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 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 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 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 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 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. Define 푓: 푋 → 퐼 by 푓(푡) = 퐵), since 푋 − {푝} is a normal proper subspace, so there are 푔(푡) at 푡 ≠ 푝 and 푓(푝) = 0. For each 휀 > 0, then 푝 푝 two disjoint open sets 푈퐴 and 푈퐵 in 푋 − {푝}, thus there are 푓−1((휀, 1]) = 푔−1((휀, 1]) and 푓−1([0, 휀)) = 푔−1([0, 휀)) ∪ 푝 two open sets 푈퐴 and 푈퐵 in 푋 such that 푈퐴 = 푈퐴 ∪ 푋 − {푝} {푝} are open sets in 푋. and 푈푝 = 푈 ∩ 푋 − {푝} containing 퐴 and 퐵, respectively, 표 퐵 퐵 Case 2: 퐴 = ∅, since 푋 is regular, there are two disjoint 푝 푝 furthermore 푈 ∩ 푈 ⊆ (푈 ∩ 푈 ) ∪ {푝} ⊆ {푝}. Similarly, open sets 푈 and 푉 such that 푎 ∈ 푈 and 퐴 ⊆ 푉, thus 푈푐 is 퐴 퐵 퐴 퐵 there are two open sets 푉 and 푉 in 푋 such that 푉 ∩ 푉 ⊆ closed, 푎 ≠ 푈푐 and 푈푐표 ≠ ∅, then apply Case 1, there is a 퐴 퐵 퐴 퐵 (푉푞 ∩ 푉푞) ∪ {푞} ⊆ {푞}. Now, 퐻 = 푈 ∩ 푉 and 퐺 = 푈 ∩ continuous function 푓: 푋 → 퐼 such that 푓(푈푐) = 0 and 퐴 퐵 퐴 퐴 퐵 푓(푎) = 1, hence 푓(퐴) = 0 and 푓(푎) = 1, therefore 푋 is 푉퐵 are disjoint open sets containing 퐴 and 퐵, respectively. completely regular. Case 3: There is exactly one point 푝 in 푋 − (퐴 ∪ 퐵). Since, |푋| ≥ 4, then one of the two sets has at least two distinct points 푥 and 푦. Without loss of generality, assume that 1 The last result shows that 푇3 and all the equivalent 2 {푥, 푦} ⊆ 퐴. Since, 푋 − {푥} is a normal proper subspace, thus 1 푥 푥 properties to 푇3 like gauge space [5] are PHP. there are two disjoint open sets 푈 and 푈 in 푋 − {푥} 2 퐴 퐵 containing 퐴 − {푥} and 퐵, respectively. Moreover, there exist 푈 and 푈 open in 푋 such that 푈푥 = 푈 ∩ 푋 − {푥} and 푈 = Based on Al-Bsoul [2] 푇퐷 is PHP without proof. Next, we 퐴 퐵 퐴 퐴 퐵 푈푥 ∩ 푋 − {푥}, also 푈 ∩ 푈 ⊆ (푈푥 ∩ 푈푥) ∪ {푥} ⊆ {푥}. will show that 푇퐷 is PHP. 퐵 퐴 퐵 퐴 퐵 Similarly, there are two disjoint open sets 푉퐴 and 푉퐵 in 푋 − Omar Al-Odat, Adnan Al-Bsoul 15

푦 푦 {푦} such that 푉퐴 = 푉퐴 ∩ 푋 − {푦} and 푉퐵 = 푉퐵 ∩ 푋 − {푦}, so Theorem 3.2. 푇푅-space is PHP. 푦 푦 푃푟표표푓. Let 푋 be a topological space with |푋| ≥ 3 and all 푉퐴 ∩ 푉퐵 ⊆ (푉퐴 ∩ 푉퐵 ) ∪ {푦} ⊆ {푦}. Now, 퐻 = 푈퐴 ∪ 푉퐴 and proper subspaces are 푇 . Let 퐴 ⊆ 푋 and 푝, 푞 are two distinct 퐺 = 푈퐵 ∩ 푉퐵 are disjoint open sets in 푋 containing 퐴 and 퐵, 푅 respectively. Therefore, 푋 is normal.  points in 퐴. Since |푋| ≥ 3, so there is 푥 ∈ 푋 − {푝, 푞}. Moreover, 푋 − {푥} is a 푇푅 proper subspace, then there is 퐵 ⊆ 푋−{푥} 푋−{푥} Corollary 2.7. Completely normal is PHP. 퐴 such that 푝 ∈ 퐵 and 푞 ∉ 퐵 , thus 푝 ∈ 퐵 and 푞 ∉

퐵. Theorem 2.8. Perfectly normal is PHP.

