On Continuity of Functions between Vector Metric Spaces

Abstract

We introduce vectorial and topological continuities for functions defined on vector metric spaces and illustrate spaces of such functions. Also, we describe some fundamental classes of vector valued functions and extension theorems.

1. Introduction and Preliminaries

Let be a Riesz space. If every nonempty bounded above (countable) subset of has a supremum, then is called Dedekind (-) complete. A subset in is called order bounded if it is bounded both above and below. We write if is a decreasing sequence in such that . The Riesz space is said to be Archimedean if holds for every . A sequence is said to be -convergent (or order convergent) to if there is a sequence in such that and for all , where for any . We will denote this order convergence by . Furthermore, a sequence is said to be -Cauchy if there exists a sequence in such that and for all and . The Riesz space is said to be -Cauchy complete if every -Cauchy sequence is -convergent.

An operator between two Riesz spaces is positive if for all . The operator is said to be order bounded if it maps order bounded subsets of to order bounded subsets of . The operator is called -order continuous if in implies in . Every -order continuous operator is order bounded. If for all , the operator is a lattice homomorphism. For further information about Riesz spaces and the operators on Riesz spaces, we refer to [1, 2].

In [3], a vector metric space is defined with a distance map having values in a Riesz space, and some results in metric space theory are generalized to vector metric space theory. Some fixed point theorems in vector metric spaces are given in [3–7]. Actually, the study of metric spaces having value on a vector space has started by Zabrejko in [8]. The distance map in the sense of Zabrejko takes values from an ordered vector space. We use the structure of lattice with the vector metrics having values in Riesz spaces; then we have new results as mentioned above. This paper is on some concepts and related results about continuity in vector metric spaces.

Definition 1. Let be a nonempty set and let be a Riesz space. The function is said to be a vector metric (or -metric) if it satisfies the following properties:(vm1) if and only if ,(vm2) for all .Also the triple is said to be vector metric space.

Definition 2. Let be a vector metric space.(a)A sequence in is vectorially convergent (or is -convergent) to some , if there is a sequence in such that and for all . We will denote this vectorial convergence by .(b)A sequence in is called -Cauchy sequence whenever there exists a sequence in such that and for all and .(c)The vector metric space is called -complete if each -Cauchy sequence in is -convergent to a limit in .

One of the main goals of this paper is to demonstrate the properties of functions on vector metric spaces in a context more general than the continuity in metric analysis. Hence, the properties of Riesz spaces will be the tools for the study of continuity of vector valued functions. The result we get from here is continuity in general sense, an order property rather than a topological feature.

In Section 2, we consider two types of continuity on vector metric spaces. This approach distinguishes continuities vectorially and topologically. Moreover, vectorial continuity examples are given and the relationship between vectorial continuity of a function and its graph is demonstrated. In Section 3, equivalent vector metrics, vectorial isometry, vectorial homeomorphism definitions, and examples related to these concepts are given. In Section 4, uniform continuity is discussed and some extension theorems for functions defined on vector metric spaces are given. Finally, in Section 5, a uniform limit theorem on a vector metric space is given, and the structure of vectorial continuous function spaces is demonstrated.

2. Topological and Vectorial Continuity

We now introduce two types of continuity in vector metric spaces. For two elements and in a Riesz space, we write to indicate that but .

Definition 3. Let and be vector metric spaces, and let .(a)A function is said to be topologically continuous at if for every in there exists some in such that whenever and . The function is said to be topologically continuous if it is topologically continuous at each point of .(b)A function is said to be vectorially continuous at if in implies in . The function is said to be vectorially continuous if it is vectorially continuous at each point of .

Theorem 4. Let and be vector metric spaces where is Archimedean. If a function is topologically continuous, then is vectorially continuous.

Proof. Suppose that . Then there exists a sequence in such that and for all . Let be any nonzero positive element in . By topological continuity of at , there exist elements in such that implies for all . Then there exist elements in such that implies for all . Since is Archimedean, . Hence, .

Vectorially continuous functions have a number of nice characterizations.

Corollary 5. For a function between two vector metric spaces and the following statements hold.(a)If is Dedekind -complete and is vectorially continuous, then whenever .(b)If is Dedekind -complete and whenever , then the function is vectorially continuous.(c)Suppose that and are Dedekind -complete. Then, the function is vectorially continuous if and only if whenever .

Proof. (a) If , then . By the vectorial continuity of the function , there is a sequence in such that and for all . Since is Dedekind -complete, holds.
(b) Let in . Then there is a sequence in such that and for all . Since is Dedekind -complete, holds. By the hypothesis, , and so in .
(c) Proof is a consequence of (a) and (b).

Example 6. Let be a vector metric space. If and , then ; that is, the vector metric map from to is vectorially continuous. Here, is equipped with the -valued vector metric defined as for all , , and is equipped with the absolute valued vector metric .

We recall that a subset of a vector metric space is called -closed (or vector closed) whenever and imply .

Theorem 7. Let and be vector metric spaces. If a function is vectorially continuous, then for every -closed subset of the set is -closed in .

