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 $\dim_H(A) = 0$?

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    $\begingroup$ The Liouville numbers do the trick: en.wikipedia.org/wiki/Liouville_number $\endgroup$ – user940 Oct 18 '11 at 4:07
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    $\begingroup$ @Bryon: Thank you for the link. An explicit example is nice to have! $\endgroup$ – JavaMan Oct 18 '11 at 4:25

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.

  • $\begingroup$ @JDH: This a cute and strikingly simple idea. Thanks! $\endgroup$ – JavaMan Oct 18 '11 at 4:27

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