# Least cardinality of a set of points in the plane

What is the least possible cardinality $K$, of a set S of points in the plane, such that there exists a point P in the plane and an open ball B centered at P, such that for all points X in B, not all the euclidean distances from X to points of S are rational?

Can we do better then $3\leq K \leq 2^{\aleph_0}$?

What can be said about K if all the points of S must be at integer distance from eachother, can it be proved to be finite?

• Three points suffice. Take $S′=\{(0,0),(1,0)\}$. Then any point with algebraic distances to $S'$ has algebraic coordinates (it lies on the intersection of circles with algebraic coefficients). Let $S=S'\cup\{(x,y)\}$, where $1,x,y$ are algebraically independent over $\mathbb{Q}$. Mar 7, 2012 at 14:07
• Is there any set of three points that does not have this property? Mar 7, 2012 at 15:45
• @Emil Right, for the second question we have $4\leq K$, atleast if they are not colinear. Mar 7, 2012 at 16:01
• @Emil: Yes. If each of the sides of a triangle has rational length, then the set of points at rational distance from all three of its vertices is dense in the plane. Mar 7, 2012 at 16:10
• @Tony: I can’t say I see this. Is there a simple argument I am missing? Mar 7, 2012 at 16:37

Here is a summary of the information in the previous question. For the second part of your question, the author (me) conjectures that for any finite set $S$ with all rational distances, no such point $P$ exists. As I noted in the comments, this is true when $|S|=3$, proven by Almering.
It is not known if there is a point with all rational distances to the unit square. However, it is known that there are no points at rational distance from all vertices of a regular $n$-gon, except perhaps when $n=4,6,8,12,24$.