0
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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 $$\dim_H A = \dim_H \Phi^{-1}(\Phi(A)) \le \dim_H \Phi(A).$$

$\endgroup$
  • $\begingroup$ 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

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.