rioaja0006

Theorem 3.36. A subset A of X is δ gβ-closed if and only if βcl(A) ⊆ ker δ (A). Proof. Suppose A is δgβ-closed set in X such that x ∈ βcl(A). If possible, let ker δ(A), then there exists a δ-open set G in X such that A ⊆ G and x  G. Since A is δ gβ-closed in X, βcl(A)⊆G implies x  βcl(A) which is a contradiction. Conversely, let βcl(A)⊆kerδ(A) be true and G is a δ-open set containing A, then ker δ(A)⊆G which implies βcl(A)⊆G. Hence A is δgβ-closed.Let A and B be subsets of a topological space X.Then: (i) δgβcl(X) = X and δgβcl(Φ) = Φ (ii) If A ⊆ B, then δgβcl(A) ⊆ δgβcl(B). (iii) δgβcl(A) ∪ δgβcl(B) ⊆ δgβcl(A∪B). (iv) δgβcl(A∩B) ⊆ δgβcl(A) ∩ δgβcl(B). (v) If A is δ gβ closed, then δgβcl(A) = A. (vi) A ⊆ δgβcl(A) ⊂ gβcl(A) ⊂ βcl(A).