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Universitas Scientiarum

Print version ISSN 0122-7483

Univ. Sci. vol.23 no.1 Bogotá Jan./Apr. 2018

http://dx.doi.org/10.11144/javeriana.sc23-1.bosa 

Artículos

Between Open Sets and Semi-Open Sets

Intermedio entre conjuntos abiertos y semiabiertos

Intermédio entre conjuntos abertos e semiaberto

Samer Al Ghour1  , Kafa Mansur1  ; Juan Carlos Salcedo-Reyes, Edited by1 

1Jordan University of Science and Technology, Faculty of Science and Arts, Department of Mathematics and Statistics 22110, Irbid, Jordan, 3030. algore@just.edu.jo 3104880610

Abstract

We introduce and investigate ω s-open sets as a new class of sets which lies strictly between open sets and semi-open sets. Then we use ω s-open sets to introduce ω s-continuous functions as a new class of functions between continuous functions and semi-continuous functions. We give several results and examples regarding our new concepts. In particular, we obtain some characterizations of ω s-continuous functions.

Key words: Semi-open set; ω-open set; Semi-continuous function

Resumen

Introducimos e investigamos los conjuntos ω s -abiertos como una nueva clase de conjuntos que se ubica estrictamente entre los conjuntos abiertos y semi-abiertos. Usamos los conjuntos ω s -abiertos para introducir las funciones ω s -continuas como un nuevo tipo de funciones que se encuentran entre las funciones continuas y semicontinuas. Proporcionamos varios resultados y ejemplos relacionados con nuestros nuevos conceptos. En particular, obtenemos algunas caracterizaciones de las funciones ω s -continuas.

Palabras-clave: Conjunto semiabierto; Conjunto ω-abierto; Función semicontinua

Resumo

Introduzimos e investigamos os conjuntos ω s-abertos como uma nova classe de conjuntos que se localiza estritamente entre os conjuntos abertos e semiabertos. Usamos os conjuntos ω s-abertos para introduzir as funções ω s-contínuas como um novo tipo de função que se encontram entre as funções contínuas e semicontinuas. Proporcionamos vários resultados e exemplos relacionados com nossos novos conceitos. Particularmente, obtemos algumas caracterizações das funções ω s-contínuas.

Palavras-Chave: Conjunto semiaberto; Conjunto ω-aberto; Função semicontínua

Introduction

Let (X, τ) be a topological space and . We will denote the complement of A in X, the closure of A, the interior of A, the exterior of A, and the relative topology on A, by X - A, , Int(A), Ext(A), and τA, respectively. In 1963, Levine [7] defined semi-open sets as a class of sets containing the open sets as follows: A is semi-open if there exists an open set U such that , this is equivalent to say that . Using semi-open sets he also generalized continuity by semi-continuity as follows: A function is semi-continuous if for all , the preimage . The complement of a semi-open set is called semi-closed[5]. A point is called a condensation point [6] of A if for every with , the set is uncountable. Hdeib [6] defined ω-closed sets and ω-open sets as follows: A is called ω-closed if it contains all its condensation points. The complement of an ω-closed set is called ω-open. The collection of all ω-open sets of a topological space (X, τ) will be denoted by τω. In [1], the author proved that (X, τω ) is a topological space and . Moreover, it was observed that A is ω-open if and only if for every x in A there is an open set U and a countable subset C such that . The ω-closure of A in (X, τ), denoted by , is the smallest ω-closed set in (X, τ) that contains A (cf. [1]). The ω-interior of A in (X, τ), denoted by Int ω(A), is the largest ω-open set in (X, τ) contained in A. The ω-exterior of A in (X, τ), denoted by Ext ω(A), is defined to be Int ω(X - A). It is clear that the ω-closure (resp. ω-interior) of A in (X, τ) equals the closure (resp. interior) of A in (X, t ω). In 2002, Al-Zoubi and Al-Nashef [2] used ω-open sets to define semi ω-open sets as a weaker form of semi-open sets as follows: A is semi ω-open if there exists an ω-open set U such that . The collection of all semi ω-open sets of a topological space (X, τ) will be denoted by SωO (X, τ). Al-Zoubi [4] used semi ω-open sets to introduce semi -continuous functions as a weaker form of -continuous functions as follows: A function is semi ω-continuous [4] if for all , the preimage . This paper is devoted to define ωs-opennes as a property of sets that is strictly weaker than openness and stronger than semi-openness as follows: A is ωs-open if there exists an open set U such that . We investigate this class of sets, and use it to study a new property of functions strictly between continuity and semi-continuity, and another new property of functions strictly between slight continuity and slight semi-continuity.

