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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 $C$?

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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).

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Well, this still doesn't answer the atomless question, but I've got a violation of the desired implication among atomic Boolean algebras.

Let $A$ consist of the finite or cofinite subsets of $\mathbb{N}$ that take $2k$ and $2k+1$ together, if at all. That is, $a\in A$ if $a\subset\mathbb{N}$ is finite or cofinite and for every $k$ we have $2k\in a\leftrightarrow 2k+1\in a$. Let $B$ be the Boolean algebra consisting of all finite or cofinite subsets of $\mathbb{N}$.

Note that $A$ is quasi-dense in $B$, since if $b$ is finite, then $b$ is contained in an interval $[0,2k+1]$ for some large $k$, and this is in $A$, and if $b$ is cofinite, then $b$ contains some final segment interval $[2k,\infty)$, which is in $A$.

Let $C$ be the algebra generated by the elements of $B$ together with the set $E$ of even numbers. Thus, every element of $C$ is the union of a finite or cofinite subset of $E$ with a finite or cofinite subset of $\mathbb{N}-E$. The algebra $B$ is dense in $C$, since the singletons are dense, and they are finite.

Finally, $A$ is not quasi-dense in $C$, because the set $E$ contains no nonzero element of $A$, as it contains no odd numbers, and is contained in no non-unital element of $A$, as the only element of $A$ containing all the even numbers is the whole of $\mathbb{N}$.


This is my original answer, which shows merely that quasi-density is not transitive.

The answer is no. For a counterexample, let $A$ have at least two atoms; let $B$ split one of those atoms, and let $C$ split both of them.

More explicitly, let $A$ be the 4-element Boolean algebra with atoms $\{0,1\}$ and $\{2,3\}$. Let $B$ be the $8$-element algebra with atoms $\{0\}$, $\{1\}$, $\{2,3\}$, and let $C$ be the full power set, with atoms $\{0\}$, $\{1\}$, $\{2\}$, $\{4\}$.

You may observe that $A$ is quasi-dense in $B$ and $B$ is quasi-dense in $C$ by inspection. But $A$ is not quasi-dense in $C$, since $\{0,2\}$ is neither above nor below any nontrivial element of $A$.

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The question was about "$B$ dense in $C$", not just quasi-dense. –  Goldstern Feb 9 '12 at 23:33
    
But I said $B$ was dense in $C$, not just quasi-dense. This was not a typo. Also, I am only really interested in the case of atomless algebras. I will put that in the original question. –  Monroe Eskew Feb 9 '12 at 23:36
    
Oh, sorry, I misread the question. –  Joel David Hamkins Feb 9 '12 at 23:57
    
I have now posted a counterexample to the implication among atomic Boolean algebras. Perhaps it could be used to construct an atomless counterexample, but I don't see it yet. –  Joel David Hamkins Feb 10 '12 at 4:46
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It's a surprisingly difficult question! –  Joel David Hamkins Feb 11 '12 at 0:16
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