Exercise 3.21

Question

Exercise 21:
    Prove the following analogue of Theorem 3.10(b): If $\{E_n\}$ is a sequence of closed nonempty and bounded sets in a {\it complete} metric space $X$, if $E_n\supset E_{n+1}$,
    and if $$\lim_{n\rightarrow\infty}\mathop{\rm diam} E_n=0,$$ then $\cup_1^\infty E_n$ consists of exactly one point.

ghostofgarborg · Accepted Answer

Pick $x_n \in E_n$ for each $n$. Since $ \{x_k\}_{k=n}^\infty \subset E_n$ and $\lim \operatorname{diam} E_n = 0$,
    $x_n$ is Cauchy. Since $X$ is complete, $x_n$ converges to a point $x$. Since $x$ is a 
    limit point for each $E_n$ and each $E_n$ is closed, $x \in E_n$ for all $n$. 
    Consequently, $ x \in \bigcap_{n=1}^\infty E_n$. There cannot be more than one point
    in this intersection, since that would contradict $\lim \operatorname{diam} E_n = 0$.

Amit Awasthi · Answer

\begin{proof}
      Construct sequence $\{x_j\}$ such that $x_j \in E_n$ for $j = n$. Because of the limit above, $\{x_j\}$ is Cauchy by definition 3.9 in Rudin. This Cauchy sequence converges to a limit point since $X$ is complete, and this limit point must be in $E_n$ since $E_n$ is closed for all $n$. Inductively, this means the subsequence of $\{x_j\}$ that starts on $n$ must be contained in $E_n$, since that is how we constructed them. Now, as $n = j \rightarrow \infty$, $x_j \rightarrow x$, $x$ being the limit point of the Cauchy sequence, and $x \in E_n$. Since we constructed $\{x_j\}$ to be contained in every $E_n$ such that $n \le j$, $x \in \bigcap^\infty_1 E_n$. And because of the limit given, $E_n$ as $n \rightarrow \infty$ only contains the one point $x$ since the diameter goes to 0.
    \end{proof}

Exercise 3.21

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Exercise 3.21

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