16
$\begingroup$

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 $\dim_H(A) = 0$?

$\endgroup$
  • 6
    $\begingroup$ The Liouville numbers do the trick: en.wikipedia.org/wiki/Liouville_number $\endgroup$ – user940 Oct 18 '11 at 4:07
  • 1
    $\begingroup$ @Bryon: Thank you for the link. An explicit example is nice to have! $\endgroup$ – JavaMan Oct 18 '11 at 4:25
15
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.