Proof. For any , there exists a sequence in such that . Since the function is vectorially continuous, . But the set is -closed, so . Then, the set is -closed.

If and are two Riesz spaces, then is also a Riesz space with coordinatewise ordering defined as for all . The Riesz space is a vector metric space, equipped with the biabsolute valued vector metric defined as for all , .

Let and be two vector metrics on which are -valued and -valued, respectively. Then the map defined as for all is an -valued vector metric on . We will call double vector metric.

Example 8. Let and be vector metric spaces.(i)Suppose that and are vectorially continuous functions. Then the function from to defined by for all is vectorially continuous with the double vector metric and the biabsolute valued vector metric .(ii)Let and be the projection maps. Any function can be written as for all where and . If is vectorially continuous, so are and since and are vectorially continuous.

Let and be vector metric spaces. Then is a vector metric space, equipped with the -valued product vector metric defined as for all , .

Consider the vector metric space . The projection maps and defined on which are -valued and -valued, respectively, are vectorially continuous. For any function from a vector metric space to , the function is vectorially continuous if and only if both and are vectorially continuous.

Example 9. Let and be vector metric spaces. Suppose that and are vectorially continuous functions. Then the function from to defined by for all , is vectorially continuous with the product vector metric and the absolute valued vector metric .

Example 10. Let and be vector metric spaces. Suppose that and are vectorially continuous functions. Then the function from to defined by for all , is vectorially continuous with the product vector metric and the biabsolute valued vector metric .

The last three examples inspire the following results.

Corollary 11. (a) If and are vectorially continuous functions, then the function from to defined by for all is vectorially continuous with the -valued double vector metric and the -valued product vector metric .
(b) Let be a Riesz space. If and are vectorially continuous functions, then the function from to defined by for all , is vectorially continuous with the -valued product vector metric and the absolute valued vector metric .
(c) If and are vectorially continuous functions, then the function from to defined by for all , is vectorially continuous with the -valued and -valued product vector metrics.

We have the next proposition for any product vector metric.

Proposition 12. Let be a sequence in and let . Then, if and only if and .

Now let us give relevance between vectorial continuity of a function and closeness of its graph.

Corollary 13. Let and be vector metric spaces and let be a function from to . For the graph of , the following statements hold.(a)The graph is -closed in if and only if for every sequence with and we have .(b)If the function is vectorially continuous, then the graph is -closed.(c)If the function is vectorially continuous at , then the induced function defined by is vectorially continuous at .

Proof. For the proof of (a), suppose that the graph is -closed. If and , then we have by Proposition 12. Hence , and so . Conversely, suppose that is a sequence in such that . By Proposition 12, and . Then , and so .
Proofs of (b) and (c) are similar to the proof of (a).

3. Fundamental Vector Valued Function Classes

Definition 14. The -valued vector metric and -valued vector metric on are said to be -equivalent if for any and any sequence in

Lemma 15. For any two -valued vector metrics and on , the following statements are equivalent.(a)There exist some in such that for all .(b)There exist two positive and -order continuous operators and from to itself such that and for all .

Proof. Let and be two operators defined as and for all . If (a) holds, then and are positive and -order continuous operators and satisfy and for all . Conversely, (a) holds since every (-)order continuous operator is order bounded [1, 1.54].

Now, we give the following result for the equivalence of vector metrics.

Theorem 16. An -valued vector metric and an -valued vector metric on are -equivalent if there exist positive and -order continuous two operators , such that for all .

Example 17. Suppose that the ordering of is coordinatewise.(a)Let and be -valued and -valued vector metrics on , respectively, defined as  where and . Consider the two operators ;   and ;   for all . Then, the operators and are positive and -order continuous, and (6) is satisfied. Hence, the metrics and are -equivalent on .(b)Let and be -valued and -valued vector metrics on , respectively, defined as  where , , and . Let and be two operators defined as and . Then, the operators and are positive and -order continuous. The condition (6) is satisfied. So, the vector metrics and are -equivalent on . On the other hand, if is another -valued vector metric on defined as  where , , and , and the operator is defined as , then the vector metrics and are -equivalent on .

Remark 18. Vectorial continuity is invariant under equivalent vector metrics.

Let us show how an isometry is defined between two vector metric spaces.

Definition 19. Let and be vector metric spaces. A function is said to be a vector isometry if there exists a linear operator satisfying the following two conditions: (i) for all ,(ii) implies for all .If the function is onto, and the operator is a lattice homomorphism, then the vector metric spaces and are called vector isometric.

Remark 20. A vector isometry is a vectorial distance preserving one-to-one function.

Example 21. Let be -valued vector metric and let be -valued vector metric on defined as where and . Consider the identity mapping on and the operator defined as for all . Then, the identity mapping is a vector isometry. So, the vector metric spaces and are vector isometric.

Definition 22. Let and be vector metric spaces. A function is said to be a vector homeomorphism if is one-to-one and vectorially continuous and has a vectorially continuous inverse on . If the function is onto, then the vector metric spaces and are called vector homeomorphic.

Remark 23. A vector homeomorphism is one-to-one function that preserves vectorial convergence of sequences.

