## Function Spaces, Hyperspaces, and Asymmetric and Fuzzy Structures

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# On the Fuzzy Number Space with the Level Convergence Topology

**Academic Editor:**Salvador Romaguera

#### Abstract

We characterize compact sets of endowed with the level convergence topology . We also describe the completion of with respect to its natural uniformity, that is, the pointwise uniformity , and show other topological properties of , as separability. We apply these results to give an Arzela-Ascoli theorem for the space of -valued continuous functions on a locally compact topological space equipped with the compact-open topology.

#### 1. Introduction

Not long after Chang and Zadeh [1] introduced the concept of fuzzy numbers with the consideration of the properties of probability functions, this topic became an active and important area of research because of the development of various theories of fuzzy numbers and their applications in fuzzy optimization, fuzzy decision making, and so forth. In this framework, the study of several types of convergence in fuzzy number spaces plays a central role (see, e.g., [2, 3]). This paper is a contribution to the study of one of these types of convergence, the so-called level convergence.

Let denote the family of all fuzzy subsets on the real numbers , that is, the set of functions . For and , the -level set of is defined by The fuzzy number space is the set of elements of satisfying the following properties:(1)is normal, that is, there exists an with ;(2) is convex, that is, for all ;(3) is upper-semicontinuous;(4) is a compact set in . The -level set of is a compact interval for each . We denote . Notice that every can be considered a fuzzy number since can be identified with the fuzzy number defined as Recall also that Goetschel and Voxman provided the following representation theorem of fuzzy numbers (see [4]).

Theorem 1.1. *Let and , . Then the pair of functions and has the following properties:*(i)* is a bounded left continuous nondecreasing function on ;*(ii)* is a bounded left continuous nonincreasing function on ;*(iii)* and are right continuous at ;*(iv)*. **Conversely, if a pair of functions and satisfies the above conditions (i)â€“(iv), then there exists a unique such that for each . *

Actually, the function is a nondecreasing function on , and the function is a nonincreasing function on . This is an easy consequence of condition (iii) of the Goetschel and Voxman representation theorem.

Since is a compact interval of for all , given two elements , and , we can consider the Hausdorff distance between and . It is well known that for compact intervals of the reals, the Hausdorff distance is given by The Hausdorff distance allows us to endow with the following metric.

*Definition 1.2 (see [4, 5]). *For ,
is a metric on . It is called the supremum metric on , and is a complete metric space.

*Definition 1.3. *We say that the net levelly converges to , if for any .

An easy characterization of level convergence is as follows.

Proposition 1.4. *The net levelly converges to if and only if and for each . *

Notice that a net -converges to if and only if converges uniformly to and converges uniformly to . Thus, the -convergence implies the -convergence. The converse fails to be true (see Example 2.1 in [6]). In [7], Fang and Huang described the topology associated with this level convergence in . They showed that is a Hausdorff, first countable topological space.

By means of the Goetschel-Voxman's representation theorem, we can regard as a subspace of the product space (which can be identified with in a canonical way). Indeed, the correspondence defines a homeomorphism from into , where denotes the pointwise topology. Thus, basic open sets in are given by for , , and .

In Section 2, we will first prove that the subset of all elements of such that and are continuous is dense in , and we will also study compact sets of , giving a characterization to be compared with the one provided in [7]. In Section 3, we will provide a description of the completion of with the pointwise uniformity , and we will show that its underlying topological space is separable, LindelĂ¶f (thus, strongly paracompact) and a Baire space. In the last section, an Arzela-Ascoli theorem for the space of -valued continuous functions will be stated.

#### 2. Two Useful Results

Our first result provides a dense subset of . We will consider the subset of given by

Theorem 2.1. * is dense in . *

*Proof. * Given a basic open set in with for , consider where is defined as
and linearly at intermediate values, and is defined as
and linearly at intermediate values. It is an easy matter to check that and that . Thus, is dense in .

By [8, Theorem 6.1], we know that the topology of pointwise convergence and the topology of uniform convergence coincide in the set of all monotone continuous functions defined on the unit interval. Taking into account these facts and properties (i) and (ii) of Theorem 1.1, it is clear that where, as usual, stands for the topology induced by the metric .

There are several characterizations of compact subsets of spaces of functions depending on the topology we endow the space with. For instance, if we deal with the topology of uniform convergence on some spaces of continuous functions, the Ascoli-Arzela's Theorem asserts that is compact if and only if is pointwise bounded, closed and equicontinuous. Dealing with the pointwise convergence topology in more general spaces of functions , it is well known that is compact if and only if is closed and pointwise bounded. Hence, it is clear that closed and pointwise bounded subsets satisfying are also equicontinuous.

*Definition 2.2 (see [5]). *It is said that is *uniformly support bounded,* if there is a constant such that for all .

As it is customary, we say that a subset of is *pointwise closed,* if is a closed subset of the product space (or equivalently, of the product space ).

