Is there an example of a countable and dense subset of $\mathbb{R}$ (or $[0,1]$) without its construction strictly relying on $\mathbb{Q}$ ?
I thought about the dyadic rational numbers, but they are already represented as fractions, e.g. $\frac{m}{2^n}$. It would be better to have a construction relying only on a set construction process without relying (explictly) on $\mathbb{Q}$ (as in the Cantor ternary set, as we take the limit of a set construction process).
On
It is well know (probably showed by Cantor) that every two countable dense linear orders without endpoints are isomorphic. In particular, every countable dense subset of $\mathbb{R}$ is isomorphic to $\mathbb{Q}$.
Edit: For a specific example, take your favorite countable and not bounded $D_0 \subset \mathbb{R} \smallsetminus \mathbb{Q}$, and defined a family $\{D_i: i \in \omega\}$ as follow:
$D_{i+1} \subset \mathbb{R} \smallsetminus \mathbb{Q}$ is the $\subset$-minimum such as for any $a, b \in D_i$ with $a<b$ there are $x, y, z \in D_{i+1}$ with $x<a<y<b<z$.
You can check that each $D_i$ is countable and $D=\bigcup_{i \in \omega}D_i$ is countable dense in $\mathbb{R}.$
Let us divide $[0,1]$ into two intervals of equal length. Let us choose any element from the first and second interval. Then divide the intervals and repeat the process, choosing numbers different from already chosen. Repeat etc.