A NEW GENERALIZATION OF SHANIN’S NOTION of R_{0} TOPOLOGICAL SPACES

M. Caldas, E. Hatir, and S. Jafari

Abstract. In this paper, we introduce and investigate some weak separation axioms by using the notions of (Λ, α)–open sets and the (Λ, α)–closure operator.

1. Introduction. In 1943, N. A. Shanin [8] offered a new weak separation axiom called R_{0} to the world of general topology. In 1961, A. S. Davis [2] rediscovered this axiom and established some properties of topological spaces endowed with it. Later on several topologists further investigated R_{0} topological spaces [3,4,5,6]. The notion of α–open set was introduced by O. Njåstad [7] in 1965. Since then it has been investigated in different respects in the literature. Quite recently Caldas, et. al. [1] introduced and studied the notions of (Λ, α)–open sets, (Λ, α)–closed sets, and the (Λ, α)–closure operator. This paper deals with some new low separation axioms by utilizing (Λ, α)–open and (Λ, α)–closed sets. There is no doubt that low separation axioms play a very important role in general topology. Indeed, there are lots of research papers which deal with different low separation axioms and also many topologists worldwide are doing research in this area. It is the aim of this paper to offer some new types of low separation axioms by using (Λ, α)–open sets and (Λ, α)–closure operator.

In this paper, by (X, τ) and (Y, σ) (or X and Y) we always mean topological spaces. Let A be a subset of X. The subset A of the topological space (X, τ) is called α–open (originally called α-sets) [7] if A ⊆ Int(Cl(Int(A))). The complement of a α–open set is called α–closed. By αO(X, τ) (respectively αC(X, τ)), we denote the family of all α–open (respectively α–closed) sets of (X, τ). Observe that α–open sets form a topology, and also α–openness does not imply openness in the underlying topology. Let C denote the standard “middle thirds” Cantor set in the unit interval [0, 1] with the standard topology, and let x ∈ C. Then, [0, 1] − {C − {x}} would be an α–open set. The intersection of all α–closed sets containing A is called the α–closure of A, denoted by Cl_{α}(A). A subset A is also α–closed if and only if A = Cl_{α}(A). A set U in a topological space (X, τ) is a α-neighborhood [7] of a point x if U contains an α–open set V such that x ∈ V.

\underline{Remark 1.1}.

(i) It is shown in [7] that αO(X, τ) is a topology on X and τ ⊂ αO(X, τ). (ii) Clearly open sets are α–open but one easily finds in the real line with the usual

topology α–open sets that are not open.

\underline{Lemma 1.2}. Let A be a subset of a topological space (X, τ).

(1) Cl_{α}(A) = ∩{F ∈ αC(X, τ) | A ⊂ F }.

(2) Cl_{α}(A) is α–closed, that is Cl_{α}(Cl_{α}(A)) = Cl_{α}(A).

\underline{Definition 1}. Let A be a subset of a topological space (X, τ). A subset Λ_{α}(A) is defined as follows: Λ_{α}(A) = ∩{O ∈ αO(X, τ) | A ⊂ O} and A is called a Λ_{α}-set if A = Λ_{α}(A) [1].

\underline{Definition 2}. Let A be a subset of a topological space (X, τ).

(i) A is called a (Λ, α)–closed set [1] if A = T ∩C, where T is a Λ_{α}–set and C is an α–closed set. The complement of a (Λ, α)–closed set is called (Λ, α)–open. We denoted the collection of all (Λ, α)–open sets (respectively (Λ, α)–closed sets) by Λ_{α}O(X, τ) (respectively by Λ_{α}C(X, τ)).

(ii) A point x ∈ X is called a (Λ, α)–cluster point of A [1] if for every (Λ, α)–open set U of X containing x we have A∩U ≠ ∅. The set of all (Λ, α)–cluster points of A is called the (Λ, α)–closure of A and is denoted by A^{(Λ,α)}.

\underline{Lemma 1.3} Let A and B be subsets of a topological space (X, τ). For the (Λ, α)–closure, the following properties hold [1].

