I'm trying to prove that $l_p ^\infty $ is complete for each $p\geq 1$ but only with the definition of $\varepsilon$-$N$.
I know that this have been proved in other posts here but I couldn't find a proof with the $\varepsilon$-$N$ method only.
Here is my try for $p=1$:
let $\{x^{(n)}\}_{n\in \mathbb{N}}$ be a Cauchy sequence in $l_1 ^\infty$. Then it is easy to show that for each fixed $i\in \mathbb{N}$, the sequence $\{x_i ^{(n)}\}_{n\in \mathbb{N}}$ converges to some $x_i\in \mathbb{R}$.
Now it is left to show that the vector $x=(x_1,x_2,\ldots)$ is in $l_1 ^\infty$ and that $x^{(n)}\longrightarrow x$.
This is the part where I can't prove only by the $\varepsilon$-$N$ method.
I know that for each $k\in \mathbb{N}$, I can define $N=\max \{N_i |1\leq i\leq k\}$
such that $\displaystyle{\sum_{i=1} ^k}|x_i ^{(n)} -x_i|<\varepsilon$ for all $n\geq N$.
What most of the proofs I saw do now is to take $k\longrightarrow \infty$ to get: $\displaystyle{\sum_{i=1} ^\infty}|x_i ^{(n)} -x_i|\leq\varepsilon$ for all $n\geq N$.
The problem with this is that $N$ depends on $k$ which means that $n$ can't be held constant when $k\longrightarrow \infty$. How can I fix this?
Once we have found the coordinstewise limit, we have to go back to the $\ell^1$-norm. For each positive $\varepsilon$, there exists an integer $N$ such that if $m,n\geqslant N$, then $$\left\lVert x^{(m)}-x^{(n)}\right\rVert_1=\sum_{i=1}^{+\infty} \left\lvert x_i^{(m)}-x_i^{(n)}\right\rvert \lt \varepsilon.$$ In particular, for any $k$, $$ \sum_{i=1}^{k} \left\lvert x_i^{(m)}-x_i^{(n)}\right\rvert \lt \varepsilon $$ provided that $m,n\geqslant N$. So $N$ does not depend on $k$.