Throughout this paper R, N, Q, and Qc, will denote the set of real numbers, the set of natural numbers, the set of rational numbers, and the set of irrational numbers, respectively. For any non-empty set X we denote by τdisc the discrete topology on X. Finally, by τu we mean the usual topology on R.

The following sequence of theorems will be useful in the sequel:

Theorem 1.1 ([3]). Let (X, τ) be a topological space and . Then

(a) If A is non-empty, then (τA)ω = (τω)A.

(b) (τω) ω = τω.

Theorem 1.2 ([2]). Let (X, τ) be a topological space. Then

(a) SO (X, τ) SωO (X, τ), and SO (X, τ) ≠ SωO (X, τ) in general.

(b) τ ω SωO (X, τ), and t ω SωO (X, τ) in general.

Theorem 1.3 ([1]). Let (X, τ) be a topological space. Then

(a) If (X, τ) is anti-locally countable, thenfor all, and Intω (A) = Int(A) for all ω-closed set A in (X, τ).

(b) If (X, τ) is locally countable, then τω is the discrete topology.

ωs -Open sets

Definition 2.1. Let A of be a subset of a topological space (X, τ). Then A is called ωs-open of (X, τ), if there exists such that and A is called ωs-closed if X - A is ωs-open. The family of all ωs-open subsets of (X, τ) will be denoted by ωs(X, τ).

Theorem 2.2. Let (X, τ) be a topological space. Then τ ω s (X, τ) SO (X, τ).

Proof. Let Aτ Take U = A. Then Uτ and . This shows that A ∈ ωs (X, τ). It follows that . Let A ∈ ωs (X, τ). Then there exists U ∈ t such that , but . Thus A ∈ SO (X, τ). This shows that .

In the following example we will see that, in general, neither of the two inclusions in Theorem 2.2 are equalities:

Example 2.3. Consider (R, τ), where . It is not difficult to check that . Thus Q ∈ SO (X, τ) - ωs(X, τ) and R - N ∈ ωs(X, τ) - τ.

Theorem 2.4. Let (X, τ) be a topological space. Then

(a) If (X, τ) is anti-locally countable, then ωs(X, τ) = SO (X, τ).

(b) If (X, τ) is locally countable, then t = ωs(X, τ).

Proof. (a) By Theorem 2.2 it is sufficient to show that SO (X, τ) ω s (X, τ). Let A ∈ SO (X, τ). Then there exists Uτ such that UAU. Since (X, τ) is anti-locally countable, then by Theorem 1.3 (a), U = U .It follows that A ∈ v, (X, τ).

(b) By Theorem 2.2 it is sufficient to show that co s (X, τ) C τ. Let us take A ∈ co s (X, τ). Then there exists U ∈ t such that U C A C U . Since (X, τ) is locally countable, then by Theorem 1.3 (b), U ω = U. It follows that A = U and hence At.

The following example shows that v-open sets and ω s-open sets are independent:

Example 2.5. Consider (R, τ) where . It is not difficult to check that . Thus [-1, ∞) ∈ ωs(X, τ) - Tω and (0, ∞) ∈ Tω - ωs(X, Τ).

Theorem 2.6. A subset A of a topological space (X, T) is ωs -open if and only if .

Proof. Necessity. Let A be ωs -open. Then there exists some U ∈ T such that . Since U A, then U = Int(U)Int(A) and so . Therefore, A .

Sufficiency. Suppose that A . Take U = Int(A). Then U ∈ T with . It follows that A is ωs-open.

