# Exposition: Complementive filters are complete lattice

(In a past version of this article I erroneously concluded that our main conjecture follows from join-closedness of .)

Let is a set. A filter (on ) is a non-empty set of subsets of such that . Note that unlike some other authors I do not require . I will denote the lattice of all filters (on ) ordered by set inclusion.

**Conjecture 1** * Let is some (fixed) filter. Let . (Obviously is a bounded lattice.) Then the set of complemented elements of the lattice (ordered by set inclusion) is a complete lattice. *

This conjecture was first formulated in this blog post and suggested as a polymath project in this blog post.

**1. Filter objects **

(Borrowed from this blog post)

For greater clarity I will use {*filter objects*} instead of filters. Below I will describe the properties of filter objects without exact definition and the proofs. You can look here for the formalistic behind.

I will denote the set of all filters objects on as . Filter objects are bijectively related with filters by the bijection “” from the set of filter objects to the set of filters. A filter object corresponding to principal filter generated by a set is equal to . (Thus the set of subsets of is a subset of .)

For formal definition of filter objects in the framework of ZF see here. Below we will not need the exact definition of filter objects, but only the facts that “” is a bijection from filter objects to filters and that a filter object corresponding to principal filter generated by a set is equal to .

I will define the order on the set of filter objects by the formula for every filter objects and . This order well-agrees with the order of sets on .

is a complete lattice. (See here for a proof.)

**2. Second definition **

Let is a bounded distributive lattice. Let . I will denote . Obviously is a bounded lattice.

The center of a bounded distributive lattice is by definition the set of all complemented elements of . It is a well known fact that the center is a boolean lattice.

I will call the elements {*complementive*} to .

Now our conjecture can be equivalently reformulated:

**Conjecture 2** * is a complete lattice whenever is an element of the lattice . *

If the conjecture is true it may be generalized for the case of being the set of filter objects on some lattice instead of our less general case of filters on a set.

**3. Special case of being a set **

I almost believe that our conjecture is true in general, because it is true in the special case when is a set. Trueness of this special case of our conjecture trivially follows from the following theorem:

**Theorem 3** * if . *

**Proof:** It follows from this theorem.

**4. Ways to attack our conjecture **

** 4.1. Calculating meets and joins **

To prove that a lattice is complete is enough to prove one of the following two statements: 1. it has all joins; or 2. it has all meets.

Directly calculating meets and joins for the lattice seems a complicated problem.

It could be simplified if meets or joins on coincide with meets or joins on .

For meets it is not the case. (A counter-example is simple to find in the case when is any infinite set.)

For joins it seems true that joins on coincide with joins on . However I failed to prove it.

**Proposition 4** * If our main conjecture is true, then the sublattice of the lattice is join-closed. *

**Proof:** Let . Let our main conjecture is true and thus exists. We need to prove that . Because is a sublattice of it’s enough to prove that for any

Really: Let and . Then ; ; . Thus .

** 4.2. Intersection with a set **

An other way to characterize elements of is:

**Theorem 5** * for every. *

**Proof:**

We may attempt to attack our conjecture by proving that meets or joins of filter objects of the form are also of the form .

Let . Then where .

How to calculate and ?

I conjecture:

and .

It probably may be helpful if we would calculate or . (However it simple to show that in general .)

. Sadly there are no distributive law we could apply here.

. Sadly is not necessarily a set.

**5. Consequences of our conjecture **

** 5.1. Closedness of joins on **

Above it was proved that closedness of joins of follows from our main conjecture.

** 5.2. Coinciding pseudodifference and second pseudodifference **

It seems that using our conjecture we can prove that quasidifference and second quasidifference coincide for the lattice of filter objects.

See this and this wiki pages for details about the current state of the problem about coinciding quasidifference and second quasidifference.

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