In this paper, the neural network domain with the backpropagation Levenberg-Marquardt scheme (NNB-LMS) is novel with a convergent stability and generates a numerical solution of the impact of the magnetohydrodynamic (MHD) nanofluid flow over a rotating disk (MHD-NRD) with heat generation/absorption and slip effects. The similarity variation in the MHD flow of a viscous liquid through a rotating disk is explained by transforming the original non-linear partial differential equations (PDEs) to an equivalent non-linear ordinary differential equation (ODEs). Varying the velocity slip parameter, Hartman number, thermal slip parameter, heat generation/absorption parameter, and concentration slip parameter, generates a Prandtl number using the Runge-Kutta 4th order method (RK4) numerical technique, which is a dataset for the suggested (NNB-LMS) for numerous MHD-NRD scenarios. The validity of the data is tested, and the data is processed and properly tabulated to test the exactness of the suggested model. The recommended model was compared for verification, and the estimation solutions for particular instances were assessed using the NNB-LMS training, testing, and validation procedures. A regression analysis, a mean squared error (MSE) assessment, and a histogram analysis were used to further evaluate the proposed NNB-LMS. The NNB-LMS technique has various applications such as disease diagnosis, robotic control systems, ecosystem evaluation, etc. Some statistical data such as the gradient, performance, and epoch of the model were analyzed. This recommended method differs from the reference and suggested results, and has an accuracy rating ranging from 10−09to 10−12.
Citation: Yousef Jawarneh, Humaira Yasmin, Wajid Ullah Jan, Ajed Akbar, M. Mossa Al-Sawalha. A neural networks technique for analysis of MHD nano-fluid flow over a rotating disk with heat generation/absorption[J]. AIMS Mathematics, 2024, 9(11): 32272-32298. doi: 10.3934/math.20241549
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In this paper, the neural network domain with the backpropagation Levenberg-Marquardt scheme (NNB-LMS) is novel with a convergent stability and generates a numerical solution of the impact of the magnetohydrodynamic (MHD) nanofluid flow over a rotating disk (MHD-NRD) with heat generation/absorption and slip effects. The similarity variation in the MHD flow of a viscous liquid through a rotating disk is explained by transforming the original non-linear partial differential equations (PDEs) to an equivalent non-linear ordinary differential equation (ODEs). Varying the velocity slip parameter, Hartman number, thermal slip parameter, heat generation/absorption parameter, and concentration slip parameter, generates a Prandtl number using the Runge-Kutta 4th order method (RK4) numerical technique, which is a dataset for the suggested (NNB-LMS) for numerous MHD-NRD scenarios. The validity of the data is tested, and the data is processed and properly tabulated to test the exactness of the suggested model. The recommended model was compared for verification, and the estimation solutions for particular instances were assessed using the NNB-LMS training, testing, and validation procedures. A regression analysis, a mean squared error (MSE) assessment, and a histogram analysis were used to further evaluate the proposed NNB-LMS. The NNB-LMS technique has various applications such as disease diagnosis, robotic control systems, ecosystem evaluation, etc. Some statistical data such as the gradient, performance, and epoch of the model were analyzed. This recommended method differs from the reference and suggested results, and has an accuracy rating ranging from 10−09to 10−12.
Let B be a subset of a topological space (Y,σ). The set of condensation points of B is denoted by Cond(B) and defined by Cond(B)={y∈Y: for every V∈σ with y∈V, V∩B is uncountable}. For the purpose of characterizing Lindelöf topological spaces and improving some known mapping theorems, the author in [1] defined ω-closed sets as follows: B is called an ω-closed set in (Y,σ) if Cond(B)⊆B. B is called an ω-open set in (Y,σ) [1] if Y−B is ω-closed. It is well known that B is ω-open in (Y,σ) if and only if for every b∈B, there are V∈σ and a countable subset F⊆Y with b∈V−F⊆B. The family of all ω-open sets in (Y,σ) is denoted by σω. It is well known that σω is a topology on Y that is finer than σ. Many research papers related to ω-open sets have appeared in [2,3,4,5,6,7,8] and others. Authors in [9,10,11] included ω-openness in both soft and fuzzy topological spaces. In this paper, we will denote the closure of B in (Y,σ), the closure of B in (Y,σω), the interior of B in (Y,σ), and the interior of B in (Y,σω) by ¯B, ¯Bω, Int(B), and Intω(B), respectively. B is called a semi-open set in (Y,σ) [12] if there exists V∈σ such that V⊆B⊆¯V. Complements of semi-open sets are called semi-closed sets. The family of all semi-open sets in (Y,σ) will be denoted by SO(Y,σ). Authors in [12,13,14,15] have used semi-open sets to define semi-continuity, semi-openness, irresoluteness, pre-semi-openness, and slight semi-continuity. The area of research related to semi-open sets is still hot [16,17,18,19,20,21,22,23,24,25,26,27,28]. Authors in [29] have defined ωs-open sets as a strong form of semi-open sets as follows: B is called an ωs-open set in (Y,σ) if there exists O∈σ such that O⊆B⊆¯Oω. Complements of ωs-open sets are called ωs-closed sets. The family of all ωs-open sets in (Y,σ) will be denoted by ωs(Y,σ). The intersection of all ωs-closed sets in (Y,σ) which contains B will be denoted by ¯Bωs, and the union of all ωs-open sets in (Y,σ) which contained in B will be denoted by Intωs(B). Authors in [29] have defined and investigated the class of ωs-continuity which lies strictly between the classes of continuity and semi-continuity.
In this paper, ωs-irresoluteness as a strong form of ωs-continuity is introduced. It is proved that ωs-irresoluteness is independent of each of continuity and ωs-irresoluteness. Also, ωs-openness which lies strictly between openness and semi-openness is introduced and investigated, and pre-ωs-openness which is a strong form of ωs -openness and independent of openness is introduced and investigated. Moreover, slight ωs-continuity as a new class of functions which lies between slight continuity and slightly semi-continuity is introduced and investigated. In addition to these, ωs-compactness as a new class of topological spaces that lies strictly between compactness and semi-compactness is introduced. Several implications, examples, counter-examples, characterizations, and mapping theorems are introduced. In particular, several sufficient conditions for the equivalence between our new concepts and other related concepts are given. We hope that this will open the door for a number of future related studies such as ωs -separation axioms and ωs-connectedness.
Throughout this paper, the usual topology on R will be denoted by τu.
Recall that a topological space (Y,σ) is called locally countable [30] (resp. anti-locally countable [31]) if (Y,σ) has a base consisting of countable sets (all non-empty open sets are uncountable sets).
The following results will be used in the sequel:
Proposition 1.1. [29] Let (Y,σ) be a topological space. Then
(a) σ⊆ωs(Y,σ)⊆SO(Y,σ).
(b) If (Y,σ) is locally countable, then σ=ωs(Y,σ).
