Uniformity on generalized topological spaces

Abstract

Purpose

The present article deals with the initiation and study of a uniformity like notion, captioned μ-uniformity, in the context of a generalized topological space.

Design/methodology/approach

The existence of uniformity for a completely regular topological space is well-known, and the interrelation of this structure with a proximity is also well-studied. Using this idea, a structure on generalized topological space has been developed, to establish the same type of compatibility in the corresponding frameworks.

Findings

It is proved, among other things, that a μ-uniformity on a non-empty set X always induces a generalized topology on X, which is μ-completely regular too. In the last theorem of the paper, the authors develop a relation between μ-proximity and μ-uniformity by showing that every μ-uniformity generates a μ-proximity, both giving the same generalized topology on the underlying set.

Originality/value

It is an original work influenced by the previous works that have been done on generalized topological spaces. A kind of generalization has been done in this article, that has produced an intermediate structure to the already known generalized topological spaces.

Citation

Dey, D., Mandal, D. and Mukherjee, M.N. (2022), "Uniformity on generalized topological spaces", Arab Journal of Mathematical Sciences, Vol. 28 No. 2, pp. 184-190. https://doi.org/10.1108/AJMS-03-2021-0058

1. Introduction and prerequisites

It was Császár [1] who first initiated the idea of generalized topological space. This opened up a new direction which was pursued by many mathematicians toward generalizations of many topological concepts to this new arena. A generalized topology (GT, for short) μ on a set X is a collection of subsets of X such that φ ∈ μ and arbitrary unions of members of μ belong to μ; and the ordered pair (X, μ) then stands for a generalized topological space (henceforth abbreviated as GTS). The sets in μ are called μ-open sets and their complements μ-closed sets. A GTS (X, μ) is called a strong GTS if X ∈ μ. For any subset A of a GTS (X, μ), the μ-interior i_μ(A) and μ-closure c_μ(A) of A are defined in the usual way as:

iμ(A)=⋃B⊆X:B⊆A and B∈μ and cμA=⋂B⊆X:A⊆B and X\B∈μ.

As is expected, μ-interior and μ-closure operators on a GTS (X, μ) obey the following basic properties:

1. i_μ(A) ⊆ A and A ⊆ c_μ(A), for all A ⊆ X.

2. A ⊆ B ⊆ X ⇒ i_μ(A) ⊆ i_μ(B) and c_μ(A) ⊆ c_μ(B).

3. A(⊆ X) is μ-open (μ-closed) if and only if A = i_μ(A) (resp. A = c_μ(A)).

4. i_μ(X \ A) = X \ c_μ(A), for all A ⊆ X.

The notion of uniformity is well-known for a topological space. This article is intended to initiate the study of a uniformity-like structure, termed μ-uniformity, on a generalized topological space.

In what follows in Section 2, we define μ-uniformities on a nonempty set X axiomatically and show that such a μ-uniformity induces a generalized topology on X. Although a μ-uniformity is not necessarily a uniformity. In Section 3, we also prove that a μ-uniform space satisfies a sort of complete regularity condition. Finally in Section 4, we establish that for a μ-uniform space, there exists a μ-proximity relation [2] such that the same generalized topology originates from both the structures.

We now recall the definition of uniformity on a set and some well-known relevant results thereof; related details may be found in [3].

Definition 1.1.

Let X be a non-empty set:

1. A non-void subset of X × X is called a binary relation on X.

2. The identity relation on X is called the diagonal in X × X and is denoted by Δ(X) or simply by Δ. Thus Δ = {(x, x) : x ∈ X}.

3. The inverse of a relation U, denoted by U⁻¹, is defined by U⁻¹ = {(y, x) : (x, y) ∈ U}.

4. A relation U is said to be symmetric if U = U⁻¹.

5. The composition of two relations U and V, denoted by U◦V, is defined by U◦V=(x,y):(x,z)∈U and (z, y) ∈ V, for some z∈X.

Definition 1.2.

Let X be a non-empty set. A non-void family U of subsets of X × X, is said to be a uniformity on X if the following conditions hold:

1. Δ ⊆ U, for every U∈U.

