Showing a functional sequence is not Cauchy wrt $\infty$ norm I’m trying to show that the functional sequence below is not Cauchy Sequence on $C[-1,1]$ space wrt $\|.\|_{\infty}$ which is equal to $\sup_{[-1,1]}|f_n(x)|$ but I cannot obtain something concrete for exact proof. Could you please fix my mistakes or warn me about unecessary things?
$$f_n(x) =
\begin{cases}
0,  & \text{if $-1\le x\le 0$} \\
nx, & \text{if $0\lt x\lt 1/n$} \\
1, & \text{if $1/n\le x\le 1$}
\end{cases}$$
I have used negation of Cauchy Sequence definition
$\exists \varepsilon_0 \gt 0$ $\forall N \in \mathbb N$ $\exists n,m \ge N$ $\|f_n-f_m\|_{\infty} \ge \varepsilon_0$
Let $m\gt n$ and $1/m \lt 1/n$ 
I’ve written 
$\|f_n-f_m\|_{\infty}$ = $\max\{\sup_{0\lt x\lt 1/m}|f_n-f_m|$ , $\sup_{1/m\lt x\lt 1/n}|f_n-f_m|\}$ = $\max \{\sup_{0\lt x\lt 1/m}|(n-m)x|$ , $\sup_{1/m\lt x\lt 1/n}|nx-1|\}$
I cannot find a $\varepsilon_0$
With all due respect, I think there ara many mistakes. I need your help
Thanks
 A: Let $m \ge n$. The functions $f_m$ and $f_n$ are equal on $[-1, 0] \cup \left[\frac1n, 0\right]$ so we have:
$$f_m(x) - f_n(x) = \begin{cases}
mx - nx,  & \text{if $x \in \left[0, \frac1m\right]$} \\
1-nx, & \text{if $x \in \left[\frac1m, \frac1n\right]$}\\
0, & \text{otherwise}
\end{cases}
 $$
Now we calculate
$$\|f_m - f_n\|_\infty = \max\left\{\sup_{x \in \left[0, \frac1m\right] } (m-n)x, \sup_{x \in \left[\frac1m, \frac1n\right] } (1-nx)\right\} = 1 - \frac{n}{m}$$
Therefore, if we set $m = 2n$ we get
$$\|f_{2n} -f_n\| = \frac12 \not\to 0$$
so the sequence $(f_n)_n$ cannot be Cauchy.

A more conceptual way to see that $(f_n)_n$ cannot be Cauchy is to recall that $C[-1,1]$ is a Banach space and that the uniform limit of continuous functions is itself a continuous function.

If $(f_n)_n$ were Cauchy, then by completeness of $C[-1,1]$ it would converge to an element of $C[-1,1]$. Since uniform convergence implies pointwise convergence, the only candidate for the uniform limit is the pointwise limit $\chi_{\langle 0, 1]}$. However, this is not a continuous function so the convergence cannot be uniform.
