Really I should first ask this question here on MathOverflow and only then [post it as an open problem](http://garden.irmacs.sfu.ca/?q=op/do_filters_complementive_to_a_given_filter_form_a_complete_lattice) in Open Problem Garden and [propose it as a polymath problem](http://portonmath.wordpress.com/2009/08/30/proposal-conjecture-complementive-filters/). Indeed I did the reverse and now hope that for somebody my "open problem" may be simple.


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.


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 bounded 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.


Read also [my current progress on this problem](http://portonmath.wordpress.com/2009/11/02/exposition-complementive-filters/).



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


