What happens if you strip everything but the “between” relation in metric spaces

Given a set $X$ together with an arbitrary function $[\bullet,\bullet] : X^2 \rightarrow \mathcal{P}(X),$ let's call $A \subseteq X$ closed if and only if for all $a,b \in A$, we have that $[a,b] \subseteq A$. It's easy to show that an arbitrary intersection of closed sets is closed, or in other words the collection of closed sets forms a Moore family. This gives us a closure operator on $\mathcal{P}(X)$, which makes $(X,\mathcal{P}(X))$ into a closure system. This means that the collection of closed sets forms a complete lattice given by $$\bigwedge_{i \in I} C_i := \bigcap_{i \in I} C_i, \qquad \bigvee_{i \in I} C_i := \mathrm{cl}\left(\bigcup_{i \in I} C_i\right).$$

If we furthermore assume that $[a,a] = \{a\},$ which is true in an arbitrary metric space (though I'm not sure whether this follows from your axioms), then we can prove that singletons are closed sets. Thus we're allowed to consider the expression $\{a\} \vee \{b\}$ in the aforementioned lattice. This set clearly includes $[a,b]$, and I think your axioms probably imply that these sets are equal. Ergo the Moore family alone allows us to recover the original between-relation.

I think that if you take the first three axioms you've written down, and add in the axiom $[a,a] = \{a\},$ the resulting axiom list should provide a complete characterization of closure systems $X$ in which firstly, every singleton is closed, and secondly, for all $A \subseteq X$, we have that $A$ is closed if and only if for all $a,b \in A$, we have $\{a\} \vee \{b\} \subseteq A$. But you should check this explicitly rather than taking my word for it, because right now this is just a hunch.

I'll also remark that you might learn interesting things by looking at partially ordered sets and defining $[a,b] := \{x \in X : a \leq x \leq b\} \cup \{x \in X : b \leq x \leq a\}.$

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:

1. $[x,y]\subset[x',y]$ 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 $[x,y]\subset[x',y]$.

2. $x\in[z,y]\implies[x,x]\subset[z,y]$.

   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 $[(x),(x)]=\{(x)\}$.