Verify the compactness of the set in $C[0,1]$ with metric sup: $$ F=\{x_{n}\in C[0,1]: x_{n}(t)=t^n, n\in\mathbb{N}\}$$
My attempt:
Consider $t_0\in [0,1]$.
$\forall\epsilon>0, \forall x_n\in F, \forall t\in [0,1]: |t-t_0|<\delta$, let choose $\delta=\dfrac{\epsilon}{n}$.
Then $|x_n(t)-x_n(t_0)|=|t^n-t_0^n|\leq |t-t_0|n<\epsilon$.
$\Rightarrow \sup_{t\in[0,1]}|x_n(t)-x_n(t_0)|\leq\epsilon.$
$\Rightarrow F$ is equicontinuous on $[0,1]$.
On the other hand, $\forall x_n\in F$ and $\forall t\in [0,1]$, we have $|x_n(t)|\leq 1$.
Apply Arzela-Ascoli theorem, we have $F$ is relatively compact in $C[0,1]$.
Is my proof right? And how can I prove if $F$ is compact or not? (I just proved it is relatively compact).
It is not relatively compact and it is not equi- continuous either. The pointwise limit of any subsequence is dis-continuous at $1$.
The flaw in your argument is your $\delta$ depends on $n$. That is not allowed.
Take $0<\epsilon <1-\frac 1 e$. Suppose There exists $\delta>0$ such that $|t^{n}-s^{n}|<\epsilon$ for all $n$ whenever $|t-s|<\delta$. Take $t=1-\frac 1 n, s=1$ to get $|(1-\frac 1n )^{n}-1|<\epsilon$ for all $n$ sufficienlty large. What happens when you let $n \to \infty$?