Heine-Borel theorem says that for any closed and bounded set $V$ in $\mathbb{R}^m$, if $V$ is contained in some infinite union of open balls, then there must be finitely many balls whose union can cover $V$. Those finite balls are called finite subcovers of $V$.
I am interested in bounding the number of finite subcovers over different coverings:
does there exist a constant $K$ such that, for any open-ball covering of $V$ in which each ball has same radius $r$ and center in $V$, $V$ can be covered by at most $K$ balls in the covering? ($K$ is independent of the infinite covering, can depend on $m$)
As an example, take $V$ as the closed unit ball in $\mathbb{R}^3$ centered at origin and $r=1$. We can choose different coverings by using unit balls with different centers. Say, some centers are close to origin, some close to boundary of $V$. In this case, there seems to be an upper bound of the number of subcovers needed regardless of the open covering. The general case $\mathbb{R}^m$ is not that straightforward to me and I don't know how to prove the claim if the answer is affirmative. If the problem is too hard, we can assume $V$ is a ball and $r$ is same to its radius.
Remark. It is crucial to restrict the radius of each ball and require the center to be in $V$.
- The answer is obviously negative if we do not restrict the radius. More balls will be needed if balls with smaller radius are used to form the covering. To see this, consider $V=[0,1]$. If we choose the open covering $\mathcal{O}_3 = \bigcup_\limits{a} (a,a+\frac{1}{3})$, then 4 subintervals are enough to cover $V$. But if we choose the covering $\mathcal{O}_{1000}=\bigcup_\limits{a}(a,a+\frac{1}{1000})$, then we need at least around 1000 subcovers to cover $V$. Hence we can see that if such a $K$ exists, it must be inifinity because we can always choose covering with smaller components.
- Perhaps not as trivial as above, the answer would also be negative if we do not impose the center of each ball to be in $V$. To see this, consider $V=$ closed unit disk centered at origin. We can choose the open covering to be the union of the open unit disk centered at origin and unit disks whose centers lie on the circle centered at origin with radius 1.99. If we vary 1.99 to be closer to 2, then more disks are needed to cover $V$. As 1.99 approaches 2, the number of finite subcovers needed approaches infinity (each exterior disk is almost tangential to $V$, making the boundary of $V$ barely covered).
[Update] After reading the comments, I found that for the aforementioned case ($V$ the closed unit ball and $r=1$), the answer is negative. A counter-example can be constructed in the spirit of the second example in the Remark above. Consider $V$ the closed unit ball and $r=1$. If we choose the open covering to be the union of the interior of $V$ and copies of $V$ with slightly shifted centers, then each of those shifted balls can only cover a tiny part of the boundary of $V$. We can always choose the shift to be small enough so that we need more than $n$ balls to cover $V$, for any given $n$. We can see that those counter-examples are all created by abusing the boundary of $V$. To avoid the boundary, I would ask the following on covering the interior of $V$ instead of $V$ itself:
Let $V$ be the closed ball in $\mathbb{R}^m$ centered at origin and $r=1$. Consider the family of open coverings of $V$ where each covering is the union of unit balls with centers in the interior of $V$. Does there exist a constant $K$ such that, for each covering in the family, we can use at most $K$ balls from it to cover the interior of $V$ (i.e. the open unit ball) ?
($K$ is independent of the covering)