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:

- $h$ 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

$h(a)leq b$. - $h$ is dense in the sense that every element of $B$ is the

join of elements in $h(A)$. - $h$ 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.