From the three point model we saw if we used only the axioms

From the three point model, we saw if we used only the axioms for incidence geometry, we could not prove that a line has more than two points on it. Using the betweenness axioms as well, prove that every line has at least five points on it. Give an informal argument to that every segment (and thus, every line) has infinitely many points on it.

Solution

Let l be a line.By incident axiom, there are at least two point, B,D incident to l. By betweenness axiom, there are three points, A,C,E, each collinear with B, D, such that A*B*D, B*C*D,and B*D*E

Note that all points must be incident to l, for otherwise there would be another line incident to B, D, violating the uniqueness portion of I-1.Thus l is incident to the ve points A, B, C, D, E

Assume three points (A,B,C) are on one side of L ( the line that creates the half plane) and ABC. Now assume half plane is not convex, that is, B is on the opposite side of L. Now work with Betweeness axioms until you reach contradiction.

For interior of triangle.The intersection of convex sets is convex. A triangle is an intersection of three interior of angle sets. Once you prove interior of angle is convex.

Similar proof for interior of angle except ABC but this time A is on one ray, C is on other and B is in middle. Again assume interior is not convex and find contradiction by making ray emanating from O (vertex of angle) through B. use Betweeness axioms again to find contradiction.