1. Introduction
Let
U be a universal set. A fuzzy subset
of
U is defined as a set of ordered pairs:
where
is called the
membership function of
. The set of all fuzzy subsets of
U is denoted by
.
We consider an onto function
, where
V is another universal set. This function is also called as a
crisp function. The purpose is to fuzzify the crisp function
f as a
fuzzy function ; that is, for any
, it means
. We remark that this fuzzy function is completely different from the concept of fuzzy function studied in Hajek [
1], Demirci [
2] and Höhle et al. [
3] in which the fuzzy function is treated as a fuzzy relation.
The principle for fuzzifying the crisp functions is called the
extension principle, which was proposed by Zadeh [
4,
5,
6]. In particular, if
, then
is called a real-valued function, and
induced from
f is called a
fuzzy-valued function.
For any
, the extension principle says that the membership function of
is defined by the following supremum:
Suppose that
U and
V are taken as the topological spaces such that the original crisp function
is continuous. In this paper, we shall investigate the continuity of fuzzified function
. Román-Flores et al. [
7] has studied the continuity of this fuzzified function when
U and
V are taken as the Euclidean space
. In this paper, we are going to extend these results to the case of normed spaces and normable topological vector spaces using the generalized extension principle discussed in Wu [
8].
Let
and
be two universal sets. We consider the onto crisp function
. The purpose is to fuzzify the crisp function
f as a fuzzy function
. For any two fuzzy subsets
and
of
and
, respectively, the membership function of
is defined by:
Nguyen [
9] has obtained the following result.
For
, the
-level set of
is defined and denoted by:
If the universal set
U is endowed with a topology, then the 0-level set of
is defined by:
which is the closure of the
support of
. However, if
U is not assumed to be a topological space, then the 0-level set is usually taken to be the whole set
U.
Theorem 1. (Nguyen [9]) Let be an onto crisp function defined on and let be a fuzzy function induced from f via the extension principle defined in Equation (2). For , , the following equality:holds true for each if and only if, for each , the following supremum:is attained; that is, we have: Fullér and Keresztfalvi [
10] generalized Theorem 1 by considering the
t-norm. In this case, the extension principle presented in Equation (
2) can be generalized in the following form:
since
is a
t-norm. Therefore, Theorem 1 can be generalized as follows.
Theorem 2. (Fullér and Keresztfalvi [10]) Let be an onto function defined on and let be a fuzzy function induced from f via the extension principle defined in Equation (5). For , , the following equality:holds true for each if and only if, for each , the following supremum:is attained. Fullér and Keresztfalvi [
10] also obtained the following interesting result.
Theorem 3. (Fullér and Keresztfalvi [10]) Let be locally compact topological spaces. Let for such that the membership functions are upper semicontinuous and the 0-level sets are compact subsets of for . Let be a continuous and onto crisp function and let be a fuzzy function induced from f via the extension principle defined in Equation (5). If the t-norm t is upper semicontinuous, then the following equality:holds true for each . Based on the Hausdorff space, Wu [
8] generalizes Theorems 2 and 3 to the case of generalized
t-norm
that is recursively defined by:
Let
be universal sets and let
be an onto crisp function defined on
. For
,
, the membership function of the fuzzy subset
of
V is defined by:
for each
. This definition extends the definition given in Equation (
7). In the sequel, we are going to consider the extension principle using an operator called
that is more general than
.
Let
be a function defined on
, which does not assume any extra conditions. Let
be universal sets and let
be an onto crisp function defined on
. For
,
, the membership function of the fuzzy subset
of
V is defined by:
for each
. Of course, this definition extends the definition given in Equation (
9). In this paper, we are going to investigate the continuity of this kind of fuzzy function. We also remark that the operator
is a kind of aggregation operator studied in Calvo et al. [
11] and Grabisch et al. [
12].
2. Generalized Extension Principle on Normed Spaces
Let be a normed space. Then, we see that the norm can induce a topology such that becomes a Hausdorff topological vector space in which is also called a norm topology.
