4
$\begingroup$

While reading around online earlier today, I managed to stumble onto a strange claim, that I have not been able to find or produce a proof for. The claim is that if $X$ is a compact subset of $\mathbb R^n$, which can be written as a disjoint union of some $X_i$ where $X_i$ is isometric to a dilation of $X$ by some constant $c_i$, then the Hausdorff dimension has the property that $\sum c_i^d = 1$.

I know the definition of Hausdorff dimension, and so I was able to verify that this is true for some examples that I know off the top of my head, i.e. Cantor sets in $\mathbb R$, Sierpinski fractals, etc. But I have not been able to come up with a proof.

I expect it is not a hard theorem, because a google search tells me that more technical versions of this theorem can be stated, with all kinds of overlap conditions on the $X_i$ and whatnot, but none of this is necessary, so somehow I think it should not be hard.. but I cannot figure it out.

Any help is appreciated.

$\endgroup$
4
$\begingroup$

Assume for the moment that the $d$-dimensional Hausdorff measure of $X$ is finite and nonzero when $d$ is the Hausdorff dimension of $X$ (this isn't necessarily the case, but it makes the proof harder if you don't assume it. The idea is the same, though.). Let $\mu$ be the $d$-dimensional Hausdorff measure. $\mu$ has the scaling property that $\mu(cY) = c^d\mu(Y)$, where $cY$ denotes the scaling of $Y$ by a factor of $c$. It also has the property that $\mu(A \cup B) = \mu(A) + \mu(B)$ whenever $A$ and $B$ are disjoint (I'm talking only about $\mu$-measurable sets here, of course). So $\mu(X) = \sum_i\mu(X_i)$. But $X_i = c_iX$, so $\mu(X) = \sum_ic_i^d\mu(X)$. Since we assumed that $\mu(X)$ was finite and nonzero, we can divide both sides by $\mu(X)$ and obtain $1 = \sum_ic_i^d$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ As always, I overthought the issue. Elegant. Thanks so much. $\endgroup$ – Alfred Yerger Sep 16 '16 at 15:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.