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The problem is given: source

Let $c_0$ be a space of real sequences $x = \{x_n\}_{n = 1}^\infty=0$ converging to $0$. Let $\ell^\infty$ be a set of real sequences $w = \{w_k\}^\infty_{k=0}$ furnished with the norm $\|w \|_\infty = \sup_{1 \leq n \leq \infty} |w_n|$. Prove that $c_0$ is closed in $\ell^\infty$

The pdf says

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TL;DR: Please check whether my recap of the proof is correct? As detail as possible so I have a deep understanding of what is really going on. Thank you

  1. Start with a sequence in $c_0$, our goal is to show that its limit $\omega \in \ell^\infty$ is also in $c_0$. This limit exists by completeness of $\ell^\infty$.

  2. Denote this sequence by $(x_n^k)_{k=1}^{\infty} = ( \{x^1_1,x_2^1,\dots \}, \{x_1^2, x_2^2,\dots \}, \dots)$. As the objects in $c_0$ are sequences, we have a sequence of sequences.

A quick remark: My intuition tells me that since the elements of $c_0$ all converge to $0$, it would seem to make sense to me that this sequence (of sequences) will also converge to a sequence (of sequences) to $0.$ That is $\omega = 0.$

  1. Instead of showing that the entire sequence (of sequences) $(x_n^k)_{k=1}^{\infty}$ converges to a limit (of sequences) $(\omega^{k}_n)_{k =1}^{\infty}$ (this should be a sequence of constant sequences), we show that each term of $(x_n^k)_{k=1}^{\infty}$ converges to each term of $(\omega_n^k)_{k = 1}^{\infty}$. Note that for a fixed $k$, $\omega_n^k = (\omega_1^k, \omega_2^k, \dots, )$, so the elements are numbers

  2. I am guessing $(\omega_n^k)_{k =1}^{\infty}$ is another sequence in $c_0$? I don't know why you want $(\omega_n^k)_{k =1}^{\infty}$to be in $\ell^\infty$ in the first place.

  3. I am also guessing that when they say

Take $\epsilon >0$ and $N_0 \in \mathbb{N}$ such that $\sup_{1 \leq n \leq \infty} |x_{n}^{k} - \omega_n| < \epsilon/2$ for all $k > N_0$.

They are using the convergence of $(x_n^k)_{k=1}^{\infty}$

  1. I am guessing that when they say

For each $k $, choose $N_1 \in \Bbb N$ such that $|x_n^k| < \epsilon/2$ for all $n > N_1$ and $k > N_0$

They are using the convergence of the elements in $(x_n^k)_{k=1}^{\infty}$, which all go to $0$.

Remark: As opposed to (5), they don't really know if the sequence (of sequence) $(\omega_n)$ really go to a sequence (of sequences) of $0$s. So they denote this "mystery" limit of sequence (of sequences) by $(\omega_n^k)_{k = 1}^{\infty}$

(7) I think the conclusion now is that $0 = \omega \in c_0$?

(8) Would it be more firm to make $k,n > \max \{N_1,N_1\}$

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  1. Start with a sequence in $c_0$, our goal is to show that its limit $\omega\in\ell^\infty$ is also in $c_0$. This limit exists by completeness of $\ell^\infty$

Incorrect. To prove that a set is closed, we start with a convergent sequence in that set, and show that the limit is contained in the set. The existence of limit is assumed, not obtained from completeness.

"Quick remark" is totally wrong. Take $x_n = (1+1/n,0,0,0,\dots)$. Then the limit of $x_n$ is $(1,0,0,\dots)$, not the zero sequence.

I don't know where you got $\omega_n^{(k)}$ from, or what it means; consequently I can't make any sense from your items 3 and following.


Advice: since you are getting confused by "sequence of sequences" business, change the notation to functions. A sequence of real numbers if just a function $f:\mathbb N\to\mathbb R$. So, you have a sequence of functions $f_n:\mathbb N\to\mathbb R$. They converge to some function $g$ in the sense that $\|f_n-g\|_\infty\to 0$. The goal is to prove that the function $g$ satisfies $\lim_{k\to\infty} g(k)=0$.

By the definition of limit, we must do the following: given $\epsilon>0$, find $K$ such that $|g(k) |<\epsilon$ whenever $k\ge K$. To this end, we

  • find $N$ such that $\|f_n-g\|_\infty<\epsilon/2$ whenever $n\ge N$.
  • find $K$ such that $|f_N(k)|<\epsilon/2$ whenever $k\ge K$. (This is possible since $f_N\in c_0$)
  • observe that $|g(k)|\le |f_N(k)|+|f_N(k)-g(k)| <\epsilon$ whenever $k\ge K$, as required.
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  • $\begingroup$ Um, doesn't that mean your example $1 + 1/n \not\in c_0$? Also, $\omega_n^k = (\omega_1^k, \omega_2^k, \dots, )$, the answer don't show the "$k$" in the upper script because it is believed the each term in the sequence is the same. $\endgroup$ – Hawk Sep 21 '14 at 20:55
  • $\begingroup$ I added more under the horizontal line, since you continue to confuse two different kinds of indices here (as in your previous comment). $\endgroup$ – user147263 Sep 21 '14 at 20:59
  • $\begingroup$ oh okay I think the function point of view clears up more. Is the sequence $\{ g(k) \}_{k \geq 1}$ in $c_0$ or $\ell^\infty$? I know that's not too important, but I still want to know. $\endgroup$ – Hawk Sep 21 '14 at 21:03
  • $\begingroup$ The function $g$ is originally assumed to be in $\ell_\infty$ (meaning it's bounded), but the proof shows that it's in $c_0$ (meaning that $\lim_{k\to\infty} g(k)=0$). $\endgroup$ – user147263 Sep 21 '14 at 21:06
  • $\begingroup$ So then when we started with a convergent sequence in $c_0$, is this not $ \{ g(k) \}$? Because the conclusion is that the limit $g(k) \to 0$. $\endgroup$ – Hawk Sep 21 '14 at 21:10

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