A Counterexample to the Generalized Ho-Zhao Problem

In this paper we find the answer to the open question in (Ho & Zhao, 2009), which states that we do not know whether the isomorphism of complete lattices C(P) and C(Q) implies that of the dcpo’s P and Q, where C(P) and C(Q) are the lattices of all Scott closed subsets of P and Q respectively. We proved that is not necessarily satisfied in general case.


Introducing the Problem
This paper depends on the work of (Ho & Zhao, 2009) about the nature of the order relation in the lattice of Scott-closed sets over semi-lattice.They mentioned at end of their paper that we still do not know whether the isomorphism of complete lattices () and () implies that of the dcpo's  and , so further work must be done to achieve a better understanding of the lattices of Scott-closed sets.

What is The Question?
The remained question is: Can one prove or deny the statement: () ≅ () implays  ≅  for two arbitrary directed complete partly ordered sets  and .

What We Are Proving in This Paper?
In this paper, we prove that it's not necessarily satisfied in general case, through defining two dcpo  and Ψ such that Υ ≇ Ψ and (Υ) ≅ (Ψ).

Method
At first, we give some preliminaries on directed complete partly ordered sets.

Definition
A nonempty subset  of a poset is said to be directed if any two elements in  have an upper bound in .See (Kelley, 1975, p 81) A poset  in which every directed subset  has a supremum (donate by ⋁ ) is called a directed-complete partial order, or dcpo for short.See (Abramsky & Jung, 1995, p 14)
A subset  of a poset  is said to be upper if  =↑  and said to be lower if  =↓ .See (Gierz, et al., 2003)

Definition
is upper set ii.
∈  implies   for all directed sets  ⊆ .
The collection of all Scott-open subsets of  is called the Scott topology of  and will be denoted by ().
The complement of a Scott-open set is called Scott-closed, The collection of all Scott-closed subsets of  will be denoted by ().
One can prove that a subset  ⊆  is Scott-closed if and only if the following two conditions are satisfied: i.
is lower set ii.
For any directed set  ⊆ , If  has a supremum ⋁  then ⋁  ∈ .
The upper subsets in Υ are the closed intervals ,, 1and the half opened intervals -, 1-, and every subset of Υ is directed because ≤ is a total order relation (for every two elements, one must be upper bound of the other).
Since ≤ is a total order relation, ever (directed) subset of Υ has an upper bound, so Υ is directed complete poset (dcpo).Now let us characterize the Scott-open sets of Υ, the first condition is to be upper set.
For every upper set of the form ,, 1with 0 <  < 1, we have the directed set  = ,0, , that does not have any intersection with it.
But the supremum of  is ⋁  =  ∈ ,, 1-, so those upper sets of the form ,, 1are not Scott-open, because they don't satisfy the second condition.
On other hand, for every upper set of the form  =-, 1-, every subset  satisfies that:  doesn't have any intersection with , will have a supremum ⋁  ≤ , that means ⋁  ∉ , so  is Scott-open.
On the other hand, for the upper sets of the form   ′ = *,0, -:  < +, the supremum of any set  where  has no intersection with   ′ , is ⋁  ⊆ ,0, -, this means ⋁  ∉   ′ .
Therefore, the sets of the form   ′ are Scott-open.

Conclusions
The counterexample we provided in this paper gives the answer to the open question in (Ho & Zhao, 2009).Thus in general case, we know now that the isomorphism of complete lattices () and () doesn't imply that of the dcpo's  and  , where () and () are the lattices of all Scott closed subsets of  and  respectively.