# New Types of Almost Countable Dense Homogeneous Space

Authors: Abdalla Tallafha; Ahmed Al-Rawashdeh
Journal: International Journal of Mathematics and Mathematical Sciences
 (2010)

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## Abstract

In 1972, Bennett studied the countable dense homogeneous (CDH) spaces
and in 1992, Fitzpatrick, White, and Zhou proved that every CDH
space is aT1 space. Afterward Bsoul, Fora, and Tallafha gave another
proof for the same result, also they defined the almost CDH spaces and
almost T1, T0 spaces, indeed they prove that every ACDH space is an almost T1 space. In this paper we introduce a new type of almost CDH
spaces called ACDH-1, we characterize the ACDH spaces, the almost
T0 spaces, we also give relations between different types of CDH spaces.
We define new type of almost T1 (AT1) spaces, and we study the relations between the old and new definitions. By extending the techniques
given by Tallafha, Bsoul, and Fora, we prove that every ACDH-1 is an AT1.


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## Body

## 1. Introduction

In 1920, Sierpinski introduced in  [1] the notion of homogeneous spaces, saying that a topological space X is a homogeneous space if for any x≠y in X, then there is a homeomorphism h of X such that h(x)=y. Fréchet in [2] and Brouwer in [3] observed that the n-dimensional Euclidean space ℝn has the property that if A,B are countable dense subsets of ℝn, then there is a homeomorphism h of ℝn such that h(A)=B. Afterward, in 1972 the abstract study was begun by Bennett in [4], who called such spaces the countable dense homogeneous (CDH) spaces.In 1974, Lauer defined in [5] the densely homogeneous (DH) spaces, and in 1992, Fitzpatrick et al. proved in [6] that DH and CDH spaces are T1 spaces, and afterward Tallafha et al. in [7] gave another proof for the fact that CDH spaces are T1 spaces. In [8], Fora et al. defined almost CDH (ACDH) spaces, almost T0,T1 (AT0,AT1) spaces, and they discussed the relation between such spaces.In the first part of this paper, we introduce the definitions given by Fora et al. in [8] for almost CDH spaces, almost T0,T1 spaces. We give a characterization of almost CDH spaces besides we define a new type of almost CDH (ACDH-1) spaces, Transposition Homogeneous space (TH). Also, we discuss the relation between the new type and the others.In Section4 we give a characterization of almost T0 we introduce new definition of almost T1 space AST1, and we study the relation between them and the old ones. We finally prove our main result using the idea of almost closurely ordered sets, more precisely, we show that every almost CDH space of type 1 (ACDH-1) is an AST1 space.Finally, the following abbreviations and symbols will be used throughout this paper. For a subsetA of a topological space (X,τ), we write A¯, or Cl(A) for the closure of A, for x∈X, by x¯ which denotes the closure of {x} and |A| which denotes the cardinality of A. By H(X) we mean the set of all homeomorphisms of X, Δ refers to the symmetric difference of sets, (X,τind), (X,τdis), (X,τcof), and (X,τcoc) denote the set X with the indiscrete, discrete, cofinite, and cocountable topologies, and (ℝ,τl·r),(ℝ,τr·r), and (ℝ,τu) denote ℝ, with the left ray, right ray, and the usual topologies, respectively.


