This is in some sense following up on my earlier question and the answer given by NN.

I am currently revising the paper which used the condition mentioned in my question. It was pointed out in NN's answer that the concept is an old and standard one, at least to lattice theorists -- namely, it turns out I rediscovered the notion of the "breadth" of a semilattice.

I would like to find references that I can cite for the following facts, both of which appear in my paper in order to provide examples:

  1. The breadth of a semilattice is less than or equal to its height.

  2. The breadth of a semilattice is less than or equal to its width.

These are easy to prove: but it seems likely that these should be in one of several compendious works on lattice theory or more general partially ordered sets, and I would like to pay my dues to the literature, as it were. Does anyone know of a reference stating these results explicitly?

Definitions of terms. Let $(P, \preceq)$ be a partially ordered set. The height of $P$ is the sup of the lengths of chains in $P$. The width of $P$ is the cardinality of the largest antichain in $P$ (if it is finite, it is also equal, by Dilworth's theorem, to the minimum "number" of chains in $P$ whose union is all of $P$).

Now suppose $P$ is a meet-semilattice, so that each pair $(x,y)$ of elements has a greatest lower bound $x\wedge y$. The breadth of $P$, as far as I can gather from reading fragments of various papers found by searching online, is defined as follows: let $E$ be a finite subset of $P$, let $E^{[n]}$ be the set of all elements of the form $x_1\wedge \dots \wedge x_n$, and let $$ b(E)=\min\{ n\geq 1 \mid E^{[n]}=E^{[n+1]} \} \leq 2^{|E|}. $$ Then the breadth of $P$ is the supremum of $b(E)$ as we range over all finite subsets $E\subseteq P$.

  • $\begingroup$ Too late, but Gratzer's book (2011 edition) should have useful material. $\endgroup$ – user46855 Feb 16 '14 at 22:33
  • $\begingroup$ @user46855 Thanks - I think I tried quickly checking it before I posted this question, and just found these given as exercises. But perhaps I will check again if I find time. $\endgroup$ – Yemon Choi Feb 27 '14 at 4:30

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