Generalized Pre α-Regular and Generalized Pre α-Normal Spaces in Topological Spaces

The concept of separation axioms constitutes a key role in general topology and all generalized forms of topologies. The present authors continued the study of gpα-closed sets by utilizing this concept, new separation axioms, namely gpα-regular and gpα-normal spaces are studied and established their characterizations. Also, new spaces namely gpα-T k for k = 0, 1, 2 are studied


Introduction
O.Njastad [1] introduced and defined α-open sets.Following the work on α-open sets, many topologists focused on generalization of topological concepts using semi-open and α-open sets.These sets play an important role in the generalization of continuity in topological spaces.Mashhour et al .[2,3] proposed pre-open sets.Since then, many topologists have applied these ideas to investigate the weak separation axioms, weak regularity and weak normality.
Proof: Let f: X → Y be bijective and pre-gpα open and X is gpα-T0 space.Let y1, y2 ∈ Y with y1≠y2.As f is bijective, and let x1, x2 ∈ X such that p(x1) = y1, p(x2) = y2.Then there exists gpαopen set U in X as a result x1 ∈ U, x2 ∉ U as X is gpα-T0.So, f(U) is a gpα-open set containing f(x1) but not f(x2).So, there exists gpα-open set f(U) ∈ N such that y1∈ f(U), n2∉f(U).Thus Y is gpα-T0 space.Theorem 3.7: For a topological space X, let gpα-O(X) is open under arbitrary union.The attributes listed below are equivalent: (i) X is gpα-T0, (ii) Each singleton set is gpα-closed, (iii) Each subset of X is the intersection of all gpα-open set containing it, (iv) The set {y} is the intersection of all gpα-open set containing the point x ∈ X.
Proof: (i) → (ii) Let x ∈ X where X is gpα-T0.Then for each y ∈ X such that y ≠ x, there exists gpα-open set U y containing y but not x.Therefore, y ∈ Uy ⊆ {x} c .Then {x} c = ∪{ U y : y ∈ {x} c }, that is {x} c is the union of gpα-open sets.Hence {x} is gpα-closed.(ii) → (iii) Assume that (ii) holds.Let A ⊆ X.Then for each point y ∉ A, there exists {y} c such that A⊆ {y} c where {y} c is gpα-open in X. Hence A = ∩{{y} c : y ∈ A c } and as a result, set A is the intersection of all gpα-open sets containing A. (iii) → (iv) Obvious.(iv) → (i) Assume that (4) holds.Let x, y ∈ X with x ≠ y.According to the hypothesis, there is a gpα-open set Ux, such that x ∈ Ux and y ∉ Ux, so the gpα-T0 space condition is satisfied.As a result, X is gpα-T0 space.Definition 3.8: A topological space (X, τ) is said to be a gpα-T1 space if for each pair of distinct points x, y in X, there are gpα-open sets U and V such that x ∈ U, y ∉ U and y ∈ V, x ∉ V. Example 3.9: Let X = {a1, a2, a3} and τ = {X, φ, {a1}, {a2, a3}}.Then the space (X, τ) is gpα-T1 space.Remark 3.10: Every T1-space is gpα-T1, but the reverse implication is not true.
Example 3.11: Let X = {a1, a2, a3} and τ = {X, φ, {a1}, {a2, a3}}.Then, the space X is gpα-T1 space but not T1-space.Theorem 3.12: Every gpα-T1 space is gpα-T0 space.Proof: Consider X is gpα-T1 space.Let c and d be two distinct points in X.Since X is gpα-T1 space there exist gpα-open sets U and V such that c ∈ U, d ∉ U and c ∉ V, d ∈ V. We have c ∈ U and d ∉ U. Thus, X is gpα-T0 space.Remark 3.13: Following example shows the converse of the preceding theorem is not true in general.
Proof: Let x be any point on X.Let y ∈ {x} c is true, then y ≠ x, as X is gpα-T1 and y differs from x, a gpα-open set G y must exist such that y ∈ G y but x ∉ G y.This implies, for each y ∈ {x} c , there exists gpα-open set G y such that y ∈ G y ⊆ {x} c .Hence ∪{y: y ≠ x} ⊆∪{ G y: ≠x} ⊆ {x} c .So {x} c ⊆∪{G y : y ≠ x} ⊆ {x} c .Hence, {x} c = ∪{G y : y ≠x}.As, G y is gpα-open set, the union of gpα-open sets is also a gpα-open.Thus {x} c is gpα-open in X, and {x} is gpαclosed in X.Conversely, let x, y ∈ X be gpα-closed in X and x ≠ y be gpα-closed in X.Then {x} c and {y} c are gpα-open in X, with y ∈ {x} c but x ∉ {x} c and x ∈ {y} but y ∈ {y} c .There are gpα-open sets {x} c and {y} c such that x ∈ {y} c , y ∈ {y} c and y ∈ {x} c , x ∉ {x} c .Thus X is gpα-T1 space.Corollary 3.16: Every finite subset of X is gpα-closed if and only if the space X is gpα-T1.Definition 3.17: [16] A topological space (X, τ ) is said to be Tgpα-space if every gpα-closed set is closed in (X, τ ).Theorem 3.18: Let f: X → Y is bijective, gpα-open with X is gpα-T1 and Tgpα-space, and then Y is also gpα-T1 space.
Proof: Let y1, y2 ∈ Y with y1≠y2.As f is bijective and Tgpα-space, there are unique points x1, x2 of X such that y1 = f(x1), y2 = f(x2), and hence gpα-open sets U and V such that x1 As a result, X is gpα-T1 space Definition 3.21: Let (X, τ) be a topological space.The space (X, τ) is said to be gpα-T2 if there are disjoint gpα-open sets U, V such that x ∈ U and y ∈ V for each x, y∈ X with x ≠ y.
Example 3.24: From Example 3.22, it is clear that the space X is gpα-T2 not gpα-T1 space.