Theorem 2.3 (compare [7]). * A subset of is compact if and only if is pointwise closed and uniformly support bounded. *

*Proof. * Suppose that is a compact subset of . Then is compact in , that is, is pointwise closed and pointwise bounded which implies that is pointwise closed and uniformly support bounded. Conversely, if is uniformly support bounded, is pointwise bounded. Thus, if is also pointwise closed, is a compact subset of . Since compactness is an absolute topological property, is a compact subset of .

*Example 2.4. *There are examples of -closed and uniformly support bounded subsets of which are not -compact. Consider a uniformly support bounded Cauchy sequence in that does not converge (see, e.g., [9]). Since is first countable, the set is uniformly support bounded and -closed in but it is not compact.

We also notice that there exist compact metric subsets of such that .

*Example 2.5. * Let be the sequence in defined as
extended linearly at intermediate values, and for all . It is clear that the sequence converges to with if and if . Then is a compact subset of . Since the function is not continuous, is not -compact.

#### 3. Topological and Uniform Properties of

First we prove the following.

Theorem 3.1. * is separable. *

*Proof. * We will denote by CMI (resp., CMD,) the set of all nondecreasing monotone continuous functions (resp., the set of all nonincreasing monotone continuous functions) on . It is well known that the space of all continuous real-valued functions on endowed with the topology of the uniform convergence is separable by the classical Stone-Weierstrass theorem. Separability is a hereditary property in the realm of metric spaces, so that the spaces CMI and CMD equipped with the -topology are separable spaces. Since the topologies of uniform convergence and pointwise convergence coincide on , the product space is a separable metric space and, consequently, the space is separable. The result now follows from Theorem 2.1.

The correspondence is a uniform isomorphism when we consider the natural admissible structure on and the pointwise uniform structure (or pointwise uniformity) uniform structure on which has as a base the sets () of the following form: where, for all , runs over . Throughout, we shall freely identify, without explicit mention, the uniformity with the restriction of the uniformity to . It is a well-known fact that the uniform space is not complete ([9]). Hence, our next goal is to describe its completion . Remark that a consequence of the previous theorem is the following.

Corollary 3.2. *The underlying topological space of is separable. *

Theorem 3.3. *The completion of the uniform space is the subspace of whose elements verify the following properties: *(i)* is a nonincreasing function on ; *(ii)* is a nondecreasing function on ; *(iii)*. *

*Proof. *The uniform space is a product of complete uniform spaces, so that it is complete. Hence is a complete uniform space containing as dense subset. By [10, Theorem ], can be identified with the completion of .

We move on to the description of . First notice that if , then is nonincreasing and is nondecreasing since a pointwise limit of a net of nonincreasing (resp., nondecreasing,) functions is a nonincreasing (resp., a nondecreasing,) function. Moreover, since every satisfies , condition (iii) is an easy consequence of the definition of the limit of a net. Thus, every element in verifies conditions (i)â€“(iii). The converse follows from an argument similar to the one used in Theorem 2.1.

Corollary 3.4. *The uniformity induced by the metric and the uniformity induce the same topology on the set , but the first is finer than the second. *

As we have seen above, the topology induced by the metric and the level topology coincide in . This question dealing with seems to have received almost no attention in the literature. We borrow from [11] the necessary techniques in order to obtain a characterization of subsets satisfying that . Recall that a function between two first countable spaces and is continuous if and only if for every sequence in the space . For every and , we will denote by the limit of when . First we need the following helpful result.

Lemma 3.5. * If is a -convergent sequence, then for all and , there exists such that and for all whenever . *

*Proof. * The proof for proceeds along the same lines as the proof for so that we only show our lemma for the sequence . Let be the limit point of . Since is left-continuous on and bearing in mind that is nondecreasing, there exists such that for all .

Now choose such that for every , we have and . Then, for all and all , we obtain
The proof is completed by invoking the fact that is left continuous at for each .

Theorem 3.6. *For a subset of the following conditions are equivalent:*(i)*; *(ii)*for each sequence which levelly converges to an element in , we have and for all ; *(iii)*every -convergent sequence is right equicontinuous. *

* Proof. * (i)(ii) easily follows from the definition of uniform convergence and (ii)(iii) runs along similar lines to the ones used in Lemma 3.5. To see (iii)(i), we need to prove that the identity map from onto is continuous, that is, if converges to in , then -converges to . Assume that does not -converge to . Then we can clearly assume that does not converge uniformly to . Under this assumption, we can find , an infinite sequence of natural numbers and a sequence such that . Let us suppose, with no loss of generality that the sequence converges to . Notice that has infinite many different elements because pointwise converges to . Assume that there exists an infinite subsequence of whose elements are greater than . For simplicity, we shall denote this subsequence again by . Our hypothesis and the fact that and the elements of are right continuous at allow us to find a sequence such that
which leads us to a contradiction. On the other hand, if there is an infinite subsequence of whose elements are less than , by means of Lemma 3.5, an argument similar to the previous one also permits us to obtain a contradiction. Thus, the proof is complete.

