[Math] Incidence Geometry Proof

geometrylogic

How do I show that the axioms of incidence geometry follow as theorems from the following axioms.

1) There exist exactly four lines.

2) Any two distinct lines are incident with exactly one point.

3) Each point is incident with exactly two lines

Best Answer

Concerning the axioms for Incidence geometry; see : Francis Borceux, An Axiomatic Approach to Geometry. Geometric Trilogy I (2014), page 306 :

Ax-I.1 Two distinct points are incident to exactly one line.

Ax-I.2 Each line is incident to at least two distinct points.

Ax-I.3 There exist three points not incident to the same line.

For Ax-I.1, let $P,Q$ such that $P \ne Q$, and suppose that exists $l,l'$ such that $l \ne l'$ and : $P,Q \in l \cap l'$. Then, $l$ and $l'$ are two distinct lines that are incident with two distinct points, contrary to axiom 2).

For Ax-I.2, consider a line $l$ and assume that there is at most one point $P$ such that $P \in l$. By axiom 3) there is onother line $l'$ such that $P \in l \cap l'$.

Consider a third line $l''$ (it exists by axiom 1)) : by axiom 2) there are two points $Q \in l' \cap l''$ and $R \in l'' \cap l$.

We have that $P \ne Q$, otherwise $P \in l, l', l''$, contrary to axiom 3).For the same reason, $P \ne R$ and $Q \ne R$.

Thus $R \in l$, contrary to assumption that $P$ is the only point on $l$.

Finally, the construction of the above proof gives us three distinct points that are not on the same line, i.e.Ax-I-3.

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