New Types of Almost Countable Dense Homogeneous Space

Abstract

In 1972, Bennett studied the countable dense homogeneous (CDH) spaces and in 1992, Fitzpatrick, White, and Zhou proved that every CDH space is a 𝑇₁ 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.

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 ℝ^𝑛 has the property that if 𝐴,𝐵 are countable dense subsets of ℝ^𝑛, then there is a homeomorphism ℎ of ℝ^𝑛 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 Section 4 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 subset 𝐴 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,𝜏_ind), (𝑋,𝜏_dis), (𝑋,𝜏_cof), and (𝑋,𝜏_coc) denote the set 𝑋 with the indiscrete, discrete, cofinite, and cocountable topologies, and (ℝ,𝜏_𝑙⋅𝑟),(ℝ,𝜏_𝑟⋅𝑟), and (ℝ,𝜏_𝑢) 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 condition ℎ(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 𝐴_𝑥={𝑦∈𝑋∶𝑦=𝑥}, 𝐴={𝑥∶|𝐴_𝑥|>1}, 𝐵={𝑥∶|𝑥|≥ℵ₀}, 𝐶_𝑥={𝑦∈𝑋∶𝑥∈𝑦}, and 𝐶={𝑥∶|𝐶_𝑥|≥ℵ₀}. And Let 𝐹={𝑥∈𝑋∶{𝑥}isnotaclosedset}.

Definition 2.4 (see [8]). A topological space (𝑋,𝜏) is called almost 𝑇_𝑜 (𝐴𝑇₀ ) if |𝐴|<ℵ₀. If, in addition, |𝐵|<ℵ₀, then (𝑋,𝜏) is called strong almost 𝑇_𝑜 (𝑆𝐴𝑇_𝑜).

Clearly every 𝑇_𝑜-space is 𝐴𝑇_𝑜.

Definition 2.5 (see [8]). A topological space (𝑋,𝜏) is called almost 𝑇₁ and denoted by 𝐴𝑇₁ if it is 𝑆𝐴𝑇_𝑜 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|<ℵ_𝑜. Clearly every 𝑇₁ space is 𝑆𝐴𝑇₁ and every 𝑆𝐴𝑇₁ is 𝐴𝑇₁.

Theorem 2.7 (see [8]). Every ACDH space is 𝐴𝑇₀.

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. Let 𝑥∈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 Proposition 2.10, 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≠𝜏_dis, 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=𝜏_dis, then 𝐴=∅.

The following example shows that ACDH-1 space need be a CDH space.

Example 2.12. Let 𝑋=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 subsets 𝐾₁,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 subsets 𝐾₁,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 subsets 𝐾₁,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 ℎ𝐾₁⧵𝐹₁=ℎ𝐾₁∪𝐺₁⧵ℎ𝐺₁⧵𝐺₂=𝐾₂⧵𝐹₂.(2.1) (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, ℎ𝐾₁⧵𝐺₁=ℎ𝐾₁⧵𝐹₁∪ℎ⁻¹𝐹₂=𝐾₂⧵𝐹₂.(2.2) But clearly (ℎ(𝐹₁)⧵𝐹₂)∩𝐾₂=∅, so 𝐾₂⧵𝐹₂=𝐾₂⧵𝐺₂. (iii) implies that (iv) is clear.
(iv) implies (i) Let 𝐾₁,𝐾₂ be two countable dense sets, then there are 𝐺₁,𝐺₂ finite sets and ℎ∈𝐻(𝑋) such that ℎ(𝐾₁⧵𝐺₁)=𝐾₂⧵𝐺₂.
Let 𝐹₁=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, ℎ𝐾₁∪𝐹₁==ℎℎ𝐾𝐾₁₁∪𝐺₂∩𝐾₂⧵ℎ𝐺₁∩𝐾₁=ℎ𝐾₁∪𝐺₂∩𝐾₂=𝐾₂⧵𝐺₁∪𝐾₁∩𝐺₁∪𝐺₂∩𝐾₂=𝐾₂⧵𝐺₂∪ℎ𝐾₁∩𝐺₁∪𝐺₂∩𝐾₂=𝐾₂∪ℎ𝐾₁∩𝐺₁⧵𝐾₂∪𝐹₂.(2.3)

