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

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

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

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

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 $latex D$.

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