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This is a question about the proof of Theorem 3.10(b) in baby Rudin:

Let $\{K_n\}$ be a collection of nonempty compact sets in a metric space $X$ such that for any $n \in \mathbb{N}$, $K_n \supset K_{n+1}$. If $\lim_{n \to \infty} \mathrm{diam} K_n = 0$, then $\bigcap_{n=1}^\infty K_n$ contains exactly one element.

(Where for any $E \subset X$, $\mathrm{diam} E = \sup \big\{ d(p, q) : p, q \in E \big\}$.)

If $\bigcap_{n=1}^\infty K_n$ were to contain more than one element, then we would have $\mathrm{diam} K_n \geq \mathrm{diam} \bigcap_{n=1}^\infty K_n > 0$, for every $n \in \mathbb{N}$. In Rudin's proof, he says that this contradicts the fact that $\mathrm{diam} K_n \to 0$.

But how is this a contradiction? For example, if $\mathrm{diam} K_n = \frac{1}{n}$ for every $n \in \mathbb{N}$, we would have a sequence which converges to zero, but each of whose elements which are $> 0$.

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I guess after "then $\bigcap_{n=1}^{\infty}K_n$" a "contain only an element" is missing. We get that $\operatorname{diam}K_n\geq \alpha>0$ where $\alpha:=\operatorname{diam}\bigcap_{n=1}^{\infty}K_n$ is a constant which doesn't depend on $\mathbb N$. – Davide Giraudo Mar 18 '12 at 19:37
Yes, sorry for the typo! It's fixed now. – jamaicanworm Mar 18 '12 at 20:24
but in my book there is no non-empty condition for $K_n$s!( 3rd edition). – M.K. Dadsetani Jan 8 '15 at 21:17
up vote 1 down vote accepted

I don't have "Baby Rudin" handy at the moment, but if $\bigcap_{n=1}^\infty K_n$ were to contains two distinct points, $x, y$, then $\mathrm{diam} ( \bigcap_{n=1}^\infty K_n ) \geq d (x,y) > 0$. As $\bigcap_{n=1}^\infty K_n \subseteq K_n$ for all $n$, it would follow that $\mathrm{diam} (K_n) \geq d(x,y)$ for all $n$, contradicting that $\mathrm{diam} (K_n) \rightarrow 0$.

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The point is that if $x$ and $y$ are distinct points of $\bigcap_{n\ge 1}K_n$, the distance between them is positive, say $r>0$. But since they are in the intersection of the $K_n$’s, then must be in each of the $K_n$’s. That is, for each $n\in\Bbb Z^+$ we have $x,y\in K_n$, and therefore $\operatorname{diam}K_n\ge r$. If every $K_n$ has diameter at least $r$, the sequence of $\operatorname{diam}K_n$ cannot approach $0$.

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