Uniformity on generalized topological spaces

Abstract

Purpose

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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

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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

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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

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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.

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 $\phi$ ∈ 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:

$i_{\mu }(A)=\bigcup \left\{B\subseteq X:B\subseteq A\right.$ and $\left.B\in \mu \right\}$ and $c_{\mu }A=\bigcap \left\{B\subseteq X:A\subseteq B\right.$ and $\left.X\backslash B\in \mu \right\}$.

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 $U◦V=\left\{(x,y):(x,z)\in U\right.$ and (z, y) ∈ V, for some $\left.z\in X\right\}$.

Definition 1.2.

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

1. Δ ⊆ U, for every $U\in \mathcal{U}$.

2. $U,V\in \mathcal{U}\Rightarrow U\cap V\in \mathcal{U}$.

3. $U\in \mathcal{U}$ and $V\supseteq U\Rightarrow V\in \mathcal{U}$.

4. $U\in \mathcal{U}\Rightarrow U^{-1}\in \mathcal{U}$.

5. $U\in \mathcal{U}\Rightarrow$ there exists $V\in \mathcal{U}$ such that V◦V ⊆ U.

The pair $(X,\mathcal{U})$ 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, $U(A)=\left\{x\in X:(a,x)\in U\right.$, for some $\left.a\in A\right\}$. 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 $(X,\mathcal{U})$.Result 1.4.

Let $\mathcal{U}$ 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 $U_p\in \mathcal{U}$ such that U_p(p) ⊆ G. Then τ is a topology on X.

Definition 1.5.

[4] If $(X,\mathcal{U})$ is a uniform space the topology $\tau (\mathcal{U})$ of the uniformity $\mathcal{U}$, or the uniform topology, is the family of all subsets G of X such that for each x in G there is U in $\mathcal{U}$ 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 $\beta \subseteq \mathcal{P}(X)$. 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 ${\mathcal{U}}_{\mu }$ of subsets of X × X is called a μ-uniformity on X if

1. Δ ⊆ U for every $U\in {\mathcal{U}}_{\mu }$,

2. $U\in {\mathcal{U}}_{\mu }$ and $V\supseteq U\in {\mathcal{U}}_{\mu }\Rightarrow V\in {\mathcal{U}}_{\mu }$,

3. $U\in {\mathcal{U}}_{\mu }\Rightarrow$ there exists a symmetric $V\in {\mathcal{U}}_{\mu }$ such that V◦V ⊆ U.

The pair $(X,{\mathcal{U}}_{\mu })$ is called a μ-uniform space.Result 2.4.

Let $(X,{\mathcal{U}}_{\mu })$ be a μ-uniform space, then for any $U\in {\mathcal{U}}_{\mu },U\subseteq U◦U$.

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 $(X,{\mathcal{U}}_{\mu })$ be a μ-uniform space, then for any $U\in {\mathcal{U}}_{\mu },U^{-1}\in {\mathcal{U}}_{\mu }$.

Proof. Let $U\in {\mathcal{U}}_{\mu }$. Then by axiom (iii), there exists a symmetric $V\in {\mathcal{U}}_{\mu }$ 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), $U^{-1}\in {\mathcal{U}}_{\mu }$. □Result 2.6.

Every uniform space $(X,\mathcal{U})$ 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 $U\in \mathcal{U}$, then by axiom (v) of Definition 1.2 there exists $V\in \mathcal{U}$ such that V◦V ⊆ U; we set W = V ∩ V⁻¹. By axioms (ii) and (iv) of Definition 1.2, we see that $W\in \mathcal{U}$ , and it is also clear that W is symmetric and W◦W ⊆ U. Hence, $(X,\mathcal{U})$ 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 ${\mathcal{U}}_{\mu }=\left\{U\subseteq X\times X:A\subseteq U\right.$ or $B\left.\subseteq U\right\}$. It is clear that ${\mathcal{U}}_{\mu }$ is a μ-uniformity on X. But $A\cap B=\{(a,a),(b,b),(c,c)\}\notin {\mathcal{U}}_{\mu }$, 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 $\mathcal{U}$ of subsets of X × X is called a generalized quasi uniformity (or g-quasi uniformity) on X if the following hold:

1. $\mathrm{\Delta }\subseteq U,\forall U\in \mathcal{U}$.

