# Are there sets of zero measure and full Hausdorff dimension?

I would like to ask the following:

Are there "many" sets, say in the interval $[0,1]$, with zero Lebesgue measure but with Hausdorff dimension $1$?

The motivation for this question is the dichotomy between measure and category. There are certainly dense sets with zero Lebesgue measure. But a dense set need not have positive Hausdorff dimension (for example, the rationals are dense but have zero Hausdorff dimension).

Honestly, I would already be satisfied with an answer to the following question:

Is there any set in $[0,1]$ with zero Lebesgue measure but with Hausdorff dimension $1$?

• Same question on MO: mathoverflow.net/questions/35986/…
– ocg
Feb 9, 2016 at 3:13
• @JulienGodawatta It is a better fit here, anyway. Feb 9, 2016 at 3:15

For any $r<1$, you can construct a Cantor set with Hausdorff dimension $r$ by varying the lengths of the intervals in the usual Cantor set construction. In particular, you can let $C_n\subset[0,1]$ be a Cantor set of Hausdorff dimension $1-1/n$ for each $n$. The union $C=\bigcup C_n$ then has Lebesgue measure $0$ because each $C_n$ does, but Hausdorff dimension $1$.
• @bartgol: It depends how you vary them. If you always remove a constant proportion of each of the intervals you have so far (as in the standard construction where you always remove $1/3$), you'll always get a set of measure zero, but the Hausdorff dimension will depend on the proportion. If you let the proportion you're removing get smaller and smaller fast enough, you get a Cantor set of positive measure. Feb 9, 2016 at 3:29
• If you remove the middle $r$ proportion of each interval, then at the $n$th stage of the construction you have a set of measure $(1-r)^n$, which goes to $0$ as $n\to\infty$. So as long as the proportion is always the same (even if it's smaller than $1/3$), you get a set of measure $0$. Feb 9, 2016 at 3:49
For a "naturally occurring" example, let $$b_1$$ and $$b_2$$ be positive integers $$\geq 2$$ such that no positive integer power of $$b_1$$ equals a positive integer power of $$b_2$$ (i.e. $$(b_1)^m = (b_2)^n$$ has no solution where $$m$$ and $$n$$ are positive integers). Kenji Nagasaka proved in 1979 that the set of real numbers normal to base $$b_1$$ but not normal to base $$b_2$$ is a measure zero set with Hausdorff dimension $$1.$$ See my 5 July 2002 sci.math post Numbers normal to one base but not to another base. (Note: In that post I seem to have reversed the definitions of multiplicatively dependent and multiplicatively independent.)
Actually, Nagasaka only proved the Hausdorff dimension $$1$$ part. The measure zero part follows from the long-known fact that all real numbers except for a set of measure zero are normal to every base.