Verify the compactness of the following set in $C[0,1]$ with metric sup $$F=\{x_{\alpha}\in C[0,1]: x_{\alpha}(t)=\sin\alpha t, > \alpha\in[1,2]\}.$$
My attempt:
Consider $t_0\in [0,1]$.
$\forall x_{\alpha}\in F, \forall t\in [0,1]: |t-t_0|<\delta$. Let choose $\delta=\dfrac{\epsilon}{2}.$
$\Rightarrow |x_{\alpha}(t)-x_{\alpha}(t_0)|=|\sin\alpha t-\sin\alpha t_0|\leq|\alpha t-\alpha t_0|<2.\frac{\epsilon}{2}=\epsilon$.
$\Rightarrow \sup_{t\in[0,1]}|x_{\alpha}(t)-x_{\alpha}(t_0)|\leq\epsilon$
$\Rightarrow F$ is equicontinuous on $[0,1]$.
On the other hand, $\forall x_{\alpha}\in F, \forall t\in [0,1]$, $|x_{\alpha}(t)|\leq 1$.
According Arzela-Ascoli theorem, $F$ is relatively compact in $C[0,1]$.
Is my proof correct? And how can I prove it is compact or not? Thanks in advance
Your proof of relative compactness is correct. To show that $F$ is compact we have to show that it is closed. Suppose $\sin(\alpha_n t)$ converges to some element $g$ of $C[0,1]$. Since $[1,2]$ is compact there is a subsequence $(\alpha_{n_i})$ converging to some $\alpha \in [1,2]$. But then $g(t)=\lim \sin (\alpha_{n_i}t)= \sin (\alpha t)$ so the limit function $g$ belongs to $F$.