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Are there any famous examples of fractals, or other closed sets, of cardinality continuum but Hausdorff dimension 0?

I can think of something ad hoc like a Cantor middle $\frac13$ set where the middle $\frac13$ is followed by middle $\frac24$, middle $\frac35$ etc. but I am looking for something natural that's been studied before.

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  • $\begingroup$ I think you mean a fractal removing something like 2/3, 3/4. 4/5, not 1/3, 1/5, 1/7 (I'm pretty sure the latter has positive dimension). $\endgroup$
    – Wojowu
    Oct 23, 2016 at 19:41

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One example: the set of Liouville numbers has Hausdorff dimension zero.

In number theory, a Liouville number is an irrational number $x$ with the property that, for every positive integer $n$, there exist integers $p$ and $q$ with $q > 1$ and such that $$ 0<\left|x-{\frac {p}{q}}\right|<{\frac {1}{q^{n}}}. $$ A Liouville number can thus be approximated "quite closely" by a sequence of rational numbers. In 1844, Joseph Liouville showed that all Liouville numbers are transcendental, thus establishing the existence of transcendental numbers for the first time.

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    $\begingroup$ This set is not closed. $\endgroup$ Oct 23, 2016 at 23:23
  • $\begingroup$ @ChristianRemling ... correct, the set is dense. $\endgroup$ Oct 23, 2016 at 23:28
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    $\begingroup$ What makes this set a fractal? Does this set has some sort of self-similar property? $\endgroup$ Oct 24, 2016 at 2:11
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If by a fractal you mean a self-similar set and if the corresponding iterated function system (IFS) satisfies the Open Set Condition (OSC), then the answer is no.

The Moran-Hutchinson formula gives that the Hausdorff dimension $s$ satisfies $$ \sum_{k=1}^n |r_k|^s=1, $$ where $|r_k|\in(0,1)$ is the contraction ratio of the $k$th map. Clearly, if $s=0$, then $n=1$, which is absurd.

If the OSC is not satisfied, this usually means that some images of this IFS overlap, which, intuitively, can only increase the HD of the attractor. So I'd say the answer is still no.

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Closed uncountable sets of Hausdorff dimension zero turn up from time to time in practice (one of my papers with Nikita Sidorov features one) but I think that there does not exist a set which is simultaneously famous, closed, uncountable, and of zero Hausdorff dimension. I think that to some extent the sets which we call "fractals" become famous due to having striking visual properties, but zero-dimensional sets are necessarily very sparse, and it is perhaps more difficult for such a sparse set to be picturesque.

Let me extend this answer with a more mathematically substantial observation. The precise meaning of "fractal" is notoriously slippery, but one possible interpretation of this term refers to a set such that every open neighbourhood contains an arbitrarily small conformally-distorted copy of the whole set [1]. I would like to mention the following theorem of Falconer (Dimensions and measures of quasi-self-similar sets, PAMS 1989):

Theorem: Let $(X,d)$ be a compact metric space with Hausdorff dimension $s$. Suppose there exist $a,r_0>0$ such that for every $r$-ball in $X$ with $r<r_0$ there is a function $\phi \colon X \to B$ satisfying $ard(x,y)\leq d(\phi(x),\phi(y))$ for every $x,y \in X$. Then the box-counting dimension of $X$ equals its Hausdorff dimension, and its $s$-dimensional Hausdorff measure is finite.

If a set with this property has Hausdorff dimension zero then since zero-dimensional Hausdorff measure is simply counting measure, the set must be finite. There is therefore no zero-dimensional infinite compact set which is "everywhere locally self-conformal" in the sense of the above theorem. This result generalises Nikita's answer; in particular it excludes any compact set $X$ which satisfies the self-similarity property $$X=\bigcup_{i=1}^N f_i(X)$$ where each $f_i$ is a $C^1$ conformal contraction, irrespective of whether or not separation conditions such as the Open Set Condition are satisfied.

[1] There is a lively literature on sets which contain arbitrarily small affinely-distorted but highly nonconformal copies of themselves, so this definition of "fractal" is in practice too restrictive.

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    $\begingroup$ Thanks. Your paper must be arxiv.org/abs/1107.3506v2 A paper in JEMS is already close to famous among mathematicians I suppose. $\endgroup$ Oct 24, 2016 at 13:10
  • $\begingroup$ The one you mention is a sequel to this one: arxiv.org/abs/1006.2117 - plenty of Sturmian sequences there, too. ;) $\endgroup$ Oct 24, 2016 at 18:37
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    $\begingroup$ I love this description: “simultaneously famous, closed, uncountable, and of zero Hausdorff dimension.” $\endgroup$ Apr 10, 2018 at 18:01
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There are Moran geometric constructions which give closed generalized self-similar sets of Hausdorff-dimension zero. For the general construction see chapter 5 of Yakov B. Pesin Dimension Theory in Dynamical Systems: Contemporary Views and Applications, Chicago Lectures in Mathematics, 1997. These sets have Hausdorff-dimension zero if you decrease the diameter in the construction superexponential. Ok, I admit that this construction is not such famous.

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For example, let $x_n=3^{-2^{n-1}}$ for $n=1,2,\ldots,$ and consider the following modified version of the classic Cantor middle set:

$$ \begin{aligned} F_1&=[0,1]\\ F_2&=[0,x_1]\cup[1- x_1,1]\\ F_3&=[0,x_2]\cup[x_1-x_2,x_1]\cup [1-x_1,1-x_1+x_2]\cup[1-x_2,1]\\ F_4&=[0,x_3]\cup [x_2-x_3,x_2]\cup [x_1-x_2,x_1-x_2+x_3]\cup [x_1-x_3,x_1]\cup [1-x_1,1-x_1+x_3]\cup [1-x_1+x_2-x_3,1-x_1+x_2]\cup [1-x_2,1-x_2+x_3]\cup [1-x_3,1]\\ &\vdots \end{aligned} $$

In the limit $F=\cap_{n\in\mathbb{N}}F_n,$ you will obtain a closed set with Hausdorff dimension equal to $0$ and cardinality continum.

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