Generating fuzzy sets from the families of sets

Abstract

The main purpose of this paper is to establish a mechanical procedure to determine the membership functions using the data collected from the economic and engineering problems. Determining the membership functions from the collected data may depend on the subjective viewpoint of decision makers. The mechanical procedure proposed in this paper can get rid of the subjective bias of decision makers. The concept of solid families is also proposed by regarding the sets in a family to be continuously varied. The desired fuzzy sets will be generated in the sense that its α-level sets will be identical to the sets of the original family. In order to achieve this purpose, any arbitrary families will be rearranged as the nested families by applying some suitable functions to the original families that are formulated from the collected data.

Keywords

Nested families non-normal fuzzy sets solid families

1 Introduction

In economic and engineering problems, when the fuzziness is taken into account, the observed data sometimes are fuzzified to be fuzzy numbers. This kind of fuzzification may result in many different types of fuzzy numbers depending on the methodology adopted by the decision makers, where the subjectivity via the viewpoint of decision makers may have bias from the reality. In this paper, we propose a general methodology that can get rid of the subjectivity by directly generating the related fuzzy sets based on the observed data without involving the possible biased viewpoint of decision makers. The main idea is based on the solid and nested families of sets.

Suppose that we want to measure the water level in the season of summer. Owing to the fluctuation of water level, we cannot simply say that the water level is now 10 meters. We should say that the water level is around 10 meters. Therefore, the reasonable way to model the water level is to treat it as a fuzzy interval or fuzzy number. Under this consideration, the water level should be taken as a fuzzy number $\tilde{10}$ . The problem is that the determination of membership function $ξ_{\tilde{10}}$ of $\tilde{10}$ is subjective depending on the viewpoint of decision makers. In other words, there are infinite ways to setup the membership functions. It may happen that the different membership functions can result in the different final results. Therefore, the better way is to follow a mechanical procedure to setup the membership functions, which is the main purpose of this paper. We briefly address this mechanical procedure as follows.

We assume that there are 100 days in summer. Suppose that the engineers can measure the water level two times each day. In other words, the engineers can obtain a bounded closed interval each day by setting the lower and upper bounds of this interval as the low and high water levels, respectively. More precisely, we consider 100 values of α in the unit intervals [0, 1]. Then, we can obtain a bounded closed interval $[m_{α}^{L}, m_{α}^{U}]$ , where $m_{α}^{L}$ denotes the low water level and $m_{α}^{U}$ denotes the high water level in the (100 * α)th day. For example, the value α = 0.08 means the 8th day of this summer. In this case, we obtain a family of closed intervals given by $M = {M_{α} : M_{α} = [m_{α}^{L}, m_{α}^{U}] for α = 1, \dots, 100} .$ This family cannot be nested in the sense of M_α ⊆ M_β for α > β. However, we can rearrange this family $M$ as a nested family $M^{(η)}$ given by $M^{(η)} = {M_{α}^{(η)} : M_{α}^{(η)} = M_{η (α)}},$ for some suitable function η : (0, 1] → (0, 1] such that we have the nestedness $M_{α}^{(η)} \subseteq M_{β}^{(η)}$ for α > β. The main purpose of this paper is to generate a fuzzy set $\tilde{A}$ such that its α-level set ${\tilde{A}}_{α}$ is identical to the closed interval $M_{α}^{(η)}$ . In this case, the fuzzy set $\tilde{A}$ can be used to describe the water level in summer in which the biased subjectivity raised by the decision makers can be avoided. In other words, this mechanical procedure is independent on decision makers.

In general, given an arbitrary family $M = {M_{α} : α \in (0, 1]}$ of nonempty subsets of a universal set U, where U can be $ℝ$ or $ℝ^{n}$ , the purpose of this paper is to generate a fuzzy set $\tilde{A}$ in U such that its α-level set ${\tilde{A}}_{α}$ is equal to M_α for each α ∈ (0, 1] when $M$ is a nested family in the sense of M_β ⊆ M_α for β > α. The nestedness of the family $M$ plays an important role for guaranteeing the existence of such kind of fuzzy sets. The concept of nested families was taken into account for the possibility and random set theories by referring to Alvarez [1], Baudrit et al. [2] and Dubois et al. [3].

The family $M = {M_{α} : α \in (0, 1]}$ cannot be nested in general. However, we can always rearrange the family $M$ by applying a function η : (0, 1] → (0, 1] to the original family $M$ to form a new family $M^{(η)} = {M_{α}^{(η)} \equiv M_{η (α)} : α \in (0, 1]}$ such that $M^{(η)}$ is a nested family in the sense of $M_{β}^{(η)} \subseteq M_{α}^{(η)}$ for β > α. In this case, we can also generate a fuzzy set $\tilde{A}$ in U satisfying ${\tilde{A}}_{α} = M_{α}^{(η)}$ for each α ∈ (0, 1]. More general in this paper, we shall generate a fuzzy set $\tilde{A}$ in U satisfying ${\tilde{A}}_{ζ (α)} = M_{α}^{(η)}$ for each α ∈ (0, 1], where ζ : (0, 1] → (0, 1] is an increasing and bijective function. In particular, when ζ is taken to be an identity function, we shall recover the equality ${\tilde{A}}_{α} = M_{α}^{(η)}$ .

In this paper, we do not consider M₀ in the family $M$ , since the 0-level set of a generated fuzzy set from the family $M$ does not necessarily include M₀. The concept of 0-level set is also an important issue. There are two different concepts regarding the 0-level set. The simple concern is to take the whole universal set U as the 0-level set based on the definition of α-level sets for α ∈ [0, 1] given by ${\tilde{A}}_{α} = {x \in U : ξ_{\tilde{A}} (x) \geq α} .$ However, this kind of consideration is not helpful for identifying the effective domain of fuzzy sets. Therefore, the more reasonable way is to define the 0-level set as ${\tilde{A}}_{0} = cl ({\tilde{A}}_{0 +}) = cl ({x \in U : ξ_{\tilde{A}} (x) > 0}),$ where the topological concept of closure is considered. In this case, the 0-level set ${\tilde{A}}_{0}$ can be treated as the effective domain of the membership function $ξ_{\tilde{A}}$ . More concern on the concept of 0-level set can also refer to Wu [16, 17]. On the other hand, Šešelja, Stojić and Tepavčević [11 –14] also studied the relationship between the α-level sets and fuzzy sets in which the 0-level set was not taken into account. In this paper, the 0-level set will be seriously considered. Jaballah and Saidi [8 –10] studied the existence of fuzzy set whose family of α-level sets can represent a pre-determined nested family, where the 0-level set was defined as the whole universal set U. In this paper, the 0-level set will not be assumed to be U and the arbitrary families instead of nested families will be considered.

In Section 2, we present some basic properties of non-normal fuzzy sets that will be used in the discussions. In Section 3, we propose the concept of solid family that will be used to generate a fuzzy set. In Section 4, we consider the arbitrary families that can be rearranged as the nested families by assigning some suitable function. Many interesting properties will be presented and will be used to generate the desired fuzzy sets. In Section 5, two theorems will be established to present the way of generating the desired fuzzy sets from the arbitrary families of sets.

2 Non-normal fuzzy sets

Let $\tilde{A}$ be a fuzzy set in a universal set U with membership function $ξ_{\tilde{A}}$ . The range of membership function $ξ_{\tilde{A}}$ is denoted by $R (ξ_{\tilde{A}})$ . The range $R (ξ_{\tilde{A}})$ can be a proper subset of [0, 1] with $R (ξ_{\tilde{A}}) \neq [0, 1]$ . For example, the range $R (ξ_{\tilde{A}})$ can be some disjoin union of intervals in [0, 1].

For α ∈ (0, 1], the α-level set ${\tilde{A}}_{α}$ of $\tilde{A}$ is defined by ${\tilde{A}}_{α} = {x : ξ_{\tilde{A}} (x) \geq α} .$ For α ∈ [0, 1], we also define the strong α-level set of $\tilde{A}$ as follows ${\tilde{A}}_{α +} = {x : ξ_{\tilde{A}} (x) > α} .$ When $ℝ$ is endowed with a usual topology, the 0-level set ${\tilde{A}}_{0}$ is defined to be the closure of the support of $\tilde{A}$ , i.e., ${\tilde{A}}_{0} = cl ({\tilde{A}}_{0 +})$ . Notice that if $α \notin R (ξ_{\tilde{A}})$ , we still can consider the α-level set ${\tilde{A}}_{α}$ .

Since the range $R (ξ_{\tilde{A}})$ is not necessarily equal to the whole unit interval [0, 1], it is possible that the α-level set ${\tilde{A}}_{α}$ can be an empty set for some α ∈ [0, 1]. In this case, we need to carefully treat the α-level sets for avoiding the emptiness.

Since the range $R (ξ_{\tilde{A}})$ plays an important role for the non-empty α-level sets, the following interesting equalities will be used in the further discussion: ${\tilde{A}}_{0 +} = ⋃_{α \in (0, 1]} {\tilde{A}}_{α} = ⋃_{{α \in R (ξ_{\tilde{A}}) : α > 0}} {\tilde{A}}_{α}$ (1) and ${\tilde{A}}_{0} = cl (⋃_{α \in (0, 1]} {\tilde{A}}_{α}) = cl (⋃_{{α \in R (ξ_{\tilde{A}}) : α > 0}} {\tilde{A}}_{α}) .$ We can rewrite the above equalities by considering the concept of interval range of $\tilde{A}$ .

Recall that the supremum of the set S, denoted by sup S, is not necessarily in S. Suppose that the supremum sup S is attained; that is, there exists α^* ∈ S satisfying α^* = sup S. Then it is also denoted by max S = α^* ∈ S, i.e., the maximum of S. Define $α^{*} = sup R (ξ_{\tilde{A}})$ and $I_{\tilde{A}} = {\begin{matrix} [0, α^{*}), & if the supremum sup R (ξ_{\tilde{A}}) is not attained \\ [0, α^{*}], & if the supremum sup R (ξ_{\tilde{A}}) is attained . \end{matrix}$ (2) It is clear to see $R (ξ_{\tilde{A}}) \subseteq I_{\tilde{A}}$ . We also see that ${\tilde{A}}_{α} \neq \emptyset$ for all $α \in I_{\tilde{A}}$ and ${\tilde{A}}_{α} = \emptyset$ for all $α \notin I_{\tilde{A}}$ . The interval $I_{\tilde{A}}$ presented in (2) is also called an interval range of $\tilde{A}$ . The role of interval range $I_{\tilde{A}}$ can be used to clarify ${\tilde{A}}_{α} \neq \emptyset$ for all $α \in I_{\tilde{A}}$ and ${\tilde{A}}_{α} = \emptyset$ for all $α \notin I_{\tilde{A}}$ . Since $R (ξ_{\tilde{A}}) \neq I_{\tilde{A}}$ in general, the interval $I_{\tilde{A}}$ plays an important role for considering the α-level sets. The range $R (ξ_{\tilde{A}})$ is not helpful for identifying the α-level sets.