(c) If (Y,σ) is anti-locally countable, then ωs(Y,σ)=SO(Y,σ).
Proposition 1.2. [29] Let (Y,σ) be a topological space and let B,C⊆Y. Then we have the following:
(a) If B⊆C⊆¯Bω and B∈ωs(Y,σ), then C∈ωs(Y,σ).
(b) If B∈σ and C∈ωs(Y,σ), then B∩C∈ωs(Y,σ).
Proposition 1.3. [31] Let (Y,σ) be anti-locally countable. Then for every B∈σω, we have ¯Bω=¯B.
Definition 2.1. A function g:(Y,σ)⟶(Z,γ) is called irresolute [14] (resp. ωs -continuous [29]), if for every V∈SO(Z,γ) (resp. V∈γ), g−1(V)∈SO(Y,σ) (resp. g−1(V)∈ωs(Y,σ)).
Definition 2.2. A function g:(Y,σ)⟶(Z,γ) is called ωs-irresolute, if for every V∈ωs(Z,γ), g−1(V)∈ωs(Y,σ).
The following two examples will show that irresoluteness and ωs -irresoluteness are independent concepts:
Example 2.3. Let X=R, Y={a,b}, τ={∅,R,N,Qc,N∪Qc}, and σ={∅,Y,{a},{b}}. Define f:(X,τ)⟶(Y,σ) by
f(x)={aif x∈Qcbif x∈Q. |
Since f−1({a})=Qc∈τ⊆SO(X,τ) and f−1({b})=Q∈SO(X,τ)−ωs(X,τ), then f is irresolute but not ωs-irresolute.
Example 2.4. Consider the topology σ={∅,N,{1},{2},{1,2}} on N. Define g:(N,σ)⟶(N,σ) by
g(t)={1if t=11if t=2tif t∈N−{1,2}. |
Since (N,σ) is locally countable, then by Proposition 1.1 (b) we have ωs(N,σ)=σ. Since f−1({1})={1,2}∈σ and f−1({2})=∅∈σ, then f is ωs-irresolute. However, f is not irresolute because there is ¯{2}=N−{1}∈SO(N,σ) such that f−1(N−{1})=N−{1,2}∉SO(N,σ).
Theorem 2.5. Let (Y,σ) and (Z,γ) be two anti-locally countable topological spaces. Then for any function g:(Y,σ)⟶(Z,γ) the followings are equivalent:
(a) g is irresolute.
(b) g is ωs-irresolute.
Proof. Follows from the definitions and Proposition 1.1 (c).
The following two examples will show that continuity and ωs -irresoluteness are independent concepts:
Example 2.6. Let X=R and τ={∅,R,(3,∞),{2},{2}∪(3,∞)}. Define f:(X,τ)⟶(X,τ) by
f(x)={2if x∈{2}∪(3,∞)xif x∈R−({2}∪(3,∞)). |
Since f−1((3,∞))=∅∈τ and f−1({2})={2}∪(3,∞)∈τ, then f is continuous. Since ¯(3,∞)ω=R−{2}, then R−{2}∈ωs(X,τ). Since f−1(R−{2})=R−({2}∪(3,∞))∉ωs(X,τ), then f is not ωs-irresolute.
Example 2.7. Let τdisc be the discrete topology on {a,b}. Define f:(R,τu)⟶({a,b},τdisc) by
f(x)={aif x∈(−∞,0)bif x∈[0,∞). |
Clearly that ωs({a,b},τdisc)=τdisc. Since f−1({a})=(−∞,0)∈τu⊆ωs(R,τu) and f−1({b})=[0,∞)∈ωs(R,τu), then f is ωs-irresolute. However, f is not continuous because there is {b}∈τdisc such that f−1({b})=[0,∞)∉τu.
Theorem 2.8. Let (Y,σ) and (Z,γ) be two locally countable topological spaces. Then for any function g:(Y,σ)⟶(Z,γ) the followings are equivalent:
(a) g is continuous.
(b) g is ωs-irresolute.
Proof. Follows from the definitions and Proposition 1.1 (b).
Theorem 2.9. If g:(Y,σ)⟶(Z,γ) is a continuous function such that g:(Y,σω)⟶(Z,γω) is an open function, then g is ωs-irresolute.
Proof. Let H∈ωs(Z,γ). Then there exists W∈γ such that W⊆H⊆¯Wω and so g−1(W)⊆g−1(H)⊆g−1(¯Wω). Since g:(Y,σ)⟶(Z,γ) is continuous, then g−1(W)∈σ. Since g:(Y,σω)⟶(Z,γω) is open, then g−1(¯Wω)⊆¯g−1(W)ω. Thus, g−1(W)⊆g−1(H)⊆¯g−1(W)ω, and hence g−1(H)∈ωs(Y,σ). Therefore, g is ωs-irresolute.
Theorem 2.10. Every ωs-irresolute function is ωs-continuous.
Proof. Let f:(X,τ)⟶(Y,σ) be ωs -irresolute. Let V∈σ. Then by Proposition 1.1 (a), V∈ωs(Y,σ). Since f is ωs-irresolute, then f−1(V)∈ωs(X,τ). Therefore, f is ωs -irresolute.
The function in Example 2.6 is continuous and hence ωs -continuous. Therefore, the converse of Theorem 2.10 is not true, in general.
Theorem 2.11. A function g:(Y,σ)⟶(Z,γ) is ωs-irresolute if and only if for every ωs-closed subset B of (Z,γ), g−1(B) is ωs-closed in (Y,σ).
Proof. Necessity. Assume that g is ωs-irresolute. Let B be an ωs-closed set in (Z,γ). Then Z−B∈ωs(Z,γ). Since g is ωs-irresolute, then g−1(Z−B)=Y−g−1(B)∈ωs(Y,σ). Hence, g−1(B) is ωs-closed in (Y,σ).
Sufficiency. Suppose that for every ωs-closed subset B of (Z,γ), g−1(B) is ωs -closed in (Y,σ). Let W∈ωs(Z,γ). Then Z−W is ωs-closed in (Z,γ). By assumption, g−1(Z−W)=Y−g−1(W) is ωs-closed in (Y,σ), and so g−1(W)∈ωs(Y,σ). Therefore, g is ωs-irresolute.
Theorem 2.12. A function g:(Y,σ)⟶(Z,γ) is ωs-irresolute if and only if for every H⊆Y, g(¯Hωs)⊆¯g(H)ωs.
