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I have to prove that $(l^{\infty},\|\cdot\|_{\infty})$ is Banach space and I have some difficulties. This is what I've done.

$l^\infty=\{x=\langle x_k\rangle, k\in N|\exists M>0 \ such\ that\forall k\in N |x_k|\leq M\}$
$\|x\|_\infty=\sup\{|x_k|:k\in N\}$

Let $\langle x^n:n\in N\rangle$ be a Cauchy sequence in $l^\infty$.
$x^1=\langle x^1_1,x^1_2,...,x^1_k,...\rangle$
$x^2=\langle x^2_1,x^2_2,...,x^2_k,...\rangle$
.
.
.
$x^n=\langle x^n_1,x^n_2,...,x^n_k,...\rangle$

Since it is Cauchy sequence, I know that it holds
$(\forall\epsilon>0)(\exists n_0\in N)(\forall m,n\geq n_0)\|x^n-x^m\|_\infty<\epsilon$
$(\forall\epsilon>0)(\exists n_0\in N)(\forall m,n\geq n_0)\sup\{|x^n_k-x^m_k|:k\in N\}<\epsilon$
$(\forall\epsilon>0)(\exists n_0\in N)(\forall m,n\geq n_0)(\forall k\in N)|x^n_k-x^m_k|<\epsilon$

Now I can choose an arbitrary $k_*\in N$ such that
$(\forall\epsilon>0)(\exists n_0\in N)(\forall m,n\geq n_0)|x^n_{k_*}-x^m_{k_*}|<\epsilon$

This means that $\langle x^n_{k_*}:n\in N\rangle$ is Cauchy sequence in $(R,|\cdot|)$ which is complete, so $\langle x^n_{k_*}:n\in N\rangle$ is convergent. That means that there is $x_{k_*}\in R$ such that $\lim_{n\rightarrow\infty}x^n_{k_*}=x_{k_*}$.

Since $k_*$ was arbitrary, we can do this for all $k\in N$ so we get that there exists a sequence $x=\langle x_k\rangle$ such that $\lim_{n\rightarrow\infty}x^n_k=x_k$,$\forall k\in N$.

This sequence, $x=\langle x_k\rangle$ is a candidate for a limit of a sequence $\langle x^n\rangle$.

I have to show now that $x\in l^\infty$.
I know that $l^\infty$ is vector space. Also I know that for some $n_0\in N$ $x^{n_0}\in l^\infty$.
If I write $x$ as $x=(x-x^{n_0})+x^{n_0}$ it's enough to show that $x-x^{n_0}\in l^\infty$

I know from what I wrote above that
$(\forall\epsilon>0)(\exists n_0\in N)(\forall m,n\geq n_0)(\forall k\in N)|x^n_k-x^m_k|<\epsilon$
$(\forall\epsilon>0)(\exists n_0\in N)(\forall n\geq n_0)(\forall k\in N)(\forall m\geq n_0)|x^n_k-x^m_k|<\epsilon$
$(\forall\epsilon>0)(\exists n_0\in N)(\forall n\geq n_0)(\forall k\in N)\lim_{m\rightarrow\infty}|x^n_k-x^m_k|\leq\epsilon$
$(\forall\epsilon>0)(\exists n_0\in N)(\forall n\geq n_0)(\forall k\in N)|x^n_k-\lim_{m\rightarrow\infty}x^m_k|\leq\epsilon$
$(\forall\epsilon>0)(\exists n_0\in N)(\forall n\geq n_0)(\forall k\in N)|x^n_k-x_k|\leq\epsilon$
For $\epsilon=1$ I can find $n_0$ such that $(\forall k\in N) |x^{n_0}_k-x_k|\leq 1$

Since this hold for all $k\in N$, than it holds that $\sup\{|x^{n_0}_k-x_k|:k\in N\}\leq 1$
i.e. $\|x^{n_0}_k-x_k\|_\infty=\|x_k-x^{n_0}_k\|_{\infty}<1=\epsilon$ i.e. $(x_k-x^{n_0}_k)\in l^{\infty}$

Is this correct? If it is, then I have to show that $\lim_{n\rightarrow\infty} x^n=x$ and that's all.
If it is not, please help me.

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    $\begingroup$ It's correct so far. And your last step shows basically that if $\lVert x^n - x^m\rVert_\infty \leqslant \varepsilon$ for all $n,\,m \geqslant N$, then $\lVert x^n - x\rVert_\infty \leqslant \varepsilon$ for all $n \geqslant N$. $\endgroup$ Aug 14 '13 at 19:58
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    $\begingroup$ Well, you should try to make it shorter and with that more readable. But basically, that's it, you show that each coordinate is a Cauchy sequence, hence convergent, the boundedness of Cauchy sequences shows that the sequence of limits is bounded, then you show that the sequence of limits is the limit. $\endgroup$ Aug 14 '13 at 20:04
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    $\begingroup$ Yes, I agree with Daniel, so far is correct, but you need to finish your proof. Your last argument shows that $x - x^n \in l^{\infty}$, hence $x = (x-x^n) + x^n \in l^{\infty}$ too. The last computations you did should show that $x^n \to x$. $\endgroup$ Aug 14 '13 at 20:04
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    $\begingroup$ @gov : Don't get me wrong, what you wrote is not invisible, I can read it! I'm just saying that the presentation is very heavy. You should learn to lighten it up, it would help you understand yourself better. $\endgroup$ Aug 14 '13 at 20:05
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    $\begingroup$ @gov : I'm sorry that you professor either taught you to do that or led you to think that you had to. I'm confident in saying that this is not an optimal way to write mathematics, i.e. to write plenty of lines full of repeated symbols. I am also confident that you understand mathematics, since what you wrote up there is non-trivial and needs understanding of mathematics to be done ; so I think that if you find a way to put less symbols and allow someone to read what you wrote twice as fast, you would have improved by a huge factor. =D That is what I would suggest to improve your answer above. $\endgroup$ Aug 14 '13 at 22:27

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