By referring to Wu [18], we also have ${\tilde{A}}_{0 +} = ⋃_{{α \in I_{\tilde{A}} : α > 0}} {\tilde{A}}_{α} = ⋃_{{α \in R (ξ_{\tilde{A}}) : α > 0}} {\tilde{A}}_{α} .$ (3) For α ∈ [0, 1), since ${\tilde{A}}_{β} = \emptyset$ for $β \notin I_{\tilde{A}}$ , we can show that ${\tilde{A}}_{α +} = ⋃_{β \in (α, 1]} {\tilde{A}}_{β} = ⋃_{β \in I_{\tilde{A}} \cap (α, 1]} {\tilde{A}}_{β} = ⋃_{{β \in R (ξ_{\tilde{A}}) : β > α}} {\tilde{A}}_{β} .$ (4)

Recall that $\tilde{A}$ is called a normal fuzzy set in $ℝ$ when there exists $x \in ℝ$ satisfying $ξ_{\tilde{A}} (x) = 1$ . In this case, the interval range of $\tilde{A}$ is given by $I_{\tilde{A}} = [0, 1]$ . However, the range $R (ξ_{\tilde{A}})$ is still not necessarily equal to the whole unit interval [0, 1] even though $\tilde{A}$ is normal. The following interesting results will be used in the subsequent discussion.

Proposition 2.1. Let $\tilde{A}$ be a fuzzy set in a universal U, and let $D_{α} \equiv {\tilde{A}}_{α} \ {\tilde{A}}_{α +}$ for $α \in R (ξ_{\tilde{A}})$ with α > 0. Then D_α≠ ∅ for any $α \in R (ξ_{\tilde{A}})$ with α > 0, and the support ${\tilde{A}}_{0 +}$ is the disjoin union of sets D_α given by ${\tilde{A}}_{0 +} = ⋃_{{α \in R (ξ_{\tilde{A}}) : α > 0}} D_{α} = ⋃_{{α \in R (ξ_{\tilde{A}}) : α > 0}} ({\tilde{A}}_{α} ∖ ⋃_{{β \in R (ξ_{\tilde{A}}) : β > α}} {\tilde{A}}_{β}),$ where D_α∩ D_β = ∅ for α ≠ β.

Proof. Given any $α \in R (ξ_{\tilde{A}})$ with α > 0, we have

$\begin{matrix} \emptyset \neq ξ_{\tilde{A}}^{- 1} (α) & = {x \in U : ξ_{\tilde{A}} (x) = α} = {x \in U : ξ_{\tilde{A}} (x) \geq α} \ {x \in U : ξ_{\tilde{A}} (x) > α} \\ = {\tilde{A}}_{α} \ {\tilde{A}}_{α +} = D_{α} = {\tilde{A}}_{α} ∖ ⋃_{{β \in R (ξ_{\tilde{A}}) : β > α}} {\tilde{A}}_{β} . (using (4)) . \end{matrix}$ (5) From (5), we obtain $\begin{matrix} {\tilde{A}}_{0 +} & = {x \in U : ξ_{\tilde{A}} (x) > 0} = ⋃_{{α \in R (ξ_{\tilde{A}}) : α > 0}} ξ_{\tilde{A}}^{- 1} (α) = ⋃_{{α \in R (ξ_{\tilde{A}}) : α > 0}} D_{α} \\ = ⋃_{{α \in R (ξ_{\tilde{A}}) : α > 0}} ({\tilde{A}}_{α} ∖ ⋃_{{β \in R (ξ_{\tilde{A}}) : β > α}} {\tilde{A}}_{β}) . \end{matrix}$ For α ≠ β, it is obvious that $D_{α} \cap D_{β} = {x \in U : ξ_{\tilde{A}} (x) = α} \cap {x \in U : ξ_{\tilde{A}} (x) = β} = \emptyset .$ This completes the proof. ■

3 Solid families

Let $M = {M_{α} : α \in (0, 1]}$ be a family of nonempty subsets of a universal set U. Let T be a subset of (0, 1]. Then, we can consider the subfamily $M_{T}$ of $M$ defined by $M_{T} = {M_{α} : α \in T} .$ Let η : S → [0, 1] be a function defined on S. The range of η is denoted by $R (η)$ . Then, we take $T = R (η)$ and rename the elements of $M$ to form a new family $M_{R (η)} = {M_{η (α)} : α \in S} .$ For convenience, we write $M_{α}^{(η)} \equiv M_{η (α)}$ for α ∈ S. Then, we consider the following family $M_{S}^{(η)} = {M_{α}^{(η)} : α \in S} = {M_{η (α)} : α \in S} = M_{R (η)} .$ We have the following observations.

Suppose that η is an identity function id. Then $M_{S}^{(id)} = M_{S}$ .

Suppose that η is an injective function from S into (0, 1]. Then, for each $α \in R (η)$ , there exists β ∈ S such that η (β) = α. In this case, we also write β = η^-1 (α). Therefore, if the function η is injective, we have $M_{α} = M_{η^{- 1} (α)}^{(η)}$ (6) for $α \in R (η)$ .

Example 3.1. Given any real numbers a₁, a₂, b₁, b₂ satisfying the order b₁ < a₁ < a₂ < b₂, for α ∈ [0, 1], we take $m_{α}^{L} = {\begin{matrix} α b_{1} + (1 - α) a_{1}, & if 0.1 < α \leq 0.8 \\ (1 - α) b_{1} + α a_{1}, & if 0 < α \leq 0.1 or 0.8 < α \leq 1 \end{matrix}$ and $m_{α}^{U} = {\begin{matrix} α b_{2} + (1 - α) a_{2}, & if 0.1 < α \leq 0.8 \\ (1 - α) b_{2} + α a_{2}, & if 0 < α \leq 0.1 or 0.8 < α \leq 1 . \end{matrix}$ It is clear to see $m_{α}^{L} < m_{α}^{U}$ for all α ∈ [0, 1]. Therefore, we define a closed interval M_α in $ℝ$ by $M_{α} = [m_{α}^{L}, m_{α}^{U}]$ for α ∈ (0, 1]. Now, we take S = [0.2, 0.9) and define a function η from S into (0, 1] by η (α) =1 - α. We also define ${\bar{m}}_{α}^{L} = m_{1 - α}^{L} and {\bar{m}}_{α}^{U} = m_{1 - α}^{U} for α \in [0, 1] .$ It is clear to see that ${\bar{m}}_{α}^{L} = {\begin{matrix} (1 - α) b_{1} + α a_{1}, & if 0.2 \leq α < 0.9 \\ α b_{1} + (1 - α) a_{1}, & if 0 \leq α < 0.2 or 0.9 \leq α < 1 \end{matrix}$ and ${\bar{m}}_{α}^{U} = {\begin{matrix} (1 - α) b_{2} + α a_{2}, & if 0.2 \leq α < 0.9 \\ α b_{2} + (1 - α) a_{2}, & if 0 \leq α < 0.2 or 0.9 \leq α < 1 \end{matrix}$ Therefore, we have $M_{α}^{(η)} = M_{η (α)} = M_{1 - α} = [m_{1 - α}^{L}, m_{1 - α}^{U}] = [{\bar{m}}_{α}^{L}, {\bar{m}}_{α}^{U}] for α \in S,$ where $M_{α}^{(η)} \subset M_{β}^{(η)}$ with $M_{α}^{(η)} \neq M_{β}^{(η)}$ for α > β.

In the sequel, we shall propose the concept of solid family. Let S be a subset of (0, 1] with sup S = α^*. For α ∈ S with α < α^*, we write $M_{α}^{S} = ⋃_{{β \in S : α^{*} > β \geq α}} M_{β}^{(η)} and M_{α +}^{S} = ⋃_{{β \in S : α^{*} > β > α}} M_{β}^{(η)} .$

Inspired by Proposition 2, by including the infimum inf S of S, we propose the following definition.

Definition 3.2. Let S be a subset of (0, 1] with sup S = α^* and $inf S = \bar{α}$ , and let η be a function from S into (0, 1]. Let $M = {M_{α} : α \in (0, 1]}$ be a family of nonempty subsets of a universal set U.

We say that $M$ is a solid family with respect to (η, S) when $D_{α}^{S} \equiv M_{α}^{S} \ M_{α +}^{S} \neq \emptyset$ for α ∈ S with α < α^*.

We say that $M$ is a relaxed solid family with respect to (η, S) when $D_{α}^{S} \equiv M_{α}^{S} \ M_{α +}^{S} \neq \emptyset$ for α ∈ S with $\bar{α} < α < α^{*}$ .

We also remark that if $\bar{α} \notin S$ , the concepts of solid family and relaxed solid family are equivalent.

Example 3.3. Continued from Example 3, we want to show $M_{α +}^{S} = ⋃_{{β \in S : α^{*} > β > α}} M_{β}^{(η)} = ⋃_{{β \in S : α^{*} > β > α}} [{\bar{m}}_{β}^{L}, {\bar{m}}_{β}^{U}] = ({\bar{m}}_{α}^{L}, {\bar{m}}_{α}^{U})$ If $x \in M_{α +}^{S}$ , then $x \in [{\bar{m}}_{β_{0}}^{L}, {\bar{m}}_{β_{0}}^{U}]$ for some β₀ ∈ S with β₀ > α. Since ${M_{α}^{(η)}}_{α \in S}$ is a strictly nested family in the sense of $M_{β}^{(η)} \subset M_{α}^{(η)}$ with $M_{β}^{(η)} \neq M_{α}^{(η)}$ for β > α, we have ${\bar{m}}_{α}^{L} < {\bar{m}}_{β_{0}}^{L} \leq x \leq {\bar{m}}_{β_{0}}^{U} < {\bar{m}}_{α}^{U},$ which says that $x \in ({\bar{m}}_{α}^{L}, {\bar{m}}_{α}^{U})$ . Therefore, we obtain the inclusion $M_{α +}^{S} \subseteq ({\bar{m}}_{α}^{L}, {\bar{m}}_{α}^{U})$ On the other hand, since ${\bar{m}}_{α}^{L}$ and ${\bar{m}}_{α}^{U}$ are continuous with respect to α, using the strict monotonicity of ${\bar{m}}_{α}^{L}$ and ${\bar{m}}_{α}^{U}$ with respect to α, for any $x \in ({\bar{m}}_{α}^{L}, {\bar{m}}_{α}^{U})$ , there exists β₁ ∈ S with β₁ > α such that $x \geq {\bar{m}}_{β_{1}}^{L} > {\bar{m}}_{α}^{L}$ and $x \leq {\bar{m}}_{β_{1}}^{U} < {\bar{m}}_{α}^{U}$ , which says that $x \in [{\bar{m}}_{β_{1}}^{L}, {\bar{m}}_{β_{1}}^{U}]$ , i.e., $x \in M_{α +}^{S}$ . Therefore, we obtain the desired equality. Since ${M_{α}^{(η)}}_{α \in S}$ is a strictly nested family, we can also obtain $M_{α}^{S} = ⋃_{{β \in S : α^{*} > β \geq α}} M_{β}^{(η)} = ⋃_{{β \in S : α^{*} > β \geq α}} [{\bar{m}}_{β}^{L}, {\bar{m}}_{β}^{U}] = [{\bar{m}}_{α}^{L}, {\bar{m}}_{α}^{U}],$ which implies $D_{α}^{S} = M_{α}^{S} \ M_{α +}^{S} = [{\bar{m}}_{α}^{L}, {\bar{m}}_{α}^{U}] \ ({\bar{m}}_{α}^{L}, {\bar{m}}_{α}^{U}) = {{\bar{m}}_{α}^{L}, {\bar{m}}_{α}^{U}} \neq \emptyset .$ This shows that $M$ is a solid family with respect to (η, S).

