Proposition 8.1.4.label Let $(X, d)$ be a metric space, then the following are equivalent:
- (1)
For any Cauchy sequence $\seq{x_n}\subset X$, there exists $x \in X$ such that $x = \limv{n}x_{n}$.
- (2)
For any Cauchy sequence $\seq{x_n}\subset X$, there exists a subsequence $\seq{n_k}$ and $x \in X$ such that $x = \limv{k}x_{n_k}$.
- (3)
For every Cauchy filter $\fF \subset 2^{X}$, there exists $x \in X$ such that $\fF \to x$.
Proof. (2) $\Rightarrow$ (3): Since $\fF$ is Cauchy, there exists $\seq{E_n}\subset \fF$ such that for each $n \in \natp$, $E_{n} \supset E_{n+1}$ and $\sup_{y, z \in E_n}d(y, z) \le 1/n$. For each $n \in \natp$, let $x_{n} \in E_{n}$, then there exists a subsequence $\seq{n_k}$ and $x \in X$ such that $x = \limv{n}x_{n}$. In which case, $x \in \bigcap_{n \in \natp}\overline{E_n}$. For each $n \in \natp$, $\sup_{y, z\in E_n}d(y, z) \le 1/n$, so $B_{X}(x, 2/n) \supset E_{n}$. Therefore $\fF \to x$.$\square$
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