On completeness of H-closed pospaces

In [GPR], they gave the following characterization of H-closedness for a linearly ordered pospace to be H-closed: A linearly ordered pospace X is H-closed if and only if X is a complete lattice with L ∈ cl↓L and L ∈ cl↓L for any nonempty chain L ⊆ X. In the same paper, they also gave a non-H-closed pospace such that it is a directed complete and down-complete poset with an infinite antichain and with L ∈ cl↓L and L ∈ cl↓L for any nonempty chain L of it. (In fact, they showed that the extension X of an infinite antichain by adding the minimal element and equipped with the discrete topology on X is a non-H-closed pospace.) In this paper, all topological spaces will be assumed Hausdorff. We extend the above characterization of H-closedness for linearly ordered pospaces to one for pospaces without an infinite antichain. For a set X, denote by X <ω the set of finite subsets of X. If A is a subset of a topological space X, then we denote the closure of the set A in X by cl X A or clA. By a partial order on a set X we mean a reflexive, transitive and anti-symmetric binary relation ≤ on X. A set endowed with a partial order is called a partially ordered set (or poset). Recall that a poset with a topology defined on it is called a topological partially ordered space (or pospace) if the partial order is a closed subset of X × X. A partial order ≤ is said to be continuous or closed if x y in X implies that there are open neighborhoods U and V of x and y respectively such that ↑U ∩ V = ∅ (i.e. U ∩ ↓V = ∅). A partial order ≤ on a topological space X is continuous if and only if (X, ≤) is a pospace [W]. In any pospace, ↓x and ↑x are both closed for any element x of it. A Hausdorff pospace X is said to be an H-closed pospace if X is a closed subspace of every Hausdorff pospace in which it is contained. Obviously that the notion of H-closedness is a generalization of compactness. For an element x of a poset X, of Y . For a subset S of a poset, denote by S ↑ (resp. S ↓ ) the set of upper (resp. lower) bounds of S. For elements x, y of a poset, x y means that x and y are incomparable. For an element x of a poset X, denote by I x the set of incomparable elements for x (i.e. I x = X -(↑x ∪ ↓x)). For a subset A of a poset, A is said to be a chain if A is linearly ordered, and is said to be an antichain if any distinct elements are incomparable. A maximal chain is a chain which is properly contained in no other chain. The Axiom of Choice implies the existence of maximal chains in any poset. A subset D of a poset X is (up-)directed (resp. down-directed) if every finite subset of D has an upper (resp. lower) bound in D. A poset X is said to be down-complete (resp. up-complete) if each down-directed (resp. up-directed) set S of X has S (resp. S). An up-complete poset is also called a directed complete poset or a dcpo. It is well-known that a poset X is directed complete if and only if each chain L of X has L. Now we state the main result. Theorem 1.1. Let X be a pospace without an infinite antichain. Then X is an H-closed pospace if and only if X is directed complete and down-directed such that L ∈ cl↓L and L ∈ cl↓L for any nonempty chain L ⊆ X. Note that ↑F ∪ ↓F = X for a maximal antichain F of a poset X. Moreover notice that that if X has no infinite antichain, then all subposet and all extensions of X by adding finite points have no infinite antichain neither. Lemma 2.1. Let X be a poset and x a point of X. Suppose that there is a subset Then ↑U ⊆ ↑x ∪ ↓x and ↓U ⊆ ↑x ∪ ↓x. Proof. Put X -:= ↓(F ⊔ {x}) and X + := ↑(F ⊔ {x}). Then X = X -∪ X + . Since ↑U ∩↓F = ∅, we have ↑U ∩X -= ↑U ∩↓(F ⊔{x}) = ↑U ∩↓x and so ↑(↑U ∩X -)∩X + ⊆ ↑x. Since U ∩X + ⊆ ↑x, we obtain ↑U ∩X + = (↑(U ∩X + )∪↑(U ∩X -))∩X + ⊆ ↑x∪↓x and so ↑U ⊆ ↑x ∪ ↓x. By symmetry, we obtain ↓U ⊆ ↑x ∪ ↓x. Lemma 2.2. Let X be an H-closed pospace without an infinite antichain. Then any maximal chain of X is complete. Proof. Suppose