1
$\begingroup$

A closure algebra C is a boolean algebra B together with a unary closure operator, and additional axioms, the Kuratowski axioms, that the closure operator must satisfy. (The Wikipedia article prefers the term "interior algebra".)

Since C is a boolean algebra, we may define a boolean filter F on C in the usual way, as a non-empty set of elements of C which contains:

  • the meet of x and y whenever x and y are both in F;

    the join of x and y, for any x in F and y in C.

The definition of a boolean filter makes no mention of the closure operator. This raises the question of whether there is another kind of filter, to be called a closure filter, which adds another condition, relating to the closure operator, to the definition of a boolean filter. For example, it might require that:

  • the interior of x is in F, for any x in F.

where interior is defined in terms of closure as usual. Or perhaps this is not the right choice for an additional condition.

If we then define a closure ultrafilter as a maximal closure filter, we might find that these have different properties from maximal boolean filters.

I have not been able to locate any research along these lines, and would appreciate hearing of any references.

Improve this question
$\endgroup$
3
  • $\begingroup$ My suspicion is that there will be little difference. One of the early researchers of closure operators is Cech (forgive my limited character set). I was influenced by McKenzie, McNulty, and Taylor's book Algebras, Lattices, Varieties, which has a section on closure operators. Perhaps the main concern will be whether what you form is algebraic or not. ALV may also have some useful bibliographic references for you. Gerhard "Email Me About System Design" Paseman, 2011.07.14 $\endgroup$
    – Gerhard Paseman
    Commented Jul 14, 2011 at 20:34
  • 1
    $\begingroup$ There is one way to figure out what the correct notion is. Filters and ideals on Boolean algebras are basically kernels of homomorphisms. Naturally, filters and ideals on closure algebra should be kernels in the same way. The requirement that filters are closed under the interior operation is certainly necessary, but I don't know off the top of my head if it is sufficient. $\endgroup$
    – François G. Dorais
    Commented Jul 14, 2011 at 21:31
  • $\begingroup$ François: Yes, you put it better than I did. A homomorphism on a closure algebra needs to preserve closures. But a homomorphism based on a boolean filter needn't do that, in general. So we need some other concept of filter, if we want to maintain the relationship between filters and homomorphisms. -- The book Gerhard mentions was published in 1987; I'm surprised if this hasn't come up before. $\endgroup$
    – MikeC
    Commented Jul 14, 2011 at 22:27

0

Reset to default

You must log in to answer this question.