Abstract
In this paper, the concepts of weak quasi-hypercontinuous posets and weak generalized finitely regular relations are introduced. The main results are: (1) when a binary relation ρ : X ⇀ Y satisfies a certain condition, ρ is weak generalized finitely regular if and only if (φρ(X, Y), ⊆) is a weak quasi-hypercontinuous poset if and only if the interval topology on (φρ(X, Y), ⊆) is split T2; (2) the relation ≰ on a poset P is weak generalized finitely regular if and only if P is a weak quasi-hypercontinuous poset if and only if the interval topology on P is split T2.
1 Introduction
In domain theory, the interval topology and the Lawson topology are two important "two-sided" topologies on posets. A basic problem (see [1, 2, 3, 4, 5]) is: When do the interval topology and the Lawson topology have T2 properties? In [5] (see also [3, 4]), Gierz and Lawson have discussed this problem for the Lawson topology, and proved that a complete lattice is a quasicontinuous lattice if and only if the Lawson topology is T2. However, T2 properties for the interval topology on posets have attracted a considerable deal of attention (see [6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18]). Especially, Erné [1] obtained several equivalent characterizations about T2 properties of the interval topology on posets. For a complete lattice L, Gierz and Lawson [5] proved that the interval topology on L is T2 if and only if L is a generalized bicontinuous lattice.
The regularity of binary relations was first characterized by Zareckiǐ [18]. In [18] he proved the following remarkable result: a binary relation ρ on a set X is regular if and only if the complete lattice (Φρ(X), ⊆) is completely distributive, where Φρ(X) = {ρ(A) : A ⊆ X}, ρ(A) = {y ∈ X : ∃a ∈ A with (a, y) ∈ ρ}. Further criteria for regularity were given by Markowsky [19] and Schein [20] (see also [21] and [22]). Motivated by the fundamental works relative Zareckiǐ on regular relations, Xu and Liu [23] introduced the concepts of finitely regular relations and generalized finitely regular relations, respectively. It is proved that a relation ρ is generalized finitely regular if and only if the interval topology on (Φρ(X), ⊆) is T2. Especially, in complete lattices, this condition turns out to be equivalent both to the T2 interval topology and to the quasi-hypercontinuous lattices.
In this paper, we mainly concentrate on the T2 interval topology of posets by using the regularity of binary relations. Therefore, we introduce the concepts of the split T2 interval topology on posets and weak generalized finitely regular relations. Meanwhile, in order to characterize split T2 interval topology of posets by a order structure, like the equivalence of the T2 interval topology and quasi-hypercontinuous lattices in [23], we give the notion of a weak quasi-hypercontinuous poset. It is proved that when a binary relation ρ : X ⇀ Y satisfies property M, ρ is weak generalized finitely regular if and only if (φρ(X, Y), ⊆) is a weak quasi-hypercontinuous poset if and only if the interval topology on (φρ(X, Y), ⊆) is split T2, where φρ(X, Y) = {ρ(x): x ∈ X}. For a poset P, the relation ≰ on P is weak generalized finitely regular if and only if P is a weak quasi-hypercontinuous poset if and only if the interval topology on P is split T2, which generalizes the corresponding works in [12, 16, 17].
2 Preliminaries
In this section, we recall some basic concepts needed in this paper; other non-explicitly stated elementary notions please refer to [4, 23, 24].
Let P be a poset. For all x ∈ P, A ⊆ P, let ↑ x = {y ∈ P : x ≤ y} and ↑ A = ⋃a∈A ↑ a; ↓ x and ↓ A are defined dually. A↑ and A↓ denote the sets of all upper and lower bounds of A, respectively. Let Aδ = (A↑)↓ and δ(P) = {Aδ : A ⊆ P}. To avoid ambiguities, we also denote A↑, A↓ and Aδ on P by
For two sets X and Y, we call ρ : X ⇀ Y a binary relation if ρ ⊆ X × Y. When X = Y, ρ is usually called a binary relation on X.
Definition 2.1
Let ρ : X ⇀ Y, τ : Y ⇀ Z be two binary relations. Define
τ ∘ ρ = {(x, z) : ∃y ∈ Y, (x, y) ∈ ρ, (y, z) ∈ τ}. The relation τ ∘ ρ : X ⇀ Z is called the composition of ρ and τ.
ρ–1 = {(y, x) ∈ Y × X : (x, y) ∈ ρ}.
ρc = X × Y ∖ ρ.
