Question

Given a projective plane, let u be a line in the plane. Let A be the set of all points and lines from the projective plane minus the line u and all points incident on u. Show that A is an affine plane. (This implies that the "line at infinity" is not a special line in the projective plane. Any line could serve that role.)

Answer 1

To show that the set A formed by removing a line u and all points incident on u from a projective plane is an affine plane, we need to show that A satisfies the axioms of an affine plane. A satisfies the axioms of an affine plane, which state that any two distinct points are still incident on a unique line, and any two distinct lines have at most one point in common.

Step-by-step explanation:

To show that the set A formed by removing a line u and all points incident on u from a projective plane is an affine plane, we need to show that A satisfies the axioms of an affine plane. An affine plane is a set of points and lines such that

1. For every pair of distinct points, there exists exactly one line that passes through both points.
2. Any two distinct lines have at most one point in common.

Let's consider the first axiom. In the projective plane, any two distinct points are incident on exactly one line. When we remove the line u and all points incident on u, any two distinct remaining points, say P and Q, are still incident on a unique line. This line can be obtained by joining P and Q in the projective plane, and since u was not part of the line joining P and Q, it is also a line in A. Therefore, A satisfies the first axiom.

For the second axiom, let's consider two distinct remaining lines, say l1 and l2. Any two lines in the projective plane intersect at exactly one point. If the intersection point lies on u, we remove it from A as well. Therefore, in A, l1 and l2 have at most one point in common, satisfying the second axiom of an affine plane.

Thus, we have shown that A is an affine plane.