2. $U\in \mathcal{U}$ and $U\subseteq V\Rightarrow V\in \mathcal{U}$.

3. $U\in \mathcal{U}\Rightarrow \exists V\in \mathcal{U}$ 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 ${\mathcal{U}}_{\mu }=\{V\subseteq X\times X:U\subseteq V\}$. It is clear that ${\mathcal{U}}_{\mu }$ is a g-quasi uniformity on X. Now $U\in {\mathcal{U}}_{\mu }$ but there does not exist any symmetric A ⊆ X × X in ${\mathcal{U}}_{\mu }$ such that A◦A ⊆ U. Hence $(X,{\mathcal{U}}_{\mu })$ 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 ${\mathcal{U}}_{\mu }$ 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 $U_p\in {\mathcal{U}}_{\mu }$ such that U_p(p) ⊆ G. Then τ_μ is a strong generalized topology on X.

Proof. Clearly $\phi$ ∈ τ_μ. For each p ∈ X, U(p) ⊆ X, for any $U\in {\mathcal{U}}_{\mu }$ 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 $U_p\in {\mathcal{U}}_{\mu }$ 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 ${\mathcal{U}}_{\mu }$ on X is called the generalized topology on X induced by ${\mathcal{U}}_{\mu }$ and will be denoted by $\tau ({\mathcal{U}}_{\mu })$.

Henceforth, the GTS $(X,\tau ({\mathcal{U}}_{\mu }))$ 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 $f:(X,\mu )\to (\mathbb{R},\nu )$ such that f(x) = 0 and f(A) = {1}, where ν is the generalized topology on the set $\mathbb{R}$ of reals generated by the base $\beta =\{(-\infty ,t):t\in \mathbb{R}\}\cup \{(t,\infty ):t\in \mathbb{R}\}$.

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 ${\mathcal{U}}_{\mu }$ on X such that $\mu =\tau ({\mathcal{U}}_{\mu })$. Let F be μ-closed and p ∉ F. Thus X \ F = W(say) is μ-open and p ∈ W, so there exists $U\in {\mathcal{U}}_{\mu }$ such that U(p) ⊆ W.

Now we shall show by induction that for every $n\in \mathbb{N}\cup \{0\}$, we can construct a symmetric member $U_n\in {\mathcal{U}}_{\mu }$ 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 $U_1\in {\mathcal{U}}_{\mu }$ 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 $U_n\in {\mathcal{U}}_{\mu }$ 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 $r=\frac{1}{2^{n_1}}+\frac{1}{2^{n_2}}+\cdots +\frac{1}{2^{n_m}}=\frac{p}{2^{n_m}}$, where p is some positive integer] r ∈ (0, 1], we define $V_r=U_{n_1}◦U_{n_2}◦\dots ◦U_{n_m}$, where $r={\mathrm{\Sigma }}_{i=1}^m{2}^{-n_i}$ 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 ${\mathcal{U}}_{\mu }$ and also note that V₁ = U₀. Then it can be shown that (Lemma 3.3 below)

$V_{k{2}^{-n}}\subseteq V_{k{2}^{-n}}◦U_n\subseteq V_{(k+1)2^{-n}}$ …(⋆)

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 $V_r=V_{i\cdot 2^{-n}}\subseteq V_{(i+1)2^{-n}}\subseteq \cdots \subseteq V_{j\cdot 2^{-n}}=V_s$. 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$$g(x)=\left\{\begin{array}{cc}\text{sup}\{r:x\notin V_r(p)\},\hfill & \text{for}x\ne p\hfill \\ 0,\hfill & \text{for}x=p.\hfill \end{array}\right.$$

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)} ≠ $\phi$. 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 $(\mathbb{R},\nu )$ 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 $n\in \mathbb{N}$ such that $2^n>\frac{2}{r}$. We show that U_n(x) ⊆ g⁻¹([0, t)), consequently $g^{-1}([0,t))\in \tau ({\mathcal{U}}_{\mu })=\mu$.

