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
    Commented May 28, 2012 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
    Commented May 28, 2012 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
    Commented May 28, 2012 at 20:34

1 Answer 1


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 $0$


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