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Give an example of a monotone decreasing sequence $\{E_n\}$ of closed sets in a complete metric space such that $\operatorname{diam}(E_1)<\infty, \lim_n \operatorname{diam}(E_n)>0$, and $\cap_{n=1}^\infty E_n=\emptyset$.

There is a hint accompanying this exercise: Let $\{y_j\}$ be any bounded sequence with no points of accumulation. Write it as a double sequence $\{x_{mn}\}$ and take $E_n=\{x_{mj}\,\mid 1\leq m<\infty,n\leq j<\infty\,\}$.

Attempt:

Not really considering the hint at first, my initial thought was to consider the complete space (proven earlier) $\ell^\infty(\mathbb{R})$, with the sequence $\{e_n\}_{n\geq 1}$ where $e_n=(0,0,...,1,0,...)$ with $1$ in the $n:th$ place. And then let $E_n=\{\,e_m\,\mid m\geq n\,\}$. Now, $\{E_n\}$ is a monotone decreasing sequence of sets in a complete metric space and $\operatorname{diam}(E_1)=1<\infty, \lim_n \operatorname{diam}(E_n)=1>0$, since $\rho(x,y)=1$ for any two distinct sequences in $E_n$ for any $n$. Also, $\cap E_n=\emptyset$. What is left to show is that each $E_n$ is closed:

Well, Given any $n\geq 1$, any convergent sequence is Cauchy and any Cauchy sequence of $E_n$ is eventually constant since again $\rho(x,y)=1$ for any two distinct points of $E_n$. Hence $E_n$ contains all its limit points, and is thus closed.

Question 1: I am very unsure about the above attempt; is it correct?

Question 2: I am also unsure as to what is meant by the hint: Why should I consider bounded sequences with no accumulation points? And what would be an example following the hint?

In my example, the sequences are bounded, i.e. they're in $\ell^\infty$, but all have 0 as an accumulation point. If I should consider a bounded sequence with no points of accumulation, I would need to take sequences in something other than $\mathbb{R}^n$ by Bolzano-Weierstrass right? And when they write 'write it as a double sequence', I am also unsure. Say we have $(a_1,a_2,\dots)$ and that it has no accumulation points. If I just stack copies of that sequence, and define $E_n$ as in the hint, we wouldn't have $\cap E_n=\emptyset$.

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  • $\begingroup$ Your attempt is correct. If $E$ is a subset of a metric space and $\inf \{d(a,b): a,b\in E\land a\ne b\}>0$ then $E$ is closed. A simpler example is $\Bbb R$ with the metric $d(x,y)=\min (1,|x-y|)$ and $E_n=\{m\in\Bbb N: m\ge n\}.$ $\endgroup$ Nov 17, 2021 at 14:01
  • $\begingroup$ An even simpler example is the space $\Bbb N$ with "the" discrete metric. That is $d(x,y)=1$ if $x\ne y.$ Every subset is closed and bounded...... Let $ E_n=\{m\in\Bbb N:m\ge n\}.$ $\endgroup$ Nov 17, 2021 at 14:08

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