Let
be normed spaces for
and let
be the product vector space. Since each normed space
can induce a Hausdorff topological vector space
, we can form a product topological vector space
using
for
, where
is the product topology. On the other hand, we can define a product norm on the product vector space
using the norms
. For example, for
, we can define the maximum norm:
which is shown in Kreyszig ([
13], p. 71) or the
p-norm:
which is shown in Conway ([
14], p. 72), where
and
are taken to be the same norm
for all
. We also see that the product norms
and
can induce the norm topologies
and
, respectively. We can show that
. However, in general, the norm topology generated by the product norm does not necessarily equal to the product topology
. Therefore, it is needed to investigate the generalized extension principle in the case of normed space separately.
We can simply regard
as a vector space over
. Therefore, we can define a norm to make it as a normed space
. In this case, we can induce a norm topology
. Alternatively, we can consider the product norm for the product vector space
over
. Let
be a real-valued function defined on
. In general, the product norm on
can be defined as:
Of course, the product norm is also a norm for . Based on this product norm, we can also induce a product norm topology . If we take , then we obtain the product norm , and if we take for , then we can obtain the product norm .
For
and
, the open ball in the normed space
is defined by:
For
, the open ball in the normed space
is given by:
and, by referring to Equation (
11), the open ball in the product normed space
is given by:
We have the following interesting result.
Proposition 1. Let be normed spaces for and let be the product vector space which is endowed with a norm . Given any and , if there exist such that the following inclusions hold true:and:then , where is the norm topology induced by the norm and is the product topology. If the product vector space is endowed with the product norm such that the inclusions Equations (12) and (13) are satisfied, then we also have , where is the product norm topology. Proof. Since
is the product topology for
, we recall that
if and only if, for any
, there exist open neighborhoods
of
for
such that
. Since
is an open neighborhood of
, there exists
such that
. Let
. Then, we have
for all
. By the assumption in Equation (
12), it follows that
Therefore, we conclude that
. Conversely, for any open set
and any element
, there exists
such that
. By the assumption in Equation (
13), we see that
. Since
is an open neighborhood of
for each
, we conclude that
. This completes the proof. ☐
Proposition 2. Let be normed spaces for and let be the product vector space. Then, the following statements hold true.
- (i)
We consider the normed space . Given any , if if and only if for all , then .
- (ii)
We consider the product normed space , where the product norm is defined by Equation (11). Given any , if if and only if for all , then .
Proof. It suffices to prove the case of (ii). By definition, we have:
The results follow immediately from Proposition 1 by taking . ☐
Remark 1. Note that if the product norm is taken as the maximum norm or the p-norm defined above, then the assumption in part (ii) of Proposition 2 is satisfied automatically.
Let and be two normed spaces. Recall that the function is continuous at if and only if, given any , there exists such that implies . The function f is continuous on X if and only if f is continuous at each point . Then, we have the following easy observation.
Remark 2. Let and be two normed spaces such that and are two norm topologies induced by the norms and , respectively. It is well-known that the function is continuous if and only if is continuous. The continuity of means that if , then .
Remark 3. Let and be the normed spaces for . Then three kinds of continuity for the function can be presented below.
Suppose that the product vector space is endowed with the norm . Then the function is continuous at if and only if, given any , there exists such that implies , where and are elements of . Since can induce a norm topology , Remark 2 says that is continuous if and only if is continuous.
Suppose that the product vector space is endowed with the product norm . Then the continuity of the function can be similarly realized. We also see that is continuous if and only if is continuous.
Since we can form a product topological vector space from the normed spaces for , we say that the function is continuous if and only if the function is continuous in the topological sense. Propositions 1 and 2 say that this kind of continuity will be equivalent to one of the above two continuities under some suitable conditions.
We say that is nondecreasing if and only if for all imply . This definition does not necessarily say that implies for all . In the subsequent discussion, the function will satisfy some of the following conditions:
- (a)
if and only if for all .
- (b)
For each , if and only if for all .
- (c)
is upper semicontinuous and nondecreasing.
- (d)
if any one of is zero, then .
- (e)
.
- (f)
.
Next, we are going to present the generalized extension principle on normed spaces.
