Journal of Computer Science & Computational Mathematics, Volume 8, Issue 2, June 2018 DOI: 10.20967/jcscm.2018.02.001 Topological Properties like Separation Axioms Satisfying Properly Hereditary Property Omar Al-Odat, Adnan Al-Bsoul* Department of Mathematics, Yarmouk University, Irbid - Jordan *Corresponding author email: [email protected] Abstract: In this paper we shall improve the definition of "properly 2 that the topological properties: Functionally Hausdorff, 푇1 , hereditary property" which raised previously [1] and we shall prove 3 completely regular, normal, perfectly normal and R0 are PHPs. that the topological properties: Functionally Hausdorff, 푻 ퟐ, ퟏ ퟑ completely regular, normal, perfectly normal, S2, 푺∞, locally Definition 2.1. [3] A topological space X is functionally connected, locally strongly connected, totally disconnected and Hausdorff if for any two distinct points a and b in X, there is Riesz separation axiom: TR are properly hereditary properties. a continuous function 푓: 푋 → 퐼 such that 푓(푎) = 0 and푓(푏) = Keywords: hereditary property, proper subspace, completely regular, 1. functionally Hausdorff, Riesz's separation, strongly connected. Theorem 2.1. Functionally Hausdorff is PHP. 푃푟표표푓. Let X be a topological space with |푋| ≥ 3 and all 1. Introduction proper subspaces are functionally Hausdorff. Since, all proper subspaces are functionally Hausdorff, then all proper In 1996, Arenas [1] introduced a definition of properly subspaces are T , so X is T . Assume that x and y are distinct hereditary property as: A property P of a space X is properly 1 1 points in X, since |푋| ≥ 3, so there is 푧 ∈ 푋 − {푥, 푦}. Hence, hereditary property if every proper subspace A of X has the X-{z} and X-{y} are functionally Hausdorff proper subspaces, property P, then the whole space X has the property P. The then there are continuous functions 푓 : 푋 − {푦} → 퐼 and property P is properly (closed, open, etc. respectively) 푦 hereditary property if every (closed, open, etc. respectively) 푓푧: 푋 − {푧} → 퐼 such that 푓푦(푥) = 푓푧(푥) = 1 and 푓푦(푧) = proper subspace has the property P, hence the whole space X 푓푧(푦) = 0. has the property P. If the property P is hereditary and properly Define 푓: 푋 → 퐼 by 푓(푡) = 푓푦(푡)푓푧(푡) for all 푡 ∉ {푦, 푧} and hereditary property, then it is called a strongly properly 푓(푡) = 0 otherwise. Since 푓 is a multiplication of two hereditary property. continuous functions on 푋 − {푦, 푧}, so 푓 is a continuous function on 푋 − {푦, 푧}, hence we shall show that 푓 is a All proper subspaces of the trivial topology on X= {a, b} continuous function on {푦, 푧}. Let 휀 > 0. 