# On the Fuzzy Number Space with the Level Convergence Topology

Authors: J. J. Font; A. Miralles; M. Sanchis
Journal: Journal of Function Spaces and Applications
 (2012)

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## Abstract

We characterize compact sets of𝔼1 endowed with the level convergence topology τℓ. We also describe the completion (𝔼1̂,𝒰̂) of 𝔼1 with respect to its natural uniformity, that is, the pointwise uniformity 𝒰, and show other topological properties of 𝔼1̂, as separability. We apply these results to give an Arzela-Ascoli theorem for the space of (𝔼1,τℓ)-valued continuous functions on a locally compact topological space equipped with the compact-open topology.


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## Body

## 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.LetF(ℝ) denote the family of all fuzzy subsets on the real numbers ℝ, that is, the set of functions u:ℝ→[0,1]. For u∈F(ℝ) and λ∈[0,1], the λ-level set of u is defined by
(1.1)[u]λ:={x∈ℝ:u(x)≥λ},λ∈]0,1],[u]0:=clℝ{x∈ℝ:u(x)>0}.
The fuzzy number space 𝔼1 is the set of elements u of F(ℝ) satisfying the following properties:(1)
uis normal, that is, there exists an x0∈ℝ with u(x0)=1;(2)
u is convex, that is, u(λx+(1-λ)y)≥min{u(x),u(y)} for all x,y∈ℝ,λ∈[0,1];(3)
u is upper-semicontinuous;(4)
[u]0 is a compact set in ℝ.The λ-level set [u]λ of u∈𝔼1 is a compact interval for each λ∈[0,1]. We denote [u]λ=[u-(λ),u+(λ)]. Notice that every r∈ℝ can be considered a fuzzy number since r can be identified with the fuzzy number r~ defined as
(1.2)r~(t):={1ift=r,0ift≠r.
Recall also that Goetschel and Voxman provided the following representation theorem of fuzzy numbers (see [4]).Theorem 1.1.
Letu∈𝔼1 and [u]λ=[u-(λ),u+(λ)], λ∈[0,1]. Then the pair of functions u-(λ) and u+(λ) has the following properties:(i)
u-(λ) is a bounded left continuous nondecreasing function on ]0,1];(ii)
u+(λ) is a bounded left continuous nonincreasing function on ]0,1];(iii)
u-(λ) and u+(λ) are right continuous at λ=0;(iv)
u-(1)≤u+(1).Conversely, if a pair of functions α(λ) and β(λ) satisfies the above conditions (i)–(iv), then there exists a unique u∈𝔼1 such that [u]λ=[α(λ),β(λ)] for each λ∈[0,1].Actually, the functionu-(λ) is a nondecreasing function on [0,1], and the function u+(λ) is a nonincreasing function on [0,1]. This is an easy consequence of condition (iii) of the Goetschel and Voxman representation theorem.Since[u]λ is a compact interval of ℝ for all λ∈[0,1], given two elements u,v∈𝔼1, [u]λ=[u-(λ),u+(λ)] and [v]λ=[v-(λ),v+(λ)], we can consider the Hausdorff distance d between [u]λ and [v]λ. It is well known that for compact intervals of the reals, the Hausdorff distance is given by
(1.3)d([u]λ,[v]λ)=max{|u-(λ)-v-(λ)|,|u+(λ)-v+(λ)|}.
The Hausdorff distance allows us to endow 𝔼1 with the following metric.Definition 1.2 (see [4, 5]).
Foru,v∈𝔼1,
(1.4)d∞(u,v):=supλ∈[0,1]d([u]λ,[v]λ)=supλ∈[0,1]max{|u-(λ)-v-(λ)|,|u+(λ)-v+(λ)|},
is a metric on 𝔼1. It is called the supremum metric on 𝔼1, and (𝔼1,d∞) is a complete metric space.Definition 1.3.
