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I am new to this site, so sorry if the question is stupid. I am learning fractals and my teacher gave me the following exercise.

Let $$E=\{0,0\}\cup\left\{\bigcup_{n=1}^\infty (x,1/\sqrt{n}\,):0\leq x\leq 1/\sqrt{n}\right\}$$ Find $\dim_\text{lower box}(E)$ and $\dim_\text{box}(L\cap E)$ for all lines $L$ with non-zero gradient and which do not pass through $(0,0)$.

I have done some of the work but I do not know if it is correct.

$E$ can be covered by $n(n+1)/2$ boxes of side $1/\sqrt{n}$. But then this gives $$\dim_\text{lower box}(E)\leq\lim_{n\to\infty}\frac{\log(n)+\log(n+1)-\log(2)}{\frac{1}{2}\log(n)}=4$$

I am certain this is not correct. Could someone possibly please help me? Thank you very much.

Ok, now I am told that I am supposed only to show $\dim_{lower\ box}\geq\frac{4}{3}$. Does anyone know how I do this? Thanks you.

Jane
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  • I don't think you're calculation of the number of boxes is correct. For example, when $n=3$, you need $3$ boxes to cover the set, but you calculation says you need $6$. – JSharpee Aug 20 '17 at 00:59
  • Ok thank you. I see that is true now. I do not see how to work out the right number. Do you know? – Jane Aug 20 '17 at 01:12
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    No, sorry. Maybe try using an equivalent definition of a covering? I think that the other part is simpler. Any non-horizontal (zero gradient) line not passing through zero intersects $E$ at finitely many points, so the box dimension should be zero (just cover by a finite number of boxes of side length $\epsilon$ and take the limit as $\epsilon\to0$. – JSharpee Aug 20 '17 at 10:14
  • Oh that is very good thanks you. That end bit is helpful to me. I will wait here for some help on the other part. – Jane Aug 20 '17 at 10:16

1 Answers1

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Since it is probably a homework problem, I'll just get you started.

Suppose the width of the box is $\epsilon$. For each line segment $(0,1/\sqrt n)\times\{1/\sqrt n\}$ you need at least $1/(\sqrt n \epsilon)$ boxes, and these boxes won't cover more than one line segment if $\frac1{\sqrt n}-\frac1{\sqrt{n+1}} > \epsilon$, which gives $n \lessapprox \epsilon^{-2/3}$, since $\frac1{\sqrt n}-\frac1{\sqrt{n+1}} \approx \frac 1{n^{3/2}}$.

The number of boxes this gives can be written as a sum, and this sum can be approximated by an integral.

Stephen Montgomery-Smith
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  • This is a problem for homework yes. Why is $\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\approx\frac{1}{n^{3/2}}$? I thought it was that it was less than $\frac{1}{2}\sqrt{n}$. I do not understand how it can be approximated with an integral. Is the sum $n/(n+1)/2$. – Jane Aug 23 '17 at 13:39
  • First Q. It is the distance between two line segments. – Stephen Montgomery-Smith Aug 23 '17 at 13:41
  • But is the approximation correct? I do not see how it is. Could you explain please? – Jane Aug 23 '17 at 13:43
  • Use the mean value theorem for the approximation. – Stephen Montgomery-Smith Aug 23 '17 at 14:06
  • I think this right now.

    At least $n^{-1/2}\epsilon^{-1}$ boxes of width $\epsilon$ are required to cover each line segment and will not cover more than one if $\epsilon<n^{-1/2}-(n+1)^{-1/2}\approx n^{-3/2}$. Let $N_{\epsilon}(E)$ be the smallest number of boxes needed to cover $E$. Then $N_\epsilon (E)>n^{-1/2}\epsilon^{-1}$ and so $$\dim_{lower\ box}(E)>\lim_{\epsilon\to0}\frac{\log(n^{-1/2}\epsilon^{-1})}{-\log(\epsilon)}>\lim_{\delta\to0}\frac{\frac{1}{2}\log(n)+\log(\epsilon)}{\log(\epsilon)}=1+\frac{1}{2}\lim_{n\to\infty}\frac{\log(n)}{\log(n^{3/2})}=\frac{4}{3}.$$

    Am I right?

     – Jane Aug 23 '17 at 14:13
  • No. $N_\epsilon(E) \ge \sum 1/(\sqrt{n} \epsilon)$, where the sum is over those $n$ for which $1/\sqrt n - 1/\sqrt{n+1} > \epsilon$. – Stephen Montgomery-Smith Aug 23 '17 at 23:08