Proposition 6

Let $\mathfrak{G}$ be a subbase of a filter on a set $X$. If for each subset $Y$ of $X$ we have either $Y \in \mathfrak{G}$ or $\complement Y \in \mathfrak{G}$, then $\mathfrak{G}$ is an ultrafilter on $X$.

Let $\mathfrak{F}$ be a filter containing $\mathfrak{G}$ (there is one, by hypothesis); then $\mathfrak{F}$ coincides with $\mathfrak{G}$; for if $Y \in \mathfrak{F}$ then $\complement Y \notin \mathfrak{F}$; hence $\complement Y \notin \mathfrak{G}$ and therefore $Y \in \mathfrak{G}$.

Example of an ultrafilter. The set of all subsets of a non-empty set $X$ which contain a given element $a \in X$ is an ultrafilter; for it is a filter, and if $Y$ is any subset of $X$ then either $a \in Y$ or $a \in \complement Y$. Such ultrafilters are called trivial.

Apart from this example, we shall never prove the existence of an ultrafilter (even on a countably infinite set) except by using Theorem 1 (and therefore the axiom of choice).