0
$\begingroup$

Consider the geometry given by the points of $\mathbb{R}^2$ and the lines being circles centered at points of the $x$-axis.

  1. Check to see if the incidence axioms hold(two distinct points make up a unique line and a line contains at least two points.)

Now I don't believe the first axiom holds since the intersection of two lines will contain more then $1$ point so they must be the same line however will also contain points that are not on the line. Though I'm unsure if I'm understanding it correctly.

1 Answers 1

1

How would you go about constructing the “lines” joining points? Given two points $A=(x_A,y_A)$ and $B=(x_B,y_B)$ you need to find a circle through these two points which has its center on the $x$ axis. You can either go about this by looking at the equation of a circle, or by observing that a circle of this kind will be symmetric wrt. the $x$ axis, so $A'=(x_A,-y_A')$ will be a third point on that circle. Here you have a circle defined by three points, so the line is defined. Right?

No. The above is true in general, but there are several ways how this might fail.

  1. For example, three points do not define a circle if they are collinear and if your definition of a circle does not include circles wich degenerate to lines.
  2. Three points also do not define a circle if two of them coincide. The case $A=B$ is usually excluded by the axioms, but $A'=B$ would be a problem for us, too, and this isn't covered. So you can't get a line through a point and its mirror image.
  3. There is also another critical situation where $A=A'$, i.e. $y_A=0$. You might use $B'=(x_B,-y_B)$ to still define this circle, but if both $A$ and $B$ lie on the $x$ axis you don't have a unique circle either. You can recover from this since the circle with center on the axis through these two points is still unique, but it's a special case worth considering nonetheless.

You might stand a better chance by taking mirror-image pairs of points as the “points” of your incidence geometry. Probably exclude the $x$ axis itself. Then the axioms you quoted should hold. Instead of taking pairs, you might exclude the lower half of the plane altogether; that's the same thing. Actually the geometry you get from this is essentially the Poincaré half-plane model of hyperbolic geometry, so there are several more axioms satisfied by this setup, and it's actually quite useful.