Theorem 3.25:
The intersection of all gpα-closed neighborhoods of each point of X is singleton set if and only if the topological space X is gpα-T2.
Proof: Let us say x, y ∈ X with x ≠ y.There are gpα-open sets U and V such that x ∈ U, y ∈ V and U ∩ V = φ, that is x ∈ U ⊆ X −V according to the definition 3.21.As a result, X − V is a gpα-closed neighborhood of x that excludes y.So, y does not form part of the intersection of all gpα-closed neighborhoods of x.Since y is an arbitrary point, the singleton set {x} is the intersection of all gpα-closed neighborhoods of x.In the alternative, consider x to be the intersection of all gpα-closed neighborhoods of any arbitrary point x ∈ X and y to be any arbitrary point of X such that x ≠ y.Since y does not belong to the intersection, there exists a gpα-closed neighborhood Y of x such that y ∉ Y. Then there is a gpα-open set U such that x ∈ U ⊆Y. Consequently, U and X − N are gpα-open sets such that x ∈ U and y ∈ X − Y such that U ∩ (X −Y) = φ.As a result, X is gpα-T2 space.Theorem 3.26: Let f: X → Y is gpα-c, injective and Y is T2-space.Then X is gpα-T2 space.
Proof: Consider any two distinct points of X to be x1 and x2.Because f is injective, disjoint points y1 and y2 of Y exist with y1 = f(x1) and y2 = f(x2).There exist disjoint open sets U and V such that y1 ∈ U, y2 ∈ V due to a property of T2-space.Thus, in X, x1 Remark 3.27: If f: X → Y is gpα -I, injective and Y is gpα-T2 then X is also gpα-T2 space.
Theorem 3.28: For any topological space X, the following properties are equivalent.(i) X is gpα-T2, (ii) There is gpα-open set U such that x ∈ U and y ∉ gpα-cl (U) for each distinct points x, y, (iii) For each point x∈ X, we have {x} =∩{gpα-cl(U):U is gpα-open in X such that x ∈ U}.
Proof: (i) → (ii) Let us say x, y ∈ X with x≠ y.Then there are gpα-open sets U and V with U ∩V = φ such that x ∈ U, y ∈ V exist.X-V is gpα-closed in X.Since U ∩V = φ, U ⊆ X-V is the result.So, gpα-cl (U) ⊆gpα-cl (X−V) because X-V is gpα-closed.Since y∉ X-V, implies y ∉ gpα-cl (U).(ii) → (iii) Let x, y ∈ X with x ≠ y.Then there is a gpα-open set U such that x ∈ U, y ∉ gpαcl(U) exists.Hence, y ∉ {gpα-cl (U) : U is gpα-open in X and x ∈ U} = {x}.(iii) → (i) Let x, y ∈ X with x ≠ y.Then there is a gpα-open set U such that x ∈ U, y ∉gpαcl(U) exists.Then there exists a gpα-closed V such that y ∈ V that is y ∈ X −V, where X -V is gpα-open.So, gpα-open sets U and X -V such that x ∈ U, y ∈ X − V and U ∩ (X −V) = φ.Hence X stands for gpα-T2 space.Theorem 3.29: Let X be gpα-T2, Tgpα-space and Y be a subspace of X.Then Y is also gpα-T2 space.

