Let $(c, \|.\|_{\infty})$ be the vector space of all real convergent sequence with sup norm. Is the space complete?
I know $(l^\infty, \|.\|_{\infty})$ is a Banach space and $c$ is subset of $l^\infty$. If I can show that $c$ is closed the it becomes complete. But all this is if the $c$ is known to be complete. What do you say about it? Is it complete? If not what Cauchy sequence in $c$ can you provide which is not convergent convergent with $\sup$ norm in $c$?
Best Answer
Yes, $c$ is closed. For each $n\in\mathbb N$, let $(x(n)_k)_{k\in\mathbb N}$ be an element of $c$ and suppose that $\bigl(x(n)\bigr)_{n\in\mathbb N}$ converges (in $l^\infty$) to some sequence $x=(x_n)_{n\in\mathbb N}$. For each $n\in\mathbb N$, let $l_n$ be the limit of the sequence $x(n)$. The sequence $x(n)_{n\in\mathbb N}$ is a Cauchy sequence in $l^\infty$. Therefore, $(l_n)_{n\in\mathbb N}$ is a Cauchy sequence of real numbers and so it converges to some $l\in\mathbb R$. I will prove that $\lim_{n\to\infty}x_n=l$, thereby proving that $(x_n)_{n\in\mathbb N}\in c$.
Take $\varepsilon>0$ and take $\varepsilon'\in(0,\varepsilon)$. There is some $p\in\mathbb N$ such that$$n\geqslant p\implies\bigl\|x-x(n)\bigr\|_\infty<\varepsilon'.$$Therefore, since $l=\lim_{n\to\infty}l_n$ and $l_n=\lim x(n)$,$$n\geqslant p\implies|l-l_n|\leqslant\varepsilon'<\varepsilon.$$
Since $c$ is closed in $l^\infty$ and $l^\infty$ is complete, $c$ is complete.