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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

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

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.

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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}.

$$

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