TITLE: When does Scott topology generated by specialization order induced by a sober space (X,$\tau$) equal the initial topology $\tau$?
QUESTION [7 upvotes]: Let X be a $T_{0}$ space. The specialization order ≤ on X is that if x is contained in cl{y}, then we call "x≤y". Obviously (X,≤) is a partially ordered set.
A sober space is a topological space such that every irreducible closed subset of X is the closure of exactly one point of X: that is, this closed subset has a unique generic point. It is not difficult to  see that specialization order induced by a sober space is a directed complete partial order (dcpo).
A subset O of a dcpo P is called Scott-open if it is an upper set and if it is inaccessible by directed joins, i.e. if all directed sets D with supremum in O have non-empty intersection with O. The Scott-open subsets of a dcpo P form a topology on P, the Scott topology. (https://en.wikipedia.org/wiki/Scott_continuity)
Let (X,$\tau$) be a sober space. The specialization order ≤ induced by a sober space makes (X,≤) a dcpo. Generally, the Scott topology on (X,≤) is finer than $\tau$. But when do these two topologies coincide with each other?
Two examples (sober c space and sober locally finitely compact space) 
are given in a book "Non-Hausdorff topology and Domain theory". (http://www.cambridge.org/ca/academic/subjects/mathematics/geometry-and-topology/non-hausdorff-topology-and-domain-theory-selected-topics-point-set-topology)
Can you find more examples like this? or Can you characterize this kind of sober space by simple property?

REPLY [4 votes]: Here is a partial answer. In P. T. Johnstone, "Stone Spaces", pp. 292, 294, it is shown that Scott topologies of continuous posets are precisely all completely distributive complete lattices. Recall that a poset $P$ is continuous iff it is a dcpo and moreover the join map $\operatorname{Ideals}(P)\to P$ has a left adjoint.
I don't know a nice characterization of spaces whose specialization posets are continuous, but for those the answer is thus that their topology coincides with the Scott topology of the specialization iff open sets form a completely distributive lattice. And moreover, any completely distributive complete lattice is the topology of a sober space which is at the same time the Scott topology of its specialization order.