Lemma 4.2.10. Let $X$ be a set and $\fF \subset 2^{X}$, then
There exists an ultrafilter $\fU \supset \fF$.
$\fF$ is the intersection of all ultrafilters containing it.
Proof. (1): Let
and order it by inclusion. For any chain $\mathbf{C}\subset \mathbf{U}$, let $\fB = \bigcup_{\fF \in \textbf{C}}\fF$, then $\fB$ is a filter subbase, and the filter $\fU \in \mathbf{U}$ generated by $\fB$ is an upper bound for $\mathbf{C}$. By Zorn’s lemma, $\textbf{U}$ has a maximal element.
(2): Let $E \subset X$. If there exists $F \in \fF$ with $F \cap E^{c} = \emptyset$, then $E \supset F$ and $E \in \fF$. Otherwise, $\fF \cup \bracs{E^c}$ is a filter subbase, and there exists an ultrafilter containing it by (1). In which case, there exists an ultrafilter $\fU \supset \fF$ with $E \not\in \fU$.$\square$