For a boolean algebra B, let d(B) be the least cardinality of a dense subset of B. Let A be a (non-regular) subalgebra of a boolean algebra B. Is it possible that d(A) > d(B)? What if d(B) = $\aleph\_0$?



density of boolean algebras
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 [Balcar-Franek theorem](http://www.cas.mcmaster.ca/~franek/journals/semifree_view/page1.html) 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 Balcar-Franek 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. 




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*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. 