Proposition 3.4. Let S be a subset of (0, 1] with sup S = α^* and $inf S = \bar{α}$ , and let η be a function from S into (0, 1]. Let $M = {M_{α} : α \in (0, 1]}$ be a family of nonempty subsets of a universal set U such that it is a relaxed solid family with respect to (η, S). Then the following statements hold true.

We have $D_{α}^{S} \cap D_{β}^{S} = \emptyset$ for α, β ∈ S with $\bar{α} < α < α^{*}$ , $\bar{α} < β < α^{*}$ and α ≠ β.

Suppose that α^* ∈ S. Then $M_{α^{*}}^{(η)} \cap D_{α}^{S} = \emptyset$ for α ∈ S with $\bar{α} < α < α^{*}$ .

Suppose that α^* ∉ S and there exists M^* satisfying $M^{*} \subseteq M_{α}^{(η)}$ for all α ∈ S. Then $M^{*} \cap D_{α}^{S} = \emptyset$ for α ∈ S with $\bar{α} < α$ .

Moreover, when

M

is a solid family with respect to (η, S), we can have the same results without assuming

α > \bar{α}

Proof. To prove part (i), given α, β ∈ S with $\bar{α} < α < β < α^{*}$ , we have $M_{β}^{S} \subseteq M_{α}^{S}, D_{β}^{S} \subseteq M_{β}^{S} \subseteq M_{α +}^{S} and D_{α}^{S} = M_{α}^{S} \ M_{α +}^{S},$ which says that $D_{α}^{S} \cap D_{β}^{S} = \emptyset$ for α ≠ β with $\bar{α} < α < α^{*}$ and $\bar{α} < β < α^{*}$ .

To prove part (ii), suppose that $x \in D_{α}^{S}$ for α ∈ S with $\bar{α} < α < α^{*}$ . By the definition of $D_{α}^{S}$ , we see that $x \notin M_{α +}^{S}$ , i.e., $x \notin M_{β}^{(η)}$ for all β ∈ S with β > α, which implies $x \notin M_{α^{*}}^{(η)}$ , since α^* ∈ S with α^* > α. Therefore, we obtain the desired result.

To prove part (iii), suppose that $x \in D_{α}^{S}$ for α ∈ S with $\bar{α} < α$ . It says that $x \notin M_{α +}^{S}$ , i.e., $x \notin M_{β}^{(η)}$ for all β ∈ S with β > α. Since β ∈ S, the assumption says that $M^{*} \subseteq M_{β}^{(η)}$ , which implies x ∉ M^*, since $x \notin M_{β}^{(η)}$ . Therefore, we obtain $M^{*} \cap D_{α}^{S} = \emptyset$ . This completes the proof. ■

Proposition 3.5. Let S be a subset of (0, 1] with sup S = α^* and $inf S = \bar{α}$ , and let η be a function from S into (0, 1]. Let $M = {M_{α} : α \in (0, 1]}$ be a family of nonempty subsets of a universal set U such that it is a relaxed solid family with respect to (η, S). Let T be a subset of S. Given any α ∈ S with $α > \bar{α}$ , we have $M_{α}^{(η)} ⋂ (⋃_{{β \in T : α^{*} > β}} D_{β}^{S}) = M_{α}^{(η)} ⋂ (⋃_{{β \in T : α^{*} > β \geq α}} D_{β}^{S}) .$ When $M$ is a solid family with respect to (η, S), we can have the same equality without assuming $α > \bar{α}$ .

Proof. Given any fixed $x \in M_{α}^{(η)}$ and $x \in D_{β}^{S}$ for some β ∈ T with $\bar{α} < β < α^{*}$ , we have $x \in D_{β}^{S} = ⋃_{{α \in S : α^{*} > α \geq β}} M_{α}^{(η)} ∖ ⋃_{{α \in S : α^{*} > α > β}} M_{α}^{(η)} .$ (7) If α > β, then the equality (7) says that $x \notin M_{α}^{(η)}$ , which contradicts $x \in M_{α}^{(η)}$ . Therefore, we must have α ≤ β < α^*, which implies the following inclusion $M_{α}^{(η)} ⋂ (⋃_{{β \in T : α^{*} > β}} D_{β}^{S}) \subseteq M_{α}^{(η)} ⋂ (⋃_{{β \in T : α^{*} > β \geq α}} D_{β}^{S}) .$ Another direction of inclusion is obvious. This completes the proof.■

4 Restricted nested families

In the sequel, we shall consider the concept of nested families with respect to a subset S of (0, 1] and a function η : S → [0, 1], and provide some interesting properties that will be used to study the existence of fuzzy sets.

Definition 4.1. Let S be a subset of (0, 1], and let $M = {M_{α} : α \in (0, 1]}$ be a family of nonempty subsets of a universal set U. We consider the subfamily $M_{S} = {M_{α} : α \in S}$ of $M$ .

We say that $M_{S}$ is a nested family when M_α ⊆ M_β for α, β ∈ S with β < α.

We say that $M_{S}$ is a strictly nested family when M_α ⊂ M_β with M_α ≠ M_β for α, β ∈ S with β < α.

If S = (0, 1], then we simply say that

M

is a (strictly) nested family if and only if

M_{S}

is a (strictly) nested family.

Even though $M$ is not a nested family, it is possible that there exists a subset S of (0, 1] such that $M_{S}$ is a nested subfamily.

Definition 4.2. Let S be a subset of (0, 1], and let $M = {M_{α} : α \in (0, 1]}$ be a family of nonempty subsets of a universal set U. Let η be a function from S into (0, 1].

We say that $M$ is a nested family with respect to (η, S) when $M_{α}^{(η)} \subseteq M_{β}^{(η)}$ for α, β ∈ S with β < α.

We say that $M$ is a strictly nested family with respect to (η, S) when $M_{α}^{(η)} \subset M_{β}^{(η)}$ with $M_{α}^{(η)} \neq M_{β}^{(η)}$ for α, β ∈ S with β < α.

Definition 4.3. Let S be a subset of (0, 1], and let $M = {M_{α} : α \in (0, 1]}$ be a family of nonempty subsets of a universal set U. We say that the family $M_{S} = {M_{α} : α \in S}$ has a focal set M^* when M^* ⊆ M_α for all α ∈ S. If M^* ≠ M_α for any α ∈ S, we also say that the nested family $M_{S}$ has a strict focal set M^*.

Proposition 4.4. Let S be a subset of (0, 1] with sup S = α^*, and let η be a function from S into (0, 1]. Let $M = {M_{α} : α \in (0, 1]}$ be a family of nonempty subsets of a universal set U such that it is a solid and nested family with respect to (η, S).

Suppose that α^* ∈ S. Then $M_{α^{*}}^{(η)}$ is a focal set of the family $M_{S}^{(η)} = {M_{α}^{(η)} : α \in S}$ . Given any α ∈ S with α < α^*, we also have $M_{α^{*}}^{(η)} ⋃ (⋃_{{β \in S : α^{*} > β \geq α}} D_{β}^{S}) \subseteq M_{α}^{(η)} .$ (8)

Suppose that α^* ∉ S. Then, given any α ∈ S, we have $⋃_{{β \in S : β \geq α}} D_{β}^{S} \subseteq M_{α}^{(η)} .$ (9)

Suppose that α^* ∉ S, and that the family $M_{S}^{(η)}$ has a focal set M^*. Given any α ∈ S, we have $M^{*} ⋃ (⋃_{{β \in S : β \geq α}} D_{β}^{S}) \subseteq M_{α}^{(η)} .$ (10)

Proof. To prove part (i), since $M$ is a nested family with respect to (η, S), we first have $M_{α^{*}}^{(η)} \subseteq M_{α}^{(η)}$ . Given any α ∈ S with α < α^*, since $M$ is a solid family with respect to (η, S), we have $D_{α}^{S} \subseteq M_{α}^{S} \subseteq M_{α}^{(η)} and D_{β}^{S} \subseteq M_{β}^{(η)} \subseteq M_{α}^{(η)}$ for β ∈ S with α < β. Therefore, we obtain the inclusion (8). Part (ii) can be realized from the proof of part (i).

To prove part (iii), by definition, we have $M^{*} \subseteq M_{α}^{(η)}$ for all α ∈ S, i.e., α < α^*. Since we also have $D_{α}^{S} \subseteq M_{α}^{(η)}$ and $D_{β}^{S} \subseteq M_{β}^{(η)} \subseteq M_{α}^{(η)}$ for β ∈ S with α < β, we immediately have the inclusion (10), and the proof is complete. ■

Proposition 4.5 Let S be a subset of (0, 1] with sup S = α^*, and let η be a function from S into (0, 1]. Let $M = {M_{α} : α \in (0, 1]}$ be a family of nonempty subsets of a universal set U such that $M$ is a solid and nested family with respect to (η, S).

Suppose that α^* ∈ S. Then $⋃_{β \in S} M_{β}^{(η)} = M_{α^{*}}^{(η)} ⋃ (⋃_{{β \in S : β < α^{*}}} D_{β}^{S})$ (11) if and only if, given any α ∈ S with α < α^*, $M_{α}^{(η)} = M_{α^{*}}^{(η)} ⋃ (⋃_{{β \in S : α^{*} > β \geq α}} D_{β}^{S}) .$ (12)

Suppose that α^* ∉ S. Then $⋃_{β \in S} M_{β}^{(η)} = ⋃_{β \in S} D_{β}^{S}$ if and only if, given any α ∈ S, $M_{α}^{(η)} = ⋃_{{β \in S : β \geq α}} D_{β}^{S} .$

Suppose that α^* ∉ S, and that the family $M_{S}^{(η)} = {M_{α}^{(η)} : α \in S}$ has a focal set M^*. Then $⋃_{β \in S} M_{β}^{(η)} = M^{*} ⋃ (⋃_{β \in S} D_{β}^{S})$ if and only if, given any α ∈ S, $M_{α}^{(η)} = M^{*} ⋃ (⋃_{{β \in S : β \geq α}} D_{β}^{S}) .$

Proof. To prove the sufficiency of part (i), using (8), we have the following inclusion $M_{α^{*}}^{(η)} ⋃ (⋃_{{β \in S : α^{*} > β \geq α}} D_{β}^{S}) \subseteq M_{α}^{(η)} .$ For proving another direction of inclusion, we have $\begin{matrix} M_{α}^{(η)} & = M_{α}^{(η)} ⋂ (⋃_{β \in S} M_{β}^{(η)}) \\ = M_{α}^{(η)} ⋂ (M_{α^{*}}^{(η)} ⋃ (⋃_{{β \in S : β < α^{*}}} D_{β}^{S})) (by the assumption (11)) \\ = M_{α}^{(η)} ⋂ (M_{α^{*}}^{(η)} ⋃ (⋃_{{β \in S : α^{*} > β \geq α}} D_{β}^{S})) \\ (using Proposition 3 by taking T = S), \end{matrix}$ which implies $M_{α}^{(η)} \subseteq M_{α^{*}}^{(η)} ⋃ (⋃_{{β \in S : α^{*} > β \geq α}} D_{β}^{S}) .$ For proving the necessity of part (i), we have $\begin{matrix} ⋃_{α \in S} M_{α}^{(η)} & = M_{α^{*}}^{(η)} ⋃ (⋃_{{α \in S : α < α^{*}}} M_{α}^{(η)}) \\ = M_{α^{*}}^{(η)} ⋃ (⋃_{{α \in S : α < α^{*}}} (M_{α^{*}}^{(η)} ⋃ (⋃_{{β \in S : α^{*} > β \geq α}} D_{β}^{S}))) (by the assumption (12) \\ = M_{α^{*}}^{(η)} ⋃ (⋃_{{α \in S : α < α^{*}}} D_{α}^{S}) . \end{matrix}$ Part (ii) can be realized from part (i) by using (9). Part (iii) can be similarly obtained by using (10). This completes the proof. ■

Proposition 4.6. ıLet S be a subset of (0, 1] with sup S = α^*, and let η be a function from S into (0, 1]. Let $M = {M_{α} : α \in (0, 1]}$ be a family of nonempty subsets of a universal set U such that $M$ is a solid and nested family with respect to (η, S).

