TITLE: Uncountable sets of Hausdorff dimension zero
QUESTION [19 upvotes]: Let $A \subset \mathbb{R}$ be a countable set.  It is easy to see that $A$ has Hausdorff dimension $\dim_H(A) = 0$.
Do there exist uncountable sets $A \subset \mathbb{R}$ with $\dim_H(A) = 0$?

REPLY [22 votes]: A thin version of the usual Cantor middle third set works.
The idea is just that you need to omit more than just a
third of the remaining intervals as the construction
proceeds, enough so as to force the Hausdorff dimension to
$0$.
Specifically, we construct the set in stages. At each
stage, we've omitted a "middle third" from each finite
interval remaining. Thus, at stage $n$, our set is
contained in $2^n$ many intervals of some finite length
$a_n$. In the typical middle-third construction, we have
$a_n=3^{-n}$. But in our construction here, we want
$a_n$ to be small enough that $2^na_n^{1/n}\to 0$. By this means, the Hausdorff dimension will be forced to $0$.
But the resulting set is perfect, and hence is uncountable
of size continuum.