Characterizations of Generalized Pre α -Difference Sets
In this section, new types of separation axioms are defined and studied in topological spaces called gpα-Dk for k = 0, 1, 2, and also some properties of these spaces are explained.

Definition 4.1:
A subset A of a space X is called a gpα-difference set (gpα-D Set) if there exist two gpα-open sets U and V in X with U ≠M and A = U \ V.

Definition 4.3:
A space X is said to be a gpα-D0, if there exists gpα-D set of X including x but not y or gpα-D set of X containing y but not x for any distinct points x, y in X.

Definition 4.4:
A space X is said to be gpα-D1, if there exists gpα-D set of X including x but not y and a gpα-D set of X containing y but not x for any pair of distinct points x, y in X.

Definition 4.5:
A space X is gpα-D2, if there are disjoint gpα-D sets U and V of X containing x and y for any pair of different points x, y of X. Proof: As a sample, we prove for k = 0. Assume that X is gpα-T0 space and then for each pair of distinct points x, y ∈ X, there exists a gpα-open set U with x ∈ U, y∉ U or x ∉ U, y ∈ U. From Remark 4.2 we know that every proper gpα-open set is gpα-D set.Thus U is gpα-D set in X.So for each x, y ∈ X with x≠ y there exists a gpα-D set U with x ∈U, y ∉ U or x ∉ U, y ∈ U. Hence, (X, τ) is gpα-D0 space.Similarly, if we take k = 1, 2 we may establish the result.Hence, every gpα-T1 space is gpα-D1 space and every gpα-T2 space is gpα-D2 space.
Proof: Let us take k = 2. Let (X, τ) be gpα-D2 space.Then there are gpα-D sets U and V containing x and y respectively for each x, y ∈ X with x ≠ y.It follows, for each x, y ∈ X with x ≠ y, a gpα-D set U including x but not y and gpα-D set V containing y but x is not exists, indicating that (X, τ) is gpα-D1 space.Similarly we can show that gpα-D1 space is gpα-D0 space.Proof: Since we know that every pre-T k space is gpα-Tk space and every gpα-Tk space is gpα-Dk space.So, every pre-D k space is gpα-Dk space.Definition 4.9: A space X is said to be gpα-symmetric if for each x, y in X, x ∈ gpα-cl ({y}) then y ∈ gpα-cl ({x}).Theorem 4.10: In a space X.The space X is gpα-symmetric if and only if {x} is gpα-closed, for each x ∈ X. Remark 5.2: Every gpα-regular space is regular.However, as the following example shows, the converse is not always true.
(ii)→ (i) Let F be any gpα-closed set in X that has the property x ∉ F. Then x ∈ X −F, where X − F is gpα-open and hence X − F is gpα-neighborhood of x.According to the hypothesis, there exists an open set V such that x ∈ V and cl(V ) ⊆ X − F, that is F ⊆ cl(X − V).Then cl(X − V) is an open set containing F with V ∩ cl(X − V) = φ.Hence X is gpα-regular.(ii)→ (iii) Let x ∈ X and F be any gpα-closed set in X such that x ∉ F. Then X−F is a gpαneighborhood of x.According to the hypothesis, there is an open set V such that cl (V) ⊆ M− F. As a result cl (V) ∩ F = φ.(iii) → (ii) Let x ∈ X and W be any gpα-neighborhood of m.Then there is an open set G in which x ∈G ⊆ W exists.Because X − G is gpα-closed and x ∉ X − G.Then, according to the hypothesis, there exists an open set U for which cl (U)∩ X−G = φ.Hence cl(U) ⊆ G ⊆ W.