Suppose that α^* ∈ S. Then, for any α ∈ S with α < α^*, we have the strict inclusion $M_{α^{*}}^{(η)} \underset{\neq}{\subset} M_{α}^{(η)}$ .

Suppose that α^* ∉ S, and that the family $M_{S}^{(η)} = {M_{α}^{(η)} : α \in S}$ has a focal set M^*. Then, for any α ∈ S, we have the strict inclusion $M^{*} \underset{\neq}{\subset} M_{α}^{(η)}$ ; that is to say, the family $M_{S}^{(η)}$ has a strict focal set M^*.

Proof. To prove part (i), since $M$ is a nested family with respect to (η, S) and α^* ∈ S, for any α ∈ S, we have the inclusion $M_{α^{*}}^{(η)} \subseteq M_{α}^{(η)}$ . Suppose that there exists β₀ ∈ S with β₀ < α^* satisfying $M_{α^{*}}^{(η)} = M_{β_{0}}^{(η)}$ . Then, for any β ∈ S with β > β₀, since $M_{α^{*}}^{(η)} \subseteq M_{β}^{(η)} \subseteq M_{β_{0}}^{(η)} = M_{α^{*}}^{(η)},$ it follows that $M_{α^{*}}^{(η)} = M_{β}^{(η)} = M_{β_{0}}^{(η)}$ for any β ∈ S with β > β₀, which implies $M_{β_{0} +}^{S} = ⋃_{{α \in S : β > β_{0}}} M_{β}^{(η)} = M_{α^{*}}^{(η)} = M_{β_{0}}^{(η)} .$ Since $M$ is a nested family with respect to (η, S), we also have $M_{β_{0}}^{S} = ⋃_{{α \in S : β \geq β_{0}}} M_{β}^{(η)} = M_{β_{0}}^{(η)} .$ Therefore, we obtain $D_{β_{0}}^{S} = M_{β_{0}}^{S} \ M_{β_{0} +}^{S} = \emptyset,$ which contradicts $D_{α}^{S} \neq \emptyset$ for any α ∈ S with α < α^*.

To prove part (ii), since the family $M_{S}^{(η)} = {M_{α}^{(η)} : α \in S}$ has a focal set M^*, we have the inclusion $M^{*} \subseteq M_{α}^{(η)}$ for any α ∈ S. Suppose that there exists β₀ ∈ S satisfying $M^{*} = M_{β_{0}}^{(η)}$ . Then, for any β ∈ S with β > β₀, since $M^{*} \subseteq M_{β}^{(η)} \subseteq M_{β_{0}}^{(η)} = M^{*},$ it follows that $M^{*} = M_{β}^{(η)} = M_{β_{0}}^{(η)}$ for any β ∈ S with β > β₀, which implies $M_{β_{0} +}^{S} = ⋃_{{α \in S : β > β_{0}}} M_{β}^{(η)} = M^{*} = M_{β_{0}}^{(η)} .$ The remaining proof follows from the above same argument. This completes the proof. ■

We say that S is a disjoin union of intervals in (0, 1] if S can be expressed as $S = I_{1} \cup \dots \cup I_{m}$ for m subintervals of (0, 1] satisfying I_i∩ I_j = ∅ for i ≠ j. In this paper, we allow m = 1 for the special case.

Proposition 4.7. Let S be a disjoin union of intervals in (0, 1] with sup S = α^*, and let η be a function from S into (0, 1]. Let $M = {M_{α} : α \in (0, 1]}$ be a family of nonempty subsets of a universal set U such that $M$ is a solid and nested family with respect to (η, S).

Suppose that α^* ∈ S, and that, for any α ∈ S with α < α^*, $M_{α}^{(η)} = M_{α^{*}}^{(η)} ⋃ (⋃_{{β \in S : α^{*} > β \geq α}} D_{β}^{S}) .$ (13) Then $⋂_{n = 1}^{\infty} M_{α_{n}}^{(η)} = M_{α^{*}}^{(η)}$ for α_n ↑ α^* with α_n ∈ S and α_n < α^* for all n.

Suppose that α^* ∉ S, and that, for any α ∈ S, $M_{α}^{(η)} = ⋃_{{β \in S : β \geq α}} D_{β}^{S} .$ (14) Then $⋂_{n = 1}^{\infty} M_{α_{n}}^{(η)} = \emptyset$ for α_n ↑ α^* with α_n ∈ S for all n.

Suppose that α^* ∉ S, that the family $M_{S}^{(η)} = {M_{α}^{(η)} : α \in S}$ has a focal set M^*, and that, for any α ∈ S, $M_{α}^{(η)} = M^{*} ⋃ (⋃_{{β \in S : β \geq α}} D_{β}^{S}) .$ (15) Then $⋂_{n = 1}^{\infty} M_{α_{n}}^{(η)} = M^{*}$ for α_n ↑ α^* with α_n ∈ S for all n.

Proof. To prove part (i), according to part (i) of Proposition 4, we define $N_{α} = M_{α}^{(η)} \ M_{α^{*}}^{(η)} \neq \emptyset for α \in S with α < α^{*} .$ (16) Then, we immediately have N_α ⊆ N_β for β < α < α^*. Suppose that $⋂_{n = 1}^{\infty} N_{α_{n}} \neq \emptyset$ for α_n ↑ α^* with α_n ∈ S and α_n < α^* for all n. We are going to lead to a contradiction. For x ∈ N_{α
_n} for all n, given any α ∈ S with α < α^*, since α_n ↑ α^* with α_n ∈ S and α_n < α^* for all n, there exists α_{m
₁} satisfying α^* > α_{m
₁} > α, which says that x ∈ N_{α
_{m
₁}} ⊆ N_α, i.e., x ∈ N_α. Using (16) and the assumption (13), we see that $x \in D_{β_{1}}^{S}$ for some β₁ ∈ S with α^* > β₁ ≥ α. Since S is a disjoin union of intervals and α^* ∈ S, given any β₁ ∈ S with β₁ < α^*, there exists β₂ ∈ S satisfying α^* > β₂ > β₁. We consider the following cases.

Since $M_{α^{*}}^{(η)} \cap D_{β_{1}}^{S} = \emptyset$ for β₁ ∈ S with β₁ < α^* by part (ii) of Proposition 3, it follows that $x \notin M_{α^{*}}^{(η)}$ .

For any β ∈ S with α^* > β ≥ β₂, since β₂ > β₁, it follows that β ≠ β₁. Since $D_{β_{1}}^{S} \cap D_{β}^{S} = \emptyset$ by part (i) of Proposition 3, we obtain $x \notin D_{β}^{S}$ .

Therefore, by the assumption (13), the above cases conclude that

x \notin M_{α^{*}}^{(η)} ⋃ (⋃_{{β \in S : α^{*} > β \geq β_{2}}} D_{β}^{S}) = M_{β_{2}}^{(η)} .

Since

N_{β_{2}} \subseteq M_{β_{2}}^{(η)}

, it follows that x ∉ N_{β
₂}. Since β₂ < α^* and α_n ↑ α^* with α_n ∈ S and α_n < α^* for all n, there exists α_{m
₂} such that α_{m
₂} ∈ S and α^* > α_{m
₂} > β₂, which says that N_{α
_{m
₂}} ⊆ N_{β
₂}. Therefore, we obtain x ∉ N_{α
_{m
₂}}, which contradicts x ∈ N_{α
_n} for all n. This shows that

⋂_{n = 1}^{\infty} N_{α_{n}} = \emptyset

. It is easy to see that

⋂_{n = 1}^{\infty} (M_{α_{n}}^{(η)} \ M_{α^{*}}^{(η)}) = ⋂_{n = 1}^{\infty} N_{α_{n}}

= \emptyset only if ⋂_{n = 1}^{\infty} M_{α_{n}}^{(η)} = M_{α^{*}}^{(η)}

when α_n ↑ α^* with α_n ∈ S and α_n < α^* for all n.

To prove part (ii), suppose that $⋂_{n = 1}^{\infty} M_{α_{n}}^{(η)} \neq \emptyset$ for α_n ↑ α^* with α_n ∈ S and α_n < α^* for all n. We are going to lead to a contradiction. For $x \in M_{α_{n}}^{(η)}$ for all n, given any α ∈ S with α < α^*, since α_n ↑ α^* with α_n ∈ S and α^* ∉ S, it follows that α_n < α^* for all n. Therefore, there exists α_{m
₃} satisfying α^* > α_{m
₃} > α, which says that $x \in M_{α_{m_{3}}}^{(η)} \subseteq M_{α}^{(η)}$ , i.e., $x \in M_{α}^{(η)}$ . Using the assumption (14), we see that $x \in D_{β_{3}}^{S}$ for some β₃ ∈ S with β₃ ≥ α. Since α^* ∉ S, it means that α^* > β₃. Since S is a disjoin union of intervals, there exists β₄ ∈ S satisfying α^* > β₄ > β₃. For any β ∈ S with β ≥ β₄, since β₄ > β₃, it follows that β ≠ β₃. Since $D_{β_{3}}^{S} \cap D_{β}^{S} = \emptyset$ by part (i) of Proposition 3, we obtain $x \notin D_{β}^{S}$ . Therefore, by the assumption (14), we see that $x \notin ⋃_{{β \in S : β \geq β_{4}}} D_{β}^{S} = M_{β_{4}}^{(η)} .$ Since β₄ < α^* and α_n ↑ α^* with α_n ∈ S, i.e., α_n < α^* for all n, there exists α_{m
₄} satisfying α_{m
₄} ∈ S and α_{m
₄} > β₄, which says that $M_{α_{m_{4}}}^{(η)} \subseteq M_{β_{4}}^{(η)}$ . Therefore, we obtain $x \notin M_{α_{m_{4}}}^{(η)}$ , which contradicts $x \in M_{α_{n}}^{(η)}$ for all n. This shows that $⋂_{n = 1}^{\infty} M_{α_{n}}^{(η)} = \emptyset$ .