Proof. Necessity. Suppose that g is ωs-irresolute and let H⊆Y. Then ¯g(H)ωs is ωs-closed in (Z,γ), and by Theorem 2.11, g−1(¯g(H)ωs) is ωs-closed in (Y,σ). Since H⊆g−1(¯g(H)ωs), then ¯Hωs⊆g−1(¯g(A)ωs). Thus,
g(¯Hωs)⊆g(g−1(¯g(H)ωs))⊆¯g(H)ωs. |
Sufficiency. Suppose that for every subset H⊆Y, g(¯Hωs)⊆¯g(H)ωs. We will apply Theorem 2.11 to show that g is ωs -irresolute. Let B be an ωs-closed subset of (Z,γ). Then by assumption we have g(¯g−1(B)ωs)⊆¯g(g−1(B))ωs⊆¯Bωs=B, and so
¯g−1(B)ωs⊆g−1(g(¯g−1(B)ωs))⊆g−1(B). |
Therefore, ¯g−1(B)ωs=g−1(B), and hence g−1(B) is ωs-closed in (Y,σ). This shows that g is ωs-irresolute.
Theorem 2.13. A function g:(Y,σ)⟶(Z,γ) is ωs-irresolute if and only if for every H⊆Z, ¯g−1(H)ωs⊆g−1(¯Hωs).
Proof. Necessity. Suppose that g is ωs-irresolute and let H⊆Z. Then by Theorem 2.11, g−1(¯Hωs) is ωs-closed in (Y,σ). Since g−1(H)⊆g−1(¯Hωs), then ¯g−1(H)ωs⊆g−1(¯Hωs).
Sufficiency. Suppose that for every H⊆Z, ¯g−1(H)ωs⊆g−1(¯Hωs). We will apply Theorem 2.11 to show that g is ωs-irresolute. Let B be an ωs-closed subset of (Z,γ). Then ¯Bωs=B. So by assumption, ¯g−1(B)ωs⊆g−1(B), and so ¯g−1(B)ωs=g−1(B). Therefore, g−1(B) is ωs-closed in (Y,σ).
Theorem 2.14. The composition of two ωs -irresolute functions is ωs-irresolute.
Proof. Let g:(Y,σ)⟶(Z,λ) and h:(Z,λ)⟶(M,γ) be ωs-irresolute functions. Let C∈ωs(M,γ). Since h is ωs -irresolute, then h−1(C)∈ωs(Z,λ). Since g is ωs-irresolute, then (h∘g)−1(C)=g−1(h−1(C))∈ωs(Y,σ). Therefore, h∘g is ωs-irresolute.
Definition 3.1. A function g:(Y,σ)⟶(Z,λ) is called
(a) ωs-open (resp. semi-open [13]) if for each U∈σ, f(U)∈ωs(Z,λ) (resp. f(U)∈SO(Z,λ)).
(b) pre-semi-open [14] if for each U∈SO(Y,σ), f(U)∈SO(Z,λ).
Theorem 3.2. Let g:(Y,σ)⟶(Z,λ) be a function. If for a base B of (Y,σ), g(B)∈λ for all B∈B, then g is ωs-open.
Proof. Suppose that for a base B of (Y,σ), g(B)∈ωs(Z,λ) for all B∈B. Let V∈σ−{∅}. Choose B1⊆B such that V=∪{B:B∈B1}. Then
g(V)=g(∪{B:B∈B1})=∪{g(B):B∈B1}. |
Since by assumption, g(B)∈ωs(Z,λ) for all B∈B1, then g(V)∈ωs(Z,λ).
Theorem 3.3. Every open function is ωs-open.
Proof. Let g:(Y,σ)⟶(Z,γ) be an open function and let V∈σ. Since g is open, then g(V)∈γ⊆ωs(Z,γ).
The converse of Theorem 3.3 is not true as shown in the next example:
Example 3.4. Consider the function g:(R,τu)⟶(R,τu) defined by g(y)=y2. We apply Theorem 3.2 to show that g is ωs-open. Consider the base {(c,d):c,d∈R and c<d} for (R,τu). Then for all c,d∈R with c<d we have
g((c,d))={(d2,c2)if c<d≤0(c2,d2)if 0≤c<d[0,max{c2,d2})if c<0<d, |
and so g((c,d))∈ωs(R,τu). Therefore, g is ωs-open. On the other hand, since R∈τu but g(R)=[0,∞)∉τu, then g is not open.
Theorem 3.5. If g:(Y,σ)⟶(Z,γ) is an ωs-open function such that (Z,γ) is locally countable, then g is open.
Proof. Follows from the definitions and Proposition 1.1 (b).
Theorem 3.6. Every ωs-open function is semi-open.
Proof. Let g:(Y,σ)⟶(Z,γ) be an ωs-open function and let V∈σ. Since g is ωs-open, then g(V)∈ωs(Z,γ)⊆SO(Z,γ).
The converse of Theorem 3.6 is not true as shown in the next example:
Example 3.7. Consider (R,σ) where σ={∅,N,R}. Define g:(R,σ)⟶(R,σ) by g(y)=y−1. Since g(N)={0}∪N⊆¯N=R, then, g(N)∈SO(R,σ). Also, g(∅)=∅∈SO(R,σ) and g(R)=R∈SO(R,σ). Therefore, g is semi-open. Conversely, g is not ωs-open since there is N∈σ such that g(N)={0}∪N∉ωs(R,σ).
Theorem 3.8. If g:(Y,σ)⟶(Z,γ) is a semi-open function such that (Z,γ) is anti-locally countable, then g is ωs-open.
Proof. Follows from the definitions and Proposition 1.1 (c).
Definition 3.9. A function g:(Y,σ)⟶(Z,γ) is called pre-ωs-open, if for every A∈ωs(Y,σ), g(A)∈ωs(Z,γ).
The following two examples will show that openness and pre-ωs -openness are independent concepts:
Example 3.10. Let X=R, Y={a,b,c}, τ={∅,(−∞,1),X}, and σ={∅,{a},Y}. Define f:(X,τ)⟶(Y,σ) by
f(x)={aif x∈(−∞,1)bif x=1cif x∈(1,∞). |
Since f(∅)=∅∈σ, f((−∞,1))={a}∈σ, and f(X)=Y∈σ, then f is open. Since (X,τ) is anti-locally countable, then by Proposition 1.3, ¯(−∞,1)ω=¯(−∞,1)=R. So, we have (−∞,1]∈ωs(X,τ) but f((−∞,1])={a,b}∉ωs(Y,σ)=σ. This shows that f is not pre-ωs-open.
Example 3.11. Consider (R,τ) where τ={∅,R,[1,∞)}. Define f:(R,τ)⟶(R,τ) by f(x)=x−1. Since (R,τ) is anti-locally countable, then by Proposition 1.1 (c), ωs(R,τ)=SO(R,τ)={∅}∪{H:[1,∞)⊆H}. To see that f is pre-ωs-open, let H∈ωs(R,τ)−{∅}. Then [0,∞)=f([1,∞))⊆f(H), and thus [1,∞)⊆f(H). Hence, f(H)∈ωs(R,τ). Therefore, f is pre-ωs-open. Conversely, f is not open since there is [1,∞)∈τ such that f([1,∞))=[0,∞)∉τ.
