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I consider a bounded open set $A$ in ${\mathbb R}^d$. Is the Hausdorff dimension of the boundary of $A$ at least $d-1$ ? 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.

flag	asked Sep 30 2010 at 9:45 Hugh J 353●2●5

Pete L. Clark · Accepted Answer · 2010-10-01 05:32:05Z UTC

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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 $d-1$ (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 Hurewicz-Wallman mentioned above).

edited Oct 1 2010 at 5:32

Pete L. Clark

answered Sep 30 2010 at 10:31

rpotrie
2,285●7●13

	Sorry, english is not my native language. I shall correct it. – rpotrie Sep 30 2010 at 11:21
	Thank you very much for your answer ! – Hugh J Sep 30 2010 at 11:52
	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. – Robert Bell Sep 30 2010 at 12:09
	rbell: Actually that is usually called the "inductive dimension". – Pietro Majer Sep 30 2010 at 15:42

Sergei Ivanov · Answer 2 · 2010-09-30 19:19:30Z UTC

Here is a simple proof for the Hausdorff dimension. Consider the orthogonal projection to $\mathbb R^{d-1}$. Since $A$ is bounded, the projection of the boundary contains the projection of $A$. The latter is open in $\mathbb R^{d-1}$ and hence has Hausdorff dimension $d-1$. On the other hand, the projection map does not increase Hausdorff dimension since it does not increase distances.

Hausdorff dimension of the boundary of an open set in the Euclidean space - lower bound

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