ordered geometry
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line segment^{} \PMlinkescapephrasegenerated by \PMlinkescapephraseopposite sides \PMlinkescapephraseend points^{}
1 Definition
Let $(A,B)$ be a linear ordered geometry, where $A=(P,n,I)$ is an incidence geometry, and $B$ is a strict betweenness relation. Recall that $P$ is partitioned into disjoint sets ${P}_{0},\mathrm{\dots},{P}_{n}$, where $n$ is a positive integer.
For $$, let
${B}_{i}$  $=$  $\mathrm{\{}(p,a,q)\in {P}_{0}\times {P}_{i}\times {P}_{0}\mid p,q\text{do not lie on}a,$  
$\text{and there exists a point}r\text{lying on}a\text{such that}(p,r,q)\in B\},$ 
and
$${B}_{i}(a)=\{(p,q)\mid (p,a,q)\in {B}_{i}\}.$$ 
For any $a\in {P}_{i}$, the set is symmetric^{} and antireflexive.
We say that the hyperplane^{} $a\mathrm{\in}{P}_{i}$ is
between $p$ and $q$ if $(p,q)\in {B}_{i}(a)$.
We see that ${B}_{0}=B$.
Let’s look at the case when $i=1$. If $(p,\mathrm{\ell},q)\in {B}_{1}$ where
$\mathrm{\ell}$ is a line, then $p,q$ and $\mathrm{\ell}$ necessarily lie on a common
plane $\pi $.
The above diagram seems to suggest that $\mathrm{\ell}$ “separates $\pi $ into two regions”. However, this is not true in general without the next axiom.
An ordered geometry $(A,B)$ is a linear ordered geometry such that

S1
for any three noncollinear points $p,q,r$, and any line $\mathrm{\ell}$ lying on the same plane $\pi $ generated by $p,q,r$, if $(p,q)\in {B}_{1}(\mathrm{\ell})$ and if $r$ does not lie on $\mathrm{\ell}$, then at least one of $(q,r),(r,p)\in {B}_{1}(\mathrm{\ell})$.
In fact, in axiom S1, it can be shown that exactly one of $(q,r)$ and $(r,p)$ is in ${B}_{1}(\mathrm{\ell})$. This axiom says that “a line lying on a plane separates the plane into two mutually exclusive subsets”.
Each subset is called an (open) half plane of the line.
A closed half plane is just the union of one of its open half planes and the line itself.
Suppose points $p,q$ and line $\mathrm{\ell}$ lie on plane $\pi $ and that $\mathrm{\ell}$ is between $p$ and $q$. Then we say that $p$ and $q$ are on the opposite sides of line $\mathrm{\ell}$. Two points are on the same side of line $\mathrm{\ell}$ if they are not on the opposite sides of $\mathrm{\ell}$. If $r$ is a third point (distinct from $p,q$) that lies on $\pi $ and not on $\mathrm{\ell}$, then according to axiom S1 above, $r$ must be on the same side of either $p$ or $q$ (but not both!). Same sidedness is an equivalence relation^{} on points of $A$.
3
An equivalent^{} characterization of axiom S1 is in the form of Pasch’s theorem.
The ten conditions or axioms (seven betweenness, two collinearity,
and one “separation^{}” axioms) are sometimes called the “order axioms”
of $A$.
It is customary, in an ordered geometry, to identify each element of
$P$ by its shadow (http://planetmath.org/IncidenceGeometry) (a subset of ${P}_{0}$), and we shall do so in this discussion. A line, for example, will then consist of points that are incident^{} with it, as opposed to an abstract element of ${P}_{1}$.
Hence, we shall also confuse the notation $\overleftrightarrow{pq}$ with ${I}_{0}(\overleftrightarrow{pq})$.
2 Remarks

•
Law of Trichotomy on a strict betweenness relation: Let $B$ be a strict betweenness relation. If $p,q,r$ are collinear, then exactly one of $(p,q,r)$, $(q,r,p)$, or $(r,p,q)\in B$.

•
In an ordered geometry, one can define familiar concepts, such as a line segment, a ray, even an angle, using the order axioms above. For example, ${B}_{p*q}$ is called the open line segment between $p$ and $q$, and is more commonly denoted by $\overline{pq}$, or $(p,q)$. A closed line segment between $p$ and $q$ is just $\{p\}\cup \overline{pq}\cup \{q\}$, denoted by $[p,q]$. From the third remark under betweenness relation, $\overline{pq}={B}_{p*q}={B}_{q*p}=\overline{qp}$. The points $p$ and $q$ are called the end points of $\overline{pq}$.

•
A ray is defined to be ${B}_{pq}$. For a more detailed discussion, see the entry on ray (http://planetmath.org/Ray).

•
$\overleftrightarrow{pq}=B(p,q)$.

•
$\overline{pq}\subset \overleftrightarrow{pq}$. The inclusion is strict, since there exists a point $r$ such that $(p,q,r)\in B$ by order axiom S1. $r$ lies on the $\overleftrightarrow{pq}$ and is clearly distinct from both $p$ and $q$.

•
Any line segment $\overline{pq}$ in an ordered geometry, in to being orderable, is linearly orderable, thanks to the Law of Trichotomy.

•
It fact, $\le $, defined on a line segment, can be extended to a linear order defined on the line that includes the segment (see the last remark above on betweenness relation). This shows that every line in an ordered geometry can be linearly ordered.
References
 1 D. Hilbert, Foundations of Geometry, Open Court Publishing Co. (1971)
 2 K. Borsuk and W. Szmielew, Foundations of Geometry, NorthHolland Publishing Co. Amsterdam (1960)
 3 R. Hartshorne, Geometry^{}: Euclid and Beyond, Springer (2000)
Title  ordered geometry 
Canonical name  OrderedGeometry 
Date of creation  20130322 15:28:21 
Last modified on  20130322 15:28:21 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  43 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 51G05 
Synonym  open interval 
Synonym  closed interval 
Synonym  interval 
Related topic  PaschsTheorem 
Defines  half plane 
Defines  side of line 
Defines  open line segment 
Defines  closed line segment 
Defines  opposite sides 
Defines  open half plane 
Defines  closed half plane 
Defines  end points 
Defines  open line segment 
Defines  closed line segment 