# If a compact set is covered by a finite union of open balls of same radii, can we always get a lesser radius?

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.

• Perhaps it's important to note that "balls" means "open balls" here Jan 22, 2018 at 16:33
• Yes, it is. Thanks.
– Myth
Jan 22, 2018 at 16:40
• The hypothesis holds for any compact $V\subset \mathbb{R}^n$, since any such set is bounded. In other words, for any $r > 0$, there exists a finite set of open balls $B_1, \dots, B_n$ of radius $r$ covering $V$. Jan 22, 2018 at 17:49
• @anomaly Reading between the lines: it seems you mention this in the hopes that we could therefore eliminate the hypothesis and get a stronger theorem. Unfortunately some of the variables bound in the hypothesis (specifically, the centers of the balls) are also mentioned in the conclusion, so we can't do that. Jan 22, 2018 at 22:00
• @DanielWagner: No, my (minor) point was just that such a cover exists for any $V$, so the first bit is a definition or construction rather than a hypothesis. Jan 22, 2018 at 22:03

Let $X$ denote the set of centers: $X = \{c_1,\ldots,c_m\}$.

The function $\phi(x) = \mathop{\rm dist} (x,X)$ is continuous on $\mathbb R^n$ and attains a maximum value on $V$ because $V$ is compact.

Note that if $x \in V$, then by definition $\phi(x) < r$. Whatever maximum it attains must be less than $r$.

Choose $s$ to lie in between this maximum and $r$.

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 $r$.

• (+1) I like this answer better than mine since it uses only the definition of compactness. Very nice argument. Jan 22, 2018 at 19:34
• However, does the finite subcover include only balls of equal radii? Jan 23, 2018 at 12:53
• @M.Herzkamp, one can always choose the maximal radius (for all of the finitely many balls) which is still smaller than $r$. Jan 23, 2018 at 13:08
• Very good! Can you edit it into your answer to make it complete? Jan 24, 2018 at 9:23