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A neural networks technique for analysis of MHD nano-fluid flow over a rotating disk with heat generation/absorption

  • Received: 03 August 2024 Revised: 12 October 2024 Accepted: 04 November 2024 Published: 14 November 2024
  • MSC : 34G20, 35A20, 35A22, 35R11

  • 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 1009to 1012.

    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 1009to 1012.



    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)={yY: for every Vσ with yVVB 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 YB is ω-closed. It is well known that B is ω-open in (Y,σ) if and only if for every bB, there are Vσ and a countable subset FY with bVFB. 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 Vσ such that VB¯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 Oσ such that OB¯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) σωs(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,CY. Then we have the following:

    (a) If BC¯Bω and Bωs(Y,σ), then Cωs(Y,σ).

    (b) If Bσ and Cωs(Y,σ), then BCωs(Y,σ).

    Proposition 1.3. [31] Let (Y,σ) be anti-locally countable. Then for every Bσω, 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 VSO(Z,γ) (resp. Vγ), g1(V)SO(Y,σ) (resp. g1(V)ωs(Y,σ)).

    Definition 2.2. A function g:(Y,σ)(Z,γ) is called ωs-irresolute, if for every Vωs(Z,γ), g1(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,NQc}, and σ={,Y,{a},{b}}. Define f:(X,τ)(Y,σ) by

     f(x)={aif xQcbif xQ.

    Since f1({a})=QcτSO(X,τ) and f1({b})=QSO(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 tN{1,2}.

    Since (N,σ) is locally countable, then by Proposition 1.1 (b) we have ωs(N,σ)=σ. Since f1({1})={1,2}σ and f1({2})=σ, then f is ωs-irresolute. However, f is not irresolute because there is ¯{2}=N{1}SO(N,σ) such that f1(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 xR({2}(3,)).

    Since f1((3,))=τ and f1({2})={2}(3,)τ, then f is continuous. Since ¯(3,)ω=R{2}, then R{2}ωs(X,τ). Since f1(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 f1({a})=(,0)τuωs(R,τu) and f1({b})=[0,)ωs(R,τu), then f is ωs-irresolute. However, f is not continuous because there is {b}τdisc such that f1({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 Hωs(Z,γ). Then there exists Wγ such that WH¯Wω and so g1(W)g1(H)g1(¯Wω). Since g:(Y,σ)(Z,γ) is continuous, then g1(W)σ. Since g:(Y,σω)(Z,γω) is open, then g1(¯Wω)¯g1(W)ω. Thus, g1(W)g1(H)¯g1(W)ω, and hence g1(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 Vσ. Then by Proposition 1.1 (a), Vωs(Y,σ). Since f is ωs-irresolute, then f1(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), g1(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 ZBωs(Z,γ). Since g is ωs-irresolute, then g1(ZB)=Yg1(B)ωs(Y,σ). Hence, g1(B) is ωs-closed in (Y,σ).

    Sufficiency. Suppose that for every ωs-closed subset B of (Z,γ), g1(B) is ωs -closed in (Y,σ). Let Wωs(Z,γ). Then ZW is ωs-closed in (Z,γ). By assumption, g1(ZW)=Yg1(W) is ωs-closed in (Y,σ), and so g1(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 HY, g(¯Hωs)¯g(H)ωs.

    Proof. Necessity. Suppose that g is ωs-irresolute and let HY. Then ¯g(H)ωs is ωs-closed in (Z,γ), and by Theorem 2.11, g1(¯g(H)ωs) is ωs-closed in (Y,σ). Since Hg1(¯g(H)ωs), then ¯Hωsg1(¯g(A)ωs). Thus,

    g(¯Hωs)g(g1(¯g(H)ωs))¯g(H)ωs.

    Sufficiency. Suppose that for every subset HY, 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(¯g1(B)ωs)¯g(g1(B))ωs¯Bωs=B, and so

    ¯g1(B)ωsg1(g(¯g1(B)ωs))g1(B).

    Therefore, ¯g1(B)ωs=g1(B), and hence g1(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 HZ, ¯g1(H)ωsg1(¯Hωs).

