8
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ Are you familiar with the compendium? $\endgroup$ Commented Aug 18, 2017 at 10:49
  • 2
    $\begingroup$ @Andrej Bauer Yes. $\endgroup$ Commented Oct 29, 2017 at 14:30
  • $\begingroup$ What is the definition of an irreducible closed subset? $\endgroup$
    – Wlod AA
    Commented Jul 22, 2018 at 4:49
  • 1
    $\begingroup$ @Wlod AA An closed set is irreducible if it can not be represented as the union of its two proper closed subsets. $\endgroup$ Commented Sep 10, 2018 at 9:16

2 Answers 2

4
$\begingroup$

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.

$\endgroup$
0
0
$\begingroup$

One such condition is: Every closed set is irreducible (for sober spaces this implies that every closed set is the closure of a singleton $\{x\}$).

$\endgroup$
3
  • 4
    $\begingroup$ Thanks for your answer. The specialization order of your condition is a chain. It's kind of trivial for me. $\endgroup$ Commented Oct 22, 2016 at 13:09
  • $\begingroup$ @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. $\endgroup$ Commented May 20, 2017 at 8:45
  • 1
    $\begingroup$ @DuchampGérardH.E. Every Hausdorff space is sober, but the only irreducible closed sets in it are singletons. $\endgroup$ Commented May 20, 2017 at 11:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.