Some applications of M-open sets in topological spaces

A mapping f: (X, τ) → (Y, σ) is called contra-continuous (Dontchev, 1996) (resp. contra α-continuous (Jafari and Noiri, 2001), contra-θ-semicontinuous, contra semicontinuous (Dontchev and Noiri, 1999), contra precontinuous (Jafari and Noiri, 2002), contra-δ-precontinuous (Ekici and Noiri, 2006), contra-a-continuous (Ekici, 2008a), contra-e-continuous (Ekici, 2008a, Ghosh and Basu, 2012), contra-e∗-continuous(Ekici, 2008a), contra-b-continuous (Ekici, 2004e), contra β-continuous (Caldas and Jafari, 2001), contra-δ-semicontinuous (Ekici, 2004c)) if, f⁻¹(V) is closed (resp. α-closed, θ-semiclosed, semiclosed, preclosed, δ-preclosed, a-closed, e-closed, e∗-closed, b-closed, β-closed, δ-semiclosed) in X for each open set in Y.

A mapping f: (X, τ) → (Y, σ) is called almost contra-θ-continuous (resp. almost contra-θ-semicontinuous, almost contra continuous, almost contra-super-continuous (Ekici, 2004f), almost contra δ-semicontinuous (Ekici, 2004d), almost contra-precontinuous (Ekici, 2004a), almost contra δ-precontinuous (Ekici, 2004b), almost contra-α-continuous (Baker, 2011), almost contra-a-continuous (Ekici, 2007) almost contra-e-continuous (Ekici, 2007), almost contra-e∗-continuous (Ekici, 2007), almost contra-γ-continuous (Ekici, 2005), almost contra-β-continuous, almost contra semicontinuous) if, f⁻¹(V) is θ-closed (resp. θ-semiclosed, closed, δ-closed, δ-semiclosed, preclosed, δ-preclosed, α-closed, a-closed, e-closed, e∗-closed, γ-closed, β-closed, semiclosed) in X for each regular open set of Y.

Let A be a subset of space (X, τ). Then:

(i) the kernel of A (Mrsevic, 1986) is given by ker(A) = ∩{U∈τ: A ⊆ U},

(ii) the M-boundary of A (El-Maghrabi and Al-Juhani, 2011) is given by M-b(A) = M-cl(A)/M-int(A)

Lemma 2.1 Jafari and Noiri (1999)

The following properties are holds for two subsets A, B of a topological space (X, τ):

(i) x ∈ ker(A) if and only if A ∩ F ≠ ϕ, for any closed set F of X containing x,

(ii) A ⊆ ker(A) and A = ker(A), if A is open in X,

(iii) If A ⊆ B, then ker(A) ⊆ ker(B).

A topological space (X, τ) is said to be:

(i) Urysohn (Singal and Mathur, 1969) if, for each two distinct points x, y of X, there exist two open sets U and V such that x ∈ U, y ∈ V and cl(U) ∩ cl(V) = ϕ,

(ii) ultra Hausdorff (Staum, 1974) if, for each two distinct points x, y of X, there exist two closed sets U and V such that x ∈ U, y ∈ V and U ∩ V = ϕ,

(iii) ultra normal (Staum, 1974) if for each pair of non-empty disjoint closed sets can be separated by disjoint clopen sets.

(iv) weakly Hausdorff (Soundararajan, 1971) if each element of X is the intersection of regular closed sets of X,

(v) strongly S-closed (Joseph and Kwack, 1980) (resp. S-closed (Dontchev, 1996), S-Lindelöff (Ekici, 2004a), countably S-closed (Dlaska et al., 1994) if for closed (resp. regular closed, regular closed, countably regular closed) cover of X has a finite (resp. finite, countable, finite) subcover.

A mapping f: (X, τ) → (Y, σ) is called contra-M-continuous, if f⁻¹(U) ∈ MC(X), for every open set U of Y.

According to Definitions 2.1, 3.1, the implications between these types of functions are given by the following diagram.

Let X = {a, b, c, d} with τ = {X, ϕ, {a}, {b}, {a, b}, {a, b, c}, {a, b, d}} and Y = {a, b, c} with σ = {Y, ϕ, {a}, {b}, {a, b}}. Hence a mapping f: (X, τ) → (Y, σ) which defined by f(a) = f(c) = c, f(b) = a and f(d) = b is contra-e-continuous but not contra-M-continuous. Since, f⁻¹({a, b}) = {b, d} is not M-closed of X.

