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!

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    $\begingroup$ 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? $\endgroup$ – John Hughes Dec 6 '18 at 15:33
  • $\begingroup$ Yes, you are right. Thank you $\endgroup$ – Just_a_newbie Dec 6 '18 at 16:07

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