Let $(P,L,\varepsilon)$ be a plane with finitely many points (i.e $P$ is finite) Assume in addition to the axioms of incidence that
- for each $Q \in P$ and $l \in L$ with $Q \not\varepsilon l$ there is exactly one $a \in L$ with $Q ε a$ such that $X \not\varepsilon a$ for all $X \not\varepsilon l$ (“$l$ and $a$ do not intersect”)
- for each $l,k \in L$ there is $p \in P$ such that $p \not\varepsilon l$ and $p \not\varepsilon k$.
Prove that all $l \in L$ have the same number of points, i.e. that for $l,k \in L$, $\#\{Q \in P \;|\; Q ε l\} = \# \{Q \in P \;|\; Q \varepsilon k\}$
Everyone I've asked in college can't seem to answer it. I was wondering if any of you could help. Thanks.
did you have the parallel transport in class? let $g,g'\in L$ be two arbitrary straight lines, then we know that there exists at least one point $P\in g$ s.t. $P\not\in g'$ and $Q\in g'$ s.t. $Q\not\in g$. This follows from the fact, that on each straight line there are at least 2 points, and we assume without loss of generality $g\neq g'$, else the problem would be trivial.
Now define $f\in L$ s.t. $P,Q\in f$. Now we know that there exists for every point $A\in g$ a unique parallel straight line $f'$ with $f'\cap f=\emptyset$ (your first point). and obviously this straight line has intersection points with $g$ and $g'$. Now you can define the map $\phi_f:g \rightarrow g'$, which takes a point from $g$ and maps it to the point in $g'$ along a parallel of $f$. By construction this map is surjective and injective (you can check it if you want, it's not hard) and hence bijective. This means that $g$ and $g'$ have the same amount of points.