In $C[0,1]$ with $$\|f\| := \max_{[0,1]} |f(x)|$$ is $f_n (x) = x^n$, for $n=1,\dots, \infty$, a Cauchy sequence?
Suppose I have a big positive natural number N, and any m , n > N. Then I need to study $$ \|f_m - f_n\| = \|x^m - x^n\| = \max_{[0,1]} |x^m - x^n| $$ Then how can I know whether it is small than $\epsilon$ or not?
No, it is not a Cauchy sequence. If it was, then there would be some $N\in\Bbb N$ such that if $m,n\geqslant N$, then $\max|x^m-x^n|\leqslant\frac12$. But then, since the $\lim_{m\to\infty}\bigl|x^m-x^N\bigr|$ exists when $x\in[0,1)$, we have$$\lim_{m\to\infty}\bigl|x^m-x^N\bigr|\leqslant\frac12;\quad\text{in other words,}\quad x^N\leqslant\frac12,$$for every $x\in[0,1)$, since in this interval we have $\lim_{x\to\infty}x^m=0$. But this is impossible, since $\lim_{x\to1}x^N=1$.