This question seems obvious, but I'm not secure of my proof.
If a compact set $V\subset \mathbb{R^n}$ is covered by a finite union of open balls of common radii $C(r):=\bigcup_{i=1}^m B(c_i,r)$, then is it true that there exists $0<s<r$ such that $V\subseteq C(s)$ as well? The centers are fixed.
I believe this statement is true and this is my attempt to prove it:
Each point of $v\in V$ is an interior point of least one ball (suppose its index is $j_v$), that is, there exists $\varepsilon_v>0$ such that $B(v,\varepsilon_v)\subseteq B(c_{j_v},r)$, so $v\in B(c_{j_v},r-\varepsilon_v)$. Lets consider only the greatest $\varepsilon_v$ such that this holds. Then defining $\varepsilon:=\inf\{\varepsilon_v\mid v\in V\}$ and $s=r-\varepsilon$ we get $V\subseteq C(s)$.
But why is $\varepsilon$ not zero? I thought that considering the greatest $\varepsilon_v$ was important, but still couldn't convince myself.
I would appreciate any help.
Best Answer
Replace each open ball $B_i$ of radius $r$ in the cover by the union of concentric open balls of radii strictly smaller than $r$. You get an infinite cover of $V$. By compactness there is a finite subcover. By construction the radii are smaller than before. Finally we choose the maximal radius (for all of the finitely many balls) which is still smaller than $r$.