Hausdorff dimension of Cantor set

Question

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?

Answer 1

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 $\mathrm{diam}(I_{j})<(1+\delta)\mathrm{diam}(E_{j})$. 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.)

Answer 2

It is a fact in general (i.e. true in any metric space and Hausdorff measures of any dimension) that you can assume your covering sets to be open or closed. See Federer. For closed version, it is easier: diameter of $\bar{S}$ and diameter of $S$ are equal, so, if a collection of $S$'s cover your set and each has diam less than $\delta$, then you can instead consider the collection of $\bar{S}$'s, which again have diam bounded by $\delta$ ans still cover your set.

For open version, you need some sacrifice! At a cost of arbitrarily small $\sigma$, you can enlarge every set of diam less than $\delta$ to an open one with diam less than $(1+\sigma)\delta$. The latter can be used to estimate $\mathcal{H}^s_{(1+\sigma)\delta}$ within $(1+\sigma)^s$ accuracy of $\mathcal{H}^s_{\delta}$. Since for Hausdorff measure $\mathcal{H}^s$, you will send $\delta \to 0$, and $(1+\sigma)\delta$ will as well, your sacrifices will not affect the ultimate measure. (However, as expected, for one fixed $\delta$, $\mathcal{H}^s_\delta$ can be different if you only allow open coverings versus all coverings.)