We finish this section by giving some topological properties of and the underlying topological space of the uniform space . Recall that a topological space is said to be *hemicompact* if it has a sequence of compact subsets such that every compact subset of the space lies inside some compact set in the sequence. The space is said to be *LindelĂ¶f,* if every open cover has a countable subcover.

Theorem 3.7. * is a hemicompact space. *

*Proof. *Let denote the set of all such that . It is clear that these sets are pointwise closed and pointwise bounded so that they are compact subsets of . Since is a closed subset of , they are compact subsets of . Now, given a compact subset of , there exists a constant such that . Thus, . Thus, if , . This completes the proof.

Hemicompact spaces are LindelĂ¶f, and every LindelĂ¶f space is strongly paracompact ([10, Theorem ]) so that we conclude the following.

Corollary 3.8. * is a LindelĂ¶f space. In particular, it is strongly paracompact and, consequently, a normal space. *

Recall that a (strongly) paracompact space is DieudonnĂ© complete, that is, the finest uniformity on is complete. Thus, the natural uniformity on is not complete but the finest uniformity is.

We close the section by showing that both and are Baire spaces, that is, the Baire category theorem holds in and in : for each sequence of open dense subsets of (resp., of ) the intersection is a dense set.

Theorem 3.9. * and are Baire spaces. *

*Proof. * Since a topological space that contains a dense subspace which is a Baire space is a Baire space itself ([10, (b)]), it suffices to prove that is a Baire space. To see this, first recall that the topologies and coincide in . Next, notice that the space is a metric complete space and apply that metric complete spaces are Baire spaces.

#### 4. An Arzela-Ascoli Theorem for

In this section, will denote a locally compact topological space. We shall assume that the space of -valued continuous functions is endowed with the compact-open topology and given a subset of , we shall write for any .

As in [12], we have the following.

Theorem 4.1. * Let . Then is compact if and only if *(i)* is closed; *(ii)*For any , is pointwise closed and uniformly support-bounded; *(iii)* is evenly equicontinuous. *

When we deal with , the definition of evenly continuous can be translated as follows.

*Definition 4.2. *A subset of is said to be *evenly equicontinuous* if for any , , and , there exists an open neighborhood of and such that if and
then
for all .

For the sake of completeness, we give the proof of Theorem 4.1 following the same pattern as in [12].

*Proof of Theorem 4.1. *
We have the following.*Necessity.* Let us suppose that is compact in the compact-open topology of .(i) is closed since is Hausdorff as is so ([7]).(ii) It is apparent that, for any , the map defined to be is continuous. Hence, since is compact, then is compact in for each . Therefore, by Theorem 2.3, is, for any , pointwise closed and uniformly support bounded.(iii) Fix , , and . Let such that . Let such that
?for all . It is apparent that is closed and, consequently, it is compact. Let us consider the continuous map defined from to . Then
?which is to say that the compact subset is included in . By Wallace theorem ([10, Theorem ]), there exists an open neighborhood, , of such that for all and all , that is,
*Sufficiency*. Let us first check that , the closure of in the pointwise convergence topology, is also evenly equicontinuous. Fix , , and . Then, by the evenly equicontinuity of , there exists a neighborhood of and such that if and
then
for all . Let us take such that
We must check
for all . To this end, let us choose a net in which converges pointwise to . Hence, eventually,
Consequently, by the evenly equicontinuity of ,
for all , which yields the desired fact.

As in the necessity case, let us consider the map defined from to . We claim that is continuous. To this end, fix , , and . Since is evenly equicontinuous, there exists an open neighborhood of and such that if , then , where
Since it is also clear that is open in and in open in , then is continuous.

Similarly, by the evenly equicontinuity of , it can be deduced that if , then .

Next, we shall check that the closure of in the pointwise convergence topology, , coincides with the closure of in the compact-open topology, . It suffices to check that, in this context, the topology of pointwise convergence is larger than the compact-open topology. Thus, it is well known that, given a compact subset of , and , the set given by
is a member of a subbase for the compact-open topology in .

Fix . Then , which is compact in , belongs to , where
From the previous paragraphs, is open. Hence, by Wallace theorem ([10, Theorem ]), there exists an open neighborhood, , of in the pointwise convergence topology such that , which is to say that . This means that the topology of pointwise convergence is larger than the compact-open topology.

Finally, we claim that is compact in the compact-open topology. Indeed, we shall check that it is compact in the topology of pointwise convergence, but we have already checked that both topologies coincide on . To this end, we can consider as a closed subset of the product . By Tychonoffâ€™s theorem, is compact in the topology of pointwise convergence since, by Theorem 2.3, is compact for every . Thus, is compact and, since by hypothesis, is closed, then is compact, as was proved.

#### Acknowledgments

This research is supported by the Spanish Ministry of Science and Education (Grant nos. MTM2011-22457 and MTM2011-23118) and by Bancaixa (Project P11B2011-30).

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Copyright Â© 2012 J. J. Font et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.