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It's a problem from the book "Topology of Metric Spaces", written by Kumaresan:

"Show that the set $\textbf{c}$ of convergent sequences in the Normed Linear Space of all bounded real sequences under the sup norm $\|\|_\infty$ is complete. Hint: Enough to show that $\textbf{c}$ is closed. If x = $(x_n)$ is a limit point of $\textbf{c}$, it suffices to show that $x$ is Cauchy."

I got a little bit confused here... What I tried to do:

I know that to show that a metric space is complete, I have to show that every Cauchy sequence is convergent there.

About the hint: I don't understand why it is enought to show that $\textbf{c}$ is closed, and what does it mean to say that $(x_n)$ is a limit point of $\textbf{c}$? Does that mean that there is a squence $(y_k) \in \textbf{c}$ such that $(y_k) \to (x_n)$?

If $(y_k) \to (x_n)$, given $\epsilon > 0$, there are $n_0, k_0 \in \mathbb{N}$ such that $$\|y_k-x_n\|_\infty<\epsilon \forall k,n \geq \max\{n_0, k_0\}=m_0$$

As we have that $(y_n) \in \textbf{c}$, we have that $(y_n) \to y \in \mathbb{R}$. Because of the uniqueness of the limit, $(x_n) \to y$, so $(x_n) \in \textbf{c}$, therefore, $\textbf{c}$ is closed.

Now, let $(y_{n_k}) \in \textbf{c}$ be a Cauchy sequence. Well, $(y_{n_k})=(y_{n_1},y_{n_2},\dots,y_{n_j},\dots)$ is a sequence of sequences. Knowing that $(y_{n_k})$ is Cauchy's, we know that, given $\epsilon > 0, \exists j_0 \in \mathbb{N}$ such that $\forall m, p \geq j_0$

$$\|y_{n_m}-y_{n_p}\|_\infty < \epsilon$$

I don't know what to do from here. My idea is to show that, since every term $y_{n_j}$, of the sequence $(y_{n_k})$, belongs to $\textbf{c}$, therefore, every term $y_{n_j}$ is convergent, we have that $(y_{n_k})$ converges for a convergent sequence $(y_n)$. So, since $(y_n)$ is convergent, $(y_n) \in \textbf{c}$, we have that every Cauchy sequence in $\textbf{c}$ is convergent, meaning that $\textbf{c}$ is complete.

But I don't know how to do that, and I don't know if I understood the problem correctly, so I don't know if what I did before is correct...

I'd be really grateful if someone could give a hint, a solution, a coment about anything I wrote here! :)

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    $\begingroup$ If it is closed, then this means it contains all of its limit points, but then if $x=(x_n)_{n\in \mathbb{N}}$ is Cauchy then it converges (see this -u.arizona.edu/~mwalker/econ519/Econ519LectureNotes/… - for verification that Cauchy implies convergent in your case). But that's exactly what we need to do... show it converges because that would show $any$ limit point is convergent (because any limit point is Cauchy). So just show any limit point is Cauchy! Does that help? Any questions? $\endgroup$ – Squirtle Jun 30 '14 at 2:59
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    $\begingroup$ Also, look up the definition of a limit point because your definition isn't quite right but its worth noting that a limit point doesn't even need to be part of a set. $\endgroup$ – Squirtle Jun 30 '14 at 3:18
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    $\begingroup$ @Carol: Notice that the points in c are themselves sequences, and then you have a sequence of sequences that (if it converges, it)converges to a sequence. Notice also that a closed subset of a complete space is complete, and that a Cauchy sequence of sequences is one where $|| (x_n)_m-(y_n)_k ||_{\infty} < \epsilon $ , for $m,k >j$ for some positive integer $j$. $\endgroup$ – user99680 Jun 30 '14 at 3:23
  • $\begingroup$ @Squirtle I read the link you sent, but I still don't understand why Cauchy implies convergent in this case... However, I think I understood why it's enough to show that $\textbf{c}$ is closed. $\textbf{c}$ is a subset of the set A of all bounded sequences. A is complete, right? So, if $\textbf{c}$ is closed, it means $\textbf{c}$ is complete. Is that correct? $\endgroup$ – Anna Jun 30 '14 at 4:28
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    $\begingroup$ Oh I'm sorry, I feel absolutely terrible, I had a dyslexia moment the link I sent showed every convergent sequence is Cauchy. Still it's not horribly difficult to prove the converse. In response to your comment about what you call $A$, please read the following post: (is the set of all bounded sequences complete) $\endgroup$ – Squirtle Jun 30 '14 at 4:45
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1)First notice that a closed subset of a complete space is closed. Another way of understanding closed sets is that a closed set contains all its limit points. Equivalently, a subset $S$ is closed, if every convergent sequence in $S$ has its limit in $S$.

2) An element $(x_n)$ is a limit point of a set $S$, if every neighborhood of $(x_n)$ intersects $S$. It is not always the case that if $(x_n)$ is a limit point, that there exists a sequence converging to $(x_n)$; this is true for 1st-countable spaces (which includes metric spaces).

3)Now, try to show c is closed by showing that if $(x_n)$ is a limit point of c, then $(x_n)$ is in $S$, or that, if you have a convergent sequence in c, then its limit is also in c. The triangle inequality should help you there.

A general observation is that you can show the subspace $c$ is closed, by showing that its complement is open: consider a point $(x_n)$ in $\mathbb R^{\mathbb R} -c$ , i.e., a sequence of Reals that does not converge . Then show that there is an $\epsilon>0$ , so that the ball $B((x_n), \epsilon)$ (ball given in the $||.||_{\infty}$ metric ), so that no sequence in that ball converges. This should not be too hard; if a sequence $(x_n)$does not converge and $Sup|x_n-y_n| < \epsilon$ , then I think it is not too hard to show that $(y_n)$ does not converge either.

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  • $\begingroup$ When you say "if you have a convergent sequence in c, then its limit is also in c", what you mean? The sequences in c are real sequences, so they converge for some real number... But real numbers are not in c, right? Because c is a set of sequences... hehe sorry, I'm confused $\endgroup$ – Anna Jun 30 '14 at 4:32
  • $\begingroup$ @Carol: Please use the @ to write a comment, to make sure I get the pingback. I mean that the limit of a Cauchy sequence in c would also be a convergent sequence of Real numbers. So you have a Cauchy sequence of convergent sequence, and its limit should be a convergent sequence ( in the $||.||_{\infty}$ norm) itself. $\endgroup$ – user99680 Jun 30 '14 at 5:02
  • $\begingroup$ Sorry! I thought that since my coment was in your answer, you'd be notified. About your answer, thank you, I understood it now. $\endgroup$ – Anna Jun 30 '14 at 13:24
  • $\begingroup$ I don't understand oO I write "@user99680" in the beginning of the coment, but when I send, it desappears, and I tried more than once! $\endgroup$ – Anna Jun 30 '14 at 13:26
  • $\begingroup$ @Carol: No problem, enjoy the futbol cup. $\endgroup$ – user99680 Jun 30 '14 at 16:43

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