TITLE: What happens if you strip everything but the “between” relation in metric spaces
QUESTION [49 upvotes]: Given a metric space $(X,d)$ and three points $x,y,z$ in $X$, say that $y$ is between $x$ and $z$ if $d(x,z) = d(x,y) + d(y,z)$, and write $[x,z]$ for the set of points between $x$ and $z$.
Obviously, we have

$x,z\in[x,z]$;
$[x,z]=[z,x]$;
$y \in [x,z]$ implies $[x,y] \subseteq [x,z]$;
$w,y \in [x,z]$ implies: $w\in [x,y]$ iff $y \in [w,z]$.

My question:
Has the family of objects with such an axiomatic “interval structure” $[\bullet,\bullet]:X \times X \to \mathcal{P}(X)$ satisfying approximately the above conditions been studied somewhere?

REPLY [9 votes]: This is more of an extended remark than a full answer.
My message is:
The property $[x,x]=\{x\}$ does not necessarily hold, but you can always make it hold by taking a natural quotient.
Your axioms don't imply $[x,x]=\{x\}$, so you have to add it if you want it.
A pseudometric (e.g. a seminorm) will satisfy your axioms, and $[x,x]$ is the set of points distance zero from $x$, and this set can have any size.
In a pseudometric space you can take the quotient by the relation $x\sim y\iff d(x,y)=0$ and you get a metric space.
It turns out the same works with the interval systems.
We can define a relation $\sim$ on $X$ so that $x\sim y\iff x\in[y,y]$.
This is an equivalence relation:

Reflexivity:
Axiom 1.
Symmetry:
By axioms 1 and 2 $x,y\in[y,x]$.
Thus axiom 4 says $x\in[y,y]\iff y\in[x,x]$.
Transitivity:
Suppose $x\sim y\sim z$.
By axiom 3 $x\sim y$ and $y\sim z$ imply $[y,x]\subset [y,y]$ and $[z,y]\subset [z,z]$.
By axioms 2 and 3 $[y,y]\subset[z,y]$.
Thus $x\in[y,x]\subset[y,y]\subset[z,y]\subset[z,z]$.

Then we can take the quotient space $X/{\sim}$.
To avoid confusion, let's denote by $(x)$ the quotient set of $x\in X$.
(In fact, $(x)=[x,x]$, but I find it cleaner to keep the separate notation.)
Then declaring $(x)\in[(y),(z)]\iff x\in[y,z]$ defines an interval structure on $X/{\sim}$.
To see this, we need to check two things:

$[x,y]\subset[x',y]$ when $x\sim x'$.
(The reverse inclusion and varying both $x$ and $y$ follow easily.)
Proof: Since $x\in[x',x']\subset[x',y]$, we have indeed $[x,y]\subset[x',y]$.
$x\in[z,y]\implies[x,x]\subset[z,y]$.
Proof: Since $x\in[z,y]$ and $x\in[x,y]$, applying axiom 3 twice gives $[x,x]\subset[x,y]\subset[z,y]$.

In the quotient structure we evidently have $[(x),(x)]=\{(x)\}$.