Theorem 4. Let be an onto crisp function defined on and let be a fuzzy function induced from f via the extension principle defined in Equation (10). Assume that and are taken to be the normed spaces for , and that the product vector space is endowed with a norm such that the inclusions Equations (12) and (13) are satisfied. We also assume that the following supremum:is attained for each . Then, the following equality:holds true for each . The results for the 0-level sets are given below. If we further assume that condition (a) for is satisfied, then: If we further assume that the function is continuous and that condition (a) for is satisfied, then:
If the product vector space is endowed with the product norm such that the inclusions Equations (12) and (13) are satisfied, then we also have the same results. The assumptions satisfying the inclusions Equations (12) and (13) are not needed when we say that the function is continuous directly in topological sense without considering the norm and the product norm . Proof. Since a normed space can induce a Hausdorff topological space, the results follow immediately from Remarks 2 and 3, Proposition 1 and Wu ([
8], Theorem 5.1). ☐
By referring to Remark 1, from part (ii) of Proposition 2, we see that if the product norm is taken as the maximum norm or the p-norm , then Theorem 4 is applicable for these norms.
Theorem 5. Let be an onto crisp function defined on and let be a fuzzy function induced from f via the extension principle defined in Equation (10). Suppose that the following supremum:is attained for each , and that that condition (b) for is satisfied. Then, for each , we have the following equalities. Let and be now taken to be the normed spaces for , and let the product vector space be endowed with a norm such that the inclusions Equations (12) and (13) are satisfied. If we further assume that the function is continuous, then we also have the following equality: If the product vector space is endowed with the product norm such that the inclusions Equations (12) and (13) are satisfied, then we also have the same results. The assumptions satisfying the inclusions Equations (12) and (13) are not needed when we say that the function is continuous directly in topological sense without considering the norm and the product norm . Proof. Since a normed space can induce a Hausdorff topological space, the results follow immediately from Remarks 2 and 3, Proposition 1 and Wu ([
8], Theorem 5.2). ☐
Theorem 6. Let and be normed spaces for , and let the product vector space be endowed with a norm such that the inclusions Equations (12) and (13) are satisfied. Let for all such that the membership functions are upper semicontinuous and the 0-level sets are compact subsets of for all . Let be a continuous and onto crisp function, and let be a fuzzy function induced from f via the extension principle defined in Equation (10). Suppose that conditions (c) and (d) for are satisfied. Then, the following statements hold true. - (i)
The membership function is upper semicontinuous.
- (ii)
For each , we have the following equality: For the 0-level sets, if we further assume that conditions (a) for is satisfied, then: Under this further assumption, the α-level sets of are closed and compact subsets of V for all .
- (iii)
If we further assume that conditions (b) for is satisfied, then:for each .
If the product vector space is endowed with the product norm such that the inclusions Equations (12) and (13) are satisfied, then we also have the same results. The assumptions satisfying the inclusions Equations (12) and (13) are not needed when we say that the function is continuous directly in topological sense without considering the norm and the product norm . Proof. Since a normed space can induce a Hausdorff topological space, the results follow immediately from Remarks 2 and 3, Proposition 1 and Wu ([
8], Theorem 5.3 and Corollary 5.1). ☐
Since we are going to consider the concept of convexity, we need to impose the vector addition and scalar multiplication upon the universal set
U. Therefore, the universal set
U is taken as a vector space. For any fuzzy subset
of
U, we say that
is
convex if and only if each
-level set
is a convex subset of
U for each
. It is well-known that:
that is, the membership function
is quasi-concave.
Let U be a vector space over which is endowed with a topology and let be a fuzzy subset of U. Since , we see that if is convex, then its 0-level set is also a convex subset of U.
Definition 1. Let U be a vector space over which is endowed with a topology τ. We denote by the set of all fuzzy subsets of U such that each satisfies the following conditions:
is normal, i.e., for some ;
is convex;
the membership function is upper semicontinuous;
the 0-level set is a compact subset of U.
For , we see that, for each , the -level set is a compact and convex subset of U. Each element of is called a fuzzy element. If , then each element of is called a fuzzy number. In addition, if , then each element of is called a fuzzy vector. Moreover, for any fuzzy number , we see that each of its -level sets is a closed, bounded and convex subset of , i.e., a closed interval in .
Theorem 7. Let and be normed spaces for , and let the product vector space be endowed with a norm such that the inclusions Equations (12) and (13) are satisfied. Let be a continuous and onto crisp function. We also assume that f is linear in the sense of: Let be a fuzzy function induced from f via the extension principle defined in Equation (10). Suppose that conditions (a), (c), (d), (e) and (f) for are satisfied. For , , the following statements hold true. - (i)
We have . The α-level sets are compact, convex and closed subsets of V for all .
- (ii)
For each , we have the following equality:For the 0-level sets, we also have: - (iii)
If we further assume that conditions (b) for is satisfied, then:for each .
If the product vector space is endowed with the product norm such that the inclusions Equations (12) and (13) are satisfied, then we also have the same results. The assumptions satisfying the inclusions Equations (12) and (13) are not needed when we say that the function is continuous directly in topological sense without considering the norm and the product norm . Proof. Since a normed space can induce a Hausdorff topological space, the results follow immediately from Remarks 2 and 3, Proposition 1 and Wu ([
8], Theorem 6.1 and Corollary 6.1). ☐
The linearity in Theorem 7 can be replaced by assuming the convexity.
Theorem 8. Let and be normed spaces for , and let the product vector space be endowed with a norm such that the inclusions Equations (12) and (13) are satisfied. Let be a continuous and onto crisp function and let be a fuzzy function induced from f via the extension principle defined in Equation (10). We further assume that is a convex subset of V for any convex subsets of , . Suppose that conditions (b), (c), (d) and (e) for are satisfied. For , , the following statements hold true. - (i)
We have . The α-level sets are compact, convex and closed subsets of V for all .
- (ii)
For each , we have the following equality:For the 0-level sets, we also have:
If the product vector space is endowed with the product norm such that the inclusions Equations (12) and (13) are satisfied, then we also have the same results. The assumptions satisfying the inclusions Equations (12) and (13) are not needed when we say that the function is continuous directly in topological sense without considering the norm and the product norm . Proof. Since a normed space can induce a Hausdorff topological space, the results follow immediately from Remarks 2 and 3, Proposition 1 and Wu ([
8], Theorem 6.2). ☐
Remark 4. For , the following functions:satisfy the conditions of Theorem 8. Let be a normed space. We recall that the norm can induce a norm topology such that becomes a Hausdorff topological vector space. Conversely, if is a topological vector space, we say that X is normable if and only if there exists a norm which can induce a norm topology such that . A subset Y of a topological vector space is called bounded if and only if, for each neighborhood N of , where is the zero element of X, there is a real number r such that . We have the following interesting results.
Proposition 3. (Kelley and Namioka ([15], p. 44)) The following statements hold true. - (i)
A topological Hausdorff vector space is normable if and only if there is a bounded convex neighborhood of 0.
- (ii)
A finite product of normable spaces is normable.
From part (ii) of Proposition 3, we see that Theorems 4–8 are still valid if the normed spaces are replaced by the normable topological vector spaces. However, if we consider the continuity of the function
instead of the continuity of the function
, then the assumption satisfying the inclusions Equations (
12) and (
13) is not needed, since, in this case, we can just apply Wu ([
8], Theorems 5.1, 5.2, 5.3, 6.1 and 6.2) by considering the product topology
instead of the norm topology
or the product norm topology
.
3. Continuity of Fuzzified Functions
Let
be a normed space. We denote by
the family of all compact subsets of
X in the sense of norm topology
induced by the norm
. Let
A and
B be any two compact subsets of
X. We can define the Hausdorff metric
between
A and
B as follows:
If X is a normable topological vector space, then we can also define the Hausdorff metric . Now we have the following simple observation.
Lemma 1. The function defined by is continuous for any given .
Proof. We have:
The continuity follows immediately from the above inequality. ☐
We need some useful properties from topological space.
Proposition 4. The following statements hold true.
- (i)
(Royden ([16], p. 158)) If X is a Hausdorff space, then a compact subset of X is closed. - (ii)
(Royden ([16], p. 158)) If f is a continuous function from the topological space X to another topological space Y, then the image is a compact subset of Y when K is a compact subset of X. - (iii)
(Royden ([16], p. 161)) If f is an upper semicontinuous real-valued function defined on X, then f assumes its maximum on a compact subset of X. If f is a lower semicontinuous real-valued function defined on X, then f assumes its minimum on a compact subset of X. - (iv)
(Royden ([16], p. 158)) Let X be a topological space and let K be a compact subset of X. If A is a closed subset of X and is also a subset of K, then A is also a compact subset of X. - (v)
(Royden ([16], p. 166)) (Tychonoff’s Theorem) Let be n topological spaces and let be compact subsets of for . Then the product is a compact subset of the product topological space , where is the product topology for .
