A study of uniformities on the space of uniformly continuous mappings

Ankit Gupta; Abdulkareem Saleh Hamarsheh; Ratna Dev Sarma; Reny George

doi:10.1515/math-2020-0110

Article Open Access

A study of uniformities on the space of uniformly continuous mappings

Ankit Gupta , Abdulkareem Saleh Hamarsheh , Ratna Dev Sarma and Reny George

Published/Copyright: December 13, 2020

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$Open Mathematics$

From the journal Open Mathematics Volume 18 Issue 1

Abstract

New families of uniformities are introduced on U C ( X , Y ) , the class of uniformly continuous mappings between X and Y, where ( X , U ) and ( Y , V ) are uniform spaces. Admissibility and splittingness are introduced and investigated for such uniformities. Net theory is developed to provide characterizations of admissibility and splittingness of these spaces. It is shown that the point-entourage uniform space is splitting while the entourage-entourage uniform space is admissible.

Keywords: uniform space; function space; splittingness; admissibility; uniformly continuous mappings

MSC 2010: 54C35; 54A20

1 Introduction

The function space C ( X , Y ) , where X , Y are topological spaces, can be equipped with various interesting topologies. Properties of these topologies vis-a-vis that of X and Y have been an active area of research in recent years. In [1], it is shown that several fundamental properties hold for a hyperspace convergence τ on C ( X , $ ) at X if and only if they hold for τ ⇑ on C ( X , ℝ ) at origin (where $ is the Sierpinski topology and τ ⇑ is the convergence on C ( X , ℝ ) determined by τ ). In [2], function space topologies are introduced and investigated for the space of continuous multifunctions between topological spaces. In [3], some conditions are discussed under which the compact-open, Isbell or natural topologies on the set of continuous real-valued functions on a space may coincide. Properties of c-compact-open topology on the C ( X ) such as metrizability, separability, and second countability have been discussed in [4]. Function space topologies over the generalized topological spaces, defined by Császár, are introduced and studied in [5]. Their dual topologies have been investigated in [6]. Similarly, the space C ( X ) , the space of continuous mappings from X to ℝ , where X is a completely regular or a Tychonoff space, has also been studied by several researchers in recent years [7,8]. It is well known that every metric space has a uniformity induced by its metric, but not every uniform space is metrizable. Similarly, every uniformity induces a topology. But not every topological space is uniformizable. The metric topology of a space can be derived purely from the properties of the induced uniform space via its uniform topology. In this sense, uniformities are positioned between metric spaces and topological structures. Hence, it is natural to look for similar studies for uniform spaces also. However, not much literature is available so far regarding function space uniformities over uniform spaces. In most of these studies, the topologies and not the uniformities of the underlying spaces are considered. For example, in [9], a quasi-uniformity is formed over a topological space X. In [10], fuzzy topology on X is considered for the space C f ( X , Y ) . Similar is the case in [11,12]. None of these studies relate to uniformities over uniform spaces, per se.

In the present paper, we provide a study of the possible uniform structures on the space of uniformly continuous mappings between uniform spaces. We have verified the existence of two such families, namely, entourage-entourage uniformities and point-entourage uniformities, for the space of uniformly continuous mappings. Our present study is centered around developing the well-known topological concepts of function spaces such as admissibility and splittingness for the function space uniformities over uniform structures. Unlike in [13–15], net theory has been used as a tool in our study. For this purpose, we have introduced the concept of pairwise Cauchy nets for uniformities. This has helped us develop a net theoretic characterization for uniform continuity between uniformities. All the concepts introduced and studied in Section 3 are new, although similar concepts do exist for function space topologies between topological spaces. It is found that a uniformity on U C ( Y , Z ) is splitting if and only if every pair of nets in U C ( Y , Z ) is pairwise Cauchy whenever it is continuously Cauchy. On the other hand, a uniformity is admissible if and only if every pair of nets in U C ( Y , Z ) is continuously Cauchy whenever it is pairwise Cauchy. While the point-entourage uniformity is splitting, entourage-entourage uniformity is found to be admissible. Several examples are provided to explain the theory developed in the paper. The successful application of net theory in the entire investigation testifies that like in topology, net theory is an effective tool for uniformities too. We have concluded the present work with some open questions for future work.

2 Preliminaries

A uniform space is a non-empty set with a uniform structure on it. A uniform structure (or a uniformity) on a set X is a collection of subsets of X × X satisfying certain conditions. More precisely, we have the following definition.

Definition 2.1

[16,17] A uniform structure or uniformity on a non-empty set X is a family U of subsets of X × X satisfying the following properties:

(2.1.1) if U ∈ U , then Δ X ⊆ U ;

where Δ X = { ( x , x ) ∈ X × X for all x ∈ X } ;

(2.1.2) if U ∈ U , then U − 1 ∈ U ,

where U − 1 is called the inverse relation of U and is defined as:

U − 1 = { ( x , y ) ∈ X × X | ( y , x ) ∈ U } ;

(2.1.3) if U ∈ U , then there exists some V ∈ U such that V ∘ V ⊆ U ,

where the composition U ∘ V = { ( x , z ) ∈ X × X | for some y ∈ X , ( x , y ) ∈ V , and ( y , z ) ∈ U } ;

(2.1.4) if U , V ∈ U , then U ∩ V ∈ U ;

(2.1.5) if U ∈ U and U ⊆ V ⊆ X × X , then V ∈ U .

The pair ( X , U ) is called a uniform space and the members of U are called entourages.

Remark 2.1

There are two more approaches to define uniformity on a set. One of them [18] uses a certain specification of a system of coverings on X. The other is via a system of pseudo-metrics. The one we have provided here is originally due to Weil [19]. This definition centers around the idea of closeness of points of X. In metric spaces, the metric defines the closeness between points. However, in topological spaces, we can only talk of a point being arbitrarily close to a set (i.e., being in the closure of the set). In a uniform space, as defined above, the closeness of points x and y is equivalent to the ordered pair ( x , y ) belonging to some entourage.

Definition 2.2

[20] A subfamily ℬ of a uniformity U is called a base for U if each member of U contains a member of ℬ .

In view of (2.1.5), a base is enough to specify the corresponding uniformity unambiguously as the uniformity consists of just the supersets of members of ℬ . Also, every uniformity possesses a base.

Definition 2.3

[20] A subfamily S of a uniformity U is called subbase for U if the family of finite intersections of members of S is a base for U .

The finite intersection of the members of a subbase generates a base. A uniformity is obtained by taking the collection of the supersets of the members of its base.

Remark 2.2

The aforementioned definitions of base and subbase are similar to those of a topology. In fact, if we replace uniformity U by a topology τ in Definitions 2.2 and 2.3, we get the definitions of base and subbase of topology τ on X. As in topology, these definitions help us to restrict our study to a smaller collection of subsets.

The conditions under which a collection of subsets of X × X becomes a base (respectively, a subbase) of a uniformity on X are provided in the following theorems.

Theorem 2.4

[20] A non-empty family U of subsets of X × X is a base for some uniformity for X if and only if the aforementioned conditions (2.1.1)–(2.1.4) hold.

Theorem 2.5

[20] A non-empty family U of subsets of X × X is a subbase for some uniformity for X if and only if the aforementioned conditions (2.1.1)–(2.1.3) hold.

In particular, the union of any collection of uniformities for X forms a subbase for a uniformity for X. The fact that a subbase (respectively, a base) uniquely defines a uniformity is being utilized in this paper for defining new uniformities.

Remark 2.3

The aforementioned two results provide us simplified methods to check whether a given collection of subsets of X × X qualifies to generate a uniformity on X.

