1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ Shooting at straws here, but I suppose skinny cantor sets are less common than the fat cantor sets of SVC because the fat cantor set's main interest is the fact that they are constructively similar to the canonical cantor set but it has positive measure while the cantor sets measure is zero. $\endgroup$ – MCT Dec 8 '15 at 5:30
  • $\begingroup$ I googled "Hausdorff dimension Cantor set", and the 4th hit was this paper arxiv.org/pdf/1312.1406.pdf, which covers your example and many more. $\endgroup$ – Lee Mosher Dec 8 '15 at 17:04
  • $\begingroup$ @LeeMosher That's an interesting paper, but it seems too advanced for me at the moment! Thanks for finding it and skimming it -- I didn't even know what I was looking for! $\endgroup$ – Andrey Kaipov Dec 9 '15 at 7:11
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.