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, $x\notin V_{k{2}^{-n}}(p)\Rightarrow k\cdot 2^{-n}\in \{r:x\notin V_r(p)\}\Rightarrow s=\text{sup}\{r:x\notin V_r(p)\}\ge k\cdot 2^{-n}$, which is a contradiction. So $x\in V_{k{2}^{-n}}(p)\Rightarrow (p,x)\in V_{k{2}^{-n}}$. Also for y ∈ U_n(x) we get (x, y) ∈ U_n. Hence, $(p,y)\in V_{k{2}^{-n}}◦U_n\subseteq V_{(k+1)2^{-n}}$, by (a), and so $y\in V_{(k+1)2^{-n}}(p)$, and hence g(y) ≤ (k + 1)2⁻ⁿ. Therefore, $g(y)-s\le (k+1)2^{-n}-(k-1)2^{-n}=\frac{2}{2^n}<r=t-s$ i.e. g(y) < t ⇒ y ∈ g⁻¹([0, t)). Hence, U_n(x) ⊆ g⁻¹([0, t)), so $g^{-1}([0,t))\in \tau ({\mathcal{U}}_{\mu })=\mu$.

Next, for g⁻¹((t, 1]), let x ∈ g⁻¹((t, 1]), then g(x) = s > t ≥ 0. Let r = s − t > 0 and $n\in \mathbb{N}$ so that $2^n>\frac{2}{r}$. 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 $y\in V_{k{2}^{-n}}(p)$ (in fact otherwise, $y\notin V_{k{2}^{-n}}(p)\Rightarrow g(y)\ge k\cdot 2^{-n}$). Therefore $(p,y)\in V_{k{2}^{-n}}$ and since y ∈ U_n(x), (x, y) ∈ U_n and hence, as U_n is symmetric, (y, x) ∈ U_n. Thus $(p,x)\in V_{k{2}^{-n}}◦U_n\subseteq V_{(k+1)2^{-n}}$ [by (⋆)]. So, $x\in V_{(k+1)2^{-n}}(p)$. Consequently, g(x) ≤ (k + 1)2⁻ⁿ. Now $g(x)-t\le (k+1)2^{-n}-(k-1)2^{-n}=\frac{2}{2^n}<r\Rightarrow s-t<r$, a contradiction to the equality.

Hence U_n(x) ⊆ g⁻¹((t, 1]), so $g^{-1}((t,1])\in \tau ({\mathcal{U}}_{\mu })=\mu$. 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 $V_{k\cdot 2^{-n}}\subseteq V_{k\cdot 2^{-n}}◦U_n\subseteq V_{(k+1)2^{-n}}$ 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 $V_{k\cdot 2^{-n}}=V_{k\cdot 2^{-n}}◦\mathrm{\Delta }\subseteq V_{k\cdot 2^{-n}}◦U_n$ is always true, it remains only to prove $V_{k\cdot 2^{-n}}◦U_n\subseteq V_{(k+1)2^{-n}}$, 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 $V_{(k+1)\cdot 2^{-n}}=V_{m\cdot 2^{-(n-1)}}◦U_n=V_{k\cdot 2^{-n}}◦U_n$, 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 $V_{m\cdot 2^{-(n-1)}}◦U_{n-1}\subseteq V_{(m+1)\cdot 2^{-(n-1)}}$. …(*)

Since U_n◦U_n ⊆ U_n−1, it implies that $V_{k\cdot 2^{-n}}◦U_n=V_{m\cdot 2^{-(n-1)}+2^{-n}}◦U_n=V_{m\cdot 2^{-(n-1)}}◦U_n◦U_n\subseteq V_{m\cdot 2^{-(n-1)}}◦U_{n-1}$ and by using (*) we get $V_{k\cdot 2^{-n}}◦U_n\subseteq V_{(m+1)\cdot 2^{-(n-1)}}=V_{(k+1)\cdot 2^{-n}}$. 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 $(X,\mathcal{U})$, there is a result that a uniformity always induces a proximity on X which generates the same topology as is induced by $\mathcal{U}$ 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 $\mathcal{P}(X)$ of a set X is called a μ-proximity on X if δ_μ satisfies the following axioms:

1. Aδ_μB iff $B{\delta }_{\mu }A,\forall A,B\in \mathcal{P}(X)$

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 $(X,{\mathcal{U}}_{\mu })$ be a μ-uniform space. Then for A, B ⊆ X, U(A) ∩ U(B) ≠ $\phi$, for all $U\in {\mathcal{U}}_{\mu }$ if and only if U(A) ∩ B ≠ $\phi$ for all $U\in {\mathcal{U}}_{\mu }$.