In fact, Theorem 2.5 is used in this paper in Lemmas 3.4 and 3.5 to establish the existence of the point-entourage uniformity and the entourage-entourage uniformity on U C ( X , Y ) and U C ( Y , Z ) , respectively.

Definition 2.6

[20] Let ( X , U ) and ( Y , V ) be two uniform spaces. A mapping f : X → Y is called uniformly continuous if for each V ∈ V , there exists U ∈ U such that f 2 [ U ] ⊂ V , where f 2 : X × X → Y × Y is a map corresponding to f defined as f 2 ( x , x ′ ) = ( f ( x ) , f ( x ′ ) ) for ( x , x ′ ) ∈ X × X .

In other words, f : X → Y is uniformly continuous if for each V = V 1 × V 2 ∈ V , there exists U = U 1 × U 2 ∈ U such that f ( U 1 ) × f ( U 2 ) ⊆ V 1 × V 2 .

The collection of all uniformly continuous functions from X to Y is denoted by U C ( X , Y ) .

3 The main results

The development of this section is as follows. In Section 3.1, we first define pairwise Cauchy nets and then use them to characterize uniform continuity. In Section 3.2, we establish the existence of uniformities in U C ( X , Y ) . Two such uniformities on U C ( X , Y ) are point-entourage and entourage-entourage uniformities, respectively. Next we define admissibility and splittingness for such uniformities on U C ( X , Y ) . Net theory has been extensively used to provide alternative characterizations for these notions. Finally, we prove that point-entourage uniformity on U C ( X , Y ) is splitting, while entourage-entourage uniformity is admissible.

Here, it may be mentioned that ℐ -Cauchy nets and ℐ -convergence using ideal of the directed sets were introduced in [21] for uniform spaces. However, convergence of ℐ -nets was defined there using open sets of the corresponding topology. In our paper, we are using nets without any such restrictions. The convergence defined here is purely in terms of uniformity. We have not come across any results in the literature resembling the net-theoretic characterization of uniform continuity provided here. Splittingness and admissibility have been studied by several authors for the function space topologies [13,17,21]. In [16], it has been proved that point-open topology is splitting and open-open topology is admissible. In this section, we extend these notions to uniformities on the space of uniformly continuous mappings. We provide characterizations of these notions using net theory. We also provide examples of splitting and admissible uniformities. The successful development of net theory and its effective applications here have established that like in topology, and net theory is an useful tool for studying uniform structures.

3.1 Uniformly continuous mappings and net theory

Definition 3.1

Let ( X , U ) be a uniform space. Two nets { x n } n ∈ D 1 and { y m } m ∈ D 2 in ( X , U ) , where D i are directed sets, are called pairwise Cauchy if { ( x n , y m ) } ( n , m ) ∈ D 1 × D 2 is eventually contained in each entourage U ∈ U , that is, for each U ∈ U , ( x n , y m ) ∈ U for all n ≥ n 0 and m ≥ m 0 for some ( n 0 , m 0 ) ∈ D 1 × D 2 .

For brevity, we simply say that { ( x n , y m ) } ( n , m ) ∈ D 1 × D 2 is pairwise Cauchy to indicate that { x n } n ∈ D 1 and { y m } m ∈ D 2 are pairwise Cauchy nets with respect to the uniformity concerned.

Now we provide a characterization for uniformly continuous mappings.

Proposition 3.2

Let ( X , U ) be a uniform space and { ( x n , y m ) } ( n , m ) ∈ D 1 × D 2 be a pair of nets. Then { ( x n , y m ) } ( n , m ) ∈ D 1 × D 2 is pairwise Cauchy if and only if the pair of nets { ( y m , x n ) } ( m , n ) ∈ D 2 × D 1 is pairwise Cauchy.

Proof

Let { ( x n , y m ) } ( n , m ) ∈ D 1 × D 2 be a pairwise Cauchy nets. Let U ∈ U be any entourage. Then there exists U − 1 ∈ U . Since the pair of nets { ( x n , y m ) } ( n , m ) ∈ D 1 × D 2 is pairwise Cauchy, ( x n , y m ) ∈ U − 1 eventually. Thus, ( y m , x n ) ∈ U eventually. Hence, the pair of nets { ( y m , x n ) } ( m , n ) ∈ D 2 × D 1 is pairwise Cauchy.

Converse is true obviously.□

Here, it may be mentioned that pairwise Cauchy nets remain pairwise Cauchy if finitely many members of the pair are replaced by other elements. In other words, the results related to pairwise Cauchy nets will remain valid if the pair of nets is eventually Cauchy.

In our next theorem, we provide an equivalent criterion for uniform continuity.

Proposition 3.3

Let ( X , U ) and ( Y , V ) be two uniform spaces. Then f : ( X , U ) → ( Y , V ) is uniformly continuous if and only if the image of every pairwise Cauchy nets in X is again pairwise Cauchy in Y.

Proof

Let f be uniformly continuous and { ( x n , y m ) } ( n , m ) ∈ D 1 × D 2 be pairwise Cauchy nets in X. Let V ∈ V be any entourage. Since f is uniformly continuous, there exists U ∈ U such that f 2 [ U ] ⊂ V . As ( x n , y m ) ∈ U eventually, we have f 2 ( x n , y m ) ∈ f 2 [ U ] ⊂ V , eventually. That is, ( f ( x n ) , f ( y m ) ) ∈ V eventually. Hence, { ( f ( x n ) , f ( y m ) ) } n × m ∈ D 1 × D 2 , the image of { ( x n , y m ) } ( n , m ) ∈ D 1 × D 2 is also pairwise Cauchy.

Conversely, let the image of every pairwise Cauchy nets be pairwise Cauchy. Let if possible, f be not uniformly continuous. Then there exists an entourage V ∈ V such that there is no entourage U ∈ U with f 2 [ U ] ⊂ V . Hence for each U ∈ U , we have f 2 [ U ] ⊈ V . Thus, for each entourage U ∈ U , there exists a pair ( x u , y u ) ∈ U such that f 2 ( x u , y u ) = ( f ( x u ) , f ( y u ) ) ∉ V . Now the collection of all entourages U ∈ U forms a directed set under the relation ≥ , which is defined by “ U ≥ V implies U ⊂ V .” Now we show that { ( x u , y u ) } u ∈ U is pairwise Cauchy in X but its image is not pairwise Cauchy in Y.

Let U 0 ∈ U . For U ≥ U 0 , that is, U ⊂ U 0 , we have ( x u , y u ) ∈ U ⊂ U 0 . Hence, ( x u , y u ) ∈ U 0 for all U ≥ U 0 . Thus, { ( x u , y u ) } u ∈ U is pairwise Cauchy. Now consider entourage V ∈ V , we have ( f ( x u ) , f ( y u ) ) ∉ V , for each u ∈ U . Hence, the image of { ( x u , y u ) } u ∈ U is not pairwise Cauchy. Thus, we got a contradiction. Therefore, f is uniformly continuous.□

3.2 Uniformity over uniform spaces

We now define a uniformity on U C ( X , Y ) in the following way:

Let ( X , U ) and ( Y , V ) be two uniform spaces. For V ∈ V and x ∈ X , we define:

( x , V ) = { ( f , g ) ∈ U C ( X , Y ) × U C ( X , Y ) | ( f ( x ) , g ( x ) ) ∈ V } .

Let S p , V = { ( x , V ) | x ∈ X , V ∈ V } .

Lemma 3.4

S p , V forms a subbase for a uniformity over U C ( X , Y ) .

Proof

By Theorem 2.5, it is enough to show that S p , V satisfies conditions (2.1.1)–(2.1.3). We proceed as follows:

Δ = { ( f , f ) | f ∈ U C ( X , Y ) } ⊂ ( x , V ) .

