On the relation between completeness and H-closedness of pospaces without infinite antichains
We study the relation between completeness and H-closedness for topological partially ordered spaces. In general, a topological partially ordered space with an infinite antichain which is even directed complete and down-directed complete, is not H-closed. On the other hand, for a topological partial...
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irk-123456789-1522962019-06-11T01:25:09Z On the relation between completeness and H-closedness of pospaces without infinite antichains Yokoyama, T. We study the relation between completeness and H-closedness for topological partially ordered spaces. In general, a topological partially ordered space with an infinite antichain which is even directed complete and down-directed complete, is not H-closed. On the other hand, for a topological partially ordered space without infinite antichains, we give necessary and sufficient condition to be H-closed, using directed completeness and down-directed completeness. Indeed, we prove that {a pospace} X is H-closed if and only if each up-directed (resp. down-directed) subset has a supremum (resp. infimum) and, for each nonempty chain L ⊆ X, ⋁ L∈ cl ↓ L and ⋀L ∈ cl ↑ L. This extends a result of Gutik, Pagon, and Repovs [GPR]. 2013 Article On the relation between completeness and H-closedness of pospaces without infinite antichains / T. Yokoyama // Algebra and Discrete Mathematics. — 2013. — Vol. 15, № 2. — С. 287–294. — Бібліогр.: 3 назв. — англ. 1726-3255 2010 MSC:Primary 06A06, 06F30; Secondary 54F05, 54H12. http://dspace.nbuv.gov.ua/handle/123456789/152296 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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We study the relation between completeness and H-closedness for topological partially ordered spaces. In general, a topological partially ordered space with an infinite antichain which is even directed complete and down-directed complete, is not H-closed. On the other hand, for a topological partially ordered space without infinite antichains, we give necessary and sufficient condition to be H-closed, using directed completeness and down-directed completeness. Indeed, we prove that {a pospace} X is H-closed if and only if each up-directed (resp. down-directed) subset has a supremum (resp. infimum) and, for each nonempty chain L ⊆ X, ⋁ L∈ cl ↓ L and ⋀L ∈ cl ↑ L. This extends a result of Gutik, Pagon, and Repovs [GPR]. |
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Yokoyama, T. On the relation between completeness and H-closedness of pospaces without infinite antichains Algebra and Discrete Mathematics |
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Yokoyama, T. |
| author_sort |
Yokoyama, T. |
| title |
On the relation between completeness and H-closedness of pospaces without infinite antichains |
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On the relation between completeness and H-closedness of pospaces without infinite antichains |
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On the relation between completeness and H-closedness of pospaces without infinite antichains |
| title_fullStr |
On the relation between completeness and H-closedness of pospaces without infinite antichains |
| title_full_unstemmed |
On the relation between completeness and H-closedness of pospaces without infinite antichains |
| title_sort |
on the relation between completeness and h-closedness of pospaces without infinite antichains |
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Інститут прикладної математики і механіки НАН України |
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2013 |
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http://dspace.nbuv.gov.ua/handle/123456789/152296 |
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On the relation between completeness and H-closedness of pospaces without infinite antichains / T. Yokoyama // Algebra and Discrete Mathematics. — 2013. — Vol. 15, № 2. — С. 287–294. — Бібліогр.: 3 назв. — англ. |
| series |
Algebra and Discrete Mathematics |
| work_keys_str_mv |
AT yokoyamat ontherelationbetweencompletenessandhclosednessofpospaceswithoutinfiniteantichains |
| first_indexed |
2025-07-13T02:46:11Z |
| last_indexed |
2025-07-13T02:46:11Z |
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1837498133773287424 |
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Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 15 (2013). Number 2. pp. 287 – 294
c© Journal “Algebra and Discrete Mathematics”
On the relation between completeness and
H-closedness of pospaces without infinite
antichains
Tomoo Yokoyama
Communicated by V. V. Kirichenko
Abstract. We study the relation between completeness and
H-closedness for topological partially ordered spaces. In general, a
topological partially ordered space with an infinite antichain which is
even directed complete and down-directed complete, is not H-closed.
