Limit theory of isv -functions and its application for rough sets

Abstract

In this paper, limit theory of set-valued functions defined on an interval (for short, isv-functions) is preliminarily established. Firstly, the concept of isv-functions is introduced. Secondly, limits of isv-functions are proposed and their properties are obtained. Thirdly, point-wise continuity of isv-functions and continuous isv-functions are discussed. Finally, an application of this theory for rough sets is given.

Keywords

limit continuity rough set

1 Introduction

It is well-known that calculus theory is the foundation of modern science. Limits of functions are its basic concepts, which play an important role in the process of development [9].

A set-valued function defined on an interval (for short, A isv-function) is the function that the input is a number and the output is a set. isv-functions are widespread. For example, considering a fuzzy set A on the universe U, the θ-level set (or θ-cut set) A_θ of A is an isv-function on [0, 1], the θ-strong level set (or θ-strong cut set) $A_{\dot{θ}}$ of A is an isv-function on [0, 1), where A_θ = {u ∈ U : A (u) ≥ θ} and $A_{\dot{θ}} = {u \in U : A (u) > θ}$ .

Rough set theory, proposed by Pawlak [12], is a mathematical tool for the study of intelligent systems characterized by insufficient and incomplete information. After thirty years development, this theory has been successfully applied to machine learning, intelligent systems, inductive reasoning, pattern recognition, mereology, image processing, signal analysis, knowledge discovery, decision analysis, expert systems and many other fields [12 –15]. The basic structure of rough set theory is an approximation space. Based on it, lower and upper approximations can be induced. Through these rough approximations, knowledge hidden in information systems may be revealed and expressed in the form of decision rules [13 –15].

Pawlak’s rough set model is based on the completeness of available information, and ignores the incompleteness of available information and the possible existence of statistical information. This model for extracting rules in uncoordinate decision information systems often seems incapable. These have motivated many researchers to investigate probabilistic generalization of rough set theory and provide new rough set models for the study of uncertain information system.

Probabilistic rough set model is probabilistic generalization of rough set theory. In probabilistic rough set model, probabilistic rough approximations are dependent on parameters. Probabilistic rough approximations are also isv-functions. Researching the infinite change trend or the limit state of these isv-functions accordance with parameters is helpful for the study of probabilistic rough sets.

Since cut sets of a fuzzy set and probabilistic rough approximations are both isv-functions, we may attempt to study the infinite change trend or the limit state of isv-functions. It is worth mentioning that there is no systematic research and summary for limits of isv-functions although the limit though of isv-functions has formed in [16, 17].

In general, most of uncertain mathematical theories can only deal with uncertainty problems of discreteness. If limit theory of isv-functions is established, then these theories may be used to solve uncertainty problems of continuity. The purpose of this paper is to establish preliminarily limit theory of isv-functions so that some uncertain mathematical theories may be used to solve uncertainty problems of continuity.

The remaining part of this paper is organized as follows. In Section 2, we recall some basic concepts about limits of s-sequences and rough sets. In Section 3, we introduce isv-functions and related notions. In Section 4, we propose the concept of limits of isv-functions and obtain their properties. In Section 5, we discuss the continuity of isv-functions including point-wise continuity of isv-functions and continuous isv-functions. In Section 6, we give an application of the proposed limit theory for rough sets. Section 7 summarizes this paper.

2 Preliminaries

In this section, we recall some basic concepts about limits of s-sequences, rough sets and isv-functions.

Throughout this paper, U denotes the universe which may be an infinite set, 2^U denotes the family of all subsets of U, R denotes the set of all real numbers, N denotes the set of all natural numbers, I denotes the interval in R, and Y^X denotes the family of all functions from X to Y.

2.1 Limits of s-sequences

Definition 2.1. ([2, 8]). Let U be the universe. If for each n ∈ N, E_n ∈ 2^U, then {E_n} is called a s-sequence in U. Define $\underset{n \to \infty}{lim^{¯}} E_{n} = {u \in U : {n \in N : u \in E_{n}} is infinite},$ $\underset{n \to \infty}{lim_{_}} E_{n} = {u \in U : {n \in N : u \notin E_{n}} is finite} .$ If $\underset{n \to \infty}{lim_{_}} E_{n} = \underset{n \to \infty}{lim^{¯}} E_{n} = E$ , then {E_n : n ∈ N} is called to has the limit E, which is denoted by $lim_{n \to \infty} E_{n}$ , i.e., $lim_{n \to \infty} E_{n} = E$ ; If $\underset{n \to \infty}{lim_{_}} E_{n} \neq \underset{n \to \infty}{lim^{¯}} E_{n}$ , then {E_n : n ∈ N} is said to have no limit.

Obviously, $\underset{n \to \infty}{lim_{_}} E_{n} \subseteq \underset{n \to \infty}{lim^{¯}} E_{n}$ .

Example 2.2. Suppose $E_{n} = [- 1, 1 - \frac{(- 1)^{n}}{n}] (n \in N)$ . Then $\underset{n \to \infty}{lim^{¯}} E_{n} = [0, 2], \underset{n \to \infty}{lim_{_}} E_{n} = [0, 1] .$

Thus {E_n : n ∈ N} has no limit.

Proposition 2.3. ([2, 8]). Let {E_n : n ∈ N} be a s-sequence in U.

(1) $\underset{n \to \infty}{lim^{¯}} E_{n} = ⋂_{n = 1}^{\infty} ⋃_{k = n}^{\infty} E_{k}$ .

(2) $\underset{n \to \infty}{lim_{_}} E_{n} = ⋃_{n = 1}^{\infty} ⋂_{k = n}^{\infty} E_{k}$ .

Proposition 2.4. ([2, 8]). Let {E_n : n ∈ N} be a s-sequence in U.

(1) If {E_n}↑, then $lim_{n \to \infty} E_{n} = ⋃_{n = 1}^{\infty} E_{n}$ .

(2) If {E_n}↓, then $lim_{n \to \infty} E_{n} = ⋂_{n = 1}^{\infty} E_{n}$ .

2.2 Rough sets

Let R be an equivalence relation on the universe U. Then the pair (U, R) is called a Pawlak approximation space. Based on (U, R), one can define the following two rough approximations: $\underline{R} (X) = {u \in U : [u]_{R} \subseteq X},$

$\bar{R} (X) = {u \in U : [u]_{R} \cap X \neq \emptyset} .$

Then $\underline{R} (X)$ and $\bar{R} (X)$ are called the Pawlak lower approximation and the Pawlak upper approximation of X, respectively.

The boundary region of X, defined by the difference between these rough approximations, that is ${Bnd}_{R} (X) = \bar{R} (X) - \underline{R} (X)$ .

A set is rough if its boundary region is not empty; Otherwise, it is crisp. Thus, X is rough if $\underline{R} (X) \neq \bar{R} (X)$ .

Definition 2.5. ([16, 17]). Let U be a finite universe. Then a function P : 2^U → [0, 1] is called a probability measure on U, if P (U) =1 and P (A ∪ B) = P (A) + P (B) whenever A∩ B = ∅.

If P is a probability measure on U, A, B ∈ 2^U and P (B) >0, then $P (A | B) = \frac{P (A \cap B)}{P (B)}$ is called the conditional probability of the event A when the event B occurs.

Definition 2.6. ([16, 17]). Let U be a finite universe, R an equivalence relation on U and P a probability measure on U. Then the pair (U, R, P) is called a probabilistic approximate space. Based on (U, R, P), the lower and upper approximation of X, are respectively denoted by ${\underline{PI}}_{θ} (X)$ and ${\bar{PI}}_{α} (X)$ , are defined as follows: ${\underline{PI}}_{θ} (X) = {u \in U : P (X | [u]) \geq θ},$ ${\bar{PI}}_{α} (X) = {u \in U : P (X | [u]) > α),$ where 0 ≤ α < θ ≤ 1.

Theorem 2.7. ([16, 17]). Let (U, R, P) be a probabilistic approximate space. Then the following properties hold.

(1) ${\underline{PI}}_{θ} (\emptyset) = {\bar{PI}}_{θ} (\emptyset) = \emptyset$ , ${\underline{PI}}_{θ} (U) = {\bar{PI}}_{θ} (U) = U$ ;

(2) ${\underline{PI}}_{θ} (X) \subseteq {\bar{PI}}_{θ} (X)$ ;

(3) ${\underline{PI}}_{θ} (U - X) = U - {\bar{PI}}_{1 - θ} (X)$ , ${\bar{PI}}_{θ} (U - X) = U - {\underline{PI}}_{1 - θ} (X)$ ;

(4) If X ⊆ Y, then

${\underline{PI}}_{θ} (X) \subseteq {\underline{PI}}_{θ} (Y)$ , ${\bar{PI}}_{θ} (X) \subseteq {\bar{PI}}_{θ} (Y)$ ;

(5) If 0 < θ₁ < θ₂ ≤ 1, 0 ≤ α₁ < α₂ < 1, then ${\underline{PI}}_{θ_{2}} (X) \subseteq {\underline{PI}}_{θ_{1}} (X)$ , ${\bar{PI}}_{α_{2}} (X) \subseteq {\bar{PI}}_{α_{2}} (X)$ .

Theorem 2.8. ([16, 17]). Let (U, R, P) be a probabilistic approximate space. Then for 0 < γ < 1, X ∈ 2^U,

(1) $lim_{θ ↑ γ} {\underline{PI}}_{θ} (X) = ⋂_{θ \in (0, γ)} {\underline{PI}}_{θ} (X) = {\underline{PI}}_{γ} (X)$ ,

$lim_{θ ↓ γ} {\underline{PI}}_{θ} (X) = ⋃_{θ \in (γ, 1]} {\underline{PI}}_{θ} (X) = {\bar{PI}}_{γ} (X)$ ;

(2) $lim_{θ ↑ γ} {\bar{PI}}_{θ} (X) = ⋂_{θ \in [0, γ)} {\bar{PI}}_{θ} (X) = {\underline{PI}}_{γ} (X)$ ,

$lim_{θ ↓ γ} {\bar{PI}}_{θ} (X) = ⋃_{θ \in (γ, 1)} {\bar{PI}}_{θ} (X) = {\bar{PI}}_{γ} (X)$ .

Although the limit though of isv-functions has formed in Theorem 2.6, there is no systematic research and summary for limits of isv-functions. Thus, limit theory of isv-functions deserves deeply study so that rough set theory can be used to solve uncertainty problems of continuity.

3 isv-functions

Definition 3.1. ([1]). Let X and Y be two sets. If for each u ∈ X there exists a corresponding set f (u) ∈2^Y, then f (u) or f is called a set-valued function from X to Y. We denote this by $f : X ⇉ Y .$

Definition 3.2. Let θ be a real variable and U a set. If f : I ⇉ U, then f (θ) or f is called a set-valued function defined on I (for short, a isv-function on I).

In this case, f (θ₀) is called the function value of θ₀; θ and f are called independent variable and the corresponding rule, respectively; I is called the domain of f, f (I) = {f (θ) : θ ∈ I} is called the range of f.

In this paper, (2^U) ^I denotes the family of all isv-functions on I.

The research in this paper is different from general set-valued function’s research. Set-valued function’s research considers that I is necessarily not the interval in R, and I and U are the same kinds of things which often give topologies or distances (see [1]). But this paper’s research is similar to the research of limit theory of functions, we look that I and U are different kinds of things. That is say, I is considered the domain which provides the range of the independent variable θ, U may be a arbitrary set which provides the corresponding subset when the independent variable θ picks a value and is not endowed the topology.

A soft set is a very broad concept. In fact, a set-valued function can be seen as a soft set. But they deal with problems from different angles. The former considers problems from the angle that the function value of a set-valued function is a set. The latter considers problems from the angle of parameterization.

In this section, we give some operations and types of isv-functions by simulating general set-valued functions and Li’s paper (see [11]).

Definition 3.3. Let f (θ) and g (θ) be two isv-functions on I.

(1) f (θ) and g (θ) are called equal, if f (θ) = g (θ) for each θ ∈ I. We write f (θ) = g (θ) or f = g.

(2) f (θ) is called a sub-function of g (θ), if f (θ) ⊆ g (θ) for each θ ∈ I. We write f (θ) ⪯ g (θ) or f ⪯ g.

(3) f (θ) is called a super-function of g (θ), if f (θ) ⊇ g (θ) for each θ ∈ I. We write f (θ) ≼ g (θ) or f ≼ g.

Obviously, $f = g \Leftrightarrow f ⪯ g, f ≼ g .$

Definition 3.4. Let f (θ) be an isv-function on I. f (θ) is called constant, if ∃ X ∈ 2^U, f (θ) = X for each θ ∈ I. We denote f (θ) by $\bar{X} (θ)$ or f by $\bar{X}$ .

