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).
Let is some (fixed) filter. Let . Obviously is a lattice.
I will call complementive such filters that:
- is a complemented element of the lattice .
Conjecture The set of complementive filters ordered by inclusion is a complete lattice.
Remark In general the set of complementive filters is not a complete sublattice of .