Theorem 5.6: A gpα-regular is hereditary property.
Proof: Let X be a gpα-regular space and Y ⊂ X.Let x∈ Y and F be a gpα-closed set in Y with x ∉ F.Then, there exists gpα-closed set A in X with F=Y ∩ A and x ∉ A. Therefore, x ∈ X and A is gpα-closed in X with x ∉ A. Then, there are open sets G and H such that x ∈ G, A⊆H and G∩H = φ.Here, Y∩G and Y∩H are open sets in Y. Also, we have x ∈ G and x ∈ Y that is x There are disjoint open sets f −1 (U) and f −1 (V) such that A ⊆ f −1 (U) and B ⊆ f −1 (V) for any two disjoint gpαclosed sets A and B in X. Hence X is gpα-normal.Remark 6.9:If f: X → Y be gpα-I, bijective, open map from a gpα-normal space X onto the space Y. Then Y is gpα-normal.

Generalized Pre α -Closed graphs
We say that a function f: X → Y has a closed graph if the graph of a function f, that is the set {(x, y) ∈ X × Y: y − f(x), x ∈ X} is a closed subset of the product X × Y. Proof: Let f be strongly gpα-closed graph.Then, for each x ∈ X and y ∈ Y with y ≠f(x), there are gpα-open sets U and V containing x and y respectively such that (U × V) ∩ G(f) = φ, that is for each x ∈ X and y ∈ Y with y ≠f(x).Hence f(U) ∩ V = φ.Conversely, let (x, y) ∉ G(f).Then there are gpα-open sets U and V each having x and y such that f(U) ∩ V = φ.Then we have f(x) ≠y for each x ∈ X and y ∈ Y. Hence (U × V) ∩ G (f) = φ.Thus, p has a strongly gpα-closed graph.Theorem 7.3: Let f: X → Y be injective with strongly gpα-closed graph G(f).Then X is gpα-T1 space.Proof: Let x, y ∈ X with x ≠y that is f (x) ≠ f (y).Then (x, f(y)) ∈ X × Y -G(f).As G (f) is strongly gpα-closed graph, there exist gpα-open sets U and V containing x and f(y) respectively such that f(U)∩V = φ, and so y ≠ U. Similarly there exist gpα-open sets Q and R containing y and f(x) with f (Q) ∩ R = φ, so x ∈ Q.Hence, for each x, y ∈ X with x ≠ y, there exists a gpαopen set U containing x but not y and gpα-open set Q containing y but not x.Hence, X is gpα-T1 space.Theorem 7.4: If f: X → Y is surjective with strongly gpα-closed graph G (f), then Y is gpα-T1 space.
Proof: Let us say x, y ∈ Y where x ≠ y.There exists o ∈ X with f(o) = y since f is surjective.As a result of Lemma 7.2 (o, x) ∈G(f).There are gpα-open sets U and V containing o and x ,respectively with f(U)∩V = φ, so y ∉ V .Similarly, there exists w ∈ X with f (w) = y and hence (w, x) ∈ G (f). Then there are gpα-open sets Q and R, each containing w and y, such that f (Q) ∩ Y = φ and x ∉ R. As a result, there exist gpα-open sets V and R with x ∈ V and y ∉ V and x ∉R and y ∈ R for any x, y ∈ Y with x ≠ y.Hence Y stands for gpα-T1 space.Theorem 79: If f: X → Y is injective, gpα-c with strongly gpα-closed graph G (f) and Y is gpα-T2 space.Then X is gpα-T2 space.Proof: Let x, y ∈ X with x ≠ y.There are disjoint open sets U and V in Y with f(x) ∈ U, f(y) ∈ V since Y is gpα-T2 space.Then, because f is gpα-c, f −1 (U) and f −1 (V ) are gpα-open sets in X containing x and y respectively.As a result, f −1 (U) ∩ f −1 (V ) = φ.Therefore, for each x, y ∈ M with x ≠ y, there exist gpα-open sets f −1 (U) and f −1 (V) containing x and y respectively such that f −1 (U) ∩ f −1 (V) = φ.Thus x ≠ y is gpα-T2 space.

