Do filters complementive to a given filter form a complete lattice?

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

Let \mathcal{A}\in\mathscr{F} is some (fixed) filter. Let D=\{\mathcal{X}\in\mathscr{F} | \mathcal{X}\supseteq \mathcal{A}\}. Obviously D is a lattice.

I will call complementive such filters \mathcal{C} that:

  1. \mathcal{C}\in D;
  2. \mathcal{C} is a complemented element of the lattice D.

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 D.

One comment

  1. Pingback: Exposition: Complementive filters are complete lattice « Victor Porton's Math Blog

Leave a Reply