that there is a maximal chain L which is non-complete. Then there is a subset S of L such that either L S or L S does not exist. We may assume that L S does not exist. If S ↑ = ∅, then let X := X ⊔ {∞} be the extension with the maximal element ∞. Define a topology τ on X by an open subbase τ ∪ { X -↓F | F ∈ X <ω }. Then X is an embedded subspace which is not closed. We will show that X is a pospace. For any element x of X, if x / ∈ max X, then the fact that L S does not exist implies that there is an element y > x of X and a finite subset F of X such that y ∈ F and F is a maximal antichain. Then Thus X is a pospace and so X is not H-closed, which contradicts. Thus S ↑ = ∅. Then L (L \ ↓S) does not exist. Let A := ↑(L \ ↓S) and B := ↓(L ∩ ↓S). Then X is an embedded subspace which is not closed. Thus it suffices to show the following claim, which induces that X is not an H-closed pospace. Claim 1. X is a pospace. Indeed, let x ∈ X be any element. If x ∈ A, then x > α. Since L (L \ ↓S) does not exist, there is an element y of L \ ↓S such that α < y < x. Then there is a finite subset E of A such that y ∈ E and E is a maximal chain. Since (E -{y}) ⊆ I y , we have that then the symmetry of pospace implies that there are a finite subset E of X , an open neighborhood U := X -↓E of α and an open neighborhood V := X -↑E of x such that ↓V ∩ U = ∅. Otherwise x α. Since L is a maximal chain, either ↑x L \ ↓S or ↓x L ∩ ↓S. First, consider the case ↑x ⊇ L \ ↓S. So ↓x L ∩ ↓S. Since x α and ↓x L ∩ ↓S, there is an element β < α such that β x. Let E be a finite subset of X such that E ⊔ {x, β} is a maximal chain. Since ↑x ⊇ L \ ↓S, we have Second, consider the case the case ↑x L \ ↓S. By symmetry, we may assume that ↓x L ∩ ↓S. Since x α and ↓x L \ ↓S, there is an element β > α such that β x. Since x α and ↓x L ∩ ↓S, there is an element γ < α such that γ x. Since x α, there is a finite subset E of X such that E ⊔ {x, α} is a maximal chain. Then U := X -(↑x ∪ ↓x) is an open neighborhood of α and ) is an open neighborhood of x. Since E ⊔{x, α} is a maximal antichain and V ⊆ X -(↑(E ⊔{α})∪↓(E ⊔{α})), Lemma 2.1 implies that ↑V ⊆ ↑x ∪ ↓x and ↓V ⊆ ↑x ∪ ↓x. Since U ∩ (↑x ∪ ↓x) = ∅, we obtain ↑V ∩ U = ∅ and ↓V ∩ U = ∅. Lemma 2.3. Let X be an H-closed pospace without an infinite antichain. Then X is directed complete and down-complete. Proof. Let L be any infinite chain of X. Then min(L ↑ ) = min((↓L) ↑ By Lemma 2.2, we obtain min(L ↑ ) = ∅. Since X has no infinite antichain, we have min(L ↑ ) is a nonempty finite subset. Put {x 1 , . . . , x n } = min(L ↑ ). Since X has no infinite antichain, there is a finite subset K such that K ⊇ min(L ↑ is a maximal antichain. If min(L ↑ ) is not a single set (i.e. n > 1), then x i x j for any distinct pair i = j. For any i = j, since X is a pospace and x i x j , there are open neighborhoods of Claim 2. X is a pospace. Indeed, let x be any element of X. Let G be a finite subset of I α such that G⊔{α} is a maximal antichain in X. If x = x i for some i, then let U := X-(↑K∪↑G∪↓G) ⊆ ↓α be an open neighborhood of α. Since ↓α = ↓ X L ⊔ {α}, we have ↓U ∩ U i = ∅. If x ∈ ↑α -{x 1 , . . . , x n }, then let U := X -↑K be an open neighborhood of α and V := X -↓K ⊆ ↑K an open neighborhood of α. Now ↓U ∩ V = ∅. If x ∈ ↓α, then there is an element y ∈ L such that x < y < α. Now there is a finite subset F of X such that y ∈ F and F ∪ {y} is a maximal antichain. Then F ⊆ I y -↑y. Hence Therefore X is not H-closed, which contradicts. Thus min(L ↑ ) is a single set and so L exists. Notice that the symmetry of pospace implies that the dual statement of Lemma 2.3 holds (i.e. An H-closed pospace without an infinite antichain is down complete). Lemma 2.4. Let X be an H-closed pospace without an infinite antichain. For any nonempty chain L ⊆ X, we have L ∈ cl↓L. Proof. Suppose that there is a chain L ⊆ X such that L / ∈ cl↓L. Put a := L. Let X := X ⊔ {α} be an extension of X by ↑α = {α} ⊔ ↑a and ↓α = {α} ⊔ ↓L. Define a topology τ on X by an open subbase τ X ∪ { X -(↑F ∪ ↓E) | F, E ∈ X <ω }, where τ X is the topology of X. Claim 3. X is a pospace. Indeed, let x be an element of X. first we consider the case x ∈ ↑α. Since X has no infinite antichain, there is a finite subset F of X such that F ⊔ {a} is a maximal antichain. Since ↑α ∪ ↓α ⊆ ↑a ∪ ↓a, we obtain F ⊔ {α} is an antichain. is an open neighborhood of a. Then ↓U ∩V ⊆ ↓a-↓α. We may assume that ↓U ∩ V = ∅. Then ↓a -↓α = ∅. Let E be a maximal antichain of ↓a-(↓α⊔{a}). Note that E ∩(↑α∪↓α) = ∅. Since X has no infinite antichain, we obtain Second, we consider the case x < α. Since a = L / ∈ L, there are an element y ∈ X and F ∈ X <ω such that x < y < α and F ⊔ {y} is a maximal chain. Then Finally, we consider the case x α. If x a, then there is a finite subset E of X such that E ⊔ {x, a} is a maximal chain. Thus U := X -(↑x ∪ ↓x) is an open neighborhood of α and V := X -(↑(E ⊔ {a}) ∪ ↓(E ⊔ {a})) is an open neighborhood of x. Since E ⊔ {x, a} is a maximal antichain, Lemma 2.1 implies that ↑V ⊆ ↑x∪↓x and ↓V ⊆ ↑x∪↓x. Since U ⊆ ↓α\(↑x∪↓x), we obtain ↑V ∩U = ∅ and ↓V ∩ U = ∅. Otherwise x < a. Since x α and a = L / ∈ L, there is an element β ∈ L such that β x. Let E be a finite subset of X such that E ⊔ {x, β} is a maximal antichain. Then U := X -(↓x ∪ ↑x) is an open neighborhood of α and ) is an open neighborhood of x. Since E ⊔ {x, β} is a maximal antichain, Lemma 2.1 implies that ↑V ⊆ ↑x ∪ ↓x and ↓V ⊆ ↑x ∪ ↓x. Since U ⊆ ↓α \ (↑x ∪ ↓x), we obtain ↑V ∩ U = ∅ and ↓V ∩ U = ∅. This completes the proof of this claim. By the definition of τ , X is a proper subspace and clX = X. Therefore X is not an H-closed pospace. Notice that the symmetry of pospace implies that the dual statement of Lemma 2.3 holds (i.e. An H-closed pospace without an infinite antichain is down complete). Now we show the another direction. Lemma 2.5. Let X be a pospace without an infinite antichain. Suppose that X is directed complete and down-complete such that L ∈ cl↓L and L ∈ cl↓L for any nonempty chain L ⊆ X. Then X is an H-closed pospace. Proof. Suppose there is a non-H-closed pospace X without an infinite antichain such that X is directed complete and down-complete, and L ∈ cl↓L and L ∈ cl↓L for any chain L ⊆ X. Then there is an embedding from X to a pospace X such that X is a dense proper subspace. Fix any element x ∈ X -X. Since X is directed complete and down-complete , X = ↓ X max X X = ↑ X min X X. Since X has no infinite antichain, we have that max X X and min X X are finite subsets. If all elements of X are incomparable to x, then x / ∈ ↓ max X X. Since ↓ max X X is X-closed, we have that x / ∈ clX, which contradicts to the density of X. Thus there is a comparable element α of X to x. By the symmetry of pospace, we may assume that α < x. If x / ∈ ↓ max X X, then the fact ↓ max X X ⊇ X implies that x / ∈ clX, which is impossible. Thus there is an element ω ∈ X such that x < ω. Since X is embedded into X, we have that A ′ := ↓x ∩ X and B ′ := ↑x ∩ X are X-closed. Let A := { X L | L = ∅ ⊆ A ′ is a chain} and B := { X L | L = ∅ ⊆ B ′ is a chain}. Since X is directed complete, we obtain max X A and min X B are nonempty such that ↓ X max X A ⊇ A and ↑ X max X B ⊇ B. Claim 4. x / ∈ ↓ max X A. Indeed, suppose that there is an element y ∈ max X A such that x < y. By the definition of A ′ , we have y / ∈ A ′ . Hence there is a chain L ⊆ A ′ such that y = X L / ∈ L. By the assumption, y ∈ cl X L ⊆ cl X A ′ = A ′ , which is impossible. By the symmetry of pospace, x / ∈ ↑ min X B. Let I := {y ∈ X | x y}. Since X has no infinite antichain, there is a finite subset F of I such that I ⊆ ↑F ∪ ↓F . Since X = A ′ ⊔ B ′ ⊔ I, ↓ X max X A ⊇ A ′ and ↑ X max X B ⊇ B ′ , we have X ⊆ ↓ max X A ∪ ↑ min X B ∪ ↑F ∪ ↓F . Since x / ∈ ↓ max X A ∪ ↑ min X B ∪ ↑F ∪ ↓F , we obtain that x / ∈ clX, which is impossible. Theorem 1.1 is induced by Lemma 2.3, 2.4, their dual statements, and Lemma 2.5.