(1) A ⊂ A^{(Λ,α)}.

(2) A^{(Λ,α)} = ∩{F | A ⊂ F and F is (Λ, α)–closed}.

(3) If A ⊂ B, then A^{(Λ,α)} ⊂ B^{(Λ,α)}.

(4) A is (Λ, α)–closed if and only if A = A^{(Λ,α)}.

(5) A^{(Λ,α)} is (Λ, α)–closed.

2. Sober Λ_{α}-R_{0} Topological Spaces. A. S. Davis [2] introduced the notion of R_{0}–axiom which in some aspects is more natural than the T_{0}–axiom. In this section we introduce the concept of sober Λ_{α}–R_{0} topological space and we show that sober Λ_{α}–R_{0} and R_{0} are independent of each other.

\underline{Lemma 2.1}. Let A be a subset of a space X. Then the following hold.

(1) If A is α–closed, then A is (Λ, α)–closed.

(2) If A is (Λ, α)–closed, then A = Λ_{α}(A) ∩ A^{(Λ,α)}.

(3) If A_{i} is (Λ, α)–closed for each i ∈ I, then ∩_{i∈I}A_{i} is (Λ, α)–closed.

(4) If A_{i} is (Λ, α)–open for each i ∈ I, then ∪_{i∈I}A_{i} is (Λ, α)–open.

\underline{Proof}. (1) It is sufficient to observe that A = X ∩ A where the whole set X is a Λ_{α}-set.

(2) Let A be (Λ, α)–closed, then there exists a Λ_{α}-set T and a α–closed set C such that A = T ∩ C. By A ⊂ T, we have A ⊂ Λ_{α}(A) ⊂ Λ_{α}(T) = T, and also by A ⊂ C, A ⊂ A^{(Λ,α)} ⊂ C^{(Λ,α)} = C. Now, A ⊂ Λ_{α}(A) ∩ A^{(Λ,α)} ⊂ T ∩ C = A. Hence, A = Λ_{α}(A) ∩ A^{(Λ,α)}.

(3) Suppose that A_{i} is (Λ, α)–closed for each i ∈ I. Then, for each i ∈ I there exists a Λ_{α}–set T_{i} and a α–closed set C_{i} such that A_{i} = T_{i} ∩ C_{i}. Now, ⋂_{i∈I} A_{i} = ⋂_{i∈I}(T_{i} ∩ C_{i}) = (⋂_{i∈I} T_{i}) ∩ (⋂_{i∈I} C_{i}). By Lemma 2.4 of [1], ⋂_{i∈I} T_{i} is a Λ_{α}-set and ⋂_{i∈I} C_{i} is α–closed. This shows that ⋂_{i∈I} A_{i} is (Λ, α)–closed.

(4) Follows from (3).

\underline{Definition 3}. Let (X, τ) be a topological space, A ⊂ X and x ∈ X. Then

(i) The Λ_{α}–kernel of A, denoted by Λ_{α}Ker(A), is defined to be the set Λ_{α}Ker(A) = ∩{G ∈ Λ_{α}O(X, τ) | A ⊂ G}.

(ii) < x >= {x}^{(Λ,α)} ∩ Λ_{α}Ker({x}).

\underline{Lemma 2.2}. Let (X, τ) be a topological space and x ∈ X. Then Λ_{α}Ker(A) = {x ∈ X | {x}^{(Λ,α)} ∩ A ≠ ∅}.

\underline{Proof}. Let x ∈ Λ_{α}Ker(A) and suppose \{x\}^{(Λ,α)} ∩ A = ∅. Hence, x ∉ X − {x}^{(Λ,α)} which is a (Λ, α)–open set containing A. This is absurd, since x ∈ Λ_{α}Ker(A). Consequently, {x}^{(Λ,α)} ∩ A ≠ ∅. Next, let x such that {x}^{(Λ,α)} ∩ A ≠ ∅ and suppose that x ∉ Λ_{α}Ker(A). Then, there exists a (Λ, α)– open set D containing A and x ∉ D. Let y ∈ {x}^{(Λ,α)} ∩ A. Hence, D is a (Λ, α)- neighborhood of y which does not contain x. But this contradicts x ∈ Ker_{α}(A) and the claim follows.