Proposition 5. Let the function be continuous. Let and be the Hausdorff metrics defined by and , respectively. Then the function defined by:is uniformly continuous. Proof. For any
, i.e.,
A and
B are compact subsets of
, since
f is continuous, by part (ii) of Proposition 4, we see that
and
are also compact subsets of
Y, i.e.,
. We need to show that, for any
, there exists
such that, for any
,
implies
. By definition, for any
and
,
implies:
By Lemma 1 and part (iii) of Proposition 4, we see that the above infimum are attained, i.e.,
and
for some
and
. Since
f is continuous at
and
, we have
and
. Since
and
are any elements of
A and
B, respectively, we have:
and:
Since
C and
D are compact subsets of
Y, there exist
and
such that:
From Equations (
16) and (
17), we obtain:
which implies
. This completes the proof. ☐
Let be a normable topological vector space. Then, we can topologize by considering the Hausdorff metric . This topological space is denoted by .
Proposition 6. Let and be two normable topological vector spaces. Let and be the topologies generated by the Hausdorff metrics and , respectively. If the function is continuous, then the function defined by is uniformly continuous; that is, the function is uniformly continuous.
Proof. The result follows immediately from Proposition 5. ☐
To study the continuity of fuzzified function, we are going to consider the space
instead of
, which is defined below. Let
U be a topological space. We denote by
the set of all fuzzy subsets of
U such that, for each
, its
-level sets
are compact subsets of
U for all
. Let
be a normed space. Then, we can define the Hausdorff metric
on
. For any
, we see that
for all
. Therefore, for any
, we can define a distance
between
and
as:
Then, we can show that forms a metric space. Based on this metric , we can induce a topology that is called a metric topology for .
Let
be normed spaces for
and let
be the product vector space that is endowed with the norm
or the product norm
. In this case, we can define a Hausdorff metric
on
based on the norm
or the product norm
. We write
. The element of
can be realized as
, where
for
. We also write
for all
. Using the Tychonoff’s theorem in part (v) of Proposition 4, we see that
for all
; that is, the product set
is a compact subset of the product space
. For any
, we define
between
and
as follows:
Then, we can also show that forms a metric space. Based on this metric , we can induce a metric topology for . Now, if we assume that are normable topological vector spaces and let be the product vector space, then part (ii) of Proposition 3 says that is also a normable topological vector space. Therefore, we still can define a Hausdorff metric on . In this case, we can also obtain the metric space .
Let
be topological spaces such that
is an onto crisp function. Then, we can induce a fuzzy function
from
f via the extension principle defined in Equation (
10). However, even
f is continuous, we cannot always induce a fuzzy function
, where the spaces
and
, instead of
and
, are considered for
. In fact, we can just induce a fuzzy function
, where the range is the space
. In other words, given
for
, we can just have
, and we cannot always guarantee
. The purpose is to present the sufficient conditions to guarantee this desired result. First of all, we need a useful lemma.
Lemma 2. (Wu [8]) Let be topological spaces and let V be a Hausdorff space. Let be a continuous and onto crisp function defined on and let be a fuzzy function induced from f via the extension principle defined in Equation (10). Assume that such that its membership function is upper semicontinuous and each 0-level set of is a compact subset of for all . Suppose that conditions (c) and (d) for are satisfied. Then, the supremum given in Equation (14) is attained. Proposition 7. Let be Hausdorff spaces, and let be a continuous and onto crisp function. Suppose that conditions (b), (c) and (d) for are satisfied. If for all , then .
Proof. Since the
-level sets
of
are compact subsets of
for all
and
, part (i) of Proposition 4 says that
are also closed subsets of
for all
and
. In other words, the membership functions
are upper semicontinuous for all
. Therefore, applying Lemma 2, we see that the supremum given in Equation (
14) is attained for each
. According to Wu ([
8], Theorem 5.2), since
f is continuous, we have:
for each
, which is a compact subset of
V by part (ii) of Proposition 4 and the Tychonoff’s theorem in part (v) of Proposition 4, i.e.,
. This completes the proof. ☐
Under the assumptions of Proposition 7, we can indeed induce a fuzzy function based on the spaces and , instead of the spaces and for . Now, if we take and to be the normed spaces, , then we need to apply Theorem 5 to induce the fuzzy function . Now we have the following result.