Example 3.5. $\bar{\emptyset}$ and $\bar{U}$ are two constant isv-functions.

Definition 3.6. Let f (θ) and g (θ) be two isv-functions on I.

(1) (f ∩ g) (θ)=f (θ) ∩ g (θ) for each θ ∈ I.

(2) (f ∪ g) (θ)=f (θ) ∪ g (θ) for each θ ∈ I.

(3) (f - g) (θ)=f (θ) - g (θ) for each θ ∈ I.

(4) f^c (θ) = U - f (θ) for each θ ∈ I.

(5) (f × g) (θ)=f (θ) × g (θ) for each θ ∈ I.

Obviously, $f^{c} = \bar{U} - f .$

Definition 3.7. Let f (θ) be an isv-function on I.

(1) If θ₁ < θ₂ implies f (θ₁) ⊂ f (θ₂) (resp . , f (θ₁) ⊃ f (θ₂)) , then f (θ) is called strictly increasing (resp., strictly decreasing) on I.

(2) If θ₁ < θ₂ implies f (θ₁) ⊆ f (θ₂) (resp . , f (θ₁) ⊇ f (θ₂)) , then f (θ) is called increasing (resp., decreasing) on I.

(3) If ∀ θ ∈ I, f (θ) ⊆ f (θ₀) (θ₀ ∈ I), then f (θ₀) is called the maximum value of f (θ) on I.

(4) If ∀ θ ∈ I, f (θ) ⊇ f (θ₀) (θ₀ ∈ I), then f (θ₀) is called the minimum value of f (θ) on I.

Definition 3.8. Let f (θ) be an isv-function on I.

(1) f (θ) is called perfect, if f : I → 2^U is surjective.

(2) f (θ) is called partition, if {f (θ) : θ ∈ I} forms a partition of U.

(3) f (θ) is called to have no kernel, if ⋂_θ∈If (θ) =∅.

Definition 3.9. Let f (θ) be an isv-function on I.

(1) f (θ) is called topological, if {f (θ) : θ ∈ I} is an Alexandrov topology on U.

(2) f (θ) is called full, if ⋃_θ∈If (θ) = U.

(3) f (θ) is called keeping intersection, if ∀ I₁ ⊆ I, ∃ α ∈ I, ⋂_θ∈I₁f (θ) = f (α).

(4) f (θ) is called keeping union, if ∀ I₁ ⊆ I, ∃ α ∈ I, ⋃_θ∈I₁f (θ) = f (α).

Clearly,

(1) f (θ) is keeping intersection and keeping union, and ∅, U ∈ {f (θ) : θ ∈ I} ⇔ {f (θ) : θ ∈ I} is a topology on U;

(2) f (θ) is keeping intersection and keeping union ⇒ {f (θ) : θ ∈ I} ∪ {∅ , U} is a topology on U.

Proposition 3.10. Let f (θ) be an isv-function on I.

(1) If f (θ) is topological, then f (θ) is full, keeping intersection and keeping union.

(2) If f (θ) is partition, then f (θ) is full.

(3) f (θ) is perfect if and only if {f (θ) : θ ∈ I} is a discrete topology on U.

(4) f (θ) is having no kernel if and only if f^c(θ) is full.

Proof. Obviously.□

Example 3.11. Let U = {u₁, u₂, u₃, u₄, u₅}, I = [0, 1). f (θ) is defined as follows: $f (θ) = {\begin{matrix} {u_{1}, u_{2}, u_{5}}, & if θ \in [0, \frac{1}{4}), \\ \emptyset, & if θ \in [\frac{1}{4}, \frac{1}{2}), \\ {u_{1}, u_{2}}, & if θ \in [\frac{1}{2}, \frac{3}{4}), \\ U, & if θ \in [\frac{3}{4}, 1) . \end{matrix}$ Then f (θ) is topological. But f (θ) is neither perfect nor partition.

Example 3.12. Let U = {u₁, u₂, u₃, u₄, u₅}, I = [0, 1). f (θ) is defined as follows: $f (θ) = {\begin{matrix} {u_{1}, u_{2}, u_{5}}, & if θ \in [0, \frac{1}{4}), \\ {u_{1}, u_{2}}, & if θ \in [\frac{1}{4}, \frac{1}{2}), \\ {u_{3}}, & if θ \in [\frac{1}{2}, \frac{3}{4}), \\ {u_{3}, u_{4}}, & if θ \in [\frac{3}{4}, 1) . \end{matrix}$

Note that {u₁, u₂, u₅} ∩ {u₃} = ∅ ≠ f (θ) (∀ θ ∈ I) . Then f (θ) does not preserve intersection.

Example 3.13. Let U = {u₁, u₂, u₃, u₄, u₅}, I = [0, 1). f (θ) is defined as follows: $f (θ) = {\begin{matrix} {u_{1}}, & if θ \in [0, \frac{1}{4}), \\ {u_{1}, u_{4}}, & if θ \in [\frac{1}{4}, \frac{1}{2}), \\ {u_{1}, u_{3}, u_{4}}, & if θ \in [\frac{1}{2}, \frac{3}{4}), \\ U, & if θ \in [\frac{3}{4}, 1) . \end{matrix}$ Then f (θ) is full, keeping intersection and keeping union. But f (θ) is not topological.

Example 3.14. Let U = {u₁, u₂, u₃, u₄, u₅}, I = [0, 1). f (θ) is defined as follows: $f (θ) = {\begin{matrix} {u_{1}, u_{2}}, & if θ \in [0, \frac{1}{4}), \\ {u_{5}}, & if θ \in [\frac{1}{4}, \frac{1}{2}), \\ {u_{3}}, & if θ \in [\frac{1}{2}, \frac{3}{4}), \\ {u_{4}}, & if θ \in [\frac{3}{4}, 1) . \end{matrix}$ Then f (θ) is partition. But f (θ) is neither topological nor perfect.

Example 3.15. Let U = {u₁, u₂, u₃, u₄, u₅}, I = [0, 1). f (θ) is defined as follows: $f (θ) = {\begin{matrix} {u_{1}, u_{2}, u_{5}}, & if θ \in [0, \frac{1}{4}), \\ \emptyset, & if θ \in [\frac{1}{4}, \frac{1}{2}), \\ {u_{3}}, & if θ \in [\frac{1}{2}, \frac{3}{4}), \\ {u_{3}, u_{4}}, & if θ \in [\frac{3}{4}, 1) . \end{matrix}$ Then f (θ) is full and keeping intersection. But ${u_{1}, u_{2}, u_{5}} \cup {u_{3}} = {u_{1}, u_{2}, u_{3}, u_{5}} \neq f (θ)$ (∀ θ ∈ I) . Thus f (θ) does not preserve union.

Example 3.16. Let U = {u₁, u₂, u₃, u₄, u₅}, I = [0, 1). f (θ) is defined as follows: $f (θ) = {\begin{matrix} {u_{1}}, & if θ \in [0, \frac{1}{4}), \\ {u_{2}}, & if θ \in [\frac{1}{4}, \frac{1}{2}), \\ {u_{1}, u_{2}}, & if θ \in [\frac{1}{2}, \frac{3}{4}), \\ U, & if θ \in [\frac{3}{4}, 1) . \end{matrix}$ Then f (θ) is full and keeping union. But ${u_{1}} \cap {u_{2}} = \emptyset \neq f (θ) (\forall θ \in I) .$ Thus f (θ) is not keeping intersection.

Example 3.17. Let U = {u₁, u₂, u₃, u₄, u₅}, I = [0, 1). f (θ) is defined as follows: $f (θ) = {\begin{matrix} {u_{1}, u_{2}, u_{3}}, & if θ \in [0, \frac{1}{2}), \\ {u_{4}, u_{5}}, & if θ \in [\frac{1}{2}, 1) . \end{matrix}$

Then {f (θ) : θ ∈ I} ∪ {∅ , U} is a topology on U. But ${u_{1}, u_{2}, u_{3}} \cap {u_{4}, u_{5}} = \emptyset \neq f (θ) (\forall θ \in I),$ ${u_{1}, u_{2}, u_{3}} \cup {u_{4}, u_{5}} = U \neq f (θ) (\forall θ \in I) .$ Thus f (θ) is neither keeping intersection nor keeping union.

From Examples 3.10, 3.11, 3.12, 3.13, 3.14 and 3.15, we have the following relationships (see Figs. 1 and 2).

Fig. 1

Relation 1 between isv-functions on I.

Fig. 2

Relation 2 between isv-functions on I.

4 Limit theory of isv-functions

4.1 Definitions of limits of isv-functions

Let θ₀ ∈ R, δ > 0. Denote $U (θ_{0}, δ) = {θ : | θ - θ_{0} | < δ},$ $U^{0} (θ_{0}, δ) = {θ : 0 < | θ - θ_{0} | < δ} .$ Then U (θ₀, δ) is called δ neighborhood of θ₀, U⁰ (θ₀, δ) is called δ neighborhood of θ₀ having no the heart, θ₀ is the center of the neighborhood, δ is the radius of the neighborhood.

U⁺ (θ₀, δ) = [θ₀, θ₀ + δ) is called the δ right neighborhood of θ₀,

U^- (θ₀, δ) = (θ₀ - δ, θ₀] is called the δ left neighborhood of θ₀.

Obviously, U (θ₀, δ) = (θ₀ - δ, θ₀ + δ) = U⁺ (θ₀, δ) ∪ U^- (θ₀, δ).

Let f (θ) be an isv-function on I. For θ₀ ∈ I, u ∈ U, denote $[u]_{f} = {θ \in I - {θ_{0}} : u \in f (θ)},$ $(u)_{f} = {θ \in I - {θ_{0}} : u \notin f (θ)} .$

Remark 4.1. (1) [u] _f ∪ (u) _f = I - {θ₀} , [u] _f ∩ (u) _f = ∅ .

(2) [u] _f ∩ [u] _g = [u] _f∩g, [u] _f ∪ [u] _g = [u] _f∪g .

(3) (u) _f ∩ (u) _g = (u) _f∪g, (u) _f ∪ (u) _g = (u) _f∩g .

(4) [u] _{f
^c} = (u) _f, (u) _{f
^c} = [u] _f .

Definition 4.2. Let f (θ) be an isv-function on I. Given θ₀ ∈ I.

(1) $\underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) = {u \in U : \forall δ > 0, [u]_{f} \cap U^{+} (θ_{0}, δ) is infinite},$ which is called the over-right limit of f (θ) as θ → θ₀ (or the over limit of $f (θ) as θ \to θ_{0}^{+})$ ;

(2) $\underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) = {u \in U : \exists δ > 0, (u)_{f} \cap U^{+} (θ_{0}, δ) is finite},$ which is called the under-right limit of f (θ) as θ → θ₀ (or the under limit of $f (θ) as θ \to θ_{0}^{+})$ .

(3) $\underset{θ \to θ_{0}^{-}}{lim^{¯}} f (θ) = {u \in U : \forall δ > 0, [u]_{f} \cap U^{-} (θ_{0}, δ) is infinite},$ which is called the over-left limit of f (θ) as θ → θ₀ (or the over limit of $f (θ) as θ \to θ_{0}^{-})$ .

(4) $\underset{θ \to θ_{0}^{-}}{lim_{_}} f (θ) = {u \in U : \exists δ > 0, (u)_{f} \cap U^{-} (θ_{0}, δ) is finite},$ which is called the under-left limit of f (θ) as θ → θ₀ (or the under limit of $f (θ) as θ \to θ_{0}^{-})$ .

Theorem 4.3. Let f (θ) be an isv-function on I. Given θ₀ ∈ I.

(1) $\underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ)$ = {u ∈ U : ∀ δ > 0, [u] _f ∩ U⁺ (θ₀, δ) ≠ ∅} $= {u \in U : \forall n \in N, [u]_{f} \cap U^{+} (θ_{0}, \frac{1}{n}) \neq \emptyset}$ .(2) $\underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ)$ = {u ∈ U : ∃ δ > 0, (u) _f ∩ U⁺ (θ₀, δ) = ∅} $= {u \in U : \exists n \in N, (u)_{f} \cap U^{+} (θ_{0}, \frac{1}{n}) = \emptyset}$ .(3) $\underset{θ \to θ_{0}^{-}}{lim^{¯}} f (θ)$ = {u ∈ U : ∀ δ > 0, [u] _f ∩ U^- (θ₀, δ) ≠ ∅} $= {u \in U : \forall n \in N, [u]_{f} \cap U^{-} (θ_{0}, \frac{1}{n}) \neq \emptyset}$ .(4) $\underset{θ \to θ_{0}^{-}}{lim_{_}} f (θ)$ = {u ∈ U : ∃ δ > 0, (u) _f ∩ U⁺ (θ₀, δ) = ∅} $= {u \in U : \exists n \in N, (u)_{f} \cap U^{-} (θ_{0}, \frac{1}{n}) = \emptyset}$ .