푃푟표표푓. Let 푋 be a topological space with |푋| ≥ 4 and let all It is easy to see that 푆 is PHP. The next Theorem show that proper subspaces are perfectly normal. Assume that 퐴 and 퐵 1 푆 is PHP.  are disjoint closed sets in 푋. 2

Case 1: 퐴표 ≠ ∅ or 퐵표 ≠ ∅. Without loss of generality, Theorem 3.3. 푆 is PHP. assume that 퐴표 ≠ ∅, then there is 푝 ∈ 퐴 has a neighborhood 2 푃푟표표푓. Let 푋 be a topological space with |푋| ≥ 3 such that 푈 ⊆ 퐴. Since, 푋 − {푝} is perfectly normal, hence there is a 0 all proper subspaces are 푆 . Assume that 푝 ≠ 푞 in 푋 with 푝 ∉ continuous function 푔: 푋 − {푝} → 퐼 such that 푔−1(0) = 퐴 − 2 {푞}, since |푋| ≥ 3, then there is 푥 ∈ 푋 − {푝, 푞}. {푝} and 푔−1(1) = 퐵. Define 푓: 푋 → 퐼 by 푓(푡) = 푔(푡) for all 푥 ≠ 푝 and 푓(푝) = 0. For each 휀 > 0, then 푓−1((휀, 1]) = Case 1: 푥 ∉ {푞}, hence there are disjoint open sets 푈푝 and 푔−1([휀, 1)) and 푓−1([0, 휀)) = 푔−1([0, 휀)) ∪ {푝} are open 푈푞 in 푋 − {푥} containing 푝 and 푞, respectively. Also, there are sets in 푋. disjoint open sets 푉푥 and 푉푞 in 푋 − {푝} containing 푥 and 푞, Case 2: 퐴표 = ∅ and 퐵표 = ∅, since 푋 is normal, there are respectively. Thus, there are open sets 푈1, 푈2, 푉1 and 푉2 in 푋 two disjoint open sets 푈 and 푉 in 푋 such that 퐴 ⊆ 푈 and 퐵 ⊆ such that 푈푝 = 푈1 ∩ 푋 − {푥}, 푈푞 = 푈2 ∩ 푋 − {푥}, 푉푥 = 푉1 ∩ 표 푉, thus 푈푐 is closed with 푈푐 ≠ ∅ and 퐴 ∩ 푈푐 = ∅, then 푋 − {푥} and 푉푞 = 푉2 ∩ 푋 − {푥}. Now, 푈 = 푈1 and 푉 = 푈2 ∩ apply Case 1, there are continuous functions 푓1, 푓2: 푋 → 퐼 such 푉2 are disjoint open sets containing 푝 and 푞, respectively. −1 푐 −1 −1 −1 that 푓1 (0) = 푈 , 푓1 (1) = 퐴, 푓2 (0) = 퐵 and 푓2 (1) = Case 2: 푥 ∈ {푞}, then 푝 ∉ {푥}, since 푋 − {푥} and 푋 − {푞} 푐 푉 , hence there is a continuous function 푓: 푋 → 퐼 defined by are proper 푆2-subspaces, then there are disjoint open sets 푈푝 −1 푓(푡) = (푓1(푡) + 푓2(푡)/2) such that 푓 (1) = 퐴 and and 푈푞 in 푋 − {푥} containing 푝 and 푞, respectively, and −1( ) 푓 0 = 퐵, therefore 푋 is perfectly normal.  disjoint open sets 푉푝 and 푉푥 in 푋 − {푞} containing 푝 and 푥, respectively. Hence, there are open sets 푈1, 푈2, 푉1 and 푉2 in 3. Riesz's Separation Axiom and Related 푋 such that 푈푝 = 푈1 ∩ 푋 − {푥}, 푈푞 = 푈2 ∩ 푋 − {푥}, 푉푝 = Axioms 푉1 ∩ 푋 − {푥} and 푉푥 = 푉2 ∩ 푋 − {푥}. Now, 푈 = 푈1 ∩ 푉1 and 푉 = 푈 are disjoint open sets containing 푝, 푞, respectively.  