## 2. Almost CDH Spaces

Fora et al. in [8] defined almost and strong almost countable dense homogeneous spaces.Definition 2.1 (see [8]).
A space(X,τ) is called an almost countable dense homogeneous (ACDH) space if it is a separable space and for any two countable dense subsets K1,K2, there are two finite subsets F1,F2, K1∩F1=K2∩F2=∅ and h∈H(X) such that h(K1∪F1)=K2∪F2 and h(F1)∩F2=∅. In addition, if |F1|=|F2|, then (X,τ) is called a strong ACDH space which is denoted by SACDH.Remark 2.2.
In the previous definition the conditionh(F1)∩F2=∅ is redundant since if there exist finite sets B1 and B2 such that K1∩B1=K2∩B2=∅ and h(K1∪B1)=K2∪B2, then we choose the finite sets F1 and F2 as follows F1=B1∖h-1(C),F2=B2∖C, where C=h(B1)∩B2.Theorem 2.3 (see [8]).
If X is countable and(X,τ) is SACDH, then τ is the discrete topology.Let(X,τ) be any topological space. For x∈X, let Ax={y∈X:y¯=x¯}, A={x:|Ax|>1}, B={x:|x¯|≥ℵ0}, Cx={y∈X:x∈y¯}, and C={x:|Cx|≥ℵ0}. And Let F={x∈X:{x}isnotaclosedset}.Definition 2.4 (see [8]).
A topological space(X,τ) is called almost To (AT0 ) if |A|<ℵ0. If, in addition, |B|<ℵ0, then (X,τ) is called strong almost To (SATo).Clearly everyTo-space is ATo.Definition 2.5 (see [8]).
A topological space(X,τ) is called almost T1 and denoted by AT1 if it is SATo and |C|<ℵ0, that is, A,B,andC are all finite sets.Definition 2.6 (see [8]).
A topological space(X,τ) is called strong almost T1 space (SAT1) if |F|<ℵo. Clearly every T1 space is SAT1 and every SAT1 is AT1.Theorem 2.7 (see [8]).
Every ACDH space isAT0.Theorem 2.8 (see [8]).
Let(X,τ) be ACDH space. If |X|≤ℵ0, then (X,τ) is AT1.Now let us define a new type of almost CDH spaces.Definition 2.9.
A topological space(X,τ) is called an almost CDH of type 1 (ACDH-1) if it is separable space and there exists a finite subset F such that, for any two countable dense subsets A and B of X, there exists h∈H(X) such that h(A∖F)=B∖F. Note that from now on, we will refer to F by a related finite set.Clearly every finite space is an almost CDH of type 1. Moreover, we have the following result.Proposition 2.10.
If(X,τ) is ACDH-1, |X|≤ℵ0, and F is a related finite set, then for all x∉F, we have {x} which is an open set in X.Proof.
Letx∈X∖F. If {x} is not an open set in X, then X∖{x}¯=X, therefore there is h∈H(X) such that h(X∖F)=(X∖{x})∖F which is a contradiction.As an application of Proposition2.10, we have the following corollary.Corollary 2.11.
If(X,τ) is an ACDH-1, |X|≤ℵ0, then there exists the smallest related finite set A.Proof.
Let(X,τ) be a nondiscrete ACDH-1 space and A={x∈X:{x} not open}. As τ≠τdis, A≠ϕ. Also by Proposition 2.10, A⊆F, for all related finite sets F, so A is finite. If K1,K2 are two dense sets and h∈H(X), then X∖A⊆K1∩K2 and h(X∖A)=X∖A, hence h(K1∖A)=h(X∖A)=X∖A=K2∖A; therefore A is the smallest finite related set. Moreover, if τ=τdis, then A=∅.The following example shows that ACDH-1 space need be a CDH space.Example 2.12.
LetX=N and β={{1,2},{3,4},{5},{6},…}. Let A,B be two dense subsets of X. Then {5,6,7,…}⊆A∩B let F={1,2,3,4}, hence (X,τ(β)) is ACDH-1 space which is not a CDH space as it is not a T1 space.Now let prove the following characterizations of ACDH spaces.Theorem 2.13.