\underline{Lemma 2.3}. If A, B ⊂ X, then

(1) A ⊂ B implies Λ_{α}Ker(A) ⊂ Λ_{α}Ker(B).

(2) Λ_{α}Ker(Λ_{α}Ker(A)) = Λ_{α}Ker(A).

\underline{Lemma 2.4}. Let (X, τ) be a topological space and x, y ∈ X. Then y ∈ Λ_{α}Ker({x}) if and only if x ∈ {y}^{(Λ,α)}.

\underline{Proof}. Let y ∉ Λ_{α}Ker(\{x\}). Then there exists a (Λ, α)–open set V containing x such that y ∉ V. Hence, x ∉ {y}^{(Λ,α)}. The converse is similarly shown.

A subset B_{x} of a topological space X is said to be (Λ, α)-neighborhood of a point x ∈ X if there exists a (Λ, α)–open set U such that x ∈ U ⊂ B_{x}.

\underline{Lemma 2.5}. A subset of a topological space X is (Λ, α)–open in X if and only if it contains a (Λ, α)–neighborhood of each of its points.

\underline{Proposition 2.6}. If (X, τ) is a topological space and A ⊂ X. Then

(1) Λ_{α}Ker(A) = {x ∈ X/{x}^{(Λ,α)} ∩ A ≠ ∅}.

(2) For each x ∈ X, Λ_{α}Ker(< x >) = Λ_{α}Ker({x}).

(3) For each x ∈ X, {< x >}^{(Λ,α)} = {x}^{(Λ,α)}.

(4) For each (Λ, α)–open set U ⊂ X, if x ∈ U then < x >⊂ U.

(5) For each (Λ, α)–closed set F ⊂ X, if x ∈ F then < x >⊂ F.

\underline{Proof}. (1) Let x ∈ Λ_{α}Ker(A) and suppose \{x\}^{(Λ,α)} ∩ A = ∅. Then, x ∉ X\{x}^{(Λ,α)} which is a (Λ, α)–open set containing A. This is impossible, since x ∈ Λ_{α}Ker(A). Consequently, {x}^{(Λ,α)} ∩ A ≠ ∅. Next, let x ∈ X such that {x}^{(Λ,α)} ∩ A ≠ ∅ and suppose that x ∉ Λ_{α}Ker(A). Then, there exists a (Λ, α)– open set U containing A and x ∉ U. Let y ∈ {x}^{(Λ,α)} ∩ A. Hence, U is a (Λ, α)- neighborhood of y which does not contain x. But this contradicts x ∈ Λ_{α}Ker(A).

(2) Follows easily from Definition 2 and Lemma 2.4.

(3) The proof is quite similar to that of (2).

(4) Since x ∈ U and U is a (Λ, α)–open set, we have that Λ_{α}Ker({x}) ⊂ U. Hence, < x >⊂ U.

(5) Since x ∈ F and F is a (Λ, α)–closed set, we have that < x >= {x}^{(Λ,α)} ∩ Λ_{α}Ker({x}) ⊂ {x}^{(Λ,α)} ⊂ F.

\underline{Lemma 2.7}. The following statements are equivalent for any points x and y in a topological space (X, τ).

(1) Λ_{α}Ker({x}) ≠ Λ_{α}Ker({y}).

(2) {x}^{(Λ,α)} ≠ {y}^{(Λ,α)}.

\underline{Proof}. (1) → (2): Suppose that Λ_{α}Ker(\{x\}) ≠ Λ_{α}Ker(\{y\}), then there exists a point z in X such that z ∈ Λ_{α}Ker({x}) and z ∉ Λ_{α}Ker({y}). From z ∈ Λ_{α}Ker({x}) it follows that {x} ∩ {z}^{(Λ,α)} ≠ ∅ which implies x ∈ {z}^{(Λ,α)}. By z ∉ Λ_{α}Ker({y}), we have {y} ∩ {z}^{(Λ,α)} = ∅. Since x ∈ {z}^{(Λ,α)} , {x}^{(Λ,α)} ⊂ {z}^{(Λ,α)} and {y} ∩ {x}^{(Λ,α)} = ∅. Therefore, it follows that {x}^{(Λ,α)} ≠ {y}^{(Λ,α)}. Now Λ_{α}Ker({x}) ≠ Λ_{α}Ker({y}) implies that {x}^{(Λ,α)} ≠ {y}^{(Λ,α)}.