Theorem 9. Let and be the normed spaces for , and let the product vector space be endowed with a norm such that the inclusions Equations (12) and (13) are satisfied. We assume that is a continuous and onto crisp function. Suppose that conditions (b), (c) and (d) for are satisfied. Let be a fuzzy function induced from f via the extension principle defined in Equation (10). Then, the fuzzy function is continuous; that is, is continuous. If the product vector space is endowed with the product norm such that the inclusions Equations (12) and (13) are satisfied, then we also have the same results. The assumptions satisfying the inclusions Equations (12) and (13) are not needed when we say that the function is continuous directly in topological sense without considering the norm and the product norm . Proof. Since a normed space can induce a Hausdorff topological space, from Remarks 2 and 3, Proposition 1 and the arguments of Proposition 7, Lemma 2 says that the supremum given in Equation (
14) is attained for each
. Therefore, from Theorem 5, we have:
for each
. From the arguments of Proposition 7, we can induce a fuzzy function
. Suppose that
for
satisfy
as
. We need to show:
as
. For
, Proposition 6 says that, for any
, there exists
such that:
Since
as
, there exists
such that:
for
, where
and
denote the
-level sets of
and
, respectively, which also says that
for all
. Therefore, according to Equation (
19), we have
for all
. Applying Lemma 2, we see that the supremum given in Equation (
14) is attained for each
. From Theorem 5, we also see that, for each
,
and
. Therefore, we obtain:
for
. This completes the proof. ☐
Theorem 10. Let be normable topological vector spaces. We also assume that V is a Hausdorff space. Let be a continuous and onto crisp function. Suppose that conditions (b), (c) and (d) for are satisfied. Let be a fuzzy function induced from f via the extension principle defined in Equation (10). Then, the fuzzy function is continuous; that is, is continuous. Proof. From Lemma 2 and Wu ([
8], Theorem 5.2), we see that, for each
,
and
. Since the normable topological vector spaces are also Hausdorff spaces, the result follows immediately from the same arguments of Theorem 9.
Example 1. To apply Theorem 10, we take and for . Let be a continuous and onto function, where the continuity is based on the usual topologies and . Suppose that conditions (b), (c) and (d) for are satisfied. For convenient, we write: Let be a fuzzy function induced from f via the extension principle defined in (10). Then the fuzzy function is continuous. Example 2. Continued from Example 1, we consider a continuous and onto function defined by . Now we take . Then, conditions (b), (c) and (d) are satisfied automatically. In this case, the fuzzy function is continuous, where the membership function of for is given by: Let U be a topological space. We denote by the set of all fuzzy subsets of U such that, for each , its membership function is upper semicontinuous and the 0-level set is a compact subset of U.
Remark 5. We have the following observations.
Let U be a topological space. For any , by the upper semicontinuity of , we see that are closed subsets of U for all . Since for all and is a compact subset of U, part (iv) of Proposition 4 says that are also compact subsets of U for all . Therefore we conclude that .
Let U be a Hausdorff space. For any , we see that are compact subsets of U for all . From Proposition 4 (i), the α-level sets are also closed subsets of U for all , i.e., the membership function is upper semicontinuous. Therefore we conclude that .
If U is taken as a normable topological vector space or is taken as the normed space, then , since the normable topological vector space and normed space are also the Hausdorff spaces.
Let
be topological spaces and let
V be a Hausdorff space. Let
be a continuous and onto crisp function. In general, we cannot induce a fuzzy function
from
f via the extension principle defined in Equation (
10). The following proposition presents the sufficient conditions to guarantee this property.
Proposition 8. Let be topological spaces and let V be a Hausdorff space. Let be a continuous and onto crisp function. Suppose that conditions (c) and (d) for are satisfied. If for all , then .
Proof. The result follows immediately from Wu ([
8], Theorem 5.3). ☐
Under the assumptions of Proposition 8, we can induce a fuzzy function . We also have the following interesting observations.
Remark 6. We remark that Theorems 9 and 10 are still valid if and are replaced by and for all , since the third observation of Remark 5 says that and for all .
Theorems 9 and 10 present the continuity of fuzzified functions based on the spaces and for . In the sequel, we are going to investigate the continuity of fuzzified functions based on the spaces and for .