If(X,τ) is a separable space, then the following are equivalent(i)
(X,τ) is an ACDH space(ii)
For any two countable dense subsetsK1,K2, there exist two finite subsets F1,F2 of K1,K2, respectively, and h∈H(X) such that h(K1∖F1)=K2∖F2.(iii)
For any two countable dense subsetsK1,K2, there exist two equipotent finite subsets F1,F2, and h∈H(X) such that h(K1∖F1)=K2∖F2.(iv)
For any two countable dense subsetsK1,K2, there exist two finite subsets F1,F2, and h∈H(X) such that h(K1∖F1)=K2∖F2.Proof.
(i) implies (ii) Suppose that(X,τ) is ACDH space. Let K1,K2 be two countable dense sets. Then there exist two finite sets G1,G2 and h∈H(X) such that G1∩K1=G2∩K2=G2∩h(G1)=∅ and h(K1∪G1)=K2∪G2. Let F1=h-1(G2)andF2=h(G1), so F1,F2 are two finite sets and
(2.1)h(K1∖F1)=(h(K1∪G1)∖h(G1))∖G2=K2∖F2. (ii) implies (iii) Let K1,K2 be two countable dense sets, then there exist two finite sets F1,F2, and h∈H(X) such that F1⊆K1,F2⊆K2, and h(K1∖F1)=K2∖F2. Let G1=F1∪h-1(F2) and G2=h(F1)∪F2. Clearly |G1|=|G2|. Moreover,
(2.2)h(K1∖G1)=h(K1∖F1∪h-1(F2))=K2∖F2.
But clearly (h(F1)∖F2)∩K2=∅, so K2∖F2=K2∖G2. (iii) implies that (iv) is clear.
(iv) implies (i) LetK1,K2 be two countable dense sets, then there are G1,G2 finite sets and h∈H(X) such that h(K1∖G1)=K2∖G2.
LetF1=h-1(G2∩K2)∖(G1∩K1) and F2=h(G1∩K1)∖(G2∩K2). Then F1,F2 are two finite sets. If t∈F1∩K1, then t∈K1∖G1 and h(t)∈G2∩K2 which gives a contradiction. Similarly, F2∩K2=∅. Moreover,
(2.3)h(K1∪F1)=h(K1)∪[(G2∩K2)∖h(G1∩K1)]=h(K1)∪(G2∩K2)=h((K1∖G1)∪(K1∩G1))∪(G2∩K2)=K2∖G2∪h(K1∩G1)∪(G2∩K2)=K2∪(h(K1∩G1)∖K2)=K2∪F2.Consequently, we have the following result.Corollary 2.14.
Every ACDH-1 space is an ACDH space.Theorem 2.15.
Let(X,τ) be a separable space. Then (X,τ) is an SACDH space if and only if for every two countable dense sets K1,K2, there exist two equipotent finite subsets F1,F2 of K1,K2, respectively, and h∈H(X) such that h(K1∖F1)=K2∖F2.Proof.
LetK1,K2 be two countable dense sets, so there exist h∈H(X), G1⊆K1, G2⊆K2, and |G1|=|G2|<ℵ0 such that h(K1∖G1)=K2∖G2. Let F1=h-1(G2)∖K1 and F2=h(G1)∖K2. It is clear that F1∩K1=F2∩K2=∅. Also,
(2.4)|F1|=|h-1(G2)∖K1|=|h-1(G2)∖(K1∩h-1(G2))|=|h-1(G2)|-|K1∩h-1(G2)|=|G1|-|(h(K1∖G1)∪h(G1))∩G2|=|G1|-|h(G1)∩G2|=|G1|-|G2∩h(G1)∪((K2∖G2)∩h(G1))|=|h(G1)|-|K2∩h(G1)|=|F2|.
Moreover,
(2.5)h(K1∪F1)=h(K1)∪h(F1)=h(K1)∪(G2∖h(K1))=h(K1)∪G2=h((K1∖G1)∪G1)∪G2=K2∪h(G1)=K2∪(h(G1)∖K2)=K2∪F2.
Hence, (X,τ) is an SACDH space.
Conversely, assume that(X,τ) is an SACDH space. Let K1,K2 be two countable dense sets, therefore, there are two finite sets G1,G2 with |G1|=|G2| and h∈H(X) such that h(K1∪G1)=K2∪G2; moreover, K1∩G1=G2∩K2=h(G1)∩G2=∅. Let F1=h-1(G2) and F2=h(G1). Then |F1|=|F2|, moreover, F1⊆K1 and F2⊆K2. We need to prove that h(K1∖F1)=K2∖F2. Now, h(K1∖F1)=h(K1∖h-1(G2))=h(K1)∖G2. Claim that h(K1)∖G2=K2∖F2. If x∈h(K1)∖G2, then x∈(K2∪G2)∖G2=K2. As K1∩G1=∅, x∉h(G1)=F2, so x∈K2∖F2. To prove the other inclusion, suppose that x∈K2∖F2=K2∖h(G1)=h(h-1(K2)∖G1). Therefore, x=h(t), for some t∈h-1(K2)∖G1, and then t∈(K1∪G1)∖G1 so t∈K1. Also t∉h-1(G2), hence x∈h(K1)∖G2.Consequently, we have the following result.Corollary 2.16.
Every ACDH-1 space is SACDH.