ρ(A) = {y ∈ Y : ∃x ∈ A with (x, y) ∈ ρ}, we call it the image of A under a binary relation ρ. Instead of ρ({x}), we write ρ(x) for short.
Φρ(X, Y) = {ρ(A) : A ⊆ X}.
φρ(X, Y) = {ρ(x) : x ∈ X}.
φy = {ρ(u) ∈ φρ(X, Y) : y ∉ ρ(u)}.
Clearly, φρ(X, Y) ⊆ Φρ(X, Y), and (Φρ(X, Y), ⊆) is a complete lattice in which the join operation ∨ is the set union operator ∪. But in general (φρ(X, Y), ⊆) is not a complete lattice. For example, let X = {x1, x2, x3} and Y = {y1, y2, y3}. Define a relation ρ = {(x1, y1), (x1, y2), (x2, y2), (x2, y3), (x3, y1), (x3, y3)}. Then ρ(x1) = {y1, y2}, ρ(x2) = {y2, y3} and ρ(x3) = {y1, y3}. It is easy to see that there is no lease upper bound of ρ(x1), ρ(x2) in (φρ(X, Y), ⊆).
Definition 2.2
[12] Let P be a poset and x ∈ P, A ⊆ P.
Define a relation ≺ on P by A ≺P x iff x ∈ intυ(P) ↑ A. Without causing confusion, we write A ≺ x for short.
P is called quasi-hypercontinuous if for all x ∈ P, ↑ x = ⋂ {↑ F : F is finite and F ≺ x} and {F ∈ P(<ω) : F ≺ x} is directed.
A complete lattice which is quasi-hypercontinuous as a poset is called a quasi-hypercontinuous lattice (see [12]). In [12], it has been proved that L is a quasi-hypercontinuous lattice if for all x ∈ L, and U ∈ υ(L) with x ∈ U, there exists F ∈ L(<ω) such that x ∈ intυ(L) ↑ F ⊆ ↑ F ⊆ U.
Definition 2.3
[12] A binary relation ρ : X ⇀ Y is called generalized finitely regular, ∀(x, y) ∈ ρ, ∃{u1, u2, …, un} ∈ X(<ω) and {v1, v2, …, vm} ∈ Y(<ω) such that
(ui, y) ∈ ρ, (x, vj) ∈ ρ (i = 1, 2, …, n; j = 1, 2, …, m), and
∀ {s1, s2, …, sm} ∈ X(<ω), {t1, t2, …, tn} ∈ Y(<ω), if (ui, ti) ∈ ρ (i = 1, 2, …, n), (sj, vj) ∈ ρ (j = 1, 2, …, m), then ∃ (l, k) ∈ {1, 2, …, m} × {1, 2, …, n} such that (sl, tk) ∈ ρ.
Theorem 2.4
[12] Let ρ : X ⇀ Y be a binary relation. Then the following conditions are equialent:
ρ is generalized finitely regular;
(Φρ(X, Y), ⊆) is a quasi-hypercontinuous lattice.
Definition 2.5
[12] Let τ and δ be two topologies on a poset P. α = τ ∨ δ is called split T2 or split Hausdorff about τ and δ, if for any x, y with x ≰ y, there exists (U, V) ∈ τ × δ such that x ∈ U, y ∈ V with U ∩ V = ∅. We call it split T2 internal topology on a poset P, if the internal topology θ(P) is split T2 about υ(P) and ω(P).
In [12, 24], it is pointed that split T2 is strictly stronger than T2 property.
3 Weak generalized finitely regular relations
In this section, we consider the split T2 interval topology of posets by using the regularity of binary relations, and obtain that the relation ≰ on a poset P is weak generalized finitely regular if and only if P is a weak quasi-hypercontinuous poset if and only if the interval topology on P is split T2.
Definition 3.1
A poset P is called weak quasi-hypercontinuous, if ↑ x = ⋂ {↑ F : F ∈ P(<ω), F ≺ x} for all x ∈ P.
In contrast to quasi-hypercontinuous posets, a weak quasi-hypercontinuous poset need not be the case that the set {F ∈ P(<ω) : F ≺ x} is directed. Clearly, P is a quasi-hypercontinuous poset ⇒ P is weak quasi-hypercontinuous, and If P is a sup-semilattice, then they are equivalent.
