5.4 Cauchy Filters
Definition 5.4.1 (Small). Let $(X, \fU)$ be a uniform space, $V \in \fU$, and $A \subset X$, then $A$ is $V$-small if $A \times A \subset V$.
Lemma 5.4.2. Let $(X, \fU)$ be a uniform space, $V \in \fU$, and $A, B \subset X$ such that:
$A, B$ are $V$-small.
$A \cap B \ne \emptyset$.
then $A \cup B$ is $V \circ V$-small.
Proof. Since $V \circ V \supset V$, it is sufficient to consider the case where $x \in A$ and $y \in B$. By assumption (2), there exists $z \in A \cap B$. By assumption (1), $(x, z), (z, y) \in V$, so $(x, y) \in V \circ V$.$\square$
Definition 5.4.3 (Cauchy Filter). Let $(X, \fU)$ be a uniform space and $\fF \subset 2^{X}$ be a filter on $X$, then $\fF$ is Cauchy if for every $V \in \fU$, there exists $E \in \fF$ such that $E$ is $V$-small.
Proposition 5.4.4 (Cauchy Criterion). Let $(X, \fU)$ be a uniform space and $\fF \subset 2^{X}$ be a convergent filter, then $\fF$ is Cauchy.
Proof. Let $x \in X$ such that $\cn(x) \subset \fF$ and $V \in \fU$. By Lemma 5.1.9, assume without loss of generality that $V \in \fU$. Since $\cn(x) \subset \fF$, then $V(x) \in \fU$ and $V(x) \times V(x) \subset V$.$\square$
Definition 5.4.5 (Cauchy Continuous). Let $X, Y$ be uniform spaces and $f: X \to Y$, then $f$ is Cauchy continuous if for any Cauchy filter base $\fB \subset 2^{X}$, $f(\fB) \subset 2^{Y}$ is Cauchy.
Proposition 5.4.6. Let $(X, \fU)$ and $(Y, \mathfrak{V})$ be uniform spaces and $f \in UC(X; Y)$, then $f$ is Cauchy continuous.
Proof. Let $V \in \mathfrak{V}$, then there exists $V' \in \fU$ such that $(f(x), f(y)) \in V$ whenever $(x, y) \in V'$. For any Cauchy filter base $\fB \subset 2^{X}$, there exists $E \in \fB$ such that $E \times E \subset V'$, so $f(E) \times f(E) \subset V$.$\square$
Definition 5.4.7 (Minimal Cauchy Filter). Let $X$ be a uniform space and $\fF \subset 2^{X}$ be a Cauchy filter, then $\fF$ is minimal if it is minimal with respect to inclusion.
Proposition 5.4.8. Let $(X, \fU)$ be a uniform space and $\fF \subset 2^{X}$ be a Cauchy filter, then:
There exists a unique minimal Cauchy filter $\fF_{0}$ such that $\fF_{0} \subset \fF$.
If $\fB \subset \fF$ is a base of $\fF$, and $\fV \subset \fU$ is a fundamental system of symmetric entourages, then $\mathfrak{M}= \bracs{V(M): V \in \fV, M \in \fB}$ is a base for $\fF_{0}$.
Proof. Since $\fV$ and $\fB$ are both non-empty, $\mathfrak{M}$ is non-empty as well. Let $V_{1}, V_{2} \in \fV$ and $M_{1}, M_{2} \in \fB$. By (F2) of $\fV$ and $\fB$, there exists $V_{0} \in \fV$ and $M_{0} \in \fB$ such that $V_{0} \subset V_{1} \cap V_{2}$ and $M_{0} \subset M_{1} \cap M_{2}$. So $V_{0}(M_{0}) \subset V_{1}(M_{1}) \cap V_{2}(M_{2})$, and $\mathfrak{M}$ satisfies (FB1). Given that $\emptyset \not\in \fV, \fB$, $\emptyset \not\in \mathfrak{M}$, so $\mathfrak{M}$ satisfies (FB2). By Proposition 4.2.3, $\mathfrak{M}$ is a filter base for
To see that $\mathfrak{M}$ is a Cauchy filter base, let $V \in \fV$. By (U2), there exists $W \in \fV$ such that $W \circ W \circ W \subset V$. Since $\fB$ is a Cauchy filter base, there exists $M \in \fB$ such that $M \times M \subset W$. Let $(x, y) \in W(M) \times W(M)$, then there exists $(p, q) \in M \times M$ such that $(p, x), (q, y) \in W$. As $M \times M$ $W$, $(x, y) \in W \circ W \circ W$ by symmetry of $W$ and Lemma 5.1.11. Therefore $W(M) \times W(M) \subset V$ and $\mathfrak{M}$ is a Cauchy filter base.
Finally, let $\fF' \subset \fF$ be a Cauchy filter. Let $M \in \fB$ and $V \in \fV$, then there exists $N \in \fF'$ such that $N \times N \subset V$. Since $N \in \fF$, $M \cap N \ne \emptyset$, and there exists $y \in M \cap N$. For any $x \in N$, $N \times N \subset V$ implies that $x \in V(y)$. Hence $N \subset V(M)$ and $\fF' \supset \fF_{0}$.$\square$
Proposition 5.4.9. Let $(X, \fU)$ be a uniform space and $\fF \subset 2^{X}$ be a minimal Cauchy filter, then $\fF$ admits a base consisting of open sets.
Proof. Let $\fV \subset \fU$ be the set of all symmetric, open entourages. By Proposition 5.1.14, $\fV$ is a fundamental system of entourages. By Proposition 5.4.8, $\mathfrak{M}= \bracs{U(M)| U \in \fV, M \in \fF}$ is a base for $\fF$.
Let $V \in \fU$, then there exists $U \in \fV$ such that $U \subset V$. For any $M \in \fF$, $U(M)$ is open by Lemma 5.1.15. Thus $\mathfrak{M}$ consists of open sets.$\square$
Proposition 5.4.10. Let $(X, \fU)$ be a uniform space, then:
For each $x \in X$, $\cn(x)$ is a minimal Cauchy filter.
For each Cauchy filter $\fF \subset 2^{X}$ and $x \in X$, $x$ is an accumulation point of $\fF$ if and only if $\fF$ converges to $x$.
For any $x \in X$, filter $\fF \subset 2^{X}$ converging to $x$, and Cauchy filter $\mathfrak{G}\subset \fF$, $\mathfrak{G}$ converges to $x$.
Proof. (1): Let $\fF = \bracs{E \subset X: x \in E}$, then $\fF$ is a filter generated by the filter base $\fB = \bracs{\bracs{x}}$. By Proposition 5.4.8, $\bracs{U(x): U \in \fU}= \cn(x)$ is the unique minimal filter contained in $\fF$.
(2): Let $x \in X$ be an accumulation point of $\fF$, then $U(x) \cap E \ne \emptyset$ for all $U \in \fU$ and $E \in \fF$. By Definition 4.2.7, there exists a filter $\fF' \supset \fF \cup \cn(x)$. Since $\fF$ is a Cauchy filter, so is $\fF'$. By minimality of $\cn(x)$, $\fF \supset \cn(x)$.
(3): Since $\fF$ converges to $x$, $x$ is a cluster point of $\fF$ and $\mathfrak{G}$. By (2), $x$ is also a limit point of $\mathfrak{G}$.$\square$