# Covering property of complete distributive lattices

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 $a\leq b \in L$.)

• Is your lattice completely distributive? (i.e. do arbitrary joins distribute over arbitrary meets?) – Ilya Bogdanov Sep 29 '17 at 9:19
• Thanks for the question! No - just finitely distributive, not necessarily completely – Dominic van der Zypen Sep 29 '17 at 12:14

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.
• $\mathbb Z$ is not a complete lattice. As far as I understand from OP's comment, the author means a complete lattice which is (finitely) distributive... – Ilya Bogdanov Sep 29 '17 at 20:03
• @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 – Dominic van der Zypen Oct 1 '17 at 6:19