## 3. T-Homogeneous Spaces

A transposition onX is a permutation on X which exchanges the places of two elements x,y, while leaving all the other elements unchanged. Now we will define the Transposition-Homogeneous (TH) spaces, and we will show that every TH SACDH space is a CDH space.Definition 3.1.
A space(X,τ) is called Transposition-Homogeneous (TH) space if every transposition on X is a homeomorphism.Proposition 3.2.
A space(X,τ) is (TH) if and only if, for any two finite subsets F1,F2 with the same cardinality and for every h∈H(X), there exists h′∈H(X) such that(1)
h′(F1)=F2,(2)
h′(x)=h(x), for every x∉F1∪h-1(F2),(3)
h′(h-1(F2))=h(F1).Proof.
Let(X,τ) be a TH space and F a finite subset of X. Let σ be a permutation which fixes X∖F, clearly σ is a composition of finite transpositions which is a homeomorphism. Let F1,F2 be two finite subsets with the same cardinality and h∈H(X), clearly |F1∖h-1(F2)|=|h-1(F2)∖F1| and F=(F1∖h-1(F2))⊔(h-1(F2)∖F1) are a disjoint union of finite sets which is finite. Let f:F→F be a bijection such that f(F1∖h-1(F2))=h-1(F2)∖F1. For each x∈F1∖h-1(F2), let σx be a transposition on X which transposes x and f(x). Now let σ be the finite composition of the transpositions σx,x∈F1∖h-1(F2). Then σx(y)=y for all y∈X∖F, σ(F1)=h-1(F2), and σ(h-1(F2))=F1. Now h′=h∘σ is the required function since h′(x)=h(x), for every x∈X∖F.
The converse is obvious by choosingh to be the identity and F1={x}, F2={y}.Example 3.3.
One can show that the spaces(X,τind), (X,τdis), (X,τcof), and (X,τcoc) are all TH spaces. However, the spaces (ℝ,τl.r) and (ℝ,τu) are not TH spaces.The following example shows that ACDH-1, TH space need be CDH space, hence ACDH TH space need be a CDH space.Example 3.4.
LetX be such that 1<|X|<ℵ0, with the indiscrete topology. The space (X,τ) is a TH-space and it is also ACDH-1, but it is not a CDH space as it is not a T1 space.Theorem 3.5.
If(X,τ) is SACDH, TH space, then (X,τ) is a CDH space.Proof.
IfA, B are two countable dense subsets of X, then there exist two finite subsets F1,F2 of A,B, respectively with |F1|=|F2| and there is h∈H(X) such that h(A∖F1)=B∖F2. As X is a TH space, there exists h′∈H(X) such that h′(F1)=F2, h′(x)=h(x), for all x∉F1∪h-1(F2) and h′(h-1(F2))=h(F1). To show that h′(A)=B we show first that A∩h-1(F2)⊆F1. Suppose that there is x∈(A∩h-1(F2))∖F1. Then h(x)∈B∖F2 which gives a contradiction. Hence h′(A∖F1)=h(A∖F1)=B∖F2, so h′(A)=h′((A∖F1)∪F1)=(B∖F2)∪F2=B.Let(X,τ) be an ACDH-1 space and 𝔉={F:Fisarelatedset} and define D={A⊆X:AisacountabledensesubsetofX}, and also we define the relation ~ on D by: A~B if and only if there is F∈𝔉 such that |A∩F|=|B∩F|. Let D1={(A,B)∈D×D:A~B}.Now we have the following result.Theorem 3.6.
If(X,τ) is an ACDH-1, TH space, and (A,B)∈D1, then there is h∈H(X) such that h(A)=B.Proof.
Suppose that(A,B)∈D1, so there are F∈𝔉 and h∈H(X) such that h(A∖F)=B∖F with |A∩F|=|B∩F|. If A∩F=∅, then we are done. In general let F1=F∩A and F2=F∩B, therefore |F1|=|F2|. As X is a TH space, there exists h′∈H(X) such that h′(F1)=F2andh′(x)=h(x), for all x∉F1∪h-1(F2) and h′(h-1(F2))=h(F1), also A∩h-1(F2)⊆F. Now h′(A∖F1)=h′(A∖F)=h(A∖F)=B∖F=B∖F2, so h′(A)=h′((A∖F1)∪F1)=(B∖F2)∪F2=B.Consequently, we have the following Corollary.Corollary 3.7.
If(X,τ) is an ACDH-1, TH space, and D1=D×D, then (X,τ) is a CDH space.


