Compact closure in $C([0,2])$

Question

a) Does the closure of $\left\{f_n(x)=\sin(x^n):n=1,2,3\dots\right\}$ form the a compact subset of $C([0,2])?$

b) Does the closure of $\left\{f_n(x)=\sin(x^\frac1n):n=1,2,3\dots\right\}$ form the a compact subset of $C([0,2])?$

I think yes for a) because it is uniformly bounded and for equicontinuous $$\begin{align}|f_n(x)-f_n(y)|&=|\sin(x^n)-\sin(y^n)|\\ &\le|x^n-y^n|\\&=|x-y||x^{n-1}+x^{n-2}y+...+y^{n-1}|\\ &\le M|x-y|\lt \epsilon \end{align}$$

Isn't it correct? please, if I'm wrong then correct me.

I think b) is not true but i don't have right counterexample.

Your help will be appreciated..

Answer 1

Fact: the uniform limit of a sequence of continuous functions is continuous.

The problem is that the constant $M$ depends on $n$. If $f_{n_k}$ is a uniformly convergent subsequence, then for any $a\in [0,1)$, we have $f_{n_k}(a)\to 0$ while $f_{n_k}(1)=\sin(1)$. We conclude that the potential limit function is not continuous, hence from the "fact" we reach a contradiction.

For b), the only candidate is the function which takes the value $\sin(1)$ for $a\in (0,2]$ and $0$ at $0$, which is not continuous.