@ Appl. Gen. Topol. 23, no. 2 (2022), 243-253 doi:10.4995/agt.2022.16720 © AGT, UPV, 2022 A Urysohn lemma for regular spaces Ankit Gupta a and Ratna Dev Sarma b a Department of Mathematics, Bharati College, (University of Delhi), Delhi 110058, India (ankitsince1988@yahoo.co.in) b Department of Mathematics, Rajdhani College (University of Delhi), Delhi 110015, India (ratna sarma@yahoo.com) Communicated by S. Garćıa-Ferreira Abstract Using the concept of m-open sets, M-regularity and M-normality are introduced and investigated. Both these notions are closed under ar- bitrary product. M-normal spaces are found to satisfy a result similar to Urysohn lemma. It is shown that closed sets can be separated by m-continuous functions in a regular space. 2020 MSC: 54A05; 54D10; 54D15; 54C08. Keywords: regularity; normality; M-normality; M-regularity; Urysohn lemma. 1. Introduction Nowadays topological approaches are being investigated in various diverse field of science and technology such as computer graphics, evolutionary theory, robotics[4, 9, 10] etc. to name a few. In a finite topological space, the intersec- tion of all open neighbourhoods of a point p is again an open neighbourhood of p, which is the smallest one. It is called the minimal neighbourhood of p. However, in general framework, we define the minimal open sets or m-open sets as the ones obtained by taking the arbitrary intersections of the open sets. They have been studied in the recent past by several researchers [2, 6]. In this paper, we further use them for construction of new notions in topology, namely M-regularity and M-normality. These two notions are distinct from the already existing notions of regularity and normality, and are found to have Received 23 November 2021 – Accepted 05 May 2022 http://dx.doi.org/10.4995/agt.2022.16720 https://orcid.org/0000-0003-0760-4467 https://orcid.org/0000-0002-4181-3732 A. Gupta and R. D. Sarma several interesting properties. Both of them are closed under arbitrary product. M-normal spaces are found to exhibit Urysohn lemma type property. Finally it is shown that even in regular spaces, disjoint pair of closed sets can be sep- arated by mappings, the so called m-continuous mappings. In that sense, the last result of the paper may be treated as Urysohn lemma for regular spaces. Here it may be mentioned that the classical Urysohn lemma is unprovable in ZF[1, 5]. The usual proof of Urysohn lemma in Kelley [8] uses the axioms of dependent choice to successfully select open sets separating previously chosen sets [1]. Similar choices have been made in our proof also, making it valid only in ZFC. 2. Preliminaries Definition 2.1 ([6]). Let (X,τ) be a topological space. A set A ⊆ X is called m-open if A can be expressed as intersection of a subfamily of open sets. The complement of an m-open set is called an m-closed set. The collection of m-open sets of a topological space (X,τ) is denoted by M. Clearly, every open set is m-open. In a finite space, open sets are the only m-open sets. The following example gives an idea about the abundance of m-open sets. Example 2.2. Let X = N be the set of natural numbers, equipped with the co-finite topology. Then every subset of X is m-open. Proposition 2.3 ([6]). For a topological space (X,τ), we have the following results: (i) ∅,X ∈M; (ii) M is closed under arbitrary union; (iii) M is closed under arbitrary intersection. Definition 2.4. [6] Let (X,τ) and (Y,µ) be two topological spaces. Then a function f : X → Y is said to be m-continuous at a point x ∈ X if for every open neighbourhood V of f(x) there exists an m-open set U containing x in X such that f(U) ⊆ V . A function f : X → Y is said to be m-continuous [6] if it is m-continuous at each point x of X. Since every open set