20
$\begingroup$

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.

$\endgroup$
6
  • 3
    $\begingroup$ Perhaps it's important to note that "balls" means "open balls" here $\endgroup$ Jan 22, 2018 at 16:33
  • $\begingroup$ Yes, it is. Thanks. $\endgroup$
    – Myth
    Jan 22, 2018 at 16:40
  • $\begingroup$ 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 $V$. $\endgroup$
    – anomaly
    Jan 22, 2018 at 17:49
  • $\begingroup$ @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. $\endgroup$ Jan 22, 2018 at 22:00
  • $\begingroup$ @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. $\endgroup$
    – anomaly
    Jan 22, 2018 at 22:03

2 Answers 2

37
$\begingroup$

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 $r$.

$\endgroup$
4
  • $\begingroup$ (+1) I like this answer better than mine since it uses only the definition of compactness. Very nice argument. $\endgroup$
    – Umberto P.
    Jan 22, 2018 at 19:34
  • $\begingroup$ However, does the finite subcover include only balls of equal radii? $\endgroup$
    – M.Herzkamp
    Jan 23, 2018 at 12:53
  • 4
    $\begingroup$ @M.Herzkamp, one can always choose the maximal radius (for all of the finitely many balls) which is still smaller than $r$. $\endgroup$ Jan 23, 2018 at 13:08
  • $\begingroup$ Very good! Can you edit it into your answer to make it complete? $\endgroup$
    – M.Herzkamp
    Jan 24, 2018 at 9:23
35
$\begingroup$

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 $r$.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .