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Join of oblique products ★★

Author(s): Porton

Conjecture   $ \left( \mathcal{A} \ltimes \mathcal{B} \right) \cup \left( \mathcal{A} \rtimes \mathcal{B} \right) = \mathcal{A} \times^{\mathsf{\ensuremath{\operatorname{RLD}}}} \mathcal{B} $ for every filter objects $ \mathcal{A} $, $ \mathcal{B} $.

Keywords: filter; oblique product; reloidal product

Posted by porton
updated August 9th, 2011
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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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