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 .
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