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As everyone knows, the standard middle-thirds Cantor set is constructed by dividing the interval into three equal parts, removing the middle one, then applying the same procedure to the remaining two intervals, etc.

The resulting set has Hausdorff dimension $s=\log 2/\log 3$, in view of Hutchinson's formula: $$ \sum_{i=1}^m r_i^s=1, $$ where in our case $m=2, r_1=r_2=1/3$.

Now, assume that $m=2$ but that on level $n$ we remove the middle-thirds interval whose relative measure is not exactly $1/3$ but $1/3+\delta_n$, where $\delta_n\to0$ sufficiently fast. (It may also depend on a position of the interval but there is always a uniform upper bound which tends to 0 sufficiently fast.)

Is it still true that the Hausdorff dimension of such a set is $\log2/\log3$?

This is actually a `toy question', since in my set-up the corresponding iterated function system is infinite countable. However, Hutchinson's formula works for such IFS just as well (with $m=\infty$), so I'm sure the conclusion should be the same. If it helps, the uniform upper bound for the $\delta_n$ in my case is a double exponent, i.e., $\frac1n \log (-\log \delta_n)\to \text{const}$ as $n\to+\infty$.

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2 Answers 2

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Since you remove "more", it should be more or less clear that $\text{HD}\; C_\delta \le \text{HD}\; C$ (where $C$ is the standard Cantor set and $C_\delta$ is the "perturbation" you describe). For proving the opposite inequality it is enough to exhibit a measure on $C_\delta$ whose Hausdorff dimension is equal to $\text{HD}\; C$. This is the usual uniform measure $m$ on $C_\delta$ (i.e., the one whose value is $1/2^n$ on each of the rank $n$ intervals), because if $\delta_n\to 0$ fast enough, then $$ \frac{\log m B(x,r)}{\log r} \to \log 2/\log 3 $$ for $m$-a.e. $x\in C_\delta$.

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Thanks. Yes, introducing measures looks like a good approach here. – Nikita Sidorov Jan 13 '13 at 17:38
With your fast convergence of $\delta_n$ the measure $m$ has also the property that $m(B(x,r)) \le cr^s$ for all $x \in \mathbb{R}^2$ (with $s = \log 2/ \log 3$). Therefore by Frostman's lemma $\mathcal{H}^s(C_\delta)>0$. For $\dim_\mathcal{H}(C_\delta) = s$ it is enough to assume $\delta_n \to 0$ and $\delta_n \in [0,2/3)$. No fast convergence is needed for this. – Tapio Rajala Jan 14 '13 at 6:28
You are right - the dimension formula I use also holds just under the assumption $\delta_n\to 0$. – R W Jan 14 '13 at 12:38

Just a quick note before having to move on, but when thinking about IFSs which approached self-similarity the papers of Igudesman have been useful. He has a notion of lacunary self-similar sets which (without checking details) looks like your set up. The main paper I am thinking of is the one referred to here.

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This paper is about lacunary sets which are generalizations of self-similar sets. I am interested in perturbed self-similar sets, so I don't quite see how his result could possibly help me. – Nikita Sidorov Jan 13 '13 at 17:37
Nikita, in the quick reading of your question that I had time for this morning it looked to me like your perturbed self-similar sets could be treated as lacunary self-similar sets. If that is not the case then such things happen and nothing but electrons have been wasted. – BSteinhurst Jan 14 '13 at 4:23

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