**Def**: A ** projective space** is an ordered pair (

Note that the only difference between the axioms for a projective space and those for a projective plane is axiom PS2. This axiom, which has also been called Pasch's Axiom (but this attribution is incorrect and is losing favor), is a clever way of permitting an extension to higher dimensions. Every projective plane is a projective space, since the projective plane axiom P2 (every pair of lines intersect) implies Veblen's Axiom, but the converse does not hold. We will provide other examples later.

**Def**: A ** subspace** of a projective space (or

**Def**: The ** span** (or

Note that a span of a set is a subspace, since the intersection of any number of subspaces is a subspace.

We now define a plane in a projective space, but we will have to prove that this definition gives the same object that we have previously called a projective plane.

**Def**: A ** plane** in a projective space is the span of a set of 3 noncollinear points.

**Proposition 22.1**: The plane <P,Q,R> consists of the points which are on lines through P which intersect the line QR.

*Pf*: Let N be the set of points which are on lines through P which intersect the line QR. If X is a point in N, and suppose the line PX intersects QR in the point S. Since <P,Q,R> is a subspace containing Q and R, it contains all the points on the line QR, in particular the point S. Since P and S are in <P,Q,R>, all the points on PS are in <P,Q,R>, in particular X. Thus, N <P,Q,R>. We will now show that N is a subspace. Let S and T be two points in N. If P is on the line ST, then by the definition of N, all the points on PS = ST are in N. So, we may assume that ST does not contain P. Since S and T are in N, there exist points U and V on QR with U on PS and V on PT. Now, let X be any point of ST other than S or T. Now the lines US and VT meet at P, so by Veblen's axiom, the lines ST and UV (= QR) meet at a point Y. Also, the lines PU and XY (= ST) meet at the point S, so again by Veblen's axiom, PX and UY must meet. But, UY = QR, so X is on a line through P which meets QR. That means that X is in N. Thus, the entire line ST is contained in N and so, N is a subspace. Since N clearly contains P, Q and R, and is a subspace, it must contain the span of P,Q and R. So, we have <P,Q,R> N, and so N = <P,Q,R>.

**Theorem 22.2:** Two distinct lines in a projective space intersect if and only if they are in the same plane (*coplanar*).

*Pf*: Suppose that lines *l* and *m* intersect in a point P. By PS3, there are points Q on *l* and R on *m*, neither equal to P. P,Q and R are not collinear by PS1. The plane <P,Q,R> contains all the points of *l* and *m* since it is a subspace.

Now suppose that the lines *l* and *m* are contained in the plane <P,Q,R>. By PS3, we can find points A and B on *l* and C and D on *m*. Since these 4 distinct points lie in the plane <P,Q,R>, we know, by Proposition 22.1, that they lie on lines through P which meet the line QR. Let U, V, S and T be the points of QR at which the lines PA, PB, PC and PD meet QR respectively. Since lines AU and BV meet at P, by PS2 we have that AB ( = *l*) meets UV (= QR) at a point X. Now, PA meets SX (=QR) at U, so PS meets AX (= *l*) at some point, say W. Also, PB meets TX (=QR) at V, so PT meets BX (= *l*) at some point, say Y. Finally, since CW meets DY at P, we have that CD (= *m*) meets WY (= *l*) at some point.

**Corollary 22.3**: Any plane in a projective space is a projective plane.

Proposition 22.1 is really a special case of a more general result which we present now.

**Theorem 22.4**: If *M* is a subspace and Q a point not in *M*, then the span <*M*,Q> is the union of all the points on all the lines joining Q to some point of *M*.

*Pf*: Let *U* = the union of all the points on all the lines joining Q to some point of *M*. It is easy to see that *M* Q *U* <*M*,Q>. Thus, to prove the result we need to show that *U* is a subspace. Suppose then that B and C are points of *U*. By the definition of *U*, there are points B' and C' in *M* with B on QB' and C on QC'. Now, let A be any point (other than B or C) on the line BC. The points Q, A, B' and C' are all in the plane <Q,B,C>, so the lines QA and B'C' are in this plane, since a plane is a subspace. By Theorem 22.2, these lines meet at a point F, which being on the line B'C', is in *M*. Hence, A is in *U*, and so, the entire line BC is in *U*. That is, *U* is a subspace.