First I'll give some definitions.

Hilbert's plane axioms of incidence: We consider a set $P$ (plane) and a family $\mathcal{L}$ (a family of lines) with axioms:

  1. For any two distinct points $a,b$ there exists exactly one line $L$ such that $a,b\in L$.
  2. For any line $L$ there exist two distinct points $a,b$ such that $a,b\in L$.
  3. There exist three distinct points not lying on one line.

Next let $B_{\mathbb{R}}\subset\mathbb{R}\times\mathbb{R}\times\mathbb{R}$ be standard (strict) betweenness relation on $\mathbb{R}$ i.e. $$B_{\mathbb{R}}(abc):\iff\left(a<b<c \vee c<b<a\right)$$

My definition of affine plane is as follows:

We say that $(P,\mathcal{L},B)$ is an affine plane whenever

  1. $(P,\mathcal{L})$ is a model of 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.
  5. For any line $L$ and point $p\notin L$ there is exactly one line parallel to $L$ passing through $p$.


Is this definition standard or equivalent to standard definitions of affine plane in terms of lines and ternary betweenness relation? And what is more important to me: Is this axiomatization categorical?

I had an idea to prove that my affine plane is isomorphic to $\mathbb{R}^2$. The idea was to pick two intersecting lines $L_1,L_2$ and isomorphisms $\xi_1:L_1\rightarrow\mathbb{R},\xi_2:L_2\rightarrow\mathbb{R}$ and to map any point $p$ on a plane to $(\xi_1(\pi_1(p)),\xi_2(\pi_2(p)))\in\mathbb{R}^2$ where $\pi_1,\pi_2$ are parallel projections to $L_1,L_2$. I think I know how to prove that this mapping is a bijection, yet I don't know how to prove it preserves betweenness. Not even sure it does.

  • $\begingroup$ The set of planes is not used in your axioms of incidence. $\endgroup$ – Julien Narboux Mar 7 at 17:06
  • $\begingroup$ Which version of Pasch do you assume ? is it a version which implies that the dimension of the space is two ? Otherwise, I don't see in your list any axiom forcing the dimension 2. $\endgroup$ – Julien Narboux Mar 7 at 17:09
  • $\begingroup$ These incidence axioms don't involve the set of planes, just one plane $P$ which is actually the whole (2-dimensional) space. And yes, it is exactly Pasch axiom which forces the geometry to be 2-dimensional. $\endgroup$ – Kulisty Mar 7 at 20:05

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