Proving the Liouville Numbers are uncountable

Question

The definition I'm given for the Liouville numbers is just that $x$ is Liouville if for every $N > 0$ there exists integers $p,q \geq 2$ such that $$\left|x-\frac{p}{q}\right| < \frac{1}{q^N}.$$

I don't have too many high powered theorems to work with, except for things like the equivalence between Zorn, AoC and Well Ordering, Schroeder-Bernstein and a few other "bread and butter" set theory results. Is there a way to do this with any of those tools, or an explicit bijection between $\mathbb{R}$ (or some other set whose uncountability is easily verified) and the Liouvilles, or something else that uses some of those bread and butter set theory results?

Thanks.

Answer 1

Any number of the form $$\sum_{n=1}^\infty {a_n\over 3^{n!}}$$ where $a_n\in \{1,2\}$ is a Liouville number. The cardinality of the set of such numbers is continuum.

Answer 2

Consider the continued fraction expansion of an irrational number $x$: $$x =[a_0; a_1, a_2, a_3, \dots] = a_0 + \cfrac1{a_1 + \cfrac1{a_2 + \cfrac1{a_3 + \dots}}}$$ If we truncate the continued fraction of $x$ to the rational number $p_n/q_n = [a_0; a_1, \dots, a_n]$, the error in the approximation $|x - p_n/q_n|$ decreases as $a_{n+1}$ increases. By choosing $a_{n+1}$ to be large enough, the error can be made arbitrarily small.

(Not much theory of continued fractions is necessary to see this; all we need to know is that the function $[a_0; a_1, \dots, a_n + z]$ is a rational function of $z$ that approaches $p_n/q_n$ as $z \to 0$, and that $x = [a_0; a_1, \dots, a_n + z]$ where $z = [0; a_{n+1}, a_{n+2}, \dots]$.)

We can construct uncountably many Liouville numbers by a continued fraction approximation that begins with, say, $a_0 = 1$, and continues as follows:

• Choose the odd terms by letting $a_{2k+1}$ be the least positive integer such that $|x - p_{2k}/q_{2k}| < (1/q_{2k})^k$.
• Choose the even terms to be arbitrary positive integers.