# When does Scott topology generated by specialization order induced by a sober space (X,$\tau$) equal the initial topology $\tau$?

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?

• Are you familiar with the compendium? – Andrej Bauer Aug 18 '17 at 10:49
• @Andrej Bauer Yes. – Zhenchao Lyu Oct 29 '17 at 14:30
• What is the definition of an irreducible closed subset? – Wlod AA Jul 22 '18 at 4:49
• @Wlod AA An closed set is irreducible if it can not be represented as the union of its two proper closed subsets. – Zhenchao Lyu Sep 10 '18 at 9:16

One such condition is: Every closed set is irreducible (for sober spaces this implies that every closed set is the closure of a singleton $\{x\}$).
• @ZhenchaoLyu If I follow the logic skeleton of the discussion, I find that your comment is "out of the road" because, if I am correct $$(sober) \Longrightarrow DvdZ_{Condition}$$ ($DvdZ_Condition$ means Dominic van der Zypen condition). Of course it is a pure form remark, we must go into details. – Duchamp Gérard H. E. May 20 '17 at 8:45
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.