Theorem 3.12. Let (Y,σ) and (Z,γ) be locally countable, and let g:(Y,σ)⟶(Z,γ) be a function. Then the followings are equivalent:
(a) g is open.
(b) g is pre-ωs-open.
Proof. Follows from the definitions and Proposition 1.1 (b).
The following two examples will show that pre-semi-openness and pre-ωs-openness are independent concepts:
Example 3.13. Consider the function g as in Example 3.7. It is not difficult to see that SO(R,σ)={∅}∪{H:N⊆H} and ωs(R,σ)=σ. To see that g is pre-semi-open, let H∈SO(R,σ)−{∅}. Then N⊆H and so N⊆N∪{0}=g(N). Thus, g(N)∈SO(R,σ). This shows that g is pre-semi-open. Conversely, g is not pre-ωs-open since there is N∈ωs(R,σ) such that g(N)={0}∪N∉ωs(R,σ).
Example 3.14. Let X=Y={1,2,3,4}, τ={∅,X,{1,2},{1},{2}}, and σ={∅,Y,{2,3,4},{1,2},{1},{2}}. Let f:(X,τ)⟶(Y,σ) be the identity function. It is clear that f is open. Since (X,τ) and (Y,σ) are locally countable, then by Theorem 3.12, f is pre-ωs-open. Conversely, f is not pre-semi-open because there is {1,3}∈SO(X,τ) such that f({1,3})={1,3}∉SO(Y,σ).
Theorem 3.15. Let (X,τ) and (Y,σ) be anti-locally countable topological spaces and let f:(X,τ)⟶(Y,σ) be a function. Then f is pre-semi-open if and only if f is pre-ωs-open.
Proof. Follows from definitions and Proposition 1.1 (c).
Theorem 3.16. Every pre-ωs-open function is ωs -open. Proof. Let f:(X,τ)⟶(Y,σ) be a pre-ωs-open function and let U∈τ⊆ωs(X,τ). Since f is pre-ωs-open, then f(U)∈ωs(Y,σ).
The function f in Example 3.10 is open but not pre-ωs-open, and by Theorem 3.3, f is ωs-open. Therefore, the converse of the implication in Theorem 3.16 is not true, in general.
Theorem 3.17. For a function g:(Y,σ)⟶(Z,λ), the followings are equivalent:
(a) g is ωs-open.
(b) g−1(¯Bωs)⊆¯g−1(B) for every B⊆Z.
(c) Int(g−1(B))⊆g−1(Intωs(B)) for every B⊆Z.
Proof. (a) ⟹ (b): Suppose that g is ωs-open and let B⊆Z. Let y∈g−1(¯Bωs). To show that y∈¯g−1(B), let V∈σ such that y∈V. Then g(y)∈g(V). Since g is ωs-open, then g(V)∈ωs(Z,λ). Since g(y)∈g(V)∩¯Bωs, then g(V)∩B≠∅. Choose t∈V such that g(t)∈B. Then t∈V∩g−1(B), and hence V∩g−1(B)≠∅. It follows that y∈¯g−1(B).
(b) ⟹ (a): Suppose that g−1(¯Bωs)⊆¯g−1(B) for every B⊆Z, and suppose to the contrary that g is not ωs-open. Then there exists V∈σ such that g(V)∉ωs(Z,λ) and so, Z−g(V) is not ωs-closed. Thus, there exists z∈g(V)∩¯Z−g(V)ωs. Choose y∈V such that z=g(y). Then y∈g−1(¯Z−g(V)ωs). By assumption, we have
g−1(¯Z−g(V)ωs)⊆¯g−1(Z−g(V))=¯Y−g−1(g(V))⊆¯Y−V=Y−V. |
Therefore, y∈Y−V. But y∈V, a contradiction.
(b) ⟹ (c): Suppose that g−1(¯Bωs)⊆¯g−1(B) for every B⊆Z. Let B⊆Z. Then by (b),
g−1(Intωs(B))=g−1(Z−¯Z−Bωs)=Y−g−1(¯Z−Bωs)⊇Y−¯g−1(Z−B)=Y−¯Y−g−1(B)=Int(g−1(B)). |
(c) ⟹ (b): Suppose that Int(g−1(B))⊆g−1(Intωs(B)) for every B⊆Z. Let B⊆Z. Then by (c),
g−1(¯Bωs)=g−1(Z−Intωs(Z−B))=Y−g−1(Intωs(Z−B))⊆Y−Int(g−1(Z−B))=Y−Int(Y−g−1(B))=¯g−1(B). |
Theorem 3.18. For a function g:(Y,σ)⟶(Z,λ), the followings are equivalent:
(a) g is pre-ωs-open.
(b) g−1(¯Bωs)⊆¯g−1(B)ωs for every B⊆Z.
(c) Intωs(g−1(B))⊆g−1(Intωs(B)) for every B⊆Z.
Proof. (a) ⟹ (b): Suppose that g is pre-ωs-open and let B⊆Z. Let y∈g−1(¯Bωs). To show that y∈¯g−1(B)ωs, let C∈ωs(Y,σ) such that y∈C. Then g(y)∈g(C). Since g is pre-ωs-open, then g(C)∈ωs(Z,λ). Since g(y)∈g(C)∩¯Bωs, then g(C)∩B≠∅. Choose t∈C such that g(t)∈B. Then t∈C∩g−1(B) and hence C∩g−1(B)≠∅. It follows that y∈¯g−1(B)ωs.
(b) ⟹ (a): Suppose that g−1(¯Bωs)⊆¯g−1(B)ωs for every B⊆Z, and suppose to the contrary that g is not pre-ωs-open. Then there is C∈ωs(Y,σ) such that g(C)∉ωs(Z,λ) and so Z−g(C) is not ωs-closed. And so there exists z∈g(C)∩¯Z−g(C)ωs. Choose y∈C such that z=g(y). Then y∈g−1(¯Z−g(C)ωs). By assumption we have
g−1(¯Z−g(C)ωs)⊆¯g−1(Z−g(C))ωs=¯Y−g−1(g(C))ωs⊆¯Y−Cωs=Y−C. |
Therefore, y∈Y−C but y∈C, a contradiction.
(b) ⟹ (c): Suppose that g−1(¯Bωs)⊆¯g−1(B)ωs for every B⊆Z. Let B⊆Z. Then by (b),
g−1(Intωs(B))=g−1(Z−¯Z−Bωs)=Y−g−1(¯Z−Bωs)⊇Y−¯g−1(Z−B)ωs=Y−¯Y−g−1(B)ωs=Intωs(g−1(B)). |
(c) ⟹ (b): Suppose that Intωs(g−1(B))⊆g−1(Intωs(B)) for every B⊆Z. Let B⊆Z. Then by (c),
g−1(¯Bωs)=g−1(Z−Intωs(Z−B))=Y−g−1(Intωs(Z−B))⊆Y−Intωs(g−1(Z−B))=Y−Intωs(Y−g−1(B))=¯g−1(B)ωs. |
Theorem 3.19. If g:(Y,σ)⟶(Z,λ) is ωs-continuous such that g:(Y,σω)⟶(Z,λω) is open, then g:(Y,σ)⟶(Z,λ) is ωs-irresolute.
