Proof that the Hausdorff Dimension of Liouville Numbers is zero.

Question

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?

Answer 1

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.

Answer 2

It occurred to me that you might find some very detailed and explicit Hausdorff dimension calculations in a Masters (or undergraduate Honors) Thesis, so I googled each of the phrases "Masters Thesis" and "Honors Thesis" with the phrase "Hausdorff dimension" and found the following:

David C Seal, An Introduction to Fractals and Hausdorff Measures, Senior Honors Thesis (under Davar Khoshnevisan), University of Utah, 2006, iii + 21 pages.

David Worth, Construction of Geometric Outer-Measures and Dimension Theory, Master of Science Thesis (under Terry Loring), University of New Mexico, December 2006, xi + 77 pages.

Kristine A. Roinestad, Geometry of Self-Similar Sets, Master of Science Thesis (under Peter Haskell), Virginia Polytechnic Institute and State University, 8 May 2007, v + 63 pages. [See Chapter 2, pp. 4-10).

Answer 3

I will try to give a self-contained proof based on the sketch presented by the OP. I think there are some details which should be addressed more precisely. Moreover, the inequality required for $n$ is not sufficient in my opinion. I have tried to stick to the notation used by the OP.

Preliminary notions. We recall the definition of Hausdorff dimension. For a subset $U$ of a metric space (e.g. real line), define $\operatorname{diam}(U)$ to be the diameter of $U$, i.e. the supremum of distances between any two elements of $U$. If $U$ would be an interval, that is just its length. Now for a set $X$, for $s\ge 0$ (which would take the role of a dimension) and $\varepsilon>0$ define $$H_\varepsilon^s(X)=\inf\left \{\sum_{i=1}^\infty (\operatorname{diam} U_i)^s: X\subseteq\bigcup_{i=1}^\infty U_i,\ \operatorname{diam} U_i<\varepsilon\right \}.$$ One observes that this is nonincreasing as a function of $\varepsilon$ (smaller $\varepsilon$ - less covers $\{U_n\}_{n\ge 1}$). Thus we define $$\displaystyle\mathcal{H}^s(X)=\sup_{\varepsilon>0}H_\varepsilon^s(X)=\lim_{\varepsilon\to 0}H_\varepsilon^s(X).$$ This turns out to be a measure (defined at least on the Borel $\sigma$-algebra (the one containing all open sets)); it is called $s$-dimensional Hausdorff measure. One may observe that $\mathcal{H}^s(X)$ is nonincreasing as a function of $s$ with values in $[0,\infty]$, and may obtain at most one finite nonzero value (consider for example subset of $[0,1]$). Define the Hausdorff dimension as the $s$ for which we switch from measure $\infty$ to measure $0$, i.e. $$\dim_{\operatorname{H}}{(X)}=\inf\{s\ge 0: \mathcal{H}^s(X)=0\}.$$

To the problem. We want to show that the set of Liouville numbers in $[0,1]$ has dimension $0$. Denote this set by $L$. Recall that by definition $x\in L $ iff for each positive integer $n$ there exist infinitely many positive integers $p,q$, such that $$\left|x-\frac{p}{q}\right|<\frac{1}{q^n}.$$ Pay attention, the usual definition requires existence of at least one such pair $p,q$; it is easy to observe that if there exists at least one pair for each $n$, then there exist infinitely many pairs for each $n$.

So let us unpack what it means for a set to have dimension $0$. Going back to the above definitions (in reverse order) we see that it suffices to show that for any $s>0$ holds $\mathcal{H}^s(X)=0$. This in turn reduces to showing that for $\varepsilon>0$ holds $H_\varepsilon^s(X)=0$. This means that $$\inf\left \{\sum_{i=1}^\infty (\operatorname{diam} U_i)^s: X\subseteq\bigcup_{i=1}^\infty U_i,\ \operatorname{diam} U_i<\varepsilon\right \}=0$$ Finally, unwrapping the infimum we may summarise the task to be done as follows:

For any $s>0$, $\varepsilon>0$ and $\delta>0$ there exists a countable collection $\{U_i\}_{i\ge 1}$ such that: $\displaystyle X\subset \bigcup_{i\ge 1}U_i$, $\operatorname{diam}U_i<\varepsilon$ for all $i$ and $$ \sum_{i=1}^\infty (\operatorname{diam} U_i)^s<\delta.$$

So fix $s>0$, $\varepsilon>0$ and $\delta>0$. We may assume $s<1$. Fix $\displaystyle n>\frac{3}{s},\ q_0>\max\left\{\frac{2}{\varepsilon},\frac{2}{\delta}\right\}$ and consider $$I_{p,q}:=\left(\frac{p}{q}-\frac{1}{q^n},\frac{p}{q}+\frac{1}{q^n}\right).$$ One observes that (recalling the modified, still equivalent defintion, we stated) $$L\subset \bigcup_{q>q_0}\bigcup_{p=1}^{q-1}I_{p,q}$$ Moreover $\displaystyle\operatorname{diam}I_{p,q}=\frac{2}{q^n}<\frac{2}{q_0}<\varepsilon.$ Finally we estimate $$\sum_{q>q_0}\sum_{p=1}^{q-1}(\operatorname{diam}I_{p,q})^s\le\sum_{q>q_0}\frac{2^s}{q^{ns}}q\le 2\sum_{q>q_0}\frac{1}{q^{ns-1}}\le 2\sum_{q>q_0}\frac{1}{q^{2}}\le \frac{2}{q_0}<\delta.$$ In the above we used the inequality $$\sum_{k>n}\frac{1}{k^2}\le \frac{1}{n}$$ which could be proven, for example, by estimating the sum from above by $\displaystyle \int_n^\infty\frac{1}{x^2}dx$.

Thus the proof is finished.

Comment. Why is this result interesting, apart from finding explicitly the Hausdorff dimension of a particular set? Regarding Lebesgue measure, it is not obvious whether there exists an uncountable set with measure zero. Lioville numbers serve as such example. However, Liouville numbers even serve as an example of an uncountable set with Hausdorff dimension zero, and this is much stronger. For example, another common reference of an uncountable set with measure $0$ is the Cantor set. Yet it has positive Hausdorff dimension ($\ln 2/\ln 3$). Thus the set of Lioville numbers, is in a sense, way smaller than the Cantor set. Yet it is still uncountable. One could also say that Liouville numbers are example of the smallest possible uncountable sets among the sets with measure zero. You can look at this MSE thread for another construction of such sets. Thanks to Dave L. Renfro, for reminding me of another aspect of Liouville numbers. They are of second Baire category, or $G_\delta$ dense, which one might consider "large" in the sense of topology. Hence Lioville numbers are big in sense of cardinality and topology (actually $G_\delta$-dense in $[0,1]$ implies uncountable), yet they are among the smallest in the sense of measure and dimension.