This follows from the definition of S p , V .
For every ( x , V ) ∈ S p , V , ( x , V ) − 1 ∈ S p , V .

Since V ∈ V , V − 1 ∈ V . We claim that ( x , V ) − 1 = ( x , V − 1 ) .

Let ( f , g ) ∈ ( x , V ) − 1 , then ( g , f ) ∈ ( x , V ) . Thus, we have ( g ( x ) , f ( x ) ) ∈ V . Hence, ( f ( x ) , g ( x ) ) ∈ V − 1 . Therefore, ( f , g ) ∈ ( x , V − 1 ) and hence ( x , V ) − 1 ⊂ ( x , V − 1 ) . On the same line, one can prove that ( x , V − 1 ) ⊂ ( x , V ) − 1 . Hence, ( x , V − 1 ) = ( x , V ) − 1 .
For every ( x , V ) ∈ S p , V , there exists some A ∈ S p , V such that A ∘ A ⊂ ( x , V ) .

Let ( x , V ) ∈ S p , V . For V ∈ V there exists V ′ ∈ V such that V ′ ∘ V ′ ⊂ V . Now, we claim that for ( x , V ′ ) ∈ S p , V we have ( x , V ′ ) ∘ ( x , V ′ ) ⊂ ( x , V ) .

Let ( f , h ) ∈ ( x , V ′ ) ∘ ( x , V ′ ) . Then there exists g ∈ U C ( X , Y ) such that ( f , g ) , ( g , h ) ∈ ( x , V ′ ) , that is, ( f ( x ) , g ( x ) ) ⊂ V ′ and ( g ( x ) , h ( x ) ) ⊂ V ′ . Thus, we have ( f ( x ) , g ( x ) ) ∘ ( g ( x ) , h ( x ) ) ⊂ V ′ ∘ V ′ ⊂ V . Hence, ( f ( x ) , h ( x ) ) ∈ V which implies ( f , h ) ∈ ( x , V ) . Thus, ( x , V ′ ) ∘ ( x , V ′ ) ⊆ ( x , V ) ∈ S p , V . Therefore, ( f , h ) ∈ ( x , V ) .

Hence, S p , V forms a subbase for a uniformity on U C ( X , Y ) .□

Uniformity generated by this subbase is called the point-entourage uniformity for U C ( X , Y ) and is denoted by U p , V .

Example 3.1

Let X = Z be the set of integers. The p-adic uniform structure on Z , for a given prime number p, is the uniformity V generated by the subsets Z n of Z × Z , for n = 1 , 2 , 3 , … , where Z n is defined as:

Z n = { ( k , m ) | k ≡ m mod p n } .

Consider the family of subsets

U ε = { ( x , y ) | | x − y | < ε }

of ℝ × ℝ for ε > 0 . The uniform structure generated by the subsets U ε for ε > 0 is called the Euclidean uniformity of ℝ . Specifically, a subset D of ℝ × ℝ is an entourage if U ε ⊂ D for some ε > 0 .

Now, we consider, for x ∈ ℝ and n ∈ N ,

( x , Z n ) = { ( f , g ) ∈ U C ( X , Y ) × U C ( X , Y ) | f ( x ) , g ( x ) ∈ Z n } .

Let S p , V = x , Z n | x ∈ ℝ , Z n ∈ V . It can be easily verified that S p , V satisfies (2.1.1) to (2.1.3) of Definition 2.1. Thus, S p , V forms a subbase for a uniformity over U C ( ℝ , Z ) which is a point-entourage uniformity for U C ( ℝ , Z ) . Here, structure of the entourage in the uniformity generated by the subbase S p , V is the collection of the pair of uniformly continuous functions from ℝ to Z , ( f , g ) such that f ( x ) − g ( x ) is divisible by p n , for some given x ∈ ℝ and n ∈ ℕ .

Now, let ( Y , V ) and ( Z , U ) be two uniform spaces. Let V ∈ V be any symmetric entourage, that is, V = V − 1 . For U ∈ U , we define:

( V , U ) = { ( f , g ) ∈ U C ( Y , Z ) × U C ( Y , Z ) | ( f ( V 1 ) , g ( V 2 ) ) ⊆ U } ∪ { ( f , f ) | f ∈ U C ( Y , Z ) } ,

where V = V 1 × V 2 .

Consider S V , U = { ( V , U ) | V ∈ V , U ∈ U , V is symmetric } .

Lemma 3.5

S V , U forms a subbase for a uniformity over U C ( Y , Z ) .

Proof

By Theorem 2.5, it is enough to show that S V , U satisfies conditions (2.1.1)–(2.1.3). We proceed as follows:

Δ = { ( f , f ) | f ∈ U C ( Y , Z ) } ⊂ ( V , U ) .

This follows from the definition of S V , U .
For every ( V , U ) ∈ S V , U , ( V , U ) − 1 ∈ S V , U .

Let ( f , g ) ∈ ( V , U ) , which implies ( f ( V 1 ) , g ( V 2 ) ) ⊆ U , where V = V 1 × V 2 . Since V ∈ V and U ∈ U , V − 1 and U − 1 belong to V and U , respectively.

We claim that ( V , U ) − 1 = ( V − 1 , U − 1 ) .

Let ( f , g ) ∈ ( V , U ) − 1 , then ( g , f ) ∈ ( V , U ) . Thus, we have ( g ( V 1 ) , f ( V 2 ) ) ⊆ U . Hence, ( f ( V 2 ) , g ( V 1 ) ) ⊆ U − 1 . Therefore, ( f , g ) ∈ ( V − 1 , U − 1 ) and hence ( V , U ) − 1 ⊂ ( V − 1 , U − 1 ) . On the same line, one can prove that ( V − 1 , U − 1 ) ⊂ ( V , U ) − 1 . Hence, ( V − 1 , U − 1 ) = ( V , U ) − 1 .
For every ( V , U ) ∈ S V , U , there exists some A ∈ S V , U such that A ∘ A ⊂ ( V , U ) .

Let ( V , U ) ∈ S V , U . For U ∈ U there exists U ′ ∈ U such that U ′ ∘ U ′ ⊂ U and similarly, there exists U ″ ∈ U such that U ″ ∘ U ″ ⊆ U ′ and hence U ″ ∘ U ″ ∘ U ″ ∘ U ″ ⊂ U . Now, we claim that for ( V , U ″ ) ∈ S V , U we have ( V , U ″ ) ∘ ( V , U ″ ) ⊂ ( V , U ) .

Let ( f , h ) ∈ ( V , U ″ ) ∘ ( V , U ″ ) . Then there exists g ∈ U C ( Y , Z ) such that ( f , g ) , ( g , h ) ∈ ( V , U ″ ) , that is, ( f ( V 1 ) , g ( V 2 ) ) ⊂ U ″ and ( g ( V 1 ) , h ( V 2 ) ) ⊂ U ″ , where V = V 1 × V 2 . Since ( g , g ) ∈ ( V , U ″ ) , where V is a symmetric entourage, we have ( g ( V 1 ) , g ( V 2 ) ) ⊂ U ″ and hence ( g ( V 2 ) , g ( V 1 ) ) ⊂ U ″ .

Thus, we have ( f ( V 1 ) , g ( V 2 ) ) ∘ ( g ( V 2 ) , g ( V 1 ) ) ∘ ( g ( V 1 ) , g ( V 1 ) ) ∘ ( g ( V 1 ) , h ( V 2 ) ) ⊂ U ″ ∘ U ″ ∘ U ″ ∘ U ″ ⊂ U . Hence, ( f ( V 1 ) , h ( V 2 ) ) ∈ U which implies ( f , h ) ∈ ( V , U ) . Thus, ( V , U ″ ) ∘ ( V , U ″ ) ⊆ ( V , U ) .