Conclusions
The class of generalized closed sets has an important role in general topology.In this work we introduced and study new types of separation axioms, namely gpα-Tk (k = 0, 1, 2) spaces, using gpα-closed sets.Several characterizations and properties of these concepts are provided.Their implication with other separation axioms have also been examined and emphasized which extends the future scope of normal and regular topological spaces.Also, we have investigated some results on gpα-Difference sets and gpα-closed graphs.
U and V are open sets in X. Since, f(U) and f(V ) are gpα-open subsets in Y, f is gpα-open.Thus, in Y there are gpα-open sets f(U) and f(V ) such that y1 ∈ f(U), y2∉ f(U) and y2 ∈ f(V ), m1 ∉ f(V ).Thus Y is gpα-T1 space.Remark 3.19: If p: M → N is gpα -I, injective and N is gpα-T1 then M is also gpα-T1 space.Theorem 3.20: Let p: M → N be gpα-c, injective function and N is T1-space.Then M is gpα-T1 space.Proof: Let x1, x2 ∈ X where x1≠ x2.Then y1, y2 of Y exist, with y1 ≠y2 such that f(x1) = y1 and p(m2) = n2.As Y is T1, there are gpα-open sets U and V in Y so that y1 ∈ U, y2 ∉U and y1 ∉V

:
If x, y ∈ Y then x, y ∈ X as Y ⊂ X.Then gpα-open sets U and V such that U ∩ V = φ.Since X is Tgpα-space, U and V are open sets in X while U ∩ Y and V ∩ Y are open sets in Y. Therefore, U ∩ Y and V ∩ Y are gpα-open sets in Y.In addition, we have x ∈ U and x ∈Y, implying that x ∈ U ∩ Y.In a similar way y ∈ V ∩ Y. Consider (U ∩ Y) ∩ (V ∩ Y)=Y ∩ (U ∩ V )= Y ∩ φ = φ.Thus for each x, y ∈ Y with x ≠ y, there exist disjoint gpα-open sets U ∩ Y and V ∩ Y such that x∈ U ∩ Y and y ∈ V ∩ Y have (U ∩ Y) ∩ (V ∩ Y) = φ.Hence, Y is gpα-T2 space.

Remark 4 . 2 :
Every proper gpα-open set is gpα-D set.Proof: Assume A is a proper gpα-open subset of X.Put U = A and V = φ.Then A = U \V with U ≠ X.Hence, A is gpα-D set.

Theorem 5 . 4 :Theorem 5 . 5 :
Let X is gpα-regular, and Y is a gpα-closed and open subset of X.Then a subspace Y is gpα-regular.Proof: Let F be any gpα-closed subset of Y and y ∈ Y − F. Then F is gpα-closed set in X, by the Theorem 3.21 of [18].As X is gpα-regular, then there exist disjoint open sets U and V such that y ∈ U and F ⊆V.So U ∩ Y and V ∩ Y are disjoint open sets in the subspace Y, such that y ∈ U ∩ Y and F ⊆ V ∩ Y. Thus Y is gpα-regular.Consider the topological space (X, τ).The following statements are identical in this case: (i) X is gpα-regular.(ii) There exists an open set U of x such that cl (U) ⊆ W, for each m ∈ M and each gpα-open neighborhood W of m. (iii) There exists an open set V such that cl (V) ∩ F = φ, for each point x ∈ X and each gpαclosed set F such that x ∉ F. Proof: (i) → (ii) Allow W to be any gpα-open neighborhood of x in X.Then there is a gpαopen set G and W such that x ∈ G ⊆ W, and because X − G is gpα-closed and x ∈ X − G, there must exist open sets U and V such that X

