# Why is it necessary to first reduce our case to a finite $J_{2} \in \mathbb N$ to show completeness of $\ell^{\infty}$

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.

• Because we dont know if $\sup\limits_{j \in \mathbb N}\vert x_{j}^{(n)}-x_{j}\vert$ is finite number. – Red shoes Jun 6 at 22:42
• 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 $n$. – Kavi Rama Murthy Jun 6 at 23:30
• If $N_\epsilon$ doesn't depend on $J_2$ then $J_2$ is useless. – reuns Jun 7 at 0:35