# Complete Metric Space if and only if

Let $$(X,d)$$ be a metric space. Let $$\mathcal{C}$$ be the set of all collections $$\{O_i\}_{i=1}^\infty$$ of non-empty closed subsets such that \begin{align*} &(a) O_{n+1}\subset O_n \forall n \\ &(b) \lim\operatorname{diam} (O_n) = 0 \ as \ n \to \infty \end{align*}

Prove that $$X$$ is complete if and only if $$\forall C \in \mathcal{C}$$ \begin{align*} \bigcap_{A\in C} A \not= \emptyset \end{align*} For the $$if$$ part: For every $$n$$, choose $$x_n \in O_n$$. Then since $$O_{n+1}\subset O_n$$, the set $$\{x_n,x_{n+1},x_{n+2},\cdots\}\subset O_n$$. Since $$\lim\operatorname{diam}(O_n) = 0$$, for any $$\epsilon > 0$$ choose a natural number $$N$$ so that $$\operatorname{diam}(O_n)<\epsilon$$ for $$n\geq N$$. This means that for any $$n,m \geq N$$, $$|x_n-x_m| \leq\operatorname{diam}(\{x_{N},x_{N+1},\dots\}) \leq\operatorname{diam}(O_N) < \epsilon$$. So $$\{x_n\}$$ is a Cauchy sequence. By completeness of $$X$$, $$\{x_n\}$$ converges to a point, $$a$$. By $$O_n$$ being closed it must contain $$\{x_n,x_{n+1},\dots\}$$. Thus, for any given $$D\in\mathcal{C}$$ we get \begin{align*} a\in \bigcap_{A \in D} A. \end{align*} hence non-empty.

Also, I am not sure how to start the other direction. I think you need to look at the tails of each sequence

Consider the Cauchy sequence $$\{x_n\}$$. Define $$F_n :=\overline{ \{x_m \vert m \geq n\}}$$.Note that $$F_n$$ is closed and satisfies the given two conditions. Let $$\mathcal{F} = \{F_n\}_{n=1}^\infty$$. Then $$F=\bigcap_{F_n\in \mathcal {F} } F_n \not= \emptyset$$. Take $$x\in F$$ and prove that $$x$$ indeed is the limit of the sequence $$\{x_n\}$$. You may have to use the fact that $$d(A)= d(\overline {A})$$, where $$d(A)$$ is the diameter of the set $$A$$ and $$\overline {A}$$ is the closure of $$A$$.