uniform convergence of $x^n$ on (0,a) for 0<a<1

Question

this is a question of my school previous years sample questions. I feel like somehow it is not correct? what do you think?
define $f_n(x)=x^n$
a) fix $0<a<1$. Show that seq $\{f_n(x)\}$ converges uniformly on (0,a).

I believe this sequence of functions does NOT converge uniformly on $(0,a)$ since let $\epsilon=\frac{1}{4}$ and let $n_k=k$ and let $\{x_k\}=\{(\frac{1}{2})^{\frac{1}{k}}\}$ then $|f_n(x_k)-f(x_k)|=\frac{1}{2}>\epsilon$

b)is there any sequence $\{f_{n_k}\}$converging uniformly on $(0,1)$

the logic of this problem is not clear to me. if in part a, we prove the whole sequence is convergent, then any subsequence of that is also convergent. I do not know how to answer these two questions and I appreciate any help.

Answer 1

$0< x^{n} <a^{n}$ for all $x \in (0,a)$ and $a^{n} \to 0$as $n \to \infty$. Hence $x^{n}$ does converge to $0$ uniformly. To be explicit $|f_n(x)| <\epsilon$ for all $x$ whenver $n >\frac {\log (\epsilon)} {\log a}$ (Observe that $\log (\epsilon)$ and $\log a$ are both negative numbers).

Part (b): The convergence is not uniform on $(0,1)$ since $f_n(1-\frac 1 n)=(1-\frac 1 n)^{n} \to \frac 1 e$.