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Let $E$ be a Banach lattice. We consider finitely generated elements in the class of all closed sublattices of $E$. Let $F$ be a closed sublattice $E$. Then a subset $G \subseteq F$ is called generator of $F$ if $$ F = \bigcap \{ \hat F \subseteq E : \hat G \subseteq \hat F, \, \hat F \text{ is a closed sublattice of } E \}. $$ We call a closed sublattice $F$ of $E$ finitely generated if $F$ has a generator $G$ with finite cardinality. The minimum cardinality of a generator of $F$ is called the rank of $F$.

Question: Is the rank operator monotone? More precisely, let $F_1$ and $F_2$ be finitely generated closed sublattices of $E$ with $F_1 \subseteq F_2$. Does $\operatorname{rank}(F_1) \leq \operatorname{rank}(F_2)$ always hold?

This seems to be the case in the Banach lattice of continuous functions from some compact Hausdorff space $K$ into the reals or the complex numbers. However, the only known proof that I know relies heavily on the Gelfand-Kakutani representation theorem for AM-spaces. I am not sure how this can proof can be generalised.

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  • $\begingroup$ To add to the original question: If $E$ is finitely generated, do we know whether every closed sublattice is also finitely generated? Intuitively I reckon this is true, but have no way to show this. $\endgroup$
    – Julian Hölz
    Commented Apr 4 at 21:19
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    $\begingroup$ In the big formula, the last $F$ has to be $\hat{F}$, right? There is a big paper Drewnowski, Wnuk - On finitely generated vector sublattices. Might be relevant $\endgroup$
    – erz
    Commented Apr 4 at 22:58
  • $\begingroup$ The typo is corrected. I did not know of this paper. Thank you $\endgroup$
    – Julian Hölz
    Commented Apr 5 at 7:55

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