# Proving that a point is the result of only two lines intersecting and a line is the result of only two points being aligned

Let $$S_0$$ be a set of four points in the real projective plane such that any three points of $$S_0$$ are not aligned. Let $$L_0 := \emptyset$$. For every integer $$n \ge 1$$, we define the following:

• If n is odd, $$S_n := S_{n - 1}$$ and $$L_n := \{p \vee q | p, q \in S_n, p \ne q\}$$.
• If n is even, $$L_n := L_{n - 1}$$ and $$S_n := \{l \cap l'|l, l' \in L_n, l \ne l'\}$$.

Is it true that for every $$p \in S_n, p \notin S_{n - 1}$$ (suppose $$n \ge 1$$) there exist two and only two lines in $$L_n$$ that contain p? Analogously, is it true that for every $$l \in L_n$$, $$l \notin L_{n - 1}$$ there exist two and only two points in $$S_n$$ contained in $$l$$?.

So far I have been able to prove that if $$p \in S_n$$, $$p \notin S_{n - 1}$$ with $$n \ge 1$$, then all the lines in $$L_n$$ that contain $$p$$ (except for maybe one) are in $$L_{n - 1} \setminus L_{n - 2}$$. That is because if two lines in $$L_n$$ that contain $$p$$ were not in $$L_{n - 1} \setminus L_{n - 2}$$, then they would be in $$L_{n - 2}$$, but that would mean that $$p \in S_{n - 1}$$ (!). I have also proved the equivalent result for $$l \in L_n$$, $$l \notin L_{n - 1}$$, however I do not know how to move from here. Any advice is appreciated!

• Can you make clear what your $\vee$ notation means? Is it just the "unique line containing $p$ and $q$"? If so, then in the definitions of $L_n$ and $S_n$ in the two bullets, should you have $p \ne q$ in the first, and a corresponding condition in the second? – John Hughes Dec 6 at 15:33
• Yes, you are right. Thank you – Just_a_newbie Dec 6 at 16:07