# Why can we cover $\mathbb R^N$ with open balls of radius $r$ such that each point is in at most $N + 1$ balls?

If $$N \geq 3$$, why can we cover $$\mathbb R^N$$ with open balls of a fixed radius $$r$$ such that each point is in at most $$N + 1$$ balls?

This is a claim in a proof of Lions' Vanishing Lemma, as presented in Willem's Minimax Theorems (Lemma 1.21). Probably very simple but I am not able to write a proper proof.

• @PaulFrost Doesn't matter - the two versions of the result are trivially equivalent. – David C. Ullrich Aug 1 '20 at 18:06
• @DaniloGregorin: No. The centres of $3$ balls are coplanar, so imagine the balls resting on a flat surface. You can rest a fourth ball in the hollow that they form, and there will be a volume immediately below it that is not covered by any of the four. It’s fairly clear intuitively that there is no way to cover that volume without using a ball that intersects all $3$ of the original balls. – Brian M. Scott Aug 1 '20 at 18:42
• @DavidC.Ullrich Equivalent - yes. Trivially? – Paul Frost Aug 1 '20 at 23:55
• @DavidC.Ullrich Doubling the radius could increase the number of balls intersecting at some points. Even adding a small $\epsilon$ to the radius may not work, if there's no lower bound on the pairwise distances between non-intersecting closed balls. – aschepler Aug 2 '20 at 14:03
• @DCao But regular simplices don't tesselate in $\mathbb{R}^n$ when $n>2$. (math.stackexchange.com/a/3087543/2236) – aschepler Aug 3 '20 at 23:45

## 1 Answer

user125932 mentioned in a comment that this seems to be an open problem, since it would imply that the covering density of any $$n$$-dimensional ball is at most $$n+1$$. As of 2018, it still seems that nobody can prove a better upper bound on that covering density than $$Θ( n · \log n )$$; see here and here. In particular, the first linked paper explicitly states that as $$n→∞$$, unit balls can cover $$\mathbb{R}^n$$ with density $$\big(\frac12+o(1)\big)\ n\ln n$$, as Corollary 2.

• I think that you are answering a different question. If I understood correctly, OP asks about the maximal number of balls at any given point, while your answer treats the question of the average number of balls at a given point, and the latter number is obviously smaller than the former. – Louis Hainaut Aug 13 '20 at 8:36
• @LouisHainaut: But that inequality is obviously enough to demonstrate that the question is an open problem, which is exactly what I stated in my post. (I said nothing about the maximum depth of the covering.) – user21820 Aug 13 '20 at 9:03
• for those that are interested, it is known that a covering exists in which every point is contained in at most $O(n \log n)$ balls -- this was proved by Erdős and Rogers here, refining Rogers' $O(n \log n)$ upper bound on the covering density. – user125932 Aug 14 '20 at 6:29
• @user125932: Nice, thanks for your comment! – user21820 Aug 14 '20 at 6:31