EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 16, No. 2, 2023, 1228-1235 ISSN 1307-5543 – ejpam.com Published by New York Business Global Locally Compact Spaces with Defects Mohmmad Zailai Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O.Box 80200 Jeddah 21589, Saudi Arabia Abstract. We call a topological space X a locally compact space with defects if all points in X possess compact neighborhoods except for some points. We investigate this weaker version of local compactness. We show that for x ∈ X• if the partition of singletons of X\(X• ∪ (U\U)) is locally finite, where U ̸= X is an open neighborhood of x, then X is a Tychonoff space. Let X be a T1c locally compact space with defects such that each x ∈ X• has an open neighborhood U such that U is a union of pairwise disjoint compact subsets ⋃ s∈S Fs. Then, we show that if the family {Fs}s∈S is locally finite except for a finite number of points, then X is a Tychonoff space. 2020 Mathematics Subject Classifications: 54D45 Key Words and Phrases: Compact, locally compact, defects, Tychonoff 1. Introduction A T1 space X is said to be locally compact if every point x ∈ X possesses a compact neighborhood, i.e, an open neighborhood such that its closure is a compact subspace. In this paper we introduce a weaker version of local compactness, which we call local compactness with defects. A T1 space X is locally compact with defects if each point of the space has a compact neighborhood except for some points. We denote by X• the set of points of X which do not have compact neighborhoods. Points of X• are called defects. A space X is said to be scattered if it contains no non-empty subset which is dense-in- itself. It is proved in [4] that for a Tychonoff space X the set X• is closed. We extend this result and show that for any space, the set of defects is a closed subset. We use that result to show that for any T1 topological space, if the set of defects is not empty then the space is not scattered. All compact subspaces in this paper are assumed to be closed and T2. We denote by T1c a T1 space such that each compact subspace is closed. This is a space which lies between T1 and T2 spaces. ℵ0 stands for a cardinality of a countable set. N stands for the set of all natural numbers. By a T3 1 2 space we mean a Tychonoff space. K stands for the sorgenfrey line, i.e., the space generated by the base B = {[x, y)}, where x, y are real numbers such that x < y, and y is a rational number. For more details about locally compact spaces, see [1]. More details about points that do not have compact neighborhoods, which we call defects, can be found in [3], [4] and [5]. DOI: https://doi.org/10.29020/nybg.ejpam.v16i2.4767 Email address: mzailai@kau.edu.sa (M. Zailai) https://www.ejpam.com 1228 © 2023 EJPAM All rights reserved. M. Zailai / Eur. J. Pure Appl. Math, 16 (2) (2023), 1228-1235 1229 2. Local Compactness with Defects Definition 2.1. Let X be a topological space and let F be any property. If Y ⊂ X is the set of points which do not satisfy the property F , i.e., only X\Y has the property F then we call the topological space X a space with defects of type F . Remark 1. This paper concerns about defect of type local compactness. Throughout this paper we write l.c.w.d. for a space with defects of type local compactness. Definition 2.2. A T1 space X is a space with defects of type local compactness, l.c.w.d., if all points have compact neighborhoods except for some points. We denote by X• the set of points which do not possess compact neighborhoods. Remark 2. It is clear that if X• = ϕ then X is a locally compact space. We always assume that X• ̸= ϕ unless stated otherwise. Example 1. [6, 118 page 137] Denote T the graph of the function h(t) = sin(1/t) where 0 < t ≤ 1, as a subset of the Euclidean space R2 with the relative topology. The set T ⋆ = {(0, 0)} ∪ T is not locally compact since the point (0, 0) has no compact neighborhood. Therefore, the topological space T ⋆ is a locally compact space with defect of type local