This conjecture has a seemingly trivial case when $latex \mathcal{A}$ is a principal filter. When I attempted to prove this seemingly trivial case I stumbled over a looking simple but yet unsolved problem:

Let $latex U$ is a set. A filter (on $latex U$) $latex \mathcal{F}$ is by definition a non-empty set of subsets of $latex U$ such that $latex A,B\in\mathcal{F} \Leftrightarrow A\cap B\in\mathcal{F}$. Note that unlike some other authors I do not require $latex \varnothing\notin\mathcal{F}$. I will denote $latex \mathscr{F}$ the lattice of all filters (on $latex U$) ordered by set inclusion. (I skip the proof that $latex \mathscr{F}$ is a lattice).

**Conjecture** The set of principal filters on a set $latex U$ is the center of the lattice of all filters on $latex U$.

Note that by *center* of a (distributive) lattice I mean the set of all its complemented elements.

I did a little unsuccessful attempt to solve this problem before I’ve put it into this blog. I will think about this more. You may also attempt to solve this open problem for me.

This problem is solved in

http://portonmath.wordpress.com/2009/10/31/principal-filters-are-center-solved/