(2) → (1): Suppose that {x}^{(Λ,α)} ≠ {y}^{(Λ,α)}. Then there exists a point z in X such that z ∈ {x}^{(Λ,α)} and z ∉ {y}^{(Λ,α)}. It follows that there exists a (Λ, α)–open set containing z and therefore x but not y, namely, y ∉ Λ_{α}Ker({x}) and thus, Λ_{α}Ker({x}) ≠ Λ_{α}Ker({y}).

\underline{Definition 4}. A topological space (X, τ) is said to be sober Λ_{α}–R_{0} if

∩_{x∈X}{x}^{(Λ,α)} = ∅.

\underline{Theorem 2.8}. A topological space (X, τ) is sober Λ_{α}–R_{0} if and only if Λ_{α}Ker({x}) ≠ X for every x ∈ X.

\underline{Proof}. Suppose that the space (X, τ) is sober Λ_{α}–R_{0}. Assume that there is a point y in X such that Λ_{α}Ker({y}) = X. Then y ∉ O which O is some proper (Λ, α)–open subset of X. This implies that y ∈ ∩_{x∈X}{x}^{(Λ,α)}. But this is a contradiction. Now assume that Λ_{α}Ker({x}) ≠ X for every x ∈ X. If there exists a point y in X such that y ∈ ∩_{x∈X}{x}^{(Λ,α)}, then every (Λ, α)–open set containing y must contain every point of X. This implies that the space X is the unique (Λ, α)–open set containing y. Hence, Λ_{α}Ker({y}) = X which is a contradiction. Therefore, (X, τ) is sober Λ_{α}–R_{0}.

Recall that a topological space (X, τ) is said to be R_{0} [2] if every open set contains the closure of each of its singletons.

\underline{Example 2.9}. Let (X, τ) be a topological space such that X = \{a, b, c\} and τ = {∅, {a}, X}. Observe that (X, τ) is sober Λ_{α}–R_{0}, but it is not R_{0}.

\underline{Example 2.10}. Let (X, τ) with τ = \{∅, X\}. Clearly (X, τ) is not sober Λ_{α}–R_{0}, but it is R_{0}.

Examples 2.9 and 2.10 show that sober Λ_{α}–R_{0} and R_{0} are independent of each other.

\underline{Definition 5}. A function f: X → Y is always called Λ_{α}–closed if the image of every (Λ, α)–closed subset of X is (Λ, α)–closed in Y.

\underline{Theorem 2.11}. If f: X → Y is an injective always Λ_{α}–closed function and X is sober Λ_{α}–R_{0}, then Y is sober α–R_{0}.

\underline{Proof}. Straightforward.

\underline{Theorem 2.12}. If the topological space X is sober Λ_{α}–R_{0} and Y is any topological space, then the product X × Y is sober Λ_{α}–R_{0}.

\underline{Proof}. By showing that ∩_{(x,y)∈X×Y} \{x, y\}^{(Λ,α)} = ∅ we are done. We have

∩_{(x,y)∈X×Y} {x, y}^{(Λ,α)} ⊆ ∩_{(x,y)∈X×Y} {x}^{(Λ,α)} × {y}^{(Λ,α)} = ∩_{x∈X}{x}^{(Λ,α)} × ∩_{y∈Y} {y}^{(Λ,α)} ⊆ ∅ × Y = ∅.

3. Λ_{α}–R_{0} Topological Spaces. In this section (Λ, α)–open sets and (Λ, α)– closure operator are used to define a new separation axiom analogous to R_{0}- axiom and we obtain several characterizations of it.

