**Definition:** Let $B$ be a boolean algebra. Say $X \subseteq B$ is *quasi-dense* in $B$ if for all $b \in B$, there is $x \in X \setminus$ { $0,1$ } such that either $x \leq b$ or $b \leq x$.


**Question:** Suppose $A \subseteq B \subseteq C$ are atomless boolean algebras, $A$ is quasi-dense in $B$, and $B$ is **dense** in $C$. Does it follow that $A$ is quasi-dense in $C$?



Quasi-dense subsets of boolean algebras
Here is a way to construct an atomless version of Joel´s counterexample: 


Let $A\_0 \subseteq A\_1 \subseteq A\_2$ be the algebras in Joel´s example (in his notation $A \subseteq B \subseteq C$) and let $X\_0$,$X\_1$ and $X\_2$ be the corresponding Stone spaces. So $X\_0$ and $X\_1$ are both just a converging sequence and $X\_2$ consists of two converging sequences (with different limit points). These spaces have isolated points, reflecting the fact that the algebras have atoms. So the idea now is to consider the (Stone) spaces $Z\_i=X\_i \times 2^\omega$ and their corresponding algebras of clopen subsets $B\_i=Cl(Z\_i)$ for $i \in 3$. These algebras are now atomless (since their Stone spaces have no isolated points). 


The inclusion maps $A\_0 \subseteq A\_1$ and $A\_1 \subseteq A\_2$ induce surjective continuous functions $\pi\_1:X\_1 \to X\_0$ and $\pi\_2:X\_2 \to X\_1$ respectively, which we can use to produce the (also continuous and surjective) functions: $\pi\_1 \times id\_{2^\omega}: Z\_1 \to Z\_0$ and $\pi\_2 \times id\_{2^\omega}:Z\_2 \to Z\_1$. These in turn induce "inclusions" $B\_0 \subseteq B\_1 \subseteq B\_2$. Arguments analogous to those in Joel´s answer can be used to show that: 


**1)** $B\_0$ is quasi-dense in $B\_1$ (here you use that any clopen set in $Z\_1$ is a finite union of boxes). 


**2)** $B\_1$ is dense in $B\_2$ (using that any clopen set in $Z\_2$ contains a box), and


**3)** $B\_0$ is not quasi-dense in $B\_2$ (here a problematic clopen subset of $Z\_2$ would be $K \times 2^\omega$ where $K$ is one of the two converging sequences).