1 1 1 2 are Ti and the whole space is not Ti for i=0, , , 1, 1 , 1 , 2, Case 1: 푡 = 푦, so there is an open set 푈 in 푋 − {푧} 4 2 3 3 1 1 containing 푦 such that 푓 (푈) ⊆ [0, 휀), since 푋 is 푇 , then {푧} 2 , 3, 3 , 4, 5, 6. Also, Al-Bsoul [2, Example 2.1] introduced 푧 1 2 2 is a closed set in 푋, moreover 푋 − {푧} is an open set in 푋, so a topological space where every proper subspace is 푇1 and the 푈 is open in 푋, also 푓(푢) = 푓푦(푢)푓푧(푢) < 푓푧(푢) < 휀 for any 4 whole space is not 푇1. So, we shall rewrite the definition of 푢 ≠ 푦 and 푓(푦) = 0 < 휀, then푓(푈) ⊆ [0, 휀). 4 Case 2: 푡 = 푧, so similarly for 푡 = 푦. proper hereditary property as the next definition. Thus, 푓 is a continuous function satisfy 푓(푥) = 1 and 푓(푦) = 0, therefore 푋 is functionally Hausdorff. Definition 1.1. A property P is properly hereditary property denoted by PHP if there exists a cardinal number 휇 such that Definition 2.2. [4] A space 푋 is 푅 -space if any open set is a given any topological space X with |푋| ≥ 휇 and all proper 0 union of closed sets. subspaces have the property P, then the whole space has the property P. It is easy to see that 푋 is 푅 -space if for each open set 푈 in 0 푋 and 푎 in 푈, there is a closed set 퐴 such that 푎 ∈ 퐴 ⊆ 푈. In this paper, |A| denotes the cardinal number of the set A and I denotes the topological space ([0,1],휏푢). Theorem 2.2. 푅0-space is PHP. 푃푟표표푓. Let 푋 be a topological space with |푋| ≥ 3 and all 2. Properties Looks like Separation Axioms proper subspaces are 푅0. Given an open set 푈 in 푋. Arenas [1] proved that Ti for all i=0, 1, 2, 3 are properly Case 1: 푈 = 푋, we are done. hereditary properties. In 2003, Al-Bsoul [2] showed that some Case 2: 푈 ⊂ 푋, then there is 푥 ∉ 푈. Let 푝 ∈ 푈. Since, |푋| ≥ 1 1 1 1 3, then there is 푦 ∈ 푋 − {푥, 푝} of non-familiar separation axioms: 푇 , 푇 , T푇1 , 푇2 are 4 2 3 2 1. 푦 ∈ 푈. Since, 푋 − {푦} is a proper 푅 -subspace, properly hereditary properties. In this section, we shall show 0 hence there is a closed set 퐹푦 in 푋 − {푦} such that 푝 ∈ 퐹푦 ⊆ 푈 − {푦}, so there is a closed set 퐹 ⊆ 푋 such 14 Topological Properties like Separation Axioms Satisfying Properly Hereditary Property that 퐹푦 = 퐹 ∩ 푋 − {푦}, then 푝 ∈ 퐹 ⊆ 퐹푦 ∪ {푦} ⊆ (푈 − {푦}) ∪ {푦} ⊆ 푈. A set 퐴 is locally closed if 퐴 − 퐴 is closed. A topological 2. 푦 ∉ 푈, since 푋 − {푥} and 푋 − {푦} are proper space 푋 is 푇퐷 if each singleton point is locally closed. subspaces, then there are closed sets 퐹푥 and 퐹푦 in 푋 − {푥} and 푋 − {푦}, respectively, such that 푝 ∈ 퐹푥 ⊆ 푈 Theorem 2.5. 