$l^{\infty}$ is a Banach Space wrt sup norm (Functional Analysis) 
$l^{\infty} := \{x=(x_n) : \sup_{n\in \mathbb N}|x_n| \lt \infty\} $
$\|x\|_{\infty} = \sup_{n \in \mathbb N} |x_n|$

Show that $l^{\infty}$ is a Banach Space (Complete) with respect to > $\|x\|_{\infty}$
If I wish a space is Banach, I must show Cauchy Sequences are convergent in this space.
I cannot find clear answers for questions below :
1) I have prior problem about writing sequences in Cauchy definition. Do I have to use sequences of sequences in $l^{\infty}$ like $\|x_{n_k}-x_{n_l}\|_{\infty} \lt \varepsilon$ ,$\forall k,l \ge N_{\varepsilon}$ or can I write Cauchy definition for only elements of $l^{\infty}$ like $\|x_{n}-x_{m}\|_{\infty} \lt \varepsilon$ , $\forall m,n \ge N_{\varepsilon}$ and if I can how should I use it? (Please fix me if there is any mistake)
2) Let $y_k$ is limit sequence of $x_{n_k}$ (or $x_n$, I am not sure about writing sequences in Cauchy definition), how should I show $y_k \in l^{\infty}$, I think it is easy but I couldn't realize :(
Thanks a lot in advance
 A: Hint:
Suppose $(X_i)_i$ be a Cauchy sequence with elements $X_i\in l^{\infty}$ that converges to $A$. We need to show that $A\in l^\infty$, i.e., $\sup_{n\in\Bbb N}|A_n|\lt\infty$
Then, given $\epsilon\gt 0$, there exists $N\in\Bbb N$ such that for all $i\geq N$, we have $||X_i-A||_\infty\lt\epsilon$
Now, $$||X_i-A||_\infty=\sup_{n\in\Bbb N}|X_{in}-A_n|\lt\epsilon$$
This implies that $|X_{in}-A_n|\lt\epsilon~\forall~n\in\Bbb N$
By the reverse triangle inequality, we have,
$$||X_{in}|-|A_n||\leq|X_{in}-A_n|\lt\epsilon$$
which shows that $|A_n|\in B_\epsilon(|X_{in}|)$ where $B_r(x)$ is the open ball of radius $r$ and center $x$
Can you now use this and the fact that $X_i\in l^\infty$ to conclude that $|A_n|\lt\infty~\forall~n\in\Bbb N$ and hence $||A||_\infty=\sup_{n\in\Bbb N}A_n\lt\infty$ ?
A: You seem to be confused with the indices. 
You can bravely introduce any notation which is convenient for yourself, e.g. writing the sequences as functions: $x_n=k\mapsto x_n(k)$.


*

*The second choice is surely good, as this is the definition: $\|x_n-x_m\|$ can be stressed to be arbitrary small for big enough $n,m$ indices.

*From where this limit $y$ is coming from? It's just the aim to prove that $x_n$ will have a limit. 
Well, the $x_n$'s converge pointwise, as for each fixed $k$, the sequence $(x_n(k))_n$ is Cauchy. 
It will indeed give an $y(k)$, and you also have to prove $\|y-x_n\|\to 0$.
To find a bound for $y$, fix an $\varepsilon$ and belonging $N$. Then $|x_n(k)|\le \|x_N\|+\varepsilon$ for all $n\ge N$ and all $k$.