## 4. Almost CDH Spaces, and New Separation Axioms

We know that almost CDH space is not aT0 space. In this section we will give a characterization of almost T0 space, also we will give a new definition of almost T1 space.Theorem 4.1.
Let(X,τ) be a topological space. Then (X,τ) is AT0 space if and only if there exists a finite subset F of X, such that, for all x≠y and {x,y}∩F=∅, there is an open set containing only one of x, y. We will refer to F by a related finite set F.Proof.
Assume that(X,τ) is an AT0 space. Let F=A. Then for all x≠y with {x,y}∩F=∅, we have that |Ax|=1 and |Ay|=1, hence y∉x¯ or x∉y¯. Conversely, suppose that there exists a finite set F such that for all x≠y with {x,y}∩F=∅, there is an open set containing only one of x, y. If |A| is an infinite, then there exists a denumerable subset of A, say {x1,x2,…}, and |Axn|>1, for all n∈ℕ, so there exist yn∈Axn and yn≠xn for all n∈ℕ. Therefore there is n0∈ℕ such that xn0≠yn0 are both not in F and x¯n0=y¯n0, which gives a contradiction.Definition 4.2.
A space(X,τ) is called an almost strong T1 (AST1) space  if there is a finite subset F of X such that, for all x≠y and {x,y}∩F=∅, there are two open subsets u1,u2 of X, such that x∈u1∖u2 and y∈u2∖u1. F is called the related finite set.One may easily prove the following proposition.Proposition 4.3.
Let(X,τ) be a topological space. If (X,τ) is an AST1 space, then for all x∉F, we have x¯⊆{x}∪F; where F is a related finite set. Conversely, if there is a finite set F such that for all x∉F, x¯⊆{x}∪F, then (X,τ) is an AST1.In the following results we show that the new separation axiomAST1 is stronger than the one defined by Fora et al. in [8].Proposition 4.4.
EveryAST1 space is AT1 space.Proof.
Suppose that(X,τ) is an AST1 space, and let F be a related finite set. By Theorem 4.1 it is an AT0, therefore, |A|<ℵ0. By Proposition 4.3, for all x∉F, we have x¯⊆{x}∪F, so that B⊆F. If |C|≥ℵ0, then there is xn∈C such that for all m≠n, xn≠xm and |Cxn|≥ℵ0, for all n∈ℕ. Let n1 be such that xn1∉F. Therefore |Cxn1|≥ℵ0, then there is t∈Cxn1∖F, so xn1∈t¯, which gives a contradiction, and hence the proposition is proved.The following example shows that the converse of the previous proposition need not be true.Example 4.5.
LetX=ℕ,:β={{1},{1,2},{3},{3,4},{5},{5,6},…} so β is a base for some topology on X. Note that for n∈ℕ, we have
(4.1)n¯={{n,n+1};nisodd,{n};niseven.
Therefore, for allx∈ℕ, Ax={x}, hence A=∅, B=∅ as for all n∈ℕ, |n¯|≤2. Now for n∈ℕ(4.2)Cn={{n-1,n};niseven,{n};nisodd.
Therefore, C=∅, and then (X,τ(β)) is AT1 space. Let F be any finite subset of X. Let m=sup:F, as 2m+2∈Cl{2m+1} and 2m+2,2m+1 are both not in F, (X,τ(β)) is not an AST1 space.One may easily prove the following proposition.Proposition 4.6.
EverySAT1 space is AST1 space.Fitzpatrick et al. proved in [6] that every CDH space is a T1 space. Indeed, Tallafha et al. in [7] gave us another proof for the same argument by using the idea of closurely ordered sets. Now, we will prove that every ACDH-1 space is AST1 space by using the idea of almost closurely ordered sets.Definition 4.7 (see [7]).
Let(X,τ) be a topological space. A countable subset K of X is said to have the closurely ordered property if there exists a numeration of K, say K={x1,x2,…} such that for all n≥2, xn∉Cl{x1,x2,…,xn-1}. The numeration {x1,x2,…} is called closurely ordered countable set.Definition 4.8 (see [7]).