We say that the net{uk}k∈D⊂𝔼1 levelly converges to u∈𝔼1, if limkd([uk]λ,[u]λ)=0 for any λ∈[0,1].An easy characterization of level convergence is as follows.Proposition 1.4.
The net{uk}k∈D⊂𝔼1 levelly converges to u∈𝔼1 if and only if limkuk+(λ)=u+(λ) and limkuk-(λ)=u-(λ) for each λ∈[0,1].Notice that a net{uk}k∈D⊂𝔼1d∞-converges to u∈𝔼1 if and only if limkuk+ converges uniformly to u+ and limkuk- converges uniformly to u-. Thus, the d∞-convergence implies the τℓ-convergence. The converse fails to be true (see Example 2.1 in [6]). In [7], Fang and Huang described the topology τℓ associated with this level convergence in 𝔼1. They showed that (𝔼1,τℓ) is a Hausdorff, first countable topological space.By means of the Goetschel-Voxman's representation theorem, we can regard(𝔼1,τℓ) as a subspace of the product space ℝ[0,1]×ℝ[0,1] (which can be identified with (ℝ×ℝ)[0,1] in a canonical way). Indeed, the correspondence u∈𝔼1↪i(u-,u+) defines a homeomorphism from (𝔼1,τℓ) into (ℝ[0,1]×ℝ[0,1],τp), where τp denotes the pointwise topology. Thus, basic open sets in τℓ are given by
(1.5)U(u,{λ1,…,λn},ϵ):={v∈𝔼1:max1≤i≤n{|v+(λi)-u+(λi)|,|v-(λi)-u-(λi)|}<ϵ},
for {λ1,…,λn}⊂[0,1], u∈𝔼1, and ϵ>0.In Section2, we will first prove that the subset of all elements u of 𝔼1 such that u- and u+ are continuous is dense in 𝔼1, and we will also study compact sets of (𝔼1,τℓ), giving a characterization to be compared with the one provided in [7]. In Section 3, we will provide a description of the completion (𝔼1^,𝒰^) of 𝔼1 with the pointwise uniformity 𝒰, and we will show that its underlying topological space is separable, Lindelöf (thus, strongly paracompact) and a Baire space. In the last section, an Arzela-Ascoli theorem for the space of (𝔼1,τℓ)-valued continuous functions will be stated.


## 2. Two Useful Results

Our first result provides a dense subset of(𝔼1,τℓ). We will consider the subset ℂ1 of 𝔼1 given by
(2.1)ℂ1:={u∈𝔼1∣u-andu+are continuousfunctions}.Theorem 2.1.
ℂ
1 is dense in (𝔼1,τℓ).Proof.
Given a basic open setU(u,{λ1,…,λn},ϵ) in (𝔼1,τℓ) with λi<λi+1 for i=1,2,…,n-1, consider (u1,u2)∈ℂ1 where u1 is defined as
(2.2)u1(λ):={u-(0)ifλ=0,u-(λi)ifλ=λi,i=1,2,…,n,u-(1)ifλ=1
and linearly at intermediate values, and u2 is defined as
(2.3)u2(λ):={u+(0)ifλ=0,u+(λi)ifλ=λi,i=1,2,…,n,u+(1)ifλ=1
and linearly at intermediate values. It is an easy matter to check that (u1,u2)∈ℂ1 and that (u1,u2)∈U(u,{λ1,…,λn},ϵ). Thus, ℂ1 is dense in (𝔼1,τℓ).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 τd∞∣ℂ1=τℓ∣ℂ1 where, as usual, τd∞ stands for the topology induced by the metric d∞.There are several characterizations of compact subsets of spaces of functionsℱ depending on the topology we endow the space with. For instance, if we deal with the topology of uniform convergence τu on some spaces of continuous functions, the Ascoli-Arzela's Theorem asserts that K⊂(ℱ,τu) is compact if and only if K is pointwise bounded, closed and equicontinuous. Dealing with the pointwise convergence topology τp in more general spaces of functions ℱ, it is well known that K⊂(ℱ,τp) is compact if and only if K is closed and pointwise bounded. Hence, it is clear that closed and pointwise bounded subsets A⊂ℱ satisfying τp=τu are also equicontinuous.Definition 2.2 (see [5]).