푇퐷 is PHP. and 푝 ∈ 퐹푦 ⊆ 푈. Moreover, there are closed sets 퐹1 푃푟표표푓. Let 푋 be a topological space with |푋| ≥ 3 such that all proper subspaces are 푇 . Assume that 푥 ∈ 푋. and 퐹2 in 푋 such that 퐹푥 = 퐹1 ∩ 푋 − {푥} and 퐹푦 = 퐷 퐹2 ∩ 푋 − {푦}. Define 퐹 = 퐹1 ∩ 퐹2, so 푝 ∈ 퐹 = 퐹1 ∩ Case 1: If {푥} = 푋, since |푋| ≥ 3, so there are 푎, 푏 ∈ {푥} − 퐹2 ⊆ (퐹푥 ∪ {푥}) ∩ (퐹푦 ∪ {푦}) = 퐹푥 ∩ 퐹푦 ⊆ 푈. {푥}. So, 푋 − {푎} and 푋 − {푏} are proper 푇퐷 subspaces, then Therefore, the whole space 푋 is 푅0-space. {푥} − {푎, 푥} and {푥} − {푏, 푥} are closed sets in {푥} − {푎} and {푥} − {푏} respectively. Hence, there are closed sets of 퐹푎 and Definition 2.3. A space 푋 is 푇 2 if every compact subset of 푋 1 퐹푏 in 푋 such that 푋 − {푥, 푎} = 퐹푎 ∩ (푋 − {푎}) and 푋 − 3 is closed. {푥, 푏} = 퐹푏 ∩ 푋 − {푏} 1. If 푎 ∈ 퐹푎 or 푏 ∈ 퐹푏, then 퐹푎 = 푋 − {푥} = {푥} − {푥} 1 Definition 2.4. [2] A space 푋 is 푇1 if every convergent or 퐹푏 = 푋 − {푥} = {푥} − {푥}, so {푥} − {푥} is closed 3 sequence converges to unique limit point. in 푋. 2. If 푎 ∉ 퐹푎 and 푏 ∉ 퐹푏, then 퐹푎 = 푋 − {푎, 푥} and 퐹푏 = 2 1 푋 − {푏, 푥}, thus 퐹푎 ∪ 퐹푏 = 푋 − {푎, 푥} ∪ 푋 − It is easy to show that every 푇1 is 푇1 . Last idea will help 3 3 {푏, 푥} = 푋 − ({푎, 푥} ∩ {푏, 푥}) = 푋 − {푥}, so {푥} − 2 us to show that 푇1 is properly hereditary property. 3 {푥} is closed in 푋. Case 2: If {푥} ≠ 푋, then {푥} is a proper subspace, so {푥} − 2 Theorem 2.3. 푇1 is PHP. 3 {푥} is closed in {푥}, thus it is closed in 푋. 푃푟표표푓. Let 푋 be a topological space with |푋| ≥ 3. Assume 2 that all proper subspaces of 푋 are 푇1 , then each proper Definition 2.5. [3] A space 푋 is perfectly normal if for each 3 pair of disjoint closed sets 퐴 and 퐵 in 푋 there is a continuous 1 1 subspaces are 푇1 , hence 푋 is 푇1 . Suppose that 퐹 is a −1 −1 3 3 function 푓: 푋 → 퐼 such that 퐴 = 푓 (0) and 퐵 = 푓 (1). A nonclosed compact set in 푋, then there exists 푝 ∉ 퐹 and 푝 ∈ space 푋 is 푇6 if 푋 is a 푇1 perfectly normal space. 퐵푑(퐹), so there is a net (푥휆) in 퐹 such that 푥휆 → 푝, since (푥휆) is a net in compact subspace 퐹, then (푥휆) has a convergent Al-Bsoul proved that 푇4, 푇5 and 푇6 are properly hereditary subnet (푥 ) in 퐹 to 푞 ∈ 퐹, also 푋 is 푇 1, then 푝 = 푞, implies 휆훽 1 properties. Next Theorems show that normal, completely 3 2 normal and perfectly normal are properly hereditary 퐹 is closed, therefore 푋 is 푇1 . 3 properties. Theorem 2.4. The completely regular is PHP. Theorem 2.6. Normality is PHP. 푃푟표표푓. Let 푋 be a topological space with |푋| ≥ 3 such that 푃푟표표푓. Let 푋 be a topological space with |푋| ≥ 4 and all all proper subspaces are completely regular. Assume that 퐴 be proper subspaces are normal space. Assume that 퐴 and 퐵 are a closed set in 푋 and 푎 ∉ 퐴 disjoint closed sets in 푋 표 Case 1: 퐴 ≠ ∅, then there is 푝 ∈ 퐴 such that 푝 has a Case 1: 푋 = 퐴 ∪ 퐵, thus 퐴 and 퐵 are clopen sets, hence 푋 is neighborhood 푈표 ⊆ 퐴. Since, 푋 − {푝} is completely regular, normal. hence there is a continuous function 푔: 푋 − {푝} → 퐼 such that Case 2: There are at least two points 푝 and 푞 in 푋 − (퐴 ∪ 푔(퐴 − {푝}) = 0 and 푔(푎) = 1.