Let $n$ be a positive integer.



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



Hausdorff dimension: subset of $\mathbb{R}^n$ vs. boundary of this subset
"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 $\ge n-1$.