To prove part (iii), since α^* ∉ S, we see that α < α^* for all α ∈ S. By part (ii) of Proposition 4, we have $M^{*} \underset{\neq}{\subset} M_{α}^{(η)}$ for all α ∈ S. Let $N_{α} = M_{α}^{(η)} \ M^{*} \neq \emptyset$ for α ∈ S. Then, we immediately have N_α ⊆ N_β for β < α with α, β ∈ S. Suppose that $⋂_{n = 1}^{\infty} N_{α_{n}} \neq \emptyset$ for α_n ↑ α^*, and that x ∈ N_{α
_n} for all n. Given any α ∈ S, i.e., α < α^*, since α_n ↑ α^*, there exists m₅ satisfying m₅ > α, which says that x ∈ N_{α
_{m
₅}} ⊆ N_α, i.e., $x \in N_{α} = M_{α}^{(η)} \ M^{*}$ and x ∉ M^*. By the assumption (15), we see that $x \in D_{β_{5}}^{S}$ for some β₅ ∈ S with β₅ ≥ α. Since α^* ∉ S, it means that α^* > β₅. Since S is a disjoin union of intervals, there exists β₆ ∈ S satisfying α^* > β₆ > β₅. For any β ∈ S with β ≥ β₆, since β₆ > β₅, it follows that β ≠ β₅. Since $D_{β_{5}}^{S} \cap D_{β}^{S} = \emptyset$ by part (i) of Proposition 3, it follows that $x \notin D_{β}^{S}$ . Since x ∉ M^*, using the assumption (15), we obtain $x \notin M^{*} ⋃ (⋃_{{β \in S : β \geq β_{6}}} D_{β}^{S}) = M_{β_{6}}^{(η)} .$ Since $N_{β_{6}} \subseteq M_{β_{6}}^{(η)}$ , it follows that x ∉ N_{β
₆}. Since α_n ↑ α^* and β₆ < α^*, there exists α_{m
₆} such that α_{m
₆} > β₆, which says that N_{α
_{m
₆}} ⊆ N_{β
₆}. Therefore, we obtain x ∉ N_{α
_{m
₆}}, which contradicts x ∈ N_{α
_n} for all n. This shows that $⋂_{n = 1}^{\infty} N_{α_{n}}^{(η)} = \emptyset$ . It is easy to see that $⋂_{n = 1}^{\infty} (M_{α_{n}}^{(η)} \ M^{*}) = \emptyset if and only if ⋂_{n = 1}^{\infty} M_{α_{n}}^{(η)} = M^{*}$ when α_n ↑ α^* with α_n ∈ S for all n. This completes the proof. ■

The converse of Proposition 2 can be obtained by providing some mild assumptions. We first present some useful inclusions

Proposition 4.8 Let S be a subset of (0, 1], and let η be a function from S into (0, 1]. Let $M = {M_{α} : α \in (0, 1]}$ be a family of nonempty subsets of a universal set U such that $M$ is a solid and nested family with respect to (η, S). We have the following properties.

Suppose that α^* ∈ S. Then, for any α ∈ S with α < α^*, we have the strict inclusion $M_{α^{*}}^{(η)} \underset{\neq}{\subset} M_{α +}^{S}$ .

Proof. To prove part (i), for any α ∈ S with α < α^*, since $M_{α^{*}}^{(η)} \subseteq M_{γ}^{(η)}$ for all γ ∈ S, we have $M_{α^{*}}^{(η)} \subseteq ⋃_{{γ \in S : γ > α}} M_{γ}^{(η)} = M_{α +}^{S} .$ (17) We want to show $M_{α^{*}}^{(η)} \neq M_{α +}^{S}$ . Suppose that $M_{α^{*}}^{(η)} = M_{α +}^{S}$ . We are going to lead to a contradiction. Since sup S = α^*, for any α ∈ S with α < α^*, according to the concept of supremum, there exists γ ∈ S such that α^* > γ > α. Therefore, we obtain $M_{α^{*}}^{(η)} \subseteq M_{γ}^{(η)} \subseteq ⋃_{{β \in S : β > α}} M_{β}^{(η)} = M_{α +}^{S} = M_{α^{*}}^{(η)},$ i.e., $M_{α^{*}}^{(η)} = M_{γ}^{(η)}$ . Since $M$ is a nested family with respect to (η, S), we also have $M_{γ}^{S} = ⋃_{{β \in S : β \geq γ}} M_{β}^{(η)} = M_{γ}^{(η)} .$ Since $M_{α^{*}}^{(η)} \subseteq M_{γ +}^{S}$ from (17), we obtain $D_{γ}^{S} = M_{γ}^{(S)} \ M_{γ +}^{S} = M_{γ}^{(η)} \ M_{γ +}^{S} \subseteq M_{γ}^{(η)} \ M_{α^{*}}^{(η)} = \emptyset,$ which contradicts $D_{α}^{S} \neq \emptyset$ for α ∈ S with α < α^*. Therefore, we have the desired strict inclusion.

To prove part (ii), by the definition of focal set, we have $M^{*} \subseteq M_{γ}^{(η)}$ for all γ ∈ S. Therefore, for any α ∈ S, we have $M^{*} \subseteq M_{α +}^{S}$ . We want to show $M^{*} \neq M_{α +}^{S}$ . Suppose that $M^{*} = M_{α +}^{S}$ . We are going to lead to a contradiction. Since sup S = α^* and α^* > α, there exists γ ∈ S such that γ > α. Since α^* ∉ S, it follows that α < γ < α^*. Therefore, we obtain $M^{*} \subseteq M_{γ}^{(η)} \subseteq ⋃_{{β \in S : β > α}} M_{β}^{(η)} = M_{α +}^{S} = M^{*},$ i.e., $M^{*} = M_{γ}^{(η)}$ . The remaining proof can follow from the arguments of part (i) by changing the role of $M_{α^{*}}^{(η)}$ as M^*. This completes the proof. ■

Theorem 4.9. Let S be a disjoin union of intervals in (0, 1] with sup S = α^*, and let η be a function from S into (0, 1]. Let $M = {M_{α} : α \in (0, 1]}$ be a family of nonempty subsets of a universal set U such that the following conditions are satisfied:

$M$ is a solid and nested family with respect to (η, S);

$⋂_{n = 1}^{\infty} M_{α_{n}}^{(η)} \subseteq M_{α}^{(η)}$ when α_n ↑ α for α < α^*.

We have the following properties.

Suppose that α^* ∈ S. Then, for any α ∈ S with α < α^*, $M_{α}^{(η)} = M_{α^{*}}^{(η)} ⋃ (⋃_{{β \in S : α^{*} > β \geq α}} D_{β}^{S})$ (18) if and only if $⋂_{n = 1}^{\infty} M_{α_{n}}^{(η)} = M_{α^{*}}^{(η)}$ for α_n ↑ α^* with α_n ∈ S and α_n < α^* for all n.

Suppose that α^* ∉ S. Then, for any α ∈ S, $M_{α}^{(η)} = ⋃_{{β \in S : β \geq α}} D_{β}^{S}$ (19) if and only if $⋂_{n = 1}^{\infty} M_{α_{n}}^{(η)} = \emptyset$ for α_n ↑ α^* with α_n ∈ S for all n.

Suppose that α^* ∉ S, and that the family $M_{S}^{(η)}$ has a focal set M^*. Then, for any α ∈ S, $M_{α}^{(η)} = M^{*} ⋃ (⋃_{{β \in S : β \geq α}} D_{β}^{S})$ (20) if and only if $⋂_{n = 1}^{\infty} M_{α_{n}}^{(η)} = M^{*}$ for α_n ↑ α^* with α_n ∈ S for all n.

Proof. To prove part (i), the sufficiency follows from part (i) in Proposition 2 immediately. To prove the necessity, assume that $⋂_{n = 1}^{\infty} M_{α_{n}}^{(η)} = M_{α^{*}}^{(η)}$ for α_n ↑ α^* with α_n ∈ S and α_n < α^* for all n. From part (i) of Proposition 4, we have the following inclusion $M_{α^{*}}^{(η)} ⋃ (⋃_{{β \in S : α^{*} > β \geq α}} D_{β}^{S}) \subseteq M_{α}^{(η)} .$ To prove another direction of inclusion, for any α ∈ S with α < α^*, according to part (i) of Proposition 4, we define $N_{α} = M_{α}^{(η)} \ M_{α^{*}}^{(η)} \neq \emptyset .$ (21) We want to show $(⋃_{{β \in S : β > α}} M_{β}^{(η)}) ∖ M_{α^{*}}^{(η)} = ⋃_{{β \in S : α^{*} > β > α}} N_{β} .$ (22) If $x \in M_{β}^{(η)}$ for some β ∈ S with β > α and $x \notin M_{α^{*}}^{(η)}$ , then it means that $x \in M_{β}^{(η)}$ for some β ∈ S with α^* > β > α and $x \notin M_{α^{*}}^{(η)}$ ; that is to say, x ∈ N_β for some β ∈ S with α < β < α^*. Therefore, we obtain the inclusion $(⋃_{{β \in S : β > α}} M_{β}^{(η)}) ∖ M_{α^{*}}^{(η)} \subseteq ⋃_{{β \in S : α^{*} > β > α}} N_{β} .$ To prove another direction of inclusion, if y ∈ N_β for some β ∈ S with α^* > β > α, then $y \in M_{β}^{(η)}$ for some β ∈ S with α^* > β > α and $y \notin M_{α^{*}}^{(η)}$ , which also says that $y \in M_{β}^{(η)}$ for some β ∈ S with β > α and $y \notin M_{α^{*}}^{(η)}$ . This shows the equality (22). From (21), it is easy to we see that, for α_n ↑ α^* with α_n ∈ S and α_n < α^*, we have $⋂_{n = 1}^{\infty} M_{α_{n}}^{(η)} = M_{α^{*}}^{(η)} if and only if ⋂_{n = 1}^{\infty} N_{α_{n}} = \emptyset .$ (23)

