How can I show that $\Phi=\{(t\mapsto t^n):n \in \mathbb{N}\}\subseteq C([0,1),\mathbb{R})$ is equicontinuous, that is
$\forall \epsilon>0: \exists \delta>0: \forall f \in \Phi: \forall x,y \in [0,1):|x-y|<\delta \Rightarrow |f(x)-f(y)| < \epsilon$
?
On
The sequence isn't equicontinuous on $C([0,1),\mathbb{R})$.
Let $\epsilon = 1/2$.
Give me your best arbitrary $\delta$.
Then pick $x_0 \in (1-\delta,1)$. Now pick $n$ so big that $x_0^n<1/4$. This is possible since
$$ \lim_{n \to \infty} t^n = 0 \qquad t<1$$ Then pick $x_1 \in (1-\delta,1)$ such that $x_1^n > 3/4$. This is always possible since
$$ \lim_{t \to 1} t^n = 1 \qquad \text{for any } n$$ Then by construction $|x_1-x_0|<\delta$ but $|x_1^n-x_0^n|>1/2$.
This proves the opposite of your claim, i.e. there exists some $\epsilon$, such that for all $\delta$'s, I can find $x,y,f$ such that $|x-y|<\delta$ but $|f(x)-f(y)|>\epsilon$.
The main problem is the endpoint $x=1$. Write $f_n(x) = x^n$. Taking $x_n = (1-1/n)$ we find that $$|f_n(x_n)-f(1)| = 1 - (1-1/n)^n \rightarrow 1-\mathrm{e}^{-1} \neq 0.$$ Equicontinuity would imply that this sequence tends to zero. Thus, $(f_n)_{n \in \mathbb{N}}$ cannot be equicontinuous.