Proof of $CFE \implies BPI$

Question

(CFE): Every filter of closed sets can be extended to a maximal one.

(BPI): Every Boolean algebra contains a prime ideal.

I am reading Herrlich's and Stepran's paper "Maximal filters, continuity and choice principles" where on page 700 it is stated that $CFE \implies BPI$ obviously holds.

Unfortunately, it's not obvious to me. My first idea was that it might hold that a maximal closed filter is prime. But I don't see how to prove it. To recall the definitions, a filter $F$ is prime iff $x \lor y \in F$ implies $x \in F$ or $y \in F$ where $\lor$ denotes the join/supremum. In a partial order of closed sets the order $A \le B$ would be defined by $A \subseteq B$ (or not?). Then $A \lor B$ would be $A \cup B$ and it is not clear to me how $A \cup B \in F$ implies that either $A \in F$ or $B\in F$. And I think the following provides a counter example: Let $X$ be a topological space endowed with the trivial topology. Let $A,B$ be any two non-empty disjoint subsets with $A \cup B = X$. Then $F = \{A \cup B\}$ is a maximal filter on $X$ that is not prime since neither $A$ nor $B$ are in it.

Hence my question: would someone post a proof of the obvious implication $CFE \implies BPI$? Thanks for your help.

Answer 1

I was making my life too complicated in earlier versions.

Note, first, that $\mathsf{BPI}$ is equivalent (under $\mathsf{ZF}$) to the "Ultrafilter Theorem": Every filter of subsets of a set $S$ can be extended to an ultrafilter. (See Jech, The Axiom of Choice, Theorem 2.2, p.17.)

If $\mathcal{F}$ is any filter of subsets of a set $S$, give $S$ the discrete topology. As $\mathcal{F}$ is now a filter of closed subsets of $S$, by $\mathsf{CFE}$ it can be extended to a maximal filter $\mathcal{G}$ of closed sets. Since we have taken the discrete topology, it is easy to show that $\mathcal{G}$ is actually an ultrafilter on $S$ extending $\mathcal{F}$.

Answer 2

Your idea is correct, but we have to work only in the context of Boolean algebras, because that is what $\sf BPI$ gives us.

In this context prime filters are maximal filters. Suppose $B$ is a Boolean algebra, $F\subseteq B$ is a prime filter, and $a\in B$ is any element. Then $a+(-a)\in F$, and therefore either $a\in F$ or $-a\in F$. So we cannot extend $F$, otherwise there would be some $a\in B$ such that $a\notin F$ and $-a\notin F$ either.

Now the implication should be more obvious. Assume $\sf CFE$, and let $B$ be a Boolean algebra. Take $F$ to be a filter on $B$, in order to use $\sf CFE$ we need to find a topological space such that $B$ is isomorphic to its closed sets, but that's easy if we think of $B$ with its order as a topological space.

Answer 3

Theorem $2$ of Kyriakos Keremedis and Eleftherios Tachtsis, On the extensibility of closed filters in $T_1$ spaces and the existence of well orderable filter bases, Commentationes Mathematicae Universitatis Carolinae, Vol. 40 (1999), No. 2, 343-353, is that $CFE_0$ is equivalent to $AC$, where $CFE_0$ is the statement that every closed filter in a $T_0$ space extends to a maximal closed filter. The proof is given in its entirety. Since we know that $AC\to BPI$, this establishes the result.