Proof. Suppose that g:(Y,σ)⟶(Z,λ) is ωs-continuous with g:(Y,σω)⟶(Z,λω) is open. Let G∈ωs(Z,λ), choose W∈λ such that W⊆G⊆¯Wω. Thus, we have g−1(W)⊆g−1(G)⊆g−1(¯Wω). Since g:(Y,σ)⟶(Z,λ) is ωs -continuous, then g−1(W)∈ωs(Y,σ). Since g:(Y,σω)⟶(Z,λω) is open, then g−1(¯Wω)⊆¯g−1(W)ω. Since we have g−1(W)⊆g−1(G)⊆¯g−1(W)ω with g−1(W)∈ωs(Y,σ), then by Proposition 1.2 (a), g−1(G)∈ωs(Y,σ). This ends the proof.
Theorem 3.20. If g:(Y,σ)⟶(Z,λ) is an ωs-open function such that g:(Y,σω)⟶(Z,λω) is a continuous function, then g:(Y,σ)⟶(Z,λ) is pre-ωs-open.
Proof. Suppose that g:(Y,σ)⟶(Z,λ) is ωs-open with g:(Y,σω)⟶(Z,λω) is continuous. Let H∈ωs(Y,σ), then we find V∈σ such that V⊆H⊆¯Vω. Thus, we have g(V)⊆g(H)⊆g(¯Vω). Since g:(Y,σ)⟶(Z,λ) is ωs-open, then g(V)∈ωs(Z,λ). By continuity of g:(Y,σω)⟶(Z,λω), g(¯Vω)⊆¯g(V)ω. Since we have g(V)⊆g(H)⊆¯g(V)ω with g(V)∈ωs(Z,λ), then by Proposition 1.2 (a), g(V)∈ωs(Z,λ). This ends the proof.
As defined in [32], a function g:(Y,σ)⟶(Z,λ) is ω-continuous if for each W∈λ, g−1(W)∈σω.
Theorem 3.21. Let g:(Y,σ)⟶(Z,λ) be pre-ωs-open and ωs-irresolute such that (Z,λ) is semi-regular and dense in itself, then g is ω-continuous.
Proof. Suppose to the contrary that g is not ω -continuous. Then there is W∈λ such that g−1(W)∉σω. So, there exists y∈g−1(W)−Intω(g−1(W)). Since g(y)∈W and (Z,λ) is semi-regular, then there is a regular open set M of (Z,λ) such that g(y)∈M⊆W. Since Int(¯Mω)⊆Int(¯M)=M, then by Theorem 2.16 of [29], M is ωs -closed. Since g is ωs-irresolute, then by Theorem 2.11, g−1(M) is ωs-closed, and so Y−g−1(M)∈ωs(Y,σ). Since g−1(M)⊆g−1(W), then Intω(g−1(M))⊆Intω(g−1(W)). Since y∉Intω(g−1(W)), then y∉Intω(g−1(M)), and so
y∈Y−Intω(g−1(M))=¯Y−g−1(M)ω. |
Thus, by Proposition 1.2 (a), we have (Y−g−1(M))∪{y}∈ωs(Y,σ) with y∉Y−g−1(M). Put S=g((Y−g−1(M))∪{y})=g(Y−g−1(M)))∪{g(y)}. Since g is pre-ωs-open, then S∈ωs(Z,λ). So by Proposition 1.2 (b), we have S∩M∈ωs(Z,λ). Since g(Y−g−1(M)))⊆Z−M, then we have
g(y)∈S∩M⊆((Z−M)∪{g(y)})∩M={g(y)}, |
and thus S∩M={g(y)}. Therefore, {g(y)}∈ωs(Z,λ). Thus, there exists O∈λ such that O⊆{g(y)}⊆¯Oω, and hence O={g(y)}. This implies that {g(y)}∈λ. But by assumption (Z,λ) is dense in itself, a contradiction.
The condition '(Z,λ) is dense in itself' in Theorem 3.21 cannot be dropped as our next example shows:
Example 3.22. Take f as in Example 2.7. Then f is ωs -irresolute and pre-ωs-open. Also, ({a,b},τdisc) is semi-regular. On the other hand, since {b}∈τdisc but f−1({b})=[0,∞)∉(τu)ω, then f is not ω -continuous.
Theorem 3.23. Let g:(Y,σ)⟶(Z,λ) be injective, pre-ωs-open, and ωs -irresolute such that (Z,λ) is semi-regular, then g is ω -continuous.
Proof. Suppose to the contrary that g is not ω -continuous. Then there is W∈λ such that g−1(W)∉σω. So, there exists y∈g−1(W)−Intω(g−1(W)). Since g(y)∈W and (Z,λ) is semi-regular, then there is a regular open set M of (Z,λ) such that g(y)∈M⊆W. Since Int(¯Mω)⊆Int(¯M)=M, then by Theorem 2.16 of [29], M is ωs -closed. Since g is ωs-irresolute, then by Theorem 2.11, g−1(M) is ωs-closed, and so Y−g−1(M)∈ωs(Y,σ). Since g−1(M)⊆g−1(W), then Intω(g−1(M))⊆Intω(g−1(W)). Since y∉Intω(g−1(W)), then y∉Intω(g−1(M)), and so
y∈Y−Intω(g−1(M))=¯Y−g−1(M)ω. |
Thus, by Proposition 1.2 (a), we have (Y−g−1(M))∪{y}∈ωs(Y,σ) with y∉Y−g−1(M). Put S=g((Y−g−1(M))∪{y})=g(Y−g−1(M)))∪{g(y)}. Since g is pre- ωs-open, then S∈ωs(Z,λ), and by Proposition 1.2 (b) we have S∩M∈ωs(Z,λ). Since g(Y−g−1(M)))⊆Z−M, then we have
g(y)∈S∩M⊆((Z−M)∪{g(y)})∩M={g(y)}, |
and thus S∩M={g(y)}. Therefore, {g(y)}∈ωs(Z,λ). Since g is ωs-irresolute and injective, then g−1({g(y)})={y}∈ωs(Y,σ). So, there exists V∈σ such that V⊆{y}⊆¯Vω, and hence {y}∈σ. Since y∈g−1(W) and {y}∈σ⊆σω, then y∈Intω(g−1(W)), a contradiction.
Example 3.23 shows that the condition 'injective' in Theorem 3.23 cannot be dropped.
Definition 3.24. A function g:(Y,σ)⟶(Z,λ) is called ωs-closed if for each ωs -closed set C of (Y,σ), g(C) is ωs-closed set in (Z,λ).