Theorem 11. Let and be normed spaces for , and let the product vector space be endowed with a norm such that the inclusions Equations (12) and (13) are satisfied. We assume that is a linear, continuous and onto crisp function. Suppose that conditions (b), (c), (d), (e) and (f) for are satisfied. Let be a fuzzy function induced from f via the extension principle defined in Equation (10). Then the fuzzy function is continuous; that is, is continuous. If the product vector space is endowed with the product norm such that the inclusions Equations (12) and (13) are satisfied, then we also have the same results. The assumptions satisfying the inclusions Equations (12) and (13) are not needed when we say that the function is continuous directly in topological sense without considering the norm and the product norm . Proof. From Remarks 2 and 3 and part (iii) of Theorem 7, we see that, for each
,
The result follows immediately from the same arguments of Theorem 9. ☐
Theorem 12. Let be normable topological vector spaces. We also assume that V is a Hausdorff space. Let be a linear, continuous and onto crisp function. Suppose that conditions (b), (c), (d), (e) and (f) for are satisfied. Let be a fuzzy function induced from f via the extension principle defined in Equation (10). Then the fuzzy function is continuous; that is, is continuous. Proof. From Wu ([
8], Corollary 6.1), we see that the equalities in (
20) are satisfied for each
. The result follows immediately from the same arguments of Theorem 9. ☐
The linearity in Theorems 11 and 12 can be replaced by assuming the convexity.
Theorem 13. Let and be normed spaces for , and let the product vector space be endowed with a norm such that the inclusions Equations (12) and (13) are satisfied. Let be a continuous and onto crisp function. We further assume that is a convex subset of V for any convex subsets of , . Suppose that conditions (b), (c), (d) and (e) for are satisfied. Let be a fuzzy function induced from f via the extension principle defined in Equation (10). Then the fuzzy function is continuous; that is, is continuous. If the product vector space is endowed with the product norm such that the inclusions Equations (12) and (13) are satisfied, then we also have the same results. The assumptions satisfying the inclusions Equations (12) and (13) are not needed when we say that the function is continuous directly in topological sense without considering the norm and the product norm . Proof. From Remarks 2 and 3 and part (ii) of Theorem 8, we see that the equalities in Equation (
20) are satisfied for each
. The result follows immediately from the same arguments of Theorem 9. ☐
Theorem 14. Let be normable topological vector spaces. We also assume that V is a Hausdorff space. Let be a continuous and onto crisp function. We further assume that is a convex subset of V for any convex subsets of , . Suppose that conditions (b), (c), (d) and (e) for are satisfied. Let be a fuzzy function induced from f via the extension principle defined in Equation (10). Then, the fuzzy function is continuous; that is, is continuous. Proof. From Wu ([
8], Theorem 6.2), we see that the equalities in (
20) are satisfied for each
. The result follows immediately from the same arguments of Theorem 9. ☐
4. Applications to Fuzzy Topological Vector Spaces
Before introducing the concept of fuzzy vector space, we need to consider the fuzzy addition and fuzzy scalar multiplication. Let
U be a vector space over
. For any
, the membership function of
is defined by:
Let
. The membership function of
is defined by:
If
is taken to be the crisp number
with value
, i.e.,
then, we simply write
as
with the membership function given by:
We have the following interesting observation.
Remark 7. Suppose that satisfies the following conditions:for any . Then, we have If we take , then conditions in Equation (21) are satisfied automatically. The following result will be used in the further discussion.
Proposition 9. (Wu [8]) Let U be a Hausdorff topological vector space over . Suppose that conditions (b), (c), (d) and (e) for are satisfied. Then, we have the following results. - (i)
If , then and for any .
- (ii)
If and , then for any .
- (iii)
If we take to be a nonnegative or nonpositive fuzzy number, then .
- (iv)
For , we have , where is a crisp number with value λ.
- (v)
If , then .
Definition 2. Let U be a vector space over and let be a subset of . We say that is a fuzzy vector space over if and only if and for any and . In other words, is closed under the fuzzy addition and scalar multiplication.
Proposition 10. Let U be a Hausdorff topological vector space over . Then, the following statements hold true.
- (i)
Suppose that conditions (b), (c), (d) and (e) for are satisfied. Then is a fuzzy vector space over .
- (ii)
Suppose that conditions (c) and (d) for are satisfied. Then is a fuzzy vector space over .
Proof. Part (i) follows immediately from parts (i) and (iv) of Proposition 9. Part (ii) follows immediately from Proposition 8 and the arguments of Proposition 9. ☐
If we consider instead of , then we can introduce another concept of fuzzy vector space in which we consider the so-called fuzzy scalar multiplication instead of scalar multiplication.
