Suppose $f_n:\mathbb{R}\to\mathbb{R}$ is defined by $f_n(x)=x^n$ where $n\in\mathbb{N}$, and $D=[0,1)$. I wish to show that $f_n(x)=x^n$ does not converge to $f(x)=0$ uniformly as $n\to\infty$, i.e.
$\exists\epsilon>0,\forall N\in\mathbb{N}, \exists x\in D \text{ s.t. } \exists n\ge N \text{ s.t. } |f_n(x)-f(x)|\ge\epsilon$
My attempt:
Let $\epsilon=1$. Suppose $N\in\mathbb{N}$. Choose... I need some help on deciding what to choose for $x$ and $n$ to make the above statement true.
Suppose $f_n \rightarrow 0$ uniformly on $[0,1).$ Then (taking $\varepsilon = 1/2$) it should exists $N \in \mathbb{N}$ such that $$x^n = |f_n(x)-0| < \frac{1}{2} \qquad \text{for all } x \in [0,1) \text{ and } n \geq N.$$ But now take $n = N$ and $x$ as the $N$th-root of $1/2$ to get a contradiction.