Compactly generated complete lattice is Heyting Algebra Let $L$ be a lattice. $a\in L$ is said to be compact if whenever $a\leq \bigvee A$ for some $A\subseteq L$, then there is some finite subset $B\subseteq A$ such that $a\leq \bigvee B$. A lattice is compactly generated if each $a\in L$ is the supremum of a set of compact elements.
Now let $L$ be a distributive lattice, which is complete and compactly generated. I am asked to show this lattice $L$ must be a Heyting algebra.
We know that a complete distributive lattice is a Heyting algebra if and only if the infinite distributivity law holds, that is:
$$a\wedge\bigvee_{i\in I}b_i = \bigvee_{i\in I} (a\wedge b_i)$$
I have tried fiddling around with the definitions, that is, rewriting all $a,b_i$ as suprema of compact elements, but without any result yet. How can I see this claim to be true?
 A: Complete compactly generated lattices are usually called algebraic. So we need to prove that every algebraic distributive lattice satisfies the infinite distributive law. The inequality from right to left is trivial.
Let $x = a \wedge \bigvee_{i \in I} b_i$. Since $L$ is algebraic we have $x = \bigvee C$ for some set $C$ of compact elements of $L$. For each $c \in C$ we have $c \leqslant \bigvee C = x = a \wedge \bigvee_{i \in I} b_i$ and hence $c \leq a$ and $c \leq \bigvee_{i \in I} b_i$. By the compactness of $c$ we have $c \leq b_{i_1} \vee \dots \vee b_{i_n}$ for some $i_1, \dots i_n \in I$. Hence $$c \leq a \wedge (b_{i_1} \vee \dots \vee b_{i_n}) = (a \wedge b_{i_1}) \vee \dots \vee (a \wedge b_{i_n}) \leq \bigvee_{i \in I} (a \wedge b_i).$$
Finally, we have
$$a \wedge \bigvee_{i \in I} b_i = x = \bigvee C \leq \bigvee_{i \in I} (a \wedge b_i),$$
where the last inequality holds since for each $c \in C$ we have $c \leq  \bigvee_{i \in I} (a \wedge b_i)$.
Note that for arbitrary (not necessarily distributive) algebraic lattice $L$ this law holds in the weaker form, namely, when $\{b_i \mid i \in I\}$ is a directed subset of $L$. See the related question.