    Proof. Necessity. Suppose that g is ωs-irresolute and let HZ. Then by Theorem 2.11, g1(¯Hωs) is ωs-closed in (Y,σ). Since g1(H)g1(¯Hωs), then ¯g1(H)ωsg1(¯Hωs).

    Sufficiency. Suppose that for every HZ, ¯g1(H)ωsg1(¯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, ¯g1(B)ωsg1(B), and so ¯g1(B)ωs=g1(B). Therefore, g1(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 Cωs(M,γ). Since h is ωs -irresolute, then h1(C)ωs(Z,λ). Since g is ωs-irresolute, then (hg)1(C)=g1(h1(C))ωs(Y,σ). Therefore, hg 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 Uσ, f(U)ωs(Z,λ) (resp. f(U)SO(Z,λ)).

    (b) pre-semi-open [14] if for each USO(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 BB, then g is ωs-open.

    Proof. Suppose that for a base B of (Y,σ), g(B)ωs(Z,λ) for all BB. Let Vσ{}. Choose B1B such that V={B:BB1}. Then

    g(V)=g({B:BB1})={g(B):BB1}.

    Since by assumption, g(B)ωs(Z,λ) for all BB1, 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 Vσ. Since g is open, then g(V)γωs(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,dR and c<d} for (R,τu). Then for all c,dR with c<d we have

    g((c,d))={(d2,c2)if c<d0(c2,d2)if 0c<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 Rτu 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 Vσ. 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)=y1. Since g(N)={0}N¯N=R, then, g(N)SO(R,σ). Also, g()=SO(R,σ) and g(R)=RSO(R,σ). Therefore, g is semi-open. Conversely, g is not ωs-open since there is Nσ such that g(N)={0}Nωs(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 Aωs(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)=Yσ, 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)=x1. 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 Hωs(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:NH} and ωs(R,σ)=σ. To see that g is pre-semi-open, let HSO(R,σ){}. Then NH and so NN{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 Nωs(R,σ) such that g(N)={0}Nωs(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 Uτωs(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) g1(¯Bωs)¯g1(B) for every BZ.

    (c) Int(g1(B))g1(Intωs(B)) for every BZ.

    Proof. (a) (b): Suppose that g is ωs-open and let BZ. Let yg1(¯Bωs). To show that y¯g1(B), let Vσ such that yV. 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 tV such that g(t)B. Then tVg1(B), and hence Vg1(B). It follows that y¯g1(B).

    (b) (a): Suppose that g1(¯Bωs)¯g1(B) for every BZ, and suppose to the contrary that g is not ωs-open. Then there exists Vσ such that g(V)ωs(Z,λ) and so, Zg(V) is not ωs-closed. Thus, there exists zg(V)¯Zg(V)ωs. Choose yV such that z=g(y). Then yg1(¯Zg(V)ωs). By assumption, we have

    g1(¯Zg(V)ωs)¯g1(Zg(V))=¯Yg1(g(V))¯YV=YV.

    Therefore, yYV. But yV, a contradiction.

    (b) (c): Suppose that g1(¯Bωs)¯g1(B) for every BZ. Let BZ. Then by (b),

    g1(Intωs(B))=g1(Z¯ZBωs)=Yg1(¯ZBωs)Y¯g1(ZB)=Y¯Yg1(B)=Int(g1(B)).

    (c) (b): Suppose that Int(g1(B))g1(Intωs(B)) for every BZ. Let BZ. Then by (c),

    g1(¯Bωs)=g1(ZIntωs(ZB))=Yg1(Intωs(ZB))YInt(g1(ZB))=YInt(Yg1(B))=¯g1(B).

    Theorem 3.18. For a function g:(Y,σ)(Z,λ), the followings are equivalent:

    (a) g is pre-ωs-open.

    (b) g1(¯Bωs)¯g1(B)ωs for every BZ.

    (c) Intωs(g1(B))g1(Intωs(B)) for every BZ.

    Proof. (a) (b): Suppose that g is pre-ωs-open and let BZ. Let yg1(¯Bωs). To show that y¯g1(B)ωs, let Cωs(Y,σ) such that yC. 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 tC such that g(t)B. Then tCg1(B) and hence Cg1(B). It follows that y¯g1(B)ωs.

