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

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?

• 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 '10 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 '10 at 22:12

"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 $\ge n-1$.

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

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.

• I don't understand your "outer measure" comment. Not all outer measures have this property? – Gerald Edgar Aug 9 '10 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 '10 at 19:56
• @Gerald, I'm not understanding your concern. Jonas supplied more details. What property are you referring to? – Ryan Budney Aug 9 '10 at 21:45
• 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 '10 at 21:56
• @Ryan: Jonas explained it. – Gerald Edgar Aug 11 '10 at 12:28