I know that by continuity transfer property we can show above very easily. But I wanted to it by definition.
My attempt:
Uniform convergence $f_n(x)\to f(x)$ on $E$ means$$\forall \epsilon>0 ,\exists N\in \mathbb N ,\forall x\in E |f_n(x)-f(x)|<\epsilon ,\forall n>N.$$
Negation of above statement is $\exists \epsilon>0 ,\forall N\in \mathbb N \exists x\in E \exists n>N |f_n(x)-f(x)|>\epsilon $ .
I wanted to use negation to show that $x^n $ is not uniform convergent .
What $\epsilon $ I have to choose so that I get answer?
Any help will be appreciated
You can take $\varepsilon=\frac12$, for instance.
If the sequence converged uniformly, then it would converge pointwise. But$$\lim_{n\to\infty}x^n=\begin{cases}0&\text{ if }x<0\\1&\text{ otherwise.}\end{cases}$$So, define$$\begin{array}{rccc}f\colon&[0,1]&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}0&\text{ if }x<0\\1&\text{ otherwise.}\end{cases}\end{array}$$
Now, take any $N\in\mathbb N$. If it was true that $(\forall x\in[0,1]):\bigl\lvert x^N-f(x)\bigr\rvert<\frac12$, that would imply that $x^N<\frac12$ if $x\in[0,1)$. But this cannot be true, since $x^N$ is continuous on $[0,1]$ and it maps $1$ into $1$. So, by the intermediate value theorem, it cannot jump from $1$ to a number smaller that $\frac12$.
In the previous paragraph, you can avoid any reference to continuity by noting that $\frac1{\sqrt[N]2}\in[0,1]$ and that $\left(\frac1{\sqrt[N]2}\right)^N=\frac12$.