Show that $f_n(x) = x^n$ defines a sequence of equicontinuous functions... 
Show that $f_n(x) = x^n$ defines a sequence of equicontinuous  functions on the interval (0, 1) that does not admit a subsequence that converges uniformly on (0, 1).

Definitions
Let E be a set formed by real functions defined in $X\subseteq R$. We say that E is equicontinuous in $x_0 \in X$ if: Given $\epsilon \gt 0$, $\exists \delta \ge 0$ such that $x \in X, |x-x_0| \lt \delta$. $|f(x)-f(x_0)| \lt \epsilon$, $\forall f \in E$
We say that E is equicontinuous if E is equicontinuous throughout $x_0 \in E$.
A sequence $f_n x \to R$ is equicontinuous in $x_0$ if $E=$ {$fn: n \in N$} is equicontinuous in $x_0$: Given $\epsilon$, $\exists \delta$ such that $x \in X$, $|x-x_0 |\lt \delta$. $|f_n(x)-f_n(x_0)| \gt \epsilon$, $\forall n$
I can't think how to do this exercise. The sequence doesn't converge, right? When n goes to zero the sequence goes to 1 and when n goes to 1 the sequence goes to x, I don't think I understand. Can someone help me, thanks!
 A: Let $x_0\in (0,1)$ and let $\epsilon>0.$ Choose $\delta_0>0$ such that $x_0+\delta_0<1.$ Then $(x_0+\delta_0)^n\to 0$ as $n\to \infty.$ Thus there exists $N$ such that $n>N$ implies $2(x_0+\delta_0)^n <\epsilon.$
Since $f_1,\dots f_N$ are continuous at $x_0,$ there exist  $\delta_1,\dots, \delta_N >0$ such that $n\le N$ and $|x-x_0|<\delta_n$ implies $|f_n(x)-f_n(x_0)|<\epsilon.$
Now set $\delta=\min (\delta_0,\delta_1,\dots,\delta_N).$ If $n\le N$ and $|x-x_0|<\delta,$ then clearly $|f_n(x)-f_n(x_0)|<\epsilon.$ If $n>N$ and $|x-x_0|<\delta,$ then $|x-x_0|<\delta_0,$ and
$|f_n(x)-f_n(x_0)|= |x^n-x_0^n| \le x^n+x_0^n< 2(x_0+\delta_0)^n<\epsilon.$
It follows that $|x-x_0|<\delta$ implies $|f_n(x)-f_n(x_0)|<\epsilon$ for all $n.$ Therefore $\{f_n\}$ is equicontinuous at $x_0$ as desired.
Finally, the sequence $x,x^2,x^3,\dots $ is practically "the" canonical example of a sequence converging to $0$ pointwise on $(0,1),$ but not uniformly. This will also be true for any subsequence $f_{n_k}.$ I'll leave it here for now, but ask if you have questions.
