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Projective Plane Axioms:

  • 1) A line lies on at least two points
  • 2) Any two distinct points have exactly one line in common
  • 3) Any two distinct lines have at least one point on common
  • 4) There is a set of four distinct points, no three are collinear.

Proof: By PA4, there is a set of four distinct points, no three collinear. By definition of collinear this means that we can create a set of four points where no three lie on the same line. We know by PA3 that any two distinct lines have one point in common. Since no three points lie on the same line, the third point, P3, must lie on a second line, L2 different from the first line, L1, containing points P1 and P2. Then we must draw a third line, L3, connecting P3 to P1. Now, consider a fourth point, P4. We construct lines L4, L5 and L6 such that PA are met.

I'm not sure if I'm going about this the correct way or if there's a better way to think about this. Any help is appreciated. Thanks!

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Prove that if there are four distinct lines, no three are concurrent.

In the Fano plane, for example, there are $7$ distinct lines, and every point has three lines that are concurrent at any given point, so this does not appear to be true, as written.

Perhaps you mean There exist four distinct lines, no three of which are concurrent.($\ast$)

We know by PA 4 that in a set of four points, no three are collinear.

That is not the axiom. It should read There exists a set of four points, no three of which are collinear.

Using this last axiom, we can prove ($\ast$) by saying that in the dual plane, there exist four lines, no three of which are concurrent. But the projective plane is self-dual, so they are in fact inside the original plane.

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