Main Theme of Question: I aim to establish the equivalence of definitions of $\dim_B F$ (i.e. the box-counting dimension) for $F\subset\mathbb R^n$. Following is an extract from Kenneth Falconer's Fractal Geometry. The author proves (1) $\Leftrightarrow$ (4) and (1) $\Leftrightarrow$ (5). I need help understanding these proofs, and coming up with proofs of other equivalences.
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 $$ with centres in $F$.
The author gives some sample proofs of equivalence, namely (1) $\Leftrightarrow$ (4) and (1) $\Leftrightarrow$ (5) as I mentioned earlier.
(1) $\Longleftrightarrow$ (4): Write $N_(F)$ for the smallest number of sets of diameter $$ that can cover $F$. Let $N'_(F)$ be the number of $$-mesh cubes that intersect $F$; since these cubes obviously provide a collection of $N'_(F)$ sets of diameter $\sqrt n$ that cover $F$, $$N_{\sqrt n}(F) \le N'_(F)$$
$$[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]$$ where $m_i \in\mathbb Z$ for every $1 \le i \le n$, is called a $\delta$-mesh (or a $\delta$-grid) in $\mathbb R^n$.
On the other hand, any set of diameter at most $$ is contained in $3^n$ mesh cubes of side $$ (by choosing a cube containing some point of the set together with its neighboring cubes), so $$N'_(F) \le 3^nN_(F)$$
Q1. I'm unable to understand the inequality above. Where does the $3^n$ come from? To visualize, I decided to put $n=1$ and think about $\mathbb R$. Consider a set $A\subset\mathbb R$ such that $\text{diam}\ A \le \delta$. Mesh cubes in $\mathbb R$ are just the intervals. $\text{diam}\ A \le \delta$ tells us that $|x-y| \le \delta$ for every $x,y\in A$. It is clear that $A$ can have non-zero overlap with at most three $\delta$-meshes in $\mathbb R$. How do I generalize to $\mathbb R^n$?
Combining these inequalities and dividing by $-\log\delta$, $$\frac{\log N_{√n}(F)}{− \log(√n) + \log √n} \le \frac{\log N'_(F)}{-\log\delta} \le \frac{\log 3^n + \log N_(F)}{-\log\delta}$$ so taking lower limits as $ → 0$, $$\underline{\lim}_{\delta\to 0}\frac{\log N_{}(F)}{− \log()} \le \underline{\lim}_{\delta\to 0}\frac{\log N'_(F)}{-\log\delta} \le \underline{\lim}_{\delta\to 0}\frac{\log N_(F)}{-\log\delta}$$ with the other terms disappearing in the limit. Thus, the definition of lower box dimension is the same working with either $N_(F)$ or $N'_(F)$. Taking upper limits, we get a similar conclusion for upper box dimension.
(1) $\Longleftrightarrow$ (5): As before let $N_(F)$ be the smallest number of sets of diameter $$ that can cover $F$. Let $N''_(F)$ be the largest possible number of disjoint balls of radius $$ with centers in $F$ and let $B_1, . . . , B_{N''_\delta(F)}$ be such a collection of balls. If $x$ belongs to $F$, then $x$ must be within distance $$ of one of the $B_i$, otherwise the ball of center $x$ and radius $$ can be added to form a larger collection of disjoint balls. Thus, the $N''_ (F)$ balls concentric with the $B_i$ but of radius $2$ (and diameter $4$) cover $F$, giving $$N_{4}(F) \le N'' _(F)$$
Q2. I didn't quite follow the last part - where does $N_{4}(F) \le N''_(F)$ come from?
Suppose now that $B_1, . . . , B_{N''(F)}$ are disjoint balls of radii $$ with centers in $F$. Let $U_1, . . . ,U_k$ be any collection of sets of diameter at most $$ which cover $F$. Since the $U_j$ must cover the centers of the $B_i$, each $B_i$ must contain at least one of the $U_j$. As the $B_i$ are disjoint, there are at least as many $U_j$ as $B_i$. Hence, $$N''_ (F) \le N_(F)$$ Just as in the previous case, on taking logarithms of these inequalities, dividing by $−\log $ and taking the limit, we see that the values of $\dim_B F$, $\overline{\dim}_B F$, $\underline{\dim}_B F$ are unaltered if $N_(F)$ is replaced by this $N''_\delta(F)$.
My attempt at (1) $\Longleftrightarrow$ (2):
Let $N_\delta(F)$ be as defined above, and let $N^c_\delta(F)$ denote the smallest number of closed balls of radius $\delta$ that cover $F$. Let these closed balls be $B_1, B_2, ..., B_{N^c_\delta(F)}$. For every $1 \le i \le N^c_\delta(F)$, $\text{diam}\ B_i \le \delta$. So, $B_1, B_2, ..., B_{N^c_\delta(F)}$ is also a cover of $F$ consisting of sets with diameter at most $\delta$. Thus, $N_\delta(F)$ can at most be $N^c_\delta(F)$. Hence,
$$N_\delta(F) \le N^c_\delta(F)$$
To get another inequality, let $S_1, S_2, ..., S_{N_\delta(F)}$ be sets of diameter at most $\delta$ covering $F$. The closure of these sets, i.e. $\overline{S_1}, \overline{S_2}, ..., \overline{S_{N_\delta(F)}}$ also covers $F$. We know that for every set $A$, $\text{diam}\ A = \text{diam}\ \overline A$, so $\overline{S_1}, \overline{S_2}, ..., \overline{S_{N_\delta(F)}}$ is a cover of $F$ consisting of closed sets of diameter at most $\delta$. Now, I claim that every closed set of diameter at most $\delta$ is contained in some closed ball of radius at most $2\delta$. To see this, suppose not. Then, there is some closed set $A$ with $\text{diam}\ A \le \delta$ and there exist $x,y\in A$ such that $|x-y| > 2\delta$. This is a contradiction, so $$N_\delta^c(F) \le N_{2\delta}(F)$$
Will this work?
My attempt at (1) $\Longleftrightarrow$ (3):
Let $N_\delta(F)$ be as defined earlier, and let $N_\delta^d(F)$ be the smallest number of cubes of side $\delta$ that cover $F$. Suppose cubes $C_1, C_2, \ldots, C_{N_\delta^d(F)}$ cover $F$. Then, $\text{diam}\ C_i = \delta\sqrt n$ for every $1\le i \le N_\delta^d(F)$. Hence,
$$N_{\delta\sqrt n}(F) \le N_\delta^d(F)$$
To get another inequality, let $S_1, S_2,\ldots, S_{N_\delta(F)}$ be sets of diameter at most $\delta$ covering $F$. My intuition says that we should be able to find cubes $C_1, C_2, \ldots, C_{N_{\delta}(F)}$ of side length $\frac{\delta}{\sqrt n}$ such that the sets $S_i$ are contained in $C_i$ for every $n$. This would probably give $$N^d_{\delta/{\sqrt n}}(F) \le N_\delta(F)$$ but I'm not sure since I haven't shown this rigorously yet. Please let me know how to complete this.
Could you please help me understand and complete the above proofs of equivalence? I suspect the overall process remains the same, i.e. establishing inequalities as in the above cases, taking the logarithm on both sides, dividing by $-\log\delta$ and finally taking the limit as $\delta\to 0$. The crucial step is establishing the right inequalities, and I would appreciate help with the same.
Thank you!
References:
Kenneth Falconer, Fractal Geometry: Mathematical Foundations and Applications