# On the Hausdorff dimension of a Cantor set

In what follows I refer to this paper by Orevkov.

I am writing a paper on this, so if somebody is interested we could consider to write a joint paper.

Consider a sequence $$R=\{R_n\}_n$$ of strictly increasing positive real numbers such that $$R_n\to\infty$$.

Take any $$a=\{a_n\}_n\subset\Bbb C$$ such that $$R_{n-1}<|a_n| and define the following polydiscs in $$\Bbb C^2$$: $$B_n=\{|z|\le R_n\}^2.$$

Define $$g_n(z)=\frac{\epsilon_n}{z-a_n};$$ and $$f_n(x,y)=\left(x,y+g_n(x)\right)\;\;\mbox{n odd}\\ f_n(x,y)=\left(x+g_n(y),y\right)\;\;\mbox{n even}$$ for $$(x,y)\in\overline{\Bbb C^2}$$ which is the one point compactification of $$\Bbb C^2$$.

Define then $$\gamma_n(z)=f_n\circ\dots\circ f_1(z,0)\;.$$ The sequence $$\epsilon=\{\epsilon_n\}_n$$ is defined to go to $$0$$ fast enough such that $$\gamma_n$$ converges.

Set $$A_n=\gamma_n(\overline{\Bbb C})=f_n(A_{n-1})$$ and consider the following open set $$\Delta_n=\gamma_n^{-1}(A_n\setminus B_n).$$ Now $$A_n\setminus B_n\subset\overline{\Bbb C\times\Bbb C}$$ has $$b_n$$ "branches" and some of them go to infinity on the first component, the other on the second. We refer to them as horizontal and vertical; correspondingly we write $$b_n=h_n+v_n$$. Creating $$\Delta_n$$, these $$b_n$$ components become $$b_n$$ (disjoint) open sets of the Riemann Sphere $$\overline{\Bbb C}$$; we write them as $$\Delta_n=\bigcup_{j=1}^{b_n}U_j^{(n)}.$$

Creating $$A_{n+1}=f_{n+1}(A_n)$$, what happens, is that the singularity of $$f_{n+1}$$ lies inside the first component for odd $$n$$, thus it depends on $$y$$ and thus it creates as many new horizontal branches as the singularity $$a_{n+1}$$ is met by the vertical components of $$A_n$$, that is, one for every $$v_n$$ components. Far from this singularity, $$f_{n+1}$$ behaves like $$f_n$$, thus just slightly moves the remaining $$b_n$$ components of $$A_n$$. Hence $$b_{n+1}=b_{n}+v_n=h_n+2v_n$$ and writing $$b_{n+1}=v_{n+1}+h_{n+1}$$, one has $$h_{n+1}=h_n+v_n$$ and $$v_{n+1}=v_n$$.

Call $$c_j^{(n)}$$ the diameter of $$U_j^{(n)}$$.

The sequence $$\{\Delta_n\}_n$$ is decreasing and it can be proved that $$C:=\bigcap_n\Delta_n$$ is a Cantor set.

Now I am investigating the Hausdorff dimension of such a Cantor set. I proved that for $$R$$ going to infinity sufficiently fast, $$\dim_H(C)$$ can be taken arbitrarily small.

I am searching now for an upperbound.

If $$T$$ is another diverging sequence and we write $$R to say $$R_n, it's clear that $$$$\label{1} \dim_H(C_T)\le\dim_H(C_R)$$$$ (we just made explicit the dependence of the Cantor set by the sequence).

Clearly we need something stronger than $$R_n for the above definition, like $$R if $$R_n=o(T_n)$$ for example; I would say, something such that the inequality $$\dim_H(C_T)<\dim_H(C_R)$$ is strict.

But this is still NOT enough to spot an upperbound.

It seems we should find some relation between the sequence $$R_n$$ and the size $$c_j^{(n)}$$.

Any hint?

Edit: As pointed out by Gerald Edgar, this Cantor set can have Hausdorff measure zero.

Maybe, since for every $$j\in\Bbb N\;\;\exists R^{(j)}=\{R_n^{(j)}\}_n$$ such that $$\dim_H(C_{R^{(j)}})\le\frac1j$$, then setting $$\tilde{R_n}:=R_n^{(n)}$$ it could be $$\dim_H(C_{\tilde R})=0$$.

But then the question is: is it possible to find some $$R$$ such that $$\dim_H(C_{R})=\alpha>0$$? What is an upper bound for $$\alpha$$?

It's clear that $$\alpha\le2$$; but for example, is it possible, for every $$\epsilon>0$$ to find $$R^{(\epsilon)}$$ such that $$\dim_H(C_{R^{(\epsilon)}})>2-\epsilon$$?

Another possibility is to conjecture that the Hausdorff measure is always zero, arguing then by contradiction.

It would be very surprising if there exists some bound $$\tilde\alpha\in(0,2)$$ such that we can find Cantor sets (built in this way of course) of $$\tilde\alpha$$ Hausdorff measure but NO Cantor set of greater Hausdorff measure. Maybe, not surprising, but very difficult, since it would require, for fixed $$R$$, to find a maximal sequence $$\epsilon$$ such that the final object is still a Cantor set, and then find out the Hausdorff dimension in function of $$R$$, and then maximize.

This would pass thru the analysis of the diameter of the $$U^{(n)}_j$$, which doesn't seem to be really friendly, so far.

• It is possible for a topological Cantor set (topological dimension zero and) also to have Hausdorff dimension zero. Feb 19, 2021 at 11:48
• Can you kindly give me some reference for this? Thank you!
– Joe
Feb 19, 2021 at 12:10
• If you take the Cantor set of all numbers whose binary expansion digits are all zero except in positions 1,4,9,.... that are squares, where they can be 0 or 1, you can show by the obvious coverings that the Hausdorff dimension is zero. Feb 19, 2021 at 18:09
• Thank you very much; I meant some general reference dealing with the stated problem (and also with a more precise statement about defining a fractal to be a topological space whose H-dim is $>$ its top. dim.)
– Joe
Feb 19, 2021 at 18:13
• Defining "fractal" by Hausdorff dimension > topological dimension. This was proposed as a "tentative definition" by Mandelbrot, but later he admitted it was not a useful definition. Then he said he preferred to leave the word "fractal" undefined. Feb 20, 2021 at 11:51