@ Appl. Gen. Topol. 22, no. 2 (2021), 355-366doi:10.4995/agt.2021.14868 © AGT, UPV, 2021 Further aspects of IK-convergence in topological spaces Ankur Sharmah and Debajit Hazarika Department of Mathematical Sciences, Tezpur University, Napam 784028, Assam, India (ankurs@tezu.ernet.in, debajit@tezu.ernet.in) Communicated by D. Georgiou Abstract In this paper, we obtain some results on the relationships between different ideal convergence modes namely, I K , I K ∗ , I, K, I ∪ K and (I ∪K)∗. We introduce a topological space namely IK-sequential space and show that the class of I K -sequential spaces contain the sequential spaces. Further I K -notions of cluster points and limit points of a func- tion are also introduced here. For a given sequence in a topological space X, we characterize the set of I K -cluster points of the sequence as closed subsets of X. 2010 MSC: 54A20; 40A05; 40A35. Keywords: I-convergence; IK-convergence; IK ∗ -convergence; I K -sequential space; I K -cluster point. 1. Introduction For basic general topological terminologies and results we refer to [5]. The ideal convergence of a sequence of real numbers was introduced by Kostyrko et al. [11], as a natural generalization of existing convergence notions such as usual convergence [5], statistical convergence [4]. It was further introduced in arbitrary topological spaces accordingly for sequences [3] and nets [2] by Das et al. The main goal of this article is to study IK-convergence which arose as a generalization of a type of ideal convergence. In this continuation we begin with a prior mentioning of ideals and ideal convergence in topological spaces. Received 01 January 2020 – Accepted 26 March 2021 http://dx.doi.org/10.4995/agt.2021.14868 A. Sharmah and D. Hazarika An ideal I on a arbitrary set S is a family I ⊂ 2S (the power set of S) that is closed under finite unions and taking subsets. Fin and I0 are two basic ideals on ω, the set of all natural numbers, defined as Fin:= collection of all finite subsets of ω and I0:= subsets of ω with density 0, we say A(⊂ ω) ∈ I0 if and only if lim supn→∞ |A∩{1,2,...,n}| n = 0. For an ideal I in P(ω), we have two additional subsets of P(ω) namely I⋆ and I+, where I⋆ := {A ⊂ ω : Ac ∈ I}, the filter dual of I and I+:= collection of all subsets not in I. Clearly, I⋆ ⊆I+. A sequence x = {xn}n∈ω is said to be I-convergent [3] to ξ, denoted by xn →I ξ, if {n : xn /∈ U} ∈ I, for all neighborhood U of ξ. A sequence x = {xn}n∈ω of elements of X is said to be I ⋆-convergent to ξ if there exists a set M := {m1 < m2 < ... < mk < ...} ∈ I ⋆ such that limk→∞ xmk = ξ. Lahiri and Das [3] found an equivalence between I and I⋆-convergences under certain assumptions. In 2011, Macaj and Sleziak [6] introduced the IK-convergence of function in a topological space, which was derived from I∗-convergence [3] by simply replacing Fin by an arbitrary ideal K. Interestingly, IK-convergence arose as an independent mode of convergence. Comparisons of IK-convergence with I-convergence [11] can be found in [1, 6, 8]. A few articles for example [9, 7] contributed to the study of IK-convergence of sequence of functions. Some of the definitions and results of [3, 6] that are used in subsequent sections are listed below. Here X is a topological space and S is a set where ideals are defined. We say that a function f : S → X is IM-convergent to a point x ∈ X