An uncountable and closed subset of the Liouville Numbers

Question

I am trying to "find" a closed and uncountable subset of the Liouville's numbers.

$x\in L$ means that for all $n\in \mathbb{N}$ exists $p,q\in \mathbb{Z}$ with $q>1$ such that $$0<\vert x-\frac{p}{q}\vert <\frac{1}{q^n}.$$

Ideas are welcome!!

Answer 1

I think the following set should work:

$$A = \left\{ \sum_{n=1}^\infty \epsilon_n 2^{-n!} : \epsilon_n \in \{0,1\} \text{ and } \epsilon_{2n-1} + \epsilon_{2n} = 1\right\}.$$

The additional restriction that the sequence of $\epsilon_n$s consists of $(0,1)$ and $(1,0)$ pairs ensures that every element of $A$ is Liouville. Furthermore I believe that every convergent sequence in $A$ has each $\epsilon_n$ eventually constant, which means the limit point also satisfies the defining property of $A$.