Index
We describe a linear space S as having the exchange property if for any two points x and y in S, and any subset X of points in S, the following holds:
In other words, x and y can be exchanged. It is sufficient to show that a linear space DOES NOT satisfy the exchange property simply by giving one example such that the above condition is not met.
Before we observe the signifigance of such a property, let's look at an example to better visualize what one of these spaces would look like.
Let's use the following example:
Let's check some examples of points and point sets for this linear space.
Continue checking all other possibilites. You will see that this linear space does indeed satisfy the exchange property.
All linear spaces, however, do not not satisfy the exchange property. See if you can find an example of a linear space that does not satisfy it.
Linear spaces that satisfy the exchange property behave nicely. For example, the concepts of linear independence and basis introduced in problem 11 suggest that it is possible to have two bases with different cardinalities for the same linear space. However if a linear space satisfies the exchange property, then any two bases will have the same number of elements (and hence the same dimension). This is a very exciting lemma! An outline of a proof is given in Batten. See if you can fill in the missing pieces by yourself.
Now, please refer back to problem 7 and the introduction of hyperplanes. A hyperplane of a fine linear space is simply a maximal, proper subspace of that linear space.
In the case where the linear space has only one line, any hyperplane is simply a point. In the case of a linear space consisting of only one point, only the empty set is a hyperplane. It seems possible, then, to conclude that hyperplanes are always of dimension one less than the space that they are in. Unfortunately this is not always true. But we propose the following lemma:
Let S be a finite near linear space with the exchange property. If H is a hyperplane of S, then dim H = dim S - 1. (for a proof, see Batten)
Let's also propose the lemma that the hyperplanes of a projective plane are the lines.
To prove this, simply show that for line l and some point 
that 
where S is the entire projective plane. Referring back to problem 10, we have shown that for any projective plane, all lines meet and each line is incident with k+1 points and each point is incident with k+1 lines. Thus if , then we know that it must lie on a line that meets l. Thus there exists a line that contains both p and a point q where q is a point on l. So 
will generate that line.
But p is incident with k+1 lines, all of which must meet l and all of which must contain both p and some point on l. Thus generates all k+1 lines that lie on p. So now we have generated line l and all k+1 lines that lie on p. We have generated k+1 points on each of the k+2 lines: (k+1)(k+2) = k2+3k+2 points all together. But we have certainly counted some points more than once. Because we have generated all k+1 lines that lie on p, we have counted p exactly k+1 many times. Thus we must subtract k because we only want to count the point p once.
Similarly, because we have generated all k+1 lines that meet l, we have counted each of the k+1 points on l twice (once on l and once on the lines that meet l). Thus we must subtract k+1 as well. So now we have
k2 + 3k + 2 - k - (k + 1) = k2 + k + 1
Thus we have k2 + k + 1 points. We showed in problem 10 that this is the total number of points in the projective plane! So if all of our points are generated by then we will generate all of the lines of S. Thus, .
In conclusion, the hyperplanes of a projective plane are just the lines.
Note: You may find it helpful to make your own illustrations when digesting a proof such as this.
Now that we have shown this, if we show that any projective plane satisfies the exchange property, then we will know that all projective planes will be of dimension 2.
This helps to explain why it is that we call the projective plane a plane. Any space that satisfies the axioms for a projective plane will be of dimension 2!
There are also 3-dimensional spaces which we call projective spaces. First, we define a triangle of a geometry to consist of three lines where each pair of lines are incident at a unique point, and the three points of incidence are non-colinear. The axioms are then as follows:
