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Disclaimer: I had probably used some kind of "extended" set of incidence axioms, which also includes planes. Even though this is part of my homework, I seriously doubt this is what I was expected to do. So, while I will try to find the exact set of incidence axioms my prof. meant, this is still a question that I'd like to have answered. Below is the set of axioms the way I found it here: https://www.imsc.res.in/~kapil/geometry/euclid/node2.html the axiom of parallels: https://www.imsc.res.in/~kapil/geometry/euclid/node4.html I will also repeat them for ease of reference:

  1. There are at least two distinct points.
  2. There is one and only one line that contains two distinct points.
  3. Every line contains at least two distinct points.
  4. There are three points that do not all lie on the same line.
  5. For any three points that do not lie on the same line there is a one and only one plane that contains them.
  6. Any plane contains at least three points.
  7. If a line lies on a plane then every point contained in the line lies on that plane.
  8. If a line contains two points which lie on a plane then the line lies on the plane.
  9. If two planes both contain a point then they also contain a line.
  10. There are at least four points that do not all lie on the same plane.

The axiom of parallels:

  • Given a line and a point outside it there is exactly one line through the given point which lies in the plane of the given line and point so that the two lines do not meet.

The problem

When I try to build the model, in a loop, I keep adding more planes (because of 5) and once I added a plane, it requires that there be four points in the plane (because of the axiom of parallels), then I get new planes (again because of 5) and the loop continues.

My homework asked whether it is necessary that such a system have 6 and then 7 lines, but it seems that it will only make sense to answer such question, if I never construct additional planes, or does it?


On the second thought, would a "cube" with lines connecting all of its points and planes drawn whenever necessary (this will give 12 planes and 36 lines) be such a model?

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By axiom 4 you have three points $A,B,C$. By 2 this gives rise to three lines $AB,AC,BC$. You still need parallels to these lines: a parallel to $AB$ through $C$, a parallel to $AC$ through $B$ and a parallel to $BC$ through $A$. All three parallels need a second point on them, because of 3. If you want to keep your plane small, choose the same point $D$ as the second point for all of these, so you have lines $AD,BD,CD$ in addition to the original three, for a total of four points and six lines. This is as small as it gets, and you can verify all the other axioms for this structure.

All of this is considered to live in a single plane, in violation of axiom 10. The other axioms talking about planes are all satisfied, although they add little of interest to the picture. If you want an affine three-space instead of only an affine plane, you need more points.

A common way to construct finite planes is building on finite fields. To obtain a structure isomorphic to the one described above, take $\mathbb F_2$ as your field, and consider all points with coordinates in $\mathbb F_2$. The points would be

$$(0,0)\qquad(1,0)\qquad(0,1)\qquad(1,1)$$

Now for every pair of these, you have a line, for a total of six lines as above.

One important thing to notice here is that you must consider the incidence structure on a purely combinatoric level. So if you were to draw the above, you might assume that the line from $(0,0)$ to $(1,1)$ might intersect the line from $(0,1)$ to $(1,0)$ in the center of the square. But that center is not a point of your plane, hence the lines do not intersect but are instead parallel.

Nevertheless, the above still resembles a square in a certain sense. So if you were indeed to create an affine space not just a plane, then the cube would be a reasonable starting point. But I'd still use $\mathbb F_2$ coordinates to get the lines and planes right. For example, the plane through $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ would be characterized by $x+y+z\equiv0$ so $(1,1,1)$ would lie on that plane as well.

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  • $\begingroup$ Thanks for the input, but I'm bothered by (10): There are at least four points that do not all lie on the same plane. I think this precludes single plane, doesn't it? But I think that the model you described is the one that I was supposed to come up with in this assignment (I just needed fewer axioms). Does this make sense? $\endgroup$ – wvxvw Feb 1 '15 at 10:54
  • $\begingroup$ @wvxvw: You are right, I missed axiom 10. Updated my answer accordingly. You have to see for yourself whether you are to construct a plane or a three-space. $\endgroup$ – MvG Feb 1 '15 at 11:37

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