Proof. (1) Put $S = \underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ),$ $T = {u \in U : \forall δ > 0, [u]_{f} \cap U^{+} (θ_{0}, δ) \neq \emptyset},$ $L = {u \in U : \forall n \in N, [u]_{f} \cap U^{+} (θ_{0}, \frac{1}{n}) \neq \emptyset} .$

Obviously, S ⊆ T ⊆ L. We only need to prove L ⊆ S.

Suppose L ⊈ S. Then L - S≠ ∅.

Pick u ∈ L - S. We have u ∉ S.

So ∃ δ₀ > 0, [u] _f ∩ U⁺ (θ₀, δ₀) is finite.

Denote $[u]_{f} \cap U^{+} (θ_{0}, δ_{0}) = {θ_{1}, θ_{2}, \dots, θ_{n}} .$

Put $θ^{*} = min {θ_{1}, θ_{2}, \dots, θ_{n}}, 0 < \frac{1}{n_{0}} < θ^{*} - θ_{0}$ . Then $0 < \frac{1}{n_{0}} < δ_{0}, [u]_{f} \cap U^{+} (θ_{0}, \frac{1}{n_{0}}) = \emptyset .$ So u ∉ L. But u ∈ L. This is a contradiction. Thus L ⊆ S.

(2) Put $P = \underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ),$ $Q = {u \in U : \exists δ > 0, (u)_{f} \cap U^{+} (θ_{0}, δ) = \emptyset},$ $K = {u \in U : \exists n \in N, (u)_{f} \cap U^{+} (θ_{0}, \frac{1}{n}) = \emptyset} .$

Obviously, K ⊆ Q ⊆ P. We only need to prove P ⊆ K. Suppose P ⊈ K. Then P - K≠ ∅. Pick u ∈ P - K. Then u ∉ K.

Now we will show that ∀ δ, B (u) ∩ U⁺ (θ₀, δ) is infinite.

In fact, suppose that ∃ δ, (u) _f ∩ U⁺ (θ₀, δ) is finite. Put $(u)_{f} \cap U^{+} (θ_{0}, δ) = {θ_{1}, θ_{2}, \dots, θ_{n}}, θ^{*} = min {θ_{1}, θ_{2}, \dots, θ_{n}}, 0 < \frac{1}{n_{0}} < θ^{*} - θ_{0} .$ Then $0 < \frac{1}{n_{0}} < δ$ , $(u)_{f} \cap U^{+} (θ_{0}, \frac{1}{n_{0}}) = \emptyset$ . So u ∈ K, But u ∉ K. This is a contradiction.

Since ∀ δ > 0, (u) _f ∩ U⁺ (θ₀, δ) is infinite, we have u ∉ P. But u ∈ P. This is a contradiction. Thus P ⊆ K.

(3) The proof is similar to (1).

(4) The proof is similar to (2).□

Example 4.4. Consider Example 3.11. Pick $θ_{0} = \frac{1}{4}$ , we have $[u_{1}]_{f} = [u_{2}]_{f} = [0, \frac{1}{4}) \cup [\frac{1}{4}, \frac{1}{2}), [u_{3}]_{f} = [\frac{1}{2}, 1),$ $[u_{4}]_{f} = [\frac{3}{4}, 1), [u_{5}]_{f} = [0, \frac{1}{4}) .$

$(u_{1})_{f} = (u_{2})_{f} = [\frac{1}{2}, 1), (u_{3})_{f} = [0, \frac{1}{4}) \cup [\frac{1}{4}, \frac{1}{2}),$ $(u_{4})_{f} = [0, \frac{1}{4}) \cup [\frac{1}{4}, \frac{3}{4}), (u_{5})_{f} = (\frac{1}{4}, 1) .$

By Theorem 4.3, $\begin{matrix} \underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) & = & {u \in U : \forall δ > 0, [u]_{f} \cap U^{+} (θ_{0}, δ) \neq \emptyset} \\ = & {u_{1}, u_{2}}; \end{matrix}$

$\begin{matrix} \underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) & = & {u \in U : \exists δ > 0, (u)_{f} \cap U^{+} (θ_{0}, δ) = \emptyset} \\ = & {u_{1}, u_{2}}; \end{matrix}$

$\begin{matrix} \underset{θ \to θ_{0}^{-}}{lim^{¯}} f (θ) & = & {u \in U : \forall δ > 0, [u]_{f} \cap U^{-} (θ_{0}, δ) \neq \emptyset} \\ = & {u_{1}, u_{2}, u_{5}}; \end{matrix}$

$\begin{matrix} \underset{θ \to θ_{0}^{-}}{lim_{_}} f (θ) & = & {u \in U : \exists δ > 0, (u)_{f} \cap U^{-} (θ_{0}, δ) = \emptyset} \\ = & {u_{1}, u_{2}, u_{5}} . \end{matrix}$

Lemma 4.5. Let f (θ) be an isv-function on I. Given θ₀ ∈ I.

(1) $\underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) = ⋂_{n = 1}^{\infty} ⋂_{θ \in (θ_{0}, θ_{0} + \frac{1}{n}) \cap I} ⋃_{α \in (θ_{0}, θ]} f (α)$ .

(2) $\underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) = ⋃_{n = 1}^{\infty} ⋃_{θ \in (θ_{0}, θ_{0} + \frac{1}{n}) \cap I} ⋂_{α \in (θ_{0}, θ]} f (α)$ .

(3) $\underset{θ \to θ_{0}^{-}}{lim^{¯}} f (θ) = ⋂_{n = 1}^{\infty} ⋂_{θ \in (θ_{0} - \frac{1}{n}, θ_{0}) \cap I} ⋃_{α \in [θ, θ_{0})} f (α)$ .

(4) $\underset{θ \to θ_{0}^{-}}{lim_{_}} f (θ) = ⋃_{n = 1}^{\infty} ⋃_{θ \in (θ_{0} - \frac{1}{n}, θ_{0}) \cap I} ⋂_{α \in [θ, θ_{0})} f (α)$ .

Proof. (1) Denote $S = \underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ), T = ⋂_{n = 1}^{\infty} ⋂_{θ \in (θ_{0}, θ_{0} + \frac{1}{n}) \cap I} ⋃_{α \in (θ_{0}, θ]} f (α) .$ To prove S = T, it suffices to show that $u \in S \Leftrightarrow$ $\forall n \in N, \forall θ \in (θ_{0}, θ_{0} + \frac{1}{n}) \cap I, \exists α \in (θ_{0}, θ], u \in f (α) .$

“⇒”. Let u ∈ S, $\forall n \in N, \forall θ \in (θ_{0}, θ_{0} + \frac{1}{n}) \cap I$ . Put δ = θ - θ₀. Then $0 < δ < \frac{1}{n} .$

Since u ∈ S, by Theorem 4.3(1), we have [u] _f ∩ U⁺ (θ₀, δ) ¬ = ∅ .

Pick α ∈ [u] _f ∩ U⁺ (θ₀, δ).

Then α ∈ [u] _f, α ∈ U⁺ (θ₀, δ).

This implies

u ∈ f (α) , θ₀ < α < θ₀ + δ = θ. Thus α ∈ (θ₀, θ].

“⇐”. ∀ n ∈ N, pick $θ \in (θ_{0}, θ_{0} + \frac{1}{n}) \cap I$ .

By the condition, ∃ α ∈ (θ₀, θ] , u ∈ f (α) . Then $α \in U^{+} (θ_{0}, \frac{1}{n}), α \in [u]_{f}$ . Thus ∀ n ∈ N, $[u]_{f} \cap U^{+} (θ_{0}, \frac{1}{n}) \neg = \emptyset$ .

By Theorem 4.3(1), u ∈ S.

(2) By (1) and Theorem 4.3(2), $u \notin \underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ)$

$\begin{matrix} \Leftrightarrow & \forall n \in N, (u)_{f} \cap U^{+} (θ_{0}, \frac{1}{n}) \neg = \emptyset \\ \Leftrightarrow & \forall n \in N, {θ \in I - θ_{0} : u \in U - f (θ)} \\ \cap U^{+} (θ_{0}, \frac{1}{n}) \neg = \emptyset \\ \Leftrightarrow & u \in ⋂_{n = 1}^{\infty} ⋂_{θ \in (θ_{0}, θ_{0} + \frac{1}{n}) \cap I} ⋃_{α \in (θ_{0}, θ]} (U - f (α)) \\ \Leftrightarrow & u \in U - ⋃_{n = 1}^{\infty} ⋃_{θ \in (θ_{0}, θ_{0} + \frac{1}{n}) \cap I} ⋂_{α \in (θ_{0}, θ]} f (α) \\ \Leftrightarrow & u \notin ⋃_{n = 1}^{\infty} ⋃_{θ \in (θ_{0}, θ_{0} + \frac{1}{n}) \cap I} ⋂_{α \in (θ_{0}, θ]} f (α) . \end{matrix}$

Hence $\underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) = ⋃_{n = 1}^{\infty} ⋃_{θ \in (θ_{0}, θ_{0} + \frac{1}{n}) \cap I} ⋂_{α \in (θ_{0}, θ]} f (α)$ .

(3) The proof is similar to (1).

(4) The proof is similar to (2).□

Lemma 4.6. Let f (θ) be an isv-function on I. Given θ₀ ∈ I.

(1) $⋂_{n = 1}^{\infty} ⋂_{θ \in (θ_{0}, θ_{0} + \frac{1}{n}) \cap I} ⋃_{α \in (θ_{0}, θ]} f (α)$ $= ⋂_{θ \in (θ_{0}, θ_{0} + 1) \cap I} ⋃_{α \in (θ_{0}, θ]} f (α) .$

(2) $⋃_{n = 1}^{\infty} ⋃_{θ \in (θ_{0}, θ_{0} + \frac{1}{n}) \cap I} ⋂_{α \in (θ_{0}, θ]} f (α)$ $= ⋃_{θ \in (θ_{0}, θ_{0} + 1) \cap I} ⋂_{α \in (θ_{0}, θ]} f (α)$ .

(3) $⋂_{n = 1}^{\infty} ⋂_{θ \in (θ_{0} - \frac{1}{n}, θ_{0}) \cap I} ⋃_{α \in [θ, θ_{0})} f (α)$ $= ⋂_{θ \in (θ_{0} - 1, θ_{0}) \cap I} ⋃_{α \in [θ, θ_{0})} f (α) .$

(4) $⋃_{n = 1}^{\infty} ⋃_{θ \in (θ_{0} - \frac{1}{n}, θ_{0}) \cap I} ⋂_{α \in [θ, θ_{0})} f (α)$ $= ⋃_{θ \in (θ_{0} - 1, θ_{0}) \cap I} ⋂_{α \in [θ, θ_{0})} f (α) .$

Proof. (1) Put $E_{n} = ⋂_{θ \in (θ_{0}, θ_{0} + \frac{1}{n}) \cap I} ⋃_{α \in (θ_{0}, θ]} f (α)$ .

Then {E_n}↑. So $⋂_{n = 1}^{\infty} E_{n} = E_{1} .$

Thus

$⋂_{n = 1}^{\infty} ⋂_{θ \in (θ_{0}, θ_{0} + \frac{1}{n}) \cap I} ⋃_{α \in (θ_{0}, θ]} f (α)$

= ⋂ _{θ∈(θ₀,θ₀+1)∩I} ⋃ _{α∈(θ₀,θ]}f (α).

(2) Put $F_{n} = ⋃_{θ \in (θ_{0}, θ_{0} + \frac{1}{n}) \cap I} ⋂_{α \in (θ_{0}, θ]} f (α) .$

Then {F_n}↓. So $⋃_{n = 1}^{\infty} F_{n} = F_{1} .$

Thus

$⋃_{n = 1}^{\infty} ⋃_{θ \in (θ_{0}, θ_{0} + \frac{1}{n}) \cap I} ⋂_{α \in (θ_{0}, θ]} f (α)$

= ⋃ _{θ∈(θ₀,θ₀+1)∩I} ⋂ _{α∈(θ₀,θ]}f (α).

(3) The proof is similar to (1).

(4) The proof is similar to (2).□

Theorem 4.7. Let f (θ) be an isv-function on I. Given θ₀ ∈ I.

