Prove that $(l^\infty,\|.\|_\infty)$ is a Banach space. $(l^\infty,\|.\|_\infty)$ is a Banach space. In the proof, $\mathbb{F}$ is either the field of complex numbers or the field of real numbers.

Proof: Let $(x^n)_{n\in\mathbb{N}}$ be a Cauchy sequence in $l^\infty$, where $x^n=(x_k^n)_{k\in\mathbb{N}}$. Consider the sequence $(x^0_k,x^1_k,\cdots,x^n_k,\cdots)$ of $k$th coordinates of the sequence $(x^n)_{n\in\mathbb{N}}$. 
Let $\epsilon>0$. Since $(x^n)_{n\in\mathbb{N}}$ is Cauchy, there exists $n_0\in\mathbb{N}$ such that $$\forall m,n>n_0,\|x^m-x^n\|_\infty<\epsilon.$$ 
Therefore for each $m,n>n_0$ we have $$|x^m_k-x^n_k|<\epsilon.$$ 
So the sequence $(x^0_k,x^1_k,\cdots,x^n_k,\cdots)$ is Cauchy. Therefore it converges to some $y_k\in\mathbb{F}\ $ ($\mathbb{F}$ is complete). 
Let $y=(y_k)_{k\in\mathbb{N}}$. Since $(x^0_k,x^1_k,\cdots,x^n_k,\cdots)$ is Cauchy it is bounded. Choose $M>0$ such that for each $n\in\mathbb{N}$, $|x^n_k|<M$. But since 
$$y_k=\lim_{n\to\infty}x_k^n,$$ we have $|y_k|\leq M$ for each $k$. Therefore, $y\in l^\infty$. 
Fix $m>n_0$. Then we have $\|x^m-x^n\|_\infty<\epsilon.$ Therefore $\|x^m-y\|_\infty<\epsilon$ as $n\to\infty$. 
Therefore for each $m>n_0,\ \|x^m-y\|_\infty<\epsilon$. Hence $(x^n)_{n\in\mathbb{N}}$ converges in $l^\infty$ and the proof is complete.

Could someone please tell me if the above proof is alright? Thanks.
 A: Your proof is very rigorous and very detailed all the way up to the point where you say

Fix $m>n_0$. Then we have $\|x^m-x^n\|_\infty<\epsilon.$ Therefore $\|x^m-y\|_\infty<\epsilon$ as $n\to\infty$. 

Now I know the inequality stands, but as you were very thorough with all your other inequalities, I think it would be nice if you wrote a little more justification for this one as well - it is not entirely obvious how the right inequality follows from the left one.

Other than that, the proof is very well written and easy to follow.
A: Rather a comment, but my reputation does not allow me to comment yet. You write 

choose $M>0$ that for each $n\in\mathbb{N}, |x^n_k|<M$

I am curios whether you have to write $M_k$, since hypothetically $M$ can depend on the choice of the sequence $(x^0_k,x^1_k,\cdots,x^n_k,\cdots)$. But then we have difficulties with proving that $y\in l^\infty$.
EDIT: the proof is more subtle than I initially thought. First we have to settle the issue 5xum mentioned. We start with
$$
\forall k\in\mathbb{N}\ \forall \epsilon > 0 \ \exists n_0:\forall n,p>n_0:|x^n_k-x^{n+p}_k|<\epsilon.
$$
This is inequality in $\mathbb{F}$, so we can let $p\to\infty$. Thus we obtain
$$
\forall k\in\mathbb{N}\ \forall \epsilon > 0 \ \exists n_0:\forall n>n_0:|x^n_k-y_k|<\epsilon.
$$
That means $\|x^n-y\|_\infty<\epsilon$, thus $(x^n)_{n\in\mathbb{N}}$ is converging to $y$.
We still have to prove that $y\in l^\infty$. But
$$
\|y\|_\infty \le \|x^n-y\|_\infty + \|x^n\|_\infty \le \epsilon + \|x^n\|_\infty < \infty
$$
A: You said:
Fix $m> n_0$. Then we have $\| x^m - x^n \|_{\infty} < \epsilon$. Therefore
\begin{eqnarray}
\| x^m - y \|_\infty < \epsilon  \quad \text{as } n \to \infty
\end{eqnarray}
I try to clean up this statement but then I have a new question after this.
\begin{eqnarray}
\lim_{n \to \infty} \| x^m - x^n \| &=& \lim_{n \to \infty} \sup_{i \in \mathbb{N}}  | x_i^m - x_i^n | \\
&=& \sup_{i \in \mathbb{N}} |  x_i^m - \lim_{n \to \infty} x_i^n| \\
&=& 
\sup_{i \in \mathbb{N}} | x_i^m - y_i | = \| x^m - y \| \le \epsilon
\end{eqnarray}
How  do you justify interchanging the order of the limit and the sup?
Note also that the limit is $\le$ and not necessarily $<$. 
Think, for example, about the sequence $(1/n)$ all terms are $>0$ but the limit is $=0$.
To justify the interchange between the sup and lim consider the following reasoning:
\begin{eqnarray*}
      \lim_{m \to \infty} \| x_n - x_m \| &=& \lim_{m \to \infty} 
      \sup_{i \in \mathbb{N}} | x_i^n-x_i^m | \\
      &=&\lim_{m \to \infty} \left ( \sup_{i \in \mathbb{N}}
      |x_i^n - x_i^1 | , \sup_{i \in \mathbb{N}}
      |x_i^n - x_i^2|, \cdots , 
      \sup_{i \in \mathbb{N}}
      |x_i^n - x_i^m|, \cdots 
      \right )
 \end{eqnarray*}
 If such a limit exists it is given by 
 \begin{eqnarray*}
 \sup_{i \in \mathbb{N}} | x_i^n - x_i^{\infty}| = \sup_{i \in \mathbb{N}}|x_i^n- y_i|.
\end{eqnarray*}
The case of $\le \epsilon$ is easy. Just asume $\epsilon/2$ above and then $\epsilon/2 < \epsilon$.
