TITLE: On well separated point sets in the plane
QUESTION [9 upvotes]: 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).

REPLY [3 votes]: 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 $G(A)=C_{2n-1}$.