Now, we are in a position to prove the following inclusion $M_{α}^{(η)} \subseteq M_{α^{*}}^{(η)} ⋃ (⋃_{{β \in S : α^{*} > β \geq α}} D_{β}^{S}) .$ For α ∈ S with α < α^* and $x \in N_{α} = M_{α}^{(η)} \ M_{α^{*}}^{(η)}$ , suppose that $x \notin D_{α}^{S}$ and $x \notin D_{β}^{S}$ for all β ∈ S with α^* > β > α. We shall lead to a contradiction. Since x ∈ N_α and $x \notin D_{α}^{S} = M_{α}^{S} \ M_{α +}^{S}$ by the hypothesis, it follows that $x \in M_{α +}^{S}$ . Using part (i) of Proposition 2, i.e., $M_{α^{*}}^{(η)} \underset{\neq}{\subset} M_{α +}^{S}$ , we also have $x \in M_{α +}^{S} \ M_{α^{*}}^{(η)}$ . Now, from (22), we obtain $x \in M_{α +}^{S} \ M_{α^{*}}^{(η)} = (⋃_{{β \in S : β > α}} M_{β}^{(η)}) ∖ M_{α^{*}}^{(η)} = ⋃_{{β \in S : α^{*} > β > α}} N_{β},$ which says that there exists β₁ ∈ S with α^* > β₁ > α such that x ∈ N_{β
₁}. Since $x \notin D_{β_{1}}^{S}$ by the hypothesis, we can similarly show that there exists β₂ ∈ S with α^* > β₂ > β₁ such that x ∈ N_{β
₂}. Also, since $x \notin D_{β_{2}}^{S}$ by the hypothesis, there exists β₃ ∈ S with α^* > β₃ > β₂ such that x ∈ N_{β
₃}. Continuing this process and argument, we can construct a strictly increasing sequence ${β_{n}}_{n = 1}^{\infty}$ contained in the open interval (α, α^*) such that x ∈ N_{β
_n} for all n, i.e., $⋂_{n = 1}^{\infty} N_{β_{n}} \neq \emptyset$ . We shall claim $α^{*} = sup_{n} β_{n}$ . Suppose that it is not true, i.e., $α < sup_{n} β_{n} = β^{*} < α^{*}$ . Then, we have $⋂_{n = 1}^{\infty} M_{β_{n}}^{(η)} \subseteq M_{β^{*}}^{(η)}$ by the second condition. Therefore, we obtain $x \in ⋂_{n = 1}^{\infty} N_{β_{n}} = ⋂_{n = 1}^{\infty} (M_{β_{n}}^{(η)} \ M_{α^{*}}^{(η)}) = (⋂_{n = 1}^{\infty} M_{β_{n}}^{(η)}) ∖ M_{α^{*}}^{(η)} \subseteq M_{β^{*}}^{(η)} \ M_{α^{*}}^{(η)} = N_{β^{*}},$ which says that x ∈ N_{β
^*} and $x \notin D_{β^{*}}^{S}$ . Using the previous arguments for the procedure of constructing the strictly increasing sequence, we can similarly show that there exists β° ∈ S with α^* > β° > β^* such that x ∈ N_β°. This says that β° must be in the sequence, and that $sup_{n} β_{n} \geq β^{\circ} > β^{*},$ which contradicts $sup_{n} β_{n} = β^{*}$ . Therefore, we must have $α^{*} = sup_{n} β_{n}$ , i.e., β_n ↑ α^* and $⋂_{n = 1}^{\infty} N_{β_{n}} \neq \emptyset$ , which also contradicts the assumption $⋂_{n = 1}^{\infty} N_{β_{n}} = \emptyset$ given in (23). Therefore, we conclude that $x \in D_{α}^{S}$ or $x \in D_{β}^{S}$ for some β ∈ S with α^* > β > α, which proves the equality (18). Part (ii) can be realized from part (i) without considering $M_{α^{*}}^{(η)}$ .

To prove part (iii), the sufficiency follows from part (iii) of Proposition 2 immediately. To prove the necessity, by part (ii) of Proposition 4, we have the following inclusion $⋃_{{β \in S : β \geq α}} D_{β}^{S} \subseteq M_{α}^{(η)} \ M^{*} .$ To prove another direction of inclusion, let $N_{α} = M_{α}^{(η)} \ M^{*}$ for α ∈ S. Then N_α≠ ∅ for α ∈ S by part (ii) of Proposition 4. Now, we have $(⋃_{{β \in S : β > α}} M_{β}^{(η)}) ∖ M^{*} = ⋃_{{β \in S : β > α}} (M_{β}^{(η)} \ M^{*}) = ⋃_{{β \in S : β > α}} N_{β} .$ (24) It is easy to see that, for α_n ↑ α^* with α_n ∈ S for all n, $⋂_{n = 1}^{\infty} M_{α_{n}}^{(η)} = M^{*} if and only if ⋂_{n = 1}^{\infty} N_{α_{n}} = \emptyset .$ (25) Now, we are in a position to prove the following inclusion $M_{α}^{(η)} \subseteq M^{*} ⋃ (⋃_{{β \in S : β \geq α}} D_{β}^{S}) .$ For α ∈ S and $x \in M_{α}^{(η)} \ M^{*}$ , suppose that $x \notin D_{α}^{S}$ and $x \notin D_{β}^{S}$ for all β ∈ S with β > α. We shall lead to a contradiction. Since $x \notin D_{α}^{S}$ by the hypothesis and $M_{α +}^{S} \ M^{*} \neq \emptyset$ by part (ii) of Proposition 2, using (24), we obtain $x \in M_{α +}^{S} \ M^{*} = (⋃_{{β \in S : β > α}} M_{β}^{(η)}) ∖ M^{*} = ⋃_{{β \in S : β > α}} N_{β},$ which says that there exists β₁ ∈ S with β₁ > α such that x ∈ N_{β
₁}. Similarly, we can construct a strictly increasing sequence ${β_{n}}_{n = 1}^{\infty}$ contained in the open interval (α, α^*) such that x ∈ N_{β
_n} for all n. We shall claim $α^{*} = sup_{n} β_{n}$ . Suppose that it is not true, i.e., $α < sup_{n} β_{n} = β^{*} < α^{*}$ . Then, we have $⋂_{n = 1}^{\infty} M_{β_{n}}^{(η)} \subseteq M_{β^{*}}^{(η)}$ by the second condition. Therefore, we obtain $x \in ⋂_{n = 1}^{\infty} N_{β_{n}} = ⋂_{n = 1}^{\infty} (M_{β_{n}}^{(η)} \ M^{*}) = (⋂_{n = 1}^{\infty} M_{β_{n}}^{(η)}) ∖ M^{*} \subseteq M_{β^{*}}^{(η)} \ M^{*} = N_{β^{*}},$ which says that x ∈ N_{β
^*} and $x \notin D_{β^{*}}^{S}$ . Using the previous arguments for the procedure of constructing the strictly increasing sequence, we can similarly show that there exists β° ∈ S with α^* > β° > β^* such that x ∈ N_β°. This says that β° must be in the sequence, and that $sup_{n} β_{n} \geq β^{\circ} > β^{*},$ which contradicts $sup_{n} β_{n} = β^{*}$ . Therefore, we must have $α^{*} = sup_{n} β_{n}$ , i.e., β_n ↑ α^* and $⋂_{n = 1}^{\infty} N_{β_{n}} \neq \emptyset$ , which also contradicts the assumption $⋂_{n = 1}^{\infty} N_{β_{n}} = \emptyset$ given in (25). Therefore, we conclude that $x \in D_{α}^{S}$ or $x \in D_{β}^{S}$ for some β ∈ S with β > α, which proves the equalities (20). This completes the proof. ■

5 Generating fuzzy sets

In the sequel, we are going to construct a fuzzy set $\tilde{A}$ in U satisfying ${\tilde{A}}_{ζ (α)} = M_{α}^{(η)}$ for each α ∈ S, where ζ is an increasing and bijective function on S. In particular, when ζ is taken to be an identity function, we have the equality ${\tilde{A}}_{α} = M_{α}^{(η)}$ .

Theorem 5.1. Let S be a subset of (0, 1] with sup S = α^*, and let η be a function from S into (0, 1]. Let $M = {M_{α} : α \in (0, 1]}$ be a family of nonempty subsets of a universal set U such that $M$ is a solid and nested family with respect to (η, S). Let ζ be an increasing and bijective function on S.

Suppose that α^* ∈ S, and that $⋃_{α \in S} M_{α}^{(η)} = M_{α^{*}}^{(η)} ⋃ (⋃_{{α \in S : α < α^{*}}} D_{α}^{S}) .$ (26) We consider two cases for defining the fuzzy sets.

When $⋃_{α \in S} M_{α}^{(η)} = U$ , we define a fuzzy subset $\tilde{A}$ of U with membership function given by $ξ_{\tilde{A}} (x) = {\begin{matrix} ζ (α), & if x \in D_{α}^{S} for α \in S with α < α^{*} \\ α^{*}, & if x \in M_{α^{*}}^{(η)} . \end{matrix}$

When $⋃_{α \in S} M_{α}^{(η)} \neq U$ , we define a fuzzy subset $\tilde{B}$ of U with membership function given by $ξ_{\tilde{B}} (x) = {\begin{matrix} ζ (α), & if x \in D_{α}^{S} for α \in S with α < α^{*} \\ α^{*}, & if x \in M_{α^{*}}^{(η)} \\ 0, & if x \notin ⋃_{α \in S} M_{α}^{(η)} . \end{matrix}$ (27)

Then, we have the following properties:

$R (ξ_{\tilde{A}}) = S$ and $R (ξ_{\tilde{B}}) = S \cup {0}$ ;

${\tilde{A}}_{ζ (α)} = {\tilde{B}}_{ζ (α)} = M_{α}^{(η)} = M_{η (α)}$ for each α ∈ S; if η is injective, then ${\tilde{A}}_{ζ (η^{- 1} (α))} = {\tilde{B}}_{ζ (η^{- 1} (α))} = M_{α}$ for each $α \in R (η)$ .

Suppose that α^* ∉ S, and that $⋃_{α \in S} M_{α}^{(η)} = ⋃_{α \in S} D_{α}^{S} .$ We consider two cases for defining the fuzzy sets.

When $⋃_{α \in S} M_{α}^{(η)} = U$ , we define a fuzzy subset $\tilde{A}$ of U with membership function given by $ξ_{\tilde{A}} (x) = ζ (α)$ for $x \in D_{α}^{S}$ and α ∈ S.

When $⋃_{α \in S} M_{α}^{(η)} \neq U$ , we define a fuzzy subset $\tilde{B}$ of U with membership function given by $ξ_{\tilde{B}} (x) = {\begin{matrix} ζ (α), & if x \in D_{α}^{S} for α \in S \\ 0, & if x \notin ⋃_{α \in S} M_{α}^{(η)} . \end{matrix}$ (28)

Then, we have the following properties:

$R (ξ_{\tilde{A}}) = S$ and $R (ξ_{\tilde{B}}) = S \cup {0}$ ;

Suppose that α^* ∉ S, that the family $M_{S}^{(η)} = {M_{α}^{(η)} : α \in S}$ has a focal set M^*, and that $⋃_{α \in S} M_{α}^{(η)} = M^{*} ⋃ (⋃_{α \in S} D_{α}^{S}) .$ We consider two cases for defining the fuzzy sets.

