Topology inducing Order We know by Munkres that any (total) ordering induces a topology and (luckily) for the real line that coincides with the euclidean topology. In fact, this construction can be carried over to any ordering (in principle even for preorders or even plain relations).
So the question arises in what extend this is still meaningful.


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*My idea -and that is my question- maybe topologies induce orderings in a fashion that is compatible to Munkres construction, that is applying both succesively gives back the ordering resp. topology.

*And, if so, would it imply that in the context of category theory orderings and topologies are basically the same. As far as I know, those concepts are very close to eachother so it would just fit inside.

*Moreover, considering nets one could turn this idea around, so that saying a topology is uniquely defined by its convergent nets would become saying a topology is given by the universal property by all chosen (convergent) functions.
 A: I don't have the answer you're looking for, but here are some obstacles to consider...
Not all topologies are orderable. Even among really "nice" spaces, there are problems. The metric topology on the circle $S^1$ is not linearly orderable. It is cyclically orderable, so maybe we should consider cyclic orders instead. However, the metric topology on the half-open interval $[0,1)$ is not cyclically orderable, even though it is linearly orderable. Already, we're having trouble finding a unified kind of order relation that reflects the topologies of one-dimensional manifolds (with boundary).
A topology can't distinguish between an order and its reverse order. So instead of considering binary linear order relations and ternary cyclic order relations, maybe we should consider ternary betweenness relations and quaternary separation relations. For example, for $a,b,c,d\in S^1$, we can define $s(a,b,c,d)$ to mean that $a$ and $c$ belong to different connected components of $S^1\setminus\{b,d\}$. However, that definition doesn't generalize very far: it's useless for totally disconnected spaces like $\mathbb Q$.
Whatever kind of order relation we try, we will need to work with the "partial" version rather than the "total" version in order to support products and coproducts.
Besides trying to find the perfect kind of order relation and the perfect kind of topological space to work with, you could probably make the task easier by asking for an adjoint to the order topology construction, rather than an inverse.
A: That is far not trivially to answer but as pointed out by @Pece finite partially ordered sets seem to induce the discrete topology which cannot uniquely give back an order. Moreover, there are in general more ways to induce a topology on a poset each of them in general not homeomorphic.
A: Alexandroff topologies are in one-to-one correspondence with preorders (see http://en.m.wikipedia.org/wiki/Alexandrov_topology)
However there is more to say for arbitrary topological spaces and this correspondence (see http://en.m.wikipedia.org/wiki/Specialization_preorder)
Besides, when dropping transitivity neighborhood systems will fail to be upwards closed within the canonical construction (proof?).
