Is there a dense subset of $\mathbb{R}^2$ with all distances being incommensurable?

Question

1. Is there a set $S$ of points on the real plane $\mathbb{R}^2$ such that:

   • there is a point belonging to $S$ in any neighborhood of every point of $\mathbb{R}^2$ (so, $S$ is dense) and
   • ratio of any two distances between points in $S$ is an irrational number?

2. Replace irrational with transcendental.

3. Replace irrational with non-period; 3'. Replace irrational with non-computable.

4. In every of previous questions, can $S$ be made uncountable?

Answer 1

1, 2, 3. Let $R$ be a cocountable subset of $\mathbb{R}$. We will construct a countable dense subset $S$ of $\mathbb{R}^2$ such that the ratio of any two distances between points in $S$ belongs to $R$. To do this, begin by placing two points $s_1, s_2 \in S$ unit distance apart. Now enumerate the disks with rational center and rational radius in $\mathbb{R}^2$ and place points $s_n$ in the interior of each such disk in turn satisfying the given condition. This is always possible because the set of all points at which $s_n$ cannot be placed is a countable union of sets of measure zero (one for each possible ratio of two distances lying outside of $S$), hence has measure zero.

The same argument together with transfinite induction should also establish 4, but I haven't thought about it too carefully.