Could you help me : If (An) is a nested descreasing sequence of non-empty closed sets in the metric space M

Question

If {A_n} is a nested decreasing sequence of non-empty closed sets in the metric space M.

i) If M is complete and $diam(A_n) \to 0$ as $n \to \infty$, how can I show that $\cap_{n=1}^{\infty}A_n$ is exactly one point.

ii) To what assertions do the sets $[n, \infty)$ provide counterexamples?

For the first part of the question, since $A_n$ is closed in M and M is complete, can I follow the way that infinite intersection of closed sets is closed and by definition of completeness any subsequence of $A_n$ will also be a convergent cauchy sequence so intersection of infinite n provide the limit (and a singleton) of sequences. Do you think this way is correct ? I do not have any idea about second part and need help.

Answer 1

Your proof doesn't really make sense. You should argue existence and uniqueness of a point in $\bigcap_nA_n$ separately:

• For existence, prove any sequence $(x_n)$ with $x_n\in A_n$ for all $n$ is Cauchy, and explain why the limit must lie in $\bigcap_nA_n$.
• For uniqueness, use the fact that if $x,y\in\bigcap_nA_n\subset A_k$ then $d(x,y)\le\operatorname{diam}(A_k)$.