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Let $B$ be a bounded open subset of $\mathbb{R}^n$ which is diffeomorphic to $\mathbb{R}^n$. (I am not sure how important the diffeomorphism is but this is the case I am interested in.) Let $C$ be its boundary.

What is the supremum of the Hausdorff dimension of sets of the form $C$ in $\mathbb{R}^n$?

I suspect it is $n$, but do not know where to find such a statement, even if it were true.

As explained here https://mathoverflow.net/questions/40593... the Hausdorff dimension of $C$ is at least $n-1$.

A look at Koch curves shows that it can easily be larger than $n-1$, see the first picture in Wikipedia: Hausdorff dimension.

Also boundaries of open sets may have Hausdorff dimension $n$, for instance the boundary of the Mandelbrot set has Hausdorff dimension $2$ everywhere locally, Theorem A in Shishikura, 1998. Though thinking about it, it does not even seem clear whether the Mandelbrot set is the closure of its interior, but this is a sidetrack. I suspect that the dimension of the boundary has more to do with all the little things attached to the body and if we were to take the large connected component of the interior of the Mandelbrot set, then the Hausdorff dimension of its boundary would be a boring $1$. So this seems to be an example that the requirement of being diffeomorphic might play a role.

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    $\begingroup$ See here for $n=2$: mathoverflow.net/questions/23960/… That's probably more than you need, you can also use Koch style curves and increase the (local) dimension as you go around $B$, so that the (global) dimension will be $n$. $\endgroup$ – Christian Remling Oct 18 '16 at 5:57
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Start with an Osgood curve $C$, a Jordan curve in $R^2$ of positive 2-dimensional measure. The curve $C$ bounds a domain $\Omega$ in $R^2$ diffeomorphic to $R^2$. Lastly, take the Cartesian product $D=\Omega\times B^{n-2}$ with the open round disk in $R^{n-2}$. The result is a bounded domain $D$ in $R^n$ diffeomorphic to $R^n$ such that $\partial D$ has positive $n$-dimensional measure.

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  • $\begingroup$ Thank you indeed. But I am not quite sure I understand. For instance in $\mathbb{R}^3$ your construction would give the set $\Omega\times (-1,1)$, so basically a cylinder with a complicated base. In the third dimension the boundary is quite regular. Why should we expect that the Hausdorff dimension of the boundary is $3$? $\endgroup$ – Fabian Wirth Oct 18 '16 at 11:17
  • $\begingroup$ @FabianWirth: In this example, $C\times [-1,1]$ already has Hausdorff dimension 3, this is just Fubini's theorem. Now use the fact that Hausdorff dimension is monotonic with respect to inclusion sets. $\endgroup$ – Moishe Kohan Oct 18 '16 at 11:21
  • $\begingroup$ OK. Thank you again. I should have checked, what is known about the Hausdorff dimension of Cartesian products. For reference: the following paper deals with this problem and solves the question I had (the statement is already on page 1): Hatano 1971 $\endgroup$ – Fabian Wirth Oct 18 '16 at 11:31

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