A countable collection𝔄 of subsets of X is said to have the closurely ordered countable property if 𝔄 can be written as 𝔄={A1,A2,…}, where An∩Cl{⋃i=1n-1Ai}=∅. The form {A1,A2,…} is called closurely ordered countable family.Theorem 4.9 (see [7]).
Let(X,τ) be a topological space and let K be any countable dense subset of X. Then there exists a countable dense subset K1 of K, such that K1 is closurely ordered countable set.Theorem 4.10 (see [7]).
Let(X,τ) be a topological space, then,(i)
ifh:X→Y is an injective open function and K has the closurely ordered property in X, then h(K) has the closurely ordered property in Y,(ii)
having closurely ordered property, is a topological property,(iii)
every subset of a set having closurely ordered property must have closurely ordered property.Now let us define the following.Definition 4.11.
A countable setK in (X,τ) is said to have the almost closurely ordered property if there is a finite set F in X such that K∖F has the closurely ordered property. If K∖F={x1,x2,…} is a closurely ordered set, then K is called almost closurely ordered set.Proposition 4.12.
If(X,τ) is an AST1 space and F is a related finite set, then each doubleton {x,y}⊆X∖F has the closurely ordered property. Conversely, in a topological space (X,τ) if there exists a finite set F all doubletons {x,y}⊆X∖F have the closurely ordered property, then (X,τ), is an AT0 space.Proof.
The first part is clear. To prove the converse, assume that there is such a finite setF. Let x,y be such that x≠y and {x,y}⊆X∖F. So x∉y¯ or y∉x¯, by Theorem 4.1 (X,τ) is an AT0.Theorem 4.13.
Every ACDH-1 space isAT0 space.Proof.
Assume that(X,τ) is an ACDH-1 space and F is a related finite set. We want to show that F is the desired set. If x≠y with {x,y}∩F=∅ and K is a countable dense subset of X, by Theorem 4.9, we may assume that K has the closurely ordered property, also by Theorem 4.10, K∖F has the closurely ordered property. Now {x,y}∪K is also a countable dense subset of X, therefore there is h∈H(X) such that h(K∖F)=(K∪{x,y})∖F. So {x,y} has the closurely ordered property, the result follows by Theorem 4.10.Theorem 4.14 (see [7]).
If(X,τ) a topological space and K is a countable dense subset, then there exists a countable collection of subsets A1,A2,… of K such that(i)
A=⋃n=1mAn⊆K:m≤ℵ0,(ii)
{A1,A2,…} is closurely ordered countable family,(iii)
eachAk has the closurely ordered property,(iv)
eachAk is either a singleton or an infinite set,(v)
A¯=K¯,(vi)
ifAk is a singleton, say {ak}, then ak∉x¯ and x∉Cl{ak}, for all x∈A∖{ak},(vii)
ifAk={a1k,a2k,…} is infinite set, then aik∈Cl{a1+ik}, for all i.It is easy to prove the following result.Proposition 4.15.
The properties (i)–(vii) in the last theorem are all preserved under homeomorphisms.We now prove the following theorem, that will be used to prove our main result.Theorem 4.16.
Let(X,τ) be an ACDH-1 space and let F be a related finite set. If x≠yand{x,y}∩F=∅ and x∈y¯, then x¯ is an infinite set.Proof.
