Proposition 4.2.3. Let $X$ be a set, $\fF \subset 2^{X}$ be a filter, and $\fB \subset \fF$ be a filter base, then:
For any $E, F \in \fB$, there exists $G \in \fB$ such that $G \subset E \cap F$.
$\emptyset \not\in \fB$.
Conversely, if $\fB \subset 2^{X}$ is a non-empty collection that satisfies (FB1) and (FB2), then $\fB$ is a base for the filter
Proof. Filter Base $\Rightarrow$ (FB1): Let $E, F \in \fB$, then $E \cap F \in \fF$. Thus there exists $G \in \fB$ such that $G \subset E \cap F$.
Filter Base $\Rightarrow$ (FB2): By (F3), $\emptyset \not\in \fB \subset \fU$.
(FB1) + (FB2) $\Rightarrow$ (F1): Let $E \in \fF$, then there exists $E_{0} \in \fB$ such that $E_{0} \subset E$. Thus for any $X \supset F \supset E$, $E_{0} \subset F \in \fF$.
(FB1) + (FB2) $\Rightarrow$ (F2): Let $E, F \in \fF$, then there exists $E_{0}, F_{0} \in \fB$ such that $E_{0} \subset E$ and $F_{0} \subset F$. By (FB1), there exists $G \in \fB$ such that $G \subset E_{0} \cap F_{0}$, then $G \subset E \cap F \in \fF$.
(FB1) + (FB2) $\Rightarrow$ (F3): Let $E \in \fF$, then there exists $E_{0} \in \fB$ such that $E_{0} \subset E$. By (FB2), $E_{0} \ne \emptyset$, so $E \ne \emptyset$ as well.
Finally, $\fB$ is a filter base for $\fF$ by definition.$\square$