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 $c<b$ 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 $h(a)leq c$,
contradiction.

($2to 3$) 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.