Consider $D\subset\Bbb R^m$ a dense subset of some open set of positive Lebesgue measure; for sake of simplicity, we can consider $D$ dense in the unit ball $B$.
Do we know something about the Hausdorff dimension of $D$?
The Hausdorff dimension measures, in some sense, how much room there is between the elements of sets; this is zero for discrete sets and maximal for full-Lebesgue measure sets, so I would say that $D$ has the same Hausdorff dimension of $B$ (since being dense, there is no room at all), which is $m$.
Is it a known result?
EDIT: as Gae S. pointed out, $\dim_H(D)=0$ is a possibility; so the question turns into: what is a sufficient condition on a dense subset to have full Hausdorff measure?