Editing Specialization (pre)order (section)

== Topologies on orders ==

The specialization order yields a tool to obtain a preorder from every topology. It is natural to ask for the converse too: Is every preorder obtained as a specialization preorder of some topology?

Indeed, the answer to this question is positive and there are in general many topologies on a set ''X'' that induce a given order ≤ as their specialization order. The [[Alexandroff topology]] of the order ≤ plays a special role: it is the finest topology that induces ≤. The other extreme, the coarsest topology that induces ≤, is the [[upper topology]], the least topology within which all complements of sets ↓''x'' (for some ''x'' in ''X'') are open.

There are also interesting topologies in between these two extremes. The finest sober topology that is order consistent in the above sense for a given order ≤ is the [[Scott topology]]. The upper topology however is still the coarsest sober order-consistent topology. In fact, its open sets are even inaccessible by ''any'' suprema. Hence any [[sober space]] with specialization order ≤ is finer than the upper topology and coarser than the Scott topology. Yet, such a space may fail to exist, that is, there exist partial orders for which there is no sober order-consistent topology. Especially, the Scott topology is not necessarily sober.