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This is stated as a problem in A Course in Point Set Topology book by J. Conway:

Let $\{E_n\}$ be a sequence of totally bounded sets. If $\operatorname{diam}E_n\to 0$ as $n\to\infty$, show that $\bigcup_{n=1}^\infty E_n$ is also totally bounded.

How can we prove this statement? I have no idea to use diameter conception to this problem (it has relation with Cauchy sequence).

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    $\begingroup$ This is problem from J.Conway A Point Set Topology book $\endgroup$
    – Analysis
    Jan 6 '15 at 17:23
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    $\begingroup$ As stated, it's obviously false, just let $E_n=\{n\}\subseteq\mathbb Z$. And why is this tagged as general topology when it's obviously about metric spaces? $\endgroup$ Jan 6 '15 at 17:27
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    $\begingroup$ It's false without some other assumption. For another example, consider $E_n = (n - \dfrac1n, n + \dfrac1n) \subset \Bbb R$. I looked up the book you are using, and found the relevant problem (1.4 exercise 5), and it is indeed stated exactly as above, so I don't know what's going on here (a misprint, perhaps). $\endgroup$
    – BaronVT
    Jan 6 '15 at 17:49
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    $\begingroup$ So, downvoters please note: this question is unanswerable as it is (because the statement is false), but not due to any error on the part of OP; the book in question does indeed have this exact wording. $\endgroup$
    – BaronVT
    Jan 6 '15 at 17:51
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    $\begingroup$ Thanks it is good counterexample. Ok i understood that there is problem in the condition of this question $\endgroup$
    – Analysis
    Jan 6 '15 at 17:52
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As stated, it's obviously false, just let $E_n=\{n\}\subseteq\mathbb Z$. -- user2345215

It's false without some other assumption. For another example, consider $E_n = \left(n - \dfrac1n, n + \dfrac1n\right) \subset \Bbb R$. I looked up the book you are using, and found the relevant problem (1.4 exercise 5), and it is indeed stated exactly as above, so I don't know what's going on here (a misprint, perhaps). -- BaronVT

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