if ∃M ∈I∗ such that the function g : S → X given by g(s) = { f(s), s ∈ M x, s /∈ M is M-convergent to x, where M is a convergence mode via ideal. If M = K∗, then f : S → X is said to be IK ∗ -convergent [6] to a point x ∈ X. Also, if M = K, then f : S → X is said to be IK-convergent [6] to a point x ∈ X. In particular, if X is a discrete space, our immediate observation is that only the I-constant functions are I-convergent, for a given ideal I, f : S → X is an I-constant function if it attains a constant value except for a set in I. It follows that I and I∗ convergence coincide for X. Thus, IK and IK ∗ -convergence modes also coincide on discrete spaces. Lemma 1.1 ([6, Lemma 2.1]). If I and K are two ideals on a set S and f : S → X is a function such that K− lim f = x, then IK − lim f = x. Remark 1.2. We say two ideals I and K satisfy ideality condition if I ∪K is an proper ideal [10]. Again, I and K satisfy ideality condition if and only if S 6= I ∪ K, for all I ∈I, K ∈K. The main results of this article are divided into 3 sections. Section 2 is devoted to a comparative study of different convergence modes for example IK, IK ∗ , I, K, I ∪K, (I ∪K)∗ etc. We justify the existence of an ideal J , © AGT, UPV, 2021 Appl. Gen. Topol. 22, no. 2 356 Further aspects of IK-convergence in topological spaces such that the behavior of IK and J -convergence coincides in Hausdorff spaces. Then in section 3, we introduce IK-sequential space and study its properties. In Section 4 we basically define IK-cluster point and IK-limit point of a function in a topological space. Here we observe that the ideality condition of I and K in IK-convergence allows to get some effective conclusions. Moreover, we characterize the set of IK-cluster points of a function as closed sets. Throughout this paper we focus on the proper ideals [10] containing Fin (S /∈I). 2. IK-convergence and several comparisons In this section, we study some more relations among different convergence modes IK, IK ∗ , I∪K, (I∪K)∗ etc. We mainly focus on IK-convergence where I∪K forms an ideal. Proposition 2.1. Let X be a topological space and f : S → X be a function. Let I,K be two ideals on S such that I∪K is an ideal. Then (i) IK ∗ − lim f = x if and only if (I∪K)∗ − lim f = x. (ii) IK − lim f = x implies I∪K− lim f = x. Proof. (i) Let f : S → X be IK ∗ -convergent to x. So, there exists a set M ∈I∗ for which the function g : S → X such that g(s) = { f(s), s ∈ M x, s /∈ M is K∗-convergent to x. So, there further exists a set N ∈K∗ for which we can consider the function h : S → X such that h(s) = { f(s), s ∈ M, s ∈ N x, s /∈ M or s /∈ N is Fin-convergent to x. Now, Let K = N∁ ∈ K, I = M∁ ∈ I (say). Then h(s) = { f(s), s ∈ (I ∪K)∁ x, s /∈ (I ∪K)∁. In essence, we can conclude f is (I∪K)∗-convergent to x. Conversely, the function f : S → X is (I ∪K)∗-convergent to x. So, there exists a set P = (I ∪ K)∁ ∈ (I ∪K)∗ for which the function h : S → X such that h(s) = { f(s), s ∈ P x, s /∈ P h(s) = { f(s), s ∈ (I ∪K)∁ x, s /∈ (I ∪K)∁ © AGT, UPV, 2021 Appl. Gen. Topol. 