# Boolean algebras and free filters generated by chains

Suppose $\mathbf{B}$ is a complete Boolean algebra with an infinite domain $B$. Suppose $\mathbf{B}$ is atomic (i.e. every element is the supremum of some set of atoms). This algebra contains the co-finite filter $\mathscr{F}_c$, which in case the set of atoms of $\mathbf{B}$ is countable is generated by a chain. This follows from the fact that we may arrange all atoms in a sequence $a_0,a_1,\ldots$ and take: $$C:=\{B-\{a_0,\ldots,a_n\}\mid n\in\omega\}$$ as a chain that generates $\mathscr{F}_c$. It is known that no free ultrafilter in $\mathbf{B}$ is generated by a chain (see this answer).

In light of the above my first question is:

(1) If $\mathbf{B}$ has a countable set of atoms, can it contain a free filter which is different from $\mathscr{F}_c$ and which is generated by a chain?

Considering an uncountable case:

(2) If $\mathbf{B}$ is atomic with uncountably many atoms, is the co-finite filter $\mathscr{F}_c$ generated by a chain?
(3) If $\mathbf{B}$ is atomic with uncountably many atoms, can it contain a free filter which is different from $\mathscr{F}_c$ and which is generated by a chain?

For Q1, take the Boolean algebra of subsets of the rational numbers, and let the filter be generated by all of the nonempty final segments of $\mathbb Q$.

For Q3, repeat the preceding with the reals in place of the rationals. For other uncountable cardinals, repeat the preceding with the rationals and reals replaced by, for example, the initial ordinal of the desired cardinality.

For Q2, the answer is no, because any chain of cofinite sets is necessarily countable; it can't contain two distinct sets whose (finite) complements have the same cardinality.

The OP requested, in a comment, more details on Q2; I hope the following clarifies it sufficiently. Consider any chain $\mathcal C$ of cofinite sets. If $A$ and $B$ are distinct elements of $\mathcal C$, their complements are distinct finite sets, with one included in the other. So these complements have distinct, finite cardinalities. That is, the function assigning to each $A\in\mathcal C$ the cardinality of its complement is a one-to-one function from $\mathcal C$ to $\mathbb N$, and therefore $\mathcal C$ is countable. Each of the countably many elements $A\in\mathcal C$ omits only finitely many atoms, so only countably many atoms are omitted by any members of $\mathcal C$. Since there are uncountably many atoms altogether, let $a$ be one that is not omitted by any $A\in\mathcal C$. Then the complement of $a$ is in the cofinite filter but is not above any member of $\mathcal C$. Therefore, $\mathcal C$ fails to generate the cofinite filter.

• In countable and continuum case - is this a canonical method, in the sense that every free filter which is not co-finite and is generated by a chain must be generated by one of the aforementioned form? – Mad Hatter Oct 5 '15 at 12:26
• I think any example has to look at least somewhat similar to the ones I suggested. Given a chain $C$ that generates a filter on a set $X$, you can define a linear pre-order (like a linear order except that distinct elements cam be $\leq$ each other) by putting $x\leq y$ when every set in $C$ that contains $x$ also contain $y$. Then extend this pre-order relation to a linear order, and notice that all elements of $C$ are final segments. It might not, however, be the case that all final segments are in the filter generated by $C$. – Andreas Blass Oct 5 '15 at 12:34
• Just to clarify the method, for $\mathbb Q$ for example. For $S\subseteq\mathbb Q$, let $U(S)$ be the set of all its upper bounds, i.e. the final segment. So the free filter $\mathscr{F}$ generated by the set $\{U(S)\mid S\subseteq\mathbb Q\}$, right? Or am I misunderstanding something? – Mad Hatter Oct 5 '15 at 12:58
• What you wrote about $U(S)$ is correct, as long as $S$ is bounded above so $U(S)$ isn't empty. (I don't want the empty set to get into my filter.) An equivalent description, in the case of $\mathbb Q$, is that the filter is generated by sets of the form $\{x\in\mathbb Q:r<x\}$ or $\{x\in\mathbb Q:r\leq x\}$for arbitrary real numbers $r$. – Andreas Blass Oct 5 '15 at 19:15
• I"ve added some more details about Q2 to my answer. – Andreas Blass Mar 15 '16 at 23:48