Cs푎́sz푎́r [6] introduced Riesz separation axiom and some 2 related Axioms. In this section, we shall show that these Theorem 3.4. 푃 -space is PHP. separation axioms are properly hereditary properties. 푅 푃푟표표푓. Let 푋 be a topological space with |푋| ≥ 3 and all proper subspaces are 푃 . Let 퐴 ⊆ 푋 and 푝, 푞 be two distinct Definition 3.1. [6] 푅 points in 퐴 such that 푝 ∈ 퐴 − {푞}. Since |푋| ≥ 3, so there is A topological space 푋 is said to be 푥 ∈ 푋 − {푝, 푞}. Since 푋 − {푥} is a proper 푃푅-subspace, so 1. 푇푅 if it is a 푇1-space and for a subset 퐴 of 푋, with 푝 푋−{푥} 푋−{푥} ′ and 푞 different elements in 퐴 , there is a subset 퐵 of there is 퐵푥 ⊆ 퐴 − {푥} such that 푝 ∈ 퐵푥 and 푞 ∉ 퐵푥 , ′ ′ 퐴 such that 푝 ∈ 퐵 and 푞 ∉ 퐵 . thus there is a closed set 퐵 ⊆ 푋 such that 퐵푥 = 퐵 ∩ 푋 − {푥} 2. 푆1 if 푝 ∉ 푞 implies 푞 ∉ 푝 for all 푝, 푞 in 푋. and 퐵 ⊆ 퐵푥 ∪ {푥}, moreover 푞 ∉ 퐵푥 ∪ {푥}, implies 푞 ∉ 퐵.  3. 푆2 if 푝 ∉ {푞} implies that 푝 and 푞 have disjoint neighborhoods for all 푝, 푞 in 푋. Theorem 3.4. 푄푅-space is PHP. 4. 푃푅 if given any two distinct points 푝 and 푞 satisfy 푃푟표표푓. Let 푋 be a topological space with |푋| ≥ 3 and all 푝 ∉ {푞} and 푝 ∈ 퐴 for any subset 퐴 ⊆ 푋, there is a proper subspaces are 푄푅. Let 퐴 ⊆ 푋 and 푝 ≠ 푞 be distinct subset 퐵 ⊆ 퐴 such that 푝 ∈ 퐵 and 푞 ∉ 퐵. points in 퐴 such that 푝 ∈ 퐴 and 푞 ∉ {푝}. Since |푋| ≥ 3, so 5. 푄푅 if given any two distinct points 푝 and 푞 satisfy there is 푥 ∈ 푋 − {푝, 푞}. Since, 푋 − {푥} is a proper 푄푅-space, 푋−{푥} 푞 ∉ {푝} and 푝 ∈ 퐴 for any subset 퐴 ⊆ 푋, there is a so there is 퐵푥 ⊆ 퐴 − {푥} such that 푝 ∈ 퐵푥 and 푞 ∈ subset 퐵 ⊆ 퐴 such that 푝 ∈ 퐵 and 푞 ∉ 퐵. 푋−{푥} 퐵푥 , thus there is a closed set 퐵 ⊆ 푋 such that 퐵푥 = 퐵 ∩ 6. 푆 if for any 푝 ∈ 퐴 where 퐴 is a subset of 푋 and 퐴 is ∞ 푋 − {푥} and 퐵 ⊆ 퐵푥 ∪ {푥}, moreover 푞 ∉ 퐵푥 ∪ {푥}, implies not compact there is a subset 퐵 ⊆ 퐴 such that 푝 ∉ 퐵 푞 ∉ 퐵.  and 퐵 is not compact. To prove the next result we shall give our next observation. Lemma 3.1. [6] Let 푋 be a topological space. 푋 is 푇푅 if given any set 퐴 ⊆ 푋 and two distinct points 푝, 푞 ∈ 퐴, there is 퐵 ⊆ Lemma 3.6. A space 푋 is compact if there exists a point 푝 in 푐 퐴 such that 푝 ∈ 퐵 and 푞 ∉ 퐵. 푋 such that each member in the {푈 : 푈 is an open set containing 푝} is compact. 16 Topological Properties like Separation Axioms Satisfying Properly Hereditary Property