22, no. 2 357 A. Sharmah and D. Hazarika is Fin-convergent to x. Lets consider the function g : S → X defined as g(s) = { f(s), s ∈ I∁ x, s /∈ I∁ for which the function h : S → X such that h(s) = { f(s), s ∈ I∁, s ∈ K∁ x, s /∈ (I ∪K)∁ is Fin-convergent to x. Consequently, f is IK ∗ -convergent to x. (ii) Let f : S → X be IK-convergent to x. So, there exists a set M ∈ I∗ for which the function g : S → X such that g(s) = { f(s), s ∈ M x, s /∈ M is K-convergent to x. Then for each Ux, neighborhood of x, we have {s : g(s) /∈ Ux} ∈ K. Accordingly, the set given by {s : f(s) /∈ Ux, s ∈ M} ∈ K. Further {s : f(s) /∈ Ux} ⊆ {s : f(s) /∈ Ux, s ∈ M}∪{s : s /∈ M}. Hence, {s : f(s) /∈Ux}∈I∪K. � Following are immediate corollaries of the above proposition provided I∪K is an ideal. Corollary 2.2. IK ∗ -convergence implies I− convergence. Corollary 2.3. IK ∗ -convergence implies K− convergence. Following results in [1] are corollaries of the above proposition. Corollary 2.4. IK-convergence implies I− convergence provided K⊆I. Corollary 2.5. IK-convergence implies K− convergence provided I ⊆K. Following diagram shows the connections between different convergence modes. I∪K←IK ←I∗ → (I∪K)∗ ≡IK ∗ →IK J In this segment we are interested to find whether there exists an ideal J such that the behavior of IK and J -convergence coincides. Recalling that a filter-base is a non empty collection closed under finite intersection, we have the following result for a given function f in X by taking an ideal-base to be complement of a filter-base. Lemma 2.6. Let I and K be two ideals on S satisfying ideality condition. f : S → X be a function on a topological space X. If J = ideal generated by (K∪J), for any J ∈I. Then f is J-convergent to x =⇒ f is IK-convergent to x. © AGT, UPV, 2021 Appl. Gen. Topol. 22, no. 2 358 Further aspects of IK-convergence in topological spaces Proof. Let f be J -convergent to x, where J= ideal generated by the ideal base (K∪ J), for any J ∈ I. Now for J = Mc, consider the function g : S → X defined as g(s) = { f(s), s ∈ M x, s /∈ M. Then, for any open set V containing x, we have {s ∈ S : g(s) /∈ V} = {s ∈ S : f(s) /∈ V, s ∈ M} ⊆{s ∈ S : f(s) /∈ V}\{s ∈ S : s /∈ M}. Since, f be J -convergent to x, that implies {s ∈ S : f(s) /∈ V}∈J . Therefore, there exists K ∈ K such that {s ∈ S : f(s) /∈ V} \ J ⊆ (K ∪ J) \ J ∈ K. Subsequently, g is K− convergent to x. Hence, f is IK-convergent to x. � Theorem 2.7 ([1, Theorem 3.1]). In a Hausdorff space X, each function f : S → X possess a unique IK-limit provided I∪K is an ideal. Theorem 2.8. Let X be a Hausdorff Space. Let f : S → X be IK-convergent to x. Then ∃ an ideal J such that x ∈ X is an IK-limit of the function f if and only if x is also a J -limit of f provided I∪K is an ideal. Proof. Let f : S → X is IK-convergent to x. So, there exists a set M ∈ I∗ such that g : S → X with g(s) = { f(s), s ∈ M x, s /∈ M is K-convergent to x. Consequently, for each neigfhborhood Ux of x. We have {s ∈ S : g(s) /∈Ux}∈K. =⇒ {s ∈ S : f(s) /∈Ux, s ∈ M}∈K. Now, let J = Mc and (K∪ J) is an ideal base provided (I ∪K) is an ideal. Now we consider J , the ideal generated by (K∪ J). Then {s ∈ S : f(s) /∈Ux}⊆{s ∈ S : f(s) /∈Ux, s ∈ M}∪{s ∈ S : s /∈ M}. Therefore, {s ∈ S : f(s) /∈Ux}∈ (K∪ J). Converse part of the proof is immediate by lemma 2.6. � The following arrow diagram exhibit the equivalence shown in theorem 2.8. K for any J∈I −−−−−−−−→J →IK fixed J∈I −−−−−−→J →I∪K Comprehensively, we may ask the following question. Problem. Whether there exists an ideal J for IK-convergence in a given non-Hausdorff topological space X such that IK ≡J -convergence? © AGT, UPV, 2021 Appl. Gen. Topol. 