For a boolean algebra B, let d(B) be the least cardinality of a dense subset of B. Let A be a (nonregular) subalgebra of a boolean algebra B. Is it possible that d(A) > d(B)? What if d(B) = $\aleph_0$?
Yes, it is possible.
Let $\mathbb{B}$ be any complete Boolean algebra with density $\aleph_0$. For example, we could use the power set Boolean algebra $\mathbb{B}=P(\mathbb{N})$, which has density $\aleph_0$ in light of the singleton sets, since every nonempty set contains a singleton.
Meanwhile, the BalcarFranek theorem asserts that every infinite complete Boolean algebra $\mathbb{B}$ contains a free subalgebra of the same size. Thus, $\mathbb{B}$ contains a free subalgebra $\mathbb{A}$ on $2^{\aleph_0}$ many generators, obtained from strongly independent sets, as the linked article explains. Since $\mathbb{A}$ is free on $2^{\aleph_0}$ generators, it must have density $2^{\aleph_0}$.
So this is an example, for which $d(\mathbb{B})=\aleph_0$ and $d(\mathbb{A})=2^{\aleph_0}$.
The same reasoning establishes:
Theorem. Every infinite complete Boolean algebra of size $\kappa$ contains a subalgebra with density $\kappa$.
Proof. The BalcarFranek theorem provides a free subalgebra of size $\kappa$, and by mapping the generators to atoms in another Boolean algebra, it is easy to see that such an subalgebra must have density $\kappa$. QED
Since there are numerous complete Boolean algebras with density smaller than their size, the theorem can be used to build numerous examples.
Addendum. In the context of forcing, it is natural to consider the case of complete Boolean algebras, for every forcing notion corresponds with a complete Boolean algebra $\mathbb{B}$, and the intermediate models of the forcing extension arise exactly from complete subalgebras of $\mathbb{B}$. A Boolean algebra $\mathbb{B}$ is complete if every subset of it has a least upper bound (and this implies one may take infinitary meets and joins). The density of the completion of a Boolean algebra (or a poset) is the same as the original density. A subalgebra $\mathbb{A}\subset\mathbb{B}$ is a complete subalgebra if infinitary meets and joins for subsets of $\mathbb{A}$, as computed in $\mathbb{B}$, exist in $\mathbb{A}$. In this setting, the density cannot go up:
Theorem. If $\mathbb{A}\subset\mathbb{B}$ is a complete subalgebra of a complete Boolean algebra $\mathbb{B}$, then $d(\mathbb{A})\leq d(\mathbb{B})$.
Proof. Suppose $D\subset \mathbb{B}$ is dense. For each $d\in D$, let $a_d$ be the meet of all the elements of $\mathbb{A}$ above $d$. This is an element of $\mathbb{A}$, and the set of all such $a_d$ is dense in $\mathbb{A}$, since for any $a\in \mathbb{A}$ there is an element $d\in D$ below $a$, and so $a_d\leq a$. Thus, we have produced a dense subset of $\mathbb{A}$ of size at most $D$, as desired. QED
Thus, when forcing, the relevant subalgebras arising from intermediate forcing extensions do respect the density.
Here is a construction that I think will work:
Take $B$ to be the Boolean algebra $P(\omega )$ with the usual operations. This has a countable dense subset. Now let $IC$ denote the collection of infinite subsets of $\omega$ with infinite complements. We will construct a family $A$ consisting of $\aleph _1$ sets, all of which are in $IC$, and such that $A$ is closed under complements, unions, and intersections; so after we throw in the nullset and total set, $A$ will be a subalgebra of $B$. It will also follow from our construction that $A$ cannot have a countable dense subset.
So here's the construction: we build $A$ inductively, in $\omega _1 $ countable stages $ (S _ \alpha ) _{ \alpha \in \omega _1 } $ which are cumulative. At each successor stage, we will add in a single new set to the previous stage, then close up under complements, unions, and intersections. So in fact there will be countably many new sets added at each successor stage. At limit stages we take the union of all previous stages. Note that each stage is closed under Boolean operations.
So we just have to say which new set $X$ we throw in at each successor stage $S_{\alpha + 1} $. We need to be sure that when we add in this new set and close up under Boolean operations, all the sets we get are still in $IC$. For this it suffices that for every $Y \in S_{\alpha}$, we have $X \cup Y \in IC$ and $X \cap Y \in IC$. In other words, $X$ and $\omega \setminus X$ must both have infinite intersection with every set in $S_{\alpha}$. And since by induction hypothesis every set in $S_{\alpha}$ is in $IC$, and there are only countably many of them, we can make our set $X$ by a routine diagonalization: for instance, we could line up pairs $(Y, n)$ for $Y \in S_{\alpha}$, $n \in \omega$ so that the pairs have ordertype $\omega$, and then go through them one by one, building initial segments of $X$ and $\omega \setminus X$ as we go to make sure that at the step $(Y, n)$, $X$ and $\omega \setminus X$ each have at least $n$ elements of $Y$ in them.
This completes the construction of $A$. Note that $A$ cannot have a countable dense subset, since any countable subset of $A$ shows up at some stage $S_{\alpha}$, and then at the next stage we added in a set which is not contained in any $Y \in S_{\alpha}$.
EDIT: A SECOND CONSTRUCTION. Here is another construction, which yields $2^{\aleph _0 }$ independent subsets of $\omega$. (Thanks to Joel for the hint about trees!) We will make a continuumsized bunch of independent reals (subsets of $\omega$), by building their initial segments in $\omega$ many stages. The idea is that we are working our way up a binary tree, level by level, and so every level is finite; and at each such level, we ensure that no nontrivial Boolean relations can hold between two reals whose initial segments at that level are different. More precisely:
We represent subsets of $\omega$ as sequences of 0's and 1's, where $n$ is in the subset iff the $n$th spot in the sequence is 1. So when we build finite initial segments of these sequences, we can specify some natural numbers to definitely be in the subset, and also specify some natural numbers to definitely NOT be in the subset. So: Start by making two different finite initial segments of reals, such that no nontrivial Boolean relations can hold between reals with those initial segments. (By 'nontrivial Boolean relation' I mean: nontrivial equation with Boolean connectives, where the only elements we have names for are the nullset, the total set, and the reals bearing the relation in question.) Since there are only finitely many possible nontrivial Boolean relations among a finite set of elements, and any particular relation can be forced to not hold by extending the initial segments in an appropriate way, we can get two different initial segments such that any two reals with these respective initial segments are already forced to not satisfy any nontrivial Boolean relations with each other. Next we fork each of these initial segments off into two longer initial segments; so now we have four initial segments, and we consider all possible nontrivial Boolean relations which could hold among these four elements. As before, extend the four segments until all such relations are forced to be false. Continue splitting segments into two, and forcing all finitely many possible Boolean relations to be false, $\omega$ many times. Then we will have a perfect binary tree of independent reals.