2. U,V∈U⇒U∩V∈U.

3. U∈U and V⊇U⇒V∈U.

4. U∈U⇒U−1∈U.

5. U∈U⇒ there exists V∈U such that V◦V ⊆ U.

The pair (X,U) is called a uniform space.

Definition 1.3.

Let U be a binary relation on X and A a non-void subset of X. Then we define, U(A)=x∈X:(a,x)∈U, for some a∈A. In particular, if A = {p}, for some p ∈ X, then U(p) = U({p}) = {x ∈ X : (p, x) ∈ U}.

Now we state some well-known results for a uniform space (X,U).

Result 1.4.

Let U be a uniformity on a non-void set X. Let a family τ of subsets of X be defined as follows: A subset G of X belongs to τ if and only if to every element p ∈ G, there corresponds some Up∈U such that U_p(p) ⊆ G. Then τ is a topology on X.

Definition 1.5.

[4] If (X,U) is a uniform space the topology τ(U) of the uniformity U, or the uniform topology, is the family of all subsets G of X such that for each x in G there is U in U such that U(x) ⊆ G.

Result 1.6.

A topological space (X, τ) is uniformizable if and only if it is completely regular.

2. μ-uniformity

Before going into the details we first state two definitions which will be required later on.

Definition 2.1.

[5] Let X be a non-empty set and β⊆P(X). Then β is called a base for a generalized topology μ on X if μ = {∪β′ : β′ ⊆ β}.

Definition 2.2.

[6] Let (X, μ) and (Y, ξ) be two generalized topological spaces. A function f : (X, μ) → (Y, ξ) is said to be μ-continuous if for any G ∈ ξ, f⁻¹(G) ∈ μ.

In [7] the concept of generalized quasi uniformity was introduced, termed as g-quasi uniformity. In the same manner, we introduce the definition of μ-uniformity as follows.

Definition 2.3.

Let X be a non-empty set. A non-void family Uμ of subsets of X × X is called a μ-uniformity on X if

1. Δ ⊆ U for every U∈Uμ,

2. U∈Uμ and V⊇U∈Uμ⇒V∈Uμ,

3. U∈Uμ⇒ there exists a symmetric V∈Uμ such that V◦V ⊆ U.

The pair (X,Uμ) is called a μ-uniform space.

Result 2.4.

Let (X,Uμ) be a μ-uniform space, then for any U∈Uμ,U⊆U◦U.

Proof.

Let (x, y) ∈ U. Then as (y, y) ∈ U[from (i)], we have (x, y) ∈ U◦U, hence U ⊆ U◦U. □

Proposition 2.5.

Let (X,Uμ) be a μ-uniform space, then for any U∈Uμ,U−1∈Uμ.

Proof. Let U∈Uμ. Then by axiom (iii), there exists a symmetric V∈Uμ such that V◦V ⊆ U. Again by Result 2.4, V ⊆ V◦V which implies V ⊆ U and so V⁻¹ ⊆ U⁻¹, i.e. V ⊆ U⁻¹ [since V is symmetric]. So by axiom (ii), U−1∈Uμ. □

Result 2.6.

Every uniform space (X,U) is a μ-uniform space.

Proof. Axioms (i) and (ii) of Definition 2.3 are obvious from the definition of uniformity given in Definition 1.2. Now for axiom (iii) of Definition 2.3, consider U∈U, then by axiom (v) of Definition 1.2 there exists V∈U such that V◦V ⊆ U; we set W = V ∩ V⁻¹. By axioms (ii) and (iv) of Definition 1.2, we see that W∈U , and it is also clear that W is symmetric and W◦W ⊆ U. Hence, (X,U) is a μ-uniform space. □

Note 2.7.

The converse of the above stated result is false i.e. a μ-uniformity on a set X need not be a uniformity on X. In fact, consider X = {a, b, c} and A = {(a, a), (b, b), (c, c), (a, b), (b, a)}, B = {(a, a), (b, b), (c, c), (c, b), (b, c)}. We set Uμ=U⊆X×X:A⊆U or B⊆U. It is clear that Uμ is a μ-uniformity on X. But A∩B={(a,a),(b,b),(c,c)}∉Uμ, which does not satisfy (ii) of Definition 1.2, and hence it is not a uniformity.

Definition 2.8.

