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Any closed subspace $V\subset {\ell}^2(\omega)$ has associated to it a subset ${\cal S}_V$ of ${\cal P}(\omega)$, call it a combinatorial Hilbert space, namely the set of all supports of all vectors in $V$.

My question: Can one give a combinatorial characterization of combinatorial Hilbert spaces?

To start things off, I'll mention obvious (or at least nearly obvious) necessary conditions on an ${\cal S}_V$:

1) Closure under countable (and thus here arbitrary) unions [countable anticipates generalization to larger ordinals than $\omega$];

2) For sets $A_1,\ldots,A_n\in {\cal S}_V$ with none contained in the union of all the others, and at least $n-1$ integers belonging to at least two $(A_i)$'s, there exists non-empty $\ C\subsetneq \bigcup A_i$ with $|C|\geq n-1$ such that $\bigcup A_i\setminus C\in {\cal S}_V$;

3) For $A,B \in {\cal S}_V$ with $\emptyset \not= B \subsetneq A$, there exists non-empty $C\subset B$ with $A\setminus C \in {\cal S}_V$.

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  • $\begingroup$ There are uncountable families of subsets of $\omega$, so your «and thus here arbitrary» needs some justification, no? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 6, 2011 at 1:52
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    $\begingroup$ @Mariano: yes, but if a subset of $\omega$ is the union of a collection of subsets of $\omega$, then it is in fact the union of countably many of them, since $\omega$ itself is countable. $\endgroup$
    – Chris Eagle
    Commented Jan 6, 2011 at 2:05
  • $\begingroup$ Now, is that provable in ZF? $\endgroup$
    – user5810
    Commented Jan 6, 2011 at 2:35
  • $\begingroup$ @Ricky No, see arxiv.org/PS_cache/arxiv/pdf/0806/0806.1957v1.pdf for example. An infinite Dedekind finite set of reals would contain no countable sequence at all. $\endgroup$
    – David Feldman
    Commented Jan 6, 2011 at 4:08
  • $\begingroup$ Is the collection of finite and cofinite subsets of $\omega$ a combinatorial Hilbert space? In the analogous question for $\ell^\infty(\omega)$ you can obviously get all subalgebras of $\mathcal P(\omega)$, but here this is not clear at all. $\endgroup$
    – Stefan Geschke
    Commented Jan 6, 2011 at 7:57

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