Disprove uniform convergence of $x^n$ where $ x\in[0,1)$

Question

So I'm looking at $f_n(x)=x^n ~~~x\in[0,1)$. It is obvious that it converges pointwise to $f(x)=0$. Is it now possible to disprove uniform convergence using the Uniform norm? This would mean I have to show $$\lim_{n\rightarrow\infty}||f-f_n||\neq0\qquad\text{and as }f=0\qquad\lim_{n\rightarrow\infty}||f_n||_D\neq0$$ With the definition of the Uniform norm: $$\lim_{n\rightarrow\infty}\sup\{x^n:x\in[0,1)\}\neq0$$ But isn't this false as $x^n$ converges to $0$ in this interval?

Or do I just misunderstood supremum, and it is actually meant the lowest value that is not reached, which would be $1$ (or at least $\neq0$?) and then everything would work out as $1\neq0$

Answer 1

Supremum is the lowest upper bound. Clearly $1$ is an upper bound of $\{x^n\mid x\in[0,1)\}$ and it's the least upper bound because $x^n\to1$ as $x$ goes to $1$. It does not have to be attained, but $x^n$ always reaches values arbitrarily close to $1$ on the interval $[0,1)$.

I suggest you look at the graph of $x^n$ for high $n$ on WolflamAlpha.

Answer 2

Hint: $$ \sup\{ x^n : x\in[0,1)\} = 1 $$ because for $r>0$, $$ \left(1-\frac rn\right)^n \ge 1-r $$ (the Bernoulli inequality)

Answer 3

$\lim_{n\to\infty}(1-\frac{1}{n})^n=\frac{1}{e}$ which contradicts uniform convergence.