3
$\begingroup$

I'm trying to understand the definition of Minkowski dimension given by Wikipedia here.

In fractal geometry, the Minkowski–Bouligand dimension, also known as Minkowski dimension or box-counting dimension, is a way of determining the fractal dimension of a set $S$ in a Euclidean space $\mathbb{R}^n$, or more generally in a metric space $(X, d)$.

It's defined as

$$\textrm{dim}(S)=\lim\limits_{\epsilon\to 0}\frac{\log N(\epsilon)}{\log(1/\epsilon)}$$ where $N(\epsilon)$ is the minimum number of balls of radius $\epsilon$ required to cover $S$.

If the above limit does not exist, one may still take the limit superior and limit inferior, which respectively define the upper box dimension and lower box dimension.

My question(s): can $S$ be anything or do we have to restrict ourselves to bounded (or even compact?) subsets of $\mathbb{R}^n$? If we don't, $N(\epsilon)$ can be infinite. In this case I guess the definition would lead us to consider that the Minkowski dimension is infinite, right? For example, $\textrm{dim}(\mathbb{R})=\infty$?

In the more general case of a metric space $(X,d)$, I have the same kind of questions. Do we have to consider special conditions on $(X,d)$ (like bounded, compact, locally compact) or on the subset $S$ of $X$?

By the way, if you have some good references about Minkowski dimension, Hausdorff measure and dimension, and related subjects...

$\endgroup$

1 Answer 1

2
$\begingroup$

You are correct - the set $S$ must be bounded for this definition to be applicable. Even if the set is bounded, there are issues with this definition. For example, it's easy to see that $\dim(S)=1$ for $S=[0,1]\cap\mathbb Q$, even though it's a countable set. Focusing on compact sets doesn't really fix the problem. If $S=\{0,1,1/2,1/3,\ldots,1/n,\ldots\}$ then $\dim(S)=1/2$, though this is a bit more work to prove.

These examples illustrate that box-counting dimension is not $\sigma$-stable. That is, we can have $$\dim\left(\bigcup_{n=1}^{\infty} E_n\right) > \sup_n \dim(E_n),$$ which is certainly an undesirable property in any notion of dimension. There is a standard way to modify the definition, namely define $$\dim_M(S) = \inf\left\{\sup_n \dim(S_n): S=\bigcup_{n=1}^{\infty} S_n\right\}.$$ That is, we consider all ways to decompose the set $S$ into countably many pieces and return the smallest possible value of the largest dimension from all decompositions. The result is not only $\sigma$-stable (alleviating the problems above) but is also applicable to unbounded sets.

Box-counting directly is still very nice for at least three reasons:

  • It is very easy to compute relative to measure theoretic definitions, like the Hausdorff dimension,
  • It provides an easy upper bound for most other dimensions,
  • It can be estimated experimentally for many natural objects.

The modified box-counting dimension, while nicer from a theoretical standpoint, loses these advantages.

$\endgroup$

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.