Show, that $(c_0, \|\cdot \|_\infty)$ with $c_0:=\{f:\mathbb{N}\to\mathbb{C}\colon \lim_{n\to\infty} f(n)=0\}$ is a banach-space.
I already showed, that $(c_0, \|\cdot \|_\infty)$ is a normed vectorspace. I am struggeling to show, that every cauchy-sequence converges.
Let $(f_k)_k\in c_0$ be a cauchy-sequence. I have to find a limit $f\in c_0$. Hence $\|f_k-f\|<\varepsilon$ for every $\varepsilon>0$
$(f_k)\in c_0$ means, that $\lim_{n\to\infty} f_k(n)=0$ for every $k\in\mathbb{N}$. Therefore there is an $N_k(\varepsilon)\in N$ such that for every $n>N(\varepsilon)$ we have $|f_k(n)|<\varepsilon$, for every $k\in\mathbb{N}$
I want to show, that $0$ is a limit of $(f_k)$. Since $(f_k)$ is a cauchy sequence we have $\|f_k(m)-f_k(m')\|_\infty<\varepsilon$ for every $m,m'>N$ and $f_k(m), f_k(m')\to 0$. Hence $|f_k(m)|<\varepsilon/2$ for every $m>N_m(\epsilon)$ and $|f_k(m')|<\varepsilon/2$ for every $m'>N_m'(\epsilon)$. Now we choose $N(\varepsilon)=\max\{N_m(\varepsilon), N_m'(\varepsilon)\}$
We proceed:
$\|f_k(m)-f_k(m')\|_\infty\leq \|f_k(m)\|_\infty+\|f_k(m'\|_\infty\\=\sup_{n\geq m}|f_k(m)|+\sup_{n\geq m'}|f_k(m')|\leq \varepsilon/2+\varepsilon/2=\varepsilon$
I doubt that this is right. Can you help me out? Is this at least a try in the right direction? Thanks in advance.