Hence, S V , U forms a subbase for a uniformity on U C ( Y , Z ) .□

The uniform space generated by the aforementioned subbase is called the entourage-entourage uniformity and it is denoted by U V , U .

Example 3.2

Now, we again consider the set of integers Z , with p-adic uniformity V and the set of real numbers ℝ with Euclidean uniformity U , defined in Example 3.1. We for given ε > 0 and n ∈ ℕ define:

( U ε , Z n ) = { ( f , g ) ∈ U C ( ℝ , Z ) × U C ( ℝ , Z ) | ( f ( V 1 ) , g ( V 2 ) ) ⊆ Z n } ∪ { ( f , f ) | f ∈ U C ( ℝ , Z ) } ,

where U ε = V 1 × V 2 .

Consider S U , V = { ( U ε , Z n ) | U ε ∈ U , Z n ∈ V } .

It is easy to verify that S U , V satisfies (2.1.1) to (2.1.3) of Definition 2.1. Hence, S U , V forms a subbase for a uniformity over U C ( ℝ , Z ) . This is an example of an entourage-entourage uniformity and it is denoted by U U , V . Here, the structure of entourage in entourage-entourage uniformity over U C ( ℝ , Z ) is the collection of pair of all uniformly continuous functions ( f , g ) from ℝ to Z such that for given ε > 0 , there always exists a natural number n ∈ ℕ such that f ( x ) − g ( y ) is divisible by p n whenever | x − y | < ε .

The above discussion clearly indicates that several uniformities do exist on U C ( X , Y ) .

Let A be a uniformity on U C ( X , Y ) , then the pair ( U C ( X , Y ) , A ) is called a uniform space over uniformly continuous mappings or uniform space over uniform continuity.

Now we introduce the notions of admissibility and splittingness for the uniform spaces over uniform continuity. Admissibility and splittingness are two very important notions in the topology of function spaces. They were introduced by Arens and Dugundji [16] and have been studied by several authors thereafter. In recent years, Georgiou, Iliadis, and others [13,14,15,17] have significantly contributed to the study of these notions. In the following, we proceed to extend the notions of splittingness and admissibility to the domain of uniformities.

Definition 3.6

Let ( Y , U ) and ( Z , V ) be two uniform spaces and let ( X , W ) be another uniform space. Then for a map g : X × Y → Z , we define g ⁎ : X → U C ( Y , Z ) by g ⁎ ( x ) ( y ) = g ( x , y ) .

The mappings g and g ⁎ related in this way are called associated maps.

Definition 3.7

Let ( Y , U ) and ( Z , V ) be two uniform spaces. A uniformity A on U C ( Y , Z ) is called

admissible if for each uniform space ( X , W ) , uniform continuity of g ⁎ : X → U C ( Y , Z ) implies uniform continuity of the associated map g : X × Y → Z ;
splitting if for each uniform space ( X , W ) , uniform continuity of g : X × Y → Z implies uniform continuity of g ⁎ : X → U C ( Y , Z ) , where g ⁎ is the associated map of g.

Now, we prove that the point-entourage uniformity defined in Example 3.1 over U C ( ℝ , Z ) is splitting.

Example 3.3

Let Y = ℝ , the set of real numbers with Euclidean uniformity U and Z = Z , the set of all integers with p-adic uniformity V be two uniform spaces. Let U C ( ℝ , Z ) be the space of all uniform continuous functions from Y to Z with point-entourage uniformity U p , V , defined in Example 3.1. Let ( X , W ) be any uniform space such that the map g : X × ℝ → Z is uniformly continuous. We have to show that the associated map g ⁎ : X → U C ( ℝ , Z ) is uniformly continuous, where g ⁎ is defined as g ⁎ ( x ) ( y ) = g ( x , y ) .

Let ( x , Z n ) be any entourage in U C ( ℝ , Z ) . Since, the map g is uniformly continuous, there exists an entourage V of X × ℝ such that g 2 [ V ] ⊆ Z n , where V = U ′ × U ε for some ε > 0 and U ′ ∈ W . We have g 2 ( U ′ × U ε ) ⊆ Z n , that is, g ( a , x ) , g ( b , y ) ∈ Z n for all ( a , b ) ∈ U ′ and ( x , y ) ∈ U ε . That is, g ( a , x ) ≡ g ( b , y ) mod p n for all ( a , b ) ∈ U ′ and ( x , y ) ∈ U ε . That is, g ⁎ ( a ) ( x ) ≡ g ⁎ ( b ) ( y ) mod p n for all ( a , b ) ∈ U ′ and ( x , y ) ∈ U ε . Since U ε ∈ U is an entourage, ( x , x ) ∈ U ε for all x ∈ ℝ . Thus, we have g ⁎ ( a ) ( x ) ≡ g ⁎ ( b ) ( x ) mod p n , which implies ( g ⁎ ( a ) , g ⁎ ( b ) ) ∈ ( x , Z n ) for all ( a , b ) ∈ U ′ . Hence, we have g 2 ⁎ [ U ′ ] ⊆ ( x , Z n ) . The associated map g ⁎ is uniformly continuous, therefore, the point-entourage uniformity over U C ( Y , Z ) is splitting.

Before proceeding further, we mention few basic results which will be used later.

Proposition 3.8

Let ( X , U ) , ( Y , V ) , and ( Z , W ) be uniform spaces and let f : ( X , U ) → ( Y , V ) and g : ( Y , V ) → ( Z , W ) be two uniformly continuous maps. Then g ∘ f : ( X , U ) → ( Z , W ) is again uniformly continuous.

Proposition 3.9

Let ( X , U ) , ( Y , V ) , and ( Z , W ) be uniform spaces and let f : ( X , U ) → ( Y , V ) be uniformly continuous. Then F : X × Z → Y × Z , defined by F ( x , z ) = ( f ( x ) , z ) is also uniformly continuous.

Proof

Let f : ( X , U ) → ( Y , V ) be any uniformly continuous function. We have to show F : X × Z → Y × Z , defined by F ( x , z ) = ( f ( x ) , z ) is uniformly continuous. Let A = V × W be any entourage in the uniformity of Y × Z , where V ∈ V and W ∈ W . Since f is uniformly continuous, there exists an entourage U ∈ U such that f 2 [ U ] ⊂ V . Thus, f 2 ( U ) × W ⊂ V × W , which implies F 2 [ U × W ] ⊂ V × W . Hence, F is uniformly continuous.□

The notion of evaluation maps of topology can be extended for uniformities also to obtain the following characterization of admissibility.

Theorem 3.10

Let ( Y , U ) and ( Z , V ) be two uniform spaces. Then a uniformity A on U C ( Y , Z ) is admissible if and only if the evaluation mapping e : U C ( Y , Z ) × Y → Z defined by e ( f , y ) = f ( y ) is uniformly continuous.

Proof

Let ( Y , U ) and ( Z , V ) be two uniform spaces and uniformity A on U C ( Y , Z ) be admissible, that is, uniform continuity of the map g ⁎ : X → U C ( Y , Z ) implies the uniform continuity of the associated map g : X × Y → Z for each uniform space ( X , W ) . We take X = U C ( Y , Z ) , then the identity map g ⁎ : U C ( Y , Z ) → U C ( Y , Z ) , where g ⁎ ( f ) = f for all f ∈ U C ( Y , Z ) is uniformly continuous. Then by the given hypothesis, the associated map g : U C ( Y , Z ) × Y → Z is also uniformly continuous. Consider, g ( f , y ) = g ⁎ ( f ) ( y ) = f ( y ) = e ( f , y ) . Thus, we have g ≡ e . Hence, the evaluation map e : U C ( Y , Z ) × Y → Z is uniformly continuous.

