Talk:Completely distributive lattice/Archive 1

Archive 1

I've started this article by copying much of the material in the "Distributivity laws for complete lattices" section of the distributivity (order theory) article here.

Please help make this article better and, in particular, add citations. I'm not a lattice theorist, but I do use lattices in my work. Wikipedia is great for info, but I need solid references to know that I'm (a) using established definitions and (b) relying on proved results. BTW, a citation for even a trivial property is typically more useful to me than a proof, since I would rarely have space in a paper for the proof. --Malcohol 09:05, 17 October 2006 (UTC)

Important: Although I've started the article, I copied much of the material from another article. I have no reference for the section on the cross-cut operator, or the property about products of chains.--Malcohol 09:25, 17 October 2006 (UTC)

The second alternative characterization seems not to be valid. Take the complete chain of all natural numbers < 7 and its set of subsets S = {{2,4}, {3,5}}. Then the intersection of S is empty, so that the join of intersection of S is the minimal element 0. On the other hand, the meet of the set of joins of S is the meet of {4,5}, which equals 4 and differs from 0. I assume you need an additional condition on S, but I don't know which one.

The first alternative characterization is valid under assumption of the Axiom of Choice (which axiom is usually taken for granted in mathematics). I have put a proof at [www.cs.rug.nl/~wim/pub/whh381.pdf] Whh7 18:38, 2 January 2007 (UTC)

I've checked your example, and you're right. I've commented out the property in the article.--Malcohol 11:22, 3 January 2007 (UTC)

I've now recovered the second alternative characterization: you have to add that S ranges over the sets of down-closed subsets of the lattice. The precise version is: A complete ordered set X is completely distributive if and only if, for every set S of down-closed subsets of X, the meet of the joins of the elements of S equals the join of the intersection of S. I've added this result to the pdf-file mentioned in my previous comment. Whh7 14:31, 3 January 2007 (UTC)