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Let $\ell^2 = \{x= (x_1,x_2,x_3,\ldots): x_n\in \mathbb C\text{ and } \sum_{n=1}^\infty |x_n|^2 < \infty\}$

and $e_n \in \ell^2 $ be the sequence whose $n$-th element is 1 and all other elements are 0.

Equip the space with $\ell_2$ with the norm $$\|x\| = \left(\sum_{n=1}^\infty |x_n|^2\right)^{1/2}$$

Then the set $S=\{e_n : n \geq 1\}$ is

  1. closed;

  2. bounded;

  3. compact;

  4. and the sequence $s=(e_n)_{n\geq 1}$ contains a convergent subsequence.

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  • $\begingroup$ $l^2$ is fine too!There was no need to edit it to $l_2$! $\endgroup$ – wannadeleteacct Aug 3 '14 at 19:32
  • $\begingroup$ Just to clarify... by $S$ do you mean the space spanned by $S$? Or do you mean the set $S$ $\endgroup$ – Mathmo123 Aug 3 '14 at 19:36
  • $\begingroup$ @Mathmo123 :Thanks... corrected it! $\endgroup$ – wannadeleteacct Aug 3 '14 at 19:42
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    $\begingroup$ @Manasi: By the way Latex provides the extra (and I think very beautiful) special character little Lebesgue '$\ell$' (\ell) in contrast to the caligraphic letter '$\mathcal{l}$'... $\endgroup$ – C-Star-W-Star Aug 3 '14 at 23:12
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My answer: only 2 is true. It is not closed as $\lim e_n$ does not exist. It is bounded as $||e_n||=1$ It is not compact as one can not find a finite sub-cover containing S. Neither does it contain a convergent subsequence as $\lim e_{n_k} $ does not exist.

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    $\begingroup$ Almost. $\lim e_n$ does not exist in $\ell_2$ either - so in fact $S$ is closed as it contains all it's limit points. I'm unconvinced by your explanation that it isn't compact - you need to find an open cover that has no finite sub cover explicitly. $\endgroup$ – Mathmo123 Aug 3 '14 at 19:39
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    $\begingroup$ You mean that as there no limit points, so it vacuously contains its limit points so S is closed! For the open cover, consider balls around each $e_n$ of radius 4 and take their union. No finite sub-cover will cover all $e_n$. Is this fine? $\endgroup$ – wannadeleteacct Aug 3 '14 at 20:00
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    $\begingroup$ That's exactly right on both parts. For the compactness, whilst your answer certainly works, I think you'd be better considering balls of radius $\frac13$ as that way, no sub cover at all will cover $S$ and that is easier to prove $\endgroup$ – Mathmo123 Aug 3 '14 at 20:01
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Every pair of orthonormal vectors $e_i\neq e_j$ has distance precisely $d(e_i,e_j)=\sqrt{2}$.

So $S$ is closed as it contains only isolated points, it is bounded as all its elements have norm one, it is not compact as it contains infinitely many isolated points, it contains a convergent subsequence the constant ones?!

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  • $\begingroup$ S is closed as there are no limit points and only isolated points as you correctly observed. Maybe you mean that S is bounded as all its elements have norm one but it is not compact. Look at the comments to my solution above: You construct an open cover by taking open balls around each $e_i$ of radius 1/4(say). None of these balls intersect. When you take finitely many balls, there is always an $e_i$ that is excluded. You can find this by considering the finite sub-cover, say ${e_s, e_p,... , e_r}$. Let $t = max[s,p,..r]$. Consider $ e_(t+1)$ which is not contained in any ball. $\endgroup$ – wannadeleteacct Aug 4 '14 at 4:17
  • $\begingroup$ There is no convergent subsequence. There are no constant ones. You can only speak of convergence in $(a_{n_i},b_{n_i},....)$ if the sequence in each ordinate converges. $0s$ will never converge to a one. Now $lim e_{n_i}$ does not exist. $\endgroup$ – wannadeleteacct Aug 4 '14 at 4:29
  • $\begingroup$ @Manasi: That was a typo (bounded rather than compact) but if you would have continued reading you might have noticed that I also answered compactness... $\endgroup$ – C-Star-W-Star Aug 4 '14 at 15:30
  • $\begingroup$ @Manasi: What is your question on subsequences precisely? You only asked for a convergent subsequence; so consider the constant sequence $(e_2,e_2,\ldots)$ then itself as a subsequence is convergent as well as all of its subsequences... Please clarify your question! $\endgroup$ – C-Star-W-Star Aug 4 '14 at 15:32
  • $\begingroup$ My last reply was based on your previous reply(before edition). Okay, it only has isolated points but as pointed out above you need to give an open cover for those isolated points and explain why you can not get a finite sub-cover- for compactness. I am afraid that I really do not get your constant sequence! Why is it constant? Is it an element of S? Here, we discuss the possibility or impossibility of the convergent (sub)sequence of S. I do not think that you follow the question. $\endgroup$ – wannadeleteacct Aug 4 '14 at 16:40

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