Def: A projective space is an ordered pair (P,L), where P is a nonempty set whose elements are called points and L is a nonempty collection of subsets of P called lines such that:
PS1: Given any two distinct points P and Q, there is one and only one line (l(P,Q)) containing them.
PS2: (Veblen's Axiom) If A, B, C and D are distinct points such that there is a point E in l(A,B) 
l(C,D), then there is a point F in l(A,C) l(B,D).
PS3: Each line contains at least three points; not all points are collinear.
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 flat) is a set of points U with the property that if two points are in U then all the points of the line determined by these points are also in U.
Def: The span (or projective closure) of a set of points S of a projective space, denoted by <S>, is the intersection of all subspaces which contain S.
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.