Definition 3.2
A binary relation ρ : X ⇀ Y is called weak generalized finitely regular, w-generalized finitely regular for short, if for any (x, y) ∈ ρ, there are {u1, u2, …, un} ∈ X(<ω) and {v1, v2, …, vm} ∈ Y(<ω) such that
(ui, y) ∈ ρ, (x, vj) ∈ ρ(i = 1, 2, …, n; j = 1, 2, …, m), and
∀s ∈ X, {t1, t2, …, tn} ⊆ Y, if (ui, ti) ∈ ρ (i = 1, 2, …, n), (s, vj) ∈ ρ (j = 1, 2, …, m), then there is a k ∈ {1, 2, …, m} such that (s, tk) ∈ ρ.
Obviously, if ρ is generalized finitely regular, then ρ is w-generalized finitely regular.
Proposition 3.3
For a binary relation ρ : X ⇀ Y, the following conditions are equivalent:
ρ is w-generalized finitely regular;
∀(x, y) ∈ ρ, ∃(U, V) ∈ X(<ω) × Y(<ω) such that
U ⊆ ρ–1(y), V ⊆ ρ(x);
∀(s, T) ∈ X × Y(<ω), if U ⊆ ρ–1(T) and V ⊆ ρ(s), then T ∩ ρ(s) ≠ ∅.
Proof
(1) ⇒ (2) For any (x, y) ∈ ρ, since ρ is w-generalized finitely regular, ∃{u1, u2, …, un} ∈ X(<ω) and {v1, v2, …, vm} ∈ Y(<ω) such that
(ui, y) ∈ ρ, (x, vj) ∈ ρ(i = 1, 2, …, n; j = 1, 2, …, m), and
∀s ∈ X, {t1, t2, …, tn} ⊆ Y, if (ui, ti) ∈ ρ (i = 1, 2, …, n), (s, vj) ∈ ρ (j = 1, 2, …, m), then ∃k ∈ {1, 2, …, m} such that (s, tk) ∈ ρ.
Let U = {u1, u2, …, un}, V = {v1, v2, …, vm}. Then (U, V) ∈ X(<ω) × Y(<ω). By the condition (a), we have that U ⊆ ρ–1(y), V ⊆ ρ(x), i.e., the condition (i) in (2) is satisfied. Now we check the condition (ii) in (2). ∀(s, T) ∈ X × Y(<ω), if U ⊆ ρ–1(T) and V ⊆ ρ(s), then ∀ i ∈ {1, 2, …, n}, ∃ti ∈ T such that (ui, ti) ∈ ρ, and ∀ j ∈ {1, 2, …, m}, (s, vj) ∈ ρ, by the condition (b), ∃k ∈ {1, 2, …, m} such that (s, tk) ∈ ρ. Thus T ∩ ρ(s) ≠ ∅.
(2) ⇒ (1) Let (x, y) ∈ ρ. By (2), ∃ (U, V) ∈ X(<ω) × Y(<ω) such that
U ⊆ ρ–1(y), V ⊆ ρ(x), and
∀(s, T) ∈ X × Y(<ω), if U ⊆ ρ–1(T) and V ⊆ ρ(s), then T ∩ ρ(s) ≠ ∅.
Let U = {u1, u2, …, un}, V = {v1, v2, …, vm}. Then by condition (i), we have that (ui, y) ∈ ρ, (x, vj) ∈ ρ (i = 1, 2, …, n; j = 1, 2, …, m). For any s ∈ X, {t1, t2, …, tn} ⊆ Y, if (ui, ti) ∈ ρ (i = 1, 2, …, n), (s, vj) ∈ ρ (j = 1, 2, …, m), let T = {t1, t2, …, tn}. Then U ⊆ ρ–1(T) and V ⊆ ρ(s). By the condition (ii), T ∩ ρ(s) ≠ ∅, i.e., ∃k ∈ {1, 2, …, m} such that (s, tk) ∈ ρ. Thus (1) holds.□
Definition 3.4
Let ρ : X ⇀ Y be a binary relation. We call ρ satisfies property M if for any y ∈ Y, φy = ∅ or φy has the greatest element, where φy = {ρ(u) ∈ φρ(X, Y) : y ∉ρ(u)}.
Example 3.5
Let E be a binary relation on a set X with reflexive and transitive. Then the relation Ec = X2 ∖ E satisfies property M.
