Defining Open Sets on a topological space Let $S$ be a set with continuously many elements, and $\_<\_:\_$ be a three-placed relation such that for any $x,y,z$ in $S$, $x<y:z$ iff $x$ is closer to $z$ than $y$ is.
Also, by definition, for any $x,y$ and a given $z$ in $S$, $\neg x<y:z \wedge \neg y<x:z$ if and only if $x \approx y:z$, or $x$ and $y$ are equally far from $z$. However, there is no distance definable between any two elements in S. S does not form a metric space.
For any given $a$ in $S$, the following holds for any $x,y,z$ in $S$:
(1) $\neg x<x:a$
(2) If $x<y:a$, then $\neg y<x:a$
(3) If $x<y:a$ and $y<z:a$, then $x<z:a$
In this case, what does it mean for a subset of S to be open?
Would the following work?
A subset $S^*$ of S is an open set iff $\forall x _{\in S^*}(\forall y_{\in S^*}((x = y)   
 \vee(\exists z(y<z:x))))$.
 A: Edit $1$. This assumes that for any fixed $p$, $\_\approx\_:p$ is an equivalence relation (which makes $\_<\_:p$ a strict total preorder).
Edit $2$. Slightly strengthening the definition so to allow the relation to give e.g. the trivial topologies.
Now, clearly a metric induces a relation as you defined (let's call it a proximity relation). For a metric space $(M,d)$ and for $x,y,z\in M$, let $x<_d y:z$ iff $d(z,x)<d(z,y)$. Let's view this as the prototypical example.
When a proximity relation is given by a metric, you probably want the metric topology to coincide with that of the relation. Here's a natural way of doing so:
Let $(X,<)$ be a set equipped with a proximity relation. For $p,x\in X$, define $B_x(p)=\{y\in X,\ y<x:p\}$ (this is like open balls for a metric space). Also let $p\in X$ be degenerate if $\forall x\in X, x\approx p:p$.
Then a set $S\subset X$ is open iff for every $p\in S$, $x\approx p:p\Rightarrow x\in S$ and when $p$ is non-degenerate, $\exists x\in X, \ x>p:p$ such that $B_x(p)\subset S$.
You can verify that the collection of open sets indeed forms a topology and that it coincides with the metric topology if the proximity relation is induced by a metric.