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. Let 𝐾₁,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, ||𝐹₁||==||||ℎℎ⁻¹₋₁𝐺₂⧵𝐾₁||=||ℎ𝐺₂⧵𝐾₁∩ℎ⁻¹𝐺₂₋₁||=||𝐺₁𝐺₂||−||𝐾₁∩ℎ⁻¹𝐺₂||=||𝐺₁||−||ℎ𝐾₁⧵𝐺₁∪ℎ𝐺₁∩𝐺₂||=||𝐺₁||−||ℎ𝐺₁∩𝐺₂||=||ℎ||𝐺−||𝐺₂∩ℎ𝐺₁∪𝐾₂⧵𝐺₂∩ℎ𝐺₁₁||=||𝐹₂||−||𝐾₂∩ℎ𝐺₁||||.(2.4) Moreover, ℎ𝐾₁∪𝐹₁==ℎℎ𝐾𝐾₁₁∪ℎ𝐹₁=ℎ𝐾₁∪𝐺₂⧵ℎ𝐾₁∪𝐺₂=ℎ𝐾₁⧵𝐺₁∪𝐺₁∪𝐺₂==𝐾𝐾₂₂∪ℎ𝐺₁=𝐾₂∪ℎ𝐺₁⧵𝐾₂∪𝐹₂.(2.5) 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 on 𝑋 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 𝜎_𝑥 be a transposition on 𝑋 which transposes 𝑥 and 𝑓(𝑥). Now let 𝜎 be the finite composition of the transpositions 𝜎_𝑥,𝑥∈𝐹₁⧵ℎ⁻¹(𝐹₂). Then 𝜎_𝑥(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 choosing ℎ to be the identity and 𝐹₁={𝑥}, 𝐹₂={𝑦}.

Example 3.3. One can show that the spaces (𝑋,𝜏_ind), (𝑋,𝜏_dis), (𝑋,𝜏_cof), and (𝑋,𝜏_coc) are all TH spaces. However, the spaces (ℝ,𝜏_𝑙.𝑟) and (ℝ,𝜏_𝑢) 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. Let 𝑋 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. If 𝐴, 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 a 𝑇₀ 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 |𝐴_𝑥|=1 and |𝐴_𝑦|=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 |𝐴_𝑥𝑛|>1, for all 𝑛∈ℕ, so there exist 𝑦_𝑛∈𝐴_𝑥𝑛 and 𝑦_𝑛≠𝑥_𝑛 for all 𝑛∈ℕ. Therefore there is 𝑛₀∈ℕ such that 𝑥_𝑛0≠𝑦_𝑛0 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 axiom 𝐴𝑆𝑇₁ is stronger than the one defined by Fora et al. in [8].

Proposition 4.4. Every 𝐴𝑆𝑇₁ 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 𝑥_𝑛∈𝐶 such that for all 𝑚≠𝑛, 𝑥_𝑛≠𝑥_𝑚 and |𝐶_𝑥𝑛|≥ℵ₀, for all 𝑛∈ℕ. Let 𝑛₁ be such that 𝑥_𝑛1∉𝐹. Therefore |𝐶_𝑥𝑛1|≥ℵ₀, then there is 𝑡∈𝐶_𝑥𝑛1⧵𝐹, so 𝑥_𝑛1∈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. Let 𝑋=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 𝑛={{𝑛𝑛,}𝑛;+1};𝑛𝑛iisseovdedn,.(4.1)
Therefore, for all 𝑥∈ℕ, 𝐴_𝑥={𝑥}, hence 𝐴=∅, 𝐵=∅ as for all 𝑛∈ℕ, |𝑛|≤2. Now for 𝑛∈ℕ𝐶_𝑛={{𝑛𝑛}−;1,𝑛};𝑛𝑛iisseovdedn.,(4.2) 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. Every 𝑆𝐴𝑇₁ 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, 𝑥_𝑛∉Cl{𝑥₁,𝑥₂,…,𝑥_𝑛−1}. The numeration {𝑥₁,𝑥₂,…} is called closurely ordered countable set.

