Density and compactness of Boolean embeddings

Question

Let A and B be Boolean algebras and $h:A\rightarrow B$ a Boolean embedding.

• If every element of $B$ can be expressed both as a join of meets and as a meet of joins of elements in $h(A)$, then the embedding $h$ is called dense.

• Assume also that $B$ is complete. If for all $S,T\subseteq A$ with $\vee h(S)\leq\wedge h(T)$, there exist finite $S ′\subseteq S$ and $T ′\subseteq T$ such that $\vee S′\leq\wedge T′$ , then the embedding $h$ is called compact.

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Is there a way to write formally the definition of a dense Boolean embedding?
Is there some intuition of what is a dense Boolean embedding?
I know there is a relation between this definition of a compact embedding and the topological compactness, but why do you have to make this definition for embeddings, and not for Boolean algebras?
Is there any relation between a dense Boolean embedding and topological density?

Answer 1

Regarding the dense embedding, perhaps this is helpful. Statement 1 can be taken as a definition of density, which makes the connection with topology by means of the lower-cone topology.

Theorem. Suppose that $A$ and $B$ are Boolean algebras and $h:A\to B$ is an embedding, meaning an injective homomorphism, an isomorphism of $A$ with a subalgebra of $B$. 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 $b\in B$ there is nonzero $a\in A$ 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. ($1\to 2$) Suppose that $h$ is dense in the sense of (1), and consider any $b\in B$. Let $A_0=\{a\in A\mid h(a)\leq b\}$. 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 $a\in A_0$ and thus $h(a)\leq c$, contradiction.

($2\to 3$) Assume $h$ is dense in the sense of (2), and consider any $b\in B$. So $\neg b=\bigvee h[A_0]$ for some $A_0\subseteq A$. By De Morgan reasoning it follows that $b=\bigwedge_{a\in A_0} \neg h(a)$, and so $b$ is the meet of $h[\{\neg a\mid a\in A_0\}]$.

($3\to 1$) Assume $h$ is dense in the sense of (3), and consider any nonzero $b\in B$. So $1\neq \neg b\in B$ and $\neg b$ is the meet of $h[A_0]$ for some set $A_0\subseteq A$. So $\neg b\leq h(a)$ for some $1\neq a\in A_0$, and consequently $0\neq h(\neg a)\leq b$, as desired for (1). $\quad\Box$

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.