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 ..



Examples of complete distributive lattices that are not Heyting algebras
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.


