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.
2 Answers
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).

$\begingroup$ Sorry, english is not my native language. I shall correct it. $\endgroup$– rpotrieSep 30, 2010 at 11:21

$\begingroup$ By definition (in Hurewicz and Wallman, if I remember correctly), the topological dimension of separable metric space $A$ is the dimension of the boundary (frontier) of $A$ plus one. Since an open set in $\mathbb{R}^d$ has dimension $d$, you are done. But if you start with the definition of topological dimension as Lebesgue's covering dimension, then you might use the theorem above. $\endgroup$ Sep 30, 2010 at 12:09

$\begingroup$ rbell: Actually that is usually called the "inductive dimension". $\endgroup$ Sep 30, 2010 at 15:42
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.

1$\begingroup$ That's great ! This is really simple. Thank you ! $\endgroup$– Hugh JSep 30, 2010 at 19:54