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Let $\mathfrak{A}$ be a bounded distributive lattice with binary meet and join $\sqcap$ and $\sqcup$.

I will denote $\partial F = \{ X\in\mathfrak{A} \mid \forall Y\in F: X\sqcap Y\ne \bot \}$ where $F$ is a filter on $\mathfrak{A}$ and $\bot$ is the least element of the lattice.

I will say that the set of filters is star-separable iff $A\ne B$ implies $\partial A \ne \partial B$ for every filters $A$ and $B$.

I my draft book it is proved that the set of filters on a boolean lattice is star-separable.

Can this result be strengthened? Particularly, is it true for all bounded distributive lattices?

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This is not true for all bounded distributive lattices. Consider the distributive lattice $\bot \to a \to \top$ and the filters $\uparrow{a}$ and $\{\top\}$.

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  • $\begingroup$ Hm, maybe it holds for all complemented lattices? $\endgroup$ – porton Apr 7 '16 at 18:29
  • $\begingroup$ nope, it does not hold for all complemented lattices. $\endgroup$ – Stu Kraji Apr 7 '16 at 21:17
  • $\begingroup$ Maybe, it holds for all Heyting or all co-Heyting lattices? $\endgroup$ – porton Apr 7 '16 at 22:11
  • $\begingroup$ Nope. It does not hold for all heyting and co-heyting algebras. Just check the example in the answer :) $\endgroup$ – Stu Kraji Apr 8 '16 at 11:38

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