Let $S$ be a non-empty set of filters on a meet-semilattice.

If our semilattice is a distributive lattice, then the supremum (on the poset of filters ordered by set-theoretic inclusion) of $S$ is the filter corresponding to the filter base generated by $\bigcup S$. You can finds a proof of such a theorem for example in my article: http://ijpam.eu/contents/2012-74-1/6/index.html

Can this statement be strengthened for a more general case than distributive lattices?

I would like to see a counter-example for the case is our semilattice is not a lattice, or better for the more specific case when it is a lattice but not a distributive lattice.


Oh, I've found an answer myself.

For every $\mathcal{A} \in S$ we have $\bigcup S \supseteq \mathcal{A}$ thus $\left[ \bigcup S \right]_{\cap} \supseteq \mathcal{A}$.

If filter $\mathcal{A} \supseteq \mathcal{X}$ for every $\mathcal{X} \in S$ then then $\mathcal{A} \supseteq \bigcup S$ then $\mathcal{A} \supseteq \left[ \bigcup S \right]_{\cap}$.

So $\left[ \bigcup S \right]_{\cap}$ is the least upper bound of $S$.


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.