Show that every line in the projective completion contains at least 3 points.

Question

We know that P is a model of incidence geometry that satisfies the Euclidean parallel property and that R is its projective completion I'm trying to show that every line in R contains at least 3 points of R.

I'm thinking there are two cases for when it is line in P and when it's line at infinity.

Is this right? Or am I approaching this all wrong?

Answer 1

What is your definition of “projective completion”? I would define projective completion as a completion which satisfies all the axioms of a projective plane, namely

1. two distinct points have a unique line joining them
2. two distinct lines have a unique point of intersection
3. there are at least four points, no three of which are collinear

At this point, you can drop the fact that you are speaking about some projective completion, and instead show that any projective plane will have at least three points on each of its lines. Without distinguishing between finite and infinite points.

I'll provide some details in a spoiler block, but encourage you to think about this first so you can try to work this out for yourself.

How do you show that? Well, take an arbitrary line $a$. Consider the points from axiom 3 which I'll call $A,B,C,D$. How many of them are incident with $a$?
1. If two of these points are incident, w.l.o.g. they are $A$ and $B$, so you have those points already. Then the line joining $C$ and $D$ will intersect $a$ in a point, which can not be $A$ or $B$ so you have to have a third point of intersection.
2. If one of these is incident, w.l.o.g. $A$, then $BC$, $CD$ and $BD$ will intersect $a$ in points which are distinct from $A$ and distinct from one another. So you have at least four points on the line.
3. If none of the points is incident, then the four points define six lines, and $a$ has to intersect all of them. Two such intersections may happen in the same point only if the lines do not have any of the four points in common. I.e. $a$ may intersect $AB$ and $CD$ in the same point, but not $AB$ and $BC$ since they already have $B$ as their point of intersection. So these six lines result in at least three distinct points of intersection.