Let X = Y = {a, b, c} with τ = {X, ϕ, {a}, {b}, {a, b}} and σ = {Y, ϕ, {a}, {b, c}}. If, f: (X, τ) → (Y, σ) which defined by the identity mapping, then f is contra-M-continuous mapping but not contra δ-precontinuous. Since, f⁻¹({a}) = {a} is not δ-preclosed set of X. Also, f is contra-M-continuous mapping but not contra θ-semicontinuous. Since, f⁻¹({b, c}) = {b, c} is not θ-semiclosed set of X.

For a mapping f: (X, τ) → (Y, σ), the following statements are equivalent:

(i) f is contra-M-continuous,

(ii) for each x ∈ X and each closed subset F of Y containing f(x), there exist U ∈ MO(X) such that x ∈ U and f(U) ⊆ F,

(iii) for every closed subset F of Y, f⁻¹(F) ∈ MO(X),

(iv) f(M-cl(A)) ⊆ ker(f(A)), for each A ⊆ X,

(v) M-cl(f⁻¹(B)) ⊆ f⁻¹(ker(B)), for each B ⊆ Y.

(i)→(ii). Let x ∈ X and F be any closed set of Y containing f(x). Then x ∈ f⁻¹(F). Hence by hypothesis, we have f⁻¹(Y/F) is M-closed in X and hence f⁻¹(F) is M-open set of X containing x. We put U = f⁻¹(F), then x ∈ U and f(U) ⊆ F.

(ii)→(iii). Let F be any closed set of Y and x ∈ f⁻¹(F). Then f(x) ∈ F. Hence by hypothesis, there exists an M-open subset U containing x such that f(U) ⊆ F, this implies that, x ∈ U ⊆ f⁻¹(F). Therefore, f⁻¹(F) = ∪{U: x ∈ f⁻¹(F)} which is M-open in X. Then f is contra-M-continuous.

(iii)→(iv). Let A be any subset of X and y ker(f(A)). Then by Lemma 2.1, there exists a closed set F of Y containing y such that f(A) ∩ F = ϕ. Hence, A ∩ f⁻¹(F) = ϕ and M-cl(A) ∩ f⁻¹(F) = ϕ. Then f(M-cl(A)) ∩ F = ϕ and y f(M-cl(A)). Therefore, f(M-cl(A)) ⊆ ker(f(A)).

(iv)→(v). Let B be any subset of Y. Then by hypothesis and Lemma 2.1, we have f(M-cl(f⁻¹(B))) ⊆ ker(f(f⁻¹(B))) ⊆ ker(B). Thus M-cl(f⁻¹(B)) ⊆ f⁻¹(ker(B)).

(v)→(i). Let V be any open subset of Y. Then by hypothesis and Lemma 2.1, M-cl(f⁻¹(V)) ⊆ f⁻¹(ker(V)) = f⁻¹(V). Therefore, f⁻¹(V) is M-closed in X. Hence, f is contra-M-continuous.

Definition 3.2 El-Maghrabi and Al-Juhani (2013a,b)

A mapping f: (X, τ) → (Y, σ) is called:

(i) M-continuous if, f⁻¹(U) ∈ MO(X), for each U ∈ σ,

(ii) M-irresolute if, f⁻¹(U) ∈ MO(X), for each U ∈ MO(Y),

(iii) pre-M-open if, f(U) ∈ MO(Y), for each U ∈ MO(X),

(iv) pre-M-closed if, f(U) ∈ MC(Y), for each U ∈ MC(X).

If a mapping f: (X, τ) → (Y, σ) is contra-M-continuous and Y is regular, then f is M-continuous.

Let x ∈ X and V be an open set of Y containing f(x). Since Y is a regular space, then there exists an open set G of Y such that f(x) ⊆ G ⊆ cl(G) ⊆ V. But, if f is contra-M-continuous, then there exists U ∈ MO(X) such that x ∈ U and f(U) ⊆ cl(G) ⊆ V. Hence, f is M-continuous. □

The composition of two contra-M-continuous mappings need not be contra-M-continuous as shown by the following example.