Theorem 2.7. Arbitrary unions of ωs-open sets in a topological space is ωs-open.

Proof. Let (X, τ) be a topological space and let . For each , there exists Uα ∈ τ such that . So, we have

, with .

It follows that .

Corollary 2.8. If is a collection of ω s -closed subsets of a topological space (X, τ), thenis ω s -closed.

The following example shows that the intersection of two ω s-open sets need not to be ω s-open in general:

Example 2.9. Consider (R, τu). Let A = [0,1], B = [1,2]. By Theorem 1.3 (a), . Thus A, B ∈ ωs(X, τ), but .

Theorem 2.10. For any topological space, the intersection of two ωs-open sets where one of them is open is also ωs-open.

Proof. Let (X, τ) be a topological space, A G t and B G ωs(X, τ). Choose a set U ∈ τ such that . Now we have , and then . This shows that, .

Corollary 2.11. For any topological space, the union of two ωs-closed sets where one of them is closed is also ωs-closed.

Theorem 2.12. Let (X, τ) be a topological space, B a non-empty subset of X and AB. Then

(a) If A ∈ ωs (X, τ), then A ∈ ωs (B, tB).

(b) If B ∈ τ and A ∈ ωs (B, τB), then A ∈ ωs (X, τ).

Proof. (a) Let A ∈ ωs(X, τ). Then there is U ∈ τ such that . Then . Note that is the closure of U in (τω)B and by Theorem 1.1 (a), it is the closure of U in B ) ω. This shows that Aω s (B, τ B).

(b) Let Bτ and A ∈ ms(B, τ B). Since A ∈ ω s (B, τ B), there is V ∈ τ B such that VAH where H is the closure of V in (B, B)m). Since Bτ, then V ∈ τ. Also,. Therefore, A ∈ ω s (X, τ).

Theorem 2.13. Let (X, τ) be a topological space. Let A ω s (X, τ) and suppose that, then B ∈ ωs(X, τ).

Proof. Since A ∈ ωs(X, τ), there exists U ∈ τ such that . Since . Therefore, we have U ∈ τ and . This shows that B ∈ ωs(X, τ).

Theorem 2.14. For any topological space (X, τ) we have that SO (X, τω) = ωs(X, τω).

Proof. By Theorem 2.2, we have ωs(X, τω)//rSO (X, τω). Conversely, let A ∈ SO (X, tm), then there exists U ∈ τω such that U AH, where H is the closure of U in (X, τ ω ). By Theorem 1.1 (b), we have ω ) ω = τ ω and so H = U .It follows that A co s (X, τ ω ).

Theorem 2.15. For any topological space (X, τ) we have the relation τ = {Int(A): Aω s (X, τ)}.

Proof. It follows because from Theorem 2.2 we have τω s (X, τ).

Theorem 2.16. A subset C of a topological space (X, τ) is ω s -closed if and only if Int ω (C)C.

Proof. Necessity. Suppose that C is ω s-closed in (X, τ). Then X - C is ω s-closed and by Theorem 2.6, X - C. So

Sufficiency. Suppose that . Then

By Theorem 2.6 it follows that X - C is ωs-open, and hence C is ωs-closed.

Definition 2.17. Let (X, τ) be a topological space and let AX.

(a) The ωs-closure of A in (X, τ) is denoted by and defined as follows:

(b) The co s-interior of A in (X, τ) is denoted by Int ωs (A) and defined as follows:

Remark 2.18. Let (X, τ) be a topological space and let A X. Then

(a) is the smallest ωs-closed set in (X, τ) containing A.

(b) A is ωs-closed in (X, τ) if and only if A = .

(c) Int ωs(A) is the largest ωs-open set in (X, τ) contained in A.

(d) A is ωs-open in (X, τ) if and only if A = Int ωs(A).

(e) x ∈ if and only if for every B ∈ ωs(X, τ) with .

(f) Int ωs(X - A).

(g) X = Int ωs(X - A) U.

(h) X - = Int ωs(X - A) and X - Int ωs(X - A) = .

Theorem 2.19. Let be an open function such that is continuous. Then for every we have .

