Consider the system [S, L, P], where S contains exactly four points A, B, C, and D, the lines are the sets with exactly two points, and the planes are sets with exactly three points. This "space" is illustrates by the following figure:

Here it should be remembered that A, B, C, and D are the only points that count. Show that there is at least one plane.

Here are the relevant Incidence Postulates in the chapter:

I-0)All lines and planes are sets of points.

I-1) Given any two different points, there is exactly one line containing them.

I-2) Given any three different noncollinear points, there is exactly one plane containing them.

I-3) If two points lie in a plane, then the line containing them lies in the plane.

I-4) If two planes interesect, then their intersection is a line.

I-5) Every line contains at least two points. S contains at least three noncollinear points. Every plane contains at least three noncollinear points. And S contains at least four noncoplanar points.

Here is my proof, but I have no idea whether or not it is correct. A push in the right direction would be greatly appreciated!

Let points A, B, C, and D in S be given. If there isn’t at least one plane, this means that any set of three points is collinear.

A, B, and C lie on a line.

A, B, and D lie on a line.

A, C, and D lie on a line.

B, C, and D lie on a line.

However, the four statements above contradict I-1, which states that given any two points, there is exactly one line containing them because each possible pair of points lies on two lines (i.e. A and B lie on line ABC and line ABD). In conclusion, there is at least one plane in the figure.

*Random question - Is there also a way to prove that there exist at least two planes?