    (b) (a): Suppose that g1(¯Bωs)¯g1(B)ωs for every BZ, and suppose to the contrary that g is not pre-ωs-open. Then there is Cωs(Y,σ) such that g(C)ωs(Z,λ) and so Zg(C) is not ωs-closed. And so there exists zg(C)¯Zg(C)ωs. Choose yC such that z=g(y). Then yg1(¯Zg(C)ωs). By assumption we have

    g1(¯Zg(C)ωs)¯g1(Zg(C))ωs=¯Yg1(g(C))ωs¯YCωs=YC.

    Therefore, yYC but yC, a contradiction.

    (b) (c): Suppose that g1(¯Bωs)¯g1(B)ωs for every BZ. Let BZ. Then by (b),

    g1(Intωs(B))=g1(Z¯ZBωs)=Yg1(¯ZBωs)Y¯g1(ZB)ωs=Y¯Yg1(B)ωs=Intωs(g1(B)).

    (c) (b): Suppose that Intωs(g1(B))g1(Intωs(B)) for every BZ. Let BZ. Then by (c),

    g1(¯Bωs)=g1(ZIntωs(ZB))=Yg1(Intωs(ZB))YIntωs(g1(ZB))=YIntωs(Yg1(B))=¯g1(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 Gωs(Z,λ), choose Wλ such that WG¯Wω. Thus, we have g1(W)g1(G)g1(¯Wω). Since g:(Y,σ)(Z,λ) is ωs -continuous, then g1(W)ωs(Y,σ). Since g:(Y,σω)(Z,λω) is open, then g1(¯Wω)¯g1(W)ω. Since we have g1(W)g1(G)¯g1(W)ω with g1(W)ωs(Y,σ), then by Proposition 1.2 (a), g1(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 Hωs(Y,σ), then we find Vσ such that VH¯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 Wλ, g1(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 Wλ such that g1(W)σω. So, there exists yg1(W)Intω(g1(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)MW. 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, g1(M) is ωs-closed, and so Yg1(M)ωs(Y,σ). Since g1(M)g1(W), then Intω(g1(M))Intω(g1(W)). Since yIntω(g1(W)), then yIntω(g1(M)), and so

    yYIntω(g1(M))=¯Yg1(M)ω.

    Thus, by Proposition 1.2 (a), we have (Yg1(M)){y}ωs(Y,σ) with yYg1(M). Put S=g((Yg1(M)){y})=g(Yg1(M))){g(y)}. Since g is pre-ωs-open, then Sωs(Z,λ). So by Proposition 1.2 (b), we have SMωs(Z,λ). Since g(Yg1(M)))ZM, then we have

    g(y)SM((ZM){g(y)})M={g(y)}

    and thus SM={g(y)}. Therefore, {g(y)}ωs(Z,λ). Thus, there exists Oλ 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 f1({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 Wλ such that g1(W)σω. So, there exists yg1(W)Intω(g1(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)MW. 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, g1(M) is ωs-closed, and so Yg1(M)ωs(Y,σ). Since g1(M)g1(W), then Intω(g1(M))Intω(g1(W)). Since yIntω(g1(W)), then yIntω(g1(M)), and so

    yYIntω(g1(M))=¯Yg1(M)ω.

    Thus, by Proposition 1.2 (a), we have (Yg1(M)){y}ωs(Y,σ) with yYg1(M). Put S=g((Yg1(M)){y})=g(Yg1(M))){g(y)}. Since g is pre- ωs-open, then Sωs(Z,λ), and by Proposition 1.2 (b) we have SMωs(Z,λ). Since g(Yg1(M)))ZM, then we have

