Here is a short question with a possibly simple and short answer:
I need an example of a complete distributive lattice that is not a Heyting algebra which should be an infinite complete lattice that does not satisfy the infinite distributivity law (in the finite world all lattices are complete). I haven't seen this matter addressed in literature yet (not even as an exercise) but it turns out that the wikipedia page on Heyting algebras (http://en.wikipedia.org/wiki/Heyting_algebra) might be giving the wrong idea in the second paragraph:
"Every Boolean algebra is a Heyting algebra when a=>b is defined as usual as \neg a v b, as is every complete distributive lattice[clarification needed] when a=>b is taken to be the supremum of the set of all c for which a ^ c \leq b."
Thank you in advance. By the way, the great book of D. Scott et al, "Continuous lattices and domains" might be giving an answer in O-45, page 39. But I didn't get there yet ..
Best Answer
I think you can get this by just dualizing a suitable complete Heyting algebra. Consider, for example, the lattice of open subsets of $\mathbb R$ ordered by $\subseteq$. This is a Heyting algebra, so it satisfies the infinite distributive law $a\wedge\bigvee_ib_i=\bigvee_i(a\wedge b_i)$, but it does not satisfy the dual law $a\vee\bigwedge_ib_i=\bigwedge_i(a\vee b_i)$. For example, let $b_i$ be the interval $(-\frac1i,\frac1i)$ and let $a=\mathbb R-\{0\}$. (Remember that the meet in this lattice is the interior of the intersection, so $\bigwedge_i(-\frac1i,\frac1i)=\varnothing$.) Therefore, the dual of this lattice, which is still complete and still distributive (because the finite distributive law implies its dual), cannot be a Heyting algebra.