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Aug 2, 2016 at 6:51 comment added Tri (concluding Professor Markowsky's comment) It is the Cartesian product of the two-element chain and the three-element chain. (iii) is not a lattice since a and b do not have a join.
Aug 2, 2016 at 6:51 comment added Tri (continuing Professor Markowsky's comment) In a Hasse diagram of a finite lattice, the join-irreducible elements are the nodes that have only one edge going downward, while the meet-irreducible elements are the nodes that have only one edge going upward from them. For the examples given: (i) has 4 join-irreducibles (a, b, c, d), 3 meet-irreducibles (c, d, e) and all maximal chains have length 3 so it is not distributive. (ii) has 3 join-irreducibles (a, b, d), 3 meet-irreducibles (a, c, d), and all maximal chains have length 3 so it is distributive.
Aug 2, 2016 at 6:50 comment added Tri (Professor George Markowsky has asked me to make the following comment.) Theorem 2 from Some Combinatorial Aspects of Lattice Theory, Proc. Univ. of Houston Lattice Theory Conf., 1973, 36-68 states: THEOREM 2. Let L be a finite lattice. The following are equivalent. (a) L has length n, satisfies the Jordan-Dedekind chain condition, has n join-irreducible elements and n meet-irreducible elements. (b) L has n join-irreducible elements, and every connected (maximal) chain between I and 0 has length n. (c) L is distributive and has n join-irreducible elements.
Jan 15, 2015 at 9:07 comment added arsmath You're completely right. My subconscious bias in favor of lattices of modules: revealed.
Jan 15, 2015 at 2:51 comment added Richard Stanley Birkhoff's theorem states that a modular lattice generated by two chains is distributive. It is not true that any lattice generated by two chains is distributive.
Jan 14, 2015 at 16:17 vote accept JohnDoe
Jan 14, 2015 at 15:50 history answered arsmath CC BY-SA 3.0