Proposition 4.2.4. Let $X, Y$ be sets, $f: X \to Y$ be a function, and $\fB \subset 2^{X}$ be a filter base, then $f(\fB) = \bracs{f(E)| E \in \fB}$ is also a filter base.
Proof. (FB1): Let $E, F \in \fB$, then there exists $G \in \fB$ such that $G \subset E \cap F$. In which case, $f(G) \subset f(E \cap F) \subset f(E) \cap f(F)$.
(FB2): For any $f(E) \in f(\fB)$, $E \ne \emptyset$ implies that $f(E) \ne \emptyset$.$\square$