Proof. Let U(A) ∩ B ≠ $\phi$. Since B ⊆ U(B) (as Δ ⊆ U), we get U(A) ∩ U(B) ≠ $\phi$ for all $U\in {\mathcal{U}}_{\mu }$.Conversely, let U(A) ∩ U(B) ≠ $\phi$ for all $U\in {\mathcal{U}}_{\mu }$ and if possible let there exist $V\in {\mathcal{U}}_{\mu }$ such that V(A) ∩ B = $\phi$. Now there exists a symmetric $W\in {\mathcal{U}}_{\mu }$ such that W◦W ⊆ V. By the given condition, W(A) ∩ W(B) ≠ $\phi$ 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 ≠ $\phi$, a contradiction. □Theorem 4.5.

For a μ-uniform space $(X,{\mathcal{U}}_{\mu })$, the relation δ_μ defined on $\mathcal{P}(X)$ by

Aδ_μB if and only if for every $U\in {\mathcal{U}}_{\mu },U(A)\cap U(B)\ne \phi$

is a μ-proximity structure on X such that $\tau ({\mathcal{U}}_{\mu })=\tau ({\delta }_{\mu })$.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 $U\in {\mathcal{U}}_{\mu }$, U(A) ∩ U(B) ≠ $\phi$. Now U(A) ⊆ U(C) and U(B) ⊆ U(D), therefore U(C) ∩ U(D) ≠ $\phi$. Hence Cδ_μD.

• (3) For all x ∈ X, x ∈ U(x) ∩ U(x), for all $U\in {\mathcal{U}}_{\mu }$ which implies U(x) ∩ U(x) ≠ $\phi$ for all $U\in {\mathcal{U}}_{\mu }$ and so {x}δ_μ{x}.

• (4) Let $A,B\in \mathcal{P}(X)$ such that A$\delta /$_μB. Then for some $U\in {\mathcal{U}}_{\mu }$, U(A) ∩ U(B) = $\phi$; we set C = U(A) and D = U(B). It is clear that A ⊆ C. We show that A$\delta /$_μ(X \ C). In fact, Aδ_μ(X \ C) ⇒ for every $V\in {\mathcal{U}}_{\mu }$, V(A) ∩ V(X \ U(A)) ≠ $\phi$. Let W be a symmetric member of ${\mathcal{U}}_{\mu }$ such that W◦W ⊆ U, then W(A) ∩ W(X \ U(A)) ≠ $\phi$ 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 A$\delta /$_μ(X \ C). Similarly, B ⊆ D and B$\delta /$_μ(X \ D), also as C ∩ D = U(A) ∩ U(B) = $\phi$, B$\delta /$_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 $\tau ({\mathcal{U}}_{\mu })=\tau ({\delta }_{\mu })$. Let A ⊆ X and x ∈ X. Then $x\in c_{\tau ({\mathcal{U}}_{\mu })}A\iff U(x)\cap A\ne \phi$, for all $U\in {\mathcal{U}}_{\mu }\iff U(x)\cap U(A)\ne \phi$, for all $U\in {\mathcal{U}}_{\mu }$ [by Lemma 4.4] $\iff \{x\}{\delta }_{\mu }A\iff x\in c_{\tau ({\delta }_{\mu })}A$ [by Proposition 4.3]. Thus, $\tau ({\mathcal{U}}_{\mu })=\tau ({\delta }_{\mu })$. □

Remark 4.6.

It is still an open problem whether a μ-proximity structure δ_μ on a set X induces a μ-uniformity ${\mathcal{U}}_{\mu }$ on X such that $\tau ({\mathcal{U}}_{\mu })=\tau ({\delta }_{\mu })$.

The authors are thankful to the referee for certain comments towards the improvement of the paper.