Heyting algebras and infinite distributive law I want to prove that "a complete lattice satisfies the infinite distributive law $a\wedge(\vee{S})=\vee\{a\wedge s|s\in S\}$ iff it is a Heyting algebra".
I proved "if" part but can't "only if" part.
I defined $a\rightarrow b$ as $(\vee \{s|a\wedge s=0\})\vee b$ and tried to prove $a\rightarrow a=1$ but failed.
 A: In the interest of having an answer available to someone who arrives to this page, I have written the following answer.
Recall that a bounded lattice $L$ is a Heyting algebra when for each $a$ and $b$ in $L$ there exists an element $a \to b$ with the property that for each $w$ in $L$, $w \leq a\to b \Leftrightarrow a \wedge w \leq b$.
Now suppose that $L$ is a complete lattice satisfying the infinite distributivity law. Let us check that the correct definition of implication is $a \to b = \vee \{s\,|\,a\wedge s \leq b\}$. If $w \leq a \to b$, then $a \wedge w \leq a \wedge (\vee \{s\,|\,a\wedge s \leq b\}) = \vee\{a\wedge s\,|\,a\wedge s \leq b\}\leq b$. Conversely, if $a \wedge w \leq b$, then trivially $w \leq a \to b$ (since $w \in \{s\,|\,a\wedge s \leq b\}$).
The fact that every complete lattice $L$ which is a Heyting algebra satisfies the infinite distributive law follows from the fact that for each $a$ in $L$, the maps defined by $s \mapsto a\wedge s$ and $s \mapsto (a \to s)$ are the left and right adjoins of a Galois connection. 
Let us recall why this is so. Suppose that $X$ and $Y$ are partially orderded sets, and that $f:X\to Y$ and $g:Y\to X$ are monotone (i.e. order preserving) maps, we say that $f$ and $g$ are part of a Galois connection when $f(x) \leq y \Leftrightarrow x \leq g(y)$. When this is the case $f$ is called the left (or lower) adjoint (of $g$) and $g$ is called the right (or upper) adjoint (of $f$). It is easy to see that the infinite distributivity law above corresponds to the fact that the left adjoints $s\mapsto a\wedge s$ preserve joins. Let us prove that left adjoints always preserve joins. Suppose that the join of $S \subseteq X$ exists in $X$ and suppose for some $w\in Y$ that for each $s \in S$, $f(s)\leq w$. Using the Galois connection we have for each $s \in S$,  $s \leq g(w)$ and hence $\vee S \leq g(w)$. Using the Galois connection again we have that $f(\vee S) \leq w$ proving that $f(\vee S)=\vee\{ f(s)\,|\,s\in S\}$.
Let us end by making several remarks:


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*The first part of this answer can also be phrased as part of a more general theory. Suppose that $f: X\to Y$ is a monotone map between partially ordered sets. If $f$ preserves arbitary joins (i.e. whichever ones exist) and for each $y$ in $Y$, the join of $\{x\,|\,x \in X, f(x) \leq y\}$ exists, then $f$ is the left adjoint of a Galois connection with right adjoint $g$ defined by $g(y) = \vee\{x\,|\,x \in X, f(x) \leq y\}$.

*Everything can be generalizer further by replacing partial orders and monotone maps by categories and functors.