[7] Let X be a nonempty set. A nonempty family U of subsets of X × X is called a generalized quasi uniformity (or g-quasi uniformity) on X if the following hold:

1. Δ⊆U,∀U∈U.

2. U∈U and U⊆V⇒V∈U.

3. U∈U⇒∃V∈U such that V◦V ⊆ U.

Remark 2.9.

It is a straightforward to observe that every μ-uniform space is also a g-quasi uniform space as defined in [7]. But the converse is not true.

Consider the set X = {a, b, c} and the subset U of X × X given by U = {(a, a), (b, b), (c, c), (a, b)}. Set Uμ={V⊆X×X:U⊆V}. It is clear that Uμ is a g-quasi uniformity on X. Now U∈Uμ but there does not exist any symmetric A ⊆ X × X in Uμ such that A◦A ⊆ U. Hence (X,Uμ) is not a μ-uniform space.

So the family of all μ-uniform spaces is coarser than the family of all g-quasi uniform spaces but finer than the collection of all uniform spaces.

Theorem 2.10.

Let Uμ be a μ-uniformity on a non-empty set X. Let a family τ_μ of subsets of X be defined by:

A subset G ∈ τ_μ if and only if for every p ∈ G, there exists some Up∈Uμ such that U_p(p) ⊆ G. Then τ_μ is a strong generalized topology on X.

Proof. Clearly φ ∈ τ_μ. For each p ∈ X, U(p) ⊆ X, for any U∈Uμ so X ∈ τ_μ.

Let G_α ∈ τ_μ, where α ∈ Λ, an index set. Let G = ⋃_α∈ΛG_α and p ∈ G. Then p ∈ G_β for some β ∈ Λ, so there exists Up∈Uμ such that U_p(p) ⊆ G_β ⊆ G. Hence, G ∈ τ_μ.

So, τ_μ is a strong generalized topology on X. □

Definition 2.11.

The generalized topology τ_μ obtained in the previous theorem from the μ-uniformity Uμ on X is called the generalized topology on X induced by Uμ and will be denoted by τ(Uμ).

Henceforth, the GTS (X,τ(Uμ)) will be called a μ-uniform space.

3. μ-uniformity and μ-complete regularity

Definition 3.1.

[2] A GTS (X, μ) is said to be μ-completely regular if for any μ-closed set A in X and for x∉A, there exists a μ-continuous function f:(X,μ)→(R,ν) such that f(x) = 0 and f(A) = {1}, where ν is the generalized topology on the set R of reals generated by the base β={(−∞,t):t∈R}∪{(t,∞):t∈R}.

Theorem 3.2.

A μ-uniformizable GTS (X, μ) is μ-completely regular.

Proof. Given that the GTS (X, μ) is μ-uniformizable, i.e. there exists a μ-uniformity Uμ on X such that μ=τ(Uμ). Let F be μ-closed and p ∉ F. Thus X \ F = W(say) is μ-open and p ∈ W, so there exists U∈Uμ such that U(p) ⊆ W.

Now we shall show by induction that for every n∈N∪{0}, we can construct a symmetric member Un∈Uμ such that U_n ⊆ U and U_n◦U_n ⊆ U_n−1 ⊆ U, when n is positive with U = U₀.

In fact, let U = U₀; then there exists a symmetric U1∈Uμ such that U₁◦U₁ ⊆ U₀, where U₁ = U₁◦Δ ⊆ U₁◦U₁ ⊆ U₀. Let U_n−1 have been constructed in this way, then there exists a symmetric Un∈Uμ such that U_n◦U_n ⊆ U_n−1 and similarly U_n = U_n◦Δ ⊆ U_n◦U_n ⊆ U_n−1 ⊆ U. So, we get a decreasing sequence {U_n : n ≥ 0} with each member being a subset of U.

Next for every diadic rational [A diadic rational number r is of the form r=12n1+12n2+⋯+12nm=p2nm, where p is some positive integer] r ∈ (0, 1], we define Vr=Un1◦Un2◦…◦Unm, where r=Σi=1m2−ni with 0 ≤ n₁ < n₂ < … < n_m; since every diadic rational number has unique expression, V_r is well-defined. We define V₀ = Δ, though it may not be in Uμ and also note that V₁ = U₀. Then it can be shown that (Lemma 3.3 below)

Vk2−n⊆Vk2−n◦Un⊆V(k+1)2−n …(⋆)

which holds for every non-negative n and all k = 0, 1, …, 2ⁿ − 1. Also for two diadic rational numbers r, s with 0 ≤ r ≤ s ≤ 1, there exists positive integer n such that r = i ⋅ 2⁻ⁿ and s = j ⋅ 2⁻ⁿ, where i, j are positive integers satisfying 0 ≤ i ≤ j ≤ 2ⁿ.

