Weak-continuity and closed graphs
... (P) For each (x, y) ф G(f), there exist open sets U c X and V c Y containing x and y, respectively, such that f(U) n Intľ(Clľ(V)) = 0. Proof. Let (x, y) ф G(f)9 then y Ф /(x). Since Уis Hausdorff, there exist disjoint open sets Vand JVcontaining y and/(x), respectively. Thus, we have Int^Clj^V)) n n ...
... (P) For each (x, y) ф G(f), there exist open sets U c X and V c Y containing x and y, respectively, such that f(U) n Intľ(Clľ(V)) = 0. Proof. Let (x, y) ф G(f)9 then y Ф /(x). Since Уis Hausdorff, there exist disjoint open sets Vand JVcontaining y and/(x), respectively. Thus, we have Int^Clj^V)) n n ...
IOSR Journal of Mathematics (IOSR-JM)
... Proof: Let x and y be any two distinct points of X. Then f x and f y are different points of Y because f is injective. Since Y is Hausdorff, there exist disjoint open sets U and V in Y containing f x and f y respectively. Since f is continuous and U ∩ V = ϕ, f −1 U and f −1 V are disjoint open sets ...
... Proof: Let x and y be any two distinct points of X. Then f x and f y are different points of Y because f is injective. Since Y is Hausdorff, there exist disjoint open sets U and V in Y containing f x and f y respectively. Since f is continuous and U ∩ V = ϕ, f −1 U and f −1 V are disjoint open sets ...
On sp-gpr-Compact and sp-gpr-Connected in Topological Spaces
... sp-gpr-closed, B is a proper non-empty subset of X which is both sp-gpr-open and sp-gpr-closed in X. Then by Theorem 3.4, X is not sp-gpr-connected. This proves the theorem. The following example shows that the converse is not true. Example 3.7: Let X = {a, b, c}, τ = {φ, {a}, {b}, {a, b}, X }. The ...
... sp-gpr-closed, B is a proper non-empty subset of X which is both sp-gpr-open and sp-gpr-closed in X. Then by Theorem 3.4, X is not sp-gpr-connected. This proves the theorem. The following example shows that the converse is not true. Example 3.7: Let X = {a, b, c}, τ = {φ, {a}, {b}, {a, b}, X }. The ...
Part II
... Proof. Write ∪i∈I Ai = U ∪ V , where U and V are open in X. Choose an index i0 . We have Ai0 ⊂ U ∪ V , and Ai0 is connected. Then all of Ai0 is in one or the other of the open sets, so we may assume Ai0 ⊂ U. Now let x ∈ ∩i∈I Ai . We have x ∈ Ai0 ⊂ U, so x ∈ U. Then, for all i ∈ I, Ai ∩ U 6= ∅. Since ...
... Proof. Write ∪i∈I Ai = U ∪ V , where U and V are open in X. Choose an index i0 . We have Ai0 ⊂ U ∪ V , and Ai0 is connected. Then all of Ai0 is in one or the other of the open sets, so we may assume Ai0 ⊂ U. Now let x ∈ ∩i∈I Ai . We have x ∈ Ai0 ⊂ U, so x ∈ U. Then, for all i ∈ I, Ai ∩ U 6= ∅. Since ...
