## General position

Prove that for a finite projective plane of sufficiently high order the generalization of the axiom for quadruples applies:

For every $$k \in \mathbb N$$ exists $$n \in \mathbb N$$ such that for any finite projective plane $$(X,\mathcal P)$$ of order at least $$n$$ there exist $$k$$ points in th general position, i.e. a set $$K \subseteq X, |K|=k$$ satisfying $$\forall P\in\mathcal P: |K\cap P|\le 2$$.

• #### Hint

First, try to construct a projective plane with a five-element set in the general position.

Then generalize the method to be possible to use mathematical induction on $$k$$.

• #### Solution

Let $$n_k$$ be the smallest number such that each projective plane of the order $$n_k$$ contains $$k$$ points in the general position. We know from the axioms of the projective plane that $$n_4 = 2$$.

First for five: If we have a projective plane of order $$n$$ and its four-element set in the general position $$K_4$$, then these four points determine $$\binom{4}{2}$$ lines. These lines have at most $$\binom{4}{2}(n+1)$$ points. (In fact, exactly $$\binom {4}{2}(n+1)-8$$, but that would only complicate our reasoning.) If the total number of points in the set $$X$$ is higher than the number of points lying on the lines specified by $$K_4$$, then there is a point $$x \in X$$ which does not lie on any lines specified by the points $$K_4$$. The set $$K_4 \cup \{x\}$$ is the wanted five-element in the general position. To get $$|X|=n^2+n+1> \binom{4}{2}(n+1)$$ it sufiifes the order $$n\ge\binom{4}{2}=6$$. Hence $$n_5 \le 6$$.

By induction we show that $$n_{k+1}\le\binom{k}{2}$$. Assume that the above estimate holds for $$k$$, and that a projective plane of order of at least $$\ binom {k} {2}$$ is given. Since $$\binom{k}{2}\ge\binom{k-1}{2}\ge n_{k}$$, according to the inductive hypothesis this projective plane contains a $$k$$-element set $$K_k$$ in general position. The points $$K_k$$ specify the $$\binom{k}{2}$$ lines. On these lines at most $$\binom{k}{2} \left(\binom{k}{2}+1 \right)$$ points may be placed. However, because the total number of points is $$\binom{k}{2}^2 + \binom{k}{2} +1$$, there is a point $$x$$ outside these lines. The set $$K_k \cup \{x \}$$ is the desired $$(k+1)$$-element set in general position.