Hence, we have Vr=Vi⋅2−n⊆V(i+1)2−n⊆⋯⊆Vj⋅2−n=Vs. Thus if 0 ≤ r ≤ s ≤ 1 and r, s are diadic rationals then V_r ⊆ V_s.

Next, we define a function g: X → [0, 1] by taking

g(x)=sup{r:x∉Vr(p)},forx≠p0,forx=p.

Since V₀ = Δ, V₀(p) = {p}. For each x( ≠ p) ∈ X, x∉V₀(p) ⇒ 0 ∈{r : x∉V_r(p)}⇒{r : x∉V_r(p)} ≠ φ. Also, r ≤ 1 ⇒{r : x∉V_r(p)} is bounded above and so its supremum exists.

Now for any point q ∈ F, i.e. q ∈ X \ W, we have q∉V₁(p), as U(p) ⊆ W and V₁ = U₀ ⊆ U. Again, q∉V₁(p) ⇒ 1 ∈{r : q∉V_r(p), r ≤ 1}⇒ g(q) = 1.

Finally, we shall show that g is μ-continuous in (X, μ). For this it is enough to show that g⁻¹([0, t)) and g⁻¹((t, 1]) are μ-open [since [0, t), (t, 1] are the basic μ-open sets of [0, 1] where t ∈ (0, 1), when it is considered as a subspace of the GTS (R,ν) defined previously]. Let x ∈ g⁻¹([0, t)), then g(x) ∈ [0, t); let us take g(x) = s then s < t ≤ 1. We set r = t − s > 0, now there exists n∈N such that 2n>2r. We show that U_n(x) ⊆ g⁻¹([0, t)), consequently g−1([0,t))∈τ(Uμ)=μ.

Now let k be the uniquely determined positive integer satisfying k − 1 ≤ s ⋅ 2ⁿ < k i.e. (k − 1)2⁻ⁿ ≤ s < k ⋅ 2⁻ⁿ, then g(x) = s < k ⋅ 2⁻ⁿ. Now, x∉Vk2−n(p)⇒k⋅2−n∈{r:x∉Vr(p)}⇒s=sup{r:x∉Vr(p)}≥k⋅2−n, which is a contradiction. So x∈Vk2−n(p)⇒(p,x)∈Vk2−n. Also for y ∈ U_n(x) we get (x, y) ∈ U_n. Hence, (p,y)∈Vk2−n◦Un⊆V(k+1)2−n, by (a), and so y∈V(k+1)2−n(p), and hence g(y) ≤ (k + 1)2⁻ⁿ. Therefore, g(y)−s≤(k+1)2−n−(k−1)2−n=22n<r=t−s i.e. g(y) < t ⇒ y ∈ g⁻¹([0, t)). Hence, U_n(x) ⊆ g⁻¹([0, t)), so g−1([0,t))∈τ(Uμ)=μ.

Next, for g⁻¹((t, 1]), let x ∈ g⁻¹((t, 1]), then g(x) = s > t ≥ 0. Let r = s − t > 0 and n∈N so that 2n>2r. We shall show that U_n(x) ⊆ g⁻¹((t, 1]). Let k be the uniquely determined positive integer satisfying (k − 1)2⁻ⁿ ≤ t < k ⋅ 2⁻ⁿ. If possible, let y ∈ U_n(x) and y∉g⁻¹((t, 1]). Then g(y) ≤ t < k ⋅ 2⁻ⁿ and so y∈Vk2−n(p) (in fact otherwise, y∉Vk2−n(p)⇒g(y)≥k⋅2−n). Therefore (p,y)∈Vk2−n and since y ∈ U_n(x), (x, y) ∈ U_n and hence, as U_n is symmetric, (y, x) ∈ U_n. Thus (p,x)∈Vk2−n◦Un⊆V(k+1)2−n [by (⋆)]. So, x∈V(k+1)2−n(p). Consequently, g(x) ≤ (k + 1)2⁻ⁿ. Now g(x)−t≤(k+1)2−n−(k−1)2−n=22n<r⇒s−t<r, a contradiction to the equality.