Let X = Y = Z = {a, b, c, d}, with topologies τ_x = {X, ϕ,{a},{b}, {a, b}}, τ_y is an indiscrete topology and τ_z = {Z, ϕ, {a, d}}. Then the identity mappings f: (X, τ_x) → (Y, τ_y) and g: (Y, τ_y) → (Z, τ_z) are contra-M-continuous mappings, but g o f is not contra-M-continuous. Since, f⁻¹({a, d}) is not M-closed of X.The next theorems give the conditions under which the composition of two contra-M-continuous mappings is also contra-M-continuous.

For two mappings f: (X, τ_x) → (Y, τ_y) and g: (Y, τ_y) → (Z, τ_z), the following properties are hold:

(i) If f is contra-M-continuous and g is continuous mappings, then g o f is contra-M-continuous,

(ii) If f is M-irresolute and g is contra-M-continuous mappings, then g o f is contra-M-continuous.

(i) Let U ∈ τ_z and g be a continuous mapping. Then g⁻¹(U) ∈ τ_y. But, f is contra-M-continuous, then (g o f)⁻¹(U) ∈ MC(X). Hence, g o f is contra-M-continuous.

(ii) Let U ∈ τ_z and g be a contra-M-continuous mapping. Then g⁻¹(U) ∈ MC(Y). But, f is M-irresolute, then (g o f)⁻¹(U) ∈ MC(X). Hence, g o f is contra-M-continuous. □

Let f: X → Y be a surjective M-irresolute and pre-M-open mapping. Then g o f: X → Z is contra-M-continuous if and only if g is contra-M-continuous.

Necessity. Obvious from Theorem 3.3.Sufficiency. Let g o f: X → Z be a contra-M-continuous mapping and F be a closed set of Z. Then (g o f)⁻¹(F) ∈ MO(X). Since f is surjective pre-M-open, then g⁻¹(F) ∈ MO(Y). Therefore, g is contra-M-continuous. □

Definition 3.3 El-Maghrabi and Al-Juhani, 2013c

A topological space (X, τ) is called:

(i) M-connected if X cannot be expressed as the union of two disjoint non-empty M-open sets of X,

(ii) M-normal if, for every pair of disjoint closed sets F₁ and F_2, there exist disjoint M-open sets U and V such that F₁ ⊆ U and F₂ ⊆ V,

(iii) M-T₁-space if for every two distinct points x, y of X, there exist two M-open sets U, V such that x ∈ U, y U and x V, y ∈ V.

(iv) M-T₂-space or M-Hausdorff space if for every two distinct points x, y of X, there exist two disjoint M-open sets U, V such that x ∈ U and y ∈ V.

If, f: (X, τ) → (Y, σ) is an injective closed and contra-M-continuous mappings, and Y is ultra normal, then X is M-normal.

Let F₁ and F₂ be two disjoint closed subsets of X. Since f is closed injection, then f(F₁) and f(F₂) are two disjoint closed subsets of Y and since Y is ultra normal space, then there exist two disjoint clopen sets U and V such that f(F₁) ⊆ U and f(F₂) ⊆ V. Hence, F₁ ⊆ f⁻¹(U) and F₂ ⊆ f⁻¹(V). Since f is injective contra-M-continuous, then f⁻¹(U) and f⁻¹(V) are two disjoint M-open sets of X. Therefore, X is M-normal. □

If, f: (X, τ) → (Y, σ) is a contra-M-continuous mapping and X is M-connected, then Y is not a discrete space.

Suppose that Y is a discrete space and U any subset of Y. Then U is open and closed set in Y. Since f is contra-M-continuous, f⁻¹(U) is M-closed and M-open in X which is a contradiction with the fact that X is M-connected. Hence, Y is not discrete space. □

If, f: (X, τ) → (Y, σ) is an injective contra-M-continuous mapping and Y is an Urysohn space, then X is M-T₂.

Let x, y ∈ X and x ≠ y. By hypothesis, f(x) ≠ f(y). Since Y is an Urysohn space, there exist two open sets U and V of Y such that f(x) ∈ U, f(y) ∈ V and cl(U) ∩ cl(V) = ϕ. Since f is contra-M-continuous, then there exist two M-open sets P and Q such that x ∈ P, y ∈ Q and f(P) ⊆ cl(U), f(Q) ⊆ cl(V). Then f(P) ∩ f(Q) = ϕ and hence, P ∩ Q = ϕ. Therefore, X is M-T₂. □

If f: (X, τ) → (Y, σ) is an injective contra-M-continuous mapping and Y is an ultra Hausdorff space, then X is M-T₂.

