# 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 $D$.