# Confused to Prove Banach Space.

Prove $$\mathcal{l}^\infty$$ for all complex sequence with $$\Vert x\Vert_\infty = \sup\limits_{i\in\mathbb{N}}\vert x_i\vert$$ is Banach space.

A Banach space is a normed linear space that is a complete metric space with respect to the metric derived from its norm.

To prove this question, let $$\{x_n\}$$ be a Cauchy sequence, $$(\forall\varepsilon>0)(\exists N\in\mathbb{N})\text{ such that }\forall{m,n>N}, \vert x_m-x_n\vert<\varepsilon.$$ So, we have $$\begin{eqnarray} \Vert x_m-x_n\Vert_\infty&=&\sup\limits_{i\in\mathbb{N}}\vert x_{m_i}-x_{n_i}\vert\\ &=&\vert x_m-x_n\vert\\ &<&\varepsilon \end{eqnarray}$$ Now I confuse to conclude that complete metric space(Cauchy sequence is convergent). Anyone can explain me how to prove Banach space?

Hints: Elements of this space are themselves sequences. So a Cauchy sequence is of the type $$(x^{n}_1,x^{n}_2,...)$$ , $$n=1,2...$$ with the property that $$\sup_i |x^{n}_i-x^{m}_i| \to 0$$ as $$n,m \to \infty$$. When this holds $$(x^{1}_i,x^{2}_i,...)$$ is a Cauchy sequence of real numbers for each $$i$$. So $$x_i=\lim_{n \to \infty} x^{n}_i$$ exists for each $$i$$. This gives a new sqequence $$(x_i)$$. Now try to show that this sequence is bounded and the original Cauchy sequence converges to this element.
There is something strange in your question, when you write about $$\lvert x_m-x_n\rvert$$. The idea is to prove that $$\ell^\infty$$ is complete and therefore to prove that every Cauchy sequence $$(x_n)_{n\in\mathbb N}$$ of elements of $$\ell^\infty$$ converges. But then you should write $$\lVert x_m-x_n\rVert$$ instead of $$\lvert x_m-x_n\rvert$$.
Anyway, if $$(x_n)_{n\in\mathbb N}$$ is a Cauchy sequence of elements of $$\ell^\infty$$, each $$x_n$$ belongs to $$\ell^\infty$$ and, in particular, it is a bounded sequence $$\bigl(x_n(k)\bigr)_{k\in\mathbb N}$$. For each $$k\in\mathbb N$$, $$\bigl(x_n(k)\bigr)_{n\in\mathbb N}$$ is a real Cauchy sequence and therefore it converges to some $$x_n\in\mathbb R$$. If $$s=(x_n)_{n\in\mathbb N}$$, prove that $$\lim_{n\to\infty}x_n=x$$.