(1) $\underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) = ⋂_{θ \in (θ_{0}, θ_{0} + 1) \cap I} ⋃_{α \in (θ_{0}, θ]} f (α)$ ;

If f (θ) is increasing, then $\underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) = ⋂_{θ \in (θ_{0}, θ_{0} + 1) \cap I} f (θ) .$

(2) $\underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) = ⋃_{θ \in (θ_{0}, θ_{0} + 1) \cap I} ⋂_{α \in (θ_{0}, θ]} f (α)$ ;

If f (θ) is decreasing, then $\underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) = ⋃_{θ \in (θ_{0}, θ_{0} + 1) \cap I} f (θ) .$

(3) $\underset{θ \to θ_{0}^{-}}{lim^{¯}} f (θ) = ⋂_{θ \in (θ_{0} - 1, θ_{0}) \cap I} ⋃_{α \in [θ, θ_{0})} f (α)$ ;

If f (θ) is decreasing, then $\underset{θ \to θ_{0}^{-}}{lim^{¯}} f (θ) = ⋂_{θ \in (θ_{0} - 1, θ_{0}) \cap I} f (θ) .$

(4) $\underset{θ \to θ_{0}^{-}}{lim_{_}} f (θ) = ⋃_{θ \in (θ_{0} - 1, θ_{0}) \cap I} ⋂_{α \in [θ, θ_{0})} f (α)$ ;

If f (θ) is increasing, then $\underset{θ \to θ_{0}^{-}}{lim_{_}} f (θ) = ⋃_{θ \in (θ_{0} - 1, θ_{0}) \cap I} f (θ) .$

Proof. This holds by Lemmas 4.5 and 4.6.□

Definition 4.8. Let f (θ) be an isv-function on I. Then for θ₀ ∈ I,

(1) If $\underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) = \underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) = S$ , then f (θ) is called to has the limit S as $θ \to θ_{0}^{+}$ (or has the right-limit S as θ → θ₀), which is denoted by $lim_{θ \to θ_{0}^{+}} f (θ)$ , i.e., $lim_{θ \to θ_{0}^{+}} f (θ) = S$ ;

If $\underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) \neq \underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ)$ , then f (θ) is said to have no limit as $θ \to θ_{0}^{+}$ (or has no the right-limit as θ → θ₀).

(2) If $\underset{θ \to θ_{0}^{-}}{lim_{_}} f (θ) = \underset{θ \to θ_{0}^{-}}{lim^{¯}} f (θ) = S$ , then f (θ) is called to has the limit S as $θ \to θ_{0}^{-}$ (or has the left-limit S as θ → θ₀), which is denoted by $lim_{θ \to θ_{0}^{-}} f (θ)$ , i.e., $lim_{θ \to θ_{0}^{-}} f (θ) = S$ ;

If $\underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) \neq \underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ)$ , then f (θ) is said to have no limit as $θ \to θ_{0}^{+}$ (or has no the left-limit as θ → θ₀).

(3) If $lim_{θ \to θ_{0}^{-}} f (θ) = lim_{θ \to θ_{0}^{+}} f (θ) = S$ , then f (θ) is called to has the limit S as θ → θ₀, which is denoted by $lim_{θ \to θ_{0}} f (θ)$ , i.e., $lim_{θ \to θ_{0}} f (θ) = S$ ;

If $lim_{θ \to θ_{0}^{-}} f (θ) \neq lim_{θ \to θ_{0}^{+}} f (θ))$ , then f (θ) is said to have no limit as θ → θ₀.

Definition 4.9. Let f (θ) be an isv-function on I. Then for θ₀ ∈ I,

(1) If $\underset{θ \to θ_{0}^{-}}{lim^{¯}} f (θ) = \underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) = S$ , then f (θ) is called to has the over-limit S as θ → θ₀, which is denoted by $\underset{θ \to θ_{0}}{lim^{¯}} f (θ)$ , i.e., $\underset{θ \to θ_{0}}{lim^{¯}} f (θ) = S$ ;

If $\underset{θ \to θ_{0}^{-}}{lim^{¯}} f (θ) \neq \underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ)$ , then f (θ) is called to has no the over-limit as $θ \to θ_{0}^{+}$ .

(2) If $\underset{θ \to θ_{0}^{-}}{lim_{_}} f (θ) = \underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) = S$ , then f (θ) is called to has the under-limit S as θ → θ₀, which is denoted by $\underset{θ \to θ_{0}}{lim_{_}} f (θ)$ , i.e., $\underset{θ \to θ_{0}}{lim_{_}} f (θ) = S$ ;

If $\underset{θ \to θ_{0}^{-}}{lim_{_}} f (θ) \neq \underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ)$ , then f (θ) is called to has no the under-limit as θ → θ₀.

(3) If $\underset{θ \to θ_{0}}{lim_{_}} f (θ) = \underset{θ \to θ_{0}}{lim^{¯}} f (θ) = S$ , then f (θ) is called to has the limit as θ → θ₀, which is denoted by $lim_{θ \to θ_{0}} f (θ)$ , i.e., $lim_{θ \to θ_{0}} f (θ) = S$ ;

If $\underset{θ \to θ_{0}}{lim_{_}} f (θ) \neq \underset{θ \to θ_{0}}{lim^{¯}} f (θ)$ , then f (θ) is said to have no limit as θ → θ₀.

Remark 4.10. The limit in Definition 4.8(3) and the limit in Definition 4.9(3) is consistent. Limit of s-sequence is a special limit of isv-function.

Proposition 4.11. For A ∈ 2^U, θ₀ ∈ I, $lim_{θ \to θ_{0}} \bar{A} = A$ .

Proof. Obviously,

$[u]_{\bar{A}} = {\begin{matrix} I - {θ_{0}}, u \in A, \\ \emptyset, u \notin A, \end{matrix}$

$(u)_{\bar{A}} = {\begin{matrix} I - {θ_{0}}, u \notin A, \\ \emptyset, u \in A . \end{matrix}$

By Theorem 4.3, $\underset{θ \to θ_{0}^{+}}{lim^{¯}} \bar{A} = {u \in U : \forall δ > 0, [u]_{\bar{A}} \cap U^{+} (θ_{0}, δ) \neq \emptyset},$

$\underset{θ \to θ_{0}^{+}}{lim_{_}} \bar{A} = {u \in U : \exists δ > 0, (u)_{\bar{A}} \cap U^{+} (θ_{0}, δ) = \emptyset} .$

Then $\underset{θ \to θ_{0}^{+}}{lim^{¯}} \bar{A} = A,$ $\underset{θ \to θ_{0}^{+}}{lim_{_}} \bar{A} = A .$

Similarly, $\underset{θ \to θ_{0}^{-}}{lim^{¯}} \bar{A} = A,$ $\underset{θ \to θ_{0}^{-}}{lim_{_}} \bar{A} = A .$

Thus $lim_{θ \to θ_{0}} \bar{A} = A .$ □

Other types of limits of isv-functions are proposed by the following definition and these limits can be discussed in a similar way.

Definition 4.12. Let f (θ) be an isv-function on (- ∞ , + ∞). Define $(1) \underset{θ \to + \infty}{lim^{¯}} f (θ) = \underset{θ \to 0^{+}}{lim^{¯}} f (\frac{1}{θ}), \underset{θ \to - \infty}{lim^{¯}} f (θ)$ $= \underset{θ \to 0^{-}}{lim^{¯}} f (\frac{1}{θ}), \underset{θ \to \infty}{lim^{¯}} f (θ) = \underset{θ \to 0}{lim^{¯}} f (\frac{1}{θ}) .$ $(2) \underset{θ \to + \infty}{lim_{_}} f (θ) = \underset{θ \to 0^{+}}{lim_{_}} f (\frac{1}{θ}), \underset{θ \to - \infty}{lim_{_}} f (θ)$ $= \underset{θ \to 0^{-}}{lim_{_}} f (\frac{1}{θ}), \underset{θ \to \infty}{lim_{_}} f (θ) = \underset{θ \to 0}{lim_{_}} f (\frac{1}{θ}) .$ $(3) lim_{θ \to + \infty} f (θ) = lim_{θ \to 0^{+}} f (\frac{1}{θ}), lim_{θ \to - \infty} f (θ)$ $= lim_{θ \to 0^{-}} f (\frac{1}{θ}), lim_{θ \to \infty} f (θ) = lim_{θ \to 0} f (\frac{1}{θ}) .$

4.2 Properties of limits of isv-functions

Proposition 4.13. For the over-right limit, the following properties hold.

(1) If f (θ) ⪯ g (θ) (∀ θ ∈ (θ₀, θ₀ + δ₀)), then $\underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) \subseteq \underset{θ \to θ_{0}^{+}}{lim^{¯}} g (θ)$ .

(2) $\underset{θ \to θ_{0}^{+}}{lim^{¯}} (f (θ) \cup g (θ)) = \underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) \cup \underset{θ \to θ_{0}^{+}}{lim^{¯}} g (θ)$ .

(3) $\underset{θ \to θ_{0}^{+}}{lim^{¯}} (U - f (θ)) = U - \underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ)$ .

(4) If $\underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) = ▵ \subset B$ , then ∃ δ > 0, ∀ θ ∈ (θ₀, θ₀ + δ), f (θ) ⊂ B.

(5) 1) $\underset{θ \to θ_{0}^{+}}{lim^{¯}} (f (θ) \times g (θ)) \subseteq \underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) \times \underset{θ \to θ_{0}^{+}}{lim^{¯}} g (θ)$ ;

2) $\underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) \times \underset{θ \to θ_{0}^{+}}{lim^{¯}} g (θ) = ⋂_{θ \in (θ_{0}, θ_{0} + 1) \cap I} ⋃_{α, γ \in (θ_{0}, θ]} (f (α) \times g (γ))$ .

Proof. (1) Denote $[u]_{f} = {θ \in I - {θ_{0}} : u \in f (θ)},$ $[u]_{g} = {θ \in I - {θ_{0}} : u \in g (θ)} .$

$\forall u \in \underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ)$ , by Theorem 4.3(1), $\forall δ > 0, [u]_{f} \cap U^{+} (θ_{0}, δ) \neg = \emptyset .$

Pick θ_δ ∈ [u] _f ∩ U⁺ (θ₀, δ).

Then u ∈ f (θ_δ), θ_δ ∈ U⁺ (θ₀, δ).

1) If δ ≤ δ₀, then θ_δ ∈ U⁺ (θ₀, δ₀). By the condition, f (θ_δ) ⊆ g (θ_δ). Then u ∈ g (θ_δ). This implies θ_δ ∈ (u) _f ∩ U⁺ (θ₀, δ). So (u) _f∩ U⁺ (θ₀, δ) ¬ = ∅.

2) If δ > δ₀, then U⁺ (θ₀, δ₀) ⊆ U⁺ (θ₀, δ).

So (u) _f ∩ U⁺ (θ₀, δ₀) ⊆ (u) _f ∩ U⁺ (θ₀, δ). Since $θ_{δ_{0}} \in (u)_{f} \cap U^{+} (θ_{0}, δ_{0}),$ we have (u) _f∩ U⁺ (θ₀, δ) ¬ = ∅.

By 1) and 2), ∀ δ> 0, (u) _f ∩ U⁺ (θ₀, δ) ¬ = ∅. By Theorem 4.3(1), $u \in \underset{θ \to θ_{0}^{+}}{lim^{¯}} g (θ)$ .

Thus $\underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) \subseteq \underset{θ \to θ_{0}^{+}}{lim^{¯}} g (θ) .$

(2) “⊇”. This holds by (1).

“⊆”. Suppose $\underset{θ \to θ_{0}^{+}}{lim^{¯}} (f (θ) \cup g (θ)) ⫋ \underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) \cup \underset{θ \to θ_{0}^{+}}{lim^{¯}} g (θ)$ . Then

$\underset{θ \to θ_{0}^{+}}{lim^{¯}} (f (θ) \cup g (θ)) - \underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) \cup \underset{θ \to θ_{0}^{+}}{lim^{¯}} g (θ) \neq \emptyset .$

Pick $u \in \underset{θ \to θ_{0}^{+}}{lim^{¯}} (f (θ) \cup g (θ)) - \underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) \cup \underset{θ \to θ_{0}^{+}}{lim^{¯}} g (θ)$ . We have

$u \in \underset{θ \to θ_{0}^{+}}{lim^{¯}} (f (θ) \cup g (θ)), u \notin \underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ)$ and $u \notin \underset{θ \to θ_{0}^{+}}{lim^{¯}} g (θ)$ .

By Theorem 4.3, ∃ δ₁, δ₂ > 0, [u] _f ∩ U⁺ (θ₀, δ₁) = ∅ , [u] _g ∩ U⁺ (θ₀, δ₂) = ∅ .

Pick δ₃=min{δ₁, δ₂}. Then [u] _f∩ U⁺ (θ₀, δ₃) = ∅ and [u] _g∩ U⁺ (θ₀, δ₃) = ∅.

It follows that

([u] _f ∪ [u] _g) ∩ U⁺ (θ₀, δ₃

= ([u] _f ∩ U⁺ (θ₀, δ₃)) ∪ ([u] _g ∩ U⁺ (θ₀, δ₃))

= ∅ .

