I want to prove $l^1$ is complete. I want to prove that any Cauchy sequence $(x_n)$ is Convergent to $(y)$ whose $y_i$ is the limit of the sequence $x_{(n,i)}$ . I want to prove this part first then I would show that $||y||$ is bounded not the other way around.
Can anyone please help me to proceed? I am really having trouble to visualize that the Cauchy sequence $(x_n)$ converges to $(y)$.
Sketch: picking up from our comments, we have $x^{(j)}_n\to y_j$ for each $j\in \mathbb N$.
Step $1$: show that $y=(y_j)\in \ell^1:$
$\|y\|:=\lim_{J\to \infty}\sum^J_{j=1}|y_j|=\lim_{J\to \infty}(\sum^J_{j=1}\lim_{n\to \infty}|x^{(j)}_n|)=\lim_{J\to \infty}\lim_{n\to \infty}(\sum^J_{j=1}|x^{(j)}_n|).$
Now, $\sum^J_{j=1}|x^{(j)}_n|\le \sum^{\infty}_{j=1}|x^{(j)}_n|=\|x_n\|\le M<\infty $ because the original sequence $(x_n)$ is Cauchy, hence bounded. So,
$\|y\|\le \lim_{n\to \infty}M=M.$
Step $2:$ Prove that $\|x_n-y\|\to 0:$
Let $\epsilon>0$ and choose an integer $N$ such that $l,n>N\Rightarrow \|x_l-x_n\|<\epsilon.$
Then, $\sum^J_{j=1}|x_n^{(j)}-x_l^{(j)}|\le \sum^{\infty}_{j=1}|x_n^{(j)}-x_l^{(j)}|=\|x_n-x_l\|<\epsilon.$
I'll leave the rest to you: keep $n>N$, fix $J$ and let $l\to \infty.$ Finally, let $J\to \infty.$