Infinitely many points in plane s.t. no point is a convex combination of other points

Let $A$ be an infinite set of points in the plane, with no three points of $A$ collinear. I want to prove that $A$ contains an infinite set $B$ such that no point of $B$ is a convex combination of other points of $B$.

1. Is it equivalent/stronger/weaker to ask for a set $B$ in which any $4$ points form a convex quadrilateral?

2. Can I prove this using Ramsey theory?

3. I know that using Ramsey theory one can prove that for any $n$, sufficiently many points in the plane contain a convex $n$-gon. Does this help to prove the infinite case?

• The answers to your questions 1. and 2. were both yes, indeed. Moreover, the infinite and finite Ramsey results are equivalent ( one implies the other), I think Ramsey originally proved the infinite case. For 3. the finite case ( and the precise number $N$ that guarantees a convex $n$-gon is again a problem of Erdos would imply directly the infinite case, without Ramsey. You can use for this the following important fact: the inverse limit (see en.wikipedia.org/wiki/Inverse_limit) of a sistem of finite sets is nonvoid. Let me know if you want to see the proof. – Orest Bucicovschi Jan 9 '16 at 18:32

Use the infinite Ramsey theorem. Consider the infinite set $A$. Color the subsets with $4$ elements of $A$ in two colors: red if the points form a convex quadrilateral, blue otherwise. There exists an infinite subset $B$ of $A$ with all subsets with $4$ elements of the same color. This color cannot be blue. Indeed, it is known that among any $5$ points in the plane, every $3$ non-colinear, $4$ of them will be the vertices of a convex quadrilateral (cf a result of Erdos) . Therefore, every $4$ points of $B$ form a convex quadrilateral, so no point is in the convex hull of the other $3$. Now, use Caratheodory theorem to show that no point of $B$ is in the convex hull of another finite subset of $B$.