It is said thatA⊂𝔼1 is uniformly support bounded, if there is a constant C>0 such that max{|u-(0)|,|u+(0)|}≤C for all u∈A.As it is customary, we say that a subsetA of 𝔼1 is pointwise closed, if A is a closed subset of the product space ℝ[0,1]×ℝ[0,1] (or equivalently, of the product space (ℝ×ℝ)[0,1]).Theorem 2.3 (compare [7]).
A subsetK of (𝔼1,τℓ) is compact if and only if K is pointwise closed and uniformly support bounded.Proof.
Suppose thatK is a compact subset of (𝔼1,τℓ). Then K is compact in ℝ[0,1]×ℝ[0,1], that is, K is pointwise closed and pointwise bounded which implies that K is pointwise closed and uniformly support bounded. Conversely, if K⊂𝔼1 is uniformly support bounded, K is pointwise bounded. Thus, if K is also pointwise closed, K is a compact subset of ℝ[0,1]×ℝ[0,1]. Since compactness is an absolute topological property, K is a compact subset of (𝔼1,τℓ).Example 2.4.
There are examples ofτℓ-closed and uniformly support bounded subsets of 𝔼1 which are not τℓ-compact. Consider a uniformly support bounded Cauchy sequence {un}n=1∞ in (𝔼1,τℓ) that does not converge (see, e.g., [9]). Since (𝔼1,τℓ) is first countable, the set K={un}n=1∞ is uniformly support bounded and τℓ-closed in (𝔼1,τℓ) but it is not compact.We also notice that there exist compact metric subsetsK of (𝔼1,τℓ) such that τd∞|K≠τℓ|K.Example 2.5.
Let{(un-,vn+)}n≥1 be the sequence in 𝔼1 defined as
(2.4)un-(λ)={1foreachλ≤12,12foreachλ>12+1n
extended linearly at intermediate values, and vn+=0 for all n≥1. It is clear that the sequence {(un-,vn+)}n≥1 converges to (u-,0) with u-(λ)=1 if λ≤1/2 and u-(λ)=1/2 if λ>1/2. Then K={(un-,vn+)}n≥1∪{(u-,0)} is a compact subset of (𝔼1,τℓ). Since the function u- is not continuous, K is not d∞-compact.


## 3. Topological and Uniform Properties of𝔼1

First we prove the following.Theorem 3.1.
(
𝔼
1
,
τ
ℓ
) 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[0,1]. It is well known that the space C([0,1]) of all continuous real-valued functions on [0,1] 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 τu-topology are separable spaces. Since the topologies of uniform convergence and pointwise convergence coincide on CMI×CMD, the product space (CMI×CMD,τp) is a separable metric space and, consequently, the space ℂ1 is separable. The result now follows from Theorem 2.1.The correspondenceu∈𝔼1↪i(u-,u+) is a uniform isomorphism when we consider the natural admissible structure 𝒰 on (𝔼1,τℓ) and the pointwise uniform structure (or pointwise uniformity) uniform structure 𝒱 on (ℝ[0,1]×ℝ[0,1],τp) which has as a base the sets U(λ1,λ2,…,λn,ε) (n≥1,ε>0) of the following form:
(3.1){((f,g),(h,t))∈(ℝ[0,1]×ℝ[0,1])×(ℝ[0,1]×ℝ[0,1]):|f(λi)-h(λi)|<ε,|g(λi)-t(λi)|<ε,i=1,2,…,n(ℝ[0,1]×ℝ[0,1])},
where, for all n≥1, (λ1,λ2,…,λn) runs over [0,1]n. Throughout, we shall freely identify, without explicit mention, the uniformity 𝒰 with the restriction of the uniformity 𝒱 to i(𝔼1). It is a well-known fact that the uniform space (𝔼1,𝒰) is not complete ([9]). Hence, our next goal is to describe its completion (𝔼1^,𝒰^). Remark that a consequence of the previous theorem is the following.Corollary 3.2.