By Remark 4.1, [u] _f∪g ∩ U⁺ (θ₀, δ₃) = ∅ .

Thus

$u \notin \underset{θ \to θ_{0}^{+}}{lim^{¯}} (f \cup g) (θ) = \underset{θ \to θ_{0}^{+}}{lim^{¯}} (f (θ) \cup g (θ)) .$ This is a contradiction.

(3) $\forall u \in \underset{θ \to θ_{0}^{+}}{lim^{¯}} (U - f (θ))$ . Then $u \in \underset{θ \to θ_{0}^{+}}{lim^{¯}} f^{c} (θ)$ .

By Theorem 4.3, ∀ δ > 0, [u] _{f
^c}∩ U⁺ (θ₀, δ) ≠ ∅.

By Remark 4.1, (u) _f∩ U⁺ (θ₀, δ) ≠ ∅.

Thus $u \in U - \underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) .$

The converse part can be proved similarly.

(4) Suppose ∀ δ > 0, ∃ θ ∈ (θ₀, θ₀ + δ), f (θ) ⊈ B or f (θ) = B.

1) If f (θ) ⊈ B, then f (θ) - B≠ ∅. Pick u ∈ f (θ) - B.

We have $u \in f (θ), u \notin B, θ \in [u]_{f} .$

Since θ ∈ (θ₀, θ₀ + δ), we have [u] _f∩ (θ₀, θ₀ + δ) ≠ ∅. Then $u \in \underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) .$

Thus u ∈ B. This is a contradiction.

2) If f (θ) = B, then ▵ - B =∅. So ∃ u∈ B, u ∉ ▵.

Since u ∈ f (θ), we have u ∈ [u] _f, [u] _f∩ (θ₀, θ₀ + δ) ≠ ∅.

So $u \in \underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) = ▵ .$ This is a contradiction.

(5) 1) Put $H_{f \times g} (θ) = ⋃_{α \in (θ_{0}, θ]} (f (α) \times g (α)) .$ By Theorem 4.7(1), $\underset{θ \to θ_{0}^{+}}{lim^{¯}} (f (θ) \times g (θ)) = ⋂_{θ \in (θ_{0}, θ_{0} + 1) \cap I} H_{f \times g} (θ) .$

$\forall (u, v) \in \underset{θ \to θ_{0}^{+}}{lim^{¯}} (f (θ) \times g (θ))$ , we have (u, v) ∈ ⋂ _{θ∈(θ₀,θ₀+1)∩I}H_f×g (θ). Since $H_{f \times g} (θ) = ⋃_{α \in (θ_{0}, θ]} (f (α) \times g (α)),$ we have ∀ θ ∈ (θ₀, θ₀ + 1) ∩ I, ∃ α_θ ∈ (θ₀, θ], (u, v) ∈ f (α_θ) × g (α_θ). It follows that u ∈ f (α_e), v ∈ g (α_θ). Then u ∈ H_f (θ) and v ∈ H_g (θ). So $u \in ⋂_{θ \in (θ_{0}, θ_{0} + 1) \cap I} H_{f} (θ) = \underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ),$ $v \in ⋂_{θ \in (θ_{0}, θ_{0} + 1) \cap I} H_{g} (e) = \underset{θ \to θ_{0}^{+}}{lim^{¯}} g (θ) .$

Thus $(u, v) \in \underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) \times \underset{θ \to θ_{0}^{+}}{lim^{¯}} g (θ)$ .

Hence $\underset{θ \to θ_{0}^{+}}{lim^{¯}} (f (θ) \times g (θ)) \subseteq \underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) \times \underset{θ \to θ_{0}^{+}}{lim^{¯}} g (θ) .$

2) $\forall (u, v) \in \underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) \times \underset{θ \to θ_{0}^{+}}{lim^{¯}} g (θ)$ , we have $u \in \underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) = ⋂_{θ \in (θ_{0}, θ_{0} + 1) \cap I} ⋃_{α \in (θ_{0}, θ]} f (α),$ $v \in \underset{θ \to θ_{0}^{+}}{lim^{¯}} g (θ) = ⋂_{θ \in (θ_{0}, θ_{0} + 1) \cap I} ⋃_{α \in (θ_{0}, θ]} g (α) .$ Then ∀ θ ∈ (θ₀, θ₀ + 1) ∩ I, ∃ α_θ, γ_θ ∈ (θ₀, θ], u ∈ f (α_e), v ∈ g (γ_θ). Then (u, v) ∈ f (α_θ) × g (γ_θ). So $(u, v) \in ⋂_{θ \in (θ_{0}, θ_{0} + 1) \cap I} ⋃_{α, γ \in (θ_{0}, θ]} (f (α) \times g (γ)) .$

The converse part can be proved similarly.

Thus $\underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) \times \underset{θ \to θ_{0}^{+}}{lim^{¯}} g (θ) = ⋂_{θ \in (θ_{0}, θ_{0} + 1) \cap I} ⋃_{α, γ \in (θ_{0}, θ]} (f (α) \times g (γ)) .$ □

Proposition 4.14. For the under-right limit, the following properties hold.

(1) If f (θ) ⪯ g (θ) (∀ θ ∈ (θ₀, θ₀ + δ₀)), then $\underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) \subseteq \underset{θ \to θ_{0}^{+}}{lim_{_}} g (θ)$ .

(2) $\underset{θ \to θ_{0}^{+}}{lim_{_}} (f (θ) \cap g (θ)) = \underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) \cap \underset{θ \to θ_{0}^{+}}{lim_{_}} g (θ)$ .

(3) $\underset{θ \to θ_{0}^{+}}{lim_{_}} (U - f (θ)) = U - \underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ)$ .

(4) If $\underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) = ▵ \supset A$ , then ∃ δ > 0, ∀ θ ∈ (θ₀, θ₀ + δ), f (θ) ⊃ A.

(5) $\underset{θ \to θ_{0}^{+}}{lim_{_}} (f (θ) \times g (θ)) = \underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) \times \underset{θ \to θ_{0}^{+}}{lim_{_}} g (θ)$ .

Proof. (1) The proof is similar to Proposition 4.13(1).

(2) “⊆”. This holds by (1).

“⊇”. Suppose $\underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) \cap \underset{θ \to θ_{0}^{+}}{lim_{_}} g (θ) ⊈ \underset{θ \to θ_{0}^{+}}{lim_{_}}$ (f (θ) ∩ g (θ)). Then $\underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) \cap \underset{θ \to θ_{0}^{+}}{lim_{_}} g (θ) - \underset{θ \to θ_{0}^{+}}{lim_{_}}$ (f (θ) ∩ g (θ)) ≠ ∅ .

Pick $u \in \underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) \cap \underset{θ \to θ_{0}^{+}}{lim_{_}} g (θ) - \underset{θ \to θ_{0}^{+}}{lim_{_}} (f (θ) \cap g (θ))$ .

We have

$u \in \underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ)$ , $u \in \underset{θ \to θ_{0}^{+}}{lim_{_}} g (θ)$ and $u \notin \underset{θ \to θ_{0}^{+}}{lim_{_}}$ (f (θ) ∩ g (θ)).

By Theorem 4.3, $\exists δ_{1}, δ_{2} > 0, (u)_{f} \cap U^{+} (θ_{0}, δ_{1}) = \emptyset, (u)_{g} \cap U^{+} (θ_{0}, δ_{2}) = \emptyset .$

Pick δ₃=min{δ₁, δ₂}. Then (u) _f∩ U⁺ (θ₀, δ₃) = ∅ , (u) _g ∩ U⁺ (θ₀, δ₃) = ∅.

It follows that

((u) _f ∪ (u) _g) ∩ U⁺ (θ₀, δ₃)

= ((u) _f ∩ U⁺ (θ₀, δ₃)) ∪ ((u) _g ∩ U⁺ (θ₀, δ₃))

= ∅ .

By Remark 4.1,

(u) _f∩g ∩ U⁺ (θ₀, δ₃) = ∅ .

Thus

$u \in \underset{θ \to θ_{0}^{+}}{lim_{_}} (f \cap g) (θ) = \underset{θ \to θ_{0}^{+}}{lim_{_}} (f (θ) \cap g (θ)) .$ This is a contradiction.

(3) $\forall u \in \underset{θ \to θ_{0}^{+}}{lim_{_}} (U - f (θ))$ . Then $u \in \underset{θ \to θ_{0}^{+}}{lim_{_}} f^{c} (θ)$ .

By Theorem 4.3, ∃ δ > 0, (u) _{f
^c}∩ U⁺ (θ₀, δ) = ∅.

By Remark 4.1, [u] _f∩ U⁺ (θ₀, δ) = ∅.

Thus $u \in U - \underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) .$

The converse part can be proved similarly.

(4) By Proposition 4.13(3), $\underset{θ \to θ_{0}^{+}}{lim^{¯}} (U - f (θ)) = U - \underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) .$

Since $\underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) = ▵ \supset A$ , we have $\underset{θ \to θ_{0}^{+}}{lim^{¯}} (U - f (θ)) \subset U - A .$

By Proposition 4.13(4), ∃ δ > 0, ∀ θ ∈ (θ₀, θ₀ + δ), U - f (θ) ⊂ U - A .

Thus $\exists δ > 0, \forall θ \in (θ_{0}, θ_{0} + δ), f (θ) \supset A .$

(5) $\forall (u, v) \in \underset{θ \to θ_{0}^{+}}{lim_{_}} (f (θ) \times g (θ))$ , by Theorem 4.7(2), $(u, v) \in ⋃_{θ \in (θ_{0}, θ_{0} + 1) \cap I} ⋂_{α \in (θ_{0}, θ]} (f (α) \times g (α)) .$

Then ∃ θ ∈ (θ₀, θ₀ + 1) ∩ I, ∀ α ∈ (θ₀, θ], (u, v) ∈ f (α) × g (α). It follows u ∈ f (α), v ∈ g (α). Then $u \in ⋃_{θ \in (θ_{0}, θ_{0} + 1) \cap I} ⋂_{α \in (θ_{0}, θ]} f (α),$ $v \in ⋃_{θ \in (θ_{0}, θ_{0} + 1) \cap I} ⋂_{α \in (θ_{0}, θ]} g (α) .$

By Theorem 4.7(2), $u \in \underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ)$ , $v \in \underset{θ \to θ_{0}^{+}}{lim_{_}} g (θ)$ . Thus $(u, v) \in \underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) \times \underset{θ \to θ_{0}^{+}}{lim_{_}} g (θ)$ .

$\forall (u, v) \in \underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) \times \underset{θ \to θ_{0}^{+}}{lim_{_}} g (θ)$ , by Theorem 4.7(2), $u \in \underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) = ⋃_{θ \in (θ_{0}, θ_{0} + 1) \cap I} ⋂_{α \in (θ_{0}, θ]} f (α),$ $v \in \underset{θ \to θ_{0}^{+}}{lim_{_}} g (θ) = ⋃_{θ \in (θ_{0}, θ_{0} + 1) \cap I} ⋂_{α \in (θ_{0}, θ]} g (α) .$ Then ∃ θ₁, θ₂ ∈ (θ₀, θ₀ + 1) ∩ I, ∀ α ∈ (θ₀, θ₁], ∀ γ ∈ (θ₀, θ₂], u ∈ f (α), v ∈ g (γ).

Put θ^* = min {θ₁, θ₂}.

Then θ^* ∈ (θ₀, θ₀ + 1) ∩ I, (θ₀, θ^*] ⊆ (θ₀, θ₁] ∩ (θ₀, θ₂].

Thus ∀ α ∈ (θ₀, θ^*], u ∈ f (α), v ∈ g (α).

It follows that (u, v) ∈ f (α) × g (α).

So $(u, v) \in ⋃_{θ \in (θ_{0}, θ_{0} + 1) \cap I} ⋂_{α, γ \in (θ_{0}, θ]} (f (α) \times g (α)) .$

By Theorem 4.7(2), $(u, v) \in \underset{θ \to θ_{0}^{+}}{lim_{_}} (f (θ) \times g (θ))$ .

Thus $\underset{θ \to θ_{0}^{+}}{lim_{_}} (f (θ) \times g (θ)) = \underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) \times \underset{θ \to θ_{0}^{+}}{lim_{_}} g (θ) .$ □

Proposition 4.15. For the over-left limit, the following properties hold.

(1) If f (θ) ⪯ g (θ) (∀ θ ∈ (θ₀ - δ₀, θ₀)), then $\underset{θ \to θ_{0}^{-}}{lim^{¯}} f (θ) \subseteq \underset{θ \to θ_{0}^{-}}{lim^{¯}} g (θ)$ .