    g(y)SM((ZM){g(y)})M={g(y)}

    and thus SM={g(y)}. Therefore, {g(y)}ωs(Z,λ). Since g is ωs-irresolute and injective, then g1({g(y)})={y}ωs(Y,σ). So, there exists Vσ such that V{y}¯Vω, and hence {y}σ. Since yg1(W) and {y}σσω, then yIntω(g1(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 xXF, there exist Uτ and Vτω such that xU, FV and UV=. As defined in [31], a function g:(Y,σ)(Z,λ) is ω-open if for each Vσ, 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 Vσ such that g(V)λω. Then we find yV such that g(y)g(V)Intω(g(V)). By ω-regularity of (Y,σ), we find Mσ such that yM¯MωV. Since ¯Mω is ωs-closed and g is ωs-closed, then Zg(¯Mω)ωs(Z,λ) with g(y)Zg(¯Mω). Since g(y)Intω(g(V)), then g(y)Intω(g(¯Mω)). Thus,

    g(y)ZIntω(g(¯Mω))=¯Zg(¯Mω)ω.

    So by Proposition 1.2 (a), (Zg(¯Mω)){g(y)}ωs(Z,λ). Set B=g1((Zg(¯Mω)){g(y)}). Since g is ωs -irresolute, then Bωs(Y,σ) and by Proposition 1.2 (b), MBωs(Y,σ). Since g is pre-ωs-open, then g(MB)ωs(Z,λ). Since

    g(y)g(MB)g(M)g(B)g(M)((Zg(¯Mω)){g(y)})={g(y)}

    then {g(y)}ωs(Z,λ). Thus, there is Kλ 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 yY and every WCO(Z,λ) with g(y)W, there exists Vσ (resp. VSO(Y,σ), Vωs(Y,σ)) such that yV 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 WCO(Z,λ), g1(W)ωs(Y,σ).

    (c) For all WCO(Z,λ), g1(W)ωsCO(Y,σ).

    Proof. (a) (b): Let WCO(Z,λ). Then for each yg1(W), g(y)W and by (a), there exists Vyωs(Y,σ) such that yVy and g(Vy)W. Thus,

    g1(W)={Vy:yg1(W)}.

    Therefore, g1(W) is a union of ωs-open sets, and hence g1(W) is ωs-open.

    (b) (c): Let WCO(Z,λ). Then ZWCO(Z,λ). Thus, by (b), g1(W)ωs(Y,σ) and g1(ZW)=Yg1(W)ωs(Y,σ). Therefore, g1(W)ωsCO(Y,σ).

    (c) (a): Let yY and WCO(Z,λ) with g(y)W. By (c), g1(W)ωsCO(Y,σ)ωs(Y,σ). Put V=g1(W). Then Vωs(Y,σ), yV, and g(V)=g(g1(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 WCO(Z,λ). Since g is slightly continuous, then g1(W)σ. So by Proposition 1.1 (a), g1(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,NQc}, and σ={,R,N,RN}. Define f :(R,τ)(R,σ) by f(x)=x. Note that CO(R,σ)=σ. Since f1(N)=Nτωs(X,τ) and f1(RN)=RNωs(X,τ), then f is slightly ωs -continuous. On the other hand, since RNCO(R,σ) but f1(RN)=RNτ, 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 WCO(Z,λ). Since g is slightly ωs-continuous, then g1(W)ωs(Y,σ). So by Proposition 1.1 (a), g1(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,NQc}, and σ={,R,Q,RQ}. Define f :(R,τ)(R,σ) by f(x)=x. Note that CO(R,σ)=σ. Since f1(Q)=QSO(X,τ) and f1(RQ)=RQτSO(X,τ), then f is slightly semi-continuous. On the other hand, since QCO(R,σ) but f1(Q)=Qωs(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 yY and let MCO(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 {zZ:(y,z)M}CO(Z,λ). Since g is slightly ωs-continuous, then {g1(z):(y,z)M}ωs(Y,σ). Moreover, y{g1(z):(y,z)M}h1(M). Hence, h1(M)ωs(Y,σ). It follows that h is slightly ωs-continuous.

    Conversely, let HCO(Z,λ). Then Y×HCO(Y×Z,γ). By slight ωs-continuity of h, h1(Y×H)ωsCO(Y,σ). Also, h1(Y×H)=g1(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 hg:(Y,σ)(W,δ) is slightly ωs -continuous.

