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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 $R_{n-1}<|a_n|<R_n$ 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 $R<T$ to say $R_n<T_n\;\;\forall n$, it's clear that \begin{equation}\label{1} \dim_H(C_T)\le\dim_H(C_R) \end{equation} (we just made explicit the dependence of the Cantor set by the sequence).

Clearly we need something stronger than $R_n<T_n$ for the above definition, like $R<T$ 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.

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    $\begingroup$ It is possible for a topological Cantor set (topological dimension zero and) also to have Hausdorff dimension zero. $\endgroup$ Feb 19, 2021 at 11:48
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    $\begingroup$ Can you kindly give me some reference for this? Thank you! $\endgroup$
    – Joe
    Feb 19, 2021 at 12:10
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    $\begingroup$ 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. $\endgroup$ Feb 19, 2021 at 18:09
  • $\begingroup$ 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.) $\endgroup$
    – Joe
    Feb 19, 2021 at 18:13
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    $\begingroup$ 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. $\endgroup$ Feb 20, 2021 at 11:51

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