(2) $\underset{θ \to θ_{0}^{-}}{lim^{¯}} (f (θ) \cup g (θ)) = \underset{θ \to θ_{0}^{-}}{lim^{¯}} f (θ) \cup \underset{θ \to θ_{0}^{-}}{lim^{¯}} g (θ)$ .

(3) $\underset{θ \to θ_{0}^{-}}{lim^{¯}} (U - f (θ)) = U - \underset{θ \to θ_{0}^{-}}{lim_{_}} f (θ)$ .

(4) If $\underset{θ \to θ_{0}^{-}}{lim^{¯}} f (θ) = ▵ \subset B$ , then ∃ δ > 0, ∀ θ ∈ (θ₀ - δ, θ₀), f (θ) ⊂ B.

(5) 1) $\underset{θ \to θ_{0}^{-}}{lim^{¯}} (f (θ) \times g (θ)) \subseteq \underset{θ \to θ_{0}^{-}}{lim^{¯}} f (θ) \times \underset{θ \to θ_{0}^{-}}{lim^{¯}} g (θ)$ .

2) $\underset{θ \to θ_{0}^{-}}{lim^{¯}} f (θ) \times \underset{θ \to θ_{0}^{-}}{lim^{¯}} g (θ)$

= ⋂ _{θ∈(θ₀-1,θ₀)∩I} ⋃ _{α,γ∈[θ,θ₀)} (f (α) × g (γ)).

Proof. The proof is similar to Proposition 4.13.□

Proposition 4.16. For the under-left limit, the following properties hold.

(1) If f (θ) ⪯ g (θ) (∀ θ ∈ (θ₀ - δ₀, θ₀)), then $\underset{θ \to θ_{0}^{-}}{lim_{_}} f (θ) \subseteq \underset{θ \to θ_{0}^{-}}{lim_{_}} g (θ)$ .

(2) $\underset{θ \to θ_{0}^{-}}{lim_{_}} (f (θ) \cap g (θ)) = \underset{θ \to θ_{0}^{-}}{lim_{_}} f (θ) \cap \underset{θ \to θ_{0}^{-}}{lim_{_}} g (θ)$ .

(3) $\underset{θ \to θ_{0}^{-}}{lim_{_}} (U - f (θ)) = U - \underset{θ \to θ_{0}^{-}}{lim^{¯}} f (θ)$ .

(4) If $\underset{θ \to θ_{0}^{-}}{lim_{_}} f (θ) = ▵ \supset A$ , then ∃ δ > 0, ∀ θ ∈ (θ₀ - δ, θ₀), f (θ) ⊃ A.

(5) $\underset{θ \to θ_{0}^{-}}{lim_{_}} (f (θ) \times g (θ)) = \underset{θ \to θ_{0}^{-}}{lim_{_}} f (θ) \times \underset{θ \to θ_{0}^{-}}{lim_{_}} g (θ)$ .

Proof. The proof is similar to Proposition 4.14.□

Corollary 4.17. Let f (θ) be an isv-function on I and A ∈ 2^U.

(1) If f (θ) ⊆ A or f (θ) ⊂ A (∀ θ ∈ (θ₀, θ₀ + δ₀)), then $\underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) \subseteq A, \underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) \subseteq A .$

(2) If f (θ) ⊆ A or f (θ) ⊂ A (∀ θ ∈ (θ₀ - δ₀, θ₀)), then $\underset{θ \to θ_{0}^{-}}{lim^{¯}} f (θ) \subseteq A, \underset{θ \to θ_{0}^{-}}{lim_{_}} f (θ) \subseteq A .$

Proof. This holds by Propositions 4.13, 4.14, 4.15 and 4.16.□

Corollary 4.18. Let f (θ) be an isv-function on I and A ∈ 2^U.

(1) If f (θ) ⊇ A or f (θ) ⊃ A (∀ θ ∈ (θ₀, θ₀ + δ₀)), then $\underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) \supseteq A, \underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) \supseteq A .$

(2) If f (θ) ⊇ A or f (θ) ⊃ A (∀ θ ∈ (θ₀ - δ₀, θ₀)), then $\underset{θ \to θ_{0}^{-}}{lim^{¯}} f (θ) \supseteq A, \underset{θ \to θ_{0}^{-}}{lim_{_}} f (θ) \supseteq A .$

Proof. This holds by Propositions 4.13, 4.14, 4.15 and 4.16.□

Theorem 4.19. For the over limit, the following properties hold.

(1) If f (θ) ⪯ g (θ) (∀ θ ∈ U⁰ (θ₀, δ₀)), then $\underset{θ \to θ_{0}}{lim^{¯}} f (θ) \subseteq \underset{θ \to θ_{0}}{lim^{¯}} g (θ)$ .

(2) $\underset{θ \to θ_{0}}{lim^{¯}} (f (θ) \cup g (θ)) = \underset{θ \to θ_{0}}{lim^{¯}} f (θ) \cup \underset{θ \to θ_{0}}{lim^{¯}} g (θ)$ .

(3) $\underset{θ \to θ_{0}}{lim^{¯}} (U - f (θ)) = U - \underset{θ \to θ_{0}}{lim_{_}} f (θ)$ .

(4) If $\underset{θ \to θ_{0}}{lim^{¯}} f (θ) = ▵ \subset B$ , then ∃ δ > 0, ∀ θ ∈ U⁰ (θ₀, δ), f (θ) ⊂ B.

(5) $\underset{θ \to θ_{0}}{lim^{¯}} (f (θ) \times g (θ)) \subseteq \underset{θ \to θ_{0}}{lim^{¯}} f (θ) \times \underset{θ \to θ_{0}}{lim^{¯}} g (θ)$ .

Proof. This follows from Propositions 4.13 and 4.15.□

Theorem 4.20. For the under limit, the following properties hold.

(1) If f (θ) ⪯ g (θ) (∀ θ ∈ U⁰ (θ₀, δ₀)), then $\underset{θ \to θ_{0}}{lim_{_}} f (θ) \subseteq \underset{θ \to θ_{0}}{lim_{_}} g (θ)$ .

(2) $\underset{θ \to θ_{0}}{lim_{_}} (f (θ) \cap g (θ)) = \underset{θ \to θ_{0}}{lim_{_}} f (θ) \cap \underset{θ \to θ_{0}}{lim_{_}} g (θ)$ .

(3) $\underset{θ \to θ_{0}}{lim_{_}} (U - f (θ)) = U - \underset{θ \to θ_{0}}{lim^{¯}} f (θ)$ .

(4) If $\underset{θ \to θ_{0}}{lim_{_}} f (θ) = ▵ \supset A$ , then ∃ δ > 0, ∀ θ ∈ U⁰ (θ₀, δ), f (θ) ⊃ A.

(5) $\underset{θ \to θ_{0}}{lim_{_}} (f (θ) \times g (θ)) = \underset{θ \to θ_{0}}{lim_{_}} f (θ) \times \underset{θ \to θ_{0}}{lim_{_}} g (θ)$ .

Proof. It can be proved by Propositions 4.14 and 4.16.□

Lemma 4.21. Let f (θ) be an isv-function on I. For θ₀ ∈ I, denote $W = {u \in U : \forall δ > 0, [u]_{f} \cap U (θ_{0}, δ) \neq \emptyset},$ $S = {u \in U : \forall δ > 0, [u]_{f} \cap U^{+} (θ_{0}, δ) \neq \emptyset},$ $T = {u \in U : \forall δ > 0, [u]_{f} \cap U^{-} (θ_{0}, δ) \neq \emptyset} .$ Then $W = S \cup T .$

Proof. Suppose W ⊈ S ∪ T. Then W - S∪ T ≠ ∅.

Pick u ∈ W - S ∪ T. Then u ∉ S, u ∉ T. So ∃ δ₁, δ₂ > 0, $[u]_{f} \cap U^{+} (θ_{0}, δ_{1}) = \emptyset, [u]_{f} \cap U^{-} (θ_{0}, δ_{2}) = \emptyset .$

Put δ^* = min {δ₁, δ₂}. Then δ^*> 0, [u] _f ∩ U⁺ (θ₀, δ^*) = ∅ , [u] _f ∩ U^- (θ₀, δ^*) = ∅.

It follows that [u] _f∩ U (θ₀, δ^*) = ∅.

Then u ∉ W. This is a contradiction.

Thus W ⊆ S ∪ T.

On the other hand, suppose S ∪ T ⊈ W, we have S∪ T - W ≠ ∅.

Pick u ∈ S ∪ T - W. Then u ∉ W.

So ∃ δ^*> 0, [u] _f ∩ U (θ₀, δ^*) = ∅.

This implies [u] _f∩ U⁺ (θ₀, δ^*) = ∅ , [u] _f ∩ U^- (θ₀, δ^*) = ∅. Then u ∉ S, u ∉ T.

So u ∉ S ∪ T. This is a contradiction.

Thus S ∪ T ⊆ W.

Hence W = S ∪ T ⊈ W.□

Theorem 4.22. Let f (θ) be an isv-function on I. Given θ₀ ∈ I.

(1) {u ∈ U : ∀ δ > 0, [u] _f ∩ U (θ₀, δ) is infinite} = {u ∈ U : ∀ δ > 0, [u] _f ∩ U (θ₀, δ) ≠ ∅} $= \underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) \cup \underset{θ \to θ_{0}^{-}}{lim^{¯}} f (θ)$ .(2) {u ∈ U : ∃ δ > 0, (u) _f ∩ U (θ₀, δ) is finite} = {u ∈ U : ∃ δ > 0, (u) _f ∩ U⁺ (θ₀, δ) = ∅} $= \underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) \cap \underset{θ \to θ_{0}^{-}}{lim_{_}} f (θ)$ .

Proof. (1) Similar to the proof of Theorem 4.3(1), we have {u ∈ U : ∀ δ > 0, [u] _f ∩ U⁺ (θ₀, δ) ≠ ∅} = {u ∈ U : ∀ δ > 0, [u] _f ∩ U⁺ (θ₀, δ) is infinite}.By Lemma 4.21, {u ∈ U : ∀ δ > 0, [u] _f ∩ U (θ₀, δ) ≠ ∅} $= \underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) \cup \underset{θ \to θ_{0}^{-}}{lim^{¯}} f (θ)$ .(2) Similar to the proof of Theorem 4.3(2), we have {u ∈ U : ∃ δ > 0, (u) _f ∩ U⁺ (θ₀, δ) = ∅} = {u ∈ U : ∃ δ > 0, (u) _f ∩ U⁺ (θ₀, δ) is finite}.By Proposition 4.13(3), $\underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) = U - \underset{θ \to θ_{0}^{+}}{lim^{¯}} (U - f (θ)) .$

By Proposition 4.15(3), $\underset{θ \to θ_{0}^{-}}{lim_{_}} f (θ) = U - \underset{θ \to θ_{0}^{-}}{lim^{¯}} (U - f (θ)) .$

By (1),

$\underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) \cap \underset{θ \to θ_{0}^{-}}{lim_{_}} f (θ)$

$= [U - \underset{θ \to θ_{0}^{+}}{lim^{¯}} (U - f (θ))] \cap [U - \underset{θ \to θ_{0}^{-}}{lim^{¯}} (U - f (θ))]$

$= U - [\underset{θ \to θ_{0}^{+}}{lim^{¯}} (U - f (θ)) \cup \underset{θ \to θ_{0}^{-}}{lim^{¯}} (U - f (θ))]$

= U - {u ∈ U : ∀ δ > 0, (u) _f ∩ U (θ₀, δ) ≠ ∅}

= {u ∈ U : ∃ δ > 0, (u) _f ∩ U (θ₀, δ) = ∅} .□

Theorem 4.23. Let f (θ) be an isv-function on I. Given θ₀ ∈ I.

(1) {u ∈ U : ∀ δ > 0, [u] _f ∩ U (θ₀, δ) is infinite} = {u ∈ U : ∀ δ > 0, [u] _f ∩ U (θ₀, δ) ≠ ∅} $= \underset{θ \to θ_{0}}{lim^{¯}} f (θ)$ .(2) {u ∈ U : ∃ δ > 0, (u) _f ∩ U⁺ (θ₀, δ) is finite} = {u ∈ U : ∃ δ > 0, (u) _f ∩ U⁺ (θ₀, δ) = ∅} $= \underset{θ \to θ_{0}}{lim_{_}} f (θ)$ .

Proof. This holds by Theorem 4.22.□

Theorem 4.24. For the right limit, the following properties hold.

(1) If f (θ) ⪯ g (θ) (∀ θ ∈ (θ₀, θ₀ + δ₀)), then $lim_{θ \to θ_{0}^{+}} f (θ) \subseteq lim_{θ \to θ_{0}^{+}} g (θ)$ .

