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Chain-meet-closed sets ★★

Author(s): Porton

Let $ \mathfrak{A} $ is a complete lattice. I will call a filter base a nonempty subset $ T $ of $ \mathfrak{A} $ such that $ \forall a,b\in T\exists c\in T: (c\le a\wedge c\le b) $.

Definition   A subset $ S $ of a complete lattice $ \mathfrak{A} $ is chain-meet-closed iff for every non-empty chain $ T\in\mathscr{P}S $ we have $ \bigcap T\in S $.
Conjecture   A subset $ S $ of a complete lattice $ \mathfrak{A} $ is chain-meet-closed iff for every filter base $ T\in\mathscr{P}S $ we have $ \bigcap T\in S $.

Keywords: chain; complete lattice; filter bases; filters; linear order; total order

Posted by porton
updated December 12th, 2009
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Do filters complementive to a given filter form a complete lattice? ★★

Author(s): Porton

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.

Keywords: complete lattice; filter

Posted by porton
updated July 31st, 2009
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