I am trying to find the lower- and upper-box dimensions (and show that they are the same) of the set $A=\{0,1,\frac{1}{4},\frac{1}{9},\ldots\}=\{\frac{1}{n^{2}}:n\in\mathbb{Z}_{\geqslant0}\}\cup\{0\}$.
My thinking: There are $k$ intervals of length $\frac{1}{k^{2}}$ at stage $k$ of the construction. So $$\dim_{B}(A)=\lim_{\varepsilon\to0}\frac{\log{N_{\varepsilon}(A)}}{-\log{\delta}}=\lim_{k\to\infty}\frac{\log{k}}{-\log{k^{2}}}=\lim_{k\to\infty}\frac{\log{k}}{2\log{k}}=0.5.$$
But this doesn't feel right. I haven't found the upper- and lower- limits, I have just kind of `done it'. Can somebody tell me if this is right? And if not, what I should do?
I also have to show that it is equal to the Hausdorff dimension, but one step at a time.
2 Answers
Computing the Box Counting Dimension
Note that showing that there is a sequence of $\varepsilon_k$ such that $\varepsilon_k \to 0$ and $$ \lim_{k\to\infty} \frac{ \log_{\varepsilon_k}(A) }{-\log(\varepsilon_k)} = \frac{1}{2} $$ is not enough to show that $$ \lim_{\varepsilon\to 0} = \frac{ \log(N_{\varepsilon}(A) }{ -\log(\varepsilon) }, $$ which is what you have done in your original computation (with $\varepsilon_k = \frac{1}{k^2}$). You need to show that the result holds for any sequence of $\varepsilon_k$ that tends to zero. This is a bit more work, and could (in general) involve the "trick" that serg_1 mentions. We can, however, do without, as shown below:
Let $\varepsilon \in (0,1]$. We require one ball of diameter $\varepsilon$ (i.e. one interval of length $\varepsilon$, or one box of side length $\varepsilon$) to cover all of the points $n^{-2}$ such that $$ \frac{1}{n^2} < \varepsilon \implies n > \frac{1}{\sqrt{\varepsilon}} = \varepsilon^{-1/2}. $$ No ball of diameter $\varepsilon$ can contain more than one of the remaining points of $A$, thus if $n \le \varepsilon^{-1/2}$, we require a ball to cover that point. As $n$ is a natural number, we will require $\lfloor \varepsilon^{-1/2} \rfloor$ additional balls to cover $A$. Hence $$ N_{\varepsilon}(A) = 1 + \lfloor \varepsilon^{-1/2} \rfloor. $$ Observe that \begin{align} \varepsilon^{-1/2} - 1 < \lfloor \sqrt{\varepsilon} \rfloor \le \varepsilon^{-1/2} &\implies \varepsilon^{-1/2} < N_{\varepsilon}(A) \le 1 + \varepsilon^{-1/2} \\ &\implies \log(\varepsilon^{-1/2}) < \log(N_{\varepsilon}(A)) \le \log(1+\varepsilon^{-1/2}) < \log(2\varepsilon^{-1/2}) \tag{1} \\ &\implies \frac{\log(\varepsilon^{-1/2})}{-\log(\varepsilon)} < \frac{\log(N_{\varepsilon}(A))}{-\log(\varepsilon)} \le \frac{\log(2)-\frac{1}{2}\log(\varepsilon^{-1/2})}{-\log(\varepsilon)} \tag{2} \\ &\implies \frac{1}{2} < \frac{\log(N_{\varepsilon}(A))}{-\log(\varepsilon)} \le \frac{1}{2} - \frac{\log(2)}{\log(\varepsilon)}. \end{align} At (1), we use the fact that $\log$ is increasing and $\varepsilon < 1 \implies \varepsilon^{-1/2} > 1$. We again use the assumption that $\varepsilon < 1$ at (2) (note the negative signs; $-\log(\varepsilon) > 0$). Taking limits as $\varepsilon \to 0$ (and so $\log(\varepsilon) \to -\infty$), we have $$ \frac{1}{2} \le \frac{\log(N_{\varepsilon}(A))}{-\log(\varepsilon)} \le \frac{1}{2}. $$ Since this limit exists, we have the desired result, namely that $$ \dim_B(A) := \frac{\log(N_{\varepsilon}(A))}{-\log(\varepsilon)} = \frac{1}{2}. $$