On the other hand, for a topological partially ordered space without
infinite antichains, we give necessary and sufficient condition to be H-
closed, using directed completeness and down-directed completeness.
Indeed, we prove that a pospace X is H-closed if and only if each
up-directed (resp. down-directed) subset has a supremum (resp.
infimum) and, for each nonempty chain L ⊆ X,
∨
L ∈ cl↓L and∧
L ∈ cl↑L. This extends a result of Gutik, Pagon, and Repovš
[GPR].
In this paper, we study the relation between completeness and H-
closedness for topological partially ordered spaces (or shortly pospaces).
Though H-closedness is a generalization of compactness, H-closedness
does not correspond with compactness for even chains and antichains
(equipped with some pospace topologies). Indeed, since the pospaces
which are antichains coincide with the Hausdorff topological spaces, we
have that H-closed non-compact topological spaces are also such pospaces
which are antichains. There is also another extremal example which is a
The author is partially supported by the JST CREST Program at Creative Research
Institution, Hokkaido University.
2010 MSC: Primary 06A06, 06F30; Secondary 54F05, 54H12.
Key words and phrases: H-closed, pospace, directed complete.
288 Relation between completeness and H-closedness
countable linearly ordered H-closed non-compact pospace (e.g. Example
4.6 in [GPR]).
In [GPR], they have shown that a linearly ordered topological semilat-
tice is H-closed if and only if it is H-closed as a topological pospace. They
also have given the following characterization of H-closedness for a lin-
early ordered pospace to be H-closed (Corollary 3.5 in [GPR]): A linearly
ordered pospace X is H-closed if and only if the following conditions hold:
(i) X is a complete set with respect to the partial order on X;
(ii) x =
∨
A for A = ↓A \ {x} implies x ∈ clA, whenever A 6= ∅ ⊆ X; and
(iii) x =
∧
B for B = ↑B \ {x} implies x ∈ clB, whenever B 6= ∅ ⊆ X.
This result can rewrite the following statement: 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.
Naturally the following question arises: Is there a similar characteri-
zation of H-closedness for topological semilattices or pospaces?
It’s easy to see that a discrete countable antichain is not H-closed as a
pospace but directed complete and down-directed complete. This means
that there is no similar characterization of H-closedness for pospaces.
However, we give the necessary and sufficient condition for pospaces
without infinite antichains to be H-closed.
In this paper, all topological spaces will be assumed Hausdorff. 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
clXA 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 = ∅ (equivalently 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, ↑ x := {y ∈ X | x ≤ y} (resp.
↓ x := {y ∈ X | y ≤ x}) is called the upset (resp. the downset) of
x. Define lx := ↑x ∪ ↓x. For a subset Y ⊆ X, ↑ Y :=
⋃
y∈Y ↑ y (resp.
T. Yokoyama 289
↓ Y :=
⋃
y∈Y ↓ y) is called the upset (resp. the downset) of Y . Define
lY := ↑Y ∪ ↓Y . For a subset S of a poset X, denote by S↑ (resp. S↓)
the set of upper (resp. lower) bounds of S. In other words, S↓ = {x ∈
X | x ≤ s for all s ∈ S} and S↑ = {x ∈ X | x ≥ s for all s ∈ S}. Define
max S := {s ∈ S | S ∩ ↑s = {s}} and min S := {s ∈ S | S ∩ ↓s = {s}}.
For elements x, y of a poset, x‖y means that x and y are incomparable.
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 (resp. antichain) is a chain (resp. antichain) which is
properly contained in no other chain (resp. antichain). 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-directed complete (resp. (up-)directed complete) if each down-
directed (resp. up-directed) set S of X has
∧
S (resp.
∨
S). An up-directed
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. Let X be a pospace without infinite antichains. Then X is
an H-closed pospace if and only if X is directed complete and down-directed
complete such that
∨
L ∈ cl↓L and
∧
L ∈ cl↑L for any nonempty chain
L ⊆ X.