is m-open. Therefore every continuous function is m- continuous. But the converse need not be true. Example 2.5. Let X = N be the set of natural numbers equipped with the co-finite topology and Y = {a,b,c} with the topology µ = {∅,{a},{b,c},Y}. Then consider a function f : X → Y defined as: f(x) =   a if x < 10, b if 10 ≤ x < 100, c otherwise. © AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 244 A Urysohn lemma for regular spaces Since every subset A ⊆ N is m-open under the co-finite topology, function f is m-continuous. But f is not a continuous function as f−1({a}) = {x ∈ N |x < 10} is not an open set in N under the co-finite topology. Theorem 2.6. Let (X,τ) and (Y,µ) be two topological spaces. Then for a function f : (X,τ) → (Y,µ), the following are equivalent: (i) f is m-continuous; (ii) inverse image of every open subset of Y is m-open; (iii) inverse image of every closed subset of Y is m-closed. Proof. (i) ⇒ (ii): Let U be any open subset of Y and let x ∈ f−1(U). Then f(x) ∈ U. Therefore there exists an m-open subset V in X such that x ∈ V and f(V ) ⊆ U. Thus x ∈ V ⊆ f−1(U), therefore f−1(U) is an m-open neighbourhood of x. Hence f−1(U) is m-open. (ii) ⇒ (iii): Let A be any closed subset of Y . Then Y �A is open and therefore f−1(Y \A) is m-open, that is, X\f−1(A) is m-open. Hence f−1(A) is m-closed. (iii) ⇒ (i): Let M be an open neighbourhood of f(x), therefore Y \M is closed, and consequently f−1(Y \M) is m-closed. Thus f−1(M) is m-open and hence x ∈ f−1(M) = N (say). Then, we have N is an m-open neighbourhood of x such that f(N) ⊆ M. � In next result, we prove that arbitrary product of m-open sets is again m- open under the product topology. Here we use the fact that “An open set V of a product topology can be realized in the form V = ∏ i∈Λ Vi, where Vi ∈ τi and Vi = Xi except for finitely many i ′s.” This can be verified using the concept of basic open sets and the fact that(⋃ i∈I Ai ) ×  ⋃ j∈J Bj  × . . . = ⋃ (i,j,...)∈I×J×... (Ai ×Bj × . . .) . We also provide below the following results, which will be used in our paper. Lemma 2.7 ([3, p. 28]). Let {Aα,β} be an arbitrary family of non-empty sets. Then we have ⋂ β (∏ α Aα,β ) = ∏ α  ⋂ β Aα,β   Lemma 2.8 ([7, p. 34]). Let {(Xα,τα)|α ∈A} be an arbitrary family of topo- logical spaces and let Aα ⊆ Xα for each α ∈A. Then we have cl (∏ α Vα ) = ∏ α (cl(Vα)). Theorem 2.9. Let (Xα,τα) be topological spaces and Uα be an m-open set in (Xα,τα). Then the product of U ′ αs is an m-open set in the product topology∏ α τα of X = ∏ α Xα. © AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 245 A. Gupta and R. D. Sarma Proof. Let {(Xα,τα)}α be a family of topological spaces and A be an m-open set in the product topology X = ∏ α Xα. Then there exists open sets Ui in X such that A = ⋂ i Ui. Since Ui is an open set in the product topology of X, therefore there exists open set Uα,i ∈ τα with Uα,i = Xα for all but finitely many α’s such that Ui = ∏ α Uα,i. Hence A = ⋂ i (∏ α Uα,i ) . Using the fact that ⋂ β (∏ α Aα,β ) = ∏ α  ⋂ β Aα,β  , in view of Lemma 2.7, we have A = ⋂ i (∏ α Uα,i ) = ∏ α (⋂ i Uα,i ) . Hence the proof. � Now, we will show that every subset of a T1-topological space (X,τ) is m- open. Theorem 2.10. Every subset of a T1-space is m-open. Proof. Let (X,τ) be a topological space, which is T1. Let A be a non-empty subset of X. Then every singleton {x}⊆ X is a closed set. Therefore, consider A = ⋂ x∈X\A X \{x}. Since every singleton is closed therefore X\{x} is an open set in X. Hence the arbitrary intersection of open sets is m-open, thus A is m-open. � 3. M-Regular Spaces Definition 3.1. A topological space (X,τ) is said to be M-regular if for each pair consisting of a point x and an m-closed set B not containing x, there exists a disjoint