In fact, for any y ∈ X, since E is reflexive, y ∉ Ec(y). Thus φy ≠ ∅. Let u ∈ X with y ∉ Ec(u), i.e., (u, y) ∈ E. Suppose that Ec(u) ⊈ Ec(y), then there is a t ∈ Ec(u) such that t ∉ Ec(y), i.e., (u, t) ∉ E and (y, t) ∈ E, we have (u, t) ∈ E since E is transitive, which contradicts (u, t) ∉ E. Thus Ec(y) is the greatest element of φy. Hence, the relation Ec satisfies property M.
Let X be a set and Y = {y}. Define a function f : X → Y by f(x) = y for any x ∈ X. Then f satisfies property M, since φy = ∅ for any y ∈ Y.
Let X, Y be two sets and g : X → Y a injective function. If |X| > 2, then g is not satisfy property M, since for any x1, x2 ∈ X, g(x1) ⊈ g(x2).
For any poset P, the relation ≤ on P is reflexive and transitive, by Example 3.5(1), we have the following corollary.
Corollary 3.6
For any poset P, the relation ≰ on P satisfies property M.
Lemma 3.7
Let ρ : X ⇀ Y be a binary relation. If ρ satisfies property M, then δ((φρ(X, Y), ⊆)) is order isomorphism to (Φρ(X, Y), ⊆).
Proof
For any A ⊆ X, define η : δ((φρ(X, Y), ⊆)) → (Φρ(X, Y), ⊆) by η(
1∘ η is order preserving. Let
2∘ ψ is order preserving. Let ρ(A) ⊆ ρ(B). We only have to show
Obviously, η ∘ ψ = id(Φρ(X,Y),⊆) and ψ ∘ η = idδ((φρ(X,Y),⊆)). All there show that δ((φρ(X, Y), ⊆)) ≅ (Φρ(X, Y), ⊆).□
From the Lemma 3.7, we can see that if ρ satisfies property M, then (Φρ(X, Y), ⊆) is the normal completion of (φρ(X, Y), ⊆).
Definition 3.8
[24] A poset P is called S-poset, if for any F, G ∈ P(<ω) ∖ {∅}, F ⊆ G↓, there exists u ∈ P such that F ⊆ ↓ u ⊆ G↓.
Lemma 3.9
[24] Let P be a sup-semilattice (inf-semilattice). Then P is an S-poset.
Lemma 3.10
Let ρ : X ⇀ Y be a relation with property M. 𝓕 ∈ φρ(X, Y)(<ω) and ρ(x) ∈ φρ(X, Y). Consider the following conditions.
𝓕 ≺Φρ(X,Y) ρ(x);
𝓕 ≺φρ(X,Y)ρ(x).
Then (1) ⇒ (2). If φρ(X, Y) is an S-poset, then they are equivalent.
Proof
(1) ⇒ (2) Let 𝓕 ≺Φρ(X,Y)ρ(x). Then there exist ρ(A1), ρ(A2), …, ρ(Am) ⊆ Φρ(X, Y) such that ρ(x) ∈ Φρ(X, Y) ∖ ↓Φρ(X,Y) {ρ(A1), ρ(A2), …, ρ(Am)} ⊆ ↑Φρ(X,Y) 𝓕. Thus for any j ∈ {1, 2, …, m}, ρ(x) ⊈ ρ(Aj), and thus there is a zj ∈ ρ(x) with zj ∉ ρ(Aj). Obviously, φzj ≠ ∅. Let Nj be the greatest of φzj. Then ρ(x) ⊈ Nj for any j ∈ {1, 2, …, m}, and thus ρ(x) ∈ φρ(X) ∖ ↓φρ(X){N1, N2, …, Nj}. Now we show that φρ(X, Y) ∖ ↓φρ(X,Y){N1, N2, …, Nj} ⊆ ↑φρ(X,Y) 𝓕. Let ρ(w) ∈ φρ(X, Y) ∖ ↓φρ(X,Y){N1, N2, …, Nj}. Then for any j ∈ {1, 2, …, m}, ρ(w) ⊈ Nj. By the definition of Nj, we have zj ∈ ρ(w), and thus ρ(w) ⊈ρ(Aj) (since zj ∉ρ(Aj)). Hence ρ(w) ∈ Φρ(X, Y) ∖ ↓Φρ(X,Y) {ρ(A1), ρ(A2), …, ρ(Am)}, it follows that ρ(w) ∈ ↑φρ(X,Y) 𝓕. Hence 𝓕 ≺φρ(X,Y)ρ(x).