compactness. Proposition 2.3. Suppose that the topological spaces X and Y are l.c.w.d. If X is home- omorphic to Y then X• and Y • have same cardinality, i.e., the number of defect points is a topological invariant. Proof. It is enough to check that the homeomorphic image of any point in X• lies in Y •. Suppose that f : X → Y is a homeomorphism. Take any point x ∈ X•. If f(x) /∈ Y • then there is a neighborhood U of y = f(x) such that the closure U is a compact subspace. Now x ∈ f−1(U) and also have f−1(U) = f−1(U) which is compact. Therefore, f−1(U) is a compact neighborhood of the point x which is a contradiction as x is a defect point. Lemma 2.4. Let X be l.c.w.d then X• is a closed subset of X. Proof. It is sufficient to that the complement X\X• is open. If X• = ϕ, then X\X• = X is closed. If X• = X, then X\X• = ϕ is closed. Now suppose X• ̸= X and X• ̸= ϕ, take any arbitrary point x ∈ X\X•. Assume that for any neighborhood of Ux of x we have that Ux ∩X• = ϕ. Since x ∈ X\X•, then there is a compact neighborhood Ux of x. Let y ∈ X•, then y does not belong to Ux. Suppose otherwise, i.e., let y ∈ Ux. then y ∈ Ux. However, that means Ux is a compact neighborhood of y. This is a contradiction as y is a defect. Proposition 2.5. If X is locally compact with defects, then X• is dense-in-itself. Proof. If X• = ϕ then it is clearly dense-in-itself. Assume that X• ̸= ϕ. Take any point x ∈ X\X•, then x is not an accumulation point of X•. Now, let x ∈ X• then the closure X•\{x} is the set X•. Therefore, the set X• contains all of its accumulation points. Hence, X• is dense-in-itself. M. Zailai / Eur. J. Pure Appl. Math, 16 (2) (2023), 1228-1235 1230 Corollary 2.6. Let X be l.c.w.d such that X• ̸= ϕ then X is not a scattered space. Lemma 2.7. [2, page 17] Suppose that the family {Ws}s∈S is locally finite, then the following ⋃ s∈S Ws = ⋃ s∈S Ws is always true. Lemma 2.8. Let X be a T1 space and suppose space that the family {Ws}s∈S is locally finite except at a finite number of points, say a1, a2, ..., an, then we have ⋃ s∈S Ws n⋃ i=1 {ai} = ⋃ s∈S Ws n⋃ i=1 {ai}. Proof. It is clearly that ⋃ s∈S Ws n⋃ i=1 {ai} ⊂ ⋃ s∈S Ws n⋃ i=1 {ai}. Now suppose that x ∈ ⋃ s∈S Ws ∪ {a1}... ∪ {an}. Let x has an open neighborhood U which intersects finitely many members of the family {Fs, {a1}, {a2}, ..., {an}}s∈S . Let S1 = {s ∈ S : U ∩Ws} be the finite indexing set. It is clear that x ̸∈ ⋃ s∈S\S1 Ws. Note that x ∈ ⋃ s∈S1 Ws n⋃ i=1 {ai} ⋃ s ̸∈S\S1 Ws. Then, we get x ∈ ⋃ s∈S1 Ws ⋃n i=1{ai} = ⋃ s∈S1 Ws ⋃n i=1 {ai}. Therefore, x ∈ ⋃ s∈S Ws ∪ {a1} ∪ {a2} ∪ ... ∪ {an}. Assume that X does not have a locally finite neighborhood, then x = ai for some i. Therefore, it is easy to see that x = ai ∈ ⋃ s∈S Ws ∪ {a1} ∪ {a2} ∪ ... ∪ {an}. Hence, ⋃ s∈S Ws n⋃ i=1 {ai} = ⋃ s∈S Ws n⋃ i=1 {ai}. Remark 3. L.c.w.d. spaces do not have to be normal. Consider product of sorgenfrey line,X = K ×K, which is locally compact i.e. X• = ϕ but it is not normal. M. Zailai / Eur. J. Pure Appl. Math, 16 (2) (2023), 1228-1235 1231 Definition 2.9. [2, page 71] Let {Bi}i∈I be a cover of the space X. Consider any family of continuous maps {gi}i∈I , where gi : Bi → Y. The maps gi are said to be compatible if for every i1, i2 of I we have gi1 |Bi1 ⋂ Bi2 = gi2 |Bi1 ⋂ Bi2 . The combination is defined as g = ⋃ i∈I gi : X → Y. Remark 4. The following two Lemmas and Theorem are used in the proofs of Theorem 2.13 and Theorem 2.15. Lemma 2.10. [2, page 17] If {Si}i∈I is a locally finite closed cover of X and {gi}i∈I is a family of compatible maps, where gi : Si → Y . Then the combination is continuous. Lemma 2.11. Let W = {Bs}s∈S ∪ {{a1}, {a2}, ...