Question about coverings of zero Hausdorff measure compact sets

Question

Consider a compact set $E\subset \mathbb{R}^n$ ad assume that ${\cal H}^{n-1}(E)=0$.

If $\epsilon>0$, there is a finite covering of $E$ made of $N_\epsilon$ open balls $B_{r_{\epsilon,k}}(x_{\epsilon,k})$ such that $\sum_{k=1}^{N_\epsilon}r_{\epsilon,k}^{n-1} \leq \epsilon$.

Indeed (I will omit the indices $_\epsilon$) for $\epsilon>0$, there is a countable family of sets $F_k$ of diameters $diam(F_j)$ such that $\sum_j diam(F_j)^{n-1}<\epsilon$. Choose open balls $B_{r_j}(y_j)$ of radius $\rho_j =diam(F_j)$ such that $B_{\rho_j}(y_j)\supset F_j$. From compactness, a finite subcovering $B_{\rho_{j_1}}(y_{j_1}), \ldots, B_{\rho_{j_N}}(y_{j_N})$ of $E$ exists such that $\sum_{k=1}^N \rho^{n-1}_{j_k} \leq \epsilon$ proving the assertion.

My question is: in the given hypoteses on $E$ and for any given $\epsilon>0$, is it possible to find a finite covering of $E$ made of balls $B_{r_\epsilon}(x_{\epsilon,k})$ with the same radius $r_\epsilon>0$, such that $\sum_{k=1}^{N_\epsilon} r_\epsilon^{n-1} \leq \epsilon$?

If considering the Lebesgue measure (i.e. ${\cal H}^n$) the assertion should be positive, what about ${\cal H}^{n-1}$ or ${\cal H}^{k}$ with $k< n$?

Or is there an explicit counterexample?

Answer 1

Let $N(r)$ be the number of $r$-balls needed to cover $E$. Then the lower Minkowski dimension of $E$ is $\liminf_{r \to 0} \log N(r)/\log(1/r)$. I claim that your assumptions imply that the lower Minkowski dimension is at most $n - 1$.

In fact, by taking $\varepsilon \leq 1$ arbitrarily small (and noticing that $r_\varepsilon \to 0$ as $\varepsilon \to 0$), it follows from your assumption that there exist arbitrarily small $r$ such that one can cover $E$ by a set of $r^{1 - n}$ $r$-balls. That is, there exist arbitrarily small $r$ such that $N(r) \leq r^{1 - n}$, which is of course equivalent to $$\liminf_{r \to 0}\frac{\log N(r)}{\log(1/r)} \leq n - 1,$$ as claimed.

(In my previous revision, I got a quantifier backwards, and incorrectly convinced myself that the upper Minkowski dimension of $E$, $\limsup_{r \to 0} \log N(r)/\log(1/r)$, is $\leq n - 1$. This would show that the packing dimension is $\leq n - 1$ as well. I am sorry for the confusion. No idea why it is claimed that the packing dimension is $0$ would come from, because it seems to be violated even by the simple example of a line segment in $\mathbb R^3$.)

Minkowski dimension is badly behaved under countable unions, unlike Hausdorff dimension. In view of this, the classic example of a compact set whose Hausdorff and Minkowski dimensions disagree is $\{1, 1/2, 1/3, \dots, 0\}$, whose Minkowski dimension is $1/2$ and whose Hausdorff dimension is $0$. This is Example 1.1.4 of Bishop and Peres' book on fractals, so let me not copy down the whole calculation here; the point is that if you want to cover $\{1, 1/2, \dots, 1/n\}$ by intervals of length $\lesssim n^{-2}$, then you need each interval to only cover a single point and therefore you need $n$ intervals.

In view of this, a counterexample to your claim is $E = \{1, 1/2, 1/3, \dots, 0\} \times C$ where $C$ is a Cantor set in $[0, 1]$ of Hausdorff dimension $\delta > 1/2$. The Hausdorff dimension of $E$ is $\delta < 1$ but the Minkowski dimension is $1/2 + \delta > 1$.

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PREVIOUS REVISION: If I'm reading your assumptions correctly, any compact set $E$ whose Hausdorff dimension is less than $n - 1$ but whose packing dimension is greater than $n - 1$ would be a counterexample. But I could be misinterpreting a quantifier.

Here's the intuition. Obviously we puny humans can't see the infinitesimal world, but things still appear $1$-dimensional, $2$-dimensional, or $3$-dimensional to us. Imagine you're nearsighted and can't tell the difference between two points at scale $r$. Let's informally say that the "dimension of $E$ at scale $r$", $s_E(r)$, is what dimension you perceive $E$ to have. Then the Hausdorff dimension is basically $\liminf_{r \to 0} s_E$ while the packing dimension is basically $\limsup_{r \to 0} s_E$. Obviously none of this is rigorous, but sometimes you hear measure theory people talking about "algorithmic fractal dimension" or "branching functions" which are both ways of making this intuition precise. Anyways, the point is we need a set which at some scales looks like it has small dimension and at other scales looks like it has big dimension.

Let $n = 2$ and start with $E_0 := [0, 1]^2$. Now let $E_1 = [1/3, 2/3] \times ([0, 1/3] \cup [2/3, 1])$, $E_2 = [4/9, 5/9] \times ([0, 1/9] \cup [2/9, 3/9] \cup [6/9, 7/9] \cup [8/9, 1])$, et cetra, so we're building a Cantor set in the line $\{1/2\} \times \mathbb R$. Keep doing this until you've built $E_{100}$, which is a union of cubes of side length $3^{-100}$. Break up each such cube into $9$ subcubes of side length $3^{-101}$ and remove the middle one, as in the construction of the Sierpinski carpet. Then do this for another $100$ stages. You get cubes of side length $3^{-200}$, in each one build a Cantor set in a line for $100$ stages, until you get cubes of side length $3^{-300}$, then in each of those build a Sierpinski carpet, and so on...

The Hausdorff dimension of this set is $\log 2/\log 3 < 1$, because when you try to compute the $s$-dimensional Hausdorff measure where $s > \log 2/\log 3$, you can bound that from above at each scale where $E$ looks like a Cantor set, and at those scales the measure is going to look arbitrarily small. But, in the definition of packing dimension, you need to be able to pack this set with balls at EVERY scale, and that includes the scales where it looks like a Cantor set. So the packing dimension of this set is $\log 8/\log 3 > 1$.