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I am kinda stuck at the idea to solve the problem below.

I know that $l^{\infty}$ is a Banach space. Further, I know the following Lemma: Let $X$ be a Banach space and $U$ a closed Subspace, then $U$ is complete.

I want to show that the subspace of convergent sequences $c$ together with the $sup$-Norm is complete by using the Lemma above.

Attempt: Let $(x^{(n)})_{n \in \mathbb{N}} $ be a sequence with elements in $c$, thus the elements of the sequence $(x^{(n)})_{n \in \mathbb{N}}=x^{(1)},x^{(2)},x^{(3)},...$ are sequences with elements in $\mathbb{R}$. I will denote $x^{(1)}=(x^{(1)}_k)_{k \in \mathbb{N}} $. To show that $c$ is complete I will assume that $(x^{(n)})_{n \in \mathbb{N}} $ converges in $l^{\infty}$ to some element $x$. This means I have to show that $x \in c$.

Looking at $x^{(n)}=(x^{(n)}_k)_{k \in \mathbb{N}} $, I will take the k-limit of these sequences, defining $\overline {x}_{n}:=lim_{k \rightarrow \infty} x^{(n)}_{k}$. Now I can take all those limits and put them to create a sequence (in $\mathbb{R}$) out of them, $\overline{x}:=\overline{x}_{1},\overline{x}_{2},\overline{x}_{3},...$.

Now, I want to show that the limit of $x^{(n)}$ is equal to $\overline{x}$.

That's how far I came on my own. I tried to search for the proof. And I found one. Ignoring notation differences, what I did til now matches the proof in this book. But I don't really understand the continuation of the proof:

Since $(\overline{x}_n)_{n \in \mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}$, the limit $\overline{x}:= lim_{n \rightarrow \infty} \overline{x}_n$ exists. Denote the sequence $x$ as $(t_m)_{m \in \mathbb{N}}$. To show that $x \in c$, it is enough to show that $lim_{m \rightarrow \infty} t_m=\overline{x}$.

I do not understand why it is enough to show that both sequences have the same limit. How does that prove that $(x^{(n)})_{n \in \mathbb{N}} $ converges to $x=\overline{x}$ ? As far as I understand it, to show that $(x^{(n)})_{n \in \mathbb{N}} $ converges to $\overline{x}$, I have to prove that $\overline{x}$ and $x$ are equal component wise.

DenisGerhardt
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  • What is $c$? the subspace of convergent sequences? – stoic-santiago Aug 02 '23 at 14:24
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    @stoic-santiago, yes with $c$ I mean the subspace of convergent sequences – DenisGerhardt Aug 02 '23 at 14:25
  • If you show that $(t_{n}){n\in\mathbb{N}}$ converges to some limit, then it follows that this sequence is in $c$ and that $c$ is closed in $l^{\infty}$. There is nothing particular about showing that it converges to the same limit. It is just to show that the sequence $(t{n})_{n\in\mathbb{N}}$ does indeed converge. – Dean Miller Aug 02 '23 at 14:32
  • @DeanMiller Oh right, I think I got it now. I don't have to show that $(t_n){n \in \mathbb{N}}$ and $(\overline{x}_n){n \in \mathbb{N}}$ are equal componentwise. The idea behind the construction of $(\overline{x}n){n \in \mathbb{N}}$ is to use it for the calculations to show that $(t_n)_{n \in \mathbb{N}}$ does in fact converge and thus is in $c$, is that correct? – DenisGerhardt Aug 02 '23 at 14:39
  • @DenisGerhardt Yes, that's correct. – Dean Miller Aug 02 '23 at 16:07
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    I prefer the following argument in order to prove completeness. The space $c$ is isometrically isomorphic to the space $C(K)$ where $K={0}\cup {{1\over n}}_{n=1}^\infty$ equipped with metric in $\mathbb{R}.$ The space $C(K)$ is complete for any compact metric space $C(K).$ The proof is similar to that of $C[0,1].$ – Ryszard Szwarc Aug 02 '23 at 20:17

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