Slightly Chunky Cantor Sets

Question

I'm familiar with the construction of so-called "fat" Cantor sets (e.g. the Volterra construction), where a Cantor-type construction is used to construct a nowhere-dense set of positive Lebesgue measure, and thus has full Hausdorf dimension. Is there a similar construction for such a set with full Hausdorf dimension, but with null Lebesgue measure? Alternatively, can it be shown that such a Cantor construction cannot be done?

Answer 1

The easiest way to construct a set Lebesgue measure zero set with Hausdorff dimension 1 is to first construct a set $C_n$ with Hausdorff dimension $1-1/n$. It sounds like you already know that this can be done via a Cantor like construction where each interval is decomposed into two sub-intervals whose length is $2^{-n/(n-1)}$ times the length of it's parent. Then, these are all null sets, so the set $$\bigcup_{n=2}^{\infty}C_n$$ satisfies your needs.

If, on the other hand, you want to mimic Cantor's construction more directly (to obtain, for example, a perfect set), then you might try to construct approximations $C_n$, where each $C_n$ consists of $2^n$ disjoint intervals of Length $$\left(\frac{1}{2}-\frac{1}{\sqrt{n}+2}\right)^n.$$

Note that $$\lim_{n\rightarrow\infty} 2^n\left(\frac{1}{2}-\frac{1}{\sqrt{n}+2}\right)^n = 0,$$ so that the intersection of all the $C_n$s is Lebesgue null. On the other hand, $$\lim_{n\to \infty } \, \frac{\left(\frac{1}{2}-\frac{1}{\sqrt {n+1}+2}\right)^{n+1}}{\left(\frac{1}{ 2}-\frac{1}{\sqrt{n}+2}\right)^n} = \frac{1}{2},$$ so we might expect get Hausdorff dimension 1. Proving that might be hard, assuming it's right. You might look at the mass distribution principle.