$\newcommand{\F}{\mathbb{F}}$ $\newcommand{\N}{\mathbb{N}}$ 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.