푃푟표표푓. Let 풰 be an open cover of 푋. Then, there is 푈푝 ∈ 풰 constant, so there exist 푎 and 푏 in 푋 such that 푓(푎) ≠ 푓(푏), 푐 containing 푝, thus 푈푝 is compact and 풱 = 풰 − {푈푝} covering hence the restriction function on the proper subspace {푎, 푏} is 푐 a non-constant continuous function, hence the proper 푈푝, so there is finite subcover {푉1, 푉2, … , 푉푛} of 풱, so the class subspace {푎, 푏} is not strongly connected, contradiction. {푉1, 푉2, … , 푉푛} ∪ {푈푝} covering 푋, therefore 푋 is compact.  Therefore, 푋 is strongly connected.  A similar observation is valid for Lindel표̈f and countably compact. Definition 4.3 [8] Space 푋 is locally strongly connected if it has a basis consisting of strongly connected open sets. Theorem 3.7. 푆 -space is PHP. ∞ Lemma (4.1) showed a feature for a connected set. It is easy 푃푟표표푓. Let 푋 be a topological space such that all proper to see that the strongly connected set has the same feature. subspaces are 푆 . Assume that 퐴 ⊆ 푋 such that 퐴 is not ∞ compact and 푝 ∈ 퐴. In each of the two Cases: 퐴 = 푋 and 퐴 ≠ Theorem 4.5. Locally strongly connected is PHP. 푋, there is 퐵 in 퐴 such that 푝 ∉ 퐵 and 퐵 is not compact 푃푟표표푓. Let (푋, 휏) be a topological space with |푋| ≥ 2 such according to Lemma (3.6).  that every proper subspace is local strongly connected. Assume that 푎 ∈ 푋 and 푈 be open in 푋 containing 푎, then 4. Connected Spaces there is 푞 ≠ 푎, since 푋 − {푞} is a proper subspace, thus there is an open strongly connected set 퐵 containing 푎 and 푎 ∈ 퐵 ⊆ In this section we shall show the properties of: locally 푈 − {푞}, implies 퐵 is strongly connected open in 푋.  connected, pathwise-connected, strongly connected, locally

strongly connected and totally disconnected are PHP. Definition 4.4. [3] 푋 is totally disconnected if the only

nonempty connected of 푋 are the one point sets. In Definition 4.1. [7] A space 푋 is locally connected if 푋 has other words, 푋 is totally disconnected if all subspaces 퐴 with base consisting of a connected set. |퐴| > 1 of 푋 are disconnected.

Lemma 4.1. Let 푋 be a topological space and 퐴 be a subspace Theorem 4.6. Totally disconnected is PHP. of 푋. If 퐵 is a connected set in 퐴, then 퐵 is connected in 푋. 푃푟표표푓. Let 푋 be a topological space with |푋| ≥ 3 and each

