Explain what it means for a metric space to be complete.

Let $X$ be a metric space. We say the subsets $A_{i}$ of $X$, with $i \in \mathbb{N}$, form a descending sequence in $X$ if $A_{1} \supset A_{2} \supset A_{3} \supset \cdots$.

Prove that the metric space $X$ is complete if and only if any descending sequence $A_{1} \supset A_{2} \supset \cdots$ of non-empty closed subsets of $X$, such that the diameters of the subsets $A_{i}$ converge to zero, has an intersection $\bigcap_{i=1}^{\infty} A_{i}$ that is non-empty.

[Recall that the diameter $\operatorname{diam}(S)$ of a set $S$ is the supremum of the set $\{d(x, y)$ : $x, y \in S\} .]$

Give examples of

(i) a metric space $X$, and a descending sequence $A_{1} \supset A_{2} \supset \cdots$ of non-empty closed subsets of $X$, with $\operatorname{diam}\left(A_{i}\right)$ converging to 0 but $\bigcap_{i=1}^{\infty} A_{i}=\emptyset$.

(ii) a descending sequence $A_{1} \supset A_{2} \supset \cdots$ of non-empty sets in $\mathbb{R}$ with $\operatorname{diam}\left(A_{i}\right)$ converging to 0 but $\bigcap_{i=1}^{\infty} A_{i}=\emptyset$.

(iii) a descending sequence $A_{1} \supset A_{2} \supset \cdots$ of non-empty closed sets in $\mathbb{R}$ with $\bigcap_{i=1}^{\infty} A_{i}=\emptyset$.