\underline{Definition 6}. A topological space (X, τ) is said to be a Λ_{α}–R_{0} space if every (Λ, α)–open set contains the (Λ, α)–closure of each of its singletons.

The next result will give the characterization of the Λ_{α}-R_{0} space in terms of the (Λ, α)–closure of points.

\underline{Theorem 3.1}. For a topological space (X, τ), the following properties are equivalent.

(1) (X, τ) is a Λ_{α}–R_{0} space.

(2) For any F ∈ Λ_{α}C(X, τ), x ∉ F implies that there exist U ∈ Λ_{α}O(X, τ), such that F ⊂ U and x ∉ U.

(3) For any F ∈ Λ_{α}C(X, τ), x ∉ F implies F ∩{x}^{(Λ,α)} = ∅.

(4) For any distinct points x and y of X, either {x}^{(Λ,α)} = {y}^{(Λ,α)} or {x}^{(Λ,α)} ∩ {y}^{(Λ,α)} = ∅.

\underline{Proof}. (1) → (2): Let F ∈ Λ_{α}C(X, τ) and x ∉ F. Then by (1) \{x\}^{(Λ,α)} ⊂ X\F. Set U = X\{x}^{(Λ,α)}, then U ∈ Λ_{α}O(X, τ), F ⊂ U and x ∉ U.

(2) → (3): Let F ∈ Λ_{α}C(X, τ) and x ∉ F. There exists U ∈ Λ_{α}O(X, τ) such that F ⊂ U and x ∉ U. Since U ∈ Λ_{α}O(X, τ), U ∩ {x}^{(Λ,α)} = ∅ and F ∩{x}^{(Λ,α)} = ∅.

(3) → (4): Assume that {x}^{(Λ,α)} ≠ {y}^{(Λ,α)} for distinct points x, y ∈ X. There exists z ∈ {x}^{(Λ,α)} such that z ∉ {y}^{(Λ,α)} (or z ∈ {y}^{(Λ,α)} such that z ∉ {x}^{(Λ,α)}). There exists V ∈ Λ_{α}O(X, τ) such that y ∉ V and z ∈ V; hence x ∈ V. Therefore, we have x ∉ {y}^{(Λ,α)}. By (3), we obtain {x}^{(Λ,α)} ∩ {y}^{(Λ,α)} = ∅. The proof for otherwise case is similar.

(4) → (1): Let V ∈ Λ_{α}O(X, τ) and x ∈ V. For each y ∉ V, x ≠ y and x ∉ {y}^{(Λ,α)}. This shows that {x}^{(Λ,α)} ≠ {y}^{(Λ,α)}. By (4), {x}^{(Λ,α)} ∩ {y}^{(Λ,α)} = ∅ for each y ∈ X\V and hence,

{x}^{(Λ,α)} ∩ ( ⋃_{y∈X\V} {y}^{(Λ,α)}) = ∅.

On the other hand, since V ∈ Λ_{α}O(X, τ) and y ∈ X\V, we have {y}^{(Λ,α)} ⊂ X\V. Therefore,

X\V = ⋃_{y∈X\V} {y}^{(Λ,α)}.

Therefore, we obtain (X\V) ∩ {x}^{(Λ,α)} = ∅ and {x}^{(Λ,α)} ⊂ V. Hence, (X, τ) is a Λ_{α}–R_{0} space.

\underline{Corollary 3.2}. A topological space (X, τ) is a Λ_{α}–R_{0} space if and only if for any x and y in X, {x}^{(Λ,α)} ≠ {y}^{(Λ,α)} implies {x}^{(Λ,α)} ∩ {y}^{(Λ,α)} = ∅.

\underline{Proposition 3.3}. A topological space (X, τ) is Λ_{α}–R_{0} if and only if for any points x and y and X, Λ_{α}Ker({x}) ≠ Λ_{α}Ker({y}) implies Λ_{α}Ker({x}) ∩ Λ_{α}Ker({y}) = ∅.

