How to prove that any line contain at least three points?

Hi i was reading a book called Symmetry and Pattern in Projective Geometry by Eric Lord, in his book the author give these axioms:

1. Any two distinct points are contained in a unique line.
2. In any plane, any two distinct lines contain a unique common point.
3. Three points that do not lie on one line are contained in a unique plane.
4. Three planes that do not contain a common line contain a unique common point.

My question is if with these axioms can i prove the statement that any line contains at least three points?

• A geometry with just two distinct points would appear to satisfy those axioms (vacuously so for #2,3,4). – dxiv Feb 26 '17 at 4:56
• I am not familiar with this book. You might have a look at the axiomatic treatment in Projective Geometry by Donald Coxeter. – DanielWainfleet Feb 27 '17 at 4:10

As far as I can see, a line with two points satisfies this system of axioms.

None of these axioms postulate the existence of noncollinear points, but that is normally a feature of axioms for the projective plane and projective $3$-space.

Perhaps the author has given these axioms in addition to some others that occurred earlier?

• Nope, he start with only these set of axioms saying: "The set P is a three-dimensional projective space if it satisfies the following axioms: " – Gonzalo Castillo Feb 26 '17 at 4:58
• @GonzaloCastillo if that's the case, then it cannot be proven (I have just given a model without even 3 points. – rschwieb Feb 26 '17 at 20:18
• @GonzaloCastillo You should put this as part of your question (that this is specifically a definition for projective 3-space). – Morgan Rodgers Feb 27 '17 at 3:45

If we're allowed to use this definition for a line in $\mathbb{R}^{3}$:

$L = \vec{a} + \lambda \vec{u}: \lambda \in \mathbb{R}$, $\vec{a}, \vec{u} \in \mathbb{R}^{3}$

Where $\vec{a}$ and $\vec{u}$ are two distinct points contained by $L$

Then by changing the value of $\lambda$ we can show that $L$ contains at least $3$ points. Although, this definition doesn't come straight from those axioms, so it seems like you can't prove the statement using only those axioms.

• Thanks for your answer. That's i am afraid of, in fact other authors include that statement (Any line contain at least three points) as an axiom. – Gonzalo Castillo Feb 26 '17 at 5:09

This is a strange definition, and not the normal set of axioms I usually see. It is possible that, for whatever reason, the author wants to include degenerate cases, such as the empty space, a single point, or a line with two points.

Maybe you can give the rest of the definition, instead of just the axioms? For example, I find it odd that planes are referred to in these axioms. Are these axioms explicitly for a projective 3-space?

One set I usually see for projective plane/space is:

1. Two points are on a unique line.
2. Let $a,b,c,d$ be four distinct points. If there is a point incident with both $\overline{ab}$ and $\overline{cd}$, then there is a point incident with both $\overline{ac}$ and $\overline{bd}$ (this basically says that if we have two lines that intersect then they determine a plane, and any two lines in that plane intersect. but without explicitly defining planes).
3. Each line is incident with at least three points.
4. There exist at least two lines (if you specifically want a projective space, require that there exist two lines with no point in common).

(Although if you want to talk more generally about a projective geometry there may be reasons you don't want to exclude the projective line. Also you sometimes want to explicitly include all of the subgeometries of each dimension as objects that are part of the geometry, instead of defining them in terms of points and lines.)