Return to Revisions

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.