Note that lF = X for a maximal antichain F of a poset X. Moreover
notice that that if X has no infinite antichains, then all subposet and all
extensions of X by adding finite points have no infinite antichain neither.
First, we show the only if part in the main theorem.
Lemma 1. Let X be a pospace without infinite antichains. Suppose that
X is directed complete and down-directed 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 infinite an-
tichains such that X is directed complete and down-directed complete,
and
∨
L ∈ cl↓L and
∧
L ∈ cl↑L for any nonempty chain L ⊆ X. Then
there is an embedding from X to a pospace X̃ such that X is a dense
proper subspace. Fix any element x ∈ X̃ \ X. Since X is directed com-
plete and down-directed complete, we have X = ↓X max X = ↑X min X.
Since X has no infinite antichains, we have that max X and min X are
290 Relation between completeness and H-closedness
finite subsets. If x /∈ ↓ max X, then the fact that ↓ max X is a closed
subset of X̃ implies that x /∈ clX, which contradicts that X is a dense
subset of X̃. Thus there is an element ω ∈ X such that x < ω. By the
symmetry of pospace, there is an element α ∈ X such that α < x. Since
X is embedded into X̃, we have that A′ := ↓x ∩ X and B′ := ↑x ∩ X
are closed subsets of X. Let A := {
∨
X L | L 6= ∅ ⊆ A′ is a chain} and
B := {
∧
X L | L 6= ∅ ⊆ B′ is a chain} be subsets of X. Since X is directed
complete and down-directed complete, we obtain max A and min B are
nonempty such that ↓X max A ⊇ A and ↑X min B ⊇ B. Next we show
that x /∈ ↓ max A. Indeed, suppose that there is an element y ∈ max A
such that x < y. By the definition of A′, we have y /∈ A′. Since y ∈ max A,
there is a chain L ⊆ A′ such that y =
∨
X L /∈ L. By the assumption,
y =
∨
X L ∈ clX↓L ⊆ clXA′ = A′, which is impossible.
By the symmetry of pospace, we have x /∈ ↑ min B. Let I := {y ∈ X |
x‖y}. Since X has no infinite antichains, there is a finite subset F of I
such that I ⊆ lF . Since X = A′ ⊔ B′ ⊔ I, the facts that ↓X max A ⊇ A′
and ↑X min B ⊇ B′, imply that X ⊆ ↓ max A ∪ ↑ min B ∪ lF and so
clX ⊆ ↓ max A ∪ ↑ min B ∪ lF . Since x /∈ ↓ max A ∪ ↑ min B ∪ lF , we
obtain that x /∈ clX, which contradicts that X is a dense subset of X̃.
Therefore X is H-closed.
Now we show the another direction. Later we assume that X is an
H-closed pospace without infinite antichains. Note that the following
three statements are equivalent for subsets U, V of a poset X: ↑U ∩V = ∅.
U ∩ ↓V = ∅. ↑U ∩ ↓V = ∅.
Lemma 2. Let X be a poset and x a point of X. Suppose that there is a
subset F of X such that the disjoint union F ⊔{x} is a maximal antichain
in X. Let U := X \ lF . Then lU ⊆ lx.
Proof. Put X− := ↓(F ⊔{x}) and X+ := ↑(F ⊔{x}). Then X = X− ∪X+.
By symmetry, it suffices to show that ↑U ⊆ lx. For any y ∈ U , if y ∈ ↑x,
then obviously ↑y ∈ ↑x. Otherwise y ∈ ↓x. Since ↑U ∩ ↓F = ∅ and y ∈ U ,
we have ↑y ∩ ↓F = ∅. Then ↑y ∩ X− ⊆ ↓x and ↑y ∩ X+ ⊆ ↑x. Thus
↑y ⊆ ↓x ∪ ↑x and so ↑U ⊆ lx.
Lemma 3. Any maximal chain of X is complete.
