I consider a bounded open set $A$ in ${\mathbb R}^d$. Is the Hausdorff dimension of the boundary of $A$ at least $d1$ ? I thought I would have found a result on this problem in any textbook about Hausdorff dimension but I failed. As you may guess I have never work with Hausdorff dimension.
If a compact set $C$ separates a connected space $X$ of dimension $d$ (with some homogeneity properties, for example, an euclidean space) in two connected components, then it must have topological dimension at least $d1$ (see for example the book by Hurewicz and Wallman on topological dimension, Theorem IV.4). In general, Hausdorff dimension exceeds (or is equal to) topological dimension (see chapter VII.2 of the book by HurewiczWallman mentioned above). 


Here is a simple proof for the Hausdorff dimension. Consider the orthogonal projection to $\mathbb R^{d1}$. Since $A$ is bounded, the projection of the boundary contains the projection of $A$. The latter is open in $\mathbb R^{d1}$ and hence has Hausdorff dimension $d1$. On the other hand, the projection map does not increase Hausdorff dimension since it does not increase distances. 

