# Proof that the normed space $(l^{\infty},\|\cdot\|_{\infty})$ is Banach space

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.
• 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 $n \geqslant N$. Aug 14 '13 at 19:58
• 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 $x^n \to x$. Aug 14 '13 at 20:04