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 τ_{ω}. In ^{[1]}, the author proved that (X, τ_{ω}
*)* 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) S

ωO(X,τ), andSO(X,τ) ≠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 *A* ∈ *τ* Take *U = A.* Then *U* ∈ *τ* 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 *U* ∈ *τ* such that *U**A**U*. 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 *A* ∈ *t.*

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

Corollary 2.8. If *is a collection of ω*
_{s}
*-closed subsets of a topological space (X*, *τ)*, *then**is 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, τ_{u}). 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 (τ_{ω})_{B} and by Theorem 1.1 (a), it is the closure of *U* in *(τ*
_{B}
*)*
_{ω}. This shows that *A* ∈ *ω*
_{s}
*(B*, *τ*
_{B}).

(b) Let *B* ∈ *τ* and *A* ∈ m_{s}(B, *τ*
_{B}). Since A ∈ *ω*
_{s}
*(B*, *τ*
_{B}), there is V ∈ *τ*
_{B} such that *V**A**H* where *H* is the closure of *V* in *(B*, *(τ*
_{B})_{m}). Since *B* ∈ *τ*, 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 *A**H*, where *H* is the closure of *U* in *(X*, *τ*
_{ω}
*).* By Theorem 1.1 (b), we have *(τ*
_{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): *A* ∈ *ω*
_{s}
*(X*, τ)}.

*Proof.* It follows because from Theorem 2.2 we have *τ**ω*
_{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 ω_{s}(A) and defined as follows:

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

(a) is the smallest ω

_{s}-closed set in (X, τ) containing 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).

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 ω_{s} continuous but not continuous. Also, Since
then *f* is semi-continuous but not ω_{s}-continuous.

**Theorem 3.4**. *Let**be a function.*

(*a*) *If (X*, *τ) is locally countable, then f is continuous if and only if f is* ω_{s}
*-continuous.*

(*b*) *If (X*, *τ) is anti-locally countable, then f is* ω_{s}
*-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 function**is* ω_{s}
*-continuous if and only if for every x ∈ X and every open set V containing f (x) there exists U ∈* ω_{s}
*(X*, *τ)*
*such that x ∈ U and f (U)**V.*

*Proof. Necessity.* Assume that is ω_{s}-continuous. Let us take *V ∈ α* with *f (x) ∈ V*. By ω_{s}-continuity, f^{-1}(V*) ∈* ω_{s}
*(X*, *τ).* Set *U = f*
^{-1}
*(V).* Then *U ∈* ω_{s}
*(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}
*∈* ω_{s}
*(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 inω_{s}(X,τ).

(

c)Inverse images of all closed subsets of (Y, σ) areω_{s}-closed in (X,τ).

. Suppose is a base for *a* such that *f*
^{-1}(B) ∈ ω_{s}
*(X*, *τ)* for every *B* ∈ . Let C be a non-empty closed subset of (Y, σ). 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 ω_{s}-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 ω_{s}-closed. □

**Theorem 3.8**. *Let**be a function. Then the following statements are equivalent:*

(

a)f isω_{s}-continuous.

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 ω_{s}-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**. *If**is* ω_{s}
*-continuous and**is continuous, then g o**is 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:

Then

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

Theorem 3.11. Let *be a family offunctions. If the function**is* ω_{s}
*-continuous, then f*
_{α}
*is* ω_{s}
*-continuous, for every α* ∈ *A.*

*Proof.* Suppose that *f* is ω_{s}-continuous and let where is the projection function on Y_{β}. Since Π_{β} is continuous, then by Theorem 3.9, *f*
_{β} is ω_{s}-continuous.

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

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

**Theorem 3.13**. *Let**be a family of functions. If**is* ω_{s}
*-continuous for some*, and if fα is continuous for all , then the function *is* ω_{s}
*-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 by**the graph function of f given by g (x) = (x*, *f (x)), for every x* ∈ *X. Then g is* ω_{s}
*-continuous if and only iff is* ω_{s}
*-continuous.*

*Proof. Necessity.* Suppose that *g* is ω_{s}-continuous. Then by Theorem 3.11, *f* is ω_{s}-continuous.

*Sufficiency.* Suppose that *f* is ω_{s}-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 ω_{s}-continuous.