New spaces and Continuity via Ω̂-closed sets

In this paper, we introduce new spaces like Ω̂ Tδ and ωTΩ̂. It turns out that the space Ω̂ Tδ coincide with semi-T1 and in ωTΩ̂-space every closed set is Ω̂-closed set and in semi-T 1 2 every Ω̂-closed set is closed in a topological space. Also we introduce some kinds of generalized continuity such as Ω̂-continuity, Ω̂-irresolute and weakly Ω̂-continuity.


Introduction
In the recent years, many new separation axioms were introduced by using generalized closed sets such as [3] T b , [12] T d , [2] semi-T 1 , semi-T 1 2 and [6] T 3 4 . The class ofΩ-closed sets [8] has been introduced recently by using the δ-closure operator. As the class ofΩ-closed sets are independent of the class of closed sets, our interest is to find the spaces in which they are interrelated. With these motivation, new separation axioms viaΩ-closed sets such asΩT δ which coincide with semi-T 1 and ω TΩ in which everyΩ-closed set is closed set are introduced and studied. Moreover, we introduce and study various types of the continuity viaΩ-closed sets. Also we investigate the role of new spaces and it's impacts on all these continuities.
Theorem 3.6. A topological space (X, τ ) isΩT δ -space if and only if every singleton set is semi closed in (X, τ ).
(ii) {x} is semi closed in X for every x ∈ X.
Proof: Since every semi-T 1 -space is semi-T 1 2 and by Theorem 3.6, everyΩT δ -space is semi-T 1 2 -space. ✷ Remark 3.9. Converse is not true in general as seen from the following example.
Proof: Necessity-Suppose that A is any closed set in (X, τ ). Then by [16], A is ω-closed set in X. By hypothesis, A isΩ-closed set in X. Sufficiency-Suppose that A is ω-closed set in X. Then by [16], cl(A) ⊆ sker(A).
Then sker(cl(A)) ⊆ sker(sker(A)) = sker(A). Since cl(A) is closed set in X, by hypothesis, cl(A) isΩ-closed set in X. Sufficiency-Suppose that A isΩ-closed set in X and there exists x ∈ cl(A) \ (A).
, a contradiction to Theorem 4.4 in [8]. In both cases, we arrive at a contradiction. Therefore, cl(A) = A and hence A is closed in X. ✷ Theorem 3.17. In a topological space (X, τ ), the following statements are equivalent.
{x} is either open or semi closed in X for every x ∈ X.
(iv) Every ω-closed set is closed in X.

4.Ω-continuity
Then f is g, δg and ω-continuous but notΩ-continuous. Remark 4.7. The pictorial representation of the above discussions and the existing results are given below. The reversible implication is never possible.

Sup Cont
ContΩ -Cont δg-Cont The following Theorem gives the conditions under which reversible implication of Theorem 4.3 holds.
Theorem 4.8. In a topological space (X, τ ), the following holds.
Proof:  (i) f is perfectly continuous.
(ii) f is completely continuous and contra continuous.
(iii) f is super continuous and contra continuous.
(v) f is ω-continuous and contra continuous.
(vi) f is gα-continuous and contra continuous.
(vii) f is pre-continuous and contra continuous.

