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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

  1. 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”)
  2. 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.

MvG
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johnfrunkis
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  • Welcome to Math SE. Please use $\LaTeX$ formatting for your math formulae, since that makes things far more readable. I did so on your post, so you can use the current source (as you see it when you press “edit”) for an idea of the syntax to use. Did you really mean the $\varepsilon$ symbol? Also have a look at In the affine plane, I am having trouble with these definitions. As far as I can see, that's essentially the same question, although stated using far less formal notation. – MvG Oct 14 '14 at 07:39

1 Answers1

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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.

Bobby
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