proper subspace is totally disconnected. Thus, each proper Arenas showed that local connectedness 푇 is properly open 1 subspace 퐴 with |퐴| > 1 is disconnected, then there exists 푎 ∈ hereditary property. Next, we will show that local connected 푋 such that 푋 − {푎} is disconnected, so there are two open sets is properly hereditary property. 푈1 and 푈2 in 푋 such that 푈1 ∩ 푈2 ⊆ {푎} Theorem 4.2. Locally connected is PHP. Case 1: 푎 ∉ (푈1 ∩ 푈2), then 퐴 = 푈1 ∪ {푎} is a proper 푃푟표표푓. Let (푋, 휏) be a topological space with |푋| ≥ 2 such disconnected subspace, so there are two disjoint open sets 푉1 that every proper subspace is locally connected. Assume that and 푉2 in 퐴 such that 푉1 ∪ 푉2 = 퐴, hence 푎 ∈ 푉1 or 푎 ∈ 푉2. 푎 ∈ 푋 and 푈 is an open set in 푋 containing 푎, then there is Without loss of generality, assume that 푎 ∈ 푉2, thus 푉1 is an 푏 ≠ 푎, and since 푋 − {푏} is proper subspace, thus there is an open set in 푈1, then 푉1 is an open set in 푋, since 푉2 is an open open connected set 퐵 in 푋 − {푏} containing 푎 such that 푎 ∈ set in 퐴, so there is an open set 퐺 in 푋 such that 푉2 = 퐺 ∩ 퐴 푐 퐵 ⊆ 푈 − {푏}. It implies that 퐵 is an open connected set in 푋 and 퐺 ∪ 푈2 ⊆ 푉2 ∪ 퐴 ∪ 푈2 = 푉2 ∪ 푈2 ⊆ 퐺 ∪ 푈2, thus 퐺 ∪ according to Lemma (4.1).  푉2 is an open set in 푋, therefore 푉2 ∪ 푈2 and 푉1 are open sets in 푋 such that 푉1 ∩ (푉2 ∪ 푈2) = ∅ and 푉1 ∪ (푉2 ∪ 푈2) = 푋. Theorem 4.3. Pathwise connected is PHP. Case 2: 푎 ∈ 푈1 ∩ 푈2, thus |푈푖| > 1 for every 푖 = 1,2, then 푃푟표표푓. Let 푋 be a topological space with |푋| ≥ 3 such that 푈1 is disconnected, equivalently there are two open sets 푉1 and all proper subspaces are pathwise connected. Assume that 푥 푉2 in 푈1 such that 푉1 ∪ 푉2 = 푈1 and 푉1 ∩ 푉2 = ∅. Since, 푈1 is and 푦 are two distinct points in 푋, since |푋| ≥ 3, there is 푎 ∈ an open set in 푋, then 푉1 and 푉2 are open sets in 푋. Without 푋 − {푥, 푦}, as 푋 − {푎} is a proper pathwise connected loss of generality, Assume that 푎 ∈ 푉2, implies 푉1 and 푉2 ∪ 푈2 subspace, hence there is a continuous function 푔: 퐼 → 푋 − {푎} are open sets in 푋 such that 푉1 ∩ (푉2 ∪ 푈2) = ∅ and 푉1 ∩ such that 푔(0) = 푥 and 푔(1) = 푦, thus the function 푓: 퐼 → 푋 (푉2 ∪ 푈2) = 푋 . So, 푋 is disconnected, therefore 푋 is totally defined by 푓(푡) = 푔(푡) for all 푡 ∈ 퐼 is continuous. Therefore, disconnected.  푋 is pathwise connected. 

Definition 4.2. [8] A space 푋 is strongly connected if any References continuous function 푓: 푋 → (ℤ, 휏푐표푓) is constant. A subset 퐴 [1] F. Arenas, “Topological Properties Preserved by Proper of 푋 is strongly connected if 퐴 is a strongly connected Subspace”, Q and A in General Topology, vol. 14, pp. subspace. 53-57, 1996. [2] A. AL-Bsoul, “Some Separation Axioms and Covering Theorem 4.4. Strongly connected is PHP. Properties Preserved by Proper Subspaces”, Q and A in 푃푟표표푓. Let 푋 be a topological space with |푋| ≥ 3 such that General Topology, vol. 21, pp. 171-175, 2003. all proper subspaces are strongly connected. Suppose that [3] S. Willard, General Topology, Addison-Wesley, 1970. 푓: 푋 → (ℤ, 휏푐표푓) be a continuous function which is not Omar Al-Odat, Adnan Al-Bsoul 17

[4] A. AL-Bsoul, N. Tahat, “Some Types of Cleavability”, [7] P. E. Long, An Introduction to General Topology, Charles Al-Manarah, vol. 4, no. 2, 1999. E. Merrill Publishing Company, Jordan Book Center [5] J. Dugundji, Topology, Allyn and Bacon, Boston, 1966. Company Limited, 1986. [6] K. Cs푎́sz푎́r, “On F. Riesz' Separation Axiom, topics in [8] Y. Wang, “A Stronger Notion of Connectedness”, topology”, Colloquia Mathematics Societatis J'anos National University of Singapore, 2010. Bolyai 8, Keszthely, pp. 173-180, 1972.