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Could you please give me some hints on (Exercise 1.7.21) of "A Course in Metric Geometry" by Burago, Burago, Ivanov.

We have a compact space $X$, which can be written as $X=\bigcup_{i=1}^n X_i$ (disjoint union), where $X_i=F_i(X)$ and $F_i$ are dilations with Lipschitz constant $c_i$. Then show that the Hausdorff dimension $d$ of $X$ satisfies $\sum_{i=1}^n c_i^d = 1$.

Let $\mu_d$ denote the $d$-dimensional Hausdorff measure, then thats how far I got:

Thanks to the corrections below I am able to show it in the case where $0<\mu_d(X)<\infty$ but I still do not know how to approach the two boundary cases. Any hints are appreciated! Thanks!

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    This is obviously wrong. Consider $X=[0,1]$, $X_1=X$, $F_1(x)=x^2$. Then the minimal Lipschitz constant is $c_1=2$. Since $d=1$ for the one-dimensional line, the claim of the problem statement is $2^1=1$.2012-11-02
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    I just read the linked original problem. It speaks of *dilations*, not of Lipschitz continuous functions.2012-11-02
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    Ah, sorry. I didn't find this notion in the sections before so I just thought it meant Lipschitz continuous. I think it makes more sense now! Thank you!2012-11-02
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    This is still wrong, unless you at least assume that the $X_i$ are disjoint. Otherwise you could just take all $F_i$ to be the identity. Also, what is $\mu_d$, and how do you know that $\mu_d(X)<\infty$?2012-11-02

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