0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – PIP Jul 6 '17 at 16:32
  • 1
    $\begingroup$ 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. $\endgroup$ – demolishka Jul 7 '17 at 15:28
0
$\begingroup$

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 $n \ge 2$, $1 \le k \le n/\log(n)$.

$\endgroup$
  • $\begingroup$ Yes, I know that it only works for bounded sets, but I'm OK with that because I'll be working with compact sets. $\endgroup$ – PIP Jun 29 '17 at 20:34
  • $\begingroup$ 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 $\varepsilon$. $\endgroup$ – PIP Jun 29 '17 at 20:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.