The complementive filters ordered by inclusion form a lattice isomorphic to the quotient of the power set of $U$ modulo the filter $\mathcal A$. So, for example, if $U=\omega$ and if $\mathcal A$ is the filter of cofinite sets, then the lattice of complementive filters would not be complete.

To establish the isomorphism, consider a complementive filter $\mathcal C$ and its complement $\mathcal C'$ in $D$. The filter generated by $\mathcal C\cup\mathcal C'$ is the join in $D$, so it must be the top element, the improper filter of all subsets of $U$. So there must be a set $Q\in\mathcal C$ whose complement $U-Q$ is in $\mathcal C'$.

I claim that $\mathcal A\cup\{Q\}$ generates $\mathcal C$. To see this, suppose not, and consider some $Z\in\mathcal C$ that is not in the filter generated by $\mathcal A\cup\{Q\}$. Then $Z\cup(U-Q)$ cannot be in $\mathcal A$ (because, if it were, then its intersection with $Q$ would be in the filter generated by $\mathcal A\cup\{Q\}$, but this intersection is included in $Z$, which is not in that filter). But, since $Z\in\mathcal C$ and $U-Q\in\mathcal C'$, the union $Z\cup(U-Q)$ is in the intersection of these two filters, which is $\mathcal A$ (because they're complements in $D$). This contradiction establishes the claim.

Thus, each complementive $\mathcal C$ in $D$ is generated by the fixed $\mathcal A$ plus one more set $Q$. It is easy to check that the filter generated by $\mathcal A\cup\{Q\}$ and the filter generated by $\mathcal A\cup\{R\}$ are equal if and only if $Q$ and $R$ represent the same element in the quotient Boolean algebra $\mathcal P(U)/\mathcal A$. Better, the filter generated by $\mathcal A\cup\{Q\}$ is included in the one generated by $\mathcal A\cup\{R\}$ if and only if the element of $\mathcal P(U)/\mathcal A$ represented by $R$ is below (in the Boolean algebra) the one represented by $Q$. Thus, the correspondence between $\mathcal C$ and ($Q$ modulo $\mathcal A$) is an order-reversing bijection between the lattice of complementive filters and the Boolean algebra $\mathcal P(U)/\mathcal A$.

Since a Boolean algebra and its dual order are isomorphic, this proves my description of the structure of the lattice of complementive filters.

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