# Strengthening of a theorem about filters vs a counter-example

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.

## 1 Answer

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 $S$.