I found that demonstration. I know one different. I need some explanation.
The demonstration:
$(a_{n})$ is a Cauchy sequence.
$|a_{k}-a_{j}|< \epsilon$ $\forall k,j>n_{\epsilon}$
We can consider $\epsilon=1$. $|a_{k}-a_{j}|< 1$ Then $a_{j}-1< a_{k}<a_{j}+1$
Now we define $m=min(a_{1},a_{2},...,a_{j}-1,a_{k},a_{j}+1)$ and $M=max(a_{1},a_{2},...,a_{j}-1,a_{k},a_{j}+1)$
$a_{k}$ with $k>j$ belongs to the interval (m,M) because $m\leqslant a_{j}-1< a_{k}<a_{j}+1\leqslant M$. In particular $m<a_{k}<M$ which ends our proof.
My doubts:
Why $k>j$ ? It should be true for any $k,j>n_{1}$
Why does it consider the minimum and maximum?
What does it happen for $k<j$?
$j$ is fixed (to define $m$ and $M$) and chosen $>n_\epsilon=n_1$ (e.g. $j=n_1+1$), and in that part of the proof (but see point 3), $k>j$, for $a_j-1< a_k<a_j+1$ to hold (but $k>n_1$ would be ok as well).
If we take only $m=a_j-1$ and $M=a_j+1$ (instead of min and max), the inequalities $m\le a_k\le M$ will only hold for all $k>n_\varepsilon$. We want them for all $k\in\mathbb N$.
for $k\le j$, we still have $m\le a_k$ because $m=\min(a_1,\dots,a_k,\dots,a_{j-1}, a_j-1)$. Similarily, we still have $a_k\le M$.