# Johnstone on constructing a complete Boolean algebra from a complete Heyting algebra

Johnstone in his Stone Spaces constructs a complete Boolean algebra from a complete Heyting algebra. The corollary of paragraph 2.6 says

Every frame is isomorphic to a subframe of a complete Boolean algebra.

The proof goes like this:

Given a frame $A$, let $B = (N(A))_{\neg\neg}$. Then $B$ is a complete Boolean algebra [...]; and by the Lemma the monomorphism $c : A \to N(A)$ factors through $B$. [...]

Here $N(A)$ is the complete Heyting algebra of nuclei of $A$, ordered pointwise. It is easy to see that $\neg\neg$ is a nucleus. I understand that the complete Boolean algebra that the corollary claims exists is $B$, and that the subframe is the image of $c$. The lemma that the proof refers to is this:

(i) For any $a \in A$, the open and closed nuclei $u(a), c(a)$ are complementary elements of $N(A)$.

(ii) The map $a \mapsto c(a)$ is a frame monomorphism $A \to N(A)$.

Note that $c(a) = a \vee (\cdot)$ and $u(a) = a \to (\cdot)$.

My problem is that I fail to understand what he means by "by the Lemma." It is unclear to me which part of the lemma the proof of the corollary refers to. Presumably, it refers to (ii); but I fail to see why the monomorphismhood is useful here. Does it not suffice to show that the range of $c$ is in $N(A)_{\neg\neg}$? (I do not know how to show it, however...)

How can this corollary be proved?

He's referring to part (i). Since $c(a)$ and $u(a)$ are complementary elements of $N(A)$, $c(a)=\neg u(a)$ and $u(a)=\neg c(a)$. Thus $c(a)=\neg u(a)=\neg\neg c(a)\in B$, and the image of $c$ is contained in $B$.