As defined in [33] a topological space (X,τ) is called ω -regular if for each closed set F in (X,τ) and x∈X−F, there exist U∈τ and V∈τω such that x∈U, F⊆V and U∩V=∅. As defined in [31], a function g:(Y,σ)⟶(Z,λ) is ω-open if for each V∈σ, g(V)∈λω.
Theorem 3.25. If g:(Y,σ)⟶(Z,λ) is ωs-closed, pre-ωs-open, and ωs -irresolute such that (Y,σ) is ω-regular, then g is ω-open.
Proof. Suppose to the contrary that there exists V∈σ such that g(V)∉λω. Then we find y∈V such that g(y)∈g(V)−Intω(g(V)). By ω-regularity of (Y,σ), we find M∈σ such that y∈M⊆¯Mω⊆V. Since ¯Mω is ωs-closed and g is ωs-closed, then Z−g(¯Mω)∈ωs(Z,λ) with g(y)∉Z−g(¯Mω). Since g(y)∉Intω(g(V)), then g(y)∉Intω(g(¯Mω)). Thus,
g(y)∈Z−Intω(g(¯Mω))=¯Z−g(¯Mω)ω. |
So by Proposition 1.2 (a), (Z−g(¯Mω))∪{g(y)}∈ωs(Z,λ). Set B=g−1((Z−g(¯Mω))∪{g(y)}). Since g is ωs -irresolute, then B∈ωs(Y,σ) and by Proposition 1.2 (b), M∩B∈ωs(Y,σ). Since g is pre-ωs-open, then g(M∩B)∈ωs(Z,λ). Since
g(y)∈g(M∩B)⊆g(M)∩g(B)⊆g(M)∩((Z−g(¯Mω))∪{g(y)})={g(y)}, |
then {g(y)}∈ωs(Z,λ). Thus, there is K∈λ such that K⊆{g(y)}⊆¯Kω. Hence, {g(y)}∈λ. Since g(y)∈g(V), then g(y)∈Int(g(V))⊆Intω(g(V)), a contradiction.
Let (Y,σ) be a topological space and let B be a subset of Y. Then B is called clopen (resp. semi-clopen, ωs-clopen) in (Y,σ) if B both open and closed (resp. semi-open and semi-closed, ωs-open and ωs-closed) in (Y,σ). Throughout this section, the family of all clopen (resp., semi-clopen, ωs -clopen) subsets of the topological space (Y,σ) will be denoted by CO(Y,σ) (resp. SCO(Y,σ), ωsCO(Y,σ)).
Definition 4.1. A function g:(Y,σ)⟶(Z,λ) is called slightly continuous [34] (resp. slightly semi-continuous [15], slightly ωs-continuous), if for every y∈Y and every W∈CO(Z,λ) with g(y)∈W, there exists V∈σ (resp. V∈SO(Y,σ), V∈ωs(Y,σ)) such that y∈V and g(V)⊆W.
As an example of a slightly continuous function that is not continuous, take the function g:(R,τu)→(R,τu) defined by g(y)=[y], where [y] is the greatest integer of y.
Theorem 4.2. For a function g:(Y,σ)⟶(Z,λ), the followings are equivalent:
(a) g is slightly ωs-continuous.
(b) For all W∈CO(Z,λ), g−1(W)∈ωs(Y,σ).
(c) For all W∈CO(Z,λ), g−1(W)∈ωsCO(Y,σ).
Proof. (a) ⟹ (b): Let W∈CO(Z,λ). Then for each y∈g−1(W), g(y)∈W and by (a), there exists Vy∈ωs(Y,σ) such that y∈Vy and g(Vy)⊆W. Thus,
g−1(W)=∪{Vy:y∈g−1(W)}. |
Therefore, g−1(W) is a union of ωs-open sets, and hence g−1(W) is ωs-open.
(b) ⟹ (c): Let W∈CO(Z,λ). Then Z−W∈CO(Z,λ). Thus, by (b), g−1(W)∈ωs(Y,σ) and g−1(Z−W)=Y−g−1(W)∈ωs(Y,σ). Therefore, g−1(W)∈ωsCO(Y,σ).
(c) ⟹ (a): Let y∈Y and W∈CO(Z,λ) with g(y)∈W. By (c), g−1(W)∈ωsCO(Y,σ)⊆ωs(Y,σ). Put V=g−1(W). Then V∈ωs(Y,σ), y∈V, and g(V)=g(g−1(W))⊆W. This shows that g is slightly ωs-continuous.
Theorem 4.3. Every slightly continuous function is slightly ωs-continuous.
Proof. Assume that g:(Y,σ)⟶(Z,λ) is slightly continuous. Let W∈CO(Z,λ). Since g is slightly continuous, then g−1(W)∈σ. So by Proposition 1.1 (a), g−1(W)∈ωs(Y,σ). Therefore, by Theorem 4.2, it follows that g is slightly ωs-continuous.
Our next example shows that the converse of Theorem 4.3 is not true, in general:
Example 4.4. Let X=Y=R, τ={∅,R,N,Qc,N∪Qc}, and σ={∅,R,N,R−N}. Define f :(R,τ)→(R,σ) by f(x)=x. Note that CO(R,σ)=σ. Since f−1(N)=N∈τ⊆ωs(X,τ) and f−1(R−N)=R−N∈ωs(X,τ), then f is slightly ωs -continuous. On the other hand, since R−N∈CO(R,σ) but f−1(R−N)=R−N∉τ, then f is not slightly continuous.
Theorem 4.5. Every slightly ωs-continuous function is slightly semi-continuous.
Proof. Assume that g:(Y,σ)⟶(Z,λ) is slightly ωs-continuous. Let W∈CO(Z,λ). Since g is slightly ωs-continuous, then g−1(W)∈ωs(Y,σ). So by Proposition 1.1 (a), g−1(W)∈SO(Y,σ). Therefore, g is slightly semi-continuous.
The converse of Theorem 4.5 is not true in general as the following example shows:
Example 4.6. Let X=Y=R, τ={∅,R,N,Qc,N∪Qc}, and σ={∅,R,Q,R−Q}. Define f :(R,τ)→(R,σ) by f(x)=x. Note that CO(R,σ)=σ. Since f−1(Q)=Q∈SO(X,τ) and f−1(R−Q)=R−Q∈τ⊆SO(X,τ), then f is slightly semi-continuous. On the other hand, since Q∈CO(R,σ) but f−1(Q)=Q∉ωs(R,τ), then f is not slightly ωs-continuous.
Theorem 4.7.(a) If g:(Y,σ)⟶(Z,λ) is slightly ωs-continuous such that (Y,σ) is locally countable, then g is continuous.
(b) If g:(Y,σ)⟶(Z,λ) is slightly semi-continuous with (Y,σ) is anti-locally countable, then g is ωs-continuous.
