I am working from Falconer's book Fractal Geometry: Mathematical Foundations and Applications (which seems to be the go-to book). I have reached the section of the modified box-counting dimensions, which are defined as
$$\underline{\dim}_{MB}(F)=\inf\left\{\sup_{i}\left\{\underline{\dim}_{B}(F_{\color{red}i})\right\}:F\subset\bigcup_{i=1}^{\infty}F_{i}\right\} \qquad \text{(lower)}$$
and
$$\overline{\dim}_{MB}(F)=\inf\left\{\sup_{i}\left\{\overline{\dim}_{B}(F_\color{red}i)\right\}:F\subset\bigcup_{i=1}^{\infty}F_{i}\right\} \qquad \text{(upper)}$$
where $\underline{\dim}_{B}$ is the lower box dimension and $\overline{\dim}_{B}$ is the upper box dimension.
I am trying to show an example of a countable set $F$ for which $\dim_{B}(F)>0$ and $\dim_{MB}(F)=0$ (after proving that $\dim_{MB}(F)=0$ for any countable set $F$).
My work so far: I have decided to use the set $F=\{0,1,\frac{1}{2},\frac{1}{3},\ldots\}$. I know that $\dim_{B}(F)=\frac{1}{2}$ (this is proven in the book), but I don't know how to calculate the modified box dimensions of $F$. Is someone able to start me off? Thank you.