Does this sequence admit a uniformly convergent subsequence on [0,1)? Let $(f_n)$ be the functions sequence difined on $[0,1],$ $$f_n(x)=x^n.$$
My question is:  
Does this sequence admit a uniformly convergent subsequence on [0,1)?  
Can any one help me on this question?
 A: There are two ways:


*

*Without Arzela-Ascoli: the uniform limit of a subsequence is a pointwise limit. The only possibility is the function which vanishes on $[0,1)$ and takes the value $1$ at $1$. But this function is not continuous. A uniform limit of continuous functions is continuous, hence no subsequence can be uniformly convergent.

*With Arzela-Ascoli: clearly, $\left\lvert f_n(x)\right\rvert\leqslant 1$ for any $n$ and any $x$, so the only thing which can fail is equi-continuity. For any $\delta$ and any $n\gt 1/\delta$, 
$$\left\lvert f_n(1)-f_n\left(1-\delta\right)   \right\rvert=1-\left(1 -\delta\right)^n \geqslant 1-\left(1 -\frac 1n\right)^n    $$
hence 
$$\sup_{\left\lvert x-y\right\rvert\lt\delta  } \left\lvert f_n(x)-f_n\left(y\right)   \right\rvert   \geqslant  1-\left(1 -\frac 1n\right)^n       $$
so that 
$$\sup_n \sup_{\left\lvert x-y\right\rvert\lt\delta  } \left\lvert f_n(x)-f_n\left(y\right)   \right\rvert   \geqslant  1-e^{-1}\gt 0.$$ 

A: Use this property to show that sequence is not uniformly convergent is$${\rm lim}\ [{\rm sup}\ \{ |  f_n(x)-f(x)|\ :\ x\in S\ \}]=1$$, where $S$ is the non zero value interval of the function.
