Proposition 6.4.12 ([Corollary 2.3.1, 2.3.2, Bou13]).label Let $(X, \fU)$ be a uniform space, then:
- (1)
For each $x \in X$, $\cn(x)$ is a minimal Cauchy filter.
- (2)
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 $x$.
- (3)
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 6.4.10, $\bracs{U(x): U \in \fU}= \cn(x)$ 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 5.2.7, 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 $\mathfrak{G}$.$\square$