Conversely, let g ⁎ : X → U C ( Y , Z ) be uniformly continuous. We define a map h : X × Y → U C ( Y , Z ) × Y defined by h ( x , y ) = ( g ⁎ ( x ) , y ) . In the light of Proposition 3.9, the map h is uniformly continuous. Since the given evaluation map e : U C ( Y , Z ) × Y → Z is uniformly continuous. Thus, the composition map e ∘ h : X × Y → Z is also uniformly continuous and e ∘ h ≡ g because e ∘ h ( x , y ) = e [ h ( x , y ) ] = e ( g ⁎ ( x ) , y ) = g ⁎ ( x ) ( y ) = g ( x , y ) . Hence, the associated map g is uniformly continuous. This completes the proof.□

The uniform space U C ( ℝ , Z ) is admissible under entourage-entourage uniformity, defined in Example 3.2.

Example 3.4

Let Y = ℝ , the set of all real numbers with the Euclidean uniformity U and Z = Z , the set of all integers with p-adic uniformity V . We have to show that the evaluation mapping e : U C ( ℝ , Z ) × ℝ → Z , defined as e ( f , y ) = f ( y ) , is uniformly continuous under the entourage-entourage uniformity over U C ( ℝ , Z ) defined in Example 3.2. Let Z n be any entourage in ( Z , V ) .

Consider an entourage V 1 = ( U ε , Z n ) and V 2 = U ε , for some ε > 0 . Then e 2 ( V 1 × V 2 ) = ( e ( f , x ) , e ( g , y ) ) = ( f ( x ) , g ( y ) ) . We have ( f , g ) ∈ ( U ε , Z n ) , thus f ( a ) , g ( b ) ∈ Z n , that is, f ( a ) ≡ f ( b ) mod p n , for all ( a , b ) ∈ Z n . Therefore, e 2 ( V 1 × V 2 ) ⊆ Z n . Thus, the evaluation map is uniformly continuous and hence U C ( ℝ , Z ) under entourage-entourage uniformity is admissible.

Before coming to the main results of this paper, we provide a small discussion on directed sets.

Let D 1 and D 2 be two directed sets. We define a uniformity U 0 on Δ = D 1 ∪ D 2 , generated by { U n 0 , m 0 | ( n 0 , m 0 ) ∈ D 1 × D 2 } , where U n 0 , m 0 = δ ∪ { ( n , m ) | ( n , m ) ≥ ( n 0 , m 0 ) } ∪ { ( m , n ) | ( m , n ) ≥ ( m 0 , n 0 ) : ( n 0 , m 0 ) ∈ D 1 × D 2 } and δ = { ( n , n ) | n ∈ Δ } , where “≥” being defined component-wise.

Lemma 3.11

Let ( Y , U ) be a uniform space and { ( y n , y m ′ } ( n , m ) ∈ D 1 × D 2 be a pair of nets in Y. Then { ( y n , y m ′ } ( n , m ) ∈ D 1 × D 2 is pairwise Cauchy if and only if the function s : Δ → Y defined by s ( n ) = y n for n ∈ D 1 , s ( m ) = y m ′ for m ∈ D 2 , n ≠ m is uniformly continuous under U 0 defined above on Δ = D 1 ∪ D 2 and U on Y. In case, D 1 ∩ D 2 ≠ ∅ , that is, n = m for some n ∈ D 1 and m ∈ D 2 , take s ( n ) = s ( m ) = y n in the above definition.

Proof

Let { ( y n , y m ′ ) } ( n , m ) ∈ D 1 × D 2 be any pairwise Cauchy nets in Y. Let U ∈ U be any entourage, then ( y n , y m ′ ) ∈ U eventually, that is, there exists ( n 0 , m 0 ) ∈ D 1 × D 2 such that ( y n , y m ′ ) ∈ U for all ( n , m ) ≥ ( n 0 , m 0 ) . That is, there exists U n 0 , m ∈ U 0 such that s 2 ( U n 0 , m 0 ) ⊂ U . Hence, s is uniformly continuous.

Conversely, let { ( y n , y m ′ ) } ( n , m ) ∈ D 1 × D 2 be any pair of nets in Y and s be a uniformly continuous mapping. Let V ∈ U be any entourage. Then there exists an entourage U n 0 , m 0 ∈ U 0 such that s 2 ( U n 0 , m 0 ) ⊂ V . Thus, s 2 ( ( n , m ) ) ∈ V for all ( n , m ) ≥ ( n 0 , m 0 ) . Hence, { ( y n , y m ′ } ( n , m ) ∈ D 1 × D 2 is pairwise Cauchy.□

In the next pair of theorems, we provide some characterizations of splittingness and admissibility of the uniform spaces over uniformly continuous mappings. We introduce the notion of continuously Cauchy nets for this purpose. The need for such notion has arisen purely out of uniformity structure of the space and has no topological or metric counterpart.

Definition 3.12

Let { ( f n , g m ) } ( n , m ) ∈ D 1 × D 2 be a pair of nets in U C ( Y , Z ) . Then { ( f n , g m ) } ( n , m ) ∈ D 1 × D 2 is said to be continuously Cauchy if for each pairwise Cauchy net { ( y k , y l ′ ) } ( k , l ) ∈ D 3 × D 4 in Y, { ( f n ( y k ) , g m ( y l ′ ) ) } ( n , m , k , l ) ∈ D 1 × D 2 × D 3 × D 4 is pairwise Cauchy in Z.

Lemma 3.13

Let ( Y , U ) and ( Z , V ) be two uniform spaces and let { ( y n , y m ′ ) } ( n , m ) ∈ D 1 × D 2 and { ( z k , z l ′ ) } ( k , l ) ∈ D 3 × D 4 be two pairwise Cauchy nets in Y and Z, respectively. Then { ( y n , z k ) , ( y m ′ , z l ′ ) } ( n , m , k , l ) ∈ D 1 × D 2 × D 3 × D 4 is pairwise Cauchy in Y × Z with respect to the product uniformity U × V and vice versa.

Proof

Let ( Y , U ) and ( Z , V ) be two uniform spaces and let { ( y n , y m ′ ) } ( n , m ) ∈ D 1 × D 2 and { ( z k , z l ′ ) } ( k , l ) ∈ D 3 × D 4 be two pairwise Cauchy nets in Y and Z, respectively. Therefore, for each U ∈ U and V ∈ V , we have ( y n , y m ′ ) ∈ U and ( z k , z l ′ ) ∈ V eventually. Hence, { ( y n , z k ) , ( y m ′ , z l ′ ) } ∈ U × V eventually. Thus, { ( y n , z k ) , ( y m ′ , z l ′ ) } ( n , m , k , l ) ∈ D 1 × D 2 × D 3 × D 4 is pairwise Cauchy in Y × Z . The converse can be proved in similar manner.□

In the remaining part of this section, we use net theory to provide further investigations about admissibility and splittingness for uniformities on U C ( Y , Z ) . In the first two theorems, we provide net-theoretic characterization for splittingness and admissibility, respectively. In the topological parlance, Arens and Dugundji were the ones to introduce the concept of continuous convergence. They have provided characterizations for splittingness and admissibility for topological function spaces, by using the concept of continuous convergence. Here, we extend the same for uniformity and use the concept of pairwise Cauchy nets to arrive at our results. Theorems 3.16 and 3.17 provide examples of splittingness and admissibility families of uniform space, respectively, on U C ( Y , Z ) .

