I'd like to prove that the power sequence
$f_n(x) = x^n$
doesn't converges uniformly on $[0,1]$, but it does on $[0,a]$ if $a < 1$.
My textbook states that a sequence of functions converges uniformly to $f(x)$ if
$ \forall \, \epsilon > 0 \,\, \exists \, \overline{n}_{\epsilon}: \forall x \in A \quad |f_n(x) - f(x)| < \epsilon$
So I found that $f_n(x)$ converges pointwise to $f(a,b) = \begin{cases} \text{$0 \leq x < 1 \implies 0$}\\ \text{$x = 1 \,\,\,\, \,\,\,\,\,\,\implies 1$} \end{cases} $
So I tried to apply the definition of uniform convergence, but I can't understand how could I find an $\epsilon > 0: \forall \, \overline{n}, \,\,\exists \, x \in [0,1], \,\,\, \exists \, n \geq \overline{n}: |f_n(x) - f(x)| \geq \epsilon$
Did I misunderstand something about the definition? Thank you in advance
In the way that I think is more constructive to think about it, uniform convergence of $f_n$ to $f$ on $A$ is equivalent to $\underset{x\in A}{\sup} \vert f_n(x)-f(x)\vert \overset{n\rightarrow \infty}{\rightarrow} 0$. So if for all $n$ there exists $x_n\in A$ such that:
$\vert f_n(x_n)-f(x_n)\vert\equiv c>0$
then you would show that there is no uniform to $f$. You have to consider just one $f$, since uniform convergence also implies point-wise convergence.
You could have also gone another route. If a sequence of continuous functions converges uniformly to $f$, then $f$ is a continuous function, which you can see that $f$ is not.