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$\newcommand{\F}{\mathbb{F}}$ $\newcommand{\N}{\mathbb{N}}$ Let $\F$ be the field of real/complex numbers and define $$c(\F)=\{s\colon\N\to\F|~\exists t\in\F~\text{such that }\lim_{i\to\infty}|s(i)-t|=0\}.$$ I want to prove that this space is complete w.r.t. the norm $\|s\|=\sup_{i\in\N}|s(i)|$.

I've found another post with a similar question and I followed the hints, but I got stuck in the proof.

This is how far I've gotten:

Let $(s_n)$ be Cauchy and fix $i\in\N$. Then the Cauchy property implies that $\{s_n(i)\}_n$ is Cauchy in $\F$. By completeness of $\F$, there is a limit, say $s(i)$. Define $s=(s(1), s(2), \ldots)$.

I have successfully shown $$\lim_{n\to\infty}\|s_n-s\|=0\implies s\in c(\F).$$ How can I prove $\lim\limits_{n\to\infty}\|s_n-s\|=0$ (straight from the definitions)?

Thanks in advance.

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The "best" proof is probably to show that $B(X)$, the set of bounded functions from a set $X$ to $\mathbb F$, is a Banach space under the uniform norm, and then to show that $c(\mathbb F)$ is a closed subspace of $\ell_\infty = B(\mathbb N)$. – kahen Sep 19 '12 at 21:42

Note that $$\begin{align} \lim_{n\to\infty}\|s_n-s\| &=\lim_{n\to\infty}\sup_{i\in\mathbb N}\left|s_n(i)-\lim_{m\to\infty}s_m(i)\right|\\ &=\lim_{n\to\infty}\sup_{i\in\mathbb N}\lim_{m\to\infty}\left|s_n(i)-s_m(i)\right|\\ &\leq\lim_{n\to\infty}\lim_{m\to\infty}\sup_{i\in\mathbb N}\left|s_n(i)-s_m(i)\right|\\ &\leq\lim_{n\to\infty}\lim_{m\to\infty}\|s_n-s_m\|=0\\ \end{align}$$ by the definition of a Cauchy sequence.

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Why is the first ``$\leq$"? – Jack Oct 1 '15 at 21:59

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