About completeness of $l^{\infty}$ with respect to sup norm Let $l^{\infty}$ be the space of all bounded sequences of real numbers $(x_n)_{n =1}^{\infty}$ with the sup norm. I have to show that $l^{\infty}$ is complete with respect to this norm.
Proof: In the proof below I am confused with the sequence $x^n = (x_1^{n},x_2^{n}\ldots )$. I am not able to visualize this sequence. How this sequence can be formed? Is $(x^{n})$ is collection of cauchy sequences?  
$x_1^{n}$,$x_2^{n}$ are different cauchy sequences they may be converging to different points so how can we assume that $(x^{n})$ will converge to fixed point $x$?   
Please help me to understand this. Any numerical example supporting this will be very much helpful.  Thanks
 A: Elements of $l^\infty$ are sequences. You have to show that a Cauchy sequence in $l^\infty$ converges.
Using a different notation, consider a Cauchy sequence $(x_n)$ in $l^\infty$. Each $x_n$ is itself a sequence; let's denote it by
$$
x_n(0), x_n(1), \dots
$$
So, if $x_n$ is the sequence $1, 1/3, 1/5,\dotsc$ (the reciprocals of the odd integers), you'd have $x_n(0)=1$, $x_n(1)=1/3$ and so on. Thus $x_n(k)$ means "the $k$-th term in the $n$-th element of the sequence we're starting with". The book page you show simply writes this as $x^n_k$.
So every $x_n$ (or $x^n$ in the book's notation) is a bounded sequence, not a Cauchy one, because it's an element of $l^\infty$. It's not at all assumed that $x_n$ converges (most elements of $l^\infty$ don't). It's the sequence (of sequences) $x_0,x_1,\dotsc$ that is a Cauchy sequence under the sup norm.
The fact that $(x_n)$ is a Cauchy sequence means that, for all $\def\eps{\varepsilon}\eps>0$, there is $N$ such that, for $n,m>N$
$$
\|x_n-x_m\|<\eps
$$
that is,
$$\sup_k|x_n(k)-x_m(k)|<\eps,$$
which implies
$$
|x_n(k)-x_m(k)|<\eps, \text{ for all } k.
$$
(it would be equivalent if we used $\le\eps$ throughout).