(2) If $lim_{θ \to θ_{0}^{+}} f (θ) = ▵$ , A ⊂ ▵ ⊂ B, then ∃ δ > 0, ∀ θ ∈ (θ₀, θ₀ + δ), A ⊂ f (θ) ⊂ B.

(3) $lim_{θ \to θ_{0}^{+}} (f (θ) \times g (θ)) \subseteq lim_{θ \to θ_{0}^{+}} f (θ) \times lim_{θ \to θ_{0}^{+}} g (θ)$ .

Proof. This follows from Propositions 4.13 and 4.14.□

Theorem 4.25. For the left limit, the following properties hold.

(1) If f (θ) ⪯ g (θ) (∀ θ ∈ (θ₀ - δ₀, θ₀)), then $lim_{θ \to θ_{0}^{-}} f (θ) \subseteq lim_{θ \to θ_{0}^{-}} g (θ)$ .

(2) If $lim_{θ \to θ_{0}^{-}} f (θ) = ▵$ , A ⊂ ▵ ⊂ B, then ∃ δ > 0, ∀ θ ∈ (θ₀ - δ, θ₀), A ⊂ f (θ) ⊂ B.

(3) $lim_{θ \to θ_{0}^{-}} (f (θ) \times g (θ)) \subseteq lim_{θ \to θ_{0}^{-}} f (θ) \times lim_{θ \to θ_{0}^{-}} g (θ)$ .

Proof. This holds by Propositions 4.15 and 4.16.□

Theorem 4.26. For the limit, the following properties hold.

(1) If f (θ) ⪯ g (θ) (∀ θ ∈ U⁰ (θ₀, δ₀)), then $lim_{θ \to θ_{0}} f (θ) \subseteq lim_{θ \to θ_{0}} g (θ)$ .

(2) If $lim_{θ \to θ_{0}^{-}} f (θ) = ▵$ , A ⊂ ▵ ⊂ B, then ∃ δ > 0, ∀ θ ∈ U⁰ (θ₀, δ₀), A ⊂ f (θ) ⊂ B.

(3) $lim_{θ \to θ_{0}} (f (θ) \times g (θ)) \subseteq lim_{θ \to θ_{0}} f (θ) \times lim_{θ \to θ_{0}} g (θ)$ .

Proof. This follows from Theorems 4.24 and 4.25.□

5 Continuity of isv-functions

5.1 Point-wise continuity of isv-functions

Definition 5.1. Let f (θ) be an isv-function on I. Given θ₀ ∈ I.

(1) f (θ) is called over-right continuous at θ₀, if $\underset{θ \to θ_{0}^{+}}{lim^{¯}} f (θ) = f (θ_{0})$ .

(2) f (θ) is called under-right continuous at θ₀, if $\underset{θ \to θ_{0}^{+}}{lim_{_}} f (θ) = f (θ_{0})$ .

(3) f (θ) is called over-left continuous at θ₀, if $\underset{θ \to θ_{0}^{-}}{lim^{¯}} f (θ) = f (θ_{0})$ .

(4) f (θ) is called under-left continuous at θ₀, if $\underset{θ \to θ_{0}^{-}}{lim_{_}} f (θ) = f (θ_{0})$ .

Definition 5.2. Let f (θ) be an isv-function on I. Given θ₀ ∈ I.

(1) f (θ) is called over-continuous at θ₀, if f (θ) is both over-left and over-right continuous at θ₀ .

(2) f (θ) is called under-continuous at θ₀, if f (θ) is both under-left and under-right continuous at θ₀ .

(3) f (θ) is called continuous at θ₀, if f (θ) is both over-continuous and under-continuous at θ₀ .

Definition 5.3. Let f (θ) be an isv-function on I. Given θ₀ ∈ I.

(1) f (θ) is called right-continuous at θ₀, if f (θ) is both over-right and under-right continuous at θ₀ .

(2) f (θ) is called left-continuous at θ₀, if f (θ) is both over-left and under-left continuous at θ₀.

(3) f (θ) is called continuous at θ₀, if f (θ) is both left-continuous and right-continuous at θ₀ .

Remark 5.4. The point-wise continuity in Definition 5.2(3) and the point-wise continuity in Definition 5.3(3) is consistent.

Denote $C^{or} (θ_{0}) = {f : f is over - right continuous at θ_{0}},$ $C^{ur} (θ_{0}) = {f : f is under - right continuous at θ_{0}},$ $C^{ol} (θ_{0}) = {f : f is over - left continuous at θ_{0}},$ $C^{ul} (θ_{0}) = {f : f is under - left continuous at θ_{0}};$ $C^{o} (θ_{0}) = {f : f is over - continuous at θ_{0}},$ $C^{u} (θ_{0}) = {f : f is under - continuous at θ_{0}};$ $C^{l} (θ_{0}) = {f : f is left - continuous at θ_{0}},$ $C^{r} (θ_{0}) = {f : f is right - continuous at θ_{0}};$ $C (θ_{0} = {f : f is continuous at θ_{0}} .$

Proposition 5.5. The following properties hold.

(1) C^o (θ₀) = C^ol (θ₀) ∩ C^or (θ₀) .

(2) C^u (θ₀) = C^ul (θ₀) ∩ C^ur (θ₀) .

(3) C^l (θ₀) = C^ol (θ₀) ∩ C^ul (θ₀) .

(4) C^r (θ₀) = C^or (θ₀) ∩ C^ur (θ₀) .

(5) C (θ₀) = C^o (θ₀) ∩ C^u (θ₀) = C^l (θ₀) ∩ C^r (θ₀) .

Proof. Obviously.□

Proposition 5.6. Let f (θ) and g (θ) be two isv-functions on I. Given θ₀ ∈ I.

(1) If f, g ∈ C^or (θ₀), then f ∪ g ∈ C^or (θ₀).

(2) If f ∈ C^or (θ₀), then f^c ∈ C^ur (θ₀).

Proof. This holds by Proposition 4.13.□

Proposition 5.7. Let f (θ) and g (θ) be two isv-functions on I. Given θ₀ ∈ I.

(1) If f, g ∈ C^ur (θ₀), then f ∩ g ∈ C^ur (θ₀).

(2) If f ∈ C^ur (θ₀), then f^c ∈ C^or (θ₀).

(3) If f, g ∈ C^ur (θ₀), then f × g ∈ C^ur (θ₀).

Proof. This follows from Proposition 4.14.□

Proposition 5.8. Let f (θ) and g (θ) be two isv-functions on I. Given θ₀ ∈ I.

(1) If f, g ∈ C^ol (θ₀), then f ∪ g ∈ C^ol (θ₀).

(2) If f ∈ C^ol (θ₀), then f^c ∈ C^ul (θ₀).

Proof. This holds by Proposition 4.15.□

Proposition 5.9. Let f (θ) and g (θ) be two isv-functions on I. Given θ₀ ∈ I.

(1) If f, g ∈ C^ul (θ₀), then f ∩ g ∈ C^ul (θ₀).

(2) If f ∈ C^ul (θ₀), then f^c ∈ C^ol (θ₀).

(3) If f, g ∈ C^ul (θ₀), then f × g ∈ C^ul (θ₀).

Proof. This follows from Proposition 4.16.□

Theorem 5.10. Let f (θ) and g (θ) be two isv-functions on I. Given θ₀ ∈ I.

(1) If f, g ∈ C^o (θ₀), then f ∪ g ∈ C^o (θ₀).

(2) If f ∈ C^o (θ₀), then f^c ∈ C^u (θ₀).

Proof. This holds by Propositions 5.6 and 5.8.□

Theorem 5.11. Let f (θ) and g (θ) be two isv-functions on I. Given θ₀ ∈ I.

(1) If f, g ∈ C^u (θ₀), then f ∩ g ∈ C^u (θ₀).

(2) If f ∈ C^u (θ₀), then f^c ∈ C^o (θ₀).

(3) If f, g ∈ C^u (θ₀), then f × g ∈ C^u (θ₀).

Proof. It can be obtained by Propositions 5.7 and 5.9.□

5.2 Continuous isv-functions

Definition 5.12. Let f (θ) be an isv-function on I.

(1) f (θ) is called over-continuous on I, if ∀ θ₀ ∈ I, f (θ) is over-continuous at θ₀.

(2) f (θ) is called under-continuous on I, if ∀ θ₀ ∈ I, f (θ) is under-continuous at θ₀.

(3) f (θ) is called left-continuous on I, if ∀ θ₀ ∈ I, f (θ) is left-continuous at θ₀.

(4) f (θ) is called right-continuous on I, if ∀ θ₀ ∈ I, f (θ) is right-continuous at θ₀.

(5) f (θ) is called continuous on I, if ∀ θ₀ ∈ I, f (θ) is continuous at θ₀.

Denote $C^{or} (θ_{0}) = {f : f is over - right continuous on I},$ $C^{ur} (θ_{0}) = {f : f is under - right continuous on I},$ $C^{ol} (θ_{0}) = {f : f is over - left continuous on I},$ $C^{ul} (θ_{0}) = {f : f is under - left continuous on I};$ $C^{o} (I) = {f : f is over - continuous on I},$ $C^{u} (I) = {f : f is under - continuous on I};$ $C^{l} (I) = {f : f is left - continuous on I},$ $C^{r} (I) = {f : f is right - continuous on I};$ $C (I) = {f : f is continuous on I} .$

Proposition 5.13. The following properties hold.

(1) C^o (I) = C^ol (I) ∩ C^or (I) .

(2) C^u (I) = C^ul (I) ∩ C^ur (I) .

(3) C^l (I) = C^ol (I) ∩ C^ul (I) .

(4) C^r (I) = C^or (I) ∩ C^ur (I) .

(5) C (I) = C^o (I) ∩ C^u (I) = C^l (I) ∩ C^r (I) .

Proof. Obviously.□

Theorem 5.14. Let f (θ) and g (θ) be two isv-functions on I.

(1) If f, g ∈ C^o (I), then f ∪ g ∈ C^o (I).

(2) If f ∈ C^o (I), then f^c ∈ C^u (I).

Proof. This holds by Theorem 5.10.□

Theorem 5.15. Let f (θ) and g (θ) be two isv-functions on I.

(1) If f, g ∈ C^u (I), then f ∩ g ∈ C^u (I).

(2) If f ∈ C^u (I), then f^c ∈ C^o (I).

(3) If f, g ∈ C^u (I), then f × g ∈ C^u (I).

Proof. This follows from Theorem 5.11.□

Theorem 5.16. Let f (θ) be an isv-function on [a, b].

(1) If f (θ) is keeping union or increasing, then f (θ) on [a, b] has the maximum value.

(2) If f (θ) is keeping intersection or decreasing, then f (θ) on [a, b] has the minimum value.

Corollary 5.17. If f (θ) is a topological isv-function on [a, b], then f (θ) on [a, b] has the maximum and minimum value.

Proof. Obviously.□

Lemma 5.18. Let f ∈ C^o (θ₀). If $lim_{n \to \infty} θ_{n} = θ_{0}$ , then $\underset{n \to \infty}{lim^{¯}} f (θ_{n}) \subseteq f (θ_{0})$ .

Proof. Since $\underset{n \to \infty}{lim^{¯}} f (θ_{n}) = ⋂_{n = 1}^{\infty} ⋃_{k = n}^{\infty} f (θ_{k})$ ,

we only need to prove that $if \forall n \in N, \exists k \geq n, u \in f (θ_{k}), then u \in f (θ_{0}) .$

$\forall δ, \exists n \in N_{1}, \frac{1}{n_{1}} < δ$ . It follows $U (θ_{0}, \frac{1}{n_{1}}) \subset U (θ_{0}, δ) .$

Since $lim_{n \to \infty} θ_{n} = θ_{0}$ , ∃ n ∈ N₂, when n > n₂ we have $θ_{n} \in U (θ_{0}, \frac{1}{n_{1}})$ .

Put n₃ = n₁ + n₂. Then ∃ k ≥ n₃, u ∈ f (θ_k). So θ_k ∈ [u] _f.

k ≥ n₃ > n₂ implies $θ_{k} \in U (θ_{0}, \frac{1}{n_{1}}) \subset U (θ_{0}, δ) .$

Then θ_k ∈ [u] _f ∩ U (θ₀, δ). So ∀ δ, [u] _f∩ U (θ₀, δ) ≠ ∅.

By Theorem 4.22, $u \in \underset{θ \to θ_{0}}{lim^{¯}} f (θ)$ .

Since f ∈ C^o (θ₀), we have $f (θ_{0}) = \underset{θ \to θ_{0}}{lim^{¯}} f (θ) .$

Hence u ∈ f (θ₀) .□

Theorem 5.19. Let f ∈ C ([a, b]).