When $⋃_{α \in S} M_{α}^{(η)} \neq U$ , we define a fuzzy subset $\tilde{B}$ of U with membership function given by $ξ_{\tilde{B}} (x) = {\begin{matrix} ζ (α), & if x \in D_{α}^{S} for α \in S \\ α^{*}, & if x \in M^{*} \\ 0, & if x \notin ⋃_{α \in S} M_{α}^{(η)} . \end{matrix}$ (29)

Then, we have the following properties:

$R (ξ_{\tilde{A}}) = S \cup {α^{*}}$ and $R (ξ_{\tilde{B}}) = S \cup {0, α^{*}}$ ;

${\tilde{A}}_{α^{*}} = {\tilde{B}}_{α^{*}} = M^{*}$ and ${\tilde{A}}_{ζ (α)} = {\tilde{B}}_{ζ (α)} = M_{α}^{(η)} = M_{η (α)}$ for each α ∈ S; if η is injective, then ${\tilde{A}}_{ζ (η^{- 1} (α))} = {\tilde{B}}_{ζ (η^{- 1} (α))} = M_{α}$ for each $α \in R (η)$ .

Moreover, for parts (i)-(iii), we have

ξ_{\tilde{A}} (x) > 0

for x ∈ U and

ξ_{\tilde{B}} (x) > 0

for

x \in ⋃_{α \in S} M_{α}^{(η)}

with

{\tilde{A}}_{0 +} = {\tilde{B}}_{0 +} = ⋃_{α \in S} M_{α}^{(η)} .

(30) If U is endowed with a topology, then

{\tilde{A}}_{0} = cl ({\tilde{A}}_{0 +}) = U = ⋃_{α \in S} M_{α}^{(η)} and {\tilde{B}}_{0} = cl ({\tilde{B}}_{0 +}) = cl (⋃_{α \in S} M_{α}^{(η)}) .

Proof. From Proposition 3, we see that the membership functions $ξ_{\tilde{A}}$ and $ξ_{\tilde{B}}$ defined in parts (i)-(iii) are well-defined. Next, we want to show ${\tilde{A}}_{ζ (α)} = {\tilde{B}}_{ζ (α)} = M_{α}^{(η)}$ for each α ∈ S. It suffices to show ${\tilde{B}}_{ζ (α)} = M_{α}^{(η)}$ for each α ∈ S. We first claim that $ξ_{\tilde{B}}^{- 1} ([ζ (α), 1]) = ⋃_{{γ \in R (ξ_{\tilde{B}}) : γ \geq ζ (α)}} ξ_{\tilde{B}}^{- 1} (γ) for each α \in S,$ (31) where $ξ_{\tilde{B}}^{- 1} ([ζ (α), 1])$ denotes the inverse image of the closed interval [ζ (α) , 1] and $ξ_{\tilde{B}}^{- 1} (γ)$ denotes the inverse image of singleton set {γ} for function $ξ_{\tilde{B}}$ . Given $x \in ξ_{\tilde{B}}^{- 1} ([ζ (α), 1])$ , we have $ξ_{\tilde{B}} (x) \geq ζ (α)$ . Let $γ = ξ_{\tilde{B}} (x) \in R (ξ_{\tilde{B}})$ . Then, we have γ ≥ ζ (α) and $x \in ξ_{\tilde{B}}^{- 1} (γ)$ , which says that $x \in ⋃_{{γ \in R (ξ_{\tilde{B}}) : γ \geq ζ (α)}} ξ_{\tilde{B}}^{- 1} (γ)$ . On the other hand, given $x \in ⋃_{{γ \in R (ξ_{\tilde{B}}) : γ \geq ζ (α)}} ξ_{\tilde{B}}^{- 1} (γ)$ , there exists $γ \in R (ξ_{\tilde{B}})$ with γ ≥ ζ (α) satisfying $x \in ξ_{\tilde{B}}^{- 1} (γ)$ , which also says that $ξ_{\tilde{B}} (x) = γ \geq ζ (α)$ . Therefore, it follows that $x \in ξ_{\tilde{B}}^{- 1} ([ζ (α), 1])$ , which proves the equality (18).

To prove part (i), since ζ is an increasing and bijective function on S, it follows that $R (ξ_{\tilde{B}}) = S \cup {0}$ (32) and ζ (α) >0 for α ∈ S, the equality (18) can be rewritten as $ξ_{\tilde{B}}^{- 1} ([ζ (α), 1]) = ⋃_{{γ \in S : γ \geq ζ (α)}} ξ_{\tilde{B}}^{- 1} (γ)$ (33) We also see that if α^* ∈ S then ζ (α^*) = α^*. Therefore, for α ∈ S with α < α^*, we have

$\begin{matrix} ⋃_{{γ \in S : α^{*} > γ \geq ζ (α)}} ξ_{\tilde{B}}^{- 1} (γ) = ⋃_{{β \in S : α^{*} > ζ (β) \geq ζ (α)}} ξ_{\tilde{B}}^{- 1} (ζ (β)) (since ζ is bijective on S) \\ = ⋃_{{β \in S : α^{*} > β \geq α}} ξ_{\tilde{B}}^{- 1} (ζ (β)) (since ζ is increasing and bijective on S) . \end{matrix}$ (34) Now, for each α ∈ S with α < α^*, we obtain

$\begin{matrix} {\tilde{B}}_{ζ (α)} & = ξ_{\tilde{B}}^{- 1} ([ζ (α), 1]) = ⋃_{{γ \in S : γ \geq ζ (α)}} ξ_{\tilde{B}}^{- 1} (γ) (by (33)) \\ = ξ_{\tilde{B}}^{- 1} (α^{*}) ⋃ (⋃_{{γ \in S : α^{*} > γ \geq ζ (α)}} ξ_{\tilde{B}}^{- 1} (γ)) \\ = ξ_{\tilde{B}}^{- 1} (α^{*}) ⋃ (⋃_{{β \in S : α^{*} > β \geq α}} ξ_{\tilde{B}}^{- 1} (ζ (β))) (by (34) \\ = M_{α^{*}}^{(η)} ⋃ (⋃_{{β \in S : α^{*} > β \geq α}} D_{β}^{S}) (by definition of ξ_{\tilde{B}} in (27)) \\ = M_{α}^{(η)} (by part (i) of Proposition 4 using assumption (26)) . \end{matrix}$ (35) We also have $\begin{matrix} {\tilde{B}}_{ζ (α^{*})} & = {\tilde{B}}_{α^{*}} = ξ_{\tilde{B}}^{- 1} ([α^{*}, 1]) = {x \in U : ξ_{\tilde{B}} (x) \geq α^{*}} \\ = {x \in U : ξ_{\tilde{B}} (x) = α^{*}} = ξ_{\tilde{B}}^{- 1} (α^{*}) = M_{α^{*}}^{(η)} . \end{matrix}$ and $\begin{matrix} {\tilde{B}}_{0 +} & = ⋃_{s \in (0, 1]} {\tilde{B}}_{s} = ⋃_{s \in R (ξ_{\tilde{B}}) \ {0}} {\tilde{B}}_{s} (by (1)) \\ = ⋃_{s \in S} {\tilde{B}}_{s} (by (32)) \\ = ⋃_{α \in S} {\tilde{B}}_{ζ (α)} (since ζ is bijective on S) \\ = ⋃_{α \in S} M_{α}^{(η)} (by (35)) . \end{matrix}$ We can similarly obtain ${\tilde{A}}_{0 +} = ⋃_{α \in S} M_{α}^{(η)},$ which proves (30). If η is injective, then $M_{α} = M_{η^{- 1} (α)}^{(η)}$ for $α \in R (η)$ by (6). Therefore, we can obtain ${\tilde{B}}_{ζ (η^{- 1} (α))} = M_{α}$ . Part (ii) can be similarly obtained from part (i) without considering $M_{α^{*}}^{(η)}$ .

To prove part (iii), since ζ (α) >0 and $R (ξ_{\tilde{B}}) = S \cup {0, α^{*}}$ , the equality (18) can be rewritten as $ξ_{\tilde{B}}^{- 1} ([ζ (α), 1]) = ⋃_{{γ \in S \cup {α^{*}} : γ \geq ζ (α)}} ξ_{\tilde{B}}^{- 1} (γ) = ξ_{\tilde{B}}^{- 1} (α^{*}) ⋃ (⋃_{{γ \in S : γ \geq ζ (α)}} ξ_{\tilde{B}}^{- 1} (γ)) .$ (36) We also have

$\begin{matrix} ⋃_{{γ \in S : γ \geq ζ (α)}} ξ_{\tilde{B}}^{- 1} (γ) = ⋃_{{β \in S : ζ (β) \geq ζ (α)}} ξ_{\tilde{B}}^{- 1} (ζ (β)) (since ζ is bijective on S) \\ = ⋃_{{β \in S : β \geq α}} ξ_{\tilde{B}}^{- 1} (ζ (β)) (since ζ is increasing and bijective on S) . \end{matrix}$ (37) Now, for each α ∈ S, we have

$\begin{matrix} {\tilde{B}}_{ζ (α)} & = ξ_{\tilde{B}}^{- 1} ([ζ (α), 1]) = ξ_{\tilde{B}}^{- 1} (α^{*}) ⋃ (⋃_{{γ \in S : γ \geq ζ (α)}} ξ_{\tilde{B}}^{- 1} (γ)) (by (36)) \\ = ξ_{\tilde{B}}^{- 1} (α^{*}) ⋃ (⋃_{{β \in S : β \geq α}} ξ_{\tilde{B}}^{- 1} (ζ (β))) (by (37)) \\ = M^{*} ⋃ (⋃_{{β \in S : β \geq α}} D_{β}^{S}) (by the definition of ξ_{\tilde{B}} in (29)) \\ = M_{α}^{(η)} (by part (iii) of Proposition 4) . \end{matrix}$ (38) We also have ${\tilde{B}}_{α^{*}} = ξ_{\tilde{B}}^{- 1} ([α^{*}, 1]) = ξ_{\tilde{B}}^{- 1} (α^{*}) = M^{*} (by definition of ξ_{\tilde{B}} in (29)) .$ (39) Finally, we can obtain $\begin{matrix} {\tilde{B}}_{0 +} & = ⋃_{s \in (0, 1]} {\tilde{B}}_{s} = ⋃_{s \in R (ξ_{\tilde{B}}) \ {0}} {\tilde{B}}_{s} (by (1)) \\ = ⋃_{s \in S \cup {α^{*}}} {\tilde{B}}_{s} = {\tilde{B}}_{α^{*}} ⋃ (⋃_{s \in S} {\tilde{B}}_{s}) = {\tilde{B}}_{α^{*}} ⋃ (⋃_{α \in S} {\tilde{B}}_{ζ (α)}) (since ζ is bijective on S) \\ = M^{*} ⋃ (⋃_{α \in S} M_{α}^{(η)}) (by (38) and (39)) \\ = ⋃_{α \in S} M_{α}^{(η)} (since M^{*} \subseteq M_{α}^{(η)} for all α \in S) . \end{matrix}$ We can similarly obtain ${\tilde{A}}_{0 +} = ⋃_{α \in S} M_{α}^{(η)},$ which proves (30). This completes the proof. ■

When S is taken to be a disjoin union of intervals in (0, 1], we can have the different sufficient conditions to guarantee the existence of fuzzy sets in U.

Theorem 5.2. Let S be a disjoin union of intervals in (0, 1] with sup S = α^*, and let η be a function from S into (0, 1] with sup S = α^*. Let $M = {M_{α} : α \in (0, 1]}$ be a family of nonempty subsets of a universal set U such that $M$ is a solid and nested family with respect to (η, S). Let ζ be an increasing and bijective function on S.

