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Let $n$ be a positive integer.

Let $S \subseteq \mathbb{R}^n$. Is the Hausdorff dimension of the boundary of $S$ always smaller than the Hausdorff dimension of $S$?

I have not found anything concerning those questions in some looked up books, I was not able to prove one of the statements, and I failed finding a counterexample. Does anybody know something about that?

flag	asked Aug 9 2010 at 19:16 Daniel Krenn 365●1●8

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If $S$ is closed and nowhere dense, then $S$ is its own boundary, so it can't be strictly less in such a case. – Joel David Hamkins Aug 9 2010 at 19:21

Maybe in your original question you meant "frontier" rather than boundary (the frontier is cl(S)-S). The answer is still negative, by the way (e.g: topologist's sine curve). – Thierry Zell Aug 9 2010 at 22:12

Gerald Edgar · Accepted Answer · 2010-08-09 19:44:18Z UTC

"Smaller" in the sense of $\le$ ... If $S$ is closed and has Hausdorff dimension $< n$, then $S$ has empty interior, so (as noted by Joel) $S$ is its own boundary, and thus we have equality for the two dimensions. And of course if (perhaps not closed) set $S$ has dimension $n$, then the boundary could have any dimension from $0$ to $n$, inclusive. If $S$ is closed and has dimension $n$, then the boundary is either empty or has dimension $\ge n-1$.

Jonas Meyer · Answer 2 · 2010-08-09 19:20:30Z UTC

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The set of rational numbers has Hausdorff dimension 0, while its boundary is the set of real numbers, with Hausdorff dimension 1.

answered Aug 9 2010 at 19:20

Jonas Meyer
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Ryan Budney · Answer 3 · 2010-08-09 19:23:05Z UTC

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For closed sets, yes, since Hausdorff dimension comes from Hausdorff measure, which is an outer measure.

For non-closed sets, no. Take the rationals in the reals. They have Hausdorf dimension 0. But their boundary is the reals, which has dimension 1.

answered Aug 9 2010 at 19:23

Ryan Budney
21.8k●1●41●102

I don't understand your "outer measure" comment. Not all outer measures have this property? – Gerald Edgar Aug 9 2010 at 19:48

@Gerald, I think he just referring to the fact that if $H^d$ denotes d-dimensional Hausdorff outer measure and $A\subseteq B\subseteq \mathbb{R}^n$, then $H^d(B)=0$ implies $H^d(A)=0$, so $\inf\{d\geq0:H^d(A)=0\}\leq\inf\{d\geq0:H^d(B)=0\}$ , along with the fact that $\partial B\subseteq B$ if $B$ is closed. – Jonas Meyer Aug 9 2010 at 19:56

@Gerald, I'm not understanding your concern. Jonas supplied more details. What property are you referring to? – Ryan Budney Aug 9 2010 at 21:45

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For the record, this answer seems to have been written independently of mine (even though technically it was posted a couple of minutes later) and it contains my answer as a proper subset. – Jonas Meyer Aug 9 2010 at 21:56

@Ryan: Jonas explained it. – Gerald Edgar Aug 11 2010 at 12:28

Hausdorff dimension: subset of $\mathbb{R}^n$ vs. boundary of this subset

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