is $x^n$ uniformly convergence on $[0,1)$

Question

So I know $f_n(x)=x^n$ is not uniformly convergence on $[0,1]$ since

\begin{equation} f_n(x)= \begin{cases} 0, & \text{if}\ x<1 \\ 1, & \text{if}\ x=1 \end{cases} \end{equation}

and $f_n(x)\longrightarrow f(x)=0$, but what about when the domain is being reduced only to $[0,1)$? Is it still not uniformly convergence? If yes or no, why?

Answer 1

Hint: If $f_n\to f=0$ uniformly, what must happen to $$\sup_{x\in[0,1)}|f_n(x)-f(x)|=\sup_{x\in[0,1)}f_n(x)$$ as $n\to\infty$? What does happen to it?