22, no. 2 359 A. Sharmah and D. Hazarika 3. IK-sequential space Recently, I-sequential space were defined by S.K. Pal [12] for an ideal I on ω. An equivalent definition was suggested by Zhou et al. [13] and further obtain that class of I-sequential spaces includes sequential spaces [5]. First, recall the notion of I-sequential spaces. Let X be a topological space and O ⊆ X is I-open if no sequence in X\O has an I-limit in O. Equivalently, for each sequence {xn : n ∈ ω}⊆ X\O with I−lim xn = x ∈ X, then x ∈ X\O. Now X is said to be an I-sequential space if and only if each I-open subset of X is open. Here we introduce a topological space namely IK-sequential space for given ideals I and K on ω. Definition 3.1. Let X be a topological space and O, A ⊆ X. Then (1) O is said to be IK-open if no sequence in X \O has an IK-limit in O. Otherwise, for each sequence {xn : n ∈ ω}⊆ X\O with I K − lim xn = x ∈ X, then x ∈ X \O. (2) A subset F ⊆ X is said to be IK-closed if X \A is IK-open in X. Remark 3.2. The following are obvious for a topological space X and ideals I and K on ω. 1. Each open(closed) set of X is IK-open(closed). 2. If A and B are IK-open (closed), then A ∪ B is IK-open (closed). 3. A topological space X is said to be an IK-sequential space if and only if each IK-open set of X is open. for I = K, each IK-sequential space coincides with a I-sequential space. Lemma 3.3. Let M1,M2 be two convergence modes in a topological space X such that M1-convergence implies M2-convergence. Then O ⊆ X is M2-open implies that O is M1-open. Proof. Let O be not M1-open in X, then ∃{xn} in (X \ O) which is M1- convergent in X. So, {xn} is (X \ O) is M2-convergent in X and hence O is not M2-open. � Corollary 3.4. Let M1,M2 be two convergence modes in X such that M1- convergence implies M2-convergence in X. Then X is a M1-sequential space implies that X is M2-sequential space. The following is an example of a topological space which is not IK-sequential space. Example 3.5. Let S = [a, b] be a closed interval with the countable comple- ment topology τcc, where a, b ∈ R. Let A be any subset of S and xn be a sequence in A, IK-convergent to x, provided I,K and I ∪K is an ideal i.e, I ∪K− lim xn = x. Consider the neighborhood U of x, be the complement of the set {xn : xn 6= x} in S. Then xn = x for all n except for a set in the ideal I ∪K. Therefore, a sequence in any set A can only I ∪K-convergent to © AGT, UPV, 2021 Appl. Gen. Topol. 22, no. 2 360 Further aspects of IK-convergence in topological spaces an element of A i.e C is I∪K-open. Thus every subset of C is IK-sequentially open. But not every subset of S is open. Hence ([a, b], τcc) is not I K-sequential. Proposition 3.6. Let X be a topological space and I1,I2,K1,K2 be ideals on S. Then the following implications hold: (1) For K1 ⊆K2 whenever U ⊆ X is I K2-open, then it is IK1-open. (2) For I1 ⊆I2 whenever U ⊆ X is I K 2 -open, then it is I K 1 -open. Proof. Let f : S → X be a function. Then by Proposition 3.6 in [6], IK1 − lim f = x =⇒ I K 2 − lim f = x. IK1 − lim f = x =⇒ IK2 − lim f = x. By lemma 3.3 we have the required results correspondingly. � Corollary 3.7. For X be topological space and I1,I2,K1,K2 be ideals on ω where I1 ⊆I2, K1 ⊆K2. Then the following observations are valid: (1) If X is IK1 -sequential, then it is I K 2 -sequential. (2) If X is IK1-sequential, then it is IK2-sequential. Theorem 3.8. In a topological space X, if O is open