# Model of ordered plane with the negation of Pasch's axiom

I am interested in finding the model of particular set of geometry axioms in which Pasch's axiom fails. First I'll give the definitions.

By ordered line I mean the set $$L$$ (line) with one three-argument relation $$B$$ whenever the following axioms are satisfied:

1. There exist $$a,b$$ such that $$a\neq b$$
2. If $$a\neq b, b\neq c, a\neq c$$, then $$B(abc)\vee B(bac)\vee B(acb)$$
3. If $$B(abc)$$, then $$a\neq c$$
4. If $$B(abc)$$, then $$B(cba)$$
5. If $$B(abc)\wedge B(acd)$$, then $$B(bcd)$$
6. If $$a\neq b$$, then there exists $$c$$ such that $$B(abc)$$
7. If $$a\neq c$$, then there exists $$b$$ such that $$B(abc)$$

I'll also give Hilbert's plane axioms of incidence: We consider a set $$P$$ (plane) and a family $$\mathcal{L}$$ of subsets of $$P$$ (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

Question. What I am looking for is the model $$(P,\mathcal{L},B)$$ such that

1. $$(P,\mathcal{L})$$ is the model of Hilbert's plane incidence axioms
2. For any line $$L\in\mathcal{L}$$, $$(L,B|_L)$$ is an ordered line
3. Whenever $$B(abc)$$, then $$a,b,c$$ are collinear
4. Pasch's axiom fails

I am aware of these two questions

A model of geometry with the negation of Pasch’s axiom?

Are there simple models of Euclid's postulates that violate Pasch's theorem or Pasch's axiom?

and the model constructed using the solutions of Cauchy's equation is also the model for my problem, but I wonder whether we can construct the model for my problem without axiom of choice (note that I don't require congruence, continuity and parallel axioms to hold)

• By the way, often people drop axioms 5 and 7 of your ordered line, since they can be deduced from Pasch's axiom. (For example, see Hilbert's axioms on Wikipedia.) In that case, if we also drop Pasch's axiom, we get weirder models, including even some finite geometries. But your axiomatization of an ordered line, I think, guarantees that the line contains infinitely many points. – Misha Lavrov Oct 30 '18 at 2:15

## 1 Answer

One possible example is coordinate geometry over the dyadic rationals: numbers of the form $$\frac{a}{2^b}$$,where $$a,b \in \mathbb Z$$.

We give this the inherited structure of the usual geometry on $$\mathbb R^2$$, but throw away all points whose $$x$$- and $$y$$-coordinates don't have this form, and throw away all lines not containing at least two of the remaining points.

Since all of your axioms were satisfied before we threw away points, the only ones we need to check are the axioms that claim something exists. (We need to check we didn't throw it away.) So:

1. There is still a line through any two points. We only threw away lines that only contained at most one dyadic rational point, such as the line $$y = \pi x$$.
2. Any line contains at least two points. If it didn't, then we threw it away!
3. If $$P,Q$$ are distinct points, there is a point $$R$$ such that $$B(P,Q,R)$$. If $$P = (x_1,y_1)$$ and $$Q = (x_2,y_2)$$, then we can take $$R = (2x_2 - x_1, 2y_2 - y_1)$$.
4. If $$P,Q$$ are distinct points, there is a point $$R$$ such that $$B(P,R,Q)$$. If $$P=(x_1,y_1)$$ and $$Q = (x_2,y_2)$$, then we can take $$R = (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$$.

Finally, Pasch's axiom is violated because some lines no longer have the intersection points they should. For example, if we take the triangle with vertices $$A = (1,0)$$, $$B = (-1,0)$$, and $$C = (0,1)$$, then the line $$y=2x$$ intersects side $$AB$$ at $$(0,0)$$ but doesn't intersect $$AC$$ or $$BC$$. (It used to intersect $$AC$$ at $$(\frac13,\frac23)$$, but then we threw away that point.)

The other common axioms this violates are:

• Completeness: we can extend this plane to a bigger one that contains it by returning to $$\mathbb R^2$$. (We don't even have any of the circle intersection properties, which are weaker than completeness.)
• The parallel postulate: lots of lines become parallel because we threw away their intersection points.
• Segment copying: we can't, for example, copy a segment of length $$1$$ onto the line $$y=x$$.

Also, depending on how some of your angle axioms are stated, they might be iffy here since this geometry doesn't have plane separation. (This will always be an issue, though, since plane separation is equivalent to Pasch's axiom!)

Another kind of example is the missing strip plane. Here, we also throw away some points: we start with $$\mathbb R^2$$ and throw away the points $$(x,y) : 0, along with vertical lines with equation $$x=c$$ where $$0. We modify the definition of the length of a line segment: if a line segment $$AB$$ lies on a line with slope $$m$$, and $$A$$ and $$B$$ are on opposite sides of the missing strip, then we decrease the length of $$AB$$ by $$\sqrt{1+m^2}$$ from the usual length.

This plane actually satisfies some notions of completeness: for example, it satisfies the line completeness axiom, since every line in the missing strip plane looks exactly like a line in the usual Cartesian plane, on its own.

Apart from Pasch's axiom (which is violated because we can draw a diagram where an intersection point ought to be in the missing strip), it only violates the parallel postulate (for similar reasons) and SAS triangle congruence (which Hilbert takes as an axiom).