Theorem 3.14

Let ( Y , U ) and ( Z , V ) be two uniform spaces. A uniformity A on U C ( Y , Z ) is splitting if and only if each pair of nets { ( f n , f m ′ ) } ( n , m ) ∈ D 1 × D 2 in U C ( Y , Z ) is pairwise Cauchy whenever it is continuously Cauchy.

Proof

Let ( X , W ) be any uniform space such that g : X × Y → Z be uniformly continuous. We have to show that the associated map g ⁎ : X → U C ( Y , Z ) is uniformly continuous. Let { ( x n , x m ′ ) } ( n , m ) ∈ D 1 × D 2 be any pairwise Cauchy nets in X. We have to show that { ( g ⁎ ( x n ) , g ⁎ ( x m ′ ) ) } ( n , m ) ∈ D 1 × D 2 is again pairwise Cauchy in U C ( Y , Z ) .

Let { ( y k , y l ′ ) } ( k , l ) ∈ D 3 × D 4 be any pairwise Cauchy net in Y. Then { ( x n , y k ) , ( x m ′ , y l ′ ) } ( n , m , k , l ) ∈ D 1 × D 2 × D 3 × D 4 is a pairwise Cauchy net in X × Y . Since g : X × Y → Z is uniformly continuous, { g ( x n , y k ) , g ( x m ′ , y l ′ ) } ( n , m , k , l ) ∈ D 1 × D 2 × D 3 × D 4 is a pairwise Cauchy net in Z. Let us define g ⁎ ( x n ) = f n and g ⁎ ( x m ′ ) = f m ′ . Then { g ( x n , y k ) , g ( x m ′ , y l ′ ) } ( n , m , k , l ) ∈ D 1 × D 2 × D 3 × D 4 = { g ⁎ ( x n ) ( y k ) , g ⁎ ( x m ′ ) ( y l ′ ) } ( n , m , k , l ) ∈ D 1 × D 2 × D 3 × D 4 = { f n ( y k ) , f m ′ ( y l ′ ) } ( n , m , k , l ) ∈ D 1 × D 2 × D 3 × D 4 is pairwise Cauchy in Z. Therefore, the pair of nets { ( f n , f m ′ ) } ( n , m ) ∈ D 1 × D 2 is continuously Cauchy. By the hypothesis, the pair of nets { ( f n , f m ′ ) } ( n , m ) ∈ D 1 × D 2 is pairwise Cauchy in U C ( Y , Z ) and hence { ( g ⁎ ( x n ) , g ⁎ ( x m ′ ) ) } ( n , m ) ∈ D 1 × D 2 is pairwise Cauchy. Therefore, g ⁎ is uniformly continuous and hence ( U C ( Y , Z ) , A ) is splitting.

In the above proof, let if possible, x n = x m ′ = x k (say) for some n ∈ D 1 and m ∈ D 2 , then we should proceed as follows.

We then define the map g ⁎ : X → U C ( Y , Z ) as g ⁎ ( x n ) = f n , g ⁎ ( x m ′ ) = f m ′ for x n ≠ x m and g ⁎ ( x k ) = f k whenever x n = x m ′ = x k . Then { g ( x n , y k ) , g ( x m ′ , y l ′ ) } ( n , m , k , l ) ∈ D 1 × D 2 × D 3 × D 4 = { g ⁎ ( x n ) ( y k ) , g ⁎ ( x m ′ ) ( y l ′ ) } ( n , m , k , l ) ∈ D 1 × D 2 × D 3 × D 4 = { f n ( y k ) , f m ′ ( y l ′ ) } ( n , m , k , l ) ∈ D 1 × D 2 × D 3 × D 4 , where f m ′ ( y l ′ ) = f k ( y l ′ ) , whenever x n = x m ′ = x k , is pairwise Cauchy net in Z as they are eventually pairwise Cauchy. Therefore, { f n ( y k ) , f m ′ ( y l ′ ) } ( n , m , k , l ) ∈ D 1 × D 2 × D 3 × D 4 is again pairwise Cauchy. Hence, the pair of nets { ( f n , f m ′ ) } ( n , m ) ∈ D 1 × D 2 is continuously Cauchy. By the hypothesis, the pair of nets { ( f n , f m ′ ) } ( n , m ) ∈ D 1 × D 2 is pairwise Cauchy in U C ( Y , Z ) and hence { ( g ⁎ ( x n ) , g ⁎ ( x m ′ ) ) } ( n , m ) ∈ D 1 × D 2 is pairwise Cauchy. Therefore, g ⁎ is uniformly continuous and hence ( U C ( Y , Z ) , A ) is splitting.

The case where x n = x m ′ for infinitely many indices, may be treated in a similar manner.

Conversely, let ( U C ( Y , Z ) , A ) be splitting and { ( f n , f m ′ ) } ( n , m ) ∈ D 1 × D 2 be any pair of nets in U C ( Y , Z ) which is continuously Cauchy. We have to show that the pair ( f n , f m ′ ) ( n , m ) ∈ D 1 × D 2 is pairwise Cauchy in ( U C ( Y , Z ) , A ) . Let U 0 be the uniformity generated on Δ , where Δ = D 1 ∪ D 2 . Then the only non-trivial pair of nets in Δ is ( n , m ) ( n , m ) ∈ D 1 × D 2 , which is pairwise Cauchy. Let S be any pairwise Cauchy net in Δ × Y . Then S = S 1 × S 2 , where S 1 and S 2 are nets in Δ × Y , where S 1 = { n , y k } ( n , k ) ∈ D 1 × D 3 and S 2 = { m , y l ′ } ( m , l ) ∈ D 2 × D 4 Then { n , m } ( n , m ) ∈ D 1 × D 2 and { y k , y l } ( k , l ) ∈ D 3 × D 4 . Then we define a map g : Δ × Y → Z as g ( n ) = f n ( y ) and g ( m , y ) = f m ′ ( y ) . Thus, g 2 ( S ) = ( g ( n , y k ) , g ( m , y l ′ ) ) ( n , m , k , l ) ∈ D 1 × D 2 × D 3 × D 4 . That is, g 2 ( S ) = ( f n ( y k ) , f m ′ ( y l ′ ) ) ( n , m , k , l ) ∈ D 1 × D 2 × D 3 × D 4 . Since the pair { ( f n , f m ′ ) } ( n , m ) ∈ D 1 × D 2 is given to be a continuously Cauchy pair, { ( g ( n , y k ) , g ( m , y l ′ ) ) } ( n , m , k , l ) ∈ D 1 × D 2 × D 3 × D 4 is pairwise Cauchy. Hence, the map g is uniformly continuous. As A is splitting, this implies g ⁎ is uniformly continuous. Since { ( n , m ) } ( n , m ) ∈ D 1 × D 2 is pairwise Cauchy, we have { ( g ⁎ ( n ) , g ⁎ ( m ) ) } ( n , m ) ∈ D 1 × D 2 is pairwise Cauchy in U C ( Y , Z ) . Now consider, g ⁎ ( n ) ( y ) = g ( n , y ) = f n ( y ) and g ⁎ ( m ) ( y ) = g ( m , y ) = f m ′ ( y ) . That is, g ⁎ ( n ) = f n and g ⁎ ( m ) = f m ′ . Hence, { ( f n , f m ′ ) } ( n , m ) ∈ D 1 × D 2 is pairwise Cauchy in ( U C ( Y , Z ) , A ) .