then O is IK-open. Proof. Let O be open and {xn} be a sequence in X\O. Let y ∈ O. Then there is a neighborhood U of y which contained in O. Hence U can not contain any term of {xn}. So y is not an I K-limit of the sequence and O is IK-open. � Theorem 3.9. In a metric space X, the notions of open and IK-open coincide. Proof. Forward implication is obvious from Theorem 3.8. Conversely, Let O be not open i.e., ∃y ∈ O such that for all neighborhood of y intersect (X \ O). Let xn ∈ (X \ O) ∩ B(y, 1 n+1 ). Then xn → y. Hence xn is IK-convergent to y. Thus O is not IK-open. � Theorem 3.10. Every first countable space is IK-sequential space. Proof. We need to prove the reverse implication. If A ⊂ X be not open. Then ∃y ∈ A such that every neighborhood of y intersects X \ A. Let {Un : n ∈ ω} be a decreasing countable basis at y (say). Consider xn ∈ (X \A)∩Un. Then for each neighborhood V of y, ∃n ∈ ω with Un ⊂ V . So, xm ∈ V,∀m ≥ n i.e xn → y. Hence K− lim xn = y. Therefore, A is not IK-open. � The following theorem about continuous mapping was also proved by Baner- jee et al. [1]. However, we have given here an alternative approach to prove. Theorem 3.11. Every continuous function preserves IK-convergence. Proof. Let X and Y be two topological spaces and c : X → Y be a continuous function. Let f : S → X be IK-convergent. So ∃M ∈I∗ such that g : S → X given by © AGT, UPV, 2021 Appl. Gen. Topol. 22, no. 2 361 A. Sharmah and D. Hazarika g(s)= { f(s), s ∈ M x, s /∈ M is K-convergent to x. Now the function c ◦ f : S → Y , the image function on Y is K-convergent to x by Theorem 3 in [3]. Hence c◦f is IK-convergent. � We now recall the definition of a quotient space. Let (X,∼) be a topological space with an equivalence relation ∼ on X. Consider the projection mapping ∏ : X → X/ ∼ (the set of equivalence classes) and taking A ⊂ X/ ∼ to be open if and only if ∏−1 (A) is open in X, we have the quotient space X/ ∼ induced by ∼ on X. Theorem 3.12. Every quotient space of an IK-sequential space X is IK- sequential. Proof. Let A ⊂ X/ ∼ be not open. Let X/ ∼ is a quotient space with an equivalence relation ∼, ∏−1 (A) is not open in X i.e., ∃ a sequence {xn} in X \ ∏−1 (A) which is IK-convergent to y ∈ ∏−1 (A). Also ∏ is continuous, hence preserves IK-convergence by Theorem 3.11. Therefore, ∏ (xn) ∈ (X/ ∼) \ A with IK-limit ∏ (y) ∈ A. So, A is not IK-open i.e., X/ ∼ is IK- sequential. � Following result is immediate via Proposition 3.4. Theorem 3.13. Every sequential space is an IK-sequential space. Recall that a topological space X is said to be of countable tightness, if for A ⊆ X and x ∈ Ā, then x ∈ C̄ for some countable subset C ⊆ A. Every sequential and I-sequential space is of countable tightness [13]. Proposition 3.14. Every IK-sequential space X is of countable tightness. Proof. Let X be an IK-sequential space and A ⊆ X. Consider [A]ω = ⋃ {B : B is a countable subset of A}. Clearly, A ⊆ [A]ω ⊆ A. We claim that, [A]ω is IK-closed in X. Consider {xn} be a sequence in [A]ω, I K-convergent to x ∈ X. Since xn ∈ [A]ω, then we can find a countable subset B of A such that xn ∈ B for all n ∈ ω. Since X be an IK-sequential space, so B is IK-closed, thus x ∈ B ⊆ [A]ω, and further [A]ω is I K-closed in X. Now, let X be an IK-sequential space and A be a subset of X. Since the set [A]ω is closed in X, and [A]ω ⊆ A ⊆ [A]ω, thus A = [A]ω. If x ∈ A, then x ∈ [A]ω, and further, there exists a countable subset C of A such that x ∈ C, i.e., X is of countable tightness. � Now we show that every IK-sequential space is hereditary with respect to IK-open (IK-closed) subspaces. First we have the following lemma. Lemma 3.15 ([13, Lemma 2.4]). Let I be an ideal on ω and xn, yn be two sequences in a topological space X such that {n ∈ ω : xn 6= yn} ∈ I. Then I− lim xn = x if and only if I− lim yn = x. © AGT, UPV, 2021 Appl. Gen. Topol. 