Calculate the Minkowski dimensions of the following objects: a point, the interval $[0,1]$, the rationals in $[0,1]$, the triadic Cantor set $C(3)$.
$\nabla \operatorname{dim}_{b o x}($ point $)=0$

$\triangleright \operatorname{dim}_{b o x}([0,1])=1$

![](https://cdn.mathpix.com/cropped/2024_05_10_da6e2ad19b1b4e3d33dcg-258.jpg?height=290&width=689&top_left_y=202&top_left_x=729)

Figure 20 - Construction of the Cantor set

$\triangleright \operatorname{dim}_{\text {box }}(\mathbb{Q} \cap[0,1])=1(\mathbb{Q} \cap[0,1]$ is dense in $[0,1])$

$\nabla \operatorname{dim}_{b o x}(C(3))=\frac{\log (2)}{\log (3)}$

It is surprising when one considers that $C(3)$ is uncountable and has a smaller dimension than $\mathbb{Q} \cap[0,1] \ldots$

To calculate the dimension of the Cantor set, we take $\epsilon_{n}=\frac{1}{3^{n}}$ and count the number of segments of length $\epsilon_{n}$ needed to cover the set. At generation $n$ of the Cantor set construction, we have $2^{n}$ segments of length $\frac{1}{3^{n}}$, and the endpoints of all these segments will be in the Cantor set. Therefore, $3^{n}$ segments of size $\epsilon_{n}$ are needed to cover the Cantor set.

We can thus write:

$$
N\left(\epsilon_{n}\right)=2^{n}
$$

and then

$$
\frac{\log \left(N\left(\epsilon_{n}\right)\right)}{\log \left(\frac{1}{\epsilon_{n}}\right)}=\frac{\log (2)}{\log (3)}
$$

The dimension of the Cantor set is therefore $\frac{\log (2)}{\log (3)}$.