The underlying topological space of(𝔼1^,𝒰^) is separable.Theorem 3.3.
The completion(𝔼1^,𝒰^) of the uniform space (𝔼1,𝒰) is the subspace of (ℝ[0,1]×ℝ[0,1],𝒱) whose elements (u,v) verify the following properties: (i)
u is a nonincreasing function on [0,1];(ii)
v is a nondecreasing function on [0,1];(iii)
u(1)≤v(1).Proof.
The uniform space(ℝ[0,1]×ℝ[0,1],𝒱) is a product of complete uniform spaces, so that it is complete. Hence clℝ[0,1]×ℝ[0,1]𝔼1 is a complete uniform space containing 𝔼1 as dense subset. By [10, Theorem 8.3.12], clℝ[0,1]×ℝ[0,1]𝔼1 can be identified with the completion of (𝔼1,𝒰).
We move on to the description ofclℝ[0,1]×ℝ[0,1]𝔼1. First notice that if (u,v)∈clℝ[0,1]×ℝ[0,1]𝔼1, then u is nonincreasing and v is nondecreasing since a pointwise limit of a net of nonincreasing (resp., nondecreasing,) functions is a nonincreasing (resp., a nondecreasing,) function. Moreover, since every (u-,u+)∈𝔼1 satisfies u-(1)≤u+(1), condition (iii) is an easy consequence of the definition of the limit of a net. Thus, every element in clℝ[0,1]×ℝ[0,1]𝔼1 verifies conditions (i)–(iii). The converse follows from an argument similar to the one used in Theorem 2.1.Corollary 3.4.
The uniformity induced by the metricd∞ and the uniformity 𝒰 induce the same topology on the set ℂ1, but the first is finer than the second.As we have seen above, the topologyτd∞ induced by the metric d∞ and the level topology τℓ coincide in ℂ1. This question dealing with 𝔼1 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 A⊂𝔼1 satisfying that τd∞|A=τp|A. Recall that a function f between   two first countable spaces X and Y is continuous if and only if f(limnxn)=limnf(xn) for every sequence {xn} in the space X. For every u∈𝔼1 and λ∈[0,1[, we will denote by u(λ+) the limit of u(t) when t→λ+. First we need the following helpful result.Lemma 3.5.
If{un}n=1∞ is a τℓ-convergent sequence, then for all ε>0 and λ0∈]0,1], there exists δ>0 such that un-(λ0)-un-(λ)<ε and un+(λ)-un+(λ0)<ε for all n∈ℕ whenever λ∈[λ0-δ,λ0].Proof.
The proof for{u+}n=1∞ proceeds along the same lines as the proof for {u-}n=1∞ so that we only show our lemma for the sequence {u-}n=1∞. Let u=(u-,u+) be the limit point of {un}n=1∞. Since u- is left-continuous on (0,1] and bearing in mind that u- is nondecreasing, there exists δ>0 such that 0≤u-(λ)-u-(λ0)<ε for all λ∈[λ0-δ,λ0].
Now choosen0 such that for every n≥n0, we have |un-(λ0-δ)-u-(λ0-δ)|<ε and |un-(λ0)-u-(λ0)|<ε. Then, for all n≥n0 and all λ∈[λ0-δ,λ0], we obtain
(3.2)0≤un-(λ0)-un-(λ)≤un-(λ0)-un-(λ0-δ)≤|un-(λ0)-u-(λ0)|+|u-(λ0)-u-(λ0-δ)|+|u-(λ0-δ)-un-(λ0-δ)|≤3ε.
The proof is completed by invoking the fact that un- is left continuous at λ0 for each n<n0.Theorem 3.6.