Proof. Suppose that there is a maximal chain L which is not 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
X̃ := X ⊔ {∞} be the extension with the greatest element ∞. Define a
T. Yokoyama 291
topology τ on X̃ by an open subbase τX ∪{X̃ \↓F | F ∈ X<ω}. Since any
neighborhood of ∞ meets S, we have ∞ ∈ clX\X. Then X is a non-closed
subset of X̃. We will show that X̃ 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
are an element y > x of X and a maximal antichain F of X containing y.
Then U := X̃ \ ↑F is an open neighborhood of x and V := X̃ \ ↓F is an
open neighborhood of ∞ such that ↓U ∩ V = ∅. Otherwise x ∈ max X.
Since S↑ = ∅, there is an element y ∈ S such that y‖x. Then there is a
maximal antichain F of X containing x and y. Let U = X̃ \ l(F \ {x})
be an open neighborhood of x and V = X̃ \ ↓F an open neighborhood of
∞. Since x ∈ max X, we have ↑x = {x, ∞}. Since F \ {x} 6= ∅, we obtain
U ⊆ ↓x. Then ↓U ∩ V = ∅. Thus X̃ is a pospace and so X is not H-closed,
which contradicts the hypothesis on X.
Thus we may assume that S↑ 6= ∅. Then
∧
L(L \ ↓S) does not exist.
Let A := L ∩ ↓S and B := L \ ↓S. Extend X to X̃ := X ⊔ {α} by
↑α := {α} ⊔ A↑ and ↓α := {α} ⊔ ↓A. Define a topology τ by an open
subbase τX ∪ {X̃ \ (↑E ∪ ↓F ) | F, E ∈ X<ω}. Since ↑α = {α} ⊔ A↑ and
since
∨
L S =
∨
L A does not exist, we have
∨
A = α. Since A is a chain
with max A = ∅ and ↑α ⊇ A↑, we obtain that any neighborhood of α
meets A. Thus X is a non-closed subset of X̃. Therefore it suffices to show
that X̃ is a pospace, which implies that X is not an H-closed pospace.
Indeed, let x ∈ X be any element. Then either x < α, x‖α, or x > α. If
x < α, then there is an element y ∈ X such that x < y < α. Let E be a
maximal antichain containing y, U = X̃ \ ↓E an open neighborhood of α,
and V = X̃ \ ↑E an open neighborhood of x. Then U ∩ ↓V = ∅. If x‖α,
then the fact that ↑α = {α}⊔A↑ implies that x /∈ A↑. Since x /∈ ↓A ⊆ ↓α,
we have that either x ≥ y or x‖y for any y ∈ A. Since x /∈ A↑ ⊆ ↑α,
there is an element y ∈ A such that x‖y. Let F be a maximal antichain
containing x and y, U = X̃ \ lx and V = X̃ \ l(F \ {x}) ⊆ lx. Then
U ∩ lV = ∅ and V is an open neighborhood of x. Since x‖α, we have
U is an open neighborhood of α. Otherwise x > α. Then x ∈ A↑. If
x /∈ min A↑, then there is an element y ∈ A↑ such that α < y < x. Let F
be a maximal antichain containing y, U = X̃ \↓F , and V = X̃ \↑F . Then
U is an open neighborhood of x and V is an open neighborhood of α such
that U ∩ ↓V = ∅. Otherwise x ∈ min A↑. Since L is a maximal chain and∧
L B does not exist, there is an element y ∈ B such that y‖x. Let F be a
maximal antichain containing x and y, U = X̃ \ l(F \ {x}), E a maximal
antichain containing α, and V = X̃ \ (l(E \ {α}) ∪ ↑F ) ⊆ (lα \ {x}) ∩ ↓F .
Obviously U is an open neighborhood of x. Since y > α, we have V is an
open neighborhood of α. Since x ∈ min A↑ and ↑α = {α} ⊔ A↑, we have
292 Relation between completeness and H-closedness
↑α ∩ ↓x = {α, x}. Since F is a maximal antichain containing y and since
α < y, we have (↑α \ {x}) ∩ ↓F ⊆ ↓(F \ {x}). Since ↓α ⊂ ↓y ⊆ ↓(F \ {x}),
we have V ⊆ ↓(F \ {x}). By the definition of U , we have ↓V ∩ U = ∅.