pair of an m-open set U and an open set V containing x and B respectively. In other words, For every pair, x and B with x /∈ B, where B is an m-closed set, there exist an m-open set U and an open set V such that x ∈ U, B ⊆ V and U ∩V = ∅. Now, we provide a characterization for M-regularity. Theorem 3.2. Let (X,τ) be a topological space. Then X is M-regular if and only if for a given point x ∈ X and an m-open neighbourhood U of x, there exists an m-open neighbourhood V of x such that x ∈ V ⊆ cl(V ) ⊆ U. Proof. Suppose that X is M-regular. Let x ∈ X and U ⊆ X, an m-open neighbourhood of x, be given. Then B = X \ U is an m-closed set disjoint from x. By the given hypothesis, there exists a disjoint pair of an m-open set V and an open set W containing x and B respectively. Thus we have, x ∈ V and X \ U = B ⊆ W , that is, x ∈ V ⊆ X \ W ⊆ U. Hence we have, x ∈ V ⊆ cl(V ) ⊆ cl(X \ W) = X \ W ⊆ U. Therefore we have, x ∈ V ⊆ cl(V ) ⊆ U. © AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 246 A Urysohn lemma for regular spaces Conversely, suppose that a point x ∈ X and an m-closed set B not containing x are given. Then U = X \ B is an m-open set containing x. Then, by the hypothesis, there exists an m-open neighbourhood V of x such that x ∈ V ⊆ cl(V ) ⊆ U. Thus we have, a disjoint pair of m-open set V and an open set X \ cl(V ) which contains x and B respectively. Therefore (X,τ) is M- regular. � With the help of following example, we show that an M-regular space need not be regular. Example 3.3. Let X = N be the set of natural numbers, equipped with the co-finite topology. Then every subset A of X is m-open. Therefore X is a M-regular space but not a regular space. But, every regular topological space is M-regular. For this, we have the following result: Theorem 3.4. Every regular space is M-regular. Proof. Let (X,τ) be a topological space which is regular. We have to show that X is M-regular. For this, let V be any m-open subset of X and let x ∈ V . Then, we have V = ⋂ j∈J Vj, where V ′ j s are open sets in X. Therefore, as x ∈ V = ⋂ j∈J Vj, we have x ∈ Vj for all j ∈ J. Since the space X is given to be regular, thus there exists open set Wj such that x ∈ Wj ⊆ cl(Wj) ⊆ Vj, for all j ∈ J. Now, consider W = ⋂ j∈J Wj an m-open set in X, we have x ∈ W ⊆ ⋂ j∈J cl(Wj) ⊆ ⋂ j∈J Vj = V . Thus we have, x ∈ W ⊆ cl(W) ⊆ V , where W is an m-open subset of X. Hence X is M-regular. � From the Theorem 2.10, one can conclude that every T1-space is m-regular. But the converse of the above statement is not true. That is, an M-regular space need not be T1. For this, we have the following example: Example 3.5. Let X = {a,b,c,d} be a non-empty set equipped with a topol- ogy τ = {∅,{a,b},{c,d},X}. Then (X,τ) is an M-regular space but it is not a T1-space. Next we will provide the decomposition of T1-space with the help of M- regularity. Theorem 3.6. Every M-regular T0-space is T1. Proof. Let (X,τ) be a topological space. Let x,y ∈ X be a pair of distinct points of an M-regular T0-space X. Let there exists an open set V in X such that x ∈ V but y /∈ V . Since every open set is m-open and the space X is given to be M-regular, therefore there exists an m-open set U in X such that © AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 247 A. Gupta and R. D. Sarma x ∈ U ⊆ cl(U) ⊆ V . Then consider, W = X \ cl(U) is open in X such that x /∈ W and y ∈ W . Thus X is T1. � From the Theorem 3.6, one can state the following: Theorem 3.7. Every T0-space is M-regular if and only if it is T1. However a T0-space need not be M-regular. Example 3.8. Let X be a Sierpiński space, that is, X = {a,b} with topology τ = {∅,{a},X}. Then (X,τ) is a T0-space but not M-regular. In our next result, we show that Hausdorffness is a sufficient condition for