(2) ⇒ (1) Suppose that 𝓕 ≺φρ(X,Y)ρ(x), then there exists {ρ(y1), ρ(y2), …, ρ(ym)} ⊆ φρ(X, Y) such that ρ(x) ∈ φρ(X, Y) ∖ ↓φρ(X,Y) {ρ(y1), ρ(y2), …, ρ(ym)} ⊆ ↑φρ(X,Y) 𝓕. Now we have to show that ρ(x) ∈ Φρ(X, Y) ∖ ↓Φρ(X,Y) {ρ(y1), ρ(y2), …, ρ(ym)} ⊆ ↑Φρ(X,Y) 𝓕. Obviously, ρ(x) ∈ Φρ(X, Y) ∖ ↓Φρ(X,Y) {ρ(y1), ρ(y2), …, ρ(ym)}. Assume that there is a ρ(A) ∈ Φρ(X, Y) ∖ ↓Φρ(X,Y) {ρ(y1), ρ(y2), …, ρ(ym)} such that ρ(A) ∉ ↑Φρ(X,Y) 𝓕. Let 𝓕 = {ρ(u1), ρ(u2), …, ρ(un)}. Then ρ(A) ⊈ ρ(yj)(j = 1, 2, …, m) and ρ(ui) ⊈ ρ(A)(i = 1, 2, …, n). Thus there exist sj ∈ A and vj ∈ ρ(sj) such that vj ∉ ρ(yj) (j = 1, 2, …, m), and ti ∈ ρ(ui) with ti ∉ ρ(A)(i = 1, 2 …, n). We can conclude that there exist k0 ∈ {1, 2, …, m} and l0 ∈ {1, 2, …, n} such that tl0 ∈ ρ(sk0). If not, then for any k ∈ {1, 2, …, m} and l ∈ {1, 2, …, n}, tl ∉ ρ(sk). Let Nl be the greatest element of φtl. Then ρ(sk) ⊆ Nl, so for any k ∈ {1, 2, …, m}, ρ(sk) is a lower bound of {N1, N2, …, Nn}. Since φρ(X, Y) is an S-poset, there exists s ∈ X such that {ρ(s1), ρ(s2), …, ρ(sm)} ⊆ ↓ ρ(s) ⊆ {N1, N2, …, Nn}↓. Thus, vj ∈ ρ(s) and ρ(s) ⊈ ρ(yj)(j = 1, 2, …, m), that is ρ(s) ∈ φρ(X, Y) ∖ ↓φρ(X,Y) {ρ(y1), ρ(y2), …, ρ(yn)}, so ρ(s) ∈ ↑φρ(X,Y) 𝓕. Thus there is a l ∈ {1, 2, …, n} such that ρ(ul) ⊆ ρ(s). Notice that tl ∈ ρ(ul), we have tl ∈ ρ(s). On the other side, since ρ(s) ∈ {N1, N2, …, Nn}↓, ρ(s) ⊆ Nl. By the definition of Nl, tl ∉ ρ(s), a contradiction. Therefore, there exist k0 ∈ {1, 2, …, m} and l0 ∈ {1, 2, …, n} such that tl0 ∈ ρ(sk0). Since ρ(sko) ⊆ ρ(A), tl0 ∈ ρ(A), which contradicts tl ∉ ρ(A) for any l ∈ {1, 2 …, n}. Hence 𝓕 ≺Φρ(X,Y)ρ(x).□
Theorem 3.11
For a binary relation ρ : X ⇀ Y with property M, consider the following conditions:
ρ is w-generalized finitely regular;
(φρ(X, Y), ⊆) is a weak quasi-hypercontinuous poset;
the interval topology on (φρ(X, Y), ⊆) is split T2;
(Φρ(X, Y), ⊆) is a quasi-hypercontinuous lattice.
Then (1) ⇔ (2) ⇔ (3) ⇐ (4). If φρ(X, Y) is an S-poset, then (1) – (4) are equivalent.
Proof
(1) ⇒ (2) For any ρ(x) ∈ φρ(X, Y), if ρ(x) ⊈ ρ(u), then there is a y ∈ ρ(x) such that y ∉ ρ(u). Since ρ is w-generalized finitely regular, there are {u1, u2, …, un} ∈ X(<ω) and {v1, v2, …, vm} ∈ Y(<ω) such that
(ui, y) ∈ ρ, (x, vj) ∈ ρ (i = 1, 2, …, n; j = 1, 2, …, m), and
∀s ∈ X, T = {t1, t2, …, tn} ⊆ Y, if (ui, ti) ∈ ρ (i = 1, 2, …, n), (s, vj) ∈ ρ (j = 1, 2, …, m), then ∃k ∈ {1, 2, …, m} such that (s, tk) ∈ ρ.