Definition 4.8 (see [7]). A countable collection 𝔄 of subsets of 𝑋 is said to have the closurely ordered countable property if 𝔄 can be written as 𝔄={𝐴₁,𝐴₂,…}, where 𝐴_𝑛∩Cl{⋃^{𝑛−1𝑖=1}𝐴_𝑖}=∅. 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)if ℎ∶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 set 𝐾 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 set 𝐹. 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 is 𝐴𝑇₀ 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=⋃^𝑚𝑛=1𝐴_𝑛⊆𝐾∶𝑚≤ℵ₀,(ii){𝐴₁,𝐴₂,…} is closurely ordered countable family,(iii)each 𝐴_𝑘 has the closurely ordered property,(iv)each 𝐴_𝑘 is either a singleton or an infinite set,(v)𝐴=𝐾,(vi)if 𝐴_𝑘 is a singleton, say {𝑎_𝑘}, then 𝑎_𝑘∉𝑥 and 𝑥∉Cl{𝑎_𝑘}, for all 𝑥∈𝐴⧵{𝑎_𝑘},(vii)if 𝐴_𝑘={𝑎^𝑘1,𝑎^𝑘2,…} is infinite set, then 𝑎^𝑘𝑖∈Cl{𝑎^𝑘1+𝑖}, 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 that 𝑥≠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=⋃^𝑚𝑛=1𝐴_𝑛⊆𝐾 and 𝐴=𝐾=𝑋. Let 𝐼={𝑖∶|𝐴_𝑖|=ℵ_𝑜}. For 𝑖∈𝐼, define 𝐵_𝑖=𝐴_𝑖⧵{𝑎^𝑖1,𝑎^𝑖2,…,𝑎^𝑖𝑁} and for 𝑖∉𝐼𝐵_𝑖=𝐴_𝑖={𝑎_𝑖}, also define 𝐵=⋃^𝑚𝑖=1𝐵_𝑖. To show that 𝐵=𝐴. If 𝑖∈𝐼, then Cl{𝑎^𝑘1,𝑎^𝑘2,…,𝑎^𝑘𝑁}⊆Cl{𝑎^𝑖𝑁+1}⊆𝐵_𝑖, therefore 𝐴_𝑖⊆𝐵_𝑖∪Cl{𝑎^𝑖1,𝑎^𝑖2,…,𝑎^𝑖𝑁}⊆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)∈𝐵_𝑖 for some 𝑖∈ℕ. If |𝐵_𝑖|=1, then 𝐵_𝑖={𝑏^𝑖} and ℎ(𝑥)=𝑏^𝑖, so 𝑏^𝑖∈ℎ(𝑦) where 𝑦∉𝐹; therefore, ℎ(𝑦)∈𝐵⧵{𝑏^𝑖} which is a contradiction by Theorem 4.14(vi). If |𝐵_𝑖|=ℵ₀ and ℎ(𝑥)=𝑏^𝑖𝑛_𝑜 for some 𝑛_𝑜>𝑁, so all 𝑏^𝑘𝑖,𝑏^𝑖2,…,𝑏^𝑖𝑛 are in ℎ(𝑥). As |ℎ(𝑥)|=|𝑥|>𝑁, which is impossible, so 𝑥 is an infinite set.

Recall that in ACDH-1 space, if 𝐾 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=⋃^𝑚𝑖=1𝐴_𝑖⊆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 above 𝐴_𝑖s 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=1𝐴_𝑖 and |𝐴_𝑖|=1.

