# Completeness/Compactness of a subset in a normed linear space

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.

Thank you

• $E$ 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 $X$? – David Mitra Dec 25 '13 at 10:08
• I do not understand why $E$ is closed.. could you please explain that – user87543 Dec 25 '13 at 10:09
• For example, it has no limit points (this follows from what I wrote earlier). – David Mitra Dec 25 '13 at 10:10
• 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... – user87543 Dec 25 '13 at 10:12
• See this post for the argument that $X$ is complete. – David Mitra Dec 25 '13 at 10:34

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 $\operatorname{diam}E=\sup\{\|x-y\|:x,y\in E\}=2$.

• Do you mind to make this community wiki? – user87543 Dec 26 '13 at 11:34
• @PraphullaKoushik,yes of course. make it:) – Haha Dec 26 '13 at 11:35
• @PraphullaKoushik,it's ok i found the button:P – Haha Dec 26 '13 at 11:36
• 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. – Martin Sleziak Dec 26 '13 at 13:36
• @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. – Josué Tonelli-Cueto Dec 26 '13 at 13:49

Hint. True, True, True, False.

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