This conjecture has a seemingly trivial case when is a principal filter. When I attempted to prove this seemingly trivial case I stumbled over a looking simple but yet unsolved problem:

Let is a set. A filter (on ) is by definition a non-empty set of subsets of such that . Note that unlike some other authors I do not require . I will denote the lattice of all filters (on ) ordered by set inclusion. (I skip the proof that is a lattice).

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

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/