Proof. Let A ∈ ωs(X, α). Then there exists U ∈ τ such that , and so . Since is open, then . Since is continuous, then . It follows that .

The condition “open function” cannot be dropped from Theorem 2.19 as shown by:

Example 2.20. Consider , where f (x) = 0 for all x ∈ R. Then it is obvious that is continuous. On the other hand, .

ωs-Continuous functions

Definition 3.1. A function is called o s-continuous, if for each V ∈ α, the preimage .

Theorem 3.2. The notions of continuity satisfy that

(a) Every continuous function is ωs-continuous.

(b) Every ωs-continuousfunction is semi-continuous.

Proof. Theorem 2.2. □

The following example will show that the converse of each of the two implications in Theorem 3.2 is not true in general:

Example 3.3. Let f, g: , with τ as in Example 2.3 and

Since τ, then f is ωs continuous but not continuous. Also, Since then f is semi-continuous but not ωs-continuous.

Theorem 3.4. Letbe a function.

(a) If (X, τ) is locally countable, then f is continuous if and only if f is ωs -continuous.

(b) If (X, τ) is anti-locally countable, then f is ωs -continuous if and only if f is semi-continuous.

Proof. (a) It is a consequence of Theorems 2.4 (a) and 3.2 (a).

(b) It is a consequence of Theorems 2.4 (b) and 3.2 (b). □

Theorem 3.5. A functionis ωs -continuous if and only if for every x ∈ X and every open set V containing f (x) there exists U ∈ ωs (X, τ) such that x ∈ U and f (U)V.

Proof. Necessity. Assume that is ωs-continuous. Let us take V ∈ α with f (x) ∈ V. By ωs-continuity, f-1(V) ∈ ωs (X, τ). Set U = f -1 (V). Then U ∈ ωs (X, τ) satisfies x ∈ U and f (U)V.

Sufficiency. Let V ∈ α. For each x ∈ f -1( V) we have f (x) ∈ V, and thus there exists U x ωs (X, τ) such that x ∈ U x, f (U x )V, and x ∈ U x f-1(V). Thus . Therefore, by Theorem 2.7, it follows f-1(V) ∈ ωs (X, τ). This shows that f is ωs-continuous.

Theorem 3.6. Let be a function. Then the following conditions are equivalent:

(a) The function f is o ¡-continuous.

(b) Inverse images of all members of a base B for a are in ωs (X, τ).

(c) Inverse images of all closed subsets of (Y, σ) are ωs -closed in (X, τ).

(d) For every AX we have.

(e) For every B Y we have.

(f) For every B Y we have.

Proof. . Obvious.

. Suppose is a base for a such that f -1(B) ∈ ωs (X, τ) for every B. Let C be a non-empty closed subset of (Y, σ). Then .

Choose such that . Then

.

By assumption f-1(B) ∈ ωs(X, τ) for every *, then by Theorem 2.7 we have X - f-1(C) ∈ ωs(X, τ), and hence f -1(C) is ωs-closed in (X, τ).

. Let A X. Then is closed in (Y, σ), and by (c) , and , and thus .

Lemma 3.7. Let (X, τ) be a topological space and let A X. Then

.

Proof. Since is ωs-closed, then by Theorem 2.15 Int">/. Therefore, , and hence . To see that , it is sufficient to show that. Therefore,

and by Theorem 2.15 it follows that is ωs-closed. □

Theorem 3.8. Letbe a function. Then the following statements are equivalent:

(a) f is ωs -continuous.

(b) For every AX we have.

(c) For every B Y we have.

Proof. . Suppose that f is ωs-continuous. Let A X. Then by Theorem 3.6 (d), . Therefore, by Lemma 3.7 we have .

. We will apply Theorem 3.6 (d). Let A X. Then by (b), we have . Also, we have always. Therefore, by Lemma 3.7 we have

. Suppose that f is ωs-continuous. Let B Y. Then by Theorem 3.6 (e), . Therefore, by Lemma 3.7 we have .

. We will apply Theorem 3.6 (e). Let B Y. Then by (c), we have . Also, we have always. Therefore, by Lemma 3.7 we have.

