Let $D$ be a set of lines in the real projective plane such that for all $d\in D$, $\bigcap (D\setminus \left\{d\right\})=\emptyset$ and let $Dot(D)$ be the set of all dots that belong to at least two members of $D$.
$D$ is complete if for any pair of dot in $Dot(D)$ the line that contain both dots of the pair is in $D$.
We note $<D>$ the smallest complete set that contains $D$.
Note that if $D$ has 4 members then there exists an homography $p$ such that $Dot(<D>) =p(Q_2)$ , where $Q_2$ is the rationnal projective plane, i.e the set of dots that homogenious coordinates are integers (indeed an homography is determined by the images of 4 lines and it is easy to see that if all members of $D$ are rational lines then $dot(<D>)$ is the rationnal projective plane)
Let's say $D$ is quasi complete if a line that joins two dots of $Dot(D)$ is either in $D$ either contains no other dot in $Dot(D)$.
If $D'\subset D$ and $|D'|=4$ , then $D\cup <D'>$ is said to be an elementary completion of $D$.
We now suppose that $D_0$ is a finite set of lines in general position, i.e. that all the $3\times |D_0|$ coeficients of their homogonious equation are algebraically independent, we then consider $D_0, D_1, .., D_k$ for some positive integer $k$ such that for all integer $i\in [1,k]$, $D_i$ is an elementary completion of $D_{i-1}$
QUESTION :
Is $D_k$ quasi complete ?
This seems very intutive to me that it is the case, it says that if some colinearity happends then it has to be either involving the completion of four lines either because we decide it in the construction [i had not enough place in the title but say that by "decide it " i mean one of the two cases I mentionned : either we are in the completion of a particular set of 4 lines, either we decide to construct lines passing by some dot that we chose on some other line and thus this dot will be on this line "by construction" ... i don't know if I'm very clear here, just trying to motivate the title...].
I have several ideas to prove this : one is by finding a contradiction with a minimal construction that would be a counterexample and the other would be on algebraic independancy considerations. But this seems to me quite fastidious as we are in two dimentions, maybe something smart involving birapport and some kind of algebraic independency would do the job....
MOTIVATION : This would answer by the negative to the weak question asked here
Indeed we could the show (quite easilly from the affirmative to present question) that in any $D$ that contains 5 lines in general position, there exists a line $d$ that has more "ordonary dot " on it than non ordinary dot in ($Dot(D) \cap d$ (ordinary dot is the intersection of exactly two lines)
If for any dot $M$ in $Dot(D)$, $charg(M)$ is the number of lines of $D$ that contain $M$ minus 3, then $charg(M)=-1$ iff $M$ is an ordinary point and the dual form of the Silvester-Gallai theorem states that : $ \Sigma_{P\in Dot(D)} charg(P)= -3 -\Delta$
with $\Delta \geq 0$ , that is $0$ iff $D$ is a simplicial aŕrangement (see comments in the mo question of the link). (1)
($\Delta$ can be seen as the number of edges that you need to addto the projective polyedron defined by $D$ in order to make all faces triangle, thus $\Delta <N$ implies there are no more than $N$ faces that are not triangles)
A consequence of this is that if a line $d$ has strictly more ordinary dots on it then other points on it, then if you remove it the sum of the chargs is greater in $D\setminus d$ then in $D$ thus it cannot be maximal in $D$ and thus $D$ cannot be a simplicial arrangement. This impossibility is a bit weaker than the one asked by the weaker question in the mo link but
In fact we can show (with the "yes" to this question post and the same argument with few minor precautions) that if the $\Delta$ in (1) is less than $|D|$ than $D$ cannot contain 5 lines in general position. And this answers no to the weak question. (because the weak question's configuration implies $\Delta \leq |D|$)