I'm learning Minkowski and Hausdorff dimensions to study Brownian motion right now, and I'm trying to understand the reasoning behind the Minkowski dimension of the set $(0,1,1/2, 1/3,\ldots)$ being $1/2$. I am not certain if my understanding of the proof of this statement is correct, so I wanted to verify on here.
So to begin, let us assume that $\epsilon < 1/n^2$. Then if we split the set into:
$$(1, 1/2, 1/3, 1/4, \ldots, 1/n)\cup (0, 1/(n+1), 1/(n+2),\ldots) = A\cup B$$
We know that there must exist precisely $n$ disjoint $\epsilon$-balls that cover $A$. Since the elements of the set $B$ are closer together than the elements in the set in $A$, then there can only be at most another $n$ $\epsilon$-balls that also cover $B$. Then the total number of $\epsilon$-balls (let us denote this value as $M(E,\epsilon)$) is $M(E,\epsilon)=\mathcal{O}(n)$. Minkowski dimension is defined to be: $$\text{Dim}(E) = \lim_{\epsilon\downarrow 0}\frac{\log(M(E, \epsilon))}{\log(1/\epsilon)}$$
Since $M(E,\epsilon) = \mathcal{O}(n)$, and $\epsilon < 1/n^2$, this implies that $n<\sqrt{1/\epsilon}$, meaning that $M(E,\epsilon) = \mathcal{O}(1/\sqrt{\epsilon})$. From here we can take the limit: $$\text{Dim}(E) = \lim_{\epsilon\downarrow 0}\frac{\log(\mathcal{O}(1/\sqrt{\epsilon}))}{\log(1/\epsilon)}$$
From here, my knowledge that the answer is $1/2$ suggests that I can somehow eliminate the big-O completely, and evaluate the limit of $\frac{\log(1/\sqrt{\epsilon})}{\log(1/\epsilon)}$ to $1/2$. Why can we get rid of the big-O like this? Or is there some flaw in my earlier reasoning?
The problem with big-Oh notation is that it only gives you an upper bound. Recall the definition:
In other words, if $x$ is very close to $a$, then $f(x)$ is "dominated by" $g(x)$. This only gives an upper bound for $|f(x)|$, not a precise estimate. In the context of the question asked, the best that we can be done is to show that there are $M$ and $\delta$ such that $0 < |\varepsilon| < \delta$ implies that $$ \frac{\log(\mathcal{O}(1/\sqrt{\varepsilon})}{\log(1/\varepsilon)} \le \frac{\log(M/\sqrt{\varepsilon})}{\log(1/\varepsilon)} = \frac{\log(M) - \frac{1}{2} \log(\varepsilon)}{\log(1) - \log(\varepsilon)} = \frac{1}{2}- \frac{\log(M)}{\log(\varepsilon)}.$$ As $\varepsilon$ goes to zero, its logarithm goes to infinity, and so the last term vanishes. It then follows that $$ \operatorname{Dim}(E) = \lim_{\varepsilon\downarrow 0} \frac{\log(M(E,\varepsilon))}{\log(1/\varepsilon)} \le \lim_{\varepsilon\downarrow 0} \left[ \frac{1}{2} - \frac{\log(M)}{\log(\varepsilon)} \right] =\frac{1}{2}, $$ assuming that the limit exists.
In order to complete the proof that $\operatorname{Dim}(E) = 1/2$, it is necessary to also obtain a lower bound. Alternatively, obtain a more explicit count of the number of $\varepsilon$-balls needed to cover the set.
In this case, why not just use the explicit upper and lower bounds already found? Fix $\varepsilon > 0$ and choose $n$ so that $$\varepsilon = \frac{1}{n^2}. $$ No two points of the set $$ \left\{ \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dotsc, \frac{1}{n} \right\} $$ can be covered by a single $\varepsilon$-ball, so $n$ $\varepsilon$-balls are needed to cover this set. The rest of $E$ is contained in the interval $[0,\frac{1}{n}]$, and this interval can be covered by fewer than $n$ balls of radius $\varepsilon$ (it can be covered by exactly $\frac{n}{2} + 1$ such balls (assuming that they are open), but this more precise estimate is a little more tedious to work with, so the sloppier estimate is fine).
This implies that it takes at least $n$, and fewer than $2n$, balls to cover all of $E$. That is, $$ n \le M(E,\varepsilon) \le 2n. $$ Thus, since $\varepsilon = \frac{1}{n^2}$ (and so $n = \varepsilon^{-1/2}$) $$ -\frac{1}{2} \log(\varepsilon) = \log(n) \le \log(M(E,\varepsilon)) \le \log\left(2n \right) = -\frac{1}{2}\log(\varepsilon) + \log(2). $$ The Minkowski–Bouligand dimension can then be obtained via the squeeze theorem: $$ \frac{1}{2} = \lim_{\varepsilon\downarrow 0} \frac{-\frac{1}{2} \log(\varepsilon)}{\log(1/\varepsilon)} \le \underbrace{\lim_{\varepsilon\downarrow 0} \frac{\log(M(E,\varepsilon))}{\log(1/\varepsilon)}}_{= \operatorname{Dim}(E)} \le \lim_{\varepsilon\downarrow 0} \frac{-\frac{1}{2}\log(\varepsilon) + \log(2)}{\log(1/\varepsilon)} = \frac{1}{2}. $$