I'm considering the problem of calculating the Hausdorff dimension of a Cantor set, according to the following lemmas:
Lemma 1
Let $C: [0, 1] \rightarrow [0, 1]$ be a Cantor staircase function. Then $C$ is $\alpha$-Holder-continuous with $\alpha=\log_{3}{2}$.
Lemma 2
If $f: [0,1] \rightarrow [0,1]$ is a $\alpha-$ Holder continuous function, then for any subset $X \subset [0, 1]$, $\dim_{H}{f(X)} \leq \frac{1}{\alpha} \dim_{H}(X)$
With the lemmas in hand we can proceed in the following way: since we know that the image of the Cantor set under $C(x)$ is $[0, 1]$, by letting $X$ to be a Cantor set and $f(X)=[0,1]$ its image under $C(x)$, we write down the following estimates:
$$ \dim_{H}{f(X)} \leq (\log_{2}{3}) \dim_{H}{X} = \log_{2}{3}$$
$$ \dim_{H}{X} \geq \log_{3}{2}$$
(since $\dim_{H}{[0,1]}=1$).
It can be found that the Hausdorff dimension of the Cantor set is precisely $\log_{3}{2}$. In order to prove it, it is reasonable to somehow derive that $\dim_{H}{X} \leq \log_{3}{2}$ or just to find a covering of a Cantor set with open balls so that the desired dimension attains.
How can I proceed in this case?
Best Answer
Yes, the way to go is to find a covering. Not necessarily with open balls: the usual definition of Hausdorff measure allows arbitrary sets. If your definition requires an open ball, observe that every set of diameter $d$ is contained in an open interval of diameter $3d$, so this doesn't really matter for computing dimension.
After $n$ stages of removing middle intervals, you have $2^n$ intervals left, each of length $3^{-n}$. This gives you a cover $\{I_{k}: k=1,\dots,2^n\}$ which satisfies $$ \sum_{k=1}^{2^n} (\operatorname{diam} I_k)^s = 1,\quad s = \log_3 2 $$ for every $n$. Hence the $s$-dimensional measure of $C$ is finite.