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.
I'm not really following your proof. Remember, you're showing that every Cauchy sequence converges. You've set up the problem by assuming you have a Cauchy sequence $(f_k)_{k \in \mathbb{N}} \in c_0$, which is good, but now you need to nominate a limit for this function, and prove it.
The first thing to notice is that the "orthogonal" sequences $(f_k(n))_{k\in \mathbb{N}}$ are all Cauchy for all $n \in \mathbb{N}$. This is because, $$|f_k(n) - f_l(n)| = |(f_k - f_l)(n)| \le \sup_{m \in \mathbb{N}} |(f_k - f_l)(m)| = \|f_k - f_l\|_\infty,$$ and $(f_k)$ is Cauchy. Since $\mathbb{R}$ is complete, there exist limits for the sequences, $f_\infty(n)$, for each $n \in \mathbb{N}$.
You now need to show two things: $f_\infty \in c_0$ and $f_k \rightarrow f_\infty$ as $k \rightarrow \infty$. Let me know if you want any more help.
EDIT: We wish to show that $f_\infty(n) \rightarrow 0$. Fix $\varepsilon > 0$, and choose $N_1$ such that $$k, l \ge N_1 \implies \|f_k - f_l\| < \frac{\varepsilon}{3}.$$ Fix $k \ge N_1$. Since $f_k(n) \rightarrow 0$ as $n \rightarrow \infty$, there exists some $N_2$ such that $$n \ge N_2 \implies |f_k(n)| < \frac{\varepsilon}{3}.$$ From the inequality before the edit, we get $$n \ge N_2 \text{ and } l \ge N_1 \implies |f_l(n)| < \frac{2\varepsilon}{3}.$$ Consider the above inequality as $l \rightarrow \infty$. We may choose $N_3$ such that $$l \ge N_3 \implies |f_l(n) - f_\infty(n)| < \frac{\varepsilon}{3},$$ so given $n \ge N_2$ and $l \ge \max \lbrace N_3, N_1 \rbrace$, we have, $$|f_\infty(n)| \le |f_\infty(n) - f_l(n)| + |f_l(n)| < \frac{\varepsilon}{3} + \frac{2\varepsilon}{3} = \varepsilon.$$ Note that $N_2$ depended only on $\varepsilon$ (via fixed $k$ and $N_1$). Therefore, $f_\infty(n) \rightarrow 0$, i.e. $f_\infty \in c_0$.
Next, we show $f_k \rightarrow f_\infty$, i.e. $\|f_k - f_\infty\|_\infty \rightarrow 0$ as $k \rightarrow \infty$. Fix $\varepsilon > 0$. Using Cauchiness of $(f_k)$ as before, fix $N_1$ such that, $$k, l \ge N_1 \implies \|f_k - f_l\|_\infty < \frac{\varepsilon}{2} \implies \forall n \in \mathbb{N}, |f_k(n) - f_l(n)| < \frac{\varepsilon}{2}.$$ Taking the limit as $l \rightarrow \infty$, we obtain, $$k \ge N_1 \implies \forall n \in \mathbb{N}, |f_k(n) - f_\infty(n)| \le \frac{\varepsilon}{2} \implies \|f_k - f_\infty\|_\infty \le \frac{\varepsilon}{2} < \varepsilon,$$ as required.