For a subsetA of 𝔼1 the following conditions are equivalent:(i)
d∞|A=τℓ|A;(ii)
for each sequence{un}n=1∞={(un-,un+)}n≥1⊂A which levelly converges to an element (u-,u+) in A, we have limnun-(λ+)=u-(λ+)   and    limnun+(λ+)=u+(λ+) for all λ∈[0,1[;(iii)
everyτℓ-convergent sequence {un}n=1∞⊂A 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 (𝔼1,τℓ) onto (𝔼1,τu) is continuous, that is, if {(un-,un+)}n≥1 converges to (u-,u+) in (𝔼1,τℓ), then {(un-,un+)}n≥1d∞-converges to (u-,u+). Assume that {(un-,un+)}n≥1 does not d∞-converge to (u-,u+). Then we can clearly assume that {un-}n≥1 does not converge uniformly to u-. Under this assumption, we can find ε>0, an infinite sequence of natural numbers n1<n2<n3<⋯ and a sequence {λnk}k∈ℕ⊂[0,1] such that |unk+(λnk)-u+(λnk)|≥3ε. Let us suppose, with no loss of generality that the sequence {λnk}k∈ℕ converges to λ0∈]0,1]. Notice that {λnk}k∈ℕ has infinite many different elements because {u-}≥1 pointwise converges to u-. Assume that there exists an infinite subsequence of {λnk}k∈ℕ whose elements are greater than λ0. For simplicity, we shall denote this subsequence again by {λnk}k∈ℕ. Our hypothesis and the fact that u- and the elements of {un-}n=1∞ are right continuous at λ0 allow us to find a sequence {nk}k such that
(3.3)|unk+(λnk)-u+(λnk)|≤|unk+(λnk)-unk+(λ0+)|+|unk+(λ0+)-u+(λ0+)|+|u+(λ0+)-u+(λnk)|≤3ε,
which leads us to a contradiction. On the other hand, if there is an infinite subsequence of {λnk}k∈ℕ whose elements are less than λ0, 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(𝔼1,τℓ) and the underlying topological space (𝔼1^,τp) of the uniform space (𝔼1^,𝒰^). 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 Lindelöf, if every open cover has a countable subcover.Theorem 3.7.
(
𝔼
1
^
,
τ
p
) is a hemicompact space.Proof.
LetKn denote the set of all u∈𝔼1^ such that |u-(0)-u+(0)|≤n. It is clear that these sets are pointwise closed and pointwise bounded so that they are compact subsets of (ℝ×ℝ)[0,1]. Since 𝔼1^ is a closed subset of (ℝ×ℝ)[0,1], they are compact subsets of (𝔼1^,τp). Now, given a compact subset K of (𝔼1^,τp), there exists a constant C>0 such that |u-(0)|,|u+(0)|≤C. Thus, |u-(0)-u+(0)|≤2C. Thus, if n>2C, K⊆Kn. This completes the proof.Hemicompact spaces are Lindelöf, and every Lindelöf space is strongly paracompact ([10, Theorem 5.3.11]) so that we conclude the following.Corollary 3.8.
(
𝔼
1
^
,
τ
p
) is a Lindelöf space. In particular, it is strongly paracompact and, consequently, a normal space.Recall that a (strongly) paracompact spaceX is Dieudonné complete, that is, the finest uniformity on X is complete. Thus, the natural uniformity on 𝔼1 is not complete but the finest uniformity is.We close the section by showing that both(𝔼1,τℓ) and (𝔼1^,τp) are Baire spaces, that is, the Baire category theorem holds in (𝔼1,τℓ) and in (𝔼1^,τp): for each sequence G1,G2,… of open dense subsets of (𝔼1,τℓ) (resp., of (𝔼1^,τp)) the intersection ⋂n≥1‍Gn is a dense set.Theorem 3.9.
(
𝔼
1
,
τ
ℓ
) and (𝔼1^,τp) are Baire spaces.Proof.
