Box Dimension Example 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.
 A: 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
$$
\dim_B(A) := \frac{\log(N_{\varepsilon}(A))}{-\log(\varepsilon)} = \frac{1}{2}.
$$
Computing the Hausdorff Dimension
The Hausdorff dimension of a countable subset of $\mathbb{R}^n$ is $0$, so we have $\dim_{H}(A) = 0$.  We'll prove the general result.  First recall that the $s$-dimensional Hausdorff measure of a set $F$ is defined to be
$$
H^s(F)
:= \lim_{\delta\to 0} H^s_\delta(F)
= \lim_{\delta\to 0} \left( \inf\left\{ \sum_{j=1}^{\infty} \mathrm{diam}(E_j)^s : \text{$\{E_j\}$ covers $E$ and $\mathrm{diam}(E_j) < \delta$} \right\} \right).
$$
The Hausdorff dimension is then defined to be
$$
\dim_{H}(F) := \inf\left\{ s\ge 0 : H^s(F) = 0 \right\} = \sup\left\{ s\ge 0 : H^s(F) = \infty \right\}.
$$
If $F$ is a countable subset of $\mathbb{R}^n$, then it can be enumerated.  So we can write $F = \{x_1, x_2, \dotsc\}$.  Fix $\delta > 0$, and for each $j=1,2,\dotsc$, let
$$ E_j = B(x_j, 2^{-(j+1)} \delta), $$
i.e. $E_j$ is the ball of radius $2^{-(j+1)}\delta$ centered at $x_j$.  Notice that


*

*the collection $\{ E_j \}$ covers all of $F$, since each point of $F$ is the center of one of the balls $E_j$, and

*$\mathrm{diam}(E_j) = 2^{-j}\delta < \delta$.


But if $s > 0$, then
$$ H^s_\delta(F)
\le \sum_{j=1}^{\infty} \mathrm{E_j}^s
= \sum_{j=1}^{\infty} 2^{-js}\delta^s
= \delta^s \sum_{j=1}^{\infty} \left( 2^{-s} \right)^j
= C \delta^s,
$$
where $C = \sum_j (2^{-s})^j$ is a finite constant, since the series is geometric with $r = 2^{-s} < 1$.  Since $s > 0$ is fixed, we take limits and obtain
$$ H^s(F)
= \lim_{\delta\to 0} H^s_\delta(F)
\le \lim_{\delta\to 0} C\delta^s
= 0.
$$
Again, this holds true for any $s>0$, from which it follows that $ \dim_{H}(F) \le 0$.  But then
$$ \dim_H(F) = 0, $$
as claimed.
A: The lower and upper box dimension calculation is similar to Example 3.5 Falconer book. Just only you replace k by $k^{2}.$
If $|U|<\frac{1}{2}$ and $k$ is the integer satisfying
$$\frac{2k-1}{k^2(k-1)^2}=\frac{1}{(k-1)^{2}}-\frac{1}{k^{2}}>\delta\geq \frac{1}{(k)^{2}}-\frac{1}{(k+1)^{2}}=\frac{2k+1}{(k+1)^2k^2}$$
Then $A$ can cover at most one of points $\{1,\frac{1}{4},...\frac{1}{k^{2}}\}.$ Thus $N_{\delta}(A)\geq k$ and hence
$$\underline{Dim}_{B}(A)=\underline{\lim}_{\delta\rightarrow 0}\frac{N_{\delta}(A)}{-log\delta}\geq {\lim}_{k\rightarrow \infty}\frac{k}{log{\frac{(k+1)^{2}k^2}{2k+1}}}=\frac{1}{3}$$
Similarly, for the upper box dimension then $k+1$ intervals of length $\delta$ cover $[0,\frac{1}{k^2}]$ leaving $k-1$ points of A which can covered by $k-1$ intervals of length $\delta.$ Thus $N_{\delta}(A)\geq 2k.$
