0

As far as I know, the Hausdorff measure of a countable subset is zero (please correct me otherwise). Is it possible to say the same about the uncountable dense subsets? Is there a general statement about their Hausdorff measure?

Xander Henderson
  • 29,772
Ozzy
  • 356
  • 3
    (1) The one dimensional Hausdorff measure of a countable set is zero. The zero dimensional Hausdorff measure is, more or less, very broadly speaking, the cardinality, so will be finite for finite sets, and infinite for infinite sets. (2) It is possible to construct a dense subset of $[0,1]$ with arbitrary one dimensional Hausdorff measure (between $0$ and $1$). For example, take $\mathbb{Q} \cup E$, where $E$ is a set of whatever measure you like. – Xander Henderson Mar 17 '23 at 18:27
  • 1
    Replacing $\mathbb Q$ with a Hausdorff dimension zero set that has cardinality continuum in every interval (such sets exist) in (2) of the comment by @Xander Henderson gives examples of $c$-dense subsets of $[0,1]$ with the same property. In fact, I guess since it appears we mostly talking about sets of Hausdorff dimension 1, we don't really need a $c$-dense Hausdorff dimension zero set -- any $c$-dense set with Hausdorff dimension less than one will do. – Dave L. Renfro Mar 18 '23 at 02:53
  • XanderHenderson and @DaveL.Renfro thank you very much for your answers, it gives me a little more insight. What I am particularly interested is quite specific actually: Dense subsets of (n-1 dimensional) boundaries of n dimensional manifolds, and under what circumstances the n-1 dimensional Hausdorff measure of those dense subsets vanish. My intuition says, given your answers, they vanish if they are countable, as their n dimensional measure is finite. – Ozzy Mar 18 '23 at 12:56

0 Answers0