Definition 3. Let U be a vector space over . Let be a subset of and be a subset of . We say that is a fuzzy vector space over if and only if and for and . In other words, is closed under the fuzzy addition and fuzzy scalar multiplication, where the fuzzy scalar multiplication should be defined in another way.
Proposition 11. Let U be a Hausdorff topological vector space over . Suppose that conditions (c) and (d) for are satisfied. If the fuzzy scalar multiplication is defined as , then is a fuzzy vector space over .
Proof. The result follows immediately from Remark 5, Proposition 8 and the arguments of Proposition 9. …
We say that the fuzzy number
is
nonnegative if and only if
for all
, and
nonpositive if and only if
for all
. We denote by
the set of all nonnegative fuzzy numbers, and by
the set of all nonpositive fuzzy numbers. Let
be a fuzzy number. We define the membership functions of
and
as follows:
and:
We see that
is a nonnegative fuzzy number and
is a nonpositive fuzzy number, since the
-level sets
and
are closed intervals for all
; that is, their membership functions
and
are upper semicontinuous (the other conditions in Definition 1 are obviously true). Furthermore, we have:
for all
. Thus
. We call
and
the
positive part and the
negative part of
, respectively.
Proposition 12. Let U be a Hausdorff topological vector space over . Suppose that conditions (b), (c), (d) and (e) for are satisfied. Then, the following statements hold true.
- (i)
Let . If the fuzzy scalar multiplication is defined as , then is a fuzzy vector space over .
- (ii)
If the fuzzy scalar multiplication is defined as:where , then is a fuzzy vector space over .
Proof. Part (i) follows immediately from part (iii) of Proposition 9. To prove part (ii), we just need to claim that for . By definition, we see that by part (iii) of Proposition 9, since . By part (i) of Proposition 9 again, we see that . This completes the proof. ☐
In the sequel, we are going to introduce the concept of the fuzzy topological vector space. Recall that, for each normed space
, we can induce a metric space
for
, where the metric
is defined in (
18). Given any
, the open ball with center
is given by:
Based on these open balls, we can induce a topological space
, where
is the so-called metric topology. Based on the topological spaces
for
, we can also form a topological space
, where
and
is the product topology. On the other hand, we can induce another metric space
by considering the norm
or the product norm
. The open ball with center
is given by:
Based on these open balls, we can also induce the metric topology for . To introduce the concept of the fuzzy topological vector space, we need to provide some suitable conditions to guarantee the equality ; that is, the product topology is equal to the metric topology.
Proposition 13. Let be normed spaces for and let be the product vector space which is endowed with a norm . Given any and:if there exist such that the following inclusions hold true:and:then the product topology is equal to the metric topology for . If the product vector space is endowed with the product norm such that the inclusions Equations (22) and (23) are satisfied, then the product topology is also equal to the metric topology. Proof. The results follow immediately from the same arguments of Proposition 1. ☐
Lemma 3. Let be normed spaces for and let be the product vector space. Then, the following statements hold true.
- (i)
We consider the normed space . Given any , assume that if and only if for all . Then, we have the following inclusions:and - (ii)
We consider the product normed space , where the product norm is defined by (11). Given any , assume that if and only if for all . Then, we also have the inclusions in Equations (24) and (25).
Proof. To prove part (i), for
, i.e.,
, we have
for all
; that is,
for all
. It suffices to consider the case of:
for all
. In this case, we see that
for all
and
. Since
is a compact subset of
, by Lemma 1 and part (iii) of Proposition 4, we see that the above infimum is attained, i.e.,
for some
, where
and
are in
. By the assumption of
, we have
for all
. Therefore, we also have:
for all
and
, which also says that:
for all
. By considering another case, we can similarly show that:
for all
. Then, we have:
which says that
for all
. Therefore, we conclude the inclusion:
On the other hand, for
,
, we have:
for all
. We consider the case of:
for all
and
. This says that
for all
,
and
. By Lemma 1 and part (iii) of Proposition 4, we see that
for some
. By the assumption of
, there exists
such that
for all
and
. Therefore, we obtain:
for all
. By considering another case, we can similarly show that:
for all
. Then, we have:
Therefore, we conclude the inclusion: