Are there sets of zero measure and full Hausdorff dimension?

Solution 1:

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 $1$.

Solution 2:

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.