Why is $\dim_{B}F \le n$ in $\mathbb R^n$? (Upper Bound on Minkowski–Bouligand dimension) (Please skip to the end for a word on notation)

For $F \subset\mathbb R^n$, where $F\ne \varnothing$, we have $$0 \le \underline\dim_B F \le \overline\dim_B F \le n$$
and hence $$\dim_B F \le n$$
where $\dim_B F$ denotes the box-counting dimension of $F$. $\underline\dim_B F$ and $\overline\dim_B F$ denote the lower and upper box-counting dimension respectively.

I need help understanding the proof of the last inequality, i.e. the $\le n$ part. The proof in the book is:

The first two inequalities are obvious; for the third, $F$ may be enclosed in a large cube $C$ so by counting $$-mesh squares $N_(F) ⩽ N_(C) ⩽ c^{−n}$ for some constant $c$.

Since $F\subset C$, $N_\delta(F) \le N_\delta(C)$ is clear. Why is $N_\delta(C) \le c\delta^{-n}$ for some $c$?

Notation:

*

*About $\dim_B F$ and $N_\delta(F)$:


The lower and upper box-counting dimensions of a subset $F$ of $ℝ^n$ are given by
$$\underline{\dim}_B F = \underline{\lim}_{\delta\to 0} \frac{\log N_\delta(F)}{-\log \delta}$$
$$\overline{\dim}_B F = \overline{\lim}_{\delta\to 0} \frac{\log N_\delta(F)}{-\log \delta}$$
and the box-counting dimension of $F$ by
$$\dim_B F = \lim_{\delta\to 0} \frac{\log N_\delta(F)}{-\log \delta}$$
(if this limit exists), where $N_\delta(F)$ is any of the following:

*

*the smallest number of sets of diameter at most $$ that cover $F$;

*the smallest number of closed balls of radius $$ that cover $F$;

*the smallest number of cubes of side $$ that cover $F$;

*the number of $$-mesh cubes that intersect $F$;

*the largest number of disjoint balls of radius $$ with centres in $F$.




*What is a $\delta$-mesh?


$$[m_1\delta, (m_1+1)\delta] \times [m_2\delta, (m_2+1)\delta] \times \ldots \times [m_n\delta, (m_n+1)\delta]$$ where $m_i \in\mathbb Z$ for every $1 \le i \le n$, is called a $\delta$-mesh (or a $\delta$-grid) in $\mathbb R^n$.


If we simply enclose $F$ in a hypercube with side $s$, that cube can be broken down into $m^n$ subcubes with side $s/m$. Then we get that
$$
\lim_{m\to\infty}\frac{\log(m^n)}{-\log(s/m)}=n
$$
This simply says that any subset of $\mathbb{R}^n$ has dimension at most $n$ (which is a mathematical justification of what seems a pretty intuitive fact).
