Abstract projective plane
An abstract projective plane
- For any two distinct points
there exists precisely one line incident with both of them. - For any two distinct lines
there exists precisely one point incident with both of them. - Each line of
is incident with at least three distinct points of . - Each point of
is incident with at least three distinct lines of .
Since ^P1 and ^P2, as well as ^P3 and ^P4, are duals of each other, the dual of any theorem following from these axioms holds. This is known as the principle of duality. The following axioms can replace both ^P3 and ^P4.
- (3a.) There are four points of
in general position, that is four points no three of which are colinear. - (3b.) At least two lines exist2, but no two lines of
cover the points of the plane, i.e. for any two lines there is a point of incident with neither line.
Proof of equivalence
Assume
satisfies ^P1, ^P2, ^P3, and ^P4. Let . By ^P4 there exist three distinct lines , and by ^P2 and ^P3 there exist distinct points , , and . each distinct from . At least one of , so without loss of generality assume . Then are in general configuration, so ^P3a holds. Now assume
satisfies ^P1, ^P2, and ^P3a, but not ^P3b, so there exist two lines covering the points of the plane. By ^P3a there exist distinct and all different from (otherwise three points would be colinear) but since cannot be on or , the assumption of not ^P3b was invalid. Hence ^P3b holds. Finally assume
satisfies ^P1, ^P2, and ^P3b. ^P4 follows immediately, since for any point there exist lines and and at least one , so . If is any line, there exists , and by ^P4 there are three distinct lines through each of which meet at a different point, hence ^P3 holds.
See Finite projective plane, and the generalizing Abstract projective space.
Footnotes
-
2020, Finite geometries, p. 1 ↩
-
Kiss and Szőnyi leave this out, but I believe without this stipulation it is possible to produce a geometry of one line and one point. ↩