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Let $(x^{(n)})_{n}\subset\ell^{\infty}$ be a Cauchy sequence (w.r.t. $\vert \vert \cdot \vert \vert _{\infty}$). Thus for any $\epsilon >0 $ there exists $N \in \mathbb N$ so that for all $n,m \geq N$ and for any $J_{1}\leq J_{2}\in \mathbb N: \sup\limits_{J_{1}\leq j \leq J_{2}}\vert x_{j}^{(m)}-x_{j}^{(n)}\vert<\epsilon$

This immediately implies that $(x_{j}^{(n)})_{n}\subset\mathbb C$ for all $j \in \mathbb N$ is a cauchy sequence and since $\mathbb C$ is complete $\lim\limits_{n \to \infty}x_{j}^{(n)}=x_{j}\in \mathbb C$ for all $j \in \mathbb N$. Define $x:=(x_{j})_{j\in \mathbb N}$ and show $\in \ell^{\infty}$. $\sup\limits_{J_{1}\leq j \leq J_{2}}\vert x_{j}\vert=\sup\limits_{J_{1}\leq j \leq J_{2}}\vert x_{j}-x_{j}^{(n)}+x_{j}^{(n)}\vert\leq \sup\limits_{J_{1}\leq j \leq J_{2}}\vert x_{j}-x_{j}^{(n)}\vert+\vert x_{j}^{(n)}\vert \leq\epsilon+\vert\vert x^{(n)}\vert \vert_{\infty}<\infty$ and since this holds for any $J_{1}\leq J_{2}\in \mathbb N$: it holds for $\vert\vert x\vert \vert_{\infty}\Rightarrow x \in \ell^{\infty}$

My question: Why have we introduced finite $J_{2}$ and $J_{1}$, rather than looking at $\sup\limits_{j \in \mathbb N}\vert x_{j}-x_{j}^{(n)}\vert<\epsilon$. Is it because we can only use the fact that $\lim\limits_{m \to \infty}x_{j}^{(m)}=x_{j}$ so that $\sup\limits_{J_{1} \leq j\leq J_{2}}\vert x_{j}^{(n)}-x_{j}\vert=\lim \limits_{m \to \infty}\sup\limits_{J_{1} \leq j\leq J_{2}}\vert x_{j}^{(n)}-x_{j}^{(m)}\vert$? In other words we cannot take out the limit when looking at $\sup\limits_{j \in \mathbb N}\vert x_{j}^{(n)}-x_{j}\vert$

Any clarification as to why finite $J_{1}, J_{2}$ are used would be excellent.

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  • $\begingroup$ Because we dont know if $\sup\limits_{j \in \mathbb N}\vert x_{j}^{(n)}-x_{j}\vert$ is finite number. $\endgroup$ – Red shoes Jun 6 at 22:42
  • $\begingroup$ The proof you have written has errors. You have not specified what $n$ is. You have to bring in $N$ to show that $x \in \ell^{\infty}$. The bound $\epsilon$ for the first term is not valid for all $n$. $\endgroup$ – Kavi Rama Murthy Jun 6 at 23:30
  • $\begingroup$ If $N_\epsilon$ doesn't depend on $J_2$ then $J_2$ is useless. $\endgroup$ – reuns Jun 7 at 0:35

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