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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?

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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.

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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 $\lim_{n\to\infty}x_n=x$.

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