Proposition 5.2.4.label Let $X, Y$ be sets, $f: X \to Y$ be a function, and $\fB \subset 2^{X}$ be a filter base, then
- (1)
$f(\fB) = \bracs{f(E)| E \in \fB}$ is also a filter base.
- (2)
If $\fB$ is an ultrafilter base, then $f(\fB)$ is also an ultrafilter base.
Proof. (1): Let $E, F \in \fB$, then there exists $G \in \fB$ such that $G \subset E \cap F$. In which case, $f(G) \subset f(E \cap F) \subset f(E) \cap f(F)$.
For any $f(E) \in f(\fB)$, $E \ne \emptyset$ implies that $f(E) \ne \emptyset$.
(2): Let $E \subset X$, then either $f^{-1}(E)$ or $f^{-1}(E^{c})$ contains an element of $\fB$. Therefore either $E$ or $E^{c}$ contains an element of $f(\fB)$.$\square$