A mapping f: (X, τ) → (Y, σ) is called weakly-M-continuous if, for each x ∈ X and each open set V of Y containing f(x), there exists U ∈ MO(X) such that x ∈ U and f(U) ⊆ cl(V).

If f: (X, τ) → (Y, σ) is a contra-M-continuous mapping, then f is weakly-M-continuous.

Let x ∈ X and V ∈ σ containing f(x). Then cl(V) is closed set in Y. Since f is contra-M-continuous, then f⁻¹(cl(V)) ∈ MO(X) and containing x. If we put U = f⁻¹(cl(V)), then f(U) ⊆ cl(V). Hence, f is weakly-M-continuous. □

The converse of Theorem 3.8 is not true as shown by the following example.

Let X = {a, b, c, d} and Y = {a, b, c} with topologies τ = {X, ϕ, {a}, {b}, {a, b}, {a, b, c}, {a, b, d}}and σ = {Y, ϕ, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}}. Then a mapping f: (X, τ) → (Y, σ) which defined by f(a) = f(c) = c, f(b) = a and f(d) = b is weakly-M-continuous mapping but, it is not contra-M-continuous. Since, f⁻¹({a, b}) = {b, d} is not M-closed of X.

A mapping f: (X, τ) → (Y, σ) is called almost-M-continuous if, for each x ∈ X and each open set V of Y containing f(x), there exists U ∈ MO(X) such that x ∈ U and f(U) ⊆ int(cl(V)), equivalently, f⁻¹(V) is M-open in X for every regular open set V of Y.

A mapping f: (X, τ) → (Y, σ) is called almost-M-continuous if and only if for each x ∈ X and each regular open set V of Y containing f(x), there exists U ∈ MO(X) containing x such that f(U) ⊆ V.

Necessity. Let V ⊆ Y be regular open set containing f(x). Then x ∈ f⁻¹(V). But f is almost-M-continuous, then f⁻¹(V) = U is regular open set of X containing x such that f(U) = f f⁻¹(V) ⊆ V. □

A mapping f: (X, τ) → (Y, σ) is called almost contra-M-continuous if, f⁻¹(V) is M-closed in X, for every regular open set V of Y.

The implication between some types of mappings of Definitions 2.2, 4.2, is given by the following diagram.

Let X = Y = {a, b, c} with topologies τ = {X, ϕ, {a}, {b}, {a, b}}andσ = {Y, ϕ, {a}, {b}, {a, b}, {b, c}}. If f: (X, τ) → (Y, σ) is the identity mapping, then f is almost contra-M-continuous mapping but not almost contra δ-precontinuous. Since,f⁻¹({a}) = {a} is not δ-preclosed in X.

Let X = Y = {a, b, c, d} with topologies τ = {X, ϕ, {a},{c},{a, b}, {a, c}, {a, b, c}, {a, c, d}} and σ = {Y, ϕ, {a}, {b}, {a, b},{a, b, c}, {a, b, d}}.If f: (X, τ) → (Y, σ) is the identity mapping, then f is almost contra-M-continuous mapping but not almost contra-θ-semicontinuous. Since, f⁻¹({a}) = {a} is not θ-semiclosed in X.

Let X = Y = {a, b, c, d} with topologies τ = {X, ϕ, {a},{b},{a, b}, {a, b, c}, {a, b, d}} and σ = {Y, ϕ, {a, b, c}, {b, c}, {a, d}, {a}}. If f: (X, τ) → (Y, σ) is the identity mapping, then f is almost contra-e-continuous mapping but not almost contra M-continuous. Since, f⁻¹({b, c}) = {b, c} is not M-closed in X.

For a mapping f: (X, τ) → (Y, σ), the following statements are equivalent:

(i) f is almost contra-M-continuous,

(ii) f⁻¹(F) is M-open in X, for every regular closed set F of Y, for each x ∈ X and each regular closed set F of Y containing f(x), there exists U ∈ MO(X) such that x ∈ U and f(U) ⊆ F,

(iii) for each x ∈ X and each regular open set V of Y not containing f(x), there exists an M-closed set K of X not containing x such that f⁻¹(V) ⊆ K.

Proof i

(i)→(ii). Let F be any regular closed set of Y. Then Y/F is regular open. By hypothesis, f⁻¹(Y/F) = X/f⁻¹(F) ∈ MC(X). Therefore, f⁻¹(F) ∈ MO(X). □

(ii)→(i). Obvious.

