Show that $\mathcal l^{\infty}$ with norm $\Vert x\Vert = \sup_{k\in \mathbb N}|x_k|$ is Banach space. Verfiying that it is a normed space was ok, but I've got a problem with completeness. There is one moment in which I cannot go further.
Suppose, that $(x^{(n)})$ is a Cauchy sequence in $\mathcal l^{\infty}$. Hence for every $\epsilon > 0$ there is a number $N$ such as, for every $m,n > N$ we have $\Vert x^{(n)}-x^{(m)} \Vert = \sup_{k\in \mathbb N}|x^{(n)}_k-x^{(m)}_k|< \epsilon$.
Which implies that for every $k\in \mathbb N$ we have $|x^{(n)}_k-x^{(m)}_k| < \epsilon$. From the completeness of $\mathbb R$ we receive that $(x^{(n)}_k)$ is convergent to some $x_k$. And what I'm trying to show is that a sequence $x = (x_1,x_2,...)$ is in $\mathcal l^{\infty}$ and $\Vert x^{(n)}-x \Vert \longrightarrow 0 $. Is is correct so far? Major problem for me is to show the boundedness of $x = (x_1,x_2,...)$.
Let $\varepsilon > 0$ be given. Since $(x^{(n)})$ is Cauchy, there exists $N \in \mathbb{N}$ such that for all $n, m \geq N$ you have $$\|x^{(n)} - x^{(m)}\|_\infty < \frac{\varepsilon}{2}.$$ Now, fix $k \in \mathbb{N}$. Then, you know that there is some $N_k \geq N$ such that $$/ |x^{(l)}_k - x_k| < \frac{\varepsilon}{2} $$ for all $l \geq N_k$.
Hence, for all $k \in \mathbb{N}$ and $n \geq N$ you get $$ |x_k^{(n)} - x_k| \leq |x_k^{(n)} - x_k^{(N_k)}| + |x_k^{(N_k)} - x_k| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon, $$ i.e. $$ \|x^{(n)} - x\|_\infty \leq \varepsilon. $$ Thus, the sequence converges in $l^\infty$ and since every convergent sequence is bounded, the claim follows.