Question

Suppose we replace Incidence Axiom 4 with the following: Given any line, there are at least three distinct points that lie on it.

What is the smallest number of points in a model for this geometry? More precisely, find a number n such that every model has at least n points and there is at least one model that has only n points, and explain why your answer is correct.

Answer 1

To construct the minimal geometry where the parallel postulate holds and where every line has at least three points, let’s again start with the bare minimum of any model: three points{A, B, C}and their associated lines AB,AC and BC(below added image, first picture). We now continue the construction by focusing first on the parallel axiom. For the line AB and the point C there has to be a line parallel to AB and containing C. We add a point D to our model and let CD be that line. Likewise, BD be the line parallel to AC and passing through B.In addition, let BD be the line through the points{B, D}(below added image, second picture).

Neither of the lines so far contains 3 points. So we add the points{E, F, G, H, I}to ensure each line has 3 points (above added image, third picture). We arrive at an inter-pretation which satisfies the parallel axiom and where each line has exactly 3points.Unfortunately it fails the first incidence axiom. For example the points G and H don’t have any lines passing through them. To remedy this we add four more lines into the picture (above added image, fourth picture). A case by case check shows that this is indeed a model of incidence geometry where the parallel axiom hold-sand where every line has precisely 3 points.