4.2 Filters
Definition 4.2.1 (Filter). Let $X$ be a set, a filter $\fF \subset 2^{X}$ is a non-empty family of sets such that:
For any $E \in \fF$ and $X \supset F \supset E$, $F \in \fF$.
For any $E, F \in \fF$, $E \cap F \in \fF$.
$\emptyset \not\in \fF$
Definition 4.2.2 (Filter Base). Let $X$ be a set, $\fF \subset 2^{X}$ be a filter, and $\fB \subset \fF$, then $\fB$ is a filter base for $\fF$ if for every $F \in \fF$, there exists $E \in \fB$ such that $E \subset F$.
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.$\square$
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 $f(E) \ne \emptyset$.$\square$
Proposition 4.2.5. Let $X, Y$ be sets, $f: X \to Y$ be a function, and $\fB \subset 2^{X}$ be a filter base. If $\emptyset \not\in f^{-1}(\fB) = \bracs{f^{-1}(E)| E \in \fB}$, then $f^{-1}(\fB)$ 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, $f^{-1}(G) \subset f^{-1}(E \cap F) = f^{-1}(E) \cap f^{-1}(F)$.
(FB2): By assumption, $\emptyset \not \in f^{-1}(\fB)$.$\square$
Definition 4.2.6 (Filter Subbase). Let $X$ be a set and $\fB_{0} \subset 2^{X}$ be a non-empty collection, then $\fB_{0}$ is a filter subbase if for any $\seqf{E_j}\subset \fB_{0}$, $\bigcap_{j = 1}^{n} E_{j} \ne \emptyset$.
Definition 4.2.7 (Generated Filter). Let $X$ be a set and $\fB_{0} \subset 2^{X}$ be a filter subbase, then there exists a filter containing $\fB_{0}$.
The smallest filter $\fF(\fB_{0}) \subset 2^{X}$ containing $\fB_{0}$ is the filter generated by $\fB_{0}$, which is given by $\fF(\fB_{0}) = \bracs{E \subset X| \exists F \in \fB: F \subset E}$, where
Proof. For any $\seqf{E_j}, \bracsn{F_j}_{1}^{m} \subset \fB_{0}$,
Thus $\fB$ satisfies (FB1). Since $\bigcap_{j = 1}^{n} E_{j} \ne \emptyset$, $\emptyset \not\in \fB$, and $\fB$ satisfies (FB2).
By Proposition 4.2.3, $\fB$ is a base for the filter $\fF = \bracs{E \subset X| \exists F \in \fB: F \subset E}$.
If $\fF' \supset \fB_{0}$ is a filter, then $\fF \supset \fB$ by (F2), and $\fF' \supset \fF$ by (F1). Thus $\fF$ is the smallest filtetr containing $\fB$.$\square$
Lemma 4.2.8. Let $X$ be a set, $\fF, \mathfrak{G}\subset 2^{X}$ be filter subbases. If for any $E \in \fF$ and $F \in \mathfrak{G}$, $E \cap F \ne \emptyset$, then there exists a filter $\fU \supset \fF \cup \mathfrak{G}$.
Proof. Let $\seqf{E_j}\subset \fF$ and $\seqf[m]{F_j}\subset \mathfrak{G}$, then $\bigcap_{j = 1}^{n} E_{j} \in \fF$ and $\bigcap_{j = 1}^{m} F_{j} \in \mathfrak{G}$. Thus their intersection is non-empty, and $\fF \cup \mathfrak{G}$ is a filter subbase. By Definition 4.2.7, there exists a filter containing $\fF \cup \mathfrak{G}$.$\square$
Definition 4.2.9 (Ultrafilter). Let $X$ be a set and $\fF \subset 2^{X}$ be a filter, then the following are equivalent:
$\fF$ is maximal with respect to inclusion.
For any $E \subset X$, either $E \in \fF$ or $E^{c} \in \fF$.
For any $\seqf{F_j}\subset X$ such that $\bigcup_{j = 1}^{n} F_{j} \in \fF$, there exists $1 \le j \le n$ such that $F_{j} \in \fF$.
If the above holds, then $\fF$ is an ultrafilter.
Proof. (1) $\Rightarrow$ (2): Let $E \subset X$ with $E^{c} \not\in \fF$, then $E \cap F \ne \emptyset$ for all $F \in \fF$. By Lemma 4.2.8, there exists a filter $\fU \supset \fF \cup \bracs{E}$. Since $\fF$ is maximal, $\bracs{E}\in \fU \subset \fF$.
(2) $\Rightarrow$ (1): Let $E \subset X$ with $E \not \in \fF$, then $E^{c} \in \fF$. Since $E \cap E^{c} = \emptyset$, there exists no filter containing $\fF \cup \bracs{E}$. Thus $\fF$ is maximal.
