Prove that $ U=\{f\in C[0,1]: f(x)\neq 0, \forall x \in [0,1]\}$ is open and find his connected components In $(C[0,1],d_\infty)$, consider $U=\{f\in C[0,1]: f(x)\neq 0, \forall x \in [0,1]\}$. Prove that $U$ is open and find his connected components.
I know that for proof the first thing, i have to show the existence of an $\epsilon\gt0$ so for all $f\in U$, $B(f,\epsilon)\subseteq U$. First of all, is that correct? Second, if it is ...how should i find that $\epsilon\gt 0$? For the connected components...i have no clue. Any suggestions?
 A: Hint:
There is no such $\epsilon$ for all $f \in U$.
You want to show that for any $f \in U$ there exists an $\epsilon$ ball $B(f,\epsilon) \subset U$. 
Note that as $f$ is continuous, $f$ is either strictly positive or strictly negative (candidate connected components?). Further, as $[0,1]$ is compact the image $f([0,1])$ is compact and $f$ attains a minimum and maximum (aka the first year calculus Extreme Value Theorem). Can you see now how to find such an $\epsilon$?
If you're still stuck, draw a graph of some $f \in U$: that's the mental image I am using in my head to see what kind of $B(f,\epsilon)$ work.
A: The connected components of $U$ are the open subsets $$U_+=\{f\in C[0,1]: f(x)\gt 0, \forall x \in [0,1]\}$$ and $$U_-=\{f\in C[0,1]: f(x)\lt 0, \forall x \in [0,1]\}$$ Indeed if $f,g\in U_+$ (resp $f,g\in U_-$), then  $tf+(1-t)g\in U_+$ (resp. $tf+(1-t)g\in U_-$) for all values of $t\in [0,1]$.
This proves that $U_+,U_-$ are both pathwise connected and thus connected.
Since they form a disjoint open partition of $U$ (notice that any $f\in U$ has constant sign since it never vanishes, so must belong to either $U_+$ or $U_-$) , they are the two connected components of $U$.
