Question

Show that every model of incidence geometry which consists of precisely 3 points is isomorphic to the 3-point plane. (8 points)

Answer 1

Two incidence planes are called      isomorphic    , if and only if there exists a bijection between the points of the two planes, and a bijection between the lines of the two planes such that incidence is preserved.

I-1 fails for either of two reasons. First, because I limited our dots to being on a sheet of paper, it is possible to draw two points for which any “line” through them would have to go off the sheet of paper. (This is the case, for instance, if the two points are at the extreme lower left and right corners.) Thus, not all pairs of distinct points lie on a line together. Alternatively, given two points near the center of the paper, it’s easy to visualize several distinct “lines” that pass through both. Thus, not all pairs of points determine a unique line.

I-2 holds, for given any “line” that you can draw on the paper, you can mark at least two points on it.

I-3 holds, but only for one particular arrangement of points (that one example is, however, enough for an existence statement): draw three points that lie on the same “childhood line” together. It is impossible to draw a circle that passes through all three of these, so the three points are indeed noncollinear in our interpretation. (It is true in “childhood geometry” that given any three noncollinear points, there is always a circle passing through all three - this can be proved using coordinate geometry or knowledge of construction techniques.)