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nested closed sets, lim diam≠0, can intersection be singleton?

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hbghlyj posted 2022-12-15 22:30 |Read mode
Last edited by hbghlyj 2022-12-17 13:07$(X,d)$ is a complete metric space. Let $\{F_n\}$ be a sequence of non-empty closed sets with $F_1\supset F_2\supset\cdots$ and $\operatorname{diam}F_n\rightarrow1$, can $\bigcap_{n=1}^{\infty}F_n$ be a singleton?

We have $\operatorname{diam} \bigcap_{n\in\Bbb{N}} F_n \le \operatorname{diam} F_m\to 1$ as $m\to\infty$
But is it possible that $\operatorname{diam} \bigcap_{n\in\Bbb{N}} F_n=0$

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Czhang271828 posted 2022-12-17 15:38
Consider $F_n=\prod_{1\leq i\leq n}\{0\}\times \prod_{j\geq n+1}[0,1]$ in $\ell^\infty$.

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original poster hbghlyj posted 2022-12-28 21:41
Czhang271828 发表于 2022-12-17 08:38
Consider $F_n=\prod_{1\leq i\leq n}\{0\}\times \prod_{j\geq n+1}[0,1]$ in $\ell^\infty$.
I see. Thanks. In this example,
For every $n$, $\operatorname{diam}F_n=1$.
$$\bigcap_{n=1}^{\infty}F_n=\left\{\prod_{i\ge1}\{0\}\right\}$$

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