Proof. (a) Follows from the definitions and Proposition 1.1 (c).
(b) Follows from the definitions and Proposition 1.1 (b).
Theorem 4.8. Let g:(Y,σ)⟶(Z,λ) be a function and let γ be the product topology of (Y,σ) and (Z,λ). Let h:(Y,σ)⟶(Y×Z,γ), where h(y)=(y,g(y)) be the graph of g. Then g is slightly ωs-continuous if and only if h is slightly ωs-continuous.
Proof. Let y∈Y and let M∈CO(Y×Z,γ) such that h(y)=(y,g(y))∈M. Then M∩({y}×Z) is a clopen set in {y}×Z which contains h(y)=(y,g(y)). Since {y}×Z is homeomorphic to Z, then {z∈Z:(y,z)∈M}∈CO(Z,λ). Since g is slightly ωs-continuous, then ∪{g−1(z):(y,z)∈M}∈ωs(Y,σ). Moreover, y∈∪{g−1(z):(y,z)∈M}⊆h−1(M). Hence, h−1(M)∈ωs(Y,σ). It follows that h is slightly ωs-continuous.
Conversely, let H∈CO(Z,λ). Then Y×H∈CO(Y×Z,γ). By slight ωs-continuity of h, h−1(Y×H)∈ωsCO(Y,σ). Also, h−1(Y×H)=g−1(H). It follows that g is slightly ωs-continuous.
Theorem 4.9. If g:(Y,σ)⟶(Z,λ) is slightly ωs-continuous and h:(Z,λ)⟶(W,δ) is slightly continuous, then h∘g:(Y,σ)⟶(W,δ) is slightly ωs -continuous.
Proof. Let M∈CO(W,δ). By slight continuity of h, g−1(M)∈CO(Z,λ). By slight ωs-continuity of g, g−1(h−1(M))=(h∘g)−1(M)∈ωsCO(Y,σ). Hence, h∘g is slightly ωs-continuous.
As defined a topological space (Y,σ) is called semi-connected if SCO(Y,σ)={∅,Y}.
Definition 4.10. A topological space (Y,σ) is called ωs-connected if ωsCO(Y,σ)={∅,Y}.
Theorem 4.11. Every ωs-connected topological space is connected.
Proof. Let (Y,σ) be ωs-connected. Then ωsCO(Y,σ)={∅,Y}. Thus, by Proposition 1.1 (a), we have {∅,Y}⊆CO(Y,σ)⊆ωsCO(Y,σ)={∅,Y}. Hence, CO(Y,σ)={∅,Y}. Therefore, (Y,σ) is connected.
The following example will show that the converse of Theorem 4.11 is not true, in general:
Example 4.12. Consider the topological space (R,τu). To see that (R,τu) is not ωs-connected, let M=(−∞,0), then M∈τu⊆ωs(R,τu). Since (0,∞)∈τu and ¯(0,∞)ω=¯(0,∞)=[0,∞)=R−M, then R−M∈ωs(R,τu). Therefore, M∈ωsCO(R,τu)−{∅,R}, and hence (R,τu) is not ωs-connected. On the other hand, (R,τu) is connected.
Theorem 4.13. Every connected locally countable topological space is ωs-connected.
Proof. Follows from the definitions and Proposition 1.1 (b).
Theorem 4.14. Every semi-connected topological space is ωs-connected.
Proof. Let (Y,σ) be semi-connected. Then SCO(Y,σ)={∅,Y}. Thus, by Proposition 1.1 (a), we have {∅,Y}⊆ωsCO(Y,σ)⊆SCO(Y,σ)={∅,Y}. Hence, ωsCO(Y,σ)={∅,Y}. Therefore, (Y,σ) is ωs-connected.
Question 4.15. Is it true that ωs-connected topological spaces are semi-connected?
The following theorem answers Question 4.15 partially:
Theorem 4.16. Every anti-locally countable ωs-connected topological space is semi-connected.
Proof. Follows from the definitions and Proposition 1.1 (c).
Theorem 4.17. A slightly ωs-continuous image of an ωs-connected space is connected.
Proof. Let g:(Y,σ)⟶(Z,λ) be surjective and slightly ωs-continuous, where (Y,σ) is ωs-connected. Suppose that (Z,λ) is not connected. Then there exists M∈CO(Z,λ)−{∅,Z}. By Theorem 4.2, g−1(M)∈ωsCO(Y,σ). Since ∅≠M≠Z and g is surjective, then ∅≠g−1(M)≠Z. Therefore, g−1(M)∈ωsCO(Y,σ)−{∅,Y} which contradicts the assumption that (Y,σ) is ωs-connected.
Definition 5.1. A topological space (Y,σ) is called ωs-compact (resp. semi-compact [36]) if for any cover A of Y with A⊆ωs(Y,σ) (resp. A⊆SO(Y,σ)), there is a finite subfamily B⊆A such that B is also a cover of Y.
Theorem 5.2. Every ωs-compact topological space is compact.
Proof. Let (Y,σ) be ωs -compact and let A be a cover of Y with A⊆σ. Then by Proposition 1.1 (a), A⊆ωs(Y,σ). Since (Y,σ) is ωs-compact, then there exists a finite subfamily B⊆A such that B is also a cover of Y. This shows that (Y,σ) is compact.
The following example will show that the converse of Theorem 5.2 is not true, in general:
Example 5.3. Consider ([0,∞),σ), where σ={∅,[0,∞)}∪{(a,∞):a≥0}. To see that ([0,∞),σ) is compact, let A be a cover of [0,∞) with A⊆σ. Then [0,∞)∈A. Choose B={[0,∞)}. Then B is a finite subfamily of A such that B is also a cover of [0,∞). This shows that ([0,∞),σ) is compact. Let A={(1,∞)∪{x}:x∈[0,1]}. Then A is a cover of [0,∞). Since (1,∞)∈σ⊆ωs([0,∞),σ) and ¯(1,∞)ω=¯(1,∞)=[0,∞), then by Proposition 1.2 (a), we have A⊆ωs([0,∞),σ). On the other hand, if B is a finite subfamily of A, then B is not a cover of [0,∞). This shows that ([0,∞),σ) is not ωs-compact.
Theorem 5.4. Let (Y,σ) be a locally countable topological space. Then (Y,σ) is ωs -compact if and only if (Y,σ) is compact.
Proof. Necessity. Follows from Theorem 5.2.
Sufficiency. Suppose that (Y,σ) is compact and let A be a cover of Y such that A⊆ωs(Y,σ). Since (Y,σ) is locally countable, then by Proposition 1.1 (b), A⊆σ. Since (Y,σ) is compact, then there exists a finite subfamily B⊆A such that B is also a cover of Y. This shows that (Y,σ) is ωs-compact.
Theorem 5.5. Every semi-compact topological space is ωs -compact.