By Theorem 7, we can develop another characterization result for vector homeomorphisms.

Theorem 24. An onto vector homeomorphism is a one-to-one function that preserves vector closed sets.

Proof. Let be an -homeomorphism. Since is a one-to-one function and its inverse is vectorially continuous, by Theorem 7 for every -closed set in , is -closed in .

The following example illustrates a relationship between vectorial equivalence and vector homeomorphism.

Example 25. Let and be two -equivalent vector metrics on . Then the vector metric spaces and are vector homeomorphic under the identity mapping. On the other hand, if any two vector metric spaces and are vector homeomorphic under a function , then the vector metrics and defined as for all are -equivalent vector metrics on .

4. Extension Theorems on Continuity

If and are vector metric spaces, , and is vectorially continuous, then we might ask whether there exists a vectorial continuous extension of . Below, we deal with some simple extension techniques.

Theorem 26. Let and be vector metric spaces, and let and be vectorially continuous functions from to . Then the set is an -closed subset of .

Proof. Let . Suppose and . Since and are vectorially continuous, there exist sequences and such that , and , for all . Then for all . So ; that is, . Hence, the set is an -closed subset of .

The following corollary is a consequence of Theorem 26.

Corollary 27. Let and be vector metric spaces and let and be vectorially continuous functions from to . If the set is -dense in , then .

Definition 28. Let and be vector metric spaces.(a)A function is said to be topological uniformly continuous on if for every in there exist some in such that for all , whenever .(b)A function is said to be vectorial uniformly continuous on if for every -Cauchy sequence the sequence is -Cauchy.

Theorem 29. Let and be vector metric spaces where is Archimedean. If a function is topological uniformly continuous, then is vectorial uniformly continuous.

Proof. Suppose that is an -Cauchy sequence. Then there exists a sequence in such that and for all and . Let be any nonzero positive element in . By topological uniform continuity of , there exist some in such that the inequality implies for all and . Then there exist elements in such that implies for all and . Since is Archimedean, . Hence, the sequence is -Cauchy.

Example 30. (a) For a vector isometry between two vector metric spaces and , the function is vectorial uniformly continuous if is positive and -order continuous.
(b) For an element in a vector metric space , the function defined by for all is vectorial uniformly continuous.
(c) For a subset of a vector metric space where is Dedekind complete, the function defined by for all is vectorial uniformly continuous.

The following theorem enables us to establish an extension property for the functions between vector metric spaces.

Theorem 31. Let be -dense subset of a vector metric space and let be an -complete vector metric space where is Archimedean. If is a topological uniformly continuous function, then has a unique vectorially continuous extension to which is also topological uniformly continuous.

Proof. Let . Then there exists a sequence in such that . Since the function is vectorial uniformly continuous on by Theorem 29, the sequence is -Cauchy in -complete vector metric space . Hence, there exists an element such that . Define an extension of on by . This extension is well-defined; that is, the value of at is independent of the particular sequence chosen -convergent to . We need to show that is topological uniformly continuous on .
Let in . Choose in such that for    implies . Let satisfy . Choose two sequences and in such that and . Then, in . Fix such that for all . Then for all . By the vectorial uniform continuity of , and are -Cauchy sequences in . Since is -complete, there exist two points and in such that and . By the definition of the function , we have and . Then in , and therefore, . This shows that is a topological uniformly continuous function on .

5. Vectorially Continuous Function Spaces

Definition 32. Let be any nonempty set and let be a vector metric space. Then a sequence of functions from to is said to be uniformly -convergent to a function , if there exists a sequence in such that and holds for all and all .

Now we give the uniform limit theorem in vector metric spaces.

Theorem 33. Let be a sequence of vectorially continuous functions between two vector metric spaces and . If is uniformly -convergent to , then the function is vectorially continuous.

Proof. Let such that in . Since is uniformly -convergent to , there is a sequence in such that and for all . For each , there is a sequence in such that and for all by the vectorial continuity of . Note that for This implies .

Let be a nonempty subset of a vector metric space . -diameter of , , is defined as . The set is called -bounded if there exists an element in such that for all . Every -bounded subset of has an -diameter whenever is Dedekind complete.

Definition 34. A function between two vector metric spaces and is called vectorial bounded if maps -bounded subsets of to -bounded subsets of .

Theorem 35. A function between two vector metric spaces and is vectorial bounded if there exists a positive operator such that for all .

Let and be the collections of all vectorially continuous and topologically continuous functions between a vector metric space and a Riesz space , respectively. By Theorem 4, whenever is Archimedean.

Theorem 36. The spaces and are Riesz spaces with the natural partial ordering defined as whenever for all .

Consider an -bounded vector metric space and a Dedekind complete Riesz space . Let be a subset of such that, for any in , there is a positive operator satisfying for all . Since the Birkhoff inequality ([1, 1.9(2)]; [2, 12.4(ii)]) holds for all , then the subset is a Riesz subspace of . By Theorem 35, every is vectorial bounded function. This argument gives us the following result.

Corollary 37. The subset described above is a vector metric space equipped with the -valued uniform vector metric defined as .

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.