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I know this is probably a easy question, but some steps in the proofs I found almost everywhere contained some parts or assumptions which I think may not be that trivial, so I would like to make it rigorous and clear enough. Here is the question:

Let $C$ be the Cantor set with delete the middle third of the interval and go on. The general Cantor can be considered similarly. We want to proof the Hausdorff dimension of $C$ is $\alpha:=\log2/\log3$. So we calculate the $d$-dimensional Hausdorff measure $H^d(C)$ for all $d$ to determine the Hausdorff dimension. Let $C(k)$ be the collection of $2^k$ intervals with length $1/3^k$ in the $k^{th}$-step of construction of Cantor set.

It is rather easy to show that $H^{\alpha}(C)<\infty$ by showing that for any covering $\{E_j\}_{j=1}^{\infty}$of $C$ the set $C(k)$ also covers $C$ for $k$ large enough, so we can bound $H^{\alpha}(C)$ from above. Which implies that the Hausdorff dimension of $C$ is less than $\alpha$.

To show the dimension is actually equal to $\alpha$, it suffices to show $H^{\alpha}(C)>0$.

Now let $\{E_j\}_{j=1}^{\infty}$ be any covering of $C$ with diameter $diam(E_j)\le \delta$ for all $j$. How do we show that $$\sum_j diam(E_j)^{\alpha}>constant$$

One author (see this link) made the following assumption: $E_j$ be open, so one can find the Lebesgue number of this covering, and when $k$ large enough, any interval in $C(k)$ will be contained in $E_j$ for some $j$. Hence one can bound the $\sum_j diam(E_j)^{\alpha}$ from below by the ones of $C(k)$.

I got confused here: First why we can assume $E_j$ to be open?

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up vote 3 down vote accepted

I can elaborate abit why it is sufficient to show it for open ones. It might be that this is not exactly what the author was going after. If you wish, I may also continue this answer into a full proof which shows that $H^{\alpha}(C)\geq \frac{1}{2}$.

Choose for starters a $\delta$-cover $\{E_{j}\}_{j=1}^{\infty}$ of $C$ with $\sum_{j=1}^{\infty}\mathrm{diam}(E_{j})^{\alpha}\leq H^{\alpha}(C)+\delta$. Then for each $j$ we may choose a closed interval $I_{j}$ with $E_{j}\subset \mathrm{int}I_{j}$ and $\mathrm{diam}(I_{j})<(1+\alpha)\mathrm{diam}(E_{j})$. Hence $\{\mathrm{int}I_{j}\}_{j=1}^{\infty}$ is an open cover of $C$, and in particular \begin{align*} H^{\alpha}(C)+\delta\geq \sum_{j=1}^{\infty}\mathrm{diam}(E_{j})^{\alpha}\geq (1+\delta)^{-\alpha}\sum_{j=1}^{\infty}\mathrm{diam}(I_{j})^{\alpha}. \end{align*} Thus to establish a lower bound for $H^{\alpha}(C)$ it suffices to establish a lower bound for $\sum_{j=1}^{\infty}\mathrm{diam}(I_{j})^{\alpha}$. (You may consider $\mathrm{int}(I_{j})$'s instead, which are open, as their diameter is the same as $I_{j}$'s.)

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great, Thanks Thomas – Sun May 30 '12 at 19:53

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