\underline{Proof}. Suppose that (X, τ) is a Λ_{α}–R_{0} space. Thus by Lemma 2.7, for any points x and y in X if Λ_{α}Ker({x}) ≠ Λ_{α}Ker({y}) then {x}^{(Λ,α)} ≠ {y}^{(Λ,α)}. Now we prove that Λ_{α}Ker({x}) ∩ Λ_{α}Ker({y}) = ∅. Assume that z ∈ Λ_{α}Ker({x}) ∩ Λ_{α}Ker({y}). By z ∈ Λ_{α}Ker({x}), it follows that x ∈ {z}^{(Λ,α)}. Since x ∈ {x}^{(Λ,α)}, by Corollary 3.2 {x}^{(Λ,α)} = {z}^{(Λ,α)}. Similarly, we have {y}^{(Λ,α)} = {z}^{(Λ,α)} = {x}^{(Λ,α)}. This is a contradiction. Therefore, we have Λ_{α}Ker({x})∩Λ_{α}Ker({y}) = ∅.

Conversely, let (X, τ) be a topological space such that for any points x and y in X, Λ_{α}Ker({x}) ≠ Λ_{α}Ker({y}) implies Λ_{α}Ker({x}) ∩ Λ_{α}Ker({y}) = ∅. If {x}^{(Λ,α)} ≠ {y}^{(Λ,α)}, then by Lemma 2.7, Λ_{α}Ker({x}) ≠ Λ_{α}Ker({y}). Therefore, Λ_{α}Ker({x}) ∩ Λ_{α}Ker({y}) = ∅ which implies {x}^{(Λ,α)} ∩ {y}^{(Λ,α)} = ∅. Because z ∈ {x}^{(Λ,α)} implies that x ∈ Λ_{α}Ker({z}), Λ_{α}Ker({x}) ∩ Λ_{α}Ker({z}) ≠ ∅. By hypothesis, we therefore have Λ_{α}Ker({x}) = Λ_{α}Ker({z}). Then z ∈ {x}^{(Λ,α)} ∩ {y}^{(Λ,α)} implies that {x}^{(Λ,α)} = {z}^{(Λ,α)} = {y}^{(Λ,α)}. This is a contradiction. Therefore, {x}^{(Λ,α)} ∩ {y}^{(Λ,α)} = ∅ and by Corollary 3.2, (X, τ) is a Λ_{α}–R_{0} space.

\underline{Proposition 3.4}. For a topological space (X, τ), the following properties are equivalent.

(1) (X, τ) is a Λ_{α}–R_{0} space.

(2) For any nonempty set A and G ∈ Λ_{α}O(X, τ) such that A ∩ G ≠ ∅, there exists F ∈ Λ_{α}C(X, τ) such that A ∩ F ≠ ∅ and F ⊂ G.

(3) Any G ∈ Λ_{α}O(X, τ), G = ∪{F ∈ Λ_{α}C(X, τ) | F ⊂ G}.

(4) Any F ∈ Λ_{α}C(X, τ), F = ∩{G ∈ Λ_{α}O(X, τ) | F ⊂ G} (i.e., F = Λ_{α}Ker(F)).

(5) For any x ∈ X, {x}^{(Λ,α)} ⊂ Λ_{α}Ker({x}).

\underline{Proof}. (1) → (2): Let A be a nonempty set of X and G ∈ Λ_{α}O(X, τ) such that A ∩ G ≠ ∅. There exists x ∈ A ∩ G. Since x ∈ G ∈ Λ_{α}O(X, τ), {x}^{(Λ,α)} ⊂ G. Set F = {x}^{(Λ,α)}, then F ∈ Λ_{α}C(X, τ), F ⊂ G and A ∩ F ≠ ∅.

(2) → (3): Let G ∈ Λ_{α}O(X, τ), then G ⊃ ∪{F ∈ Λ_{α}C(X, τ) | F ⊂ G}. Let x be any point of G. There exists F ∈ Λ_{α}C(X, τ) such that x ∈ F and F ⊂ G. Therefore, we have x ∈ F ⊂ ∪{F ∈ Λ_{α}C(X, τ) | F ⊂ G} and hence, G = ∪{F ∈ Λ_{α}C(X, τ) | F ⊂ G}.