(ii)→(iii). Let F be any regular closed set of Y containing f(x). Then by hypothesis, f⁻¹(F) ∈ MO(X) and x ∈ f⁻¹(F). Put U = f⁻¹(F), then f(U) ⊆ F.

(iii)→(ii). Let F be any regular closed set of Y and x ∈ f⁻¹(F). By hypothesis, there exist U ∈ MO(X) such that x ∈ U and f(U) ⊆ F. Hence, x ∈ U ⊆ f⁻¹(F). That implies f⁻¹(F) = ∪{U: x ∈ f⁻¹(F)} Therefore, f⁻¹(F) ∈ MO(X).

(iii)→(iv). Let V be any regular open set of Y non-containing f(x). Then Y/V is regular closed set of Y containing f(x). By (iii), there exists U ∈ MO(X) such that x ∈ U and f(U) ⊆ Y/V. Then U ⊆ f⁻¹(Y/V) ⊆ X/f⁻¹(V) and so f⁻¹(V) ⊆ X/U. Since U ∈ MO(X), then X/U = K is M-closed set of X not containing x and f⁻¹(V) ⊆ K.

(iv)→(iii). Obvious. □

The composition of two almost contra-M-continuous mappings need not be almost contra-M-continuous as shown by the following example.

Let X = Y = Z = {a, b, c, d}, with topologies τ_x = {X, ϕ,{a},{b}, {a, b}, {a, b, c}, {a, b, d}}, τ_y is an indiscrete topology and τ_z = {Z, ϕ, {a}, {b}, {a, b}, {a, d}, {a, b, c}, {a, b, d}}. Then the identity mappings f: (X, τ_x) → (Y, τ_y) and g: (Y, τ_y) → (Z, τ_z) are almost contra-M-continuous mappings, but g o f is not almost contra-M-continuous. Since, f⁻¹({a, d}) = {a, d} is not M-closed of X.

For two mappings f: (X, τ_x) → (Y, τ_y) and g: (Y, τ_y) → (Z, τ_z), the following properties are hold:

(i) If, f is a surjective pre-M-open and g o f: X → Z is almost contra-M-continuous, then g is almost contra-M-continuous.

(ii) If, f is a surjective pre-M-closed and g o f: X → Z is almost contra-M-continuous, then g is almost contra-M-continuous.

Let V ⊆ Z be regular closed set. Since, g o f is almost contra-M-continuous, then (g o f)⁻¹(V) ∈ MO(X). But, f is surjective pre-M-open, then g⁻¹(V) ∈ MO(Y). Therefore, g is almost contra-M-continuous. □(i) Obvious. □

If f: (X, τ) → (Y, σ) is an injective almost contra-M-continuous mapping and Y is weakly Hausdorff, then X is M-T₁.

Let x, y be two distinct points of X. Since f is injective, then f(x) ≠ f(y) and since Y is weakly Hausdorff, there exist two regular closed sets U and V such that f(x) ∈ U, f(y)  U and f(x)  V, f(y) ∈ V. Since f is an almost contra-M-continuous, we have f⁻¹(U) and f⁻¹(V) are M-open sets in X such that x ∈ f⁻¹(U), y  f⁻¹(U) and x  f⁻¹(V), y ∈ f⁻¹(V) and f⁻¹(U) ∩ f⁻¹(V) = ϕ. Hence X is M-T₁. □

A topological space (X, τ) is said to be:

(i) M-compact if every M-open cover of X has finite subcover,

(ii) countably M-compact if every countable cover of X by M-open sets has a finite subcover,

(iii) M-Lindelöff if every M-open cover of X has a countable subcover.

If f: (X, τ) → (Y, σ) is a surjective almost contra-M-continuous mapping, then the following statements are hold:

(i) If X is M-compact, then Y is S-closed,

(ii) If X is countably M-compact, then Y is countably S-closed,

(iii) If X is M-Lindelöff, then Y is S-Lindelöff.

Let {V_i: i ∈ I} be any regular closed cover of Y and f be almost contra-M-continuous. Then {f⁻¹(V_i): ∈ I}is M-open cover of X. But X is M-compact, there exists a finite subset I_o of I such that X = ∪{f⁻¹(V_i): i ∈ I_o}, hence Y = ∪ {f f⁻¹(V_i): i ∈ I_o} and then Y = ∪{V_i: i∈I_o}. Hence Y is S-closed.

(i) Similar to (i).

(ii) Similar to (i). □