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 $({\cal P},{\cal L})$ is 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

  1. Every line has cardinality at least 3. (Cf. Beck-Bleicher-Crowe's Excursions into Mathematics: The Millenium Edition (2000, CRC Press).)

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?

  • 1
    $\begingroup$ 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). $\endgroup$
    – Jean Marie
    Apr 14, 2021 at 20:44
  • $\begingroup$ @JeanMarie I think you meant to post this comment on the other post... $\endgroup$ Apr 14, 2021 at 22:16
  • $\begingroup$ Gossett's Discrete Mathematics with Proof (2003, Pearson) affirms in Lemma 8.3 the conjectured result: that every line has cardinality at least 3. He proves the result by first showing the "dual" of Axiom 3: 3'. There exist four lines with the property that no three of these lines have a common point. $\endgroup$ Dec 25, 2021 at 6:37

1 Answer 1


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

  • $\begingroup$ Why must there exist a point $P$ that's on neither line? $\endgroup$ Apr 14, 2021 at 22:04
  • $\begingroup$ 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 $C$". $\endgroup$ Apr 14, 2021 at 22:15
  • $\begingroup$ I still think this answer is incomplete; it seems to be assuming Axiom 4, which is what we're trying to prove. $\endgroup$ Dec 25, 2021 at 6:28
  • $\begingroup$ I see no use of Axiom 4 in my argument. We may just have to disagree about it's correctness. $\endgroup$ Dec 25, 2021 at 14:14
  • $\begingroup$ @xFioraMstr18 Re:"Why must there exist a point $P$ that's on neither line?" This result is proved via contradiction as Lemma 2.1 in Albert & Sandler's An Introduction to Finite Projective Planes (1968, Holt, Rinehart and Winston, p. 16). $\endgroup$ Dec 26, 2021 at 0:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.