> [!definition] > > Let $X$ be a [[Set|set]] and $\mathfrak B \subset \pow{X}$ be a family of subsets, let > $ > \mathfrak B_S = \bracs{E \subset X| \exists F \in \mathfrak B: E \supset F} > $ > then $\mathfrak B_S$ is a [[Filter|filter]] if and only if > 1. For every $E, F \in \mathfrak B$, there exists $G \in \mathfrak B$ such that $G \subset E \cap F$. > 2. $\fB \ne \emptyset$ and $\emptyset \not\in \fB$. > > If the above holds, then $\fB$ is a **base** for the filter $\fB_S$. Two filter bases are equivalent if they generate the same filter. > [!theorem] > > Let $\fF$ be a filter and $\fB \subset \fF$, then the following are equivalent: > 1. $\fB$ is a base for $\fF$. > 2. For every $E \in \fF$, there exists $F \in \fB$ such that $E \supset F$. > [!theorem] > > Let $X, Y$ be sets, $f: X \to Y$ be a mapping, and $\fB \subset \pow{X}$, then > > 1. If $\fB$ is a filter base, then so is $f(\fB)$. > 2. If $\fB$ is an ultrafilter base, then so is $f(\fB)$. > > Let $\fS \subset \pow{Y}$ be a filter base, then > > 3. $f^{-1}(\fS)$ is a filter base if and only if $f(X) \cap E \ne \emptyset$ for all $E \in \fS$. > > *Proof*. $(1)$: For every $E \in \fB$, $f(E) \ne \emptyset$. Since $f(E \cap F) \subset f(E) \cap f(F)$, $f(\fB)$ is a filter base. > > $(2)$: Let $E \subset Y$. If there exists $F \in \fB$ such that $f^{-1}(E) \supset F$, then $E \supset f(F)$. Otherwise, there exists $F \in \fB$ such that $f^{-1}(E^c) \supset F$, and $E^c \supset f(F)$. Thus $\fB$ is an ultrafilter base. > > $(3)$: If $f^{-1}(E) \ne \emptyset$ for all $E \in \fS$, then $f^{-1}(E \cap F) = f^{-1}(E) \cap f^{-1}(F)$ implies that $f^{-1}(\fS)$ is a filter base. If $f^{-1}(\fS)$ is a filter base, then $f^{-1}(E) \ne \emptyset$ and $f(X) \cap E \ne \emptyset$ for all $E \in \fS$.