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Let $(X,\|\cdot\|)$ be the normed linear space consisting of the sequences $a=(a_n)_{n=1}^{\infty}$, for which the corresponding series $\sum_{n=1}^{\infty} a_n$ converges absolutely, with norm $\|a\|=\sum_{n=1}^{\infty} |a_n|$.

Let $e_k\in X$ be the sequence whose $k$-th term is $1$ and all other terms are zero and let $E=\{e_k : k\in \mathbb{N}\}$.

Then which of the following are true?

  1. $E$ is complete in the norm $\|\cdot\|$.
  2. $E$ is bounded subset of $X$.
  3. $E$ is a closed subset of $X$.
  4. $E$ is a compact subset of $X$.

All I can say is :

$E$ is bounded as norm of any element in $E$ is $1$.

$E$ is not closed because $(e_k)\rightarrow(1,1,1,\dots)$ which is not an element of $E$.

I think $E$ is not complete though I can not say anything about the proof.

$E$ is not compact I guess as it is not closed though that this not sufficient/necessary.

Please help me to see if my justification for second and third bullets are sufficient and please help me to see more clearly what would first and fourth bullets be.

Thank you

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    $\begingroup$ $E$ is closed. But $(e_k)$ is not Cauchy. In fact $\Vert e_j-e_i\Vert=2$ for all $i\ne j$. So $(e_i)$ has no Cauchy subsequence. What does this tell you about the compactness of $X$? $\endgroup$ – David Mitra Dec 25 '13 at 10:08
  • $\begingroup$ I do not understand why $E$ is closed.. could you please explain that $\endgroup$ – user87543 Dec 25 '13 at 10:09
  • $\begingroup$ For example, it has no limit points (this follows from what I wrote earlier). $\endgroup$ – David Mitra Dec 25 '13 at 10:10
  • $\begingroup$ Oh my Bad... If it have limit points then i have to worry if they are in $E$ or not to check if it is closed.. If there are no limit points there is no point... a Very valid point... $\endgroup$ – user87543 Dec 25 '13 at 10:12
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    $\begingroup$ See this post for the argument that $X$ is complete. $\endgroup$ – David Mitra Dec 25 '13 at 10:34
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What David Mitra said is quite concise and true. To sum up.

1)There are only trivial Cauchy sequences in $E$. So $E$ it can be shown that it is complete.

3)Now as stated above, $E$ is closed.

4)$E$ is not compact because there is a sequence that has no convergent subsequence, since it has no Cauchy subsequences and every convergent subsequence must be a Cauchy subsequence (David Mitra).

2)Also $E$ is bounded because $\operatorname{diam}E=\sup\{\|x-y\|:x,y\in E\}=2$.

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  • $\begingroup$ Do you mind to make this community wiki? $\endgroup$ – user87543 Dec 26 '13 at 11:34
  • $\begingroup$ @PraphullaKoushik,yes of course. make it:) $\endgroup$ – Haha Dec 26 '13 at 11:35
  • $\begingroup$ @PraphullaKoushik,it's ok i found the button:P $\endgroup$ – Haha Dec 26 '13 at 11:36
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    $\begingroup$ You wrote: There are not Cauchy sequences in $E$. Certainly, a constant sequence is Cauchy. But any Cauchy sequence in this space must be eventually constant. Maybe "There are no non-trivial Cauchy sequences in $E$." would be a better wording. $\endgroup$ – Martin Sleziak Dec 26 '13 at 13:36
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    $\begingroup$ @MartinSleziak Furthermore, $E$ is complete. Thus the reasoning of $4)$ is incorrect. It is not compact because there is a sequence that has no convergent subsequence, since it has no Cauchy subsequences and every convergent subsequence must be a Cauchy subsequence. (This is David Mitra's argument.) Also, I must point out that $E$ is bounded although it is not totally bounded. These are different concepts in metric spaces. $\endgroup$ – Josué Tonelli-Cueto Dec 26 '13 at 13:49
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Hint. True, True, True, False.

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    $\begingroup$ this is not even a hint... $\endgroup$ – user87543 Dec 25 '13 at 10:42
  • $\begingroup$ The answer should help to prove, now that you know what to prove. $\endgroup$ – Yiorgos S. Smyrlis Dec 25 '13 at 10:45
  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. $\endgroup$ – Salech Rubenstein Dec 25 '13 at 11:37

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