my task is to show that every set of non-collinear points contains three non-collinear points. I can only use:
$(I1): A,B \in\mathcal{P}, A\neq B \implies \exists! g\in \mathcal{G}: A,B\in g$
$(I2):$ Every line has at least two points
$\mathcal{P}$ is the set of points and $\mathcal{G}$ is a set of lines which consists of pairs of points.
What I've done so far:
Let $M$ be a set of non-collinear points. Than exists at least one line through two points from M and another point which is not part of this line. Because of $I2$ the line consists at least of two points. So I have three non-collinear points.
This "proof" does not look correct. Can anyone give me a hint?
2026-03-27 07:49:42.1774597782