Showing that a sequence converges. Assume $x_n^k, x_n$ to be real for all $k$ and for all $n$. Suppose that I have $x_k^{(n)}\to x_k$ as $n\to \infty$ for every $k\in \mathbb N$. Then can something be said about $\sup_k|x_k^{(n)}-x_k|$?
Given any $\epsilon>0$, $(\forall k\in \mathbb N) \exists N_k\in \mathbb N, n\ge N_k\implies |x_k^{(n)}-x_k|<\epsilon/2$.
I can't do $\sup_k$ on  both sides as that would change $N_k$ as well.
Is there any way by which I can conclude that the supremum is $0$ for large $n$?
While trying to prove this, I observed the following inequality: $\sup\{|x_n|: n\in \mathbb N\}\le \sum_{j=1}^\infty|x_n|$. (Proof: If RHS is infinite then there is nothing to prove so suppose RHS to be $<\infty$. For every $n$, $|x_n|\le \sum_{j=1}^\infty |x_n|$, whence the result follows. QED)
Fixing an $\epsilon>0$, for every $k\in \mathbb N$, there exists $N_k\in \mathbb N$ such that $k\ge N_k\implies |x_k^{(n)}-x_k|\lt \frac{\epsilon}{2^{k+1}}$.
Using the inequality shown above, I'm tempted to do the following:
$\sup_k|x_k^{(n)}-x_k|\le \epsilon \sum_k\frac 1{2^{k+1}}<\epsilon.$, whence I get that $\|(x_k^{(n)})-(x_k)\|_\infty\to 0.$
But it seems wrong as $N_k$'s are varying. Is there any way to get the conclusion?
 A: It looks like you're trying to prove that $\ell^\infty(\mathbb N)$ is a complete metric space. That is, given a Cauchy sequence $z_n\in\ell^\infty$, you want to show that $z_n$ converges to an element $z\in\ell^\infty$. The candidate for $z$ is the sequence whose coordinates are the pointwise limits of the coordinates of $z_n$:
$$
z := (\lim_{n}x_1^{(n)},\lim_nx_2^{(n)},\dots).
$$
The first thing to show is that $z$ is actually a bounded sequence; I leave it to you.
Now you'd like to show that
$$
\sup_k|x^{(n)}_k-x_k|\xrightarrow{n\to\infty} 0.
$$
For a general infinite collection of convergent sequences $\{(x_k^{(n)})_{k=1}^\infty\}_n$, say with $x_k^{(n)}\xrightarrow{n\to\infty}x_k$, nothing like this can be said because of examples like $x_k^{(n)} = k/n$. However, in our situation, we also are given that the sequence $z_n$ is Cauchy. So, let $\epsilon > 0$ and find $N$ so that if $n,m\ge N$, we have
$$
\sup_k|x_k^{(n)}-x_k^{(m)}|\le\epsilon.
$$
Fix $k$ then and let $m\to\infty$ in the inequality $|x_k^{(n)}-x_k^{(m)}|\le\epsilon$ to find
$$
|x_k^{(n)}-x_k|\le\epsilon.
$$
The important point now is that this last inequality holds for any fixed $k$, so we conclude that if $n\ge N$,
$$
\sup_k|x_k^{(n)}-x_k|\le\epsilon.
$$
Together with your own proof that $z$ is actually a bounded sequence, this finishes the proof that $\ell^\infty$ is complete.