Let 𝓕 = {ρ(u1), ρ(u2), …ρ(un)}. Then 𝓕 ∈ φρ(X, Y)(<ω) and ρ(ui) ⊈ ρ(u) (i = 1, 2, …, n), thus ρ(u) ∉ 𝓕. Let Nj be the greatest element of φvj (if φvj = ∅, let Nj = ∅). Then ρ(x) ∈ φρ(X, Y) ∖ ↓φρ(X,Y){N1, N2, …, Nm} since vj ∈ ρ(x) (j = 1, 2, … m). For any ρ(s) ∈ φρ(X, Y) ∖ ↓φρ(X,Y){N1, N2, …, Nm}, ρ(s) ⊈ Nj(j = 1, 2, …, m). By the definition of Nj, vj ∈ ρ(s). If ρ(s) ∉ ↑φρ(X,Y) 𝓕, then for any i ∈ {1, 2, …, n}, ρ(ui) ⊈ ρ(s), so there is a ti ∈ ρ(ui) with ti ∉ ρ(s). By the condition (b), there is a k ∈ {1, 2, …, m} such that (s, tk) ∈ ρ, i.e., tk ∈ ρ(s), which contradicts ti ∉ ρ(s) for any i ∈ {1, 2, …, n}. Thus ρ(s) ∈ ↑φρ(X,Y) 𝓕.
All above show that ρ(x) ∈ φρ(X, Y) ∖ ↓φρ(X,Y) {N1, N2, …, Nm} ⊆ ↑φρ(X,Y) 𝓕, i.e., 𝓕 ≺φρ(X,Y)ρ(x). Note that ρ(u) ∉ ↑φρ(X,Y) 𝓕. Hence, for any ρ(x) ∈ φρ(X, Y), ↑φρ(X,Y)ρ(x) = ⋂ {↑φρ(X,Y) 𝓕 : 𝓕 ∈ φρ(X, Y)(<ω) and 𝓕 ≺φρ(X,Y)ρ(x)}. Therefore, (φρ(X, Y), ⊆) is a weak quasi-hypercontinuous poset.
(2) ⇒ (3) For any ρ(x), ρ(y) ∈ φρ(X, Y) with ρ(x) ⊈ ρ(y). By (2), there exists 𝓕 ∈ φρ(X, Y)(<ω) such that 𝓕 ≺φρ(X,Y)ρ(x) and ρ(y) ∉ ↑φρ(X,Y) 𝓕. By the definition of ≺, we have that ρ(x) ∈ intυ((φρ(X,Y),⊆)) ↑φρ(X,Y) 𝓕 ⊆ ↑φρ(X,Y) 𝓕 ⊆ φρ(X, Y) ∖ ↓φρ(X,Y)ρ(y). Let 𝓤 = intυ((φρ(X,Y),⊆)) ↑φρ(X,Y) 𝓕 and 𝓥 = φρ(X, Y) ∖ ↑φρ(X,Y) 𝓕. Then ρ(x) ∈ 𝓤 ∈ υ((φρ(X, Y), ⊆)), ρ(y) ∈ 𝓥 ∈ ω((φρ(X, Y), ⊆)) and 𝓤 ∩ 𝓥 = ∅. Hence, the interval topology on (φρ(X, Y), ⊆) is split T2;
(3) ⇒ (1) For any (x, y) ∈ ρ, let Ny be the greatest element of φy (if φy = ∅, let Ny = ∅). Then ρ(x) ⊈ Ny. Since the interval topology on (φρ(X, Y), ⊆) is split T2, there exist {ρ(x1), ρ(x2), …, ρ(xm)} ∈ φρ(X, Y)(<ω) and {ρ(u1), ρ(u2), …, ρ(un)}(<ω) such that ρ(x) ∈ φρ(X, Y) ∖ ↓φρ(X,Y) {ρ(x1), ρ(x2), …, ρ(xm)}, Ny ∈ φρ(X, Y) ∖ ↑φρ(X,Y) {ρ(u1), ρ(u2), …, ρ(un)} and ↓φρ(X,Y) {ρ(x1), ρ(x2) …, ρ(xm)} ⋃ ↑φρ(X,Y) {ρ(u1), ρ(u2), …, ρ(un)} = φρ(X, Y).
