Minkowski–Bouligand dimension

Remarks

One gets the same values if $N(A, \varepsilon)$ counts the cubes of an $\varepsilon$-mesh grid that meet $A$, or the largest number of disjoint balls of radius $\varepsilon$ centred in $A$. For bounded $A$, $\overline{\dim}_B A = \inf \set{ s : \mathcal{M}^{*s}(A) = 0 }$, where $\mathcal{M}^{*s}$ is the upper Minkowski content. Always $\dim_H A \le \underline{\dim}_B A \le \overline{\dim}_B A$, where $\dim_H$ is the Hausdorff dimension.

Example: let $C$ be the middle-thirds Cantor set. It is covered by $2^k$ intervals of length $3^{-k}$, and each set of diameter at most $3^{-k}$ meets at most two of them. So $2^{k-1} \le N(C, 3^{-k}) \le 2^k$, and the ratio tends to $\log 2 / \log 3$, so $\dim_B C = \log 2 / \log 3$.

The box dimension does not ignore countable sets. The set $\Q \cap [0, 1]$ is dense, so $N(\Q \cap [0, 1], \varepsilon) = N([0, 1], \varepsilon)$, and its box dimension is $1$. Its Hausdorff dimension is $0$.