Hence U_n(x) ⊆ g⁻¹((t, 1]), so g−1((t,1])∈τ(Uμ)=μ. Hence, g is μ-continuous and so (X, μ) is μ-completely regular. □

Lemma 3.3.

Following the same notations as in Theorem 3.2, the inclusion relation Vk⋅2−n⊆Vk⋅2−n◦Un⊆V(k+1)2−n holds for every non-negative integer n and for k = 0, 1, 2, …, 2ⁿ − 1.

Proof. This relation holds for n = 0, since for n = 0, k = 0 and V₀ = Δ so that V₀◦U₀ = U₀ = V₁. Let n > 0 and we assume that the inclusions hold for n − 1. We shall prove the inclusions for n. Since Vk⋅2−n=Vk⋅2−n◦Δ⊆Vk⋅2−n◦Un is always true, it remains only to prove Vk⋅2−n◦Un⊆V(k+1)2−n, for k = 0, 1, 2, …, 2ⁿ − 1.

If k is an even integer, say k = 2m, we have k ⋅ 2⁻ⁿ = (2m) ⋅ 2⁻ⁿ = m ⋅ 2⁻⁽ⁿ⁻¹⁾, i.e. (k + 1) ⋅ 2⁻ⁿ = m ⋅ 2⁻⁽ⁿ⁻¹⁾ + 2⁻ⁿ = (2m + 1) ⋅ 2⁻ⁿ.

It then follows from the definition of the sets V_r, given in Theorem 3.2, that V(k+1)⋅2−n=Vm⋅2−(n−1)◦Un=Vk⋅2−n◦Un, thus the inclusion is proved in this case.

If k is an odd integer, say k = 2m + 1, then k ⋅ 2⁻ⁿ = (2m + 1) ⋅ 2⁻ⁿ = m ⋅ 2⁻⁽ⁿ⁻¹⁾ + 2⁻ⁿ and (k + 1) ⋅ 2⁻ⁿ = (2m + 2) ⋅ 2⁻ⁿ = (m + 1) ⋅ 2⁻⁽ⁿ⁻¹⁾. By our induction hypothesis, we get Vm⋅2−(n−1)◦Un−1⊆V(m+1)⋅2−(n−1). …(*)

Since U_n◦U_n ⊆ U_n−1, it implies that Vk⋅2−n◦Un=Vm⋅2−(n−1)+2−n◦Un=Vm⋅2−(n−1)◦Un◦Un⊆Vm⋅2−(n−1)◦Un−1 and by using (*) we get Vk⋅2−n◦Un⊆V(m+1)⋅2−(n−1)=V(k+1)⋅2−n. Thus, the inclusion also holds for odd integers. □

Remark 3.4.

It is still an open problem whether a μ-completely regular GT is μ-uniformizable.

4. μ-uniformity and μ-proximity

In a uniform space (X,U), there is a result that a uniformity always induces a proximity on X which generates the same topology as is induced by U on X. In the following theorem, we also have a similar result for a GTS. First we state the definition of μ-proximity.

Definition 4.1.

[2] A binary relation δ_μ on the power set P(X) of a set X is called a μ-proximity on X if δ_μ satisfies the following axioms:

1. Aδ_μB iff BδμA,∀A,B∈P(X)

2. If Aδ_μB, A ⊆ C and B ⊆ D, then Cδ_μD

3. {x}δ_μ{x}, ∀x ∈ X

4. Aδ∕_μB ⇒∃ E(⊆ X) such that Aδ∕_μE and (X \ E)δ∕_μB.

Now δ_μ generates a generalized topology on X which is given below:

Proposition 4.2.

[2] Let a subset A of a μ-proximity space (X, δ_μ) be defined to be δ_μ-closed iff ({x}δ_μA ⇒ x ∈ A). Then the collection of complements of all δ_μ-closed sets so defined, yields a generalized topology μ = τ(δ_μ) on X.

Proposition 4.3.