22, no. 2 362 Further aspects of IK-convergence in topological spaces Theorem 3.16. If X is an IK-sequential space then every IK-open (IK-closed) subspaces of X is IK-sequential. Proof. Let X be an IK-sequential space. Suppose that Y is an IK-open subset of X. Then Y is also open in X. We anticipate Y to be IK-sequential space. Consider U to be IK-open in Y . Here Y is open, so we claim that U is open in X. Since X is IK-sequential space, we need to show that U is IK-open in X. Contra-positively, take U be not IK-open in X. Then, ∃{xn} in X \ U such that IK − lim xn = x (∈ U.) i.e. ∃M ∈I ∗ such that xnk →K x, where nk ∈ M and xnk ∈ X \ U. Now {nk : xnk /∈ Y } ∈K. For a point y ∈ Y \ U (assume), Now Consider a sequence {yn} such that yn = xn for n ∈ M and yn = ynk for n /∈ M where {ynk} is defined as ynk = xnk for xnk ∈ Y and ynk = y for xnk /∈ Y . Then by Lemma 3.15, {ynk} is K-convergent to x. Hence {yn} is IK-convergent to x. So U is not IK-open in Y . That is a contradiction to our assumption. Let Y be an IK-closed subset of X. Then Y is closed in X. For any IK- closed subset F of Y , it is sufficient to show that F is closed in X. Since X is an IK-sequential space, it is enough to show that F is IK-closed in X. Therefore, let {xn : n ∈ ω} be an arbitrary sequence in F with I K − lim xn = x in X. We claim that x ∈ F . Indeed, since Y is closed, we have x ∈ Y , and then it is also clear that x ∈ F since F is an IK-closed subset of Y . � Proposition 3.17. The disjoint topological sum of any family of IK-sequential spaces is IK-sequential. Proof. Let (Xα)α∈∆ be a family of I K-sequential space and X = ⊕α∈∆Xα. We claim that X is IK-sequential space. Let F be IK-closed in X. For each α ∈ ∆, Xα is closed in X i.e., Xα is I K-closed in X. Hence, F∩Xα is I K-closed in X by Remark 3.2. As (F ∩ Xα) ⊆ Xα i.e. F ∩ Xα is closed in Xα. Now F is closed in X ≡ X \F is open in X ≡∪α(Xα \F) is open in X if and only if Xα \F is open in Xα ≡ F ∩ Xα is closed in Xα. Hence F is closed in X. � 4. IK-cluster point and IK-limit point The notions I-cluster point and I-limit point in a topological space X were defined by Das et al. [3] and also characterized Cx(I), the collection of all I-cluster points of a given sequence x = {xn} in X, as closed subsets of X (Theorem 10, [3]). Here we define IK-notions of cluster point and limit points for a function in X. For I∗-convergence, I∪Fin is an ideal, thereupon I and Fin satisfy ideality condition. Moreover we assume ideality condition of I and K in IK-convergence to investigate some results. Definition 4.1. Let f : S → X be a function and I, K be two ideals on S. Then x ∈ X is called an IK-cluster point of f if there exists M ∈I∗ such that the function g : S → X defined by g(s)= { f(s), s ∈ M x, s /∈ M © AGT, UPV, 2021 Appl. Gen. Topol. 