Liouville numbers and numbers having infinite irrationality measure

Question

Let's start with the definition of a Liouville number:

A Liouville number is a real number $\xi$ such that for any $m\in\mathbb N_>0$ there exists a pair of coprime integers $(p,q)\in\mathbb Z^2$ with $q>0$ such that: $$ 0<\left\vert\xi-\frac{p}{q}\right\vert< q^{-m} $$

Then let's give the definition of irrationality measure:

The irrationality measure of $\xi\in\mathbb R$ is: $$ \mu(\xi):=\sup_{t\in \mathbb R}\left\{\text{$\exists$ infinitely many coprime couples $(p,q)\in\mathbb Z^2$ s. t. } 0<\left\vert\xi-\frac{p}{q}\right\vert< \vert q\vert^{-t}\right\} $$

It is obvious the fact that if $\mu(\xi)=+\infty$ then it must be a Liouville number. But it seems to me that it is not true that any Liouville number that irrationality measure equal to $+\infty$. Altough I don't have any example.

There are some authors (for instance Waldschmidt, Bugeaud) that as definition of Liouville numbers take exactly those numbers $\xi\in\mathbb R$ such that $\mu(\xi)=+\infty$. This makes me suspect that my previous claim is perhaps wrong. So what is the truth behind this story? Are Liouville number exactly those numbers with infinite irrational measure?

Answer 1

First, I think your first definition is slightly off since for any irrational number $\xi$ we have

$$0<\left|\xi -\frac{\lfloor \xi\rfloor}{1}\right|<1=1^{-m}$$

for all $m$. I think your definition has to be changed to $q>1$. Continuing:

You are trying to show the equivalence (or lack thereof) of two supposed definitions for Liouville numbers. You are satisfied that an irrationality measure of $+\infty$ means the number is a Liouville number, but not that being a Liouville number (according to your first definition) implies an infinite irrationality measure.

Assume by way of contradiction that we have a Liouville number $\xi$ with a finite irrationality measure $\mu$. By definition, there are only finitely many coprime-pairs $(p,q)$ such that

$$\left|\xi-\frac{p}{q}\right|<\frac{1}{q^{\chi}}$$

where $\chi=\lceil \mu\rceil +1>\mu$. Let $(P,Q)$ be given such that $Q$ is the largest amongst such $Q$. Such a $Q$ must exist since

$$\left|\xi-\frac{\lfloor\xi\rfloor}{1}\right|<1=\frac{1}{1^{\chi}}$$

Now, let $(p_n,q_n)$ be a sequence of coprime-integers such that

$$\left|\xi-\frac{p_n}{q_n}\right|<\frac{1}{q_n^n}$$

Next, we shall show that $q_n$ goes to infinity. If it didn't and was bounded by some $M\in\mathbb{N}$ then

$$\left|\xi M!-\frac{M!p_n}{q_n}\right|\geq \min\{M!\xi-\lfloor M!\xi\rfloor,\lceil M!\xi\rceil-M!\xi\}$$

since $\frac{M!p_n}{q_n}$ is always integral. But this is a finite number, which is a contradiction since

$$0\leq\lim_{n\to\infty}\frac{1}{q_n^n}\leq \lim_{n\to\infty}\frac{1}{2^n}=0$$

(note that this is the step where we use $q>1$). We conclude that $q_n\to\infty$. This implies that there exists $r$ such that $q_r>Q$. Consider what happens at $N=\max\{r,\chi\}$: we have

$$\left|\xi-\frac{p_N}{q_N}\right|<\frac{1}{q_N^N}\leq \frac{1}{q_N^\chi}$$

However, this is a contradiction since by our construction $q_N>Q$. We conclude both definitions are equivalent (after fixing the slight issue with the first one).

Answer 2

I have the answer. It should be a consequence of the Bolzano-Weierstrass theorem.

Let $\xi$ be a Liouville number, then we have a sequence of approximants $\frac{p_n}{q_n}$ such that: $$ \left\vert\xi-\frac{p_n}{q_n}\right\vert<\left(\frac{1}{q_n}\right)^n $$

By the Bolzano Weierstrass theorem the sequence $\frac{1}{q_n}$ admits a convergent subsequence $\frac{1}{q_{k_n}}\to\ell\in[0,\frac{1}{2}]$. Therefore $\left(\frac{1}{q_{k_n}}\right)^{k_n}\to 0$ and in particular the set $\left\{ \frac{1}{q_{k_n}} \right\}$ contains infinitely many distinct elements.

Now for any $t\in\mathbb R$ consider all the elements of the sequence $k_n$ such that $k_n>t$, then we have $$ \left\vert\xi-\frac{p_{k_n}}{q_{k_n}}\right\vert<\left(\frac{1}{q_{k_n}}\right)^{k_n}<{q^{-t}} $$