0
$\begingroup$

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?

|cite|improve this question
$\endgroup$
2
$\begingroup$

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

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.