(2) $\Rightarrow$ (3): Suppose that $F_{n} \not\in \fF$, then $F_{n}^{c} \in \fF$ and $F_{n}^{c} \subset \bigcup_{j = 1}^{n - 1}F_{n} \in \fF$. Proceeding inductively, there must exist $1 \le j \le n$ such that $F_{n} \in \fF$.
(3) $\Rightarrow$ (2): $E \cup E^{c} = X \in \fF$.$\square$
Lemma 4.2.10. Let $X$ be a set and $\fF \subset 2^{X}$, then
There exists an ultrafilter $\fU \supset \fF$.
$\fF$ is the intersection of all ultrafilters containing it.
Proof. (1): Let
and order it by inclusion. For any chain $\mathbf{C}\subset \mathbf{U}$, let $\fB = \bigcup_{\fF \in \textbf{C}}\fF$, then $\fB$ is a filter subbase, and the filter $\fU \in \mathbf{U}$ generated by $\fB$ is an upper bound for $\mathbf{C}$. By Zorn’s lemma, $\textbf{U}$ has a maximal element.
(2): Let $E \subset X$. If there exists $F \in \fF$ with $F \cap E^{c} = \emptyset$, then $E \supset F$ and $E \in \fF$. Otherwise, $\fF \cup \bracs{E^c}$ is a filter subbase, and there exists an ultrafilter containing it by (1). In which case, there exists an ultrafilter $\fU \supset \fF$ with $E \not\in \fU$.$\square$
Definition 4.2.11 (Convergence). Let $X$ be a topological space, $\fF \subset 2^{X}$ be a filter with base $\fB \subset 2^{X}$, and $x \in X$, then the following are equivalent:
$\fF \supset \cn(x)$.
For each ultrafilter $\fU \supset \fF$, $\fU \supset \cn(x)$.
There exists a fundamental system of neighbourhoods $\cb(x) \subset \cn(x)$ 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 $\fB$ converges to $x$.
If $A \subset X$ and $\fB \subset 2^{A}$, then $\fB$ converges to $x$ if $\fF(\fB) \supset \bracsn{U \cap A| U \in \cn(x)}$.
Proof. (1) $\Leftrightarrow$ (2): By (2) of Lemma 4.2.10.
(1) $\Rightarrow$ (3): $\cn(x)$ is a fundamental system of neighbourhoods at $x$.
(3) $\Rightarrow$ (1): For any $U \in \cn(x)$, there exists $B \in \cb(x)$ and $F \in \fB$ with $F \subset B \subset U$. In which case, $U \in \fF$.$\square$
Definition 4.2.12 (Accumulation Point). Let $X$ be a topological space, $\fF \subset 2^{X}$ be a filter with base $\fB$, and $x \in X$, then the following are equivalent:
$x \in \bigcap_{E \in \fB}\overline{E}$.
$x \in \bigcap_{E \in \fF}\overline{E}$.
There exists a fundamental system of neighbourhoods $\cb(x) \subset \cn(x)$ such that for every $E \in \cb$ and $f \in \fB$, $E \cap F \ne \emptyset$.
There exists a filter $\fU \supset \fB$ that converges to $x$.
If the above holds, then $x$ is a cluster/accumulation point of $\fB$. In particular, if $\fF$ is an ultra filter, then (6) implies that the limit points and cluster points of $\fF$ coincide.
Proof. (1) $\Rightarrow$ (3): Let $U \in \cn(x)$, then $U \cap E \ne \emptyset$ for all $E \in \fB$.
(3) $\Rightarrow$ (4): By Lemma 4.2.8, there exists a filter $\fU \supset \cb(x) \cup \fB$. Since $\cb(x)$ and $\fB$ are bases for $\cn(x)$ and $\fF$, respectively, $\fU \supset \cn(x) \cup \fF$.
(4) $\Rightarrow$ (1): Let $U \in \cn(x)$, then since $\fU \supset \fB \cup \cn(x)$, $U \cap E \ne \emptyset$ for all $E \in \fB \subset \fF \subset \fU$. Thus $x \in \bigcap_{E \in \fF}\ol{E}$.$\square$
Definition 4.2.13 (Limit). Let $X, Y$ be topological spaces, $A \subset X$, and $f: A \to Y$ be a function. For any filter base $\fB \subset 2^{A}$, if $f(\fB)$ converges to $y \in Y$, then $y = \lim_{x, \fB}f(x)$ is a limit of $f$ with respect to $\fB$.
For any $x_{0} \in \overline{A}$, let $\fF = \bracs{U \cap A| U \in \cn(x_0)}$ be the trace of $\cn(x_{0})$ on $A$, then $\fF \subset 2^{A}$ is a filter. If $f(\fF)$ converges to $y \in Y$, then
is a limit of $f$ at $y$ with respect to $A$. If $A = X$, then $x \in A$ may be omitted.