7
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ as i remember we must prove trick inequality, for example: $ \frac{1}{2}\leqslant \underline{dim}_BA\leqslant \overline{dim}_BA\leqslant \frac{1}{2}$, for details see in book: Falconer. Fractal geometry. $\endgroup$
    – serg_1
    Jun 12 '17 at 21:30
  • $\begingroup$ This looks fine to me. The upper and lower limits are usually only of interest if they are not equal. In this case, the limit does exist (as you've shown) and the upper and lower limits are both $\frac12$. (Note to self: Wiki Page) $\endgroup$ Jun 12 '17 at 21:34
  • $\begingroup$ Thank you both. @AlexisOlson, do you think you could you help me do the Hausdorff dimension please? $\endgroup$
    – JSharpee
    Jun 13 '17 at 8:06
4
$\begingroup$

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

  1. the collection $\{ E_j \}$ covers all of $F$, since each point of $F$ is the center of one of the balls $E_j$, and
  2. $\mathrm{diam}(E_j) = 2^{-j}\delta < \delta$.

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.

$\endgroup$
1
  • $\begingroup$ This is a really nice, well-explained answer. Well done and thank you :^) $\endgroup$
    – JSharpee
    Aug 6 '17 at 14:06
0
$\begingroup$

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 $N_{\delta}(A)\geq 2k.$

$\endgroup$
1
  • $\begingroup$ This is not a standalone answer, since it depends upon Falconer's book. $\endgroup$ Oct 23 '19 at 14:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.