> [!definition] > > Let $X$ be a [[Set|set]] and $\mathfrak F \subset \pow{X}$ be a family of subsets, then $\mathfrak F$ is a **filter** if: > 1. For any $E \in \mathfrak F$ and $E \subset F \subset X$, $F \in \mathfrak F$ as well. > 2. For any $E, F \in \mathfrak F$, $E \cap F \in \mathfrak F$. > 3. $\emptyset \not\in \mathfrak F$. > [!theorem] > > Let $\mathfrak S \subset \pow{X}$, then the following are equivalent: > 1. There exists a filter $\mathfrak F \supset \mathfrak S$. > 2. For every $\seqf{E_i} \in \mathfrak S$, $\bigcap_{i = 1}^nE_i \ne \emptyset$. > > If the above holds, then the smallest filter containing $\mathfrak S$ is the filter **generated by** $\mathfrak S$, and $\mathfrak S$ is a **subbase** of $\mathfrak F$. > > *Proof*. $(2) \Rightarrow (1)$: Define > $ > \mathfrak S_I = \bracs{\bigcap_{i = 1}^n E_i: \seqf{E_i} \subset \mathfrak S, n \in \nat} > $ > as the family of finite intersections of $\mathfrak S$, and > $ > \mathfrak S_F = \bracs{E \subseteq X| \exists F \in \mathfrak S_I: E \supseteq F} > $ > For any $E_1, E_2 \in \mathfrak S_F$, there exists $F_1, F_2 \in \mathfrak{S}_I$ such that $E_1 \supset F_1$, $E_2 \supset F_2$. Thus $E_1 \cap E_2 \supset F_1 \cap F_2 \in \mathfrak S_I$, and $E_1 \cap E_2 \in \mathfrak S_F$. For any $E \in \mathfrak S_F$, there exists $F \in \mathfrak S_I$ such that $E \supset F$. So for every $E' \supset E$, $E' \supset F$ and $E' \in \mathfrak S_F$ as well. Lastly, by assumption $(2)$, $\emptyset \not\in \mathfrak S_I$, and $\emptyset \not\in \mathfrak S_F$. Therefore $\mathfrak S_F$ is a filter as well. > [!theorem] > > Let $\seqi{\mathfrak F}$ be a family of filters, then the following are equivalent: > 1. There exists a filter $\mathfrak F \supset \bigcup_{i \in I}\mathfrak F_i$. > 2. For every $\seqf{i_j}$ and $E_j \in \mathfrak F_{i_j}$, $\bigcap_{j = 1}^n E_j \ne \emptyset$. > [!definition] > > Let $(X, \topo)$ be a [[Topological Space|topological space]], $\fB \subset \pow{X}$ be a filter base, $\fF$ be the filter generated by $\fB$, and $x \in X$. The following are equivalent: > > 1. $\fF$ contains the [[Neighbourhood|neighbourhood]] filter $\cn(x)$ at $x$. > 2. Every [[Ultrafilter|ultrafilter]] containing $\fF$ contains the neighbourhood filter $\cn(x)$ at $x$. > 3. There exists a [[Neighbourhood Base|neighbourhood base]] $\cb$ at $x$ such that for every $E \in \cb$, there exists $F \in \fB$ with $F \subset E$. > > If the above holds, then $x$ is a **limit point** of $\fB$ and $\fF$ **converges** to $x$. In addition, the following are equivalent: > > 4. $x \in \bigcap_{E \in \fB}\overline{E}$. > 5. $x \in \bigcap_{E \in \fF}\overline{E}$. > 6. There exists a [[Neighbourhood Base|neighbourhood base]] $\cb$ at $x$ such that for every $E \in \cb$ and $F \in \fB$, $E \cap F \ne \emptyset$. > 7. There exists a filter $\fU \supset \fB$ that converges to $x$. > > If the above holds, then $x$ is a **cluster point** of $\fB$. In particular, $(6)$ implies that the limit points and cluster points of an ultrafilter coincide. > > *Proof*. $(1) \Leftrightarrow (2)$ follows from the fact that every filter is the intersection of all ultrafilters containing it. > > $(1) \Rightarrow (3)$: $\cn(x)$ itself is a neighbourhood base. > > $(3) \Rightarrow (1)$: Let $E \in \cn(x)$, then there exists $F \in \cb$ and $G \in \fB$ such that $G \subset F \subset E$. Hence $E \in \fF$. > > $(4) \Leftrightarrow (6)$: For every $F \in \fB$, $F \cap E \ne \emptyset$ for all $E \in \cn(x)$ if and only if $x \in \overline{E}$. > > $(6) \Leftrightarrow (7)$: $(6)$ is equivalent to $\fB \cup \cn(x)$ being a filter subbase, which is equivalent to $(7)$.