Limit points and closed subsets of $\mathcal{C}([0,1])$ Let $\mathcal{C}([0,1])$ be set the of continuos real functions equipped with the norm $\Vert  f\Vert = \sup{|f(x)|}$
Let $F=\{f\in \mathcal{C}:f(x) \ge 0 \ \ \forall \ x \in [0,1]\}$
I'm trying to find all the limit points of the set $F$ and decide if F is closed.
I've previously showed that the set of positive functions is open in $\mathcal{C}([0,1])$, but I'm having trouble working this one out. Would proving that the complement is open work in this case? Working out that the set of negative functions is open seems reasonable, but that seems suspiciously simple.
I'm also having trouble figuring out what a limit point of this set would look like. Working from definition we would have that every that in each
$B_{\epsilon}(f) = \{\forall g \in \mathcal{C}([0,1]): \Vert f-g \Vert <\epsilon \}$ there is some other function. As the functions are continuous shouldn't we be able to translate the function $f$ by some small value $\delta$ and always have atleast one function in $B_{\epsilon}(f)$? So every function in $F$ should be a limit point thereof. Is there a way to prove this in a more rigorous manner?
I'm not really used to working with functions spaces and their properties as metric spaces and I haven't been able to find a good resource for their properties. Most sources I've found are too advanced. This is for a course in real analysis. If anyone could recommend a good resource that would be gladly appreciated!
 A: It has become little bit lengthy because I tried to avoid the use of sequences of functions.
We apply the general definition of a limit point:
"$f$ is a limit point of the set $F$ iff for all $\epsilon>0$ there exists a $g\in F$ with $f\neq g$ such that $\Vert f-g\Vert <\epsilon$".
Now we take an arbitrary limit point $f$ of $F$ and show that $f<0$ is impossible.
Let's assume $\epsilon>0$ and $f(z)=-\epsilon$ for an arbitrary $z\in[0,1]$. As stated above, we know that there exists a $g\in F$ with $g\neq f$ and $\Vert f-g\Vert = \sup\limits_{x\in[0,1]}|f(x)-g(x)| <\epsilon$. Note that $g(x)\geq 0$ for all $x\in[0,1]$. This leads to a contradiction because then we would have $\epsilon<|-\epsilon-g(z)|=|f(z)-g(z)|\leq\sup\limits_{x\in[0,1]}|f(x)-g(x)| <\epsilon$. So it must hold that $f(x)\geq 0$ for all $x\in[0,1]$ as $z$ was arbitrarily chosen.
Now we check that $f$ is continuous:
We know that every $g\in F$ is continuous on a closed intervall and hence uniformous continuous. So we know that for a given $\epsilon>0$ there exists a $\delta>0$ such that for all $x,y\in[0,1]$ with $|x-y|<\delta$ it holds $\vert g(x)-g(y)\vert<\epsilon$.
Let's choose an arbitrary $\epsilon>0$ and define $\epsilon':=\frac{\epsilon}{3}>0$. So there exists a $g\in F$ with $\Vert g-f\Vert <\epsilon'$. Moreover we know that there exists a $\delta>0$ such that for every $x,y\in[0,1]$ with $|x-y|<\delta$ it holds $\vert g(x)-g(y)\vert<\epsilon'$ (uniformous continuity of $g$). Let be $a\in[0,1]$ arbitrary but fixed and $x\in[0,1]$ with $|x-a|<\delta$. Then we have $$\vert f(x)-f(a)\vert \leq \vert f(x)-g(x)\vert +\vert g(x)-g(a)\vert+ \vert f(a)-g(a)\vert \leq \\ \Vert f-g\Vert +\vert g(x)-g(a)\vert+ \Vert f-g\Vert<\epsilon'+\epsilon'+\epsilon' = \epsilon.$$
So $f$ is continuous on $[0,1]$ because for an abritrary $\epsilon>0$ we have found a $\delta>0$ such that for all $x\in[0,1]$ with $|x-a|<\delta$ it holds that $|f(x)-f(a)|<\epsilon$.
So we can conclude that $f\in F$. As $f$ was arbitrarily chosen $F$ is closed.
