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Let $(X,d)$ be a complete metric space and let $C_n$ be a sequence of connected, closed sets such that $C_{n+1} \subset C_n$ for every $n \in \mathbb{N}$. Assume that $\bigcap\limits_{n =1}^\infty C_n$ consists of one single point. I would like to show that $\text{diam}(C_n)$ converges to $0$ as $n$ goes to infinity, but my attempts does not seem to work. Maybe I am wrong and we can construct a counterexample.

Anyone can help me? Thank you very much.

Eric Wofsey
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javi1996
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This is false. For instance, let $X=\ell^\infty$ and let $C_n\subseteq X$ be the set of sequences whose first $n$ coordinates are all $0$. Then each $C_n$ has infinite diameter but their intersection is $\{0\}$.

Eric Wofsey
  • 330,363
  • Nice, thank you! I think this kind of counterexamples only works if all the diameters are infinite and the result may be right for bounded sets. – javi1996 Feb 02 '20 at 22:28
  • No, you could intersect each $C_n$ with a closed ball around $0$ and it would still work. – Eric Wofsey Feb 02 '20 at 22:29