(1) Suppose f (a) ⊂ f (b). Then ∀ μ with f (a) ⊆ μ ⊆ f (b) , ∃ θ₀ ∈ [a, b], f (θ₀) = μ. Moreover, if f (a) ⊂ μ ⊂ f (b), then ∃ θ₀ ∈ (a, b), f (θ₀) = μ.

(2) Suppose f (b) ⊂ f (a). Then ∀ μ : f (b) ⊆ μ ⊆ f (a) , ∃ θ₀ ∈ [a, b], f (θ₀) = μ. Moreover, if f (b) ⊂ μ ⊂ f (a), then ∃ θ₀ ∈ (a, b), f (θ₀) = μ.

Proof. (1) It suffices to show that $if f (a) \subset μ \subset f (b), then \exists θ_{0} \in (a, b), f (θ_{0}) = μ .$

Denote E = {θ ∈ [a, b] : f (θ) ⊃ μ}. Put θ₀ = inf E. Then $\exists {θ_{n} : n \in N} \subseteq E - {θ_{0}}, lim_{n \to \infty} θ_{n} = θ_{0} .$

Since ∀ n ∈ N, f (θ_n) ⊃ μ, we have $\underset{n \to \infty}{lim^{¯}} f (θ_{n}) = ⋂_{n = 1}^{\infty} ⋃_{k = n}^{\infty} f (θ_{k}) \supseteq μ$ . Since f ∈ C^o (θ₀), by Lemma 5.18, $f (θ_{0}) \supseteq \underset{n \to \infty}{lim^{¯}} f (θ_{n}) \supseteq μ .$

Note that f (a) ⊂ μ. Then θ₀ ≠ a.

We assert θ₀ ≠ b. Suppose θ₀ = b. Since $μ \subset f (b) = lim_{θ \to b^{-}} f (θ) = \underset{θ \to b^{-}}{lim_{_}} f (θ),$ by Proposition 4.16(4), then $\exists δ, \forall θ \in (b - δ, b), f (θ) \supset μ .$

Put θ₁ ∈ (b - δ, b). Then f (θ₁) ⊃ μ. We have θ₁ ∈ E. This implies θ₁ ≥ θ₀. But θ₁ < b = θ₀. This is a contradiction.

Thus θ₀ ∈ (a, b).

We claim f (θ₀) ⊅μ . Suppose f (θ₀) ⊃ μ. Since f ∈ C^u (θ₀), we have $μ \subset f (θ_{0}) = lim_{θ \to θ_{0}} f (θ) = \underset{θ \to θ_{0}}{lim_{_}} f (θ) .$

By Theorem 4.20(4), $\exists δ, \forall θ \in U^{0} (θ_{0}, δ), f (θ) \supset μ .$

Put θ₁ ∈ (θ₀ - δ, θ₀). Then f (θ₁) ⊃ μ. We have θ₁ ∈ E. This implies θ₁ ≥ θ₀. This is a contradiction.

Note that f (θ₀) ⊇ μ . Thus f (θ₀) = μ .

(2) The proof is similar to (1).□

6 An application for rough sets

In this section, we give an application of the proposed limit theory for rough sets.

We deal with the application of the proposed limit theory for rough sets according to the following ideas: the upper and lower approximations of the concept of a probability approximation space are first seen as isv-functions on I. Then, the limits of these upper and lower approximations are established. Thus, rough set theory may be used to solve uncertainty problems of continuity by means of limits.

Definition 6.1. Let (U, R, P) be a probabilistic approximate space. For θ ∈ I, X ∈ 2^U, denote $f_{X} (θ) = {\underline{PI}}_{θ} (X), g_{X} (θ) = {\bar{PI}}_{θ} (X) .$ Then f_X (θ) and g_X (θ) are two isv-functions of X on I, which are called the isv-functions induced by the lower and upper approximation of X, respectively.

Theorem 6.2. Let (U, R, P) be a probabilistic approximate space. Then for θ₀ ∈ (0, 1) , X ∈ 2^U,

(1) 1) $\underset{θ \to θ_{0}^{+}}{lim^{¯}} f_{X} (θ) = ⋂_{θ \in (θ_{0}, 1]} ⋃_{α \in (θ_{0}, θ]} f_{X} (α)$ ;

2) $\underset{θ \to θ_{0}^{-}}{lim^{¯}} f_{X} (θ) = ⋂_{θ \in [0, θ_{0})} f_{X} (θ) = f_{X} (θ_{0})$ ;

3) $\underset{θ \to θ_{0}^{+}}{lim_{_}} f_{X} (θ) = ⋃_{θ \in (θ_{0}, 1]} f_{X} (θ) = g_{X} (θ_{0})$ ;

4) $\underset{θ \to θ_{0}^{-}}{lim_{_}} f_{X} (θ) = ⋃_{θ \in [0, θ_{0})} ⋂_{α \in [θ, θ_{0})} f_{X} (α)$ .

(2) 1) $\underset{θ \to θ_{0}^{+}}{lim^{¯}} g_{X} (θ) = ⋂_{θ \in (θ_{0}, 1]} ⋃_{α \in (θ_{0}, θ]} g_{X} (α)$ ;

2) $\underset{θ \to θ_{0}^{-}}{lim^{¯}} g_{X} (θ) = ⋂_{θ \in [0, θ_{0})} g_{X} (θ) = f_{X} (θ_{0})$ ;

3) $\underset{θ \to θ_{0}^{+}}{lim_{_}} g_{X} (θ) = ⋃_{θ \in (θ_{0}, 1]} g_{X} (θ) = g_{X} (θ_{0})$ ;

4) $\underset{θ \to θ_{0}^{-}}{lim_{_}} g_{X} (θ) = ⋃_{θ \in [0, θ_{0})} ⋂_{α \in [θ, θ_{0})} g_{X} (α)$ .

(3) 1) f_U-X (θ) = U - g_X (1 - θ),

2) g_U-X (θ) = U - f_X (1 - θ).

Proof. This holds by Theorems 2.6, 2.7 and 4.7.□

Corollary 6.3. Let (U, R, P) be a probabilistic approximate space. Then for X ∈ 2^U, $f_{X} \in C^{ol} ((0, 1)), g_{X} \in C^{ur} ((0, 1)) .$

Proof. This follows from Theorems 6.2.□

Example 6.4. Let U = {u_i : 1 ≤ i ≤ 20}, $P (X) = \frac{| X |}{| U |} (X \in 2^{U})$ , U/R = {X₁, X₂, X₃, X₄, X₅, X₆} where X₁ = {u₁, u₂, u₃, u₄, u₅}, X₂ = {u₆, u₇, u₈}, X₃ = {u₉, u₁₀, u₁₁, u₁₂}, X₄ = {u₁₃, u₁₄}, X₅ = {u₁₅, u₁₆, u₁₇, u₁₈}, X₆ = {u₁₉, u₂₀}. Put $X^{*} = {u_{6}, u_{7}, u_{8}, u_{13}, u_{17}} .$

By Example 4.6 in [16] or Example 8.1 in [17], $f_{X^{*}} (0.5) = X_{2} \cup X_{4}, g_{X^{*}} (0.5) = X_{2} .$

By Theorem 2.7, $\underset{θ \to 0 . 5^{+}}{lim_{_}} f_{X^{*}} (θ) = g_{X^{*}} (0.5) \neq f_{X^{*}} (0.5) .$

By Theorem 2.7, $\underset{θ \to 0 . 5^{-}}{lim^{¯}} g_{X^{*}} (θ) = f_{X^{*}} (θ_{0}) \neq g_{X^{*}} (0.5) .$

Thus $f_{X^{*}} \notin C^{ur} (0.5), g_{X^{*}} \notin C^{ol} (0.5) .$

This example illustrates that $f_{X^{*}} \notin C^{ur} ((0, 1)), g_{X^{*}} \notin C^{ol} ((0, 1)) .$

Example 6.5. Suppose U = {u_i : 1 ≤ i ≤ 10}. Put $P (X) = \frac{| X |}{| U |} (X \in 2^{U})$ , U/R = {X₁, X₂, X₃, X₄}, where X₁ = {u₁, u₃}, X₂ = {u₂, u₄, u₅, u₇}, X₃ = {u₆, u₈}, X₄ = {u₉, u₁₀} .

(1) Put X^* = {u₁, u₅, u₆, u₈} . Then $f_{X^{*}} (θ) = {\begin{matrix} X_{1} \cup X_{2} \cup X_{3}, & if θ \in (0, \frac{1}{4}], \\ X_{1} \cup X_{3}, & if θ \in (\frac{1}{4}, \frac{1}{2}], \\ X_{3}, & if θ \in (\frac{1}{2}, 1]; \end{matrix}$ $g_{X^{*}} (θ) = {\begin{matrix} X_{1} \cup X_{2} \cup X_{3}, & if θ \in [0, \frac{1}{4}), \\ X_{1} \cup X_{3}, & if θ \in [\frac{1}{4}, \frac{1}{2}), \\ X_{3}, & if θ \in [\frac{1}{2}, 1) . \end{matrix}$

So $\underset{θ \to 0 . 5^{+}}{lim^{¯}} f_{X^{*}} (θ) = ⋂_{θ \in (0.5, 1]} ⋃_{α \in (0.5, θ]} f_{X^{*}} (α) = X_{3} \neq X_{1} \cup X_{3} = f_{X^{*}} (0.5)$ ,

$\underset{θ \to 0 . 5^{-}}{lim_{_}} g_{X} (θ) = ⋃_{θ \in [0, 0.5)} ⋂_{α \in [θ, 0.5)} g_{X} (α) = X_{1} \cup X_{3} \neq X_{3} = g_{X^{*}} (0.5)$ .

Thus $f_{X^{*}} \notin C^{or} (0.5), g_{X^{*}} \notin C^{ul} (0.5) .$

(2) Put Y^* = {u₂, u₉, u₁₀} . Then $f_{Y^{*}} (θ) = {\begin{matrix} X_{2} \cup X_{4}, & if θ \in (0, \frac{1}{4}], \\ X_{4}, & if θ \in (\frac{1}{4}, 1] . \end{matrix}$

So $\underset{θ \to 0 . 5^{-}}{lim_{_}} f_{Y^{*}} (θ) = ⋃_{θ \in [0, 0.5)} ⋂_{α \in [θ, 0.5)} f_{Y^{*}} (α) = X_{2} \cup X_{4} \neq X_{4} = f_{Y^{*}} (0.5)$ .

Thus $f_{Y^{*}} \notin C^{ul} (0.5) .$

(3) Put $Z^{*} = U - Y^{*} .$

By Proposition 4.13(3) and Theorem 2.7,

$\begin{matrix} \underset{θ \to 0 . 5^{+}}{lim^{¯}} g_{Z^{*}} (θ) & = & \underset{θ \to 0 . 5^{+}}{lim^{¯}} (U - f_{Y^{*}} (1 - θ)) \\ = & U - \underset{θ \to 0 . 5^{+}}{lim_{_}} f_{Y^{*}} (1 - θ) \\ = & U - \underset{1 - θ \to 0 . 5^{-}}{lim_{_}} f_{Y^{*}} (1 - θ) . \end{matrix}$

Note that $\underset{θ \to 0 . 5^{-}}{lim_{_}} f_{Y^{*}} (θ) \neq f_{Y^{*}} (0.5)$ . Then by Theorem 2.7, $\underset{θ \to 0 . 5^{+}}{lim^{¯}} g_{Z^{*}} (θ) \neq U - f_{Y^{*}} (0.5) = g_{Z^{*}} (0.5) .$

Thus $g_{Z^{*}} \notin C^{or} (0.5) .$

This example illustrates that $f_{X^{*}} \notin C^{or} ((0, 1)), g_{X^{*}} \notin C^{ul} ((0, 1));$ $f_{Y^{*}} \notin C^{ul} ((0, 1)); g_{Z^{*}} \notin C^{or} ((0, 1)) .$

7 Conclusions

In this paper, limits of isv-functions have been proposed. Point-wise continuity of isv-functions and continuous isv-functions have been investigated. An application for rough sets has been given. These results will be helpful for the study of set-value analysis and rough set theory. In the future, we will further study applications of this limit theory for information science.

Footnotes

Acknowledgments

The authors would like to thank the editors and the anonymous reviewers for their valuable comments and suggestions, which have helped immensely in improving the quality of the paper. This work is supported by Special Scientific Research Project of Young Innovative Talents in Guangxi (2019AC20052), Natural Science Foundation of Guangxi (2019JJA110036, AD19245102, 2018GXNSFDA281028), Guangxi Higher Education Institutions of China (Document No.[2018] 35), Guangxi Higher Education Reform Project (2020X JJGZD17), Research Project of Institute of Big Data in Yulin (YJKY03) and Engineering Project of Undergraduate Teaching Reform of Higher Education in Guangxi (2017JGA179).

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