gn.general topology – Density and compactness of Boolean embeddings

Regarding the dense embedding, perhaps this is helpful. Statement 1 can be taken as a definition of density.

Theorem. Suppose that $$A$$ and $$B$$ are Boolean algebras and
$$h:Ato B$$ is an embedding, meaning an injective homomorphism. Then
the following are equivalent:

1. $$h$$ is dense in the sense that $$h(A)$$ is a dense subset of $$B$$,
meaning that for every nonzero $$bin B$$ there is nonzero $$ain A$$ with
$$h(a)leq b$$.
2. $$h$$ is dense in the sense that every element of $$B$$ is the
join of elements in $$h(A)$$.
3. $$h$$ is dense in the sense that every element of $$B$$ is the meet
of elements in $$h(A)$$.

Proof. ($$1to 2$$) Suppose that $$h$$ is dense in the sense of
(1), and consider any $$bin B$$. Let $$A_0={ain Amid h(a)leq b}$$. So $$h(A_0)$$ lies entirely below $$b$$, but the join of $$h(A_0)$$
must equal $$b$$, for otherwise there is some $$c which is an upper
bound of $$h(A_0)$$. In this case, $$b-c$$ is nonzero and so has some
nonzero $$h(a)leq b-c$$. So $$ain A_0$$ and thus $$h(a)leq c$$,
($$2to 3$$) Assume $$h$$ is dense in the sense of (2), and consider
any $$bin B$$. So $$neg b=bigvee h(A_0)$$ for some $$A_0subseteq A$$.
By De Morgan reasoning it follows that $$b=bigwedge_{ain A_0} neg h(a)$$, and so $$b$$ is the meet of $$h({neg amid ain A_0})$$.
($$3to 1$$) Assume $$h$$ is dense in the sense of (3), and consider
any nonzero $$bin B$$. So $$1neq neg bin B$$ and $$neg b$$ is the
meet of $$h(A_0)$$ for some set $$A_0subseteq A$$. So $$neg bleq h(a)$$ for some $$1neq ain A_0$$, and consequently $$0neq h(neg a)leq b$$, as desired for (1). $$quadBox$$
The theorem shows that a dense embedding is one whose range is dense in the lower-cone topology on $$B$$, which answers your final question.