The Dimension of a Cantor Set

Edited by Leun Kim

Let $ K_0 = [0,1]$, and $K_j$ be obtained by removing the open middle $ \alpha_j$th from each of the intervals that make up $K_{j-1}$(Here, $\{ \alpha_j \}$ is any sequence of numbers in (0,1)). The resulting limiting set $K = \bigcap_{j=1}^\infty K_j$ is called a generalized Cantor set.
Now consider the fractal dimension of $K$, which is defined by $$ \dim K = \lim_{\epsilon \to 0} \frac{\log N_{\epsilon} (K)}{- \log{\epsilon}} $$ where $N_{\epsilon} (K)$ is the smallest number of $\epsilon$-balls needed to cover $K$. Let $\epsilon >0$ be given. Then we have $$ \prod_{j=1}^n \frac{1 – \alpha_j}{2} < \epsilon \leqslant \prod_{j=1}^{n-1} \frac{1 – \alpha_j}{2} \Rightarrow N_{\epsilon} (K) \leqslant 2^n $$ $$ \prod_{j=1}^{n+1} \frac{1 – \alpha_j}{2} \leqslant \epsilon < \prod_{j=1}^n \frac{1 – \alpha_j}{2} \Rightarrow N_{\epsilon} (K) \geqslant 2^n $$ for all $n$. Thus we have $$ \dim K = \lim_{\epsilon \to 0} \frac{\log N_{\epsilon} (K)}{- \log{\epsilon}} \leqslant \lim_{n \to \infty} \frac{n \log 2}{\log \prod_{j=1}^{n-1} \frac{2}{1 – \alpha_j}} = \lim_{n \to \infty} \frac{n \log 2}{(n-1) \log 2 + \sum_{j=1}^{n-1} \log \frac{1}{1- \alpha_j}} $$ $$ \dim K = \lim_{\epsilon \to 0} \frac{\log N_{\epsilon} (K)}{- \log{\epsilon}} \geqslant \lim_{n \to \infty} \frac{n \log 2}{\log \prod_{j=1}^{n+1} \frac{2}{1 – \alpha_j}} = \lim_{n \to \infty} \frac{n \log 2}{(n+1) \log 2 + \sum_{j=1}^{n+1} \log \frac{1}{1- \alpha_j}} $$ respectively. Thus we can conclude that $$ \dim K = \lim_{n \to \infty} \frac{n \log 2}{n \log 2 + \sum_{j=1}^n \log \frac{1}{1- \alpha_j}} = \lim_{n \to \infty} \frac{n \log 2}{n \log 2 +s(n)} $$ where $ s(n) = \sum_{j=1}^n \log \frac{1}{1- \alpha_j} $. We can observe that if $ s(n) = o(n) $, then $ \dim K = 1 $, and if $ s(n) = \omega (n) $, then $ \dim K = 0 $.

Here is a simple example. Let $\alpha_j = j/(j+1)$. Then $$ \dim K = \lim_{n \to \infty} \frac{n \log 2}{n \log 2 + \log \prod_{j=1}^n (j+1)} = \lim_{n \to \infty} \frac{n \log 2}{n \log 2 + \log (n+1)!} = 0. $$ And if $\alpha_j = 1/(j+1)$, then $$ \dim K = \lim_{n \to \infty} \frac{n \log 2}{n \log 2 + \log \prod_{j=1}^n \frac{j+1}{j}} = \lim_{n \to \infty} \frac{n \log 2}{n \log 2 + \log(n+1)} = 1 $$ as we expected.

 

References.
[1] G. B. Folland, Real Analysis: Modern Techniques and Their Applications.
[2] M. Pollicott, Lectures on Fractals and Dimension Theory.

여러가지 Fractal들의 Hausdorff Dimension은 여기 를 참조!

MathJax가 좋다고 하길래 한 번 시험삼아 작성해 보았는데 편하긴 하네요. 그나저나 세미나 관련 책을 어서 빌려야 되는데 이럴수가, 도서관이 토요일인데도 문 닫았네요 ㅠㅠ

 

 
I was born and raised in Daegu, S. Korea. I majored in electronics and math in Seoul from 2007 to 2012. I've had a great interest in math since freshman year, and I studied PDE in Osaka, Japan from 2012-2014. I worked at a science museum and HUFS from 2014 in Seoul. Now I'm studying PDE in Tokyo, Japan. I also developed an interest in music, as I met a great piano teacher Oh in 2001, and joined an indie metal band in 2008. In my spare time, I enjoy various things, such as listening music, blogging, traveling, taking photos, and playing Go and Holdem. Please do not hesitate to contact me with comments, email, guestbook, and social medias.



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