Theorem 5 . 7 :Theorem 6 . 7 :Theorem 6 . 8 :
So, for each x ∈ Y and each closed set F in Y, there exists an open sets Y ∩ G and Y ∩ H such that x ∈ Y ∩ G and F ⊆ Y ∩ H with (Y ∩ G) ∩ (Y ∩ H) = φ.Hence Y is gpα-regular.If f: X → Y is gpα-I, bijective, open and M is gpα-regular, then N is also gpαregular.f −1 (A) ⊆ U and f −1 (B) ⊆ V , as f is open and bijective f(U) and f(V) are open sets in Y such that A ⊆ f(U) and B ⊆ f(V ) with f(U) ∩ f(V ) = φ.Hence Y is gpα-normal.Let X is gpα-normal, Y is gpα-closed subset of X.Then the subspace Y is gpαnormal.Proof: Let A and B be two gpα-closed sets in Y that are disjoint.Then A and B are disjoint gpα-closed sets in X.There are disjoint open sets U and V such that A⊆ U and B ⊆ V exist as X is gpα-normal.As a U ∩ Y and V ∩ Y are disjoint open subsets of the subspace Y with A ⊆ U ∩ Y and B ⊆ V ∩ Y respectively.Thus, Y is gpα-normal.If f: X → Y is pre gpα-closed, continuous, and injective, Y is gpα-normal.Then X is gpα-normal.Proof: Let A and B be any two gpα-closed disjoint sets in X. f(A) and f(B) are disjoint gpαclosed sets in Y since f is pre gpα-closed.There are disjoint open sets U and V such that f(A) ⊆ U and f

Definition 7 . 1 :Lemma 7 . 2 :
A function f: X → Y is said to be a strongly gpα-closed graph if for each {(x, y) ∈ X × Y − G(f)}, there is a gpα-open set U and a gpα-open set V in Y containing x and y respectively such that (U × V ) ∩ G(f) = φ.Where G( f ) is graph of a function f.For a function f: X → Y, the graph G(f) is said to be strongly in X × Y, if and only if there exist gpα-open sets U, V such that f(U) ∩ V= φ for each {(x, y)∈ X ×Y − G(f)}.

Theorem 7 . 5 : 4 Theorem 7 . 6 :Theorem 7 . 8 :
Let f: X → Y be bijective with strongly gpα-closed graph G(f).Then both X and Y are gpα-T1.Proof: Follows from Theorem 7.3 and Theorem 7.Let f: X → Y is gpα-c and Y is gpα-T2 space.Then G (f) is strongly gpα-closed.Proof: Let (x, y) ∉G(f), and hence (x, y) ∈ (X × Y) − G(f), as Y is gpα-T2 space, y ∉ f(x).There are two gpα-open sets U and V that have f(x) ∈ U, y ∈ V in Y and U ∩ V = φ.There is a gpαopen neighborhood W of X with f(w) ⊂ U according to gpα-open neighborhood W of X with f(w) ⊂ U according to gpα-continuity.Hence, f(W) ∩ V φ.Thus (W × V) ∩ G (f) = φ.Hence f is strongly gpα-closed graph.Definition 7.7: [16] A function f: X → Y is said to be pre gpα-open if the image of every gpαopen set in X is gpα-open in Y. Let f: X → Y be pre gpα-open surjective function with strongly gpα-closed graph G (f). Then Y is gpα-T2 space.Proof: Let y, w ∈ Y with y ≠ w.As f is surjective, there exist x, o ∈ X such that f(x) = y, f (o) = w, and so (x, w) ∉ G(f).Because G (f) is strongly gpα-closed graph, there are gpα-open sets U and V containing x and w respectively such that f(U) ∩ V = φ.However, because f(U) is gpα-open, it must include a point y.Hence, Y is gpα-T2 space.