    Proof. Let MCO(W,δ). By slight continuity of h, g1(M)CO(Z,λ). By slight ωs-continuity of g, g1(h1(M))=(hg)1(M)ωsCO(Y,σ). Hence, hg 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 Mτuωs(R,τu). Since (0,)τu and ¯(0,)ω=¯(0,)=[0,)=RM, then RMωs(R,τu). Therefore, MωsCO(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 MCO(Z,λ){,Z}. By Theorem 4.2, g1(M)ωsCO(Y,σ). Since MZ and g is surjective, then g1(M)Z. Therefore, g1(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 Aωs(Y,σ) (resp. ASO(Y,σ)), there is a finite subfamily BA 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 Aσ. Then by Proposition 1.1 (a), Aωs(Y,σ). Since (Y,σ) is ωs-compact, then there exists a finite subfamily BA 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,):a0}. To see that ([0,),σ) is compact, let A be a cover of [0,) with Aσ. 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,)σωs([0,),σ) and ¯(1,)ω=¯(1,)=[0,), then by Proposition 1.2 (a), we have Aωs([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 Aωs(Y,σ). Since (Y,σ) is locally countable, then by Proposition 1.1 (b), Aσ. Since (Y,σ) is compact, then there exists a finite subfamily BA 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 Aωs(Y,σ). Then by Proposition 1.1 (a), ASO(Y,σ). Since (Y,σ) is semi-compact, then there exists a finite subfamily BA 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}:xN{1,2}}. Then A is a cover of N. Since ¯{1}=N{2}, then ASO(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 ASO(Y,σ). Since (Y,σ) is anti-locally countable, then by Proposition 1.1 (c), Aωs(Y,σ). Since (Y,σ) is ωs-compact, then there is a finite subfamily BA 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={YH:HH}. Then A is a cover of Y and Aωs(Y,σ). Since (Y,σ) is ωs-compact, then there is a finite subfamily A1A such that A1 is also a cover of Y. Let H1={YA:AA1}. Then H1 is a finite subcollection of H such that

    H1=AA1(YA)=YAA1A=YY=.

    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 Aωs(Y,σ) and any finite subcollection of A is not a cover of Y. Let H={YA:AA}. Then H is a family of ωs-closed sets and H has the finite intersection property. So, by assumption H, and thus YHY. But

    YH=AAAY

    a contradiction.

    Definition 5.9. Let (Y,σ) be a topological space and let (xd)dD be a net in (Y,σ). A point yY is called an ωs -cluster point of (yd)dD in (Y,σ) if for every Vωs(Y,σ) with yV and every dD, there is d0D such that dd0 and yd0V.

    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)dD be a net in (Y,σ). For each dD, let Td={yd:dD and dd}. Let A={¯Tdωs:dD}. Then A is a family of ωs-closed sets.

    Claim 1. A has the finite intersection property.

    Proof of Claim 1. Let d1,d2,...,dnD. Choose d0D such that did0 for all i=1,2,...,n. Then yd0ni=1Tdini=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 ydD¯Tdωs.

    Claim 2. y is an ωs-cluster point of (yd)dD in (Y,σ).

    Proof of Claim 2. Let Vωs(Y,σ) such that yV, and let dD. Since yV¯Tdωs, then VTd, and so there exists dD such that dd and xdV. This shows that y is an ωs-cluster point of (yd)dD 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,d2Dd1d2 if and only if d2d1.

    Then (D,) is a directed set. For every dD, choose ydd. By assumption, there is an ωs-cluster point y of (yd)dD.

    Claim 3. y¯Aωs for all AA, and hence yAA¯Aωs=AAA.

    Proof of Claim 3. Let AA and Vωs(Y,σ) with yV. Let d=A, then dD. Since y is an ωs-cluster point of (yd)dD, then there is dD such that dd and ydV, say d=F. Then FA, and hence ydVA. 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 Hωs(Z,γ). Let M={g1(H):HH}. Then M is a cover of Y. Also, by ωs-irresoluteness of g, we have Mωs(Y,σ). Since (Y,σ) is ωs-compact, then there exist H1,H2,...,HnH such that ni=1g1(Hi)=Y, and so ni=1Hig(g1(ni=1Hi))=g(Y). Since g is surjective, then g(Y)=Z. Therefore, Z=ni=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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