There is Theorem 3.10(b) in baby Rudin.

If $K_n$ is a sequence of compact sets in a metric space $X$ such that $K_n\supset K_{n+1}$ and if $$\lim_{n\to \infty}\text{diam}K_n=0,$$ then $\bigcap_{n=1}^{\infty}K_n$ consists of exactly one point.

I think that all $K_n$ must be a nonempty. But why Rudin didn't write this?

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    $\begingroup$ I think maybe because the diameter is only defined for nonempty sets by the author and since each $K_n$ has a diameter their non-voidness is implicit. $\endgroup$ – Ishfaaq Aug 14 '15 at 6:36
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    $\begingroup$ @Vim: Yes I guess most authors encompass the empty set in their definition of compactness and so does Rudin. $\endgroup$ – Ishfaaq Aug 14 '15 at 6:40
  • $\begingroup$ Ishfaaq thank you very much! Rudin really defines diameter for nonempty. $\endgroup$ – ZFR Aug 14 '15 at 6:41
  • $\begingroup$ @Ishfaaq ok. And you are right. Rudin indeed only defined $\text{diam}$ for nonempty sets (Def 3.9, page 52). $\endgroup$ – Vim Aug 14 '15 at 6:43
  • $\begingroup$ @Pacman - Anytime. The book is notorious for this sort of "economical" writing. Good of you to spot it. $\endgroup$ – Ishfaaq Aug 14 '15 at 6:44

As many authors, Rudin defines diameter only for nonempty set; thus, making an assumption about $\operatorname{diam}K_n$ means that $K_n$ is implicitly assumed nonempty.

A related question: What is the best way to define the diameter of the empty subset of a metric space?


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