I have a question about Hausdorff dimensions and hope some of you can help me. I'm quite new with this topic so I hope this is not a too stupid question.

For a given $\Phi: \Omega \rightarrow \mathbb{R}^n$, a diffeomorphism and $K\subset\Omega$ compact I try to show, that the Hausdorff dimension of $A$ is the same as the Hausdorff dimension of $\Phi(A)$.

We defined the Hausdorff dimension by $\dim_{\mathcal{H}}(A):=\inf\{s\geq 0: \mathcal{H}^s(A)=0 \}$. My idea was to use that $\Phi$ is continous, due to it is a diffeomorphism. Because $\Omega \subset \mathbb{R}^n$ we know, that the compact $A$ is bounded, so $\Phi$ even is uniformly continous $\forall \epsilon>0 ~\exists \delta>0: ~\forall x,y \in A: |x-y|<\delta \Rightarrow |\Phi(x)-\Phi(y)| < \frac{\epsilon}{n} $ Furthermore we know, that for a compact $A$ there exist $C_1,...,C_n: A \subset\cup_{i=1}^{n} C_i$, a finite subcover of $A$.

So calling $s$ the Hausdorff dimension of $A$ and using the Definition we can follow: \begin{align} \mathcal{H}^s(\Phi(A))&= \lim_{\delta \rightarrow 0} \inf\Bigl\{\sum_{i=1}^{n}\left(\frac{\mathrm{diam}(\Phi(C_i))}{2}\right)^s: \mathrm{diam}(C_i)<\delta, A \subset\cup_{i=1}^{n} C_i \Bigr\}\\& \leq \lim_{\delta \rightarrow 0} \inf\Bigl\{\sum_{i=1}^{n}\left(\frac{ \mathrm{diam}(C_i)}{2} \right)^s \epsilon^s: \mathrm{diam}(C_i)<\delta, A \subset\cup_{i=1}^{n} C_i\Bigr\} = n \epsilon^s \mathcal{H}^s(A) = \tilde{\epsilon}\mathcal{H}^s(A) \end{align}

For $\epsilon \rightarrow 0$ we get $\mathcal{H}^s(\Phi(A)) \leq 0 \Rightarrow \mathcal{H}^s(\Phi(A)) = 0$

But now we still arent't finished. We still need to show, that for every $t<s$ $\mathcal{H}^t(\Phi(A)) >0$. (If I'm right?) Sadly I'm not able to show this…

Does anyone have an idea how to finish my proof or is there a better way to Show the statement?

Thanks in advance for your help!


You proved that if $A$ is compact then $H^s(A) = 0$ implies $H^s(\Phi(A)) = 0$. Note $$s > \dim_H A \implies H^s(A) = 0 \implies H^s(\Phi(A)) = 0 \implies \dim_H \Phi(A) \le s.$$ Thus $\dim_H \Phi(A) \le \dim_H A$.

Since $\Phi$ is a diffeomorphism (and in particular the image of a compact set $A$ is compact) the same argument gives you $$\dim_H A = \dim_H \Phi^{-1}(\Phi(A)) \le \dim_H \Phi(A).$$

  • $\begingroup$ Hi and thanks for your respond! The second line was what I meant (but reading my post again it seems, as if I didn't point that out very clear) so thanks for your clearification! Oooh of course. The same applies to $\Phi^{-1}$, so basically the proof is finished! Thanks a lot for your help! $\endgroup$ – pcalc Feb 22 at 10:07

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