(3) → (4): This is obvious.

(4) → (5): It follows from (4) and Lemma 1.3.

(5) → (1): Let G ∈ Λ_{α}O(X, τ) and x ∈ G. Let y ∈ Λ_{α}Ker({x}), then x ∈ {y}^{(Λ,α)} and y ∈ G. This implies that Λ_{α}Ker({x}) ⊂ G. Therefore, we obtain x ∈ {x}^{(Λ,α)} ⊂ Λ_{α}Ker({x}) ⊂ G. This shows that (X, τ) is a Λ_{α}–R_{0} space.

\underline{Corollary 3.5}. For a topological space (X, τ), the following properties are equivalent.

(1) (X, τ) is a Λ_{α}–R_{0} space.

(2) {x}^{(Λ,α)} = Λ_{α}Ker({x}) for all x ∈ X.

\underline{Proof}. (1) → (2): Suppose that (X, τ) is a Λ_{α}–R_{0} space. By Proposition 3.4, {x}^{(Λ,α)} ⊂ Λ_{α}Ker({x}) for each x ∈ X. Let y ∈ Λ_{α}Ker({x}), then {x}^{(Λ,α)} ∩ {y}^{(Λ,α)} ≠ ∅. By Corollary 3.2 we obtain {x}^{(Λ,α)} = {y}^{(Λ,α)}. Therefore, y ∈ {x}^{(Λ,α)} and hence, Λ_{α}Ker({x}) ⊂ {x}^{(Λ,α)}. This shows that {x}^{(Λ,α)} = Λ_{α}Ker({x}).

(2) → (1): Proposition 3.4.

\underline{Corollary 3.6}. Let X be a Λ_{α}–R_{0} topological space. For any x ∈ X if < x >= {x}, then {x}^{(Λ,α)} = {x}.

\underline{Proof}. It is a consequence of Corollary 3.5.

\underline{Proposition 3.7}. For a topological space (X, τ), the following properties are equivalent.

(1) (X, τ) is a Λ_{α}–R_{0} space.

(2) x ∈ {y}^{(Λ,α)} if and only if y ∈ {x}^{(Λ,α)}.

\underline{Proof}. (1) → (2): Let X be Λ_{α}–R_{0}. Let x ∈ \{y\}^{(Λ,α)} and U be any (Λ, α)– open set such that y ∈ U. Hence, Λ_{α}Ker({y}) ⊂ U. Since x ∈ {y}^{(Λ,α)} and (X, τ) is Λ_{α}–R_{0}, by Corollary 3.5, x ∈ Λ_{α}Ker({y}) ⊂ U. Therefore, every (Λ, α)–open set which contains y contains x. Hence, y ∈ {x}^{(Λ,α)}.

(2) → (1): Let U be a (Λ, α)–open set and x ∈ U. If y ∉ U, then x ∉ {y}^{(Λ,α)} and hence, y ∉ {x}^{(Λ,α)}. This implies that {x}^{(Λ,α)} ⊂ U. Hence, (X, τ) is a Λ_{α}–R_{0} space.

\underline{Proposition 3.8}. For a topological space (X, τ), the following properties are equivalent.

(1) X is a Λ_{α}–R_{0} space.

(2) < x >= {x}^{(Λ,α)} for each x ∈ X.

(3) < x > is (Λ, α)–closed for each x ∈ X.

\underline{Proof}. (1) → (2): By Corollary 3.5, \{x\}^{(Λ,α)} = Λ_{α}Ker(\{x\}) for each x ∈ X. Hence, {x}^{(Λ,α)} = {x}^{(Λ,α)} ∩ Λ_{α}Ker({x}) =< x >.

(2) → (1): Since {x}^{(Λ,α)} =< x > for each x ∈ X, we have {x}^{(Λ,α)} ⊂ Λ_{α}Ker({x}). By Proposition 3.4(5), X is Λ_{α}–R_{0}.

(2) ⟷ (3): It is consequence of Lemma 2.1.

