# Filters of Boolean algebras which are Boolean algebras

Looking at some filters generated by elements of a finite Boolean algebra I have the impression that many/most/all of them are by themselves Boolean algebras (at least I didn't stumble upon a counterexample). For example the filter generated by $\lbrace 1,2\rbrace$ and $\lbrace 1,3\rbrace$ in the power set algebra of $\lbrace 1,2,3,4\rbrace$ is isomorphic to the power set algebra of $\lbrace 1,2,3\rbrace$ - with $\lbrace 1\rbrace$ instead of $\emptyset$ being the minimal element and accordingly $\lbrace 1,3,4\rbrace$ instead of $\lbrace 3,4\rbrace$ being the negation of $\lbrace 1,2\rbrace$.

My question is:

Which filters of which Boolean algebras are by themselves Boolean algebras?

If the answer is not "all filters of all Boolean algebras" what is the smallest/simplest counterexample?

The Fréchet (cofinite) filter on $\mathbb{N}$ is not a Boolean algebra, the simplest reason being that there is no unique bottom element.
Furthermore, a filter $F$ in a Boolean algebra $B$ is itself a Boolean algebra (under the inherited join and meet operations) iff it is principal.
($\Rightarrow$) If $F \subseteq B$ is the filter $\{ x \in B : a \leq x \}$ for some $a \in B$, it is then easy to show that $F$ is a Boolean algebra with bottom element $a$. ($\Leftarrow$) If $F \subseteq B$ is a filter which is also a Boolean algebra under the inherited meet and join operations, consider the bottom element $a$ of $F$. It follows that $F = \{ x \in B : a \leq x \}$, and so is principal.
• @HansStricker: Yes, I changed "finitely generated" to principal because they are the same. If $F$ is generated by $a_1 , \ldots , a_n$, then it is also generated by their meet. – user642796 May 8 '13 at 10:35