If D 1 ∩ D 2 in non-empty and finite, then in the above discussion, we take g ( n , y ) = f n ( y ) and g ( m , z ) = f m ' ( z ) , whenever y ≠ z ∈ Y . Furthermore, if we have, n = m and y = z for some y , z ∈ Y , then we define g ( n , y ) = g ( m , z ) = f n ( y ) . Then in the above proof, g 2 ( S ) = ( f n ( y k ) , f m ′ ( y l ′ ) ) ( n , m , k , l ) ∈ D 1 × D 2 × D 3 × D 4 . Since the pair { ( f n , f m ′ ) } ( n , m ) ∈ D 1 × D 2 is continuously Cauchy, g 2 ( S ) = ( f n ( y k ) , f m ′ ( y l ′ ) ) ( n , m , k , l ) ∈ D 1 × D 2 × D 3 × D 4 is pairwise Cauchy. Hence, the proof.

If D 1 ∩ D 2 in non-empty and infinite and y = z , then the images of S 1 and S 2 under g will coincide in infinitely many places. Thus, proof becomes a trivial case of the above discussion for the converse part.□

Example 3.5

Let ( ℝ , U ) and ( Z , V ) be uniform spaces as stated in Example 3.1. It has been shown that the point-entourage uniform space ( U C ( ℝ , Z ) , U p , V ) is splitting (see Example 3.3). We now show that this uniformity satisfies the conditions stated in the above theorem.

For this, we have to show that if a pair of nets ( f n , f m ) ( n , m ) ∈ D 1 × D 2 in U C ( ℝ , Z ) × U C ( ℝ , Z ) is continuously Cauchy, then ( f n , f m ) ( n , m ) ∈ D 1 × D 2 is pairwise Cauchy under ( U C ( ℝ , Z ) , U p , V ) . Let ( y , Z n ) ∈ U p , V , for some y ∈ ℝ and for some n ∈ N , be any entourage in the point-entourage uniformity. Then, consider ( y l , y k ) ( l , k ) ∈ D 3 × D 4 = ( y , y ) , which is constant and hence is pairwise Cauchy net. Since ( f n , f m ) ( n , m ) ∈ D 1 × D 2 is assumed to be continuously Cauchy, ( f n ( y ) , f m ( y ) ) ( n , m ) ∈ D 1 × D 2 is pairwise Cauchy net in Z . Hence, for Z n ∈ V , we have ( f n ( y ) , f m ( y ) ) ∈ Z n eventually. Therefore, ( f n , f m ) ∈ ( y , Z n ) eventually. Hence, the pair of nets ( f n , f m ) ( n , m ) ∈ D 1 × D 2 is pairwise Cauchy net in ( U C ( Y , Z ) , U p , V ) . Thus, the point-entourage uniform space ( U C ( Y , Z ) , U p , ⊑ ) satisfies the conditions of Theorem 3.14.

Theorem 3.15

Let ( Y , U ) and ( Z , V ) be two uniform spaces. A uniformity A on U C ( Y , Z ) is admissible if and only if each pair of nets ( f n , f m ' ) ( n , m ) ∈ D 1 × D 2 in U C ( Y , Z ) is continuously Cauchy under A if ( f n , f m ' ) ( n , m ) ∈ D 1 × D 2 is pairwise Cauchy.

Proof

Let ( U C ( Y , Z ) , A ) be admissible and ( f n , g m ) ( n , m ) ∈ D 1 × D 2 be a pair of nets in U C ( Y , Z ) , which is pairwise Cauchy. We have to show that the pair ( f n , g m ) ( n , m ) ∈ D 1 × D 2 is continuously Cauchy in ( U C ( Y , Z ) , A ) . Let T 0 be the uniformity generated on Δ , where Δ = D 1 ∪ D 2 . The only non-trivial pair of nets in Δ , which is pairwise Cauchy, is { ( n , m ) } ( n , m ) ∈ D 1 × D 2 . Then define a map g ⁎ : Δ → U C ( Y , Z ) by g ⁎ ( n ) = f n and g ⁎ ( m ) = g m . Consider ( g ⁎ ( n ) , g ⁎ ( m ) ) ( n , m ) ∈ D 1 × D 2 = ( f n , g m ) ( n , m ) ∈ D 1 × D 2 is pairwise Cauchy. Hence, the map g ⁎ is uniformly continuous. Since ( U C ( Y , Z ) , A ) is given to be admissible, the associated map g : Δ × Y → Z is also uniformly continuous. Let ( y k , y l ′ ) ( k , l ) ∈ D 3 × D 4 be any pairwise net in Y. Therefore, { ( n , y k ) , ( m , y l ′ ) } ( n , m , k , l ) ∈ D 1 × D 2 × D 3 × D 4 is again a pairwise Cauchy net in Δ × Y . Since the map g is uniformly continuous, { g ( n , y k ) , g ( m , y l ′ ) } ( n , m , k , l ) ∈ D 1 × D 2 × D 3 × D 4 is pairwise Cauchy.

That is, { g ⁎ ( n ) ( y k ) , g ⁎ ( m ) ( y l ′ ) } ( n , m , k , l ) ∈ D 1 × D 2 × D 3 × D 4 = ( f n ( y k ) , g m ( y l ′ ) ) is pairwise Cauchy. Hence, the pair ( f n , g m ) ( n , m ) ∈ D 1 × D 2 is continuously Cauchy.

Conversely, let g ⁎ : X → U C ( Y , Z ) be uniformly continuous. We have to show that the associated map g is uniformly continuous. Let { ( x n , y m ) , ( x n ′ , y m ′ ) } ( n , m ) ∈ D 1 × D 2 be a pairwise Cauchy net in X × Y . Then { ( x n , x m ′ } ( n , m ) ∈ D 1 × D 2 and { ( y n , y m ′ ) } ( n , m ) ∈ D 1 × D 2 are pairwise Cauchy net in X and Y, respectively. Since { ( x n , x m ′ ) } ( n , m ) ∈ D 1 × D 2 is a pairwise Cauchy net in X and g ⁎ is uniformly continuous, { g ⁎ ( x n ) , g ⁎ ( x m ′ ) } ( n , m ) ∈ D 1 × D 2 is also a pairwise Cauchy net in U C ( Y , Z ) , that is, { ( f n , g m ) } ( n , m ) ∈ D 1 × D 2 is a pairwise Cauchy in U C ( Y , Z ) , where f n = g ⁎ ( x n ) and g m = g ⁎ ( x m ′ ) , respectively. Then, by the given hypothesis, the pair { ( f n , g m ) } ( n , m ) ∈ D 1 × D 2 is continuously Cauchy. Hence, for the pairwise Cauchy net { ( y n , y m ′ ) } ( n , m ) ∈ D 1 × D 2 in Y, we have { ( f n ( y n ) , g m ( y m ′ ) ) } ( n , m ) ∈ D 1 × D 2 is pairwise Cauchy in Z, that is, { g ( x n , y n ) , g ( x m ′ , y m ′ ) } ( n , m ) ∈ D 1 × D 2 is pairwise Cauchy. Hence, g is uniformly continuous. Therefore, ( U C ( Y , Z ) , A ) is admissible.□

Remark 3.1

In the above proof, if x n = x m ′ , for some n ∈ D 1 , m ∈ D 2 , we may proceed in a similar way as in Theorem 3.14. Similarly, in the converse part, if D 1 ∩ D 2 ≠ ∅ , we proceed as in Theorem 3.14.

Example 3.6

In Example 3.4, it has shown that the uniform space U C ( ℝ , Z ) is admissible under the entourage-entourage uniformity defined as in Example 3.2. Now, we show that this uniformity satisfies the conditions laid down in the above theorem.

For this, we show that if a pair of nets ( f n , f m ) ( n , m ) ∈ D 1 × D 2 ∈ U C ( ℝ , Z ) is pairwise Cauchy, then ( f n , f m ) ( n , m ) ∈ D 1 × D 2 is continuously Cauchy.

