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/