# Proof that the Hausdorff Dimension of Liouville Numbers is zero.

I am studying a proof that the set $L$ of Liouville numbers in $[0,1]$ has Hausdorff dimension zero. $L=\lbrace x\in [0,1]: \forall n\in \mathbb{N}, \exists p,q\in \mathbb{Z}, q>1, \text{and such that}\,0<\vert x-\frac{p}{q} \vert <\frac{1}{q^n}\rbrace.$ We give an $\varepsilon$-cover of $L\cap [0,1]$: Let $p,q\in \mathbb{Q}$. Define $I_{p/q}$ to be an interval centered at $\frac{p}{q}$ of length $\frac{1}{q^n}<\varepsilon.$ Then $J_n=\lbrace I_{p/q}:\frac{p}{q}\in \mathbb{Q}, q^n>q_0, q_0>\varepsilon ^{-1} \rbrace$ is our $\varepsilon$-cover. $$\sum_{p/q\in \mathbb{Q}, q>q_0}|I_{p/q}|^s\le \sum_{q>q_0}\frac{2^s}{q^{ns}}\cdot q<2\sum_{q>q_0}\frac{1}{q^{ns-1}}<\delta.$$

$(*)$ This proves $L$ has Hausdorff dimension $0$ since for $\varepsilon>0, \delta >0, \exists \varepsilon-\text{cover}$ (by choosing $q_0$ large) such that $\sum_{I\, \text{is a cover}}|I|^s<\delta \Rightarrow \text{Hdim}\le s.$ (TRUE provided $s>\frac{1}{n} \Rightarrow \text{Hdim}(L)\le \frac{1}{n} \Rightarrow \text{Hdim}(L)=0$)

Questions:

1) Why does the author sum the $|I_{p/q}|$ rather than the $J_n$?

2) $(*)$ Doesn't make any sense to me. Can someone give an example or a link to a detailed example of how to prove that the Hausdorff dimension of a set is a given number?

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You can learn Hausdorff measure and dimension in Fractal Geometry: Mathematical Foundations and Applications by Kenneth Falconer. There are beginner examples of computing fractal dimensions of sets as well as ones related to approximation of irrationals.

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