On well separated point sets in the plane - MathOverflow

On well separated point sets in the plane

2012-12-05T07:44:03Z

Let us say that a finite set $A$ in the plane is $1$-separated if:

1) it has an even number of points;

2) no open ball of diameter $1$ contains more than $|A|/2$ points.

For a $1$-separated set $A$ define $G(A)$ to be a graph where two points $x,y$ in $A$ are joined by an edge iff the distance between them is at least $1$.

Question: can one find a finite set of graphs $G _ 1,\dots,G _ n$ such that any $1$-separated set $A $ can be partitioned into non-empty $1$-separated sets $A _ 1,\dots,A _ k$ such that $G(A _ i)$ is isomorphic to one of the $G _ j$'s?

Comment: The definition makes sense on the real line (the ball of diameter $1$ is replaced by an interval of length $1$). In that case we can take $n=1$ and $G_1$ to be a graph on two vertices joined by an edge (that is, $G(A)$ contains a matching).

Answer by domotorp for On well separated point sets in the plane

2012-12-07T01:26:01Z

No, there is a counterexample. Take a circle, whose diameter is slightly larger than 1 and put |A|-1 points evenly around its boundary and the last point to its center. This set will be 1-separated, but no matter how you partition it, the part containing the center will not be 1-separated.

Answer by Alfred for On well separated point sets in the plane

2012-12-07T16:39:44Z

I think Domotorp is correct. Take a regular $(2n-1)$-gon such that its longest diagonal is 1, along with its center. Then $A$ cannot be partitioned, and $G(A)=C_{2n-1}$.