M-regularity. Theorem 3.9. Every Hausdorff space is M-regular. Proof. Let (X,τ) be a Hausdorff space. Let x and B be a pair of a point and an m-closed set such that x /∈ B. Then for every y ∈ B, we have x 6= y. Therefore by the given hypothesis, there exists a disjoint pair of open sets Uy and Vy such that x ∈ Uy and y ∈ Vy. Then consider V = ⋃ y {Vy |y ∈ B}, an open cover of B and U = ⋂ y {Uy |y ∈ B} is an m-open set containing x. Thus we have a disjoint pair consisting an open set V and an m-open set U containing B and x respectively. Therefore X is M-regular. � The converse, however need not be true. Example 3.10. Let X = {a,b,c,d} be a non-empty set equipped with a topology τ = {∅,{a,b},{c,d},X}. Then (X,τ) is an M-regular space but it is not a Hausdorff space. Our next result is on the product of M-regular spaces. Theorem 3.11. Any arbitrary product of M-regular spaces is again M-regular. Proof. Let {(Xα,τα)}α be a family of M-regular spaces and X = ∏ α Xα. Let x = (xα) ∈ X be a point and U be an m-open neighbourhood of x ∈ X. Since U is an m-open set in X, therefore U = ⋂ i Ui, where Ui is an open set in X under the product topology. Therefore, we have Ui = ∏ α Uα,i, where Uα,i ∈ τα. Hence, we have U = ⋂ i (∏ α Uα,i ) . We use the fact that ⋂ β (∏ α Aα,β ) = ∏ α  ⋂ β Aα,β   in view of Lemma 2.7. Therefore, we have U = ⋂ i (∏ α Uα,i ) = ∏ α (⋂ i Uα,i ) and x ∈ U. Hence we have xα ∈ ⋂ i Uα,i, © AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 248 A Urysohn lemma for regular spaces where ⋂ i Uα,i = Wα (say) is an m-open set in (Xα,τα). Since (Xα,τα) is M-regular, therefore there exists an m-open set Vα of (Xα,τα) such that xα ∈ Vα ⊆ clα(Vα) ⊆ Wα. Now, V = ∏ α Vα is an m-open set in X in view of Theorem 2.9. Again cl(V ) = cl (∏ α Vα ) = ∏ α clα (Vα), in view of Lemma 2.8. Thus, we have an m-open set V in X such that x ∈ V ⊆ cl(V ) ⊆ ∏ α Wα ⊆ U. Hence X is M-regular. � 4. M-Normal Spaces Definition 4.1. A topological space (X,τ) is said to be M-normal if for each disjoint pair consisting of a closed set A and an m-closed set B, there exists a disjoint pair consisting of an m-open set U and an open set V in X containing A and B respectively. Remark 4.2. An M-normal space need not be M-regular. For this, consider the Sierpiński space mentioned in the Example 3.8. The space (X,τ) is M-normal but not M-regular. In our next result, we provide a characterization for M-normality. Theorem 4.3. Let (X,τ) be a topological space. Then (X,τ) is M-normal if and only if for a given closed set C and an m-open set D such that C ⊆ D, there is an m-open set G such that C ⊆ G ⊆ cl(G) ⊆ D. Proof. Let C and D be the closed and m-open sets respectively such that C ⊆ D. Then X \D is an m-closed set such that C ∩(X \D) = ∅. Then, from the M-normality, there exist an m-open set G and an open set V such that C ⊆ G, X\D ⊆ V and G∩V = ∅. Therefore X\V ⊆ D and hence C ⊆ G ⊆ X\V ⊆ D, where X \V is a closed set. Hence C ⊆ G ⊆ cl(G) ⊆ cl(X \V ) = X \V ⊆ D. Conversely, consider D and C as closed and m-closed sets respectively such that C ∩D = ∅. Then X \C is m-open set containing D. Then by the given hypothesis, there exist an m-open set G such that D ⊆ G ⊆ cl(G) ⊆ X \ C. Thus, we have D ⊆ G, C ⊆ V and G∩V = ∅, where V = X \ cl(G), an open set. Hence X is M-normal. � From the Example 3.3, one can easily verify that M-normality doesn’t imply Normality. Here the space X is M-normal but it is not normal. In the following result, we show that every normal space is M-normal. Theorem 4.4. Every normal space is M-normal. Proof. Let (X,τ) be a topological space which is normal. We have to show that X is M-normal. For this, let A be any closed subset of X and let V be an m-open subset of X such that A ⊆ V . Then, we have V = ⋂ j∈J Vj, where V ′ j s © AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 249 A. Gupta and R. D. Sarma are open sets in X. Therefore, we have A ⊆ V = ⋂ j∈J Vj, that is, A ⊆ Vj for all j ∈ J. Since the space X is given to be normal, thus there exists open set Wj such that A ⊆ Wj ⊆ cl(Wj) ⊆ Vj, for all j ∈ J. Now, consider W = ⋂ j∈J Wj, an m-open set in X, we have A ⊆ W ⊆ ⋂ j∈J cl(Wj) ⊆ ⋂ j∈J Vj = V . Thus we have, A ⊆ W ⊆ cl(W) ⊆ V , where W is an m-open subset of X. Hence X is M-normal. � One specialty of M-normality is that it is preserved under arbitrary product. Theorem 4.5. Any arbitrary product of M-normal spaces is again M-normal. Proof. Let {(Xα,τα)}α be a family of M-normal spaces and X = ∏ α Xα. Let A ⊆ X be a closed set and U be an m-open set such that A ⊆ U. Since U is an m-open set in X, therefore U = ⋂ i Ui, where Ui is an open set in X under the product topology. Thus, we have Ui = ∏ α Uα,i, where Uα,i ∈ τα and Uα,i = Xα for all but finitely many α’s, as explained before Theorem 2.9 . Hence, we have U = ⋂ i (∏ α Uα,i ) . We use the fact that ⋂ β (∏ α Aα,β ) = ∏ α  ⋂ β Aα,β   following Lemma 2.7. Thus we have U = ⋂ i (∏ α Uα,i ) = ∏ α (⋂ i Uα,i ) . Similarly, we have A = ∏ α Aα, where Aα is a closed set in Xα. Since A ⊆ U, thus we have Aα ⊆ ⋂ i Uα,i. Let ⋂ i Uα,i = Wα, an m-open set in Xα. We have Aα ⊆ Wα and since Xα is an M-normal space, therefore, there exists an m-open set Vα such that Aα ⊆ Vα ⊆ cl(Vα) ⊆ Wα. Thus we have, ∏ α Aα ⊆ ∏ α Vα ⊆∏ α clα(Vα) ⊆ ∏ α Wα. Now, ∏ α Vα is m-open in view of Theorem 2.9. Also, cl (∏ α Vα ) = ∏ α (cl(Vα)), in view of Lemma 2.8. Hence we have V = ∏ α Vα, an m-open set in X such that A ⊆ V ⊆ cl(V ) = ∏ α (cl(Vα)) = cl (∏ α Vα ) . It follows that A ⊆ V ⊆ cl(V ) ⊆ ∏ α Wα ⊆ U. Hence X is M-normal. � Following result for M-normal spaces is analogous to the well known Urysohn lemma for normal spaces. © AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 250 A Urysohn lemma for regular spaces Theorem 4.6. Let (X,τ) be a topological space. Then for each pair of disjoint subsets A and B of X, one of which is closed and the other is M-closed, there exists an m-continuous function f on X to [0, 1] (resp. [a,b] for any real number a,b, a < b ), such that f(A) = {0} and f(B) = {1} (resp. f(A) = {a}, and f(B) = {b} ) provided X is M-normal. Proof. Let (X,τ) be an M-normal space. Let C0 and C1 be two disjoint sets, where C0 is closed and C1 is m-closed in X. Since C0 ∩ C1 = ∅, therefore C0 ⊆ X\C1. Let P be the set of all dyadic rational numbers in [0, 1]. We shall define for each p in P , an m-open set Up of X, in such a way that whenever p < q, we have cl(Up) ⊂ Uq. First we define U1 = X \ C1, an m-open set such that C0 ⊆ U1. Since X is M-normal space, by Theorem 4.3, there exists an m-open set U1/2 such that C0 ⊆ U1/2 ⊆ cl(U1/2) ⊆ U1. Similarly, there also exist another m-open sets U1/4 and U3/4 such that C0 ⊆ U1/4 ⊆ cl(U1/4) ⊆ U1/2 ⊆ cl(U1/2) ⊆ U3/4 ⊆ cl(U3/4) ⊆ U1, because cl(U1/4) is again a closed set. Continuing the process, we define Ur, for each r ∈ P such that C0 ⊆ Ur ⊆ cl(Ur) ⊆ U1 and cl(Ur) ⊆ Us whenever r < s, for r,s ∈ P . Let us define Q(x) to be the set of those dyadic rational numbers p such that the corresponding m-open sets Up contains x: Q(x) = {p | x ∈ Up} Now we define a function f : X → [0, 1] as f(x) = inf Q(x) = inf{p | x ∈ Up} Clearly, f(C0) = {0} and f(C1) = {1}. Then we show that f is the desired m-continuous function. For a given point x0 ∈ X and an open interval (c,d) in [0, 1] containing the point f(x0). We wish to find an m-open neighbourhood U of x0 such that f(U) ⊆ (c,d). Let us choose rational numbers p and q such that c < p < f(x0) < q < d. Then U = Uq \ cl(Up) = Uq ∩ (X \ cl(Up)) is the desired m-open neighbourhood of x0. Here, we will show that