Tour of Fano’s Geometry

Axioms:

1. There exists at least one line.

2. Every line of the geometry has exactly three points on it.

3. Not all points of the geometry are on the same line.

4. For each two distinct points, there exists exactly one line onboth of them.

5. Each two lines have at least one point on both of them.

A Quick Theorem

Theorem (Theorem 1.7)

Each two distinct lines have exactly one point in common.

Proof.Suppose `1 and `2 are distinct lines in Fano’s Geometry. Axiom 5says there is at least one point on both of them. If points P andP ′ are on both `1 and `2 then Axiom 4 is contradicted as there aretwo lines on both P and P ′.

What we’ve shown

Because every addition we made was strictly required to fulfill theaxioms, we’ve established two facts:

I There are at least 7 points in Fano’s Geometry.

I There are at least 7 lines in Fano’s Geometry.

In what seems to be the usual way, we can prove that there are nomore than 7 points in Fano’s Geometry by using an argument bycontradiction.

Part of a theorem

TheoremFano’s Geometry has exactly 7 points.

Proof.We’ve established that there are at least 7 points. Suppose there isan eighth, P8. By Axiom 4, there must be a line, ` that containsboth P1 and P8. Since ` must intersect the line that contains P3,P7, and P4, and that line contains no other points, it follows that `contains either P3, P7, or P4. This contradicts Axiom 4.

The rest of the theorem

TheoremFano’s Geometry has exactly 7 lines.

Proof.Suppose Fano’s geometry had a line ` that was not among the 7we’ve established. By Axiom 4, each two distinct points must haveexactly one line on both of them, thus ` can contain at most oneof these 7 points. By axiom 2, though, ` contains 3 points.Therefore, ` contains at least two points that are not among theoriginal 7. This contradicts our previous argument, that there areexactly 7 points in the geometry.