# Convergence in the space of convergent sequences


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)?

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.