Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
  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?

share|improve this question
    
+1, interesting problem! My initial guess would be to do something like, take a transcendence basis $T$ for $\mathbb{R}$ over $\mathbb{Q}$, which will be uncountable, and take $S$ to be all points in the plane whose coordinates are in $T$. Not quite sure how to show this is dense, though. –  Zev Chonoles Nov 29 '11 at 19:37
    
@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. –  Jonas Meyer Nov 29 '11 at 20:05
add comment

1 Answer

up vote 12 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.