6

For simplicity, let us suppose we have the unit rectangle $[0,1] \times [0,1]$ and the fact that it is compact. As it is compact, it is totally bounded. In the words of H. Weyl, a city is compact if it can be guarded by a finite number of arbitrarily near-sighted policemen.

A metric space is said to be totally bounded if for all epsilon, there exist a finite number of balls, all of whose centers is in the metric space, and union is a super set of the metric space. Paraphrased wiki

I am interested how would the minimum number of balls to cover out set vary with epsilon. Is there some approximate formula for it?

To get started, I found out that if epsilon is big enough, then it's only one. But, as we reduce it, there is a critical value, which is the diameter of the set (refer), below which, we need to add one at least one more ball.

Bonus: How would the answer change if we used some random compact set? Note: My argument above holds for any compact set.

Edit: I found this older question which is about the precise bounds 1

user76284
  • 6,408
Clemens Bartholdy
  • 14,222

2 Answers2

14

The answer is exactly the Minkowski dimension of the compact set.

More precisely, the Minkowski dimension is the exponent that tells us how the number of policemen required grows as they become more nearsighted: the Minkowski dimension is $d$ if there is some constant $C>0$ such that, as $\varepsilon \to 0$, the number of $\varepsilon$-sighted policement grows as $C \varepsilon^{-d}$.

The Minkowski dimension of a square city in the shape of $[0,1]\times [0,1]$ is $2$: the number of policemen needed to guard it grows with the square of $1/\varepsilon$.

If the city does not have the budget for this many policemen, perhaps it should consider carving itself out into the shape of a Sierpiński carpet. Here, because the Sierpiński carpet consists of eight copies of smaller Sierpiński carpets scaled by a factor of $\frac13$, we can show that the Minkowski dimension is only $\log_3 8 \approx 1.89$. The taxpayers may initially appreciate this proposal, though maybe not once they find out that the area of the Sierpiński carpet is $0$.

user76284
  • 6,408
Misha Lavrov
  • 159,700
  • This question comes out of ignorance, but how does this compare to the Hausdorff dimension? I believe they agree on self-similar shapes. – Malady Sep 03 '24 at 00:18
  • 8
    @Malady They're the same in many cases. I think the Minkowski dimension was the earlier definition and the Hausdorff dimension was a later refinement to patch some weird cases where the Minkowski dimension didn't do what we wanted - it doesn't behave nicely under countable unions, and in particular $[0,1]\cap \mathbb Q$ has Minkowski dimension $1$ even though it's a countable union of points. I used it here anyway because the Hausdorff dimension has a more complicated definition in which we allow our policemen to vary in their nearsightedness, and then carefully aggregate them. – Misha Lavrov Sep 03 '24 at 02:19
  • So what can we say about the value of the constant $C$ in the case of the unit square? Is it known? – Marc Sep 03 '24 at 14:19
  • The best covering in the long run is almost certainly going to be to place the policemen in a triangular lattice (though I'm only aware of the results showing that this is optimal for packings). Here, each policeman is at the center of a hexagon of area $\frac{3\sqrt3}2\varepsilon^2$, and if we take enough hexagons to cover $[0,1]^2$, there's $\frac{2}{3\sqrt3} \varepsilon^{-2}$ of them. (As $\varepsilon\to0$, we can ignore the boundary effects that come from some hexagons ending up outside the square.) – Misha Lavrov Sep 03 '24 at 14:34
  • May I ask that nobody start an edit war in the answer that I wrote over whether Weyl's "policemen" should be changed to "police officers" or not? – Misha Lavrov Dec 29 '24 at 23:30
0

A square of side $s$ can be covered by one open ball of radius $r$ iff $r > \sqrt{2}/2$. A square of side $n s$, where $n$ is a positive integer, can be tiled by $n^2$ squares of side $s$, and therefore can be covered by $n^2$ open balls of side $r$ if $r > \sqrt{2}/2$.

Robert Israel
  • 470,583