{an}} be a closed cover of X such that the family W is locally finite except for a1, a2, ..., an. Let {gs}s∈S be a family of compatible maps, where gs : Bs → Y such that all members of the family are constant of the form gs(Bs) = k except for a finite number of members. Then, the combination g = ⋃ s∈S gs n⋃ i=1 fi : X → Y is continuous, where fi : {ai} → Y is defined as fi(ai) = k for i = 1, 2, ..., n. Proof. Let F be a closed subset of Y. We Want to show that the the inverse image of F is closed in X under that map g. Assume that k ̸∈ F , then g−1(F ) = ⋃ s∈S1 gs(F ) for a finite subset S1 ⊂ S. Therefore, g−1(F ) is a finite union of closed subsets, i.e., g−1 is closed. Now, let us assume that k ∈ F. We have g−1(F ) = ⋃ s∈S\S1 g−1 s (F ) n⋃ i=1 f−1 i (F ) = ⋃ s∈S\S1 g−1 s (F ) n⋃ i=1 {ai} Therefore, by using Lemma 2.8 we have that g−1(F ) is a closed subset of X. Hence, g is continuous. Theorem 2.12. [2, page 148] Let X be a T1 space such that any x ∈ X possesses a compact neighborhood. Then for any closed subset F ⊂ X such that x ̸∈ F there exists a continuous f : X → I where f(x) = 0 and f(F ) ⊂ {1}. Theorem 2.13. Let X• be a compact subset of X. Suppose that each point x ∈ X• has an open neighborhood U ̸= X such that the partition of singletons of the complement of X• ∪ (U\U) is locally finite, then X is T3 1 2 . M. Zailai / Eur. J. Pure Appl. Math, 16 (2) (2023), 1228-1235 1232 Proof. Let x ∈ X• and take any closed subset F of X such that x ̸∈ F. Let U ̸= X be an open neighborhood of x which satisfies assumption above. Define F0 = ((U\U) ∪ (U ∩ F )) ∩X•, which is a closed subset of the closed subspace X• such that x ̸∈ F0. Therefore, there exists a map f : X• → I such that f(x) = 0 and f(F0) ⊂ {1}. Let g : U\U → I be a constant map which is defined as g(y) = 1 for any y ∈ U\U. Let us define also the following maps fs∈S : {as} → I; as 7→ 1. Now, the combination h = f ∪ g ⋃ s∈S fs : X → I is continuous such that h(x) = 0 and h(F ) ⊂ {1}. Proposition 2.14. Let X be a second countable space such that X• is a discrete subspace. If X• is compact such that each of its points satisfies the assumption in theorem 2.13, then X• is of cardinality ℵ0. Proof. First from Theorem 2.13 we have that X is a T3 1 2 space which tells us that X is a regular space. Since every second countable regular space is metrizable, then X is a metrizable space. Separability and second countability are equivalent in metrizable spaces. Hence, X is a separable space. However, we know that every closed discrete subspace of a separable normal space has cardinality ≤ ℵ0. Theorem 2.15. Let X be a T1c l.c.w.d space such that for each point x ∈ X• there exists an open neighborhood U of x such that the closure U = ⋃ s∈S Fs is a union of compact subsets. If the family {Fs}s∈S is pairwise disjoint and locally finite except for a finite number of points, then X is T3 1 2 . Proof. Let x be a defect, i.e., x ∈ X•. Let F be closed such that x ̸∈ F. Take an open neighborhood U of x such that the closure U = ⋃ s∈S Fs is a union pairwise disjoint compact subsets, where W = {Fs}s∈S is locally finite except at a1, a2, ..., an. Note that x belongs to only one member of the family W, say Fsk for some sk ∈ S. Define F0 = ((U\U) ∪ (U ∩ F )) ∩ Fsk which is a closed subset of the subspace Fsk and we have that x ̸∈ F0. Therefore, there is a map fsk : Fsk → I such that fsk(x) = 0 and fsk(F0) ⊂ {1}. Define also following constant maps fs : Fs → I, y 7→ 1; for s ̸= sk g : X\U → I, y 7→ 1. M. Zailai / Eur. J. Pure Appl. Math, 16 (2) (2023), 1228-1235 1233 Now, suppose that one of the a′is is x, say am = x. Then, we define the following maps gi : {ai} → I, ai 7→ 1 for i = 1, ..., n and i ̸= m gm : {am} → I, am 7→ 0. If all a′is are distinct from x, then we define: gi : {ai} → I, ai 7→ 1 for i = 1, ..., n First, we need to check that the map h = ⋃ s∈S fs n⋃ i=1 gi ∪ g : X → I is continuous. Let C ⊂ I be closed. If 1 ̸∈ C, then h−1(C) = f−1 sk (C) is closed or h−1(C) = f−1 sk (C) ∪ g−1 m (C) which is also closed. Assume that 1 ∈ C, then h−1(C) =⋃ s∈S f−1 s (C) ⋃n i=1 g −1 i (C) ⋃ g−1(C) which is clearly closed by using Lemma 2.8. Hence, h is continuous. It clear that h(x) = 0. Now, take y ∈ F. If y ∈ F0, then we have h(y) = 1. Assume that y ̸∈ F0, then we have two cases: • Case 1: y ̸∈ Fsk , then it is easy to see that h(y) = 1, • Case 2: y ∈ Fsk , and y ̸∈ U which cannot happen as Fsk ⊂ U. Then, we conclude that if y ∈ F and y ̸∈ Fsk . Therefore, h(y) = 1. Hence, X is a T3 1 2 space. Proposition 2.16. Let {Xs}s∈S be a collection of pairwise disjoint l.c.w.d topological spaces. If each point xs ∈ Xs has an open neighborhood U such that its closure is a union of pairwise disjoint compact subsets, then so does each point x ∈ X = ⊕s∈SXs. Example 2. (Modified Arens-Fort Space): Here we modify the Arens-fort space. Let (A, τ) be the set of all ordered pairs of N×N. We declare that all the singletons of this set are open sets except the points (0, 0), (1, 0), ..., (n, 0) for some positive integer n. Let us define open neighborhoods of each point of {(0, 0), (1, 0), ..., (n, 0)} as any set U such that {(0, 0), (1, 0), ..., (n, 0)} ⊂ U , and all but a finite number of points of each but a finite number of the sets Td = {(l, d) : l is fixed and d ∈ N}. Note that this space is not locally compact as the points (0, 0), (1, 0), ..., (n, 0) do not possess compact neighborhoods. Let us check that the point (0, 0) does not have a compact neighborhood and all other points can be verified analogously. Let U be an open neighborhood of (0, 0). Consider the following U = {{as}s∈S , V } such that each as is distinct from all the points (0, 0), (1, 0), ..., (n, 0), and V is an open neighborhood of (0, 0) which is distinct from U in the following sense. If D = {(l, d) : l is fixed and l ̸= 0} ⊂ U , then we require that D ̸⊂ V. Now, U is an open cover of the closure U which does not have a finite open subcover. For any point x ∈ X• one can take X as a neighborhood. Now, X can be written REFERENCES 1234 as a union of singletons. Clearly, each one-point set in X is compact. Also, note that the partition of singletons is locally finite except for a finite number of points. Therefore, by using Theorem 2.15 we see that X is a Tychonoff space. We can also apply Theorem 2.13 to see that this space is a Tychonoff space. Namely, X• is finite, then is compact. Observe that partition of singletons of X\(X• ∪ (X\X) is locally finite. Proposition 2.17. Let {X1, X2, ..., Xn} be a collection of l.c.w.d. topological spaces. Sup- pose that for each topological space Xi we have X• ̸= ϕ ̸= Xi. Then for X = ∏n i=1Xi, we have X• ̸= ϕ ̸= X. Proof. It is straightforward. Proposition 2.18. Let each of {Xs}s∈S be a collection of l.c.w.d spaces such each point of Xs has an open neighborhood with closure being a union of pairwise disjoint compact subsets, then so does each point of the cartesian product ∏ s∈S Xs. Proof. It is straightforward. Proposition 2.19. Let X be l.c.w.d such that any x ∈ X• has a σ-compact neighborhood. Then for any closed subspace F ⊂ X, x ∈ X• ∩ F has a σ-compact neighborhood of the subspace F , i.e., this space is hereditarily with respect to closed subspaces. Proof. Take any x ∈ X• ∩ F , then there is an open neighborhood U ⊂ X of x such that U = ⋃∞ s=1 Fs where each Fs is compact as a subset of X. Observe that U ∩ F is an open of x in F such that its closure, (U ∩ F ) ∩ F , in F σ-compact. 3. Conclusion A well-known result in general topology states that any locally compact space is a Tychonoff space. In this paper we investigate a weaker version of local compactness. Instead of assuming that all points in a space have compact neighborhoods, we allow a possibility of having some points which do not possess compact neighborhoods. We denote by X• a set of points which do not have open neighborhoods with compact closures. One of the results we obtain is that by requiring the set X• to be compact, we show that if each point x ∈ X• has an open neighborhood U ̸= X such that the partition of singletons of the complement of X• ∪ (U\U) is locally finite, then the space is a Tychonoff space. The following questions are still not answered. Could we assume that X• is locally compact instead of being compact in Theorem 2.13? Could we drop the requirement of compact subsets need to be closed in Theorem 2.15? References [1] J Dugundji. Topology. Allyn and Bacon, INC, Boston, 1972. 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