As is well known, a metric induces a topology, therefore a metric space is always a topological space. However the reverse is not true, you can have topological spaces that do not correspond to a metric (that is, you cannot find a metric that induces that topology).
Now both topology and metrics have a certain concept of "nearness". In topological spaces it's concept of neighbourhood, but that's a relatively flexible concept (as all topological concepts are, by design). In particular, in topology it doesn't make sense to say how close a certain point is to another. On the other hand, metrics has the concept of distance (after all, that's what metric spaces are all about), which is about the most rigid concept of "nearness" you can think of: Every pair of points gets a number that tells you exactly how near or far they are to each other.
Now the question crossed my mind whether there is something in between. Say, a space $C$ based on the relation "$a$ is closer to $b$ than to $c$"; lets write that $b <_a c$. Clearly, a metric fixes that relation, by $b <_a c\iff d(a,b) < d(a,c)$. Also, that relation fixes a topology, by defining an open ball with center $c$ and border point $b$ as $B_b(c) = \{x\in C: (x <_c b)\}$ and then proceeding in the usual way. Clearly for a metric space, this is the open ball around $c$ with radius $d(b,c)$.
Note that this gives automatically a T1 space, as for any points $b,c\in C$, $b\notin B_b(c)$ and $c\notin B_c(b)$.
I think a reasonable definition for such a space would be:
A set $C$ with a ternary relation $(b <_a c)$ is a "closeness space" if the following axioms are fulfilled:
- For fixed $a\in C$, the relation $x <_a y$ is a strict weak order.
- For any $b\ne a$, $a <_a b$ (a is closer to itself than any other point).
However, does this really give anything new? That is, are there any "closeness spaces" that cannot be given a metric?
Assuming the answer to that question is "yes", are there any T1 spaces that cannot be obtained from a "closeness space"?
Edit: I misremembered the definition of Hausdorff; the condition I gave is not Hausdorff, but T1. I suspect a "closeness space" is also Hausdorff, but I'm not sure yet.
Edit 2: I just noticed that for a "closeness space" derived from a metric space the topology defined by the "comparison space" open balls is not always the same as that derived from the original metric space (the reason is that there can be metric open balls that are not closeness open balls).
For example, consider $C=\{0\}\cup (1,2)$ with the usual $d(x,y) = \left|x-y\right|$. Then in the topology derived from the metric, $\{0\}$ is an open set, but in the topology derived from the corresponding "closeness space", it is not. Indeed, the closeness space derived from that is the same as the closeness space derived from $[1,2)$ if identifying $0$ with $1$, and the topology derived from that "closeness space" is the topology derived from the distance on $[1,2)$.
Of course, in that example there still exists a metric that both induces the "closeness space" and the same topology as induced by the "closeness space", so the question whether there exists one not derivable from a metric is still open.
