Definition 4.2.9 (Ultrafilter). Let $X$ be a set and $\fF \subset 2^{X}$ be a filter, then the following are equivalent:
$\fF$ is maximal with respect to inclusion.
For any $E \subset X$, either $E \in \fF$ or $E^{c} \in \fF$.
For any $\seqf{F_j}\subset X$ such that $\bigcup_{j = 1}^{n} F_{j} \in \fF$, there exists $1 \le j \le n$ such that $F_{j} \in \fF$.
If the above holds, then $\fF$ is an ultrafilter.
Proof. (1) $\Rightarrow$ (2): Let $E \subset X$ with $E^{c} \not\in \fF$, then $E \cap F \ne \emptyset$ for all $F \in \fF$. By Lemma 4.2.8, there exists a filter $\fU \supset \fF \cup \bracs{E}$. Since $\fF$ is maximal, $\bracs{E}\in \fU \subset \fF$.
(2) $\Rightarrow$ (1): Let $E \subset X$ with $E \not \in \fF$, then $E^{c} \in \fF$. Since $E \cap E^{c} = \emptyset$, there exists no filter containing $\fF \cup \bracs{E}$. Thus $\fF$ is maximal.
(2) $\Rightarrow$ (3): Suppose that $F_{n} \not\in \fF$, then $F_{n}^{c} \in \fF$ and $F_{n}^{c} \subset \bigcup_{j = 1}^{n - 1}F_{n} \in \fF$. Proceeding inductively, there must exist $1 \le j \le n$ such that $F_{n} \in \fF$.
(3) $\Rightarrow$ (2): $E \cup E^{c} = X \in \fF$.$\square$