# Minkowski Dimension of Special Cantor Set

As can be seen at the top of the page here (exercise 1), Terry Tao gives an exercise to find the Minkowski Dimension of the Quadnary Cantor Set, and of a special Quadnary Cantor Set.

The two sets are:
$$C := \left\{\sum_{i=1}^{\infty} a_i4^{-i} : a_i \in \{0,3\} \right\}$$

and

$$C’ : = \left\{\sum_{i=1}^{\infty} a_i4^{-i} : a_i \in \{0,3\} \, \, \text{if} \, \, (2k)! \leq i \leq (2k+1)! \,\, \text{and arbitrary otherwise} \right\}.$$

I managed to find the Minkowksi Dimension of the first one by noting that $C_\delta$, the $\delta$-fattening of $C$, has $\text{vol}(C_\delta)$ equal to the $n^{th}$ iteration of the Quadnary Cantor set.

For the second one, I’ve found a closed form expression for the volume if $\delta_n = 4^{-(2n)!}$, since I suspect this should be a lim sup, but I’m not sure.

The closed form I found is:

$$\text{vol}(C_{\delta_n}) = \sum_{k=1}^{n-1} \sum_{j=0}^{2k(2k)!} 2^{-(2k-1)!+2j)},$$

but I’m not sure how to even begin manipulating the expression

$$\frac{\log(\text{vol}(C_{\delta_n}))}{-(2n)! \log(4)}.$$

In particular, we are to show that the upper MD of $C’$ is $1$, while the lower MD is $\frac{1}{2}$, but it seems to me that the worst sequence we can use is $4^{-(2n+1)!}$, since this would give a volume of

$$\text{vol}(C_{\delta_n}) = \sum_{k=1}^{n} \sum_{j=0}^{2k(2k)!} 2^{-(2k-1)!+2j)}.$$

To note how I got my closed form expression, at the $i=(2k)!$ iteration, we are removing
$$2^{(2k)! + ((2k)! – (2(k-1)+1)!)}$$ intervals of length $$4^{-(2k)!}$$ note that we must remove $((2k)! – (2(k-1)+1)!)$ more intervals, since we did not remove anything for some time and also that $$(2k+1)! – (2k)! = 2k(2k)!.$$

Also it is clear that the lim inf (lower MD) is at least $\frac{1}{2}$, since $C \subset C’$, but it’s unclear why it isn’t strictly larger.

#### Solutions Collecting From Web of "Minkowski Dimension of Special Cantor Set"

I think you’re making this a bit harder than it needs to be. You don’t need an exact formula for $\text{vol}(C_{\delta_n})$, just upper and lower bounds. Furthermore, these can be quite crude – within a constant multiple.

To make this precise and to set notation, define an $\varepsilon$-mesh interval to be one of the form $(j\varepsilon,(j+1)\varepsilon)$ and define $N_{\varepsilon}(C’)$ to be the number of $\varepsilon$-mesh intervals that intersect $C’$. The upper and lower box-counting dimensions of $C’$ are then $\limsup$ and $\liminf$ as $\varepsilon\to0^{-}$ of
$$\frac{\log(N_{\varepsilon}(C’))}{\log(1/\varepsilon)}.$$
Note that I’ve chosen to use open intervals, since that will ease some further computations. If I had used closed intervals to obtain a related quantity, say $\overline{N}_{\varepsilon}$, then we could increase the value by as much as a factor of $3$. Ultimately, this doesn’t affect the dimension computation since
$$\frac{\log(N_{\varepsilon}(C’))}{\log(1/\varepsilon)}\leq \frac{\log(\overline{N}_{\varepsilon}(C’))}{\log(1/\varepsilon)} \leq \frac{\log(N_{\varepsilon}(C’))+\log{(3)}}{\log(1/\varepsilon)}$$
and that extra $\log(3)$ has no bearing on the limit. Also note that there is an easy relationship between these quantities and volume of the $\varepsilon$ fattening, as Tao shows later on that same page.

We’ll now show that the lower box-counting dimension of $C’$ is $1/2$; the other computation is similar. To do so, it is sufficient to show that
$$\liminf_{k\to\infty} \frac{\log(N_{\varepsilon_k}(C’))}{\log(1/\varepsilon_k)}\leq \frac{1}{2}$$
for $\varepsilon_k = 4^{-((2k+1)!-1)}$, as you’ve already mentioned that we know the lower box-counting dimension is at least $1/2$. To do this, we need an upper bound on $N_{\varepsilon_k}(C’)$.

First, an easy upper bound on $N_{4^{-(2k)!}}(C’)$ is $4^{(2k)!}$. Next, for each increment of $i$ over the span
$$(2k)!\leq i < (2k+1)!,$$
$N_{4^{-i}}(C’)$ increases by the factor $2$. Furthermore,
$$(2k+1)!-(2k)! = (2k)!((2k+1)-1) = 2k(2k)!.$$
Thus,
$$N_{4^{-((2k+1)!-1)}}(C’) \leq 4^{(2k)!} 2^{2k(2k)!} = 4^{(2k)!(k+1)}$$
and
$$\frac{\log(N_{4^{-((2k+1)!-1)}}(C’))}{\log\left(4^{((2k+1)!-1)}\right)} \leq \frac{(2k)!(k+1)\log(4)}{(2k+1)!\log(4) – \log(4)} \to \frac{1}{2}.$$