# Hausdorff-dimensions and self-similarity

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.

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 $1 = \sum_ic_i^d$.