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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 $\dim(A) \leq \dim(B)$?

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

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

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