Let ( y l , y k ) ( l , k ) ∈ D 3 × D 4 be a pair of Cauchy nets in ( ℝ , U ) . Thus, for any given ε > 0 , ( y l , y k ) ∈ U ε eventually. Since the pair of nets ( f n , f m ) ( n , m ) ∈ D 1 × D 2 is also a Cauchy pair, ( f n , f m ) ∈ ( U ε , Z n ) eventually. Hence, ( f n ( V 1 ) , f m ( V 2 ) ) ∈ Z n eventually, where U ε = V 1 × V 2 . Therefore, we have ( f n ( y l ) , f m ( y k ) ) ∈ Z n eventually. Hence, ( f n , f m ) ( n , m ) ∈ D 1 × D 2 is continuously Cauchy. Thus, entourage-entourage uniform space ( U C ( Y , Z ) , U V , U ) satisfies the condition laid down in Theorem 3.15.

In our next pair of theorems, we provide the existence of some uniform spaces over U C ( Y , Z ) , which satisfy the conditions of splittingness and admissibility, respectively. Using the results obtained so far, we show that every point-entourage uniform space is splitting, whereas every entourage-entourage uniform space is admissible.

Theorem 3.16

Let ( Y , V ) and ( Z , U ) be two uniform spaces. Then the point-entourage uniform space ( U C ( Y , Z ) , U p , U ) is splitting.

Proof

Let ( Y , V ) and ( Z , U ) be two uniform spaces. We have to prove that the point-entourage uniform space ( U C ( Y , Z ) , U p , U ) is splitting. For this, we have to show that if a pair of nets ( f n , f m ) ( n , m ) ∈ D 1 × D 2 in U C ( Y , Z ) × U C ( Y , Z ) is continuously Cauchy, then ( f n , f m ) ( n , m ) ∈ D 1 × D 2 is pairwise Cauchy under ( U C ( Y , Z ) , U p , U ) .

Let ( y , U ) ∈ U p , U be any entourage in point-entourage uniformity. Consider ( y l , y k ) ( l , k ) ∈ D 3 × D 4 = ( y , y ) is the pairwise constant Cauchy net. Since ( f n , f m ) ( n , m ) ∈ D 1 × D 2 is continuously Cauchy, ( f n ( y ) , f m ( y ) ) ( n , m ) ∈ D 1 × D 2 is pairwise Cauchy net in Z. Hence for U ∈ U , we have ( f n ( y ) , f m ( y ) ) ∈ U eventually. Therefore, ( f n , f m ) ∈ ( y , U ) eventually. Hence, the pair of nets ( f n , f m ) ( n , m ) ∈ D 1 × D 2 is pairwise Cauchy net in ( U C ( Y , Z ) , U p , U ) . Thus, the point-entourage uniform space ( U C ( Y , Z ) , U p , U ) is splitting.□

In the next theorem, we show that the entourage-entourage uniform space ( U C ( Y , Z ) , U V , U ) is admissible.

Theorem 3.17

Let ( Y , V ) and ( Z , U ) be two uniform spaces. Then the entourage-entourage uniform space ( U C ( Y , Z ) , U V , U ) is admissible.

Proof

Let ( Y , V ) and ( Z , U ) be two uniform spaces. We have to prove that the entourage-entourage uniform space ( U C ( Y , Z ) , U V , U ) is admissible. For this, we have to show that if a pair of nets ( f n , f m ) ( n , m ) ∈ D 1 × D 2 ∈ U C ( Y , Z ) is pairwise Cauchy provided ( f n , f m ) ( n , m ) ∈ D 1 × D 2 is continuously Cauchy.

Let ( y l , y k ) ( l , k ) ∈ D 3 × D 4 be a net of Cauchy pair in ( Y , V ) . Thus, ( y l , y k ) ∈ V eventually for all V ∈ V . Since the pair of nets ( f n , f m ) ( n , m ) ∈ D 1 × D 2 is also a Cauchy pair, ( f n , f m ) ∈ ( V , U ) eventually. Hence, ( f n ( V 1 ) , f m ( V 2 ) ) ⊆ U eventually, where V = V 1 × V 2 . Therefore, we have ( f n ( y l ) , f m ( y k ) ) ∈ U eventually. Thus, entourage-entourage uniform space ( U C ( Y , Z ) , U V , U ) is admissible.□

Does there exist any uniformity on U C ( Y , Z ) which is both admissible and splitting? The answer is yes. In the following, we provide an example to show this fact. This example also highlights applications of net-theoretic characterization of admissibility obtained in Theorem 3.15.

Let X be a non-empty set. We call a uniformity ℐ on X an indiscrete uniformity if it has only one entourage, that is, X × X . In an indiscrete uniform space, every pair of nets is a Cauchy pair.

Example 3.7

Let ( Y , U ) and ( Z , V ) be two uniform spaces, where V is the indiscrete uniformity over Z. The point-entourage uniformity, U p , V generated over the class of all uniform continuous function U C ( Y , Z ) is splitting in view of Theorem 3.14.

Now, let f n , f m ′ ( n , m ) ∈ D 1 × D 2 be pairwise Cauchy in U C ( Y , Z ) . We show that it is continuously Cauchy. Let y k , y l ′ ( k , l ) ∈ D 3 × D 4 be pairwise Cauchy nets in Y. As V is indiscrete uniformity, the pair of nets f n ( y k ) , f m ′ ( y l ′ ) is pairwise Cauchy. Therefore, f n , f m ′ is continuously Cauchy. Hence, U C ( Y , Z ) is admissible, in view of Theorem 3.15.

Remark 3.2

Consider the point-entourage uniformity over the space of all uniform functions U C ( ℝ , ℝ ) , with Euclidean uniformity. Then the convergence of the sequence of functions in the topology generated by point-entourage uniformity coincides with the point-wise convergence of the sequence of the function.

4 Conclusion

This study establishes that the uniformly continuous mappings between uniform spaces do possess interesting uniform structures. We have also shown that net theory can successfully be used in the realm of uniform spaces. The authors have not come across similar work in the literature so far in the domain of uniform spaces. The present work is expected to encourage researchers working in the field of metric spaces and topologies as uniform spaces lie between metric spaces and topological spaces. Also, one can develop the concept of dual uniformity in the line of the studies carried out in [6].

To begin with, let ( Y , U ) and ( Z , V ) be two uniform spaces, U be a uniformity on U C ( Y , Z ) . For ℋ ⊆ U C ( Y , Z ) × U C ( Y , Z ) , U ∈ U , V = V 1 × V 2 ∈ V , we define

( ℋ , U ) = { ( f 2 − 1 ( U ) , g 2 − 1 ( U ) ) | ( f , g ) ∈ ℋ } ,

S ( T ) = ( ℋ , U ) | H ∈ T , U ∈ U .

Does S ( T ) form a subbase for a uniformity on U Z ( Y ) , where

U Z ( Y ) = { f 2 − 1 ( U ) | f ∈ U C ( Y , Z ) , U ∈ U } ?

If yes, do properties of this uniformity depend on that of U on U C ( Y , Z ) and vice versa? How to define splittingness and admissibility for such spaces and how they are related to those of U on U C ( Y , Z ) ?

Acknowledgments

This research was supported by Deanship of Scientific Research at Prince Sattam bin Abdulaziz University, Al kharj, Kingdom of Saudi Arabia. The authors sincerely thank the reviewers of the paper for their valuable suggestions which helped improve the quality of the paper.

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Received: 2020-01-25

Revised: 2020-10-28

Accepted: 2020-11-02

Published Online: 2020-12-13

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/math-2020-0110

Keywords for this article

uniform space; function space; splittingness; admissibility; uniformly continuous mappings

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