Theorem 3.9. Ifis ωs -continuous andis continuous, then g ois a o ¡-continuous.

Proof. Let . Since g is continuous, then . Since f is ωs-continuous, then . □

In general, the composition of two ωs-continuous functions does not need to be ωs-continuous as the following example clarifies:

Example 3.10. Let , where

Then

Since f and g are obviously semi-continuous and is anti-locally countable, then by Theorem 3.4 (b) f and g are ωs-continuous. On the other hand, since , then g o f is not ωs-continuous.

Theorem 3.11. Let be a family offunctions. If the functionis ωs -continuous, then f α is ωs -continuous, for every αA.

Proof. Suppose that f is ωs-continuous and let where is the projection function on Yβ. Since Πβ is continuous, then by Theorem 3.9, f β is ωs-continuous.

The following example will show that the converse of Theorem 3.11 is not true in general:

Example 3.12. Define by

.

Since f and g are obviously semi-continuous, and is anti-locally countable, then by Theorem 3.4 (b) f and g are ωs-continuous. On the other hand, since , then h is not ωs-continuous.

Theorem 3.13. Letbe a family of functions. Ifis ωs -continuous for some, and if fα is continuous for all , then the function is ωs -continuous.

Proof. We will apply statement (b) of Theorem 3.6. Let A be a basic open set of , without loss of generality we may assume that A = , where is a basic open set of for all i = 0,1, ..., n. Then

By assumption for all i = 0,1, ..., n. Thus , and by Theorem 2.9, we have that . It follows that f is ωs-continuous.

Corollary 3.14. Let be a function and denote bythe graph function of f given by g (x) = (x, f (x)), for every xX. Then g is ωs -continuous if and only iff is ωs -continuous.

Proof. Necessity. Suppose that g is ωs-continuous. Then by Theorem 3.11, f is ωs-continuous.

Sufficiency. Suppose that f is ωs-continuous. Note that h(x) = (I(x), f (x)) where is the identity functions. Since the function I is continuous, then by Theorem 3.13, g is ωs-continuous.

References

[1] Al Ghour S. Certain covering properties related to paracompact-ness. Doctoral dissertation, University of Jordan, Jordan 1999. https://theses.ju.edu.jo/Origmal_Abstract/JUF0492638/JUF0492638.pdfLinks ]

[2] Al-Zoubi K, Al-Nashef B. Semi ω-open subsets, Abhath Al-Yarmouk, 11:829-838, 2002. [ Links ]

[3] Al-Zoubi K, Al-Nashef B. The topology of ω-open subsets, Al-Manarah Journal, 9:169-179, 2003. [ Links ]

[4] Al-Zoubi K. Semi ω-continuous functions, Abhath Al-Yarmouk, 12:119-131, 2003. [ Links ]

[5] Crossley SG, Hildebrand SK. Semi-closure, Texas Journal of Sciences, 22:99-112,1971. [ Links ]

[6] Hdeib H. ω-closed mappings, Revista Colombiana de Mathematicas, 16:65-78, 1982. [ Links ]

[7] Levine N. Semi-open sets and semi-continuity in topological spaces, The American Mathematical Monthly, 70:36-41, 1963. [ Links ]

Samer Al Ghour He is a professor of mathematics at Jordan University of Science Technology and obtained his PhD degree in mathematics at University of Jordan. His research interests includes homogeneity and classes of sets in the structures: Topology, Generalized Topology, Bitopology and Fuzzy Topology.

Kafa Mansur She is graduated with a master's degree in mathematics from Jordan University of Science Technology. She is working now a teacher at the Jordanian Ministry of Education.

4Funding: N.A.

5Electronic supplementary material: N.A.

Citation: Al Ghour S, Mansur K Between Open Sets and Semi-Open Sets, Universitas Scientiarum, 23 (1): 9-20, 2018. doi: 10.11144/Javeriana.SC23-1.bosa

Received: June 28, 2017; Accepted: December 2017

Conflic of interest

The authors declare that they have no conflicts of interest to disclse.

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