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Let $(L,\land,\lor)$ be a complete distributive lattice. Given $x\neq y \in L$, is there a finite set ${\cal I}$ of closed intervals in $L$ such that

  1. no member of ${\cal I}$ contains both $x$ and $y$, and
  2. $\bigcup {\cal I} = L$

?

(A closed interval in $L$ is a subset of the form $[a, b] = \{x\in L: a\leq x\leq b\}$ where $a\leq b \in L$.)

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  • $\begingroup$ Is your lattice completely distributive? (i.e. do arbitrary joins distribute over arbitrary meets?) $\endgroup$
    – Ilya Bogdanov
    Commented Sep 29, 2017 at 9:19
  • $\begingroup$ Thanks for the question! No - just finitely distributive, not necessarily completely $\endgroup$
    – Dominic van der Zypen
    Commented Sep 29, 2017 at 12:14

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I think you do mean completely distributive, not just finitely. Otherwise $\mathbb{Z}$ with its usual ordering is not a finite union of any set of closed intervals. For complete distributive lattices, let $L$ be the lattice of all measurable subsets of the unit interval $[0,1]$ modulo sets of measure 0, ordered by inclusion (up to sets of measure 0). This a complete distributive lattice (in fact, a complete boolean algebra). It is easy to see that if $L$ is written as a finite union of closed intervals, then one of these intervals is the whole lattice.

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    $\begingroup$ $\mathbb Z$ is not a complete lattice. As far as I understand from OP's comment, the author means a complete lattice which is (finitely) distributive... $\endgroup$
    – Ilya Bogdanov
    Commented Sep 29, 2017 at 20:03
  • $\begingroup$ That said, clearly +1 for a nice example (which is even more than OP asked for) --- which may seem incorrect at the first glance! $\endgroup$
    – Ilya Bogdanov
    Commented Sep 29, 2017 at 20:09
  • $\begingroup$ @richardstanley, the requirement about the intervals is equivalent to saying the interval topology is $T_2$, and $[0,1]\times[0,1]$ has $T_2$ interval topology, so I am not sure about your example $\endgroup$
    – Dominic van der Zypen
    Commented Oct 1, 2017 at 6:19
  • $\begingroup$ @DominicvanderZypen: oops, you are right. I have deleted this statement. $\endgroup$
    – Richard Stanley
    Commented Oct 2, 2017 at 16:44

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