Suppose thatx≠y,{x,y}∩F=∅ and x∈y¯. Suppose that x¯ is a finite set, then there is N∈ℕ such that |x¯|=N. Let K be any countable dense set in X, then by Theorem 4.14, there is a countable collection of subsets of K say A1,A2,… satisfing the conditions (i)–(vii). So, A=⋃n=1mAn⊆K and A¯=K¯=X. Let I={i:|Ai|=ℵo}. For i∈I, define Bi=Ai∖{a1i,a2i,…,aNi} and for i∉IBi=Ai={ai}, also define B=⋃i=1mBi. To show that B¯=A¯. If i∈I, then Cl{a1k,a2k,…,aNk}⊆Cl{aN+1i}⊆B¯i, therefore A¯i⊆B¯i∪Cl{a1i,a2i,…,aNi}⊆B¯i. Then B is a countable dense set in X and so is B∪{x,y}. Therefore, there exists h∈H(X) such that h((B∪{x,y}∖F))=B∖F. As x∉F, we have h(x)∈Bi for some i∈ℕ. If |Bi|=1, then Bi={bi} and h(x)=bi, so bi∈h(y¯) where y∉F; therefore, h(y)∈B∖{bi} which is a contradiction by Theorem 4.14(vi). If |Bi|=ℵ0 and h(x)=bnoi for some no>N, so all bik,b2i,…,bni are in h(x)¯. As |h(x)¯|=|x¯|>N, which is impossible, so x¯ is an infinite set.Recall that in ACDH-1 space, ifK is a countable dense subset of X, then by Theorem 4.14 there are countable subsets A1,A2,… of K satisfying (i)–(vii) of the pointed theorem.Moreover,A=⋃i=1mAi⊆K, A¯=K¯=X. Therefore, there is h∈H(X) such that h(A∖F)=K∖F. By Proposition 4.15, K∖F can be decomposed in the same way as A∖F.The following theorem shows that all the aboveAis are singletons.Theorem 4.17.
Let(X,τ) be an ACDH-1 space, K any countable dense subset of X, and F a related finite set. Then K∖F=⋃i=1∞Ai and |Ai|=1.Proof.
LetI={i:|Ai|=ℵ0}. If I≠∅, then Ai={a1i,a2i,…}, for some i∈I. We have a1i∈Cl{a2i} and {a1i,a2i}∩F=∅, so, by Theorem 4.16, we have Cl{a1i} which is an infinite set. Let a0i∉F with a0i∈Cl{a1i}∖{a1i}. In a similar way, let a-1i∉F with a-1i∈Cl{a0i}∖{a0i}. By the same argument, we have a sequence …,a-nia-n+1i,…,a-1i,a0i and a-ki∈Cl{a-k+1i}∖{a-k+1i}. Now we claim that for all k,n∈ℕ,a-ki≠ani. If a-ki=ani for some n≥2 and k≥0, then ani∈Cl{a1i} which contradicts the fact that {a1i,a2i,…} is a closurely ordered set. Also a1i≠a-ki for all k>0, since a1i∉Cl{a0i} and a-ki∈Cl{a0i}, so we proved our claim. Let K1=[⋃i∉IAi]∪[⋃i∈I{…,a-2ia-1i,a0i,a1i,a1i,…}]. Then X=K¯⊆K1∪F¯, hence K1∪F is a countable dense subset of X. Then there is h∈H(X) such that h(K∖F)=(K1∪F)∖F=K1. For i∉I,Ai={ai} and ai∉x¯, for all x∈(K∖F)∖{ai}. Then by Proposition 4.15, h(Ai)={aj}=Aj, j∉I. Now define i0=inf(I), therefore Ai0={a1i0,a2i0,…}. Moreover, h(a1i0)∈Aj0, for some jo∈I, where Aj0={…,a-2j0,a-1j0,⋯,a1j0,a2j0,…}. Then h(a1i0)=akj0, for some k∈ℤ. Also x∈Cl{a1i0}, where x∈K∖F and x=h-1(ak-1j0). If x∈Ar, for some r∈I, then Ar∩Cl{⋃j=1r-1Aj}≠∅ which is a contradiction, so I=ϕ.As a consequence of the previous theorem, we have the following results.Corollary 4.18.
If(X,τ) is an ACDH-1 space, F is a related finite set, and K is a countable dense subset of X, then K∖F has the closurely ordered property.Proof.
IfK is a countable dense subset of X, then by using Theorem 4.17, we have that K∖F=⋃i=1∞Ai and |Ai|=1, for all i∈ℕ. Therefore,K∖F=⋃i=1∞ai indeed, ai∉Cl{aj}, for all i≠j.Corollary 4.19.
Every ACDH-1 space is anAST1 space.Proof.
LetF be a related finite set and x≠y with {x,y}∩F=∅. If K is a countable dense subset of X, say K={x1,x2,…}, then the set K1={x,y,x1,x2,…} is also a countable dense subset of X, therefore by Corollary 4.18, we have that K1∖F has the closurely ordered property and {x,y}⊆K1∖F, therefore y∉x¯. Similarly, x∉y¯.



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*Source: 375293-2010-05-05.xml*