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Ex 3.21 If $\{E_n\}$ is a sequence of closed nonempty and bounded sets in a complete metric space $X$, if $E_{n+1}\subset E_n$ and if $$\lim_{n\rightarrow\infty} \mathbb{diam}\ E_n=0$$ then $\bigcap_{n=1}^\infty E_n$ consists of exactly one point.

my proof: Let $E=\bigcap_{n=1}^\infty E_n$, suppose there are two points $p,q$ in $E$, then, $\forall n\in \mathbb{N},E\subseteq E_n$ and $\mathbb{diam}\ E_n\geq \mathbb{diam}\ E\geq d(p,q)$, therefore there is a contradiction for $$\lim_{n\rightarrow\infty} \mathbb{diam}\ E_n=0$$ therefore, if $E$ is nonempty, $E$ has a unique point.

now show that $E$ is nonempty, Select $x_n\in E_n$, and let $$X_N:=\{x_n:n\geq N\}$$then $$X_n\in E_n,\mathbb{diam}\ E_n\geq \mathbb{diam}\ X_n,\lim_{n\rightarrow\infty} \mathbb{diam}\ X_n=0$$ therefore sequence $\{x_n\}$ is cauchy sequence, and because X is complete, $\{x_n\}$ has limit point $x$, and $E_n$ is closed ,$$\forall n\in \mathbb{N}, x\in E_n,(\because X_n\subseteq E_n)$$$$\therefore x\in E \ \blacksquare$$ Question 1. Is there any error in my proof?

Question 2. why are we need boundness of $E_n$?

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  • $\begingroup$ We need boundedness so that all sets $E_n$ have a diameter at all. Plus for unbounded closed sets there are counterexamples like $E_n=[n,\infty)$ in the reals. $\endgroup$ Dec 22, 2018 at 19:55
  • $\begingroup$ You mean, diameter should be define for only bounded set, isn't it? $\endgroup$
    – 백주상
    Dec 22, 2018 at 20:32

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For the condition that $\operatorname{diam}(E_n) \to 0$ to make sense we must have that at most finitely many $E_n$ have infinite diameter, and then we could just start the sequence at the first finite diameter set. All sets in the sequence are then bounded, with diameters tending to $0$. So the condition "forces" bounded sets, and anyway, there are counterexamples for unbounded closed sets as $E_n = [n, \infty)$ in the reals, another good reason to stay with bounded sets (and if one $E_n$ is bounded, the tail will consist of bounded sets by nestedness and we could just as well have used only bounded sets).

As to your proof, the idea is fine, but the write-up could sometimes be better.

Unicity is fine, if $p \neq q $ are both in $\cap_n E_n$, for all $n$ $\operatorname{diam}(E_n) \ge d(p,q) >0$ and the limit condition cannot hold.

Existence: just pick $x_n \in E_n$ arbitarily. If $\varepsilon >0$, pick $N$ so that $\operatorname{diam}(E_N) < \varepsilon$. Then for $n,m \ge N$, we know that $x_n, x_m \in E_{m'} \subseteq E_N$ where $m'=\max(n,m) \ge N$ because the sets are nested and thus $$d(x_n, x_m) \le \operatorname{diam}(E_{m'}) \le \operatorname{diam}(E_N) < \varepsilon$$ and $(x_n)$ is Cauchy because $\varepsilon>0$ was arbitrary.

We know by completeness that $(x_n)$ converges to some $p \in X$. Fix $n$ for now. Then for all $m \ge n$, $x_m \in E_m \subseteq E_n$ and as $E_n$ is closed and $(x_m)_{m \ge n} \to p$ also, $p \in E_n$. As $n$ was arbitary, $p \in \bigcap_n E_n$ as required. Your sets $X_n$ are superfluous.

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  • $\begingroup$ Thank you for your answer, but I have a question, I don't know how $E_n=[n,\infty)$ is counterexample of our problem, if we remove boundness condition, actually I don't know how we explain $\lim_{n\rightarrow \infty}\mathbb{diam} [n,\infty)=0$ in $\mathbb{R}$. Can you give some helps? $\endgroup$
    – 백주상
    Dec 25, 2018 at 17:14
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    $\begingroup$ @백주상 the example is meant to illustrate that we need bounded sets: the sets are closed, nested but have empty intersection. They all have infinite diameter so they don’t obey the final condition which is thus necessary. $\endgroup$ Dec 25, 2018 at 18:43

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