[2] Let (X, δ_μ) be a μ-proximity space and μ = τ(δ_μ). Then the μ-closure c_μ(A) of a set A in (X, μ) is given by c_μ(A) = {x : {x}δ_μA}.

Lemma 4.4.

Let (X,Uμ) be a μ-uniform space. Then for A, B ⊆ X, U(A) ∩ U(B) ≠ φ, for all U∈Uμ if and only if U(A) ∩ B ≠ φ for all U∈Uμ.

Proof. Let U(A) ∩ B ≠ φ. Since B ⊆ U(B) (as Δ ⊆ U), we get U(A) ∩ U(B) ≠ φ for all U∈Uμ.Conversely, let U(A) ∩ U(B) ≠ φ for all U∈Uμ and if possible let there exist V∈Uμ such that V(A) ∩ B = φ. Now there exists a symmetric W∈Uμ such that W◦W ⊆ V. By the given condition, W(A) ∩ W(B) ≠ φ and let p ∈ W(A) ∩ W(B), i.e. (a, p) ∈ W and (b, p) ∈ W for some a ∈ A, b ∈ B. Since W is symmetric, we get (a, b) ∈ W◦W ⊆ V which implies b ∈ V(a) ⊆ V(A). Thus V(A) ∩ B ≠ φ, a contradiction. □

Theorem 4.5.

For a μ-uniform space (X,Uμ), the relation δ_μ defined on P(X) by

Aδ_μB if and only if for every U∈Uμ,U(A)∩U(B)≠φ

is a μ-proximity structure on X such that τ(Uμ)=τ(δμ).

Proof.

To show that δ_μ is a μ-proximity on X we proceed in the following manner:

• (1) For A, B ⊆ X, clearly Aδ_μB iff Bδ_μA.

• (2) Let Aδ_μB with A ⊆ C and B ⊆ D, so for any U∈Uμ, U(A) ∩ U(B) ≠ φ. Now U(A) ⊆ U(C) and U(B) ⊆ U(D), therefore U(C) ∩ U(D) ≠ φ. Hence Cδ_μD.

• (3) For all x ∈ X, x ∈ U(x) ∩ U(x), for all U∈Uμ which implies U(x) ∩ U(x) ≠ φ for all U∈Uμ and so {x}δ_μ{x}.

• (4) Let A,B∈P(X) such that Aδ/_μB. Then for some U∈Uμ, U(A) ∩ U(B) = φ; we set C = U(A) and D = U(B). It is clear that A ⊆ C. We show that Aδ/_μ(X \ C). In fact, Aδ_μ(X \ C) ⇒ for every V∈Uμ, V(A) ∩ V(X \ U(A)) ≠ φ. Let W be a symmetric member of Uμ such that W◦W ⊆ U, then W(A) ∩ W(X \ U(A)) ≠ φ and so there exists p ∈ W(A) ∩ W(X \ U(A)). Therefore, there exists a ∈ A, b ∈ X \ U(A) such that (a, p) ∈ W and (b, p) ∈ W, now W being symmetric, (a, b) ∈ W◦W ⊆ U which implies b ∈ U(a) ⊆ U(A), a contradiction to the fact that b ∈ X \ U(A). Thus Aδ/_μ(X \ C). Similarly, B ⊆ D and Bδ/_μ(X \ D), also as C ∩ D = U(A) ∩ U(B) = φ, Bδ/_μC. In fact, if Bδ_μC then as C ⊆ (X \ D) that implies Bδ_μ(X \ D) [using (ii) in this proof shown above], a contradiction. Thus, we see that axiom (iv) of μ-proximity is satisfied.

Finally, we show that τ(Uμ)=τ(δμ). Let A ⊆ X and x ∈ X. Then x∈cτ(Uμ)A⇔U(x)∩A≠φ, for all U∈Uμ⇔U(x)∩U(A)≠φ, for all U∈Uμ [by Lemma 4.4] ⇔{x}δμA⇔x∈cτ(δμ)A [by Proposition 4.3]. Thus, τ(Uμ)=τ(δμ). □

Remark 4.6.

It is still an open problem whether a μ-proximity structure δ_μ on a set X induces a μ-uniformity Uμ on X such that τ(Uμ)=τ(δμ).