# Model of ordered plane which is neither isomorphic to $\mathbb{R}^2$ nor to Klein model

Let $$B_{\mathbb{R}}\subset\mathbb{R}\times\mathbb{R}\times\mathbb{R}$$ be standard (strict) betweenness relation on $$\mathbb{R}$$ i.e. $$B_{\mathbb{R}}(abc):\iff\left(a

My definition of ordered plane is as follows:

$$P$$ is a set (plane), $$\mathcal{L}\subset 2^P$$ is a family of lines and $$B\subset P\times P\times P$$ is a ternary betweenness relation. We say that $$(P,\mathcal{L},B)$$ is an ordered plane whenever

1. $$(P,\mathcal{L})$$ is a model of Hilbert's incidence axioms.
2. $$B(abc)$$ implies that $$a,b,c$$ are collinear.
3. For any line $$L$$: $$(L,B|_{L\times L\times L})$$ is isomorphic to $$(\mathbb{R},B_{\mathbb{R}})$$.
4. Pasch's axioms holds.

I know two standard models of this axioms i.e. $$\mathbb{R}^2$$ and Klein model and I'm looking for a model which is not isomorphic to any of these two.

Basically these axioms are a variant of Hilbert's neutral geometry axioms excluding congruence relations. Actually I think it's a bit more because every line is isomorphic to $$\mathbb{R}$$ and maybe there are models of Hilbert's axioms without congruence (but with continuity) in which at least one line is not isomorphic to $$\mathbb{R}$$ (This is something I am asking as well).

It is known that neutral geometry has exactly two models up to isomorphism but excluding congruence should make it have more models. Note that these axioms involve (Dedekind) continuity axiom and as a result there are no countable models such as "rational" plane.

• Have you tried Moulton's plane ? – Julien Narboux Apr 16 at 11:34