v,w} = \{A \in P: \langle A^nv,w\rangle \neq 0\mbox{ for some $n$ such that $A^nv$ is defined}\}.$$ It is easy to see that each $D_{v,w}$ is dense in $P$. Having a counterexample to the ISP is the same as having a filter of $P$ that meets each $D_{v,w}$.

My question for the set theory experts on MO is simply whether this version of the problem suggests any possible approaches?

awesome. – Asaf Karagila Jul 9 '13 at 7:48