Since ρ(x) ⊈ ρ(xj) (j = 1, 2, …, m), choose vj ∈ ρ(x) and vj ∉ ρ(xj). On the other side, ρ(ui) ⊈ Ny (i = 1, 2, …, n). By the definition of Ny, y ∈ ρ(ui) (i = 1, 2, …, n). Thus {u1, u2, …, un} and {v1, v2, …, vm} satisfy the condition (a) of Definition 3.2. ∀ s ∈ X, {t1, t2, …, tn} ⊆ Y, if (ui, ti) ∈ ρ (i = 1, 2, …, n), (s, vj) ∈ ρ (j = 1, 2, …, m), we have ρ(s) ⊈ ρ(xj) for any j ∈ {1, 2, …, m}. Thus ρ(s) ∈ ↑φρ(X,Y) {ρ(u1), ρ(u2), …, ρ(un)}, i.e., there is a k ∈ {1, 2, …, n} such that ρ(uk) ⊆ ρ(s). Notice that tk ∈ ρ(uk), thus tk ∈ ρ(s), i.e., (s, tk) ∈ ρ. Hence ρ is w-generalized finitely regular.
(4) ⇒ (1) By Theorem 2.4, ρ is generalized finitely regular. Hence (1) holds.
(2) ⇒ (4) For any ρ(A) ∈ (Φρ(X, Y), ⊆), let ρ(B) ∈ Φρ(X, Y) with ρ(A) ⊈ ρ(B). Then there is a y ∈ ρ(A) such that y ∉ ρ(B). Choose x ∈ A with y ∈ ρ(x). Let Ny be the greatest element of φy. Then ρ(x) ⊈ Ny and ρ(B) ⊆ Ny. Since (φρ(X, Y), ⊆) is weak quasi-hypercontinuous, there exists 𝓕 ∈ φρ(X, Y)(<ω) such that 𝓕 ≺φρ(X,Y)ρ(x) with Ny ∉ ↑φρ(X,Y) 𝓕. By Lemma 3.10 and ρ(B) ⊆ Ny, 𝓕 ≺Φρ(X,Y)ρ(x) ⊆ ρ(A) and ρ(B) ∉ ↑Φρ(X,Y) 𝓕. Hence (Φρ(X, Y), ⊆) is a quasi-hypercontinuous lattice.□
For any poset P, let ρ = ≰ on P. Then δ(P) is order isomorphism to (Φ≰(P), ⊆) (define δ(P) → (Φ≰(P), ⊆) by Aδ ↦ P ∖ A↑) (see [24]). Furthermore, we can check that P is order isomorphism to (φ≰(P), ⊆). In deed, define f : P → (φ≰(P), ⊆) by f(x) = P ∖ ↑ x and g : (φ≰(P), ⊆) → P by g(ρ(x)) = x. One can easily check that f, g are order preserving and f ∘ g = id(φ≰(P),⊆), g ∘ f = idP. Therefore, using Corollary 3.6 and Theorem 3.11, we have the following.
Theorem 3.12
Let P be a poset. Consider the following conditions.
P is a weak quasi-hypercontinuous poset;
the relation ≰ is w-generalized finitely regular;
for any x, y ∈ P with x ≰ y, there are {u1, u2, …, un}, {v1, v2, …, vm} ∈ P(<ω) such that
ui ≰ y, x ≰ vj (i = 1, 2, …, n; j = 1, 2, …, m), and
∀z ∈ P, ∃k ∈ {1, 2, …, n} such that uk ≤ z or ∃l ∈ {1, 2, …, m} such that z ≤ vl;
for any x, y ∈ P with x ≰ y, there exist F, G ∈ P(<ω) such that
x ∉ ↓ G, y ∉ ↑ F, and
↓ G ⋃ ↑ F = P;
θ(P) is split T2;
there is a w-generalized finitely regular relation ρ : X ⇀ Y satisfying property M such that P ≅ (φρ(X, Y), ⊆);
there is a w-generalized finitely regular relation ρ : X ⇀ X satisfying property M such that P ≅ (φρ(X), ⊆);
(δ(P), ⊆) is a quasi-hypercontinuous lattice.
Then (1) ⇔ (2) ⇔ (3) ⇔ (4) ⇔ (6) ⇔ (7)⇐ (8). If P is an S-poset, then (1) – (8) are equivalent.
Proof
(1) ⇒ (2) Let ρ = ≰ on P. By Theorem 3.11 and P ≅ (φ≰(P), ⊆).