Proof. Let 𝐼={𝑖∶|𝐴_𝑖|=ℵ₀}. If 𝐼≠∅, then 𝐴_𝑖={𝑎^𝑖1,𝑎^𝑖2,…}, for some 𝑖∈𝐼. We have 𝑎^𝑖1∈Cl{𝑎^𝑖2} and {𝑎^𝑖1,𝑎^𝑖2}∩𝐹=∅, so, by Theorem 4.16, we have Cl{𝑎^𝑖1} which is an infinite set. Let 𝑎^𝑖0∉𝐹 with 𝑎^𝑖0∈Cl{𝑎^𝑖1}⧵{𝑎^𝑖1}. In a similar way, let 𝑎^𝑖−1∉𝐹 with 𝑎^𝑖−1∈Cl{𝑎^𝑖0}⧵{𝑎^𝑖0}. By the same argument, we have a sequence …,𝑎^𝑖−𝑛𝑎^{𝑖−𝑛+1},…,𝑎^𝑖−1,𝑎^𝑖0 and 𝑎^𝑖−𝑘∈Cl{𝑎^{𝑖−𝑘+1}}⧵{𝑎^{𝑖−𝑘+1}}. Now we claim that for all 𝑘,𝑛∈ℕ,𝑎^𝑖−𝑘≠𝑎^𝑖𝑛. If 𝑎^𝑖−𝑘=𝑎^𝑖𝑛 for some 𝑛≥2 and 𝑘≥0, then 𝑎^𝑖𝑛∈Cl{𝑎^𝑖1} which contradicts the fact that {𝑎^𝑖1,𝑎^𝑖2,…} is a closurely ordered set. Also 𝑎^𝑖1≠𝑎^𝑖−𝑘 for all 𝑘>0, since 𝑎^𝑖1∉Cl{𝑎^𝑖0} and 𝑎^𝑖−𝑘∈Cl{𝑎^𝑖0}, so we proved our claim. Let 𝐾₁=[⋃_𝑖∉𝐼𝐴_𝑖]∪[⋃_𝑖∈𝐼{…,𝑎^𝑖−2𝑎^𝑖−1,𝑎^𝑖0,𝑎^𝑖1,𝑎^𝑖1,…}]. 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,𝐴_𝑖={𝑎^𝑖} and 𝑎^𝑖∉𝑥, for all 𝑥∈(𝐾⧵𝐹)⧵{𝑎^𝑖}. Then by Proposition 4.15, ℎ(𝐴_𝑖)={𝑎^𝑗}=𝐴_𝑗, 𝑗∉𝐼. Now define 𝑖₀=inf(𝐼), therefore 𝐴_𝑖0={𝑎^𝑖0₁,𝑎^𝑖0₂,…}. Moreover, ℎ(𝑎^𝑖0₁)∈𝐴_𝑗0, for some 𝑗_𝑜∈𝐼, where 𝐴_𝑗0={…,𝑎^𝑗0₋₂,𝑎^𝑗0₋₁,⋯,𝑎^𝑗0₁,𝑎^𝑗0₂,…}. Then ℎ(𝑎^𝑖0₁)=𝑎^𝑗0_𝑘, for some 𝑘∈ℤ. Also 𝑥∈Cl{𝑎^𝑖0₁}, where 𝑥∈𝐾⧵𝐹 and 𝑥=ℎ⁻¹(𝑎^𝑗0_𝑘−1). If 𝑥∈𝐴_𝑟, for some 𝑟∈𝐼, then 𝐴_𝑟∩Cl{⋃^{𝑟−1𝑗=1}𝐴_𝑗}≠∅ 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. If 𝐾 is a countable dense subset of 𝑋, then by using Theorem 4.17, we have that 𝐾⧵𝐹=⋃^∞𝑖=1𝐴_𝑖 and |𝐴_𝑖|=1, for all 𝑖∈ℕ. Therefore,𝐾⧵𝐹=⋃^∞𝑖=1𝑎^𝑖 indeed, 𝑎^𝑖∉Cl{𝑎^𝑗}, for all 𝑖≠𝑗.

Corollary 4.19. Every ACDH-1 space is an 𝐴𝑆𝑇₁ space.

Proof. Let 𝐹 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.

Acknowledgment

This paper is financially supported by the Deanship of Academic Research at the University of Jordan, Amman, Jordan.