Since a topological spaceX that contains a dense subspace which is a Baire space is a Baire space itself ([10, 3.9.J (b)]), it suffices to prove that (ℂ1,τℓ) is a Baire space. To see this, first recall that the topologies τd∞ and τℓ coincide in ℂ1. Next, notice that the space (ℂ1,d∞) is a metric complete space and apply that metric complete spaces are Baire spaces.


## 4. An Arzela-Ascoli Theorem forC(X,(𝔼1,τl))

In this section,X will denote a locally compact topological space. We shall assume that the space C(X,(𝔼1,τl)) of 𝔼1-valued continuous functions is endowed with the compact-open topology and given a subset F of C(X,(𝔼1,τl)), we shall write F[x]:={f(x):f∈F} for any x∈X.As in [12], we have the following.Theorem 4.1.
LetF⊂C(X,(𝔼1,τl)). Then F is compact if and only if  (i)
F is closed;(ii)
For anyx∈X, F[x] is pointwise closed and uniformly support-bounded;(iii)
F is evenly equicontinuous.When we deal withC(X,(𝔼1,τl)), the definition of evenly continuous can be translated as follows.Definition 4.2.
A subsetF of C(X,(𝔼1,τl)) is said to be  evenly equicontinuous if for any x0∈X, u∈𝔼1, {α1,…,αn}⊂[0,1] and ϵ>0, there exists an open neighborhood V of x0 and {β1,…,βm}⊂[0,1] such that if f∈F and
(4.1)max1≤i≤m{|f(x0)+(βi)-u+(βi)|,|f(x0)-(βi)-u-(βi)|}<ϵ,
then
(4.2)max1≤i≤n{|f(x)+(αi)-u+(αi)|,|f(x)-(αi)-u-(αi)|}<ϵ,
for all x∈V.For the sake of completeness, we give the proof of Theorem4.1 following the same pattern as in [12].Proof of Theorem4.1.
We have the following.
Necessity. Let us suppose that F is compact in the compact-open topology of C(X,(𝔼1,τl)).(i)
F is closed since C(X,(𝔼1,τl)) is Hausdorff as (𝔼1,τl) is so ([7]).(ii)
It is apparent that, for anyx∈X, the map Lx:C(X,(𝔼1,τl))→(𝔼1,τl) defined to be Lx(f):=f(x) is continuous. Hence, since F is compact, then Lx(F)=F[x] is compact in (𝔼1,τl) for each x∈K. Therefore, by Theorem 2.3, F[x] is, for any x∈K, pointwise closed and uniformly support bounded.(iii)
Fixx0∈K, u∈𝔼1, {α1,…,αn}⊂[0,1] and ϵ>0. Let {β1,…,βm}⊂[0,1] such that {α1,…,αn}⊆{β1,…,βm}. Let F'⊆F such that
(4.3)max1≤i≤m{|f(x0)+(βi)-u+(βi)|,|f(x0)-(βi)-u-(βi)|}≤ϵ,for allf∈F'. It is apparent that F' is closed and, consequently, it is compact. Let us consider the continuous map L(f,x):=f(x) defined from F×X to 𝔼1. Then
(4.4)L(F′×{x0})⊂U:={v∈𝔼1:max1≤i≤n{|v+(αi)-u+(αi)|,|v-(αi)-u-(αi)|}<ϵ},which is to say that the compact subsetF'×{x0} is included in L-1(V). By Wallace theorem ([10, Theorem 3.2.10]), there exists an open neighborhood, V, of x0 such that f(x)∈U for all f∈F' and all x∈V, that is,
(4.5)max1≤i≤n{|f(x)+(αi)-u+(αi)|,|f(x)-(αi)-u-(αi)|}<ϵ.
