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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 ..

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Fixed the typo in the WP article. –  Emil Jeřábek Jan 23 at 17:07

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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.

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The situation mentioned in Andreas Blass's answer happens for most complete Heyting algebras that one encounters. In other words, the dual of a frame is usually not a frame even though it is a distributive lattice. For example, the lattice of closed subsets of any non-discrete $T_{1}$-space is not a Heyting algebra.

A bounded distributive lattice $L$ is said to be subfit if whenever $a,b\in L$ and $a\vee c=1\Rightarrow b\vee c=1$ for all $c$, then $a\leq b$. For example, if $(X,\mathcal{T})$ is a $T_{1}$-space, then the frame $\mathcal{T}$ is subfit. Therefore subfitness is a very weak separation axiom that works particularly well in point-free topology. Therefore, most frames that people study in point-free topology are subfit.

Let $L$ be a complete lattice. Then we say that an element $a\in L$ is linear if whenever $I$ is a set and $b_{i}\in L$ for $i\in I$, then $$a\wedge\bigvee_{i\in I}b_{i}=\bigvee_{i\in I}(a\wedge b_{i}).$$

We say that an element $a\in L$ is colinear if whenever $I$ is a set and $b_{i}\in L$ for $i\in I$, then

$$a\vee\bigwedge_{i\in I}b_{i}=\bigwedge_{i\in I}(a\vee b_{i}).$$

$\mathbf{Lemma}$ Every complemented element in a complete distributive lattice is linear and colinear.

$\mathbf{Proof}$ Since the notions of linearity and colinearity are dual, we only need to show that each element on $L$ is linear. Suppose that $L$ is a complete distributive lattice and $a,c\in L$ are complemented elements. Take note that $$a\vee(a\wedge\bigvee_{i\in I}b_{i})=a=a\vee\bigvee_{i\in I}(a\wedge b_{i}).$$ Furthermore, $$c\vee(a\wedge\bigvee_{i\in I}b_{i})=(c\vee a)\wedge(c\vee\bigvee_{i\in I}b_{i})=(c\vee\bigvee_{i\in I}b_{i})$$ $$=\bigvee_{i\in I}(b_{i}\vee c)=\bigvee_{i\in I}((b_{i}\vee c)\wedge(a\vee c))=\bigvee_{i\in I}(b_{i}\wedge a)\vee c=(\bigvee_{i\in I}(b_{i}\wedge a))\vee c.$$

Therefore.

$$a\wedge\bigvee_{i\in I}b_{i}=(a\wedge c)\vee(a\wedge\bigvee_{i\in I}b_{i})= (a\vee(a\wedge\bigvee_{i\in I}b_{i}))\wedge(c\vee(a\wedge\bigvee_{i\in I}b_{i}))$$ $$=(a\vee\bigvee_{i\in I}(a\wedge b_{i}))\wedge(c\vee\bigvee_{i\in I}(a\wedge b_{i})) =(a\wedge c)\vee\bigvee_{i\in I}(a\wedge b_{i})=\bigvee_{i\in I}(a\wedge b_{i}).$$ $\mathbf{QED}$

$\mathbf{Theorem}$ Let $L$ be a complete distributive fit lattice. Then an element $a\in L$ is complemented if and only if it is colinear.

$\mathbf{Proof}$ $\rightarrow$. This direction has been proven already in the above lemma.

$\leftarrow$. Suppose $a\in L$ is colinear. Let $c=\bigwedge\{x\in L|a\vee x=1\}$. Then $a\vee c=1$ as well by colinearity. Now suppose that $a\wedge c\neq 0$. Then by subfitness, there is some $x\neq 1$ with $1=x\vee(a\wedge c)=(x\vee a)\wedge(x\vee c)$. Therefore $x\vee a=1$, so $x\geq c$, hence $x\geq x\vee c=1>x$, a contradiction. We therefore conclude that $a\wedge c=0$, so $a$ and $c$ are complements. $\mathbf{QED}$

In particular, if $L$ is a subfit frame and $L^{*}$ is the lattice with the same underlying set as $L$ but with the reverse ordering, then $L^{*}$ is a frame if and only if $L$ is a complete Boolean algebra. Also, if $X$ is a $T_{1}$-space and $\mathcal{C}$ is the lattice of closed sets of $X$, then $\mathcal{C}$ is a frame if and only if $X$ is discrete.

All these facts about frames can be found in Picado and Pultr's new book [1].

  1. Picado, Jorge, and Aleš Pultr. Frames and Locales: Topology without Points. Basel: Birkhäuser, 2012.
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