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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?

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    $\begingroup$ Two obvious observations are that the problem is absolute (Shoenfield) and that one can essentially recast your $P$ as a version of Cohen forcing (by using separability of $l^2$ carefully). Both observations seem to be folklore, but I haven't seen any reasonable way of taking advantage of them. I once attended a talk at the analysis seminar at Berkeley where the speaker began by basically describing a version of $P$, and forcing (and $\mathsf{MA}$), and then quickly hitting a wall and not getting anywhere. This was about 12 years ago, and I haven't seen any scenarios building on it since. $\endgroup$
    – Andrés E. Caicedo
    Jul 9, 2013 at 6:08
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    $\begingroup$ I suspect I was the speaker at that talk. $\endgroup$
    – Nik Weaver
    Jul 9, 2013 at 6:26
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    $\begingroup$ I have nothing intelligible to say about the question, but the comment thread is awesome. $\endgroup$
    – Asaf Karagila ♦
    Jul 9, 2013 at 7:48
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    $\begingroup$ @Sébastien: your comments in this thread have not been helpful. $\endgroup$
    – Nik Weaver
    Jul 17, 2013 at 18:19
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    $\begingroup$ @Nik: the purpose of my comments is not to be helpful for you or some experts, their purpose is to better understand your approach. Conversely, your explanations could be very helpful for me and some non-experts. Next, when my understanding will be better, maybe I could help you. $\endgroup$
    – Sebastien Palcoux
    Jul 17, 2013 at 19:03

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