Let $(X,d)$ be a metric space and for $\delta > 0$, $s \in [0, \infty)$, define

$$ \mathcal{H}_\delta^s(A) = \inf \left\lbrace \omega_s\sum_{n=1}^\infty \left( \frac{diam(E_n)}{2} \right)^s : A \subseteq \bigcup_{n=1}^\infty E_n, \, diam(E_n) \leq \delta \right\rbrace, \quad A \subseteq X.$$

Then, define the $s$-dimensional Hausdorff measure by

$$ \mathcal{H}^s(A) = \lim_{\delta \to 0} \mathcal{H}_\delta^s(A), \quad A \subseteq X. $$

Finally, the Hausdorff dimension of a subset $A$ is given by

$$ \dim(A) = \inf \{ s \in [0, \infty): \mathcal{H}^s(A) = 0\}. $$

How does one show that the Hausdorff dimension is monotonic, i.e., that $A \subseteq B$ implies $\dim(A) \leq \dim(B)$?


Hint: If $A\subseteq B$ and $\mathcal{H}^s(B) = 0$ then what is $\mathcal{H}^s(A)$?

  • $\begingroup$ Ok, I got it. Simple! $\endgroup$ – Eduardo Longa Apr 26 '18 at 20:50

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.