# Can a line in a projective plane have just two points?

Here is the definition of a projective plane from Stillwell's The Four Pillars of Geometry (2005, Springer):

Let $$\cal P$$ ("points") be a set, and let $$\cal L$$ ("lines") be a set of subsets of $$\cal P.$$ We say $$({\cal P},{\cal L})$$ a projective plane to mean

1. Any two distinct points lie on exactly one line;
2. Any two distinct lines intersect at exactly one point; and
3. There exist four points with the property that no three of these points are collinear.

Can a line have cardinality strictly less than 3? People sometimes add an axiom saying that there are at least 3 points (cf. Beck-Bleicher-Crowe's Excursions into Mathematics: The Millenium Edition (2000, CRC Press)), and I was wondering whether this "fourth axiom" is implied by the first three axioms. It is not extremely difficult to show that lines must have at least two points. Thus, we can reformulate the question as: Can a line have exactly two points?

• You have posted a similar question with 2 points instead of 1 just before. You should refer to it (or more exactly to the first issue where you have had an answer). – Jean Marie Apr 14 at 20:44
• @JeanMarie I think you meant to post this comment on the other post... – xFioraMstr18 Apr 14 at 22:16

## 1 Answer

No. Let $$A, B, C, D$$ be four points that axiom 3 guarantees. The line $$AB$$ must meet line $$CD$$ in a third point $$E$$ on each of those two lines.

It follows that every line must contain at least three points since all lines have the same number of points. To see that, suppose $$L$$ and $$M$$ are different lines. Pick some $$P$$ on neither. Then join $$P$$ to all the points on $$L$$. The intersections of those lines with $$M$$ establishes a bijection.

Almost a duplicate of Number of points on a line in a finite projective plane

• Why must there exist a point $P$ that's on neither line? – xFioraMstr18 Apr 14 at 22:04
• John Hughes seems to use his "Axiom D" in his proof when he says "Let $D$ be a point of $D$ [sic] distinct from $B$ and $C$". – xFioraMstr18 Apr 14 at 22:15