(2) ⇒ (3) Let x, y ∈ P with x ≰ y. By the definition of w-generalized finitely regular, there are {u1, u2, …, un}, {v1, v2, …, vm} ∈ P(<ω) such that
ui ≰ y, x ≰ vj (i = 1, 2, …, n; j = 1, 2, …, m), and
∀s ∈ X, {t1, t2, …, tn} ⊆ Y, if ui ≰ ti (i = 1, 2, …, n), s ≰ vj (j = 1, 2, …, m), then ∃k ∈ {1, 2, …, m} such that s ≰ tk.
For any z ∈ P, let s = z and ti = z (i = 1, 2, …, n). Then by (ii), ∃k ∈ {1, 2, …, n} such that uk ≤ z or ∃l ∈ {1, 2, …, m} such that z ≤ vl.
(3) ⇒ (1) Let x, y ∈ P with x ≰ y. By (3), there are {u1, u2, …, un}, {v1, v2, …, vm} ∈ P(<ω) such that
ui ≰ y, x ≰ vj (i = 1, 2, …, n; j = 1, 2, …, m), and
∀z ∈ P, ∃k ∈ {1, 2, …, n} such that uk ≤ z or ∃l ∈ {1, 2, …, m} such that z ≤ vl.
Let F = {u1, u2, …, un}. Then we have y ∉ ↑ F and x ∈ P ∖ ↓ {v1, v2, …, vm}. Now we show that P ∖ ↓ {v1, v2, …, vm} ⊆ ↑ F. For any z ∈ P ∖ ↓ {v1, v2, …, vm}, z ≰ vl(l = 1, 2, …, m), by (ii), there is a k ∈ {1, 2, …, n} such that uk ≤ z. Thus z ∈ ↑ F, and thus F ≺ x. Therefore P is a weak quasi-hypercontinuous poset.
(3) ⇔ (4) See [24].
(4) ⇒ (5) Let x, y ∈ P with x ≰ y. By (4) there exist F, G ∈ P(<ω) such that
x ∉ ↓ G, y ∉ ↑ F;
↓ G ⋃ ↑ F = P.
Let U = P ∖ ↓ G, V = P ∖ ↑ F. Then x ∈ U ∈ υ(P), y ∈ V ∈ ω(P) and U ∩ V = ∅, thus θ(P) is split T2.
(5) ⇒ (4) Let x, y ∈ P with x ≰ y. By (5), there are U ∈ υ(P), V ∈ ω(P) such that x ∈ U, y ∈ V with U ∩ V = ∅. Thus there exist F, G ∈ P(<ω) such that x ∈ P ∖ ↓ G ⊆ U and y ∈ P ∖ ↑ F ⊆ V. It is easy to see that F, G satisfy the conditions (a) and (b) of (4).
(2) ⇒ (6) Let X = Y = P and ρ = ≰. By Corollary 3.6.
(6) ⇒ (7) Obviously.
(7) ⇒ (1) Follows from Theorem 3.11.
(8) ⇒ (1) By P ≅ (φ≰(P), ⊆), δ(P) ≅ (Φ≰(P), ⊆) and Theorem 3.11.
(1) ⇒ (8) Let P be an S-poset. Since P ≅ (φ≰(P), ⊆) and δ(P) ≅ (Φ≰(P), ⊆), by Theorem 3.11, the condition (8) holds.□
Corollary 3.13
Let P be a poset. Then P is weak quasi-hypercontinuous if and only if Pop is weak quasi-hypercontinuous.
Corollary 3.14
Let P be a sup-semilattice. Then the following two conditions are equivalent:
P is a quasi-hypercontinuous poset;
(δ(P), ⊆) is a quasi-hypercontinuous lattice.
Follows from Corollary 3.14, we establish a necessary and sufficient condition for a quasi-hypercontinuous poset to have a normal completion which is a quasi-hypercontinuous lattice, that is, P only need to be a sup-semilattice. This condition is weaker than that appears in Theorem 5.2 of [29].
Acknowledgements
The authors would like to thank the referees for their very careful reading of the manuscript and valuable comments. This research is supported by the National Natural Science Foundation of China (No. 11661057, 11661040), the Ganpo 555 project for leading talents of Jiangxi Province and the Natural Science Foundation of Jiangxi Province (No. 20192ACBL20045), the Foundation of Education Department of Jiangxi Province (No. GJJ160660).
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© 2019 Shuzhen Luo and Xiaoquan Xu, published by De Gruyter
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