A Cantor-like set formed by removing open intervals of increasing length. The classic middle-thirds Cantor set can be generalized to a middle-$\alpha$-th Cantor set, for $0<\alpha<1$. It can be formed by removing the open middle $\alpha$-th from the unit interval, and continuing to remove the middle $\alpha$-th from each of the remaining intervals ad infinitum.
It can be shown by induction that at the $k$-th level of this construction process, there are $2^k$ disjoint sub-intervals. Further, each sub-interval at the $k$-th level has length $\left(\frac{1-a}{2}\right)^k$.
With this knowledge, we can intuitively find the Hausdorff dimension of this generalized Cantor set to be $$0 < \frac{1}{1-\lg(1-\alpha)} < 1.$$
But this was easy because we are always removing a fixed percentage of the remaining intervals at each level. That is, at the $k$-th level we always remove the middle $\alpha$-th from an interval of length $\left(\frac{1-a}{2}\right)^k$, i.e. we remove $\alpha\cdot\left(\frac{1-a}{2}\right)^k$.
A variation of this is the SVC set, where we remove smaller percentages of the remaining intervals at each level, to get a sort of fat Cantor set. 
My question is what happens if we decide to remove larger percentages of the remaining intervals at each level, to get a sort of skinny Cantor set? I find no results on Google about these kinds of Cantor sets. Specifically, I'd to find their dimension, but since I have no formal education in topology, I cannot work this out by hand. Any accessible information on them would be pretty cool.
As an example, let's say we begin with the interval $[0,1]$, and let's say we take out the open middle third so that our next level is $\left[0,\frac{1}{3}\right]\cup\left[\frac{2}{3},1\right]$. Now, instead of removing the middle third from $\left[0,\frac{1}{3}\right]$, let's say we remove the middle-$\left(\frac{1}{3}+\varepsilon\cdot\frac{1}{3}\right)$. That is, we removed an $\varepsilon>0$ percent larger interval than what we would have normally.

The motivation for this question comes from investigating the logistic map $f_r(x) = rx(1-x)$ for when $r > 4$. According to Richard Holmgren in his book A First Course in Discrete Dynamical Systems, the points inside $[0,1]$ whose $f_r$-iterations remain in $[0,1]$ forever form a Cantor set.
Upon further investigation it looks like this set is of the skinny variety described above, but I can't find the factor that we increase our removals by!
 A: In general, after $k$ removals we have $2^k$ intervals of size $\ell_k$ each. These form a cover of the set. We are interested in the values of $d$ for which the $d$-dimensional measure is zero: the infimum of such values is the dimension. So far, it looks like the measure is bounded by 
$2^k \ell_k^{d}$, which $\le 1$ if $d\ge k\log 2/|\log \ell_k|$.
If the same proportion is removed at every step, the quantity $k\log 2/|\log \ell_k|$ is independent of $k$. In general, it can vary. The definition of Hausdorff measure does not require us to have an efficient covering on every scale: it takes the infimum over all sufficiently small scales. So, the relevant quantity is 
$$
\liminf_{k\to\infty} \frac{k\log 2}{-\log \ell_k}  \tag{1}
$$
which is what the Hausdorff dimension of this set is. 
The above isn't a complete proof. It has essentially all of the proof that the dimension does not exceed the value in $(1)$. The reverse inequality is based on the mass distribution principle (the easy half of Frostman's lemma): see Lower Bound of Hausdorff Dimension of Cantor Set.
