Really I should first ask this question here on MathOverflow and only then post it as an open problem in Open Problem Garden and propose it as a polymath problem. 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.

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    $\begingroup$ You haven't included the requirement that filters are closed upwards under superset. $\endgroup$ Nov 11, 2010 at 0:22
  • $\begingroup$ Perhaps add the lattice tag? $\endgroup$ Nov 11, 2010 at 0:39
  • $\begingroup$ Does "complementive" or "complemented" here mean closed under taking complements as subsets of $U$? If so, then these "complementive filters" as you are calling them sound like the same things as Boolean subalgebras of the power set of $U$ (although I wouldn't call them filters unless you mean to include upward closure, which it sounds like you don't want to do). $\endgroup$
    – Todd Trimble
    Nov 11, 2010 at 1:42
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    $\begingroup$ @Joel and Todd: Upward closure follows from Porton's assumption about intersections, because he has a biconditional there, not just the left-to-right implication. $\endgroup$ Nov 11, 2010 at 2:07
  • $\begingroup$ Yes, Andreas. Also, I disagree with the votes to close---this is an interesting question. $\endgroup$ Nov 11, 2010 at 2:16

1 Answer 1


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.

  • $\begingroup$ What you mean by "modulo filter"? I don't know such a term. $\endgroup$
    – porton
    Nov 11, 2010 at 13:32
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    $\begingroup$ The quotient of a Boolean algebra $B$ modulo a filter $F$ is obtained from $B$ by identifying two elements $x$ and $y$ iff there is $a\in F$ such that $x\wedge a=y\wedge a$. This relation between $x$ and $y$ is a congruence relation (in the sense of universal algebra), so the quotient inherits from $B$ a Boolean algebra structure. $\endgroup$ Nov 11, 2010 at 13:45
  • $\begingroup$ Why $\omega$ modulo filter of co-finite sets is not complete? $\endgroup$
    – porton
    Nov 11, 2010 at 14:13
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    $\begingroup$ Partition $\omega$ into infinitely many infinite sets $A_n$; their images $[A_n]$ in the quotient have no supremum, for the following reason. Given any upper bound $[X]$ (so, for each $n$, $X$ contains all but finitely many elements of $A_n$), choose, for each $n$, one element $z_n$ in $X\cap A_n$, and let $Y=X-\{z_n:n\in\omega\}$. Then $[Y]$ is also an upper bound for the $[A_n]$'s and it is strictly below $[X]$ because $X-Y$ is infinite. $\endgroup$ Nov 11, 2010 at 14:31
  • $\begingroup$ @Andreas: It is expected that I will greatly simplify your proof using terminology which I define in mathematics21.org/binaries/filters.pdf Indeed do you want me to acknowledge you in my publication? $\endgroup$
    – porton
    Nov 11, 2010 at 22:50

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