Is every norm-dense order dense sublattice super order dense?

Eugene Bilokopytov — Wed, 09 Apr 2025 08:44:03 +0000

Let $F$ be a norm-dense linear sublattice of a Banach lattice $E$, which is order dense, i.e. for every $e\texttt{ > } 0$ there is $f\in F$ such that $0 \lt f\le e$. Note that since $E$ is Archimedean, order denseness implies that for every $e\ge 0$ there is $A\subset F$ such that $e=\bigvee A$.

I am wondering whether under our assumption we can get a stronger property called super order denseness (but should be called $\sigma$-order dense-ness in my opinion). Namely, is it true that for every $e\ge 0$ there is a countable $A\subset F$ such that $e=\bigvee A$ ?

Some remarks: If $F$ has a strong unit then every norm-dense sublattice is super order dense.
If $F$ has the countable supremum property, then every order dense sublattice is super order dense. This property is quite common, in particular every order continuous BL has it, and for instance every Free BL.

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Is every norm-dense order dense sublattice super order dense?