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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?

RFZ
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    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. – Ishfaaq Aug 14 '15 at 06:36
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    @Vim: Yes I guess most authors encompass the empty set in their definition of compactness and so does Rudin. – Ishfaaq Aug 14 '15 at 06:40
  • Ishfaaq thank you very much! Rudin really defines diameter for nonempty. – RFZ Aug 14 '15 at 06:41
  • @Ishfaaq ok. And you are right. Rudin indeed only defined $\text{diam}$ for nonempty sets (Def 3.9, page 52). – Vim Aug 14 '15 at 06:43
  • @Pacman - Anytime. The book is notorious for this sort of "economical" writing. Good of you to spot it. – Ishfaaq Aug 14 '15 at 06:44
  • It might also be that he uses $\supset$ to mean $\supsetneq$, which means that none of the sets are empty, because if one of the $K_n$ was empty, then all subsequent $K_m$ would be equal. But I haven't looked at the book, so I wouldn't be able to tell if this was the case. – Arthur Aug 14 '15 at 07:29

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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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