EDIT :
lemma 2 is wrong, see comment (i try to see if something in the proof can be saved)This [should] answer by the negative to the question https://mathoverflow.net/questions/155196/extending-a-line-arrangement-so-that-the-bounded-components-of-its-complement-ar in the case of the projective plane, the ideas might be used later to give stronger result
[I put it in the stack exchange because it is an attempt to an unaswered question" in MO, so I wan't to be sure that it is not wrong before posting it, I hope that this is not a behaviour that is inappropriated here on stack exchange... if there is some questions please ask in the comments, because in english I am very often missundertood, and this will help me to give a better answer in case it is correct , also if some one can help improve the formulation (or any other aspect of the proof) so that it is said in good english, I would be more than thanksfull ]
To summarize I noticed that if a set of lines is triangulating in the projective plane then $f=2(v-1)$ where $f$ is the number of faces of the (projective) planar graph induced, and $v$ the number of vertices, but when we look at classical affine triangulation we kind of notice that the cardinality of vertices is greater than that of bounded (triangle) connected components, this suggests, like it has been said in comments by David Eppstein, that there is very few example of these projective triangulation, and that can happen only if there is not "too much triangle in the middle" (middle is a ruff abreviation of "the union of bounded connected component" looked from an affine point of wiew), like the example of 3 line in generic position, there is $4$ triangle components but when we look at it in the induced affine plan, 3 of them are not bounded. Let's give a formal frame to these intuitive considerations.
Let $\mathbb P$ be the projective plane, and $\mathbb P^*$ the set of projective lines. For all finite $\mathcal E\subset \mathbb P^*$ we define $f (\mathcal E)$ to be the set of the connected component of $\mathbb P\setminus \bigcup \mathcal E$. For any $k\in f(\mathcal E)$ we denote by $V_k$ the set of its vertices. $V(\mathcal E):=\bigcup_{k\in f(\mathcal E)}V_k=\left\{l\cap m\in \mathbb P, \, (l,m)\in \mathcal E\right\}$ is the set of all vertices that lies in the picture. Let $d\in \mathbb P$ s.t $d\cap V(\mathcal E)=\emptyset$, we consider : $f_d(\mathcal E)\subset f(\mathcal E)$ to be the set of connected component that intersect $d$ (we also pose $f^d=f\setminus f_d$)
If $d$ is seen like the "line at infinity", then $\mathbb P\setminus \left\{d\right\}$ can be identified to the affine plane where elements of $f^d$ coorespound to the "bounded" connected component
Claim
if $|f_d(\mathcal E)|< |f^d(\mathcal E)|-1$ then $\mathcal E$ cannot be augmented in a triangulation.
Let's observe that such a $\mathcal E$ exists for example let's take $\mathcal E=\mathcal E_M\cup \mathcal E_N$ where $\bigcap \mathcal E_M=\left\{M\right\}\subset \mathbb P$, $\bigcap \mathcal E_N=\left\{N\right\}$, and say $|\mathcal E_M|=|\mathcal E_N|=5$. Note that we can make $|f^d|-|f_d|$ as big as we want : this might have a relevance in order to restreight the result (see note 2 at the end)
This is the concequence of the three following lemmas :
lemma 1 If $\mathcal E\subset \mathcal F\subset \mathbb P^*$, and $|\mathcal F|$ is finite and $d\cap V(\mathcal F)=\emptyset$, then : $|f^d(\mathcal F)|-|f_d(\mathcal F)|\geq |f^d(\mathcal E)|-|f_d(\mathcal E)| $
lemma 2 on same hypothesis $|V(\mathcal F)|>|f^d(\mathcal F)|$
lemma 3 if $\mathcal F$ is triangulated, (i.e elements of $f(\mathcal F)$ are triangles), then : $|f(\mathcal F)|=2(|V(\mathcal F)|-1)$
It is easy to see that the condition on lemma 3 and lemma 2 contradicts the condition of lemma 1 under the hypothesis of the claim.
The proofs of the lemma1 is an immediate induction on $|\mathcal F|$ : adding a line $l$ strcitly increase $|f^d|$, and $|f_d|$ from at most $1$ unit, indeed only the element of $f_d$ that contain $d\cap l$ is concerned by creating a new connected component.
Lemma 2 is obtain by induction on $|f^d|$, we make an element $k\in f^d$ "vanish" by making all the lines that define $k$ pass throught a single dot, then the cardinality of $V$ have decreased more than that of $|f^d|$ and we apply the induction hypothesis.
To prove lemma 3, we use the Poincaré's formula and the Euler caracteristique of $\mathbb P$ that is $1$.
Poincaré's formula tells us that $v:=|V(\mathcal F)|= U_1-U_2+U_3-....$ where $U_i=\Sigma_{P\subset f(\mathcal F),|P|=i}|\bigcap_{k\in P}V(k)|$ (+)
$U_1=3|f(\mathcal F)|:=3f$ (I use the letter $f$ for "face") $U_2=U_2'+U_2''$ where $U_2'$ is two times the cardinality of pairs of connected component that intersects in exaclty $2$ vertices (hence $U_2'=2e$ where $e$ is the total number of edges, and $U"_2$ are pairs of connected component that have just one vertex in common.
We are going to prove that $U''_2-U_3+U_4...=v$
If write $U_i:=\Sigma_{M\in V(\mathcal F)}$ where $U_i(M)$ is the number of non trivial intersection of exactly $i$ connected component that intersection is exactly $M$, we see that it is suffisant to prove that for each $M\in V(\mathcal F)$,
$U''_2(M)-U_3(M)+....+(-1)^nU_n=1$ (**) where $n$ is the number of lines that passes through $M$
$U''_2(M)=n(n-1)/2-n=-C_n^1+C_n^2$ is the numbers of pairs among $n$, from which we subtract the $n$ "consecutive pairs"
$U_i(M)=C_n^{i}$ is the i-th binomial coeficient, so we finally get (**)
Getting back to (+), we then have : $3f=2e$ (***)
In an other hand the equlity $f-e+v=1$ (+++) holds for any polyedron in $\mathbb P$ ($1$ is the Euler characteristic of $\mathbb P$) so combining (***) and (+++) we finaly get what we wanted:
$f=2(v-1)$.
Note :
Back to the claim, we show that we cannot augment $\mathcal E$ with lines that are not $d$ and such that two lines of the set of lines never intersect in $d$, but i leave the reader convince himself that this restriction can be cancelled by taking instead of $d$ some line in the neighbourhood of $d$ such that the data we need are not changing.
The same king of argument can be used to show that we can neither triangulate up to a fixed autorized number of $4$-gone (or other poygon whose cardinality and number of vertex are fixed) . Indeed the generalization of the formula $f=2(v-1)$ is $f_3+2f_4+3f_5...=2(v-1)$ where $f_i$ is the number of faces that are $i$-gons, this is not giving us the general answer to the question but gives kind of a hope, especially if we are able to some how control $|f_d(\mathcal E)|$ compare to $f(\mathcal E)$ in some sens that I'm going to investigate a little bit in the next few days.
