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...](https://mathoverflow.net/questions/40593/hausdorff-dimension-of-the-boundary-of-an-open-set-in-the-euclidean-space-lowe) 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](https://en.wikipedia.org/wiki/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](http://www.jstor.org/stable/121009?seq=1#page_scan_tab_contents). 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.



Hausdorff dimension of boundaries of open sets diffeomorphic to $\mathbb{R}^n$
Start with an [Osgood curve](https://en.wikipedia.org/wiki/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. 


