These are homework questions: Give examples of the following spaces

  1. Uncountable metric space with Hausdorff dimension 0.

  2. $\dim X=1$ with Hausdorff dimension 1 measure measure = 0.

I can't think there is any connection between countable and measure. I have a vagure idea that the first example should be somehow modified Cantor set, each time we remove an interval with length = $c_n$, with $\sum c_n=1$. But is this set still uncountable?

  • 1
    $\begingroup$ Cantor-type sets are uncountable no matter what the length of removed intervals. (It's a general theorem about compact sets without isolated points.) Hausdorff dimension is related to covering the set by small sets. In case of Cantor-type constructions, you cover the set by the intervals you have at step N. $\endgroup$ – user31373 May 28 '12 at 20:05
  • $\begingroup$ Thanks Leonid. I still confused on how to calculate the Hausdorff measure of the general cantor set, since it needs to take infimum over all coverings with diameter $\epsilon$ and then let it goes to zero. Why covered by the intervals is enough? $\endgroup$ – user17150 May 28 '12 at 20:28
  • $\begingroup$ The infimum certainly cannot be negative. Thus, if you can find coverings with arbitrarily small sum $\sum (\dots )^d$, it follows that the infimum is zero. $\endgroup$ – user31373 May 28 '12 at 20:34

If $(X,\rho)$ is a metric space, then for any subset $S$, we have $$ H_\delta^d(S):=\inf\ \{ \sum_{i=1}^\infty ({\rm diam}\ U_i)^d | S\subset \bigcup_{i=1}^\infty U_i,\ {\rm diam}\ U_i < \delta \} $$ where the infimum is over all countable covers of $S$. Then we have $H^d(S):=\lim_{\delta \rightarrow 0 } H^d_\delta (S)$, which is called the $d$-dimensional Hausdorff measure.

(1) If $C$ is Cantor set, then two intervals of length $\frac{1}{3}$ cover $C$ So $$H^d_\delta (C) \leq 2(\frac{1}{3})^d,\ \delta=\frac{1}{3} $$

Considering construction of $C$, we have $$H^d_\delta (C) \leq 2^n(\frac{1}{3})^{nd},\ \delta=\frac{1}{3^n} $$

That is dimension is $ H^d(C)=1,\ \infty,\ 0$ when $d=d_0:=\ln_32$, $d<d_0$, $d > d_0$ respectively So $1$-dimension Hausdorff measure is $0$


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.