$\begingroup$ Very nice. What you seem to be building are strongly independent sets, so that your subalgebra will be free. Perhaps you can rearrange your construction (as a tree?) to get $2^{\aleph_0}$ many independent sets? $\endgroup$ – Joel David Hamkins Aug 8 '11 at 21:17

$\begingroup$ Very good, thanks! If we want to start with an atomless algebra, then I think we can take $B$ to be the boolean completion of Cohen forcing, then fix an infinite maximal antichain in it, and use that in place of ω above. Then build the $\aleph_1$ independent elements using your same procedure, taking sums of the chosen subsets of the antichain. $\endgroup$ – Monroe Eskew Aug 8 '11 at 22:45

$\begingroup$ The process of building the new set at successor stages can also be described as taking a set which is Cohengeneric over a countable collection of dense sets $D_{(Y,n)}$, defined by the same criterion used to build $X$ by initial segments. Therefore, assuming MA, the same argument can be used to get $2^{\aleph_0}$ independent sets. $\endgroup$ – Monroe Eskew Aug 9 '11 at 5:27

$\begingroup$ Yes, building a tree works to get $2^{\aleph _0 }$ independent sets! But I ended up using a somewhat different construction; it's posted under my original answer. This construction will not, however, generalize to other cardinals; it essentially uses the fact that $w$ is strongly inaccessible. A small formatting question: how can I make a horizontal line, or something to split the above answer into two parts? I tried the usual LaTeX command but it didn't seem to work... (sorry if this is inappropriate for a comment post!) $\endgroup$ – Andy Voellmer Aug 9 '11 at 7:00

$\begingroup$ Andy, your construction is reminiscent of the following folklore construction: add a Cohen real and use its binary digits to label the nodes of the tree $2^{\lt\omega}$. It follows that any finitely many branches through this tree are mutually generic Cohen reals, since any partial labeling can be extended so as to meet any of the relevant dense sets. (Thus, adding one Cohen real creates a continuumsized family of finitelymutually generic Cohen reals in the extension.) The idea adapts to your argument, since countably many dense sets ensure independence, so the whole construction is in V. $\endgroup$ – Joel David Hamkins Aug 9 '11 at 16:27