# Clarification of Open cover

$E$ is a compact metric space. Consider a compact set $A \subset C(E)$ where $C(E)$ denotes the set of all continuous functions on $E$. Since $A$ is compact, any open cover of $A$ has a finite subcover.

My question is what does it mean to say that " consider a finite collection of open balls of a specified radius centered at the functions which cover $A$". I cannot understand what it means to say open balls centered at functions, doe sit mean for every function and every point of this function, I need to create an open ball.

All this is in context of a metric given by the sup norm

Thanks

Take for instance a function $f$ in $C([-1, 1])$. The open ball of radius $r$ centered at $f$ is the set of functions $g$ such that $|g(x) - f(x)| < r$ for all $x\in [-1, 1]$.
If the function $f$ is given by $f(x) = x^2$, say, then the open ball of radius $1$ centered at $f$ contains all functions whose graph lie inside the dotted line (right click and select "View image" / "Open image in new tab" or whatever your browser uses).
The open ball of radius $r$, centered at $f\in C(E)$ is the set $$B_r(f)=\{g\in C(E) : \sup_E|f-g|<r\}$$
If you were to look at the center of an open interval on the real number line then you have a point as your center. Think of your functions as elements or points'' in your compact set. You have a given metric, so for any open cover derived from the topology generated by the metric then you are going to find a finite subcover as you stated. It doesn't mean you have to find an open set around every function in your set, since it is compact, you just need finitely many of them and their corresponding open sets to cover your set.