Let $H$ be a separable Hilbert space, $F\subset H$ be a finite subset. Let $K=\operatorname{span}\{F\}$, which is a finite dimensional subspace of $H$. Thus, the unit ball of $K$ is compact. I want to understand the following statement in a proof:
"Let $\epsilon >0$. By compactness of the unit ball of $K$ there is a finite set $A\subset K$ which is $\epsilon$-dense in the unit ball of $K$."
My question is about the existence of such a set $A$: How too choose such a subset? My guess/attempt: Let $B_1^K(0)$ be the closed unit ball of $K$. By compactness of $B_1^K(0)$, each of its open cover has a finite subcover. Now one probably has to choose a suitable open cover, so that from the finite subcover one can extract finitely many points which form the desired set $A$. However, I don't know how to do it in detail.
Thank you.
You're on your right way. Take a cover of the closed unit ball by $\epsilon$-balls, then there is a finite subcover. This means that there is a family of finitely many balls, $B_\epsilon(x_i)$ for $i<n$, of the unit ball.
Now that means exactly that $\{x_i\mid i<n\}$ is $\epsilon$-dense, since every point on the unit ball lies in one of the $\epsilon$-balls, and therefore is closer than $\epsilon$ to some $x_i$.
As for "constructing" or "choosing" this set, this is not something you can do easily, and it will depend on $\epsilon$. You could, probably, figure out some explicit construction given from finding one in one dimensional, as a function of $\epsilon$. But this is a waste of time and effort for most abstract purposes. One exists, so we can just fix one.