Proof. Let (Y,σ) be semi-compact and let A be a cover of Y such that A⊆ωs(Y,σ). Then by Proposition 1.1 (a), A⊆SO(Y,σ). Since (Y,σ) is semi-compact, then there exists a finite subfamily B⊆A such that B is also a cover of Y. This shows that (Y,σ) is ωs-compact.
The following example will show that the converse of Theorem 5.5 is not true, in general:
Example 5.6. Consider (N,σ), where σ={∅,N,{1},{2},{1,2}}. Since {1}, {2}, and {1,2} are countable sets, then ¯{1}ω={1}, ¯{2}ω, and ¯{1,2}ω={1,2}. Thus, σ=ωs(Y,σ). This shows that (N,σ) is ωs-compact. Let A={{2}}∪{{1,x}:x∈N−{1,2}}. Then A is a cover of N. Since ¯{1}=N−{2}, then A⊆SO(Y,σ). If B is a finite subfamily of A, then ⋃B is a finite subset of N. This shows that (N,σ) is not semi-compact.
Theorem 5.7. Let (Y,σ) be an anti-locally countable topological space. Then (Y,σ) is semi-compact if and only if (Y,σ) is ωs -compact.
Proof. Necessity. Follows from Theorem 2.5.
Sufficiency. Suppose that (Y,σ) is ωs-compact and let A be a cover of Y such that A⊆SO(Y,σ). Since (Y,σ) is anti-locally countable, then by Proposition 1.1 (c), A⊆ωs(Y,σ). Since (Y,σ) is ωs-compact, then there is a finite subfamily B⊆A such that B is also a cover of Y. This shows that (Y,σ) is semi-compact.
Theorem 5.8. A topological space (Y,σ) is ωs-compact if and only if every family of ωs -closed sets which has the finite intersection property must have non-empty intersection.
Proof. Necessity. Suppose that (Y,σ) is ωs-compact, and suppose to the contrary that there exists a family H of ωs-closed such that H has the finite intersection property and ⋂H=∅. Let A={Y−H:H∈H}. Then A is a cover of Y and A⊆ωs(Y,σ). Since (Y,σ) is ωs-compact, then there is a finite subfamily A1⊆A such that A1 is also a cover of Y. Let H1={Y−A:A∈A1}. Then H1 is a finite subcollection of H such that
⋂H1=⋂A∈A1(Y−A)=Y−⋃A∈A1A=Y−Y=∅. |
This contradicts the assumption that H has the finite intersection property.
Sufficiency. Suppose that every family of ωs-closed sets which has the finite intersection property must have non-empty intersection, and suppose to the contrary that (Y,σ) is not ωs-compact. Then there is a cover A of Y such that A⊆ωs(Y,σ) and any finite subcollection of A is not a cover of Y. Let H={Y−A:A∈A}. Then H is a family of ωs-closed sets and H has the finite intersection property. So, by assumption ⋂H≠∅, and thus Y−⋂H≠Y. But
Y−⋂H=⋃A∈AA≠Y, |
a contradiction.
Definition 5.9. Let (Y,σ) be a topological space and let (xd)d∈D be a net in (Y,σ). A point y∈Y is called an ωs -cluster point of (yd)d∈D in (Y,σ) if for every V∈ωs(Y,σ) with y∈V and every d∈D, there is d0∈D such that d≤d0 and yd0∈V.
Theorem 5.10. A topological space (Y,σ) is ωs-compact if and only if every net in (Y,σ) has an ωs-cluster point.
Proof. Necessity. Suppose that (Y,σ) is ωs-compact and let (yd)d∈D be a net in (Y,σ). For each d∈D, let Td={yd′:d′∈D and d≤d′}. Let A={¯Tdωs:d∈D}. Then A is a family of ωs-closed sets.
Claim 1. A has the finite intersection property.
Proof of Claim 1. Let d1,d2,...,dn∈D. Choose d0∈D such that di≤d0 for all i=1,2,...,n. Then yd0∈n⋂i=1Tdi⊆n⋂i=1¯Tdωsi. This ends the proof that A has the finite intersection property.
Since (Y,σ) is ωs-compact, then by Claim 1 and Theorem 5.8, there exists y∈⋂d∈D¯Tdωs.
Claim 2. y is an ωs-cluster point of (yd)d∈D in (Y,σ).
Proof of Claim 2. Let V∈ωs(Y,σ) such that y∈V, and let d∈D. Since y∈V∩¯Tdωs, then V∩Td≠∅, and so there exists d′∈D such that d≤d′ and xd′∈V. This shows that y is an ωs-cluster point of (yd)d∈D in (Y,σ).
Sufficiency. Suppose that every net in (Y,σ) has an ωs-cluster point. We will apply Theorem 5.8. Let A be a family of ωs-closed sets which has the finite intersection property. Let D be the family of all finite intersections of members of A. Define the relation ≤ on D as follows:
For every d1,d2∈D, d1≤d2 if and only if d2⊆d1. |
Then (D,≤) is a directed set. For every d∈D, choose yd∈d. By assumption, there is an ωs-cluster point y of (yd)d∈D.
Claim 3. y∈¯Aωs for all A∈A, and hence y∈⋂A∈A¯Aωs=⋂A∈AA.
Proof of Claim 3. Let A∈A and V∈ωs(Y,σ) with y∈V. Let d=A, then d∈D. Since y is an ωs-cluster point of (yd)d∈D, then there is d′∈D such that d≤d′ and yd′∈V, say d′=F. Then F⊆A, and hence yd′∈V∩A. Therefore, y∈¯Aωs.
Theorem 5.11. Let g:(Y,σ)⟶(Z,γ) be ωs-irresolute and surjective. If (Y,σ) is ωs-compact, then (Z,γ) is ωs-compact.
Proof. Suppose that (Y,σ) is ωs-compact and let H be a cover of Z such that H⊆ωs(Z,γ). Let M={g−1(H):H∈H}. Then M is a cover of Y. Also, by ωs-irresoluteness of g, we have M⊆ωs(Y,σ). Since (Y,σ) is ωs-compact, then there exist H1,H2,...,Hn∈H such that n⋃i=1g−1(Hi)=Y, and so n⋃i=1Hi⊆g(g−1(n⋃i=1Hi))=g(Y). Since g is surjective, then g(Y)=Z. Therefore, Z=n⋃i=1Hi. Hence, (Z,γ) is ωs-compact.
In this paper, we introduce ωs-irresoluteness, ωs -openness, pre-ωs-openness, and slight ωs-continuity as new classes of functions. And, we define ωs-compactness as a new class of topological spaces which lies between the classes compactness and semi-compactness. Several implications, examples, counter-examples, characterizations, and mapping theorems are introduced. The following topics could be considered in future studies: (1) To define ωs-open separation axioms; (2) To define ωs-connectedness; (3) To improve some known topological results.
We declare no conflicts of interest in this paper.
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