22, no. 2 363 A. Sharmah and D. Hazarika has a K-cluster point x, i.e., {s ∈ S : g(s) ∈ Ux} /∈K. Definition 4.2. Let f : S → X be a function and I, K be two ideals on S. Then x ∈ X is called an IK-limit point of f if there exists M ∈ I∗ such that for the function g : S → X defined by g(s)= { f(s), s ∈ M x, s /∈ M has a K-limit point x. For I = K, we know the convergence modes IK ≡ I ≡ K. Hence definitions 4.1 and 4.2 generalizes the definitions of I or K-(limit point and cluster point) correspondingly. Again, for nets in a topological space I-limit points and I- cluster points coincide [2]. Therefore, IK-cluster points and IK-limit points of nets also coincide. Following the notation in [11], we denote the collection of all IK-limit points and IK-cluster points of a function f in a topological space X by Lf(I K) and Cf (I K) respectively. We observe that Cf (I K) ⊆ Cf (K) and Lf(I K) ⊆ Lf(K). We also observe that Lf (I ∗) = L(I∗), where L(I∗) denote the collection of I∗-limits of f. Lemma 4.3. If I and K be two ideal then Lf (I K) ⊆ Cf (I K). Proof. Since Lf (K) ⊆ Cf (K) for an ideal K, hence the result is immediate. � We have the following lemma provided the ideals I and K satisfy ideality condition. Lemma 4.4. Cf (I∪K) ⊆ Cf (I K). Proof. Let y be not a IK-cluster point of x = {xn}n∈ω. Then for all M ∈ I ∗ such that for the function g : S → X defined by g(s)= { f(s), s ∈ M x, s /∈ M, the set {s ∈ S : g(s) ∈ Ux}∈K. Since {s : f(s) ∈ Ux}⊆{s : g(s) ∈ Ux}∈K. i.e. {s : f(s) ∈ Ux}∈I∪K. Hence y is not a (I∪K)-cluster point of x. � Since above set inequalities signify the implication K → IK → I ∪K, We expect the following conclusion. Conjecture 4.5. Lf(I∪K) ⊆ Lf(I K). For sequential criteria in [11], we observe the following result. Theorem 4.6. Let I, K be two ideals on ω and X be a topological space. Then (i) For x = {xn}n∈ω, a sequence in X; Cx(I K) is a closed set. (ii) If (X, τ) is closed hereditary separable and there exists a disjoint se- quence of sets {Pn} such that Pn ⊂ ω, Pn /∈ I,K for all n, then for every non empty closed subset F of X, there exists a sequence x in X such that F = Cx(I K) provided I∪K is an ideal. © AGT, UPV, 2021 Appl. Gen. Topol. 22, no. 2 364 Further aspects of IK-convergence in topological spaces Proof. Consider the sequence x = {xn} in X and I, K be the two ideals on ω. (i) Let y ∈ Cx(IK); the derived set of Cx(I K). Let U be an open set containing y. It is clear that U ∩ Cx(I K) 6= φ. Let p ∈ (U ∩ Cx(I K)) i.e., p ∈ U and p ∈ Cx(I K). Now there exist a set M ∈ I∗, such that {yn}n∈ω given by yn = xn if n ∈ M and p, otherwise; we have {n ∈ ω : yn ∈ U} /∈ K. Consider the sequence {zn}n∈ω given by zn = xn if n ∈ M and y, otherwise; then {n ∈ ω : zn ∈ U} = {n ∈ ω : yn ∈ U} /∈K. Hence y ∈ Cx(I K). (ii) Being a closed subset of X, F is separable. Let S = {s1, s2, ...}⊂ F be a countable set such that S = F . Consider xn = si for n ∈ Pi. Thus we have the subsequence {kn} of {n} for which assume the sequence x = {xnk}. Let y ∈ Cx(K) (taking y 6= si otherwise if y = si for some i, then y is eventually in F). We claim Cx(K) ⊂ F . Let U be any open set containing y. Then {n : xnk ∈ U} /∈ K and hence non empty i.e., si ∈ U for some i. Therefore F ∩U is non empty, So y is a limit point of F and closedness of F gives y ∈ F . Hence Cx(K) ⊂ F . Further Cx(I K) ⊆ Cx(K) ⊂ F . Conversely, for a ∈ F and U be an open set containing a, then there exists si ∈ S such that si ∈ U. 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