4. Λ_{α}–R_{1} Topological Spaces and a Question. A. S. Davis [2] also introduced the notion of R_{1} topological spaces which is strictly weaker than T_{2}. Now we offer a new generalization of R_{1} by utilizing the notions of (Λ, α)–open sets and (Λ, α)–closure operator.

\underline{Definition 7}. A topological space (X, τ) is said to be Λ_{α}–R_{1} if for x, y in X with {x}^{(Λ,α)} ≠ {y}^{(Λ,α)}, there exists disjoint (Λ, α)–open sets U and V such that {x}^{(Λ,α)} is a subset of U and {y}^{(Λ,α)} is a subset of V.

\underline{Proposition 4.1}. If (X, τ) is Λ_{α}–R_{1}, then (X, τ) is Λ_{α}–R_{0}.

\underline{Proof}. Let U be (Λ, α)–open and let x ∈ U. If y ∉ U, then since x ∉ \{y\}^{(Λ,α)}, {x}^{(Λ,α)} ≠ {y}^{(Λ,α)} and there exists a (Λ, α)–open V_{y} such that {y}^{(Λ,α)} ⊂ V_{y} and x ∉ V_{y}, which implies y ∉ {x}^{(Λ,α)}. Thus, {x}^{(Λ,α)} ⊂ U. Hence, (X, τ) is Λ_{α}–R_{0}.

Recall that a topological space (X, τ) is said to be R_{1} [2] if for x, y in X with Cl({x}) ≠ Cl({y}), there exists disjoint open sets U and V such that Cl({x}) is a subset of U and Cl({y}) is a subset of V.

\underline{Example 4.2}. Let (X, τ) be a topological space such that X = \{a, b, c\} and τ = {∅, {a}, X}. Clearly, the family of all (Λ, α)–closed sets consists of {∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, X}. We have the following for the space (X, τ).

1) (X, τ) is Λ_{α}–R_{0}, but it is not R_{0}.

2) (X, τ) is Λ_{α}–R_{1}, but it is not R_{1}.

\underline{Question}. Characterize Λ_{α}–R_{1} spaces. Is there any example showing that a topological space is Λ_{α}–R_{0} but not Λ_{α}–R_{1}?

\underline{Acknowledgment}. The authors are grateful to the referee for his remarkable work which improved the quality of this paper.

\underline{References}

1. M. Caldas, D. N. Georgiou, and S. Jafari, “Study of α–open Sets and the Related Notions in Topological Spaces (submitted).

2. A. S. Davis, “Indexed Systems of Neighborhoods for General Topological Spaces,” American Mathematical Monthly, 68 (1961), 886–893.

3. K. K. Dube, “A Note on R_{0} Topological Spaces,” Math. Vesnik, 11 (1974), 203–208.

4. D. W. Hall, S. K. Murphy, and J. Rozycki, “On Spaces Which are Essentially T_{1}, J. Austr. Math. Soc., 12 (1974), 451–455.

5. H. Herlich, “A Concept of Nearness,” Gen. Topol. Appl., 4 (1974), 191–212.

6. S. A. Naimpally, “On R_{0}-Topological Spaces,” Ann. Univ. Sci. Budapest Eotvos, Sect. Math. 10 (1967), 53–54.

7. O. Njåstad, “On Some Classes of Nearly Open Sets,” Pacific Journal of Mathematics, 15 (1965), 961–970.

8. N. A. Shanin, “On Separation in Topological Spaces,” Dokl. Akad. Nauk. SSSR, 38 (1943), 110–113.

Miguel Caldas Departamento de Matematica Aplicada Universidade Federal Fluminense Rua Mario Santos Braga, s/n 24020-140, Niteroi, RJ Brasil email: gmamccs@vm.uff.br

Esref Hatir Department of Mathematics Education Faculty Selcuk University, 42090 Konya, Turkey email: ehatir@selcuk.edu.tr

Saeid Jafari Department of Mathematics and Physics Roskilde University, Postbox 260, 4000 Roskilde, Denmark email: sjafari@ruc.dk