Ultrafilter on a Boolean algebra

Remarks

The filters of $B$ are the nonempty order filters of $B$ seen as a lattice. For a proper filter $U$, these are equivalent. (a) $U$ is an ultrafilter. (b) For every $a \in B$, exactly one of $a$ and $\neg a$ lies in $U$. (c) $U$ is prime: $a \vee b \in U$ implies $a \in U$ or $b \in U$. Ultrafilters $U$ correspond to homomorphisms $\chi \colon B \to \set{0, 1}$, by $U = \chi^{-1}(1)$.

Example: in the power set of a set $X$, for $x \in X$, the family $\set{A \subseteq X : x \in A}$ is an ultrafilter. It is a proper filter, and for each $A$, either $x \in A$ or $x \in X \setminus A$.

Non-example: in the power set of $\set{1, 2}$, the family $F = \set{\set{1, 2}}$ is a proper filter but not an ultrafilter. It contains neither $\set{1}$ nor its complement $\set{2}$, and it lies in the larger proper filter $\set{\set{1}, \set{1, 2}}$.