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Let $(M,d)$ be a metric space and let $X_n$ be a sequence such that $X_{n+1}\subset X_n.$ Prove that if $\text{diam}(X_n)\to0$ as $x\to\infty$ and $\bigcap_{n=1}^\infty X_n\ne\varnothing,$ then $(M,d)$ is complete.
Let $x_n$ be a Cauchy sequence and consider $X_n=\overline{\{x_m:m\ge n\}},$ then $X_n$ is closed and $X_{n+1}\subset X_n,$ but I don't see how to prove that $\text{diam}(X_n)\to0,$ any ideas?
Let $x_n$ be a Cauchy sequence and consider $X_n=\overline{\{x_m:m\ge n\}},$ then $X_n$ is closed and $X_{n+1}\subset X_n,$ but I don't see how to prove that $\text{diam}(X_n)\to0,$ any ideas?