This show that X̃ is a pospace.
Lemma 4. X is directed complete.
Proof. Let L be any infinite chain of X. Then min(L↑) = min((↓L)↑).
By the Axiom of choice, there is a maximal chain L′ containing L. By
Lemma 3, there is the element x ∈ L′ ∩ min(L↑) and so min(L↑) 6= ∅ and
↑ min(L↑) = L↑. Since X has no infinite antichains, we have min(L↑) is
a nonempty finite subset. Put {x1, . . . , xn} = min(L↑). Since X has no
infinite antichains and since min L is an antichain, there is a maximal
antichain K containing {x1, . . . , xn}. Suppose that min(L↑) is not a
singleton. Then xi‖xj for any distinct pair i 6= j. The facts that X is a
pospace and xi‖xj for any i 6= j, imply that for any i = 1, . . . , n, there is
an open neighborhood Ui ⊆ lxi \ l(K \ {xi}) of xi such that Ui ∩ ↓L = ∅
and Ui ∩ lUj = ∅ for any distinct j. Extend X to X̃ := X ⊔ {α} by
↑α := ↑X{x1, . . . , xn} ⊔ {α} = L↑ ⊔ {α} and ↓α := ↓XL ⊔ {α}. Define a
topology τ on X̃ by an open subbase τX ∪ {X̃ \ (↑F ∪ ↓E) | F, E ∈ X<ω},
where τX is the topology of X. Since max L = ∅,
∨
X̃
L = α, and ↑α ⊇ L↑,
we have that any neighborhood of α meets L and so X is an embedded
subset of X̃ which is not closed. Next we show that X̃ is a pospace. Indeed,
let x be any element of X. Then either x ∈ ↑X{x1, . . . , xn}, x ≤ α, or x‖α.
Let G be a finite subset of X such that G⊔{α} is a maximal antichain in X̃.
If x = xi for some i, then let U := X̃ \(↑K ∪lG) be an open neighborhood
of α. Since ↑α \ {α} ⊆ ↑K, we have U ⊆ ↓α. Since ↓α = ↓L ⊔ {α} and
↓L ∩ Ui = ∅, we have ↓U ∩ Ui = ∅. If x ∈ ↑α \ {x1, . . . , xn}, then let
U := X̃ \ ↑K be an open neighborhood of α and V := X̃ \ ↓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 maximal antichain F
of X containing y. Then U := X̃ \ ↓F is an open neighborhood of α and
V := X̃\↑F is an open neighborhood of x such that U ∩↓V = ∅. Otherwise
x‖α. Since x /∈ ↓L ⊆ ↓α, we obtain that either x ≥ y or x‖y for any y ∈ L.
Since x /∈ L↑ ⊆ ↑α, there is an element y ∈ L such that y‖x. Then there
is a maximal antichain F of X containing x and y. Thus U := X̃ \ lx
is an open neighborhood of α and V := X̃ \ l(F \ {x}) ⊆ lx is an open
neighborhood of x. Since F is a maximal antichain, by Lemma 2, we have
lV ⊆ lx and so lV ∩ U = ∅. This show that X̃ is a pospace. Therefore
X is not H-closed, which contradicts the hypothesis of X. Thus min(L↑)
is a singleton and so
∨
L exists. Hence X is directed complete.
T. Yokoyama 293
Notice that the symmetry of pospace implies that the dual statement
of Lemma 4 holds (i.e. An H-closed pospace without infinite antichains is
down-directed complete).
Lemma 5. 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 X̃ := X ⊔ {α} be an extension of X by ↑α = {α} ⊔ ↑a
and ↓α = {α} ⊔ ↓L. Define a topology τ on X̃ by an open subbase
τX ∪ {X̃ \ (↑F ∪ ↓E) | F, E ∈ X<ω}, where τX is the topology of X.
Since any neighborhood of α meets L, we have α ∈ cl↓L \ ↓L. Thus X
is a subset of X̃ which is not closed. Next we show that X̃ is a pospace.
