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?
Replace irrational with transcendental.
Replace irrational with non-period; 3'. Replace irrational with non-computable.
In every of previous questions, can $S$ be made uncountable?
Is there a dense subset of $\mathbb{R}^2$ with all distances being incommensurable?
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0@Zev: Note that $T$ need not be dense, because for example it could be chosen to be contained in $(0,1)$, but since it is infinite its elements could be scaled by rationals to form another transcendence basis that is dense. – 2011-11-29
1 Answers
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.