x0 /∈ cl(Up). If x0 ∈ cl(Up), then for s > p, we have x0 ∈ cl(Up) ⊆ Us. Thus x0 ∈ Us for all s > p. This implies that f(x0) ≤ p, as f(x0) = Inf{s |x0 ∈ Us}. This contradicts the fact that f(x0) > p. Similarly, as f(x0) < q, therefore x0 ∈ Uq and hence U = Uq \ cl(Up) is the desired neighbourhood of x0. Hence f is an m-continuous function on X to [0, 1] with f(C0) = {0} and f(C1) = {1}. This completes the proof. � Since every closed set is m-closed, we get the following result: Corollary 4.7. Let (X,τ) be an M-normal topological space. Then for each pair of disjoint closed subsets A and B of X, there exists an m-continuous function f on X to [0, 1] (resp. [a,b] for any real number a,b, a < b ), such that f(A) = {0} and f(B) = {1} (resp. f(A) = {a}, and f(B) = {b} ). © AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 251 A. Gupta and R. D. Sarma Remark 4.8. From the proof of Theorem 4.6, it is clear that the proof is in line of the usual proof of Urysohn lemma in Kelley[8], wherein the choice function plays its role. Hence the proof is valid for ZFC. Again, it has been pointed out in [1] that the axiom of multiple choice also implies Urysohn lemma, since one can use the intersection of the finitely many separating open sets provided by MC. Essentially the same argument shows that DMC implies Urysohn lemma. Since similar working is followed in our Theorem 4.6, hence the variant of Urysohn lemma in our paper is also valid for ZF with DMC. Our next theorem provides the relation between regularity and M-normality. Theorem 4.9. Every regular space is M-normal. Proof. Let A and B be two disjoint sets such that A is closed and B is m-closed. Since A∩B = ∅, therefore B ⊆ X \A, where X \A is an open set containing the m-closed set B. Then, by the given hypothesis, for each b ∈ B ⊆ X \ A, there exists an open set Ub such that b ∈ Ub ⊆ cl(Ub) ⊆ X \A. Thus, we have a collection D = {Ub |b ∈ B} which covers B. Further, if D ∈D, then cl(D) is disjoint from A because cl(D) ⊆ X \A. Consider V = ⋃ {D |D ∈D}. Then V is an open set in X which contains B, an m-closed set. Since D lies in some Ub whose closure is disjoint from A , therefore W = ⋃ {cl(D) |D ∈D} is disjoint from A. Therefore, V = ⋃ {D |D ∈D} and X \ W are two disjoint subsets of X. Now W , being union of closed sets, is m-closed. Thus we have one open set and one m-open set containing the sets B and A respectively. Hence X is M-normal. � In view of Theorem 4.6 and 4.9, one can observe the following: Theorem 4.10. Let (X,τ) be a regular space. Then for every disjoint pair of sets consisting of a closed set A and an m-closed set B, there always exists an m-continuous mapping f from X to [a,b] such that f(A) = {a} and f(B) = {b}. The last and the final result of this section is a simple corollary of Theorem 4.10. However its importance lies in revealing the fact that even in regular spaces, closed sets can be separated by mappings, the so-called m-continuous mappings. In that sense, this result may be treated as Urysohn lemma for regular spaces. Theorem 4.11. Let (X,τ) be a regular space. Then for every disjoint pair of closed sets A and B, there always exists an m-continuous mapping f from X to [a,b] such that f(A) = {a} and f(B) = {b}. © AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 252 A Urysohn lemma for regular spaces Acknowledgements. The authors sincerely acknowledge the reviewers of the paper for their valuable suggestions. References [1] A. Blass, Injectivity, projectivity and the axiom of choice, Trans. Am. Math. Soc. 255 (1970), 31–59. [2] C. Boonpok, ξµ-sets in generalized topological spaces, Acta Math. Hungar. 96 (2012), 269–285. [3] J. Dugundji, Topology, Allyn and Bacon (1966). [4] E. 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