Example of decreasing sequence of closed sets in a complete metric space s.t. $d(E_1)<\infty, \lim_n d(E_n)>0, \cap E_n=\emptyset$

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 $$\cap E_n=\emptyset$$.

• 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 $E_n=\{m\in\Bbb N: m\ge n\}.$ Nov 17, 2021 at 14:01
• 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 $E_n=\{m\in\Bbb N:m\ge n\}.$ Nov 17, 2021 at 14:08