# Growth of the size of coverings for sets with prescribed upper Minkowski dimension

For subsets of $\mathbb{R}^n$, I want a notion of dimension $\operatorname{dim}$ verifying:

If $\operatorname{dim}(A) = d$, then there's a constant $C$, depending only on $A$, $d$ and $n$, such that for every $\varepsilon >0$ it's true that $A$ can be covered by at most $C\big ( \frac{1}{\varepsilon}\big )^d$ balls of radii less than $\varepsilon$.

I think that the upper Minkowski dimension could work, but I'm not sure. Minkowski dimensions are defined for every subset $A$ of $\mathbb{R}^n$ by $$\overline{\operatorname{dim}}_M (A)= \inf{\{ s:\limsup_{\varepsilon\downarrow0}{N(A,\varepsilon)\varepsilon^s = 0}\}}$$ and $$\underline{\operatorname{dim}}_M (A)= \inf{\{ s:\liminf_{\varepsilon\downarrow0}{N(A,\varepsilon)\varepsilon^s = 0}\}},$$ where $$N(A,\varepsilon) = \min{\{ k : A \subset \bigcup_{i=1}^{k}{B(x_i,\varepsilon)} \quad \text{for some}\; x_i \in \mathbb{R}^n\}}.$$ We call $\overline{\operatorname{dim}}_M (A)$ and $\underline{\operatorname{dim}}_M (A)$ the upper and lower Minkowsi dimension respectively.

If $\overline{\operatorname{dim}}_M (A) = d$, then for any $\delta >0$ there is a constant $C$, depending on $\delta$ and $A$, such that for any $\varepsilon >0$ the set $A$ can be covered by at most $C\big ( \frac{1}{\varepsilon}\big )^{d+\delta}$ balls of radii less than or equal to $\varepsilon$. This is similar to, but not exactly what I needed. Nevertheless, this could be fixed if the infima in the defininitions of Minkowski dimensions were minima under reasonable hypothesis.

So, my questions are:

1. When do the infima are minima?

2. If the answer is not "always", what are some counterexamples?

3. Is there some other notion of dimension that can work?

The value $N(A,\varepsilon)$ does not necessarily has growth rate of a polynomial. There are simple examples.

Take $A := \{0\} \cup \{a_{0},a_{1},a_{2},\ldots\} \subset \mathbb{R}$, where $a_{n} = e^{-n}$. It is easy too see that $N(A,a_{n}) = n+1$ and, consequently, $N(A,\varepsilon) \approx \ln (\frac{1}{\varepsilon})$, so $\overline{\dim}_{M}(A)=0$, but $N(A,\varepsilon)$ may be arbitrary large. If u take $a_{n} = n^{-\alpha} \ln^{\alpha} n$, for some $\alpha \geq 1$ u will have that $\overline{\dim}_{M}(A)=\frac{1}{\alpha}$, but $N(A,a_{n}) = n+1$ can not be majorized by $(\frac{1}{a_{n}})^\frac{1}{\alpha} = \frac{n}{\ln n}$.

The sets for which the growth rate of $N(A,\varepsilon)$ is a polynomial of $\frac{1}{\varepsilon}$ can be of distinct natures. Such a set may be a smooth manifold or a cantor-like set. The only I can say is that the property is preserved under Bi-Lipschitz homeomorphism.

• Thanks! Do you have any reference for the fact that $N(A, \varepsilon)$ is a polynomial in $\frac{1}{\varepsilon}$ for smooth manifolds and cantor-like sets?
– PIP
Jul 6 '17 at 16:32
• Any point on a smooth manifold (in my definition) has a neighborhood diffeomorphic to an $n$-dimensional open cube. For the cube it is obvious that $N(\varepsilon)$ has a polynomial growth rate. Diffeomorphism is a Bi-Lipschitz map and, consequently, we have the same growth rate of $N(\varepsilon)$ for the neighborhood. Now use compactness. For a cantor-like set this is not true in general. You can see that sets $A$ in my example for distinct sequences $a_n$ can be homeomorphic, but the discussed property may not be preserved. The same idea works for cantor-like sets. Jul 7 '17 at 15:28

If there are arbitrarily small $\epsilon > 0$ such that $A$ can be covered by at most $C (1/\epsilon)^d$ balls of radius $< \epsilon$ (and thus diameter $< 2 \epsilon$), then the Hausdorff $d$-dimensional measure of $A$ is at most $2^d C$, and so the Hausdorff dimension of $A$ is at most $d$.

On the other hand, your "dimension" will always be infinite for unbounded sets, because they can never be covered by finitely many balls. Maybe you want to restrict $A$ to be bounded?

EDIT: With your definition, countable compact sets in $\mathbb R$ need not have dimension $< 1$. For example, let $A$ consist of $0$ and the points $k/n$ for integers $n, k$ with $n \ge 2$, $1 \le k \le n/\log(n)$.

• Yes, I know that it only works for bounded sets, but I'm OK with that because I'll be working with compact sets.
– PIP
Jun 29 '17 at 20:34
• Also, I need the implication the other way round: if the dimension is at most $d$, then there is a covering with at most $C\big ( \frac{1}{\varepsilon}\big )^d$ balls of radii less than or equal to $\varepsilon$.
– PIP
Jun 29 '17 at 20:37