How to prove that $f_n(x):=x^n$ is not a Cauchy sequence in $C[0,1]$ under the norm $\|f\|= \sup|f(x)|$? 
How to prove that $f_n(x)=x^n$ is not a Cauchy sequence in $C[0,1]$ under the norm $\|f\|= \sup_{x\in [a,b]}|f(x)|$, by showing that it does not satisfy the definition of a Cauchy sequence?

 A: Note that if $f_n(x)=x^n$ is Cauchy then since $(C[0,1],||.||_\infty)$ is complete so $f_n\to f$ where $f(x)=$\begin{cases} 0 &\text{$0\le x<1$}\\ 1& x=1\end{cases} must be in $C[0,1]$ but it is not so.
Alternatively,if the sequence is Cauchy
$||f_n-f_m||=\sup_{x\in [0,1]}|f_n(x)-f_m(x)|=\sup|x^n-x^m|\to 0$;
But on Putting $n=2m ;x=\dfrac{1}{2^{2m}}$; We have $||f_n-f_m||=\dfrac{1}{4}\neq 0$
A: For a fixed $n \ge 1$ and for $m \ge n$,
$$
             \|x^n-x^m\|=\max_{x\in[0,1]}(x^n-x^m) \ge (x^n-x^m)|_{x=1/2^{1/n}}
         = \frac{1}{2}-\left(\frac{1}{2}\right)^{m/n},
$$
Hence,
$$
         \|x^n-x^m\| \ge \frac{1}{4} \mbox{ whenever } \left(\frac{1}{2}\right)^{m/n} \le \frac{1}{4},
$$
which occurs whenever $m/n > 2$, or $m > 2n$. So $\{ x^n \}_{n=1}^{\infty}$ cannot be a Cauchy sequence in $C[0,1]$.
A: Fix $N\in\mathbb{N}$. Note that:
$$ \| f_N-f_{2N}\|=\sup_{0\le x\le 1} |x|^N|1-x^N|\ge (2^{-1/N})^N|1-(2^{-1/N})^N|=2^{-1}(1-2^{-1})=1/4$$
Since $N$ was arbitrary, this shows that the definition fails with $\varepsilon=1/4$. 
A: You want to show that
$$
\forall\epsilon>0,\exists N,\forall n,m:n,m>N\implies\|x^n-x^m\|=\max_{x\in[0,1]}|x^n-x^m|<\epsilon\tag{$\star$}
$$
is not satisfied.

Proposition: Every $\epsilon\in(0,1)$ fails to satisfy $(\star)$.

Proof: Choose any $\epsilon\in(0,1)$, $N\in\mathbb{N}$ and $m>N$. Let $x_0\in(0,1)$ be such that
$$
x_0^m=\epsilon+\frac{1-\epsilon}{2}
$$
Now, since $x_0^n\to0$ as $n\to\infty$, we see that upon choosing $n>m$ large enough we obtain
\begin{align}
|x_0^n-x_0^m| &= x_0^m-x_0^n\\
&=\epsilon+\frac{1-\epsilon}{2}-x_0^n\\
&>\epsilon
\end{align}
Remark: The idea is that for fixed $x_0$ and $m$, the value $x_0^m-x_0^n$ tends to $x_0^m$. Hence all we needed was to make sure that $x_0^m>\epsilon$.