Suppose that α^* ∈ S, and that $⋂_{n = 1}^{\infty} M_{α_{n}}^{(η)} \subseteq M_{α^{*}}^{(η)} when α_{n} ↑ α^{*} with α_{n} \in S and α_{n} < α^{*} for all n .$ We consider two cases for defining the fuzzy sets.

Then, we have the following properties:

$R (ξ_{\tilde{A}}) = S$ and $R (ξ_{\tilde{B}}) = S \cup {0}$ ;

${\tilde{A}}_{ζ (α)} = {\tilde{B}}_{ζ (α)} = M_{α}^{(η)} = M_{η (α)}$ for each α ∈ S with α < α^*; if η is injective, then ${\tilde{A}}_{ζ (η^{- 1} (α))} = {\tilde{B}}_{ζ (η^{- 1} (α))} = M_{α}$ for each $α \in R (η)$ .

The interval ranges

I_{\tilde{A}}

and

I_{\tilde{B}}

\tilde{A}

and

\tilde{B}

, respectively, are given by

I_{\tilde{A}} = I_{\tilde{B}} = [0, α^{*}] .

Suppose that α^* ∉ S, and that $⋂_{n = 1}^{\infty} M_{α_{n}}^{(η)} = \emptyset when α_{n} ↑ α^{*} and α_{n} \in S for all n .$ We consider two cases for defining the fuzzy sets.

When $⋃_{α \in S} M_{α}^{(η)} = U$ , we define a fuzzy subset $\tilde{A}$ of U with membership function given by $ξ_{\tilde{A}} (x) = ζ (α)$ for $x \in D_{α}^{S}$ and α ∈ S.

When $⋃_{α \in S} M_{α}^{(η)} \neq U$ , we define a fuzzy subset $\tilde{B}$ of U with membership function given by $ξ_{\tilde{B}} (x) = {\begin{matrix} ζ (α), & if x \in D_{α}^{S} for α \in S \\ 0, & if x \notin ⋃_{α \in S} M_{α}^{(η)} . \end{matrix}$

Then, we have the following properties:

$R (ξ_{\tilde{A}}) = S$ and $R (ξ_{\tilde{B}}) = S \cup {0}$ ;

The interval ranges

I_{\tilde{A}}

and

I_{\tilde{B}}

\tilde{A}

and

\tilde{B}

are given by

I_{\tilde{A}} = I_{\tilde{B}} = [0, α^{*}) .

Suppose that α^* ∉ S, that the family $M_{S}^{(η)} = {M_{α}^{(η)} : α \in S}$ has a focal set M^*, and that $⋂_{n = 1}^{\infty} M_{α_{n}}^{(η)} \subseteq M^{*} when α_{n} ↑ α^{*} and α_{n} \in S for all n .$ We consider two cases for defining the fuzzy sets.

When $⋃_{α \in S} M_{α}^{(η)} \neq U$ , we define a fuzzy subset $\tilde{B}$ of U with membership function given by $ξ_{\tilde{B}} (x) = {\begin{matrix} ζ (α), & if x \in D_{α}^{S} for α \in S \\ α^{*}, & if x \in M^{*} \\ 0, & if x \notin ⋃_{α \in S} M_{α}^{(η)} . \end{matrix}$

Then, we have the following properties:

$R (ξ_{\tilde{A}}) = S \cup {α^{*}}$ and $R (ξ_{\tilde{B}}) = S \cup {0, α^{*}}$ ;

Moreover, for parts (i)-(iii), we have

ξ_{\tilde{A}} (x) > 0

for x ∈ U and

ξ_{\tilde{B}} (x) > 0

for

x \in ⋃_{α \in S} M_{α}^{(η)}

with

{\tilde{A}}_{0 +} = {\tilde{B}}_{0 +} = ⋃_{α \in S} M_{α}^{(η)} .

If U is endowed with a topology, then

{\tilde{A}}_{0} = cl ({\tilde{A}}_{0 +}) = U = ⋃_{α \in S} M_{α}^{(η)} and {\tilde{B}}_{0} = cl ({\tilde{B}}_{0 +}) = cl (⋃_{α \in S} M_{α}^{(η)}) .

The interval ranges

I_{\tilde{A}}

and

I_{\tilde{B}}

\tilde{A}

and

\tilde{B}

are given by

I_{\tilde{A}} = I_{\tilde{B}} = [0, α^{*}] .

Proof. To prove part (i), using the nestedness, we see that $M_{α^{*}}^{(η)} \subseteq ⋂_{n = 1}^{\infty} M_{α_{n}}^{(η)} when α_{n} ↑ α^{*} with α_{n} \in S and α_{n} < α^{*} for all n .$ Therefore, combining it with the assumption, we obtain $M_{α^{*}}^{(η)} = ⋂_{n = 1}^{\infty} M_{α_{n}}^{(η)} when α_{n} ↑ α^{*} with α_{n} \in S and α_{n} < α^{*} for all n .$ By Theorem 2* and Proposition 4, we see that the assumption of part (i) of Theorem 5 is satisfied. Therefore, the results follow from part (i) of Theorem 5 immediately. Part (ii) can be similarly obtained.

To prove part (iii), since M^* is a focal set, it follows that $M^{*} \subseteq M_{α}^{(η)}$ for α ∈ S. This also says that $M^{*} \subseteq ⋂_{n = 1}^{\infty} M_{α_{n}}^{(η)} when α_{n} ↑ α^{*} with α_{n} \in S and α_{n} < α^{*} for all n .$ Therefore, combining it with the assumption, we obtain $M^{*} = ⋂_{n = 1}^{\infty} M_{α_{n}}^{(η)} when α_{n} ↑ α^{*} with α_{n} \in S and α_{n} < α^{*} for all n .$ The remaining proof follows from the above arguments. Finally, the interval ranges $I_{\tilde{A}}$ and $I_{\tilde{B}}$ can be realized immediately from (2). This completes the proof. ■

Example 5.3. Continued from Example 3, we can obtain $\begin{matrix} M_{α^{*}} ⋃ (⋃_{{α \in S : α < α^{*}}} D_{α}^{S}) = M_{0.9} ⋃ (⋃_{0.2 \leq α < 0.9} {{\bar{m}}_{α}^{L}, {\bar{m}}_{α}^{U}}) \\ = [{\bar{m}}_{0.2}^{L}, {\bar{m}}_{0.2}^{U}] = [0.8 b_{1} + 0.2 a_{1}, 0.8 b_{2} + 0.2 a_{2}] = ⋃_{α \in S} M_{α}^{(η)} . \end{matrix}$ Therefore, according to part (iii) of Theorem 5, we can generate a fuzzy subset $\tilde{B}$ of $ℝ$ with membership function given by $\begin{matrix} ξ_{\tilde{B}} (x) & = {\begin{matrix} ζ (α), & if x \in D_{α}^{S} = {{\bar{m}}_{α}^{L}, {\bar{m}}_{α}^{U}} for 0.2 \leq α < 0.9 \\ 0.9, & if x \in M_{0.9} \\ 0, & if x \notin ⋃_{α \in S} M_{α}^{(η)} \end{matrix} \\ = {\begin{matrix} ζ (α), & if x = (1 - α) b_{1} + α a_{1} or (1 - α) b_{2} + α a_{2} for 0.2 \leq α < 0.9 \\ 0.9, & if x \in [0.1 b_{1} + 0.9 a_{1}, 0.1 b_{2} + 0.9 a_{2}] \\ 0, & if x \notin [0.8 b_{1} + 0.2 a_{1}, 0.8 b_{2} + 0.2 a_{2}] \end{matrix} \end{matrix}$ such that the following properties are satisfied.

The range of $\tilde{B}$ is given by $R (ξ_{\tilde{B}}) = [0.2, 0.9] \cup {0}$ .

For each α ∈ [0.2, 0.9), we have ${\tilde{B}}_{ζ (α)} = M_{α}^{(η)} = M_{η (α)} = M_{1 - α} = [{\bar{m}}_{α}^{L}, {\bar{m}}_{α}^{U}] = [(1 - α) b_{1} + α a_{1}, (1 - α) b_{2} + α a_{2}] .$ We also have ${\tilde{B}}_{0.9} = M_{0.9}$ . For α ∈ [0.2, 0.9), the α-level set of $\tilde{B}$ is given by $\begin{matrix} {\tilde{B}}_{α} & = M_{ζ^{- 1} (α)}^{(η)} = M_{η (ζ^{- 1} (α))} = M_{1 - ζ^{- 1} (α)} = [{\bar{m}}_{ζ^{- 1} (α)}^{L}, {\bar{m}}_{ζ^{- 1} (α)}^{U}] \\ = [(1 - ζ^{- 1} (α)) b_{1} + ζ^{- 1} (α) a_{1}, (1 - ζ^{- 1} (α)) b_{2} + ζ^{- 1} (α) a_{2}] . \end{matrix}$

The interval range

I_{\tilde{B}}

\tilde{B}

I_{\tilde{B}} = [0, 0.9]

6 Conclusions

A mechanical procedure has been proposed in this paper to generate a fuzzy set from an arbitrary family of subsets of a universal set U. This kind of generation is mainly based on the concepts of solid and nested family. The basic idea of solid family considers the sets in the family to be continuously varied. In other words, the cardinalities of the sets will be changed smoothly (continuously) rather than abruptly. Also, the idea of nested family means that the sets in the family will be shrunk gradually from the largest set. In general, an arbitrary family $M = {M_{α} : α \in (0, 1]}$ of subsets of U is not necessarily nested in the sense of M_α ⊆ M_β for β < α. However, we can always rearrange the original family $M$ to be a nested family $M^{(η)} = {M_{α}^{(η)} : M_{α}^{(η)} = M_{η (α)} for α \in (0, 1]}$ by designing a suitable function η : (0, 1] → (0, 1] such that $M_{α}^{(η)} \subseteq M_{β}^{(η)}$ for β < α.

The main purpose of this paper is to generate a fuzzy set $\tilde{A}$ in U such that the α-level set ${\tilde{A}}_{α}$ is identical to $M_{α}^{(η)}$ for each α ∈ S ⊂ [0, 1]. In this case, when η is an one-to-one function on S, we see the following families ${{\tilde{A}}_{α} : α \in S} = {M_{α} : α \in S} = {M_{α}^{(η)} : α \in S}$ are identical. In other words, we are going to generate a fuzzy set $\tilde{A}$ such that the original family $M$ is equal to the family of α-level sets of $\tilde{A}$ .

In real applications, suppose that the economic and engineering problems will be considered under fuzzy environment. Then, we shall consider the fuzzy data instead of real number data. However, we can just collect the real number data from the real world. In order to model the fuzzy data based on the real number data, this paper proposes a mechanical procedure to determine the fuzzy data by using the observed real number data to formulate the original family $M$ that can be used to generate a reasonable fuzzy data. This kind of mechanical procedure can avoid the bias of directly determining the fuzzy data raised by the subjectivity of decision makers.

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