Consider a Cantor set $E$ where the intervals at every level of the construction maintain a minimum spacing and have a finite number of intervals on each level. I have two questions regarding finding the lower bound of the Hausdorff dimension on such a set. Assume we have a collection of sets $\{U_i\}$ such that $|U|<\epsilon$ and the union of these sets provide a cover of $C$. $\mathcal{H}^s(E)$ is the Hausdorff $s$-dimensional measure.
1) The mass distribution principle that says if we have a mass distribution on $E$ and $\mu(U)<c|U|^s$ then we can claim that $\mathcal{H}^s(E)>0$, and hence claim $s$ is a lower estimate on Hdim$E$. Is it possible to apply this method when the intervals from a given level have varying lengths? If yes, how do we assign masses to the intervals? If this method does not apply here, what other methods are there for this class of sets?
2) I attended a talk where professor claimed that when establishing a lower estimate on Hdim of a Cantor set it is sufficient to show that we can find a special collection of intervals in our construction such that $\sum|I_k|<c|U|^s$ for an arbitrary $U$. Can anyone help me connect the dots between this claim and $\mathcal{H}(E)^s>0$?
Suppose F is a metric space. Also assume that there are constants $C$ and $\delta$ such that. $$(1) \quad \mu(S) \le C \cdot |S|^d $$ for a all closed sets $S$ and $d \gt 0$ Then $$H_d^{\delta}(F) \ge {{\mu(F)} \over C}$$
Proof:
Let $\bigcup_i {F_i}$ be a covering of $F$ It's easy to show that we can take $|F_i|$ to be closed. So we have... $$ (2) \quad 0 \lt \mu(F) \le \mu \left( \bigcup_i F_i \right) \le \sum_i \mu(F_i) \le C \cdot \sum_i |F_i|^d \le H_a^{\delta}(F)$$ $$ \Rightarrow dim(F) \ge d$$ where the last part of (2) is given from the assumption (1). Thus, $$H_d^{\delta}(F) \ge {{\mu(F)} \over C}$$ hence, $dim(F) \ge d$
To answer your questions. The above gives a single bound on the dimension. If the interval has varying lengths, the estimate will probably be bad. For example if a cantor fractal has one length at iteration 1 of $1/3$ and another of $1/4$, you'll have to redistribute the measure so that the $1/3$ side receives $4/7$ and the $1/4$ side receives $3/7$.