Indeed, let x be an element of X. Then either x > α, x < α, or x‖α.
If x ∈ ↑α, then there is a finite subset F of X such that F ⊔ {a} is a
maximal antichain. Since lα ⊆ la, we obtain F ⊔ {α} is an antichain.
If x = a, then U = X̃ \ (↑a ∪ lF ) ⊆ ↓a is an open neighborhood of α.
Since a /∈ clX↓L, there is an open neighborhood W ⊂ X of a such that
W ∩↓L = ∅. Then V := W \lF ⊆ la is an open neighborhood of a. Since
α /∈ V ⊆ W , we have ↓U ∩V ⊆ ↓a\(↓L⊔{α, a}) = ↓a\(↓α⊔{a}). We may
assume that ↓U ∩ V 6= ∅. Let E be a maximal antichain of ↓a \ (↓α ⊔ {a}).
Note that E ∩ lα = ∅. Since α /∈ lE, we have U ′ := U \ lE is an open
neighborhood of α. Since ↓a\↓α ⊆ lE, we have U ′ ⊆ ↓α = {α}⊔↓L. Since
W ∩ ↓L = ∅ and α /∈ W , we have ↓U ′ ∩ V = ∅. Otherwise x > a. Then
U = X̃ \↑(F ∪{a}) is an open neighborhood of α and V := X̃ \↓(F ∪{a})
is an open neighborhood of x with ↓U ∩ V = ∅. If x < α, then there are
an element y ∈ L and a maximal antichain F ∈ X<ω containing y such
that x < y < α. Then U := X̃ \ ↑F is an open neighborhood of x and
V := X̃ \ ↓F is an open neighborhood of α. Then ↓U ∩ V = ∅. Otherwise
x‖α. If x‖a, then there is a finite subset E of X such that E ⊔ {x, a} is a
maximal antichain. Thus U := X̃ \ lx is an open neighborhood of α and
V := X̃ \ l(E ⊔ {a}) is an open neighborhood of x. Since E ⊔ {x, a} is
a maximal antichain, we have lV ⊆ lx. Since U ⊆ ↓X̃ \ lx, we obtain
lV ∩ U = ∅. Otherwise x < a. Since x‖α and a =
∨
L /∈ L, there is
an element y ∈ L such that y‖x. Let E be a maximal antichain of X
containing x and y. Then U := X̃ \ lx is an open neighborhood of α and
V := X̃ \ l(E \ {x}) is an open neighborhood of x. Since E is a maximal
antichain, we have lV ⊆ lx. Since U ⊆ X̃ \lx, we obtain lV ∩U = ∅. This
shows that X̃ is a pospace. Therefore X is not an H-closed pospace.
Notice that the symmetry of pospace implies that the dual state-
ment of Lemma 5 holds (i.e. For any nonempty chain L ⊆ X, we have
294 Relation between completeness and H-closedness
∧
L ∈ cl↑L.). Therefore Theorem 1 is induced by Lemma 4, 5, their dual
statements, and Lemma 1.
Note that all H-closed pospaces which the author knows are directed
complete. Naturally the following question arises:
Question. Is there an H-closed pospace which is not directed complete?
Acknowledgments
I would like to thank Professor Dušan Repovš for informing me of
their interesting works.
References
[GPR] O. Gutik - D. Pagon - D. Repovš On Chains in H-Closed Topological Pospaces
Order (2010) 27: 69-81
[GR] O. Gutik - D. Repovš On linearly ordered H-closed topological semilattices
Semigroup Forum (2008) 77: 474-481
[W] L. E. Ward, Jr. Partially ordered topological spaces Proc. Amer. Math. Soc.
(1954) 5:1, 144-161.
Contact information
T. Yokoyama Department of Mathematics, Hokkaido Uni-
versity, Kita 10, Nishi 8, Kita-Ku, Sapporo,
Hokkaido, 060-0810, Japan
E-Mail: yokoyama@math.sci.hokudai.ac.jp
Received by the editors: 25.08.2011
and in final form 25.01.2013.
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