Proof:
(i) Since a clopen set is regular closed and open, it holds.
(ii) Since a regular closed set is δ-closed, it holds.
(v) Since a ω-closed set is gα-closed, it holds.
(vi) Since a gα-closed set is pre-closed, it holds.
(vii) If A is pre-closed and open set in (X, τ ), then cl(int(A)) ⊆ A, and A = intA.
Thus cl(A) ⊆ A and hence A is closed. Thus A is clopen in X. Therefore, every pre-continuous and contra continuous function is perfectly continuous.
✷ Remark 4.12. The composition ofΩ-continuous functions is not alwaysΩ-continuous function as seen from the following example.
for all x ∈ X is not aΩ-continuous function.
Proof: Suppose that y ∈ Y and W is any open set in Z containing g(y). Since f is surjective, there exists x ∈ X such that f (x) = y.
Let us characterizeΩ-continuous function.
is any function, then the following statements are equivalent.
(iii) The inverse image of every closed set in (Y, σ) isΩ-closed set in (X, τ ).
Sufficiency-It suffices to show that inverse image of a closed subset of Y isΩclosed subset of X. Suppose that F is any closed set in X. Then by Then f isΩ-irresolute but notΩ-continuous.
Proof: It follows from their definitions. ✷ Proof: It follows from their definitions. ✷ Composition theorems.
Theorem 5.14. The following statements are true.
Proof: They follow straight forward from their definitions. ✷ Let us prove an application ofΩ-irresolute functions as follows. Proof: Let x ∈ (X \ A).Then f (x) = g(x). Since Y is Hausdorff space, there exists two disjoint open sets U and V in Y such that f (x) ∈ U and g(x) ∈ V. Since f iŝ Ω-continuous and by [8] This implies that (X \ A) isΩ-open in (X, τ ) and hence A isΩ-closed set in (X, τ ). ✷ Theorem 5.20. Let (X, τ ) be a semi-T 1 2 -space. If f : (X, τ ) → (Y, σ) isΩcontinuous injective and Y is Hausdorff, then X is Hausdorff.
Proof: Suppose that x, y ∈ X such that x = y. Since f is injective, f (x) = f (y) and since Y is Hausdorff space, there exists two disjoint open sets U and V in Y such that f (x) ∈ U and f (y) ∈ V. Since f isΩ-continuous, f −1 (U ) and f −1 (V ) are disjointΩ-open sets in X containing x and y respectively. Since X is semi- are disjoint open sets in X containing x and y respectively. Thus, (X, τ ) is Hausdorff. ✷ Definition 5.21. A space is said to beΩ-T 2 , if for every distinct pair of points x, y ∈ X there exists disjointΩ-open sets in X such that x ∈ U, y ∈ V.
6. weaklyΩ-continuous If f is weaklyΩ continuous at each x ∈ X, then f is weaklyΩ-continuous on X. Proof: Let x ∈ X and V ∈ O(Y, f (x)). By hypothesis,their exists U ∈ΩO(X, x) such that f (U ) ⊆ V ⊆ cl(V ). Thus, f is weaklyΩ-continuous function. ✷ Remark 6.4. The converse is not possible in general, as seen from the example 6.2.
The following theorem gives the conditions under which the reversible implication of Theorem 6.3 is true.
Proof: Suppose that y ∈ Y and x ∈ {x ∈ X : f (x) = y}.Since f (x) = y in a Hausdorff space Y , there exists two disjoint open sets U and V in Y such that f (x) ∈ U and y ∈ V .Moreover,cl(U ) ∩ V = ∅ implies that y / ∈ cl(U ). Since f is weaklyΩcontinuous,there existsΩ-open set W in X containing x such that f (W ) ⊆ cl(U ). If W {x ∈ X : f (x) = y},then choose a point z ∈ W such that f (z) = y. Then y = f (z) ∈ f (W ) ⊆ cl(U ), a contradiction.Hence x ∈ W ⊆ {x ∈ X : f (x) = y} and hence {x ∈ X : f (x) = y} isΩ-open in X.Therefore, f −1 (y) = {x ∈ X : f (x) = y} isΩ-closed in X. ✷ Definition 6.14. If A ⊆ X, weaklyΩ-continuous retraction of X onto A is a weaklyΩ-continuous function f : X → A such that f (a) = a for each a ∈ A.
Theorem 6.15. If A ⊆ X, and f : X → A is a weaklyΩ-continuous retraction of a Hausdorff space X onto A, then A isΩ-closed in X. If we choose y ∈ U ∩ W ∩ A, then y ∈ A implies that y = f (y) ∈ U and hence f (y) / ∈ cl(V ). That is,there exists y ∈ W such that f (y) / ∈ cl(V ). Therefore, f (W ) cl(V ) for anyΩ-open set in X containing x, a contradiction. Thus, A isΩ-closed in X. ✷ Definition 6.16. A space (X, τ ) is called Urysohn space if for every pair of distinct points x and y in X, there exists open sets U and V in X such that x ∈ U, y ∈ V and cl(U ) ∩ cl(V ) = ∅.
Proof: Suppose that x, y ∈ X such that x = y. Since f is injective, f (x) = f (y) and since Y is Urysohn space, there exists two open sets U and V in Y such that f (x) ∈ U, f (y) ∈ V and cl(U ) ∩ cl(V ) = ∅. Since f is weaklyΩ-continuous, there existsΩ-open set U 1 and V 1 in X containing x and y respectively such that f (U 1 ) ⊆ cl(U ), f (V 1 ) ⊆ cl(V ). Moreover, U 1 ∩ V 1 = ∅. Thus, X isΩ-T 2 -space. ✷