Sufficiency. Let us first check that F-, the closure of F in the pointwise convergence topology, is also evenly equicontinuous. Fix x0∈K, u∈𝔼1, {α1,…,αn}⊂[0,1] and ϵ>0. Then, by the evenly equicontinuity of F, there exists a neighborhood V of x0 and {β1,…,βm}⊂[0,1] such that if f∈F and
(4.6)max1≤i≤m{|f(x0)+(βi)-u+(βi)|,|f(x0)-(βi)-u-(βi)|}<ϵ,
then
(4.7)max1≤i≤n{|f(x)+(αi)-u+(αi)|,|f(x)-(αi)-u-(αi)|}<ϵ,
for all x∈V. Let us take g∈F- such that
(4.8)max1≤i≤m{|g(x0)+(βi)-u+(βi)|,|g(x0)-(βi)-u-(βi)|}<ϵ.
We must check
(4.9)max1≤i≤n{|g(x)+(αi)-u+(αi)|,|g(x)-(αi)-u-(αi)|}<ϵ
for all x∈V. To this end, let us choose a net {gn} in F which converges pointwise to g. Hence, eventually,
(4.10)max1≤i≤m{|gn(x0)+(βi)-u+(βi)|,|gn(x0)-(βi)-u-(βi)|}<ϵ.
Consequently, by the evenly equicontinuity of F,
(4.11)max1≤i≤n{|gn(x)+(αi)-u+(αi)|,|gn(x)-(αi)-u-(αi)|}<ϵ
for all x∈V, which yields the desired fact.
As in the necessity case, let us consider the mapL(f,x):=f(x) defined from F-×X to 𝔼1. We claim that L is continuous. To this end, fix x0∈K, u∈𝔼1, {α1,…,αn}⊂[0,1] and ϵ>0. Since F- is evenly equicontinuous, there exists an open neighborhood V of x0 and {β1,…,βm}⊂[0,1] such that if f∈W:={f∈F-:max1≤i≤m{|f(x0)+(βi)-u+(βi)|,|f(x0)-(βi)-u-(βi)|}<ϵ}, then L(W×V)⊂U, where
(4.12)U:={v∈𝔼1:max1≤i≤n{|v+(αi)-u+(αi)|,|v-(αi)-u-(αi)|}<ϵ}.
Since it is also clear that W is open in F- and U in open in (𝔼1,τl), then L is continuous.
Similarly, by the evenly equicontinuity ofF-, it can be deduced that if f∈F-, then f∈C(X,(𝔼1,τl)).
Next, we shall check that the closure ofF in the pointwise convergence topology, F-, coincides with the closure of F in the compact-open topology, cl(F). 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 K1 of K,u0∈𝔼1, and {α1,…,αn}⊂[0,1], the set B(K1,{α1,…,αn},u0) given by
(4.13){g∈F-:max1≤i≤n{|g(x)+(αi)-u0+(αi)|,|g(x)-(αi)-u0-(αi)|}<ϵ∀x∈K1}
is a member of a subbase for the compact-open topology in C(X,(𝔼1,τl)).
Fixf∈B(K1,{α1,…,αn},u0). Then {f}×K1, which is compact in F-×K, belongs to L-1(U), where
(4.14)U:={v∈𝔼1:max1≤i≤n{|v+(αi)-u0+(αi)|,|v-(αi)-u0-(αi)|}<ϵ}.From the previous paragraphs, L-1(U) is open. Hence, by Wallace theorem ([10, Theorem 3.2.10]), there exists an open neighborhood, N, of f in the pointwise convergence topology such that N×K1∈L-1(U), which is to say that N⊂B(K1,{α1,…,αn},u0). This means that the topology of pointwise convergence is larger than the compact-open topology.
Finally, we claim thatcl(F) 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 F-=cl(F). To this end, we can consider F- as a closed subset of the product ∏x∈K‍{F[x]¯}. By Tychonoff’s theorem, ∏x∈K‍{F[x]¯} is compact in the topology of pointwise convergence since, by Theorem 2.3, F[x] is compact for every x∈K. Thus, F- is compact and, since by hypothesis, F is closed, then F is compact, as was proved.



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*Source: 326417-2012-12-06.xml*