Are principal filters the center of the lattice of filters?

This conjecture has a seemingly trivial case when \mathcal{A} is a principal filter. When I attempted to prove this seemingly trivial case I stumbled over a looking simple but yet unsolved problem:

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. (I skip the proof that \mathscr{F} is a lattice).

Conjecture The set of principal filters on a set U is the center of the lattice of all filters on U.

Note that by center of a (distributive) lattice I mean the set of all its complemented elements.

I did a little unsuccessful attempt to solve this problem before I’ve put it into this blog. I will think about this more. You may also attempt to solve this open problem for me.

2 comments

  2. Pingback: Principal filters are center – solved « Victor Porton's Math Blog

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