i am currently trying to solve an exercise that says: "Let $E=[-1,1]$ be the closed interval and let $C(E,\mathbb{R})$ be the set of continuous functions $f:E\rightarrow\mathbb{R}$, equipped with the metric $D(f,g)=\text{sup}_{t\in E}|f(t)-g(t)|$. Consider the subset $C^1(E,\mathbb{R}):=\{f\in C(E,\mathbb{R}):\text{f is cont. diff. on the interior of }E\}$
1: Is $C^1(E,\mathbb{R}) \text{ closed in }C(E,\mathbb{R})$?
2: Is the metric space $(C^1(E,\mathbb{R}),D)$ complete?"
My idea so far for the first part was the following:
Assume $C^1\text{ is closed in }C\implies C\setminus C^1\text{ is open.}$
Then $\forall f\in C\setminus C^1\exists\epsilon>0\text{ s.t. } U_\epsilon(f)\subset C\setminus C^1\Leftrightarrow \forall f\in C\setminus C^1\;\exists\epsilon>0\text{ s.t. } U_\epsilon(f)\cap C^1=\emptyset$
Right here i am struggling a bit with proving or disproving this. I just found an example like choosing $f(x)=|x|$ and $g\in U_\epsilon(f)\text{ s.t. }g(x)=\sqrt{x^2+\delta}$, then we can choose $\delta(\epsilon)$ to make $g$ fit within the "$\epsilon$-tube" around $f$, with $g\in C^1$, so we have a contradiction, which means, that $C^1$ is not closed in $C$. But to be honest, i wouldn't have come up with this idea by myself. Is there a more like "rigorous way", to prove that it is not closed?
Im very thankful for every kind of feedback and advice, of course also on mistakes i made!