Some set-valued and multi-valued contraction results in fuzzy cone metric spaces

Abstract

This paper aims to present the concept of multi-valued mappings in fuzzy cone metric spaces and prove some basic lemmas, a Hausdorff metric, and fixed point results for set-valued fuzzy cone-contraction and for multi-valued fuzzy cone-contraction mappings. We prove a fixed point theorem for multi-valued rational type fuzzy cone-contractions in fuzzy cone metric spaces. Our results extend and improve some results given in the literature.

1 Introduction

Huang et al. [1] introduced the concept of cone metric spaces by using an ordered Banach space instead of a real number set and proved some fixed point results under cone contraction conditions. After the publication of this article, a number of researchers contributed their ideas to the problems on cone metric spaces by using different contractive type mappings and spaces (see, e.g., [2–11] and the references therein).

Kramosil et al. [12] introduced a fuzzy metric space (FM-space) by using the notion of a fuzzy set and some more notions derived from the one in ordered. These researchers have compared the fuzzy metric notion with the statistical metric space and proved that both conceptions are equivalent in some cases. Later on, the modified form of the metric fuzziness was given by George et al. in [13] by using the continuous t-norm. After that, a number of authors have studied and contributed their ideas to the problems on FM-spaces. Some of their results can be found in [14–25] and the references therein.

Lopez et al. [26] introduced the Hausdorff fuzzy metric on a compact set for a given FM-space and proved some properties for a Hausdorff fuzzy metric. Kiany et al. [19] proved some fixed point results for set-valued mappings and an endpoint theorem in FM-spaces by using contraction conditions. Some other properties and fixed theorems on multi-valued mappings in FM-spaces can be found in [27–29].

The concept of a fuzzy cone metric space (FCM-space) was given by Oner et al. in [30]. They established some properties and a fuzzy cone Banach principle theorem. Some more topological properties, fixed point theorems, and common fixed point theorems in FCM-spaces can be found in [31–37].

In this paper, we introduce the concept of multi-valued mappings in FCM-spaces and prove some basic lemmas and a Hausdorff metric in FCM-spaces. Our result extends and improves the result of Kiany et al. [19] and presents a set-valued fuzzy cone contraction theorem in FCM-spaces. Moreover, we present some fixed point results via multi-valued fuzzy cone contractions in FCM-spaces by extending and improving the result of Ali et al. [27] and a rational type multi-valued fuzzy cone contraction theorem.

2 Preliminaries

Definition 2.1

([38])

A binary operation $\ast:[0,1]^{2}\to [0,1]$ is called a continuous t-norm if:

• (i) ∗ is associative, commutative, and continuous;

• (ii) $\forall a_{0},a_{1},b_{0},b_{1}\in [0,1]$, then $1\ast a_{0}=a_{0}$, while $a_{0}\ast a_{1}\le b_{0}\ast b_{1}$, whenever $a_{0}\le b_{0}$ and $a_{1}\le b_{1}$.

The basic continuous t-norms are minimum, the product and the Lukasiewicz t-norms are defined, respectively, as follows (see [38]):

$$\begin{aligned} a_{0}\ast b_{0}=\min \{a_{0},b_{0} \},\qquad a_{0}\ast b_{0}=a_{0} b_{0},\quad \text{and}\quad a_{0}\ast b_{0}= \max \{a_{0}+b_{0}-1,0 \}. \end{aligned}$$

Throughout this paper, a set of natural numbers is denoted by $\mathbb{N}$ and a real Banach space is denoted by $\mathbb{E}$. θ represents the zero element of $\mathbb{E}$.

Definition 2.2

([1])

A subset $P\subset \mathbb{E}$ is known as a cone if

• (i) $P\neq \emptyset$, closed, and $P\ne \{\theta \}$;

• (ii) If $a_{0}, b_{0}\geq 0$ and $\mu, \nu \in P$, then $a_{0} \mu + b_{0} \nu \in P$;

• (iii) If both $-\mu,\mu \in P$, then $\mu =\theta$.

A partial ordering “⪯” on $P\subset \mathbb{E}$ is defined by $\mu \preceq \nu$ if and only if $\nu - \mu \in P$. $\mu \prec \nu$ stands for $\mu \preceq \nu$ and $\mu \ne \nu$, while $\mu \ll \nu$ stands for $\nu - \mu \in \operatorname{int}(P)$ and all cones have nonempty interior.

Definition 2.3

([30])

A 3-tuple $(U,F_{m},\ast )$ is known as an FCM-space if $P\subset \mathbb{E}$ is a cone, U is an arbitrary set, ∗ is a continuous t-norm, and $F_{m}$ is a fuzzy set on $U\times U\times \operatorname{int}(P)$ satisfying the following;

• (i) $F_{m}(\mu, \nu, t)>0$, and $F_{m}(\mu,\nu,t)=1$ if and only if $\mu =\nu$;

• (ii) $F_{m}(\mu,\nu,t)=F_{m}(\nu,\mu,t)$;

• (iii) $F_{m}(\mu,\omega,t)\ast F_{m}(\omega,\nu,s)\le F_{m}(\mu,\nu,t+s)$;

• (iv) $F_{m}(\mu,\nu,.):\operatorname{int}(P)\to [0,1]$ is continuous

for all $\mu,\nu,\omega \in U$ and $t,s\in \operatorname{int}(P)$.

Definition 2.4

([30])

Let $(U,F_{m},\ast )$ be an FCM-space, $\mu \in U$, and $(\mu _{n})$ be a sequence in U. Then

• (i) $(\mu _{n})$ is said to converge to μ if, for $t\gg \theta$ and $0< r<1$, there exists $n_{1}\in \mathbb{N}$ such that

  $$\begin{aligned} F_{m}(\mu _{n},\mu,t)>1-r,\quad \forall n\ge n_{1}. \end{aligned}$$

  This can be written as $\lim_{n\to \infty }\mu _{n}=\mu$ or $\mu _{n}\to \mu$, as $n\to \infty$.

• (ii) $(\mu _{n})$ is said to be a Cauchy sequence if, for $t\gg \theta$ and $0< r<1$, there exists $n_{1}\in \mathbb{N}$ such that

  $$\begin{aligned} F_{m}(\mu _{m},\mu _{n},t)>1-r,\quad \forall m,n\ge n_{1}. \end{aligned}$$

• (iii) $(U,F_{m},*)$ is complete if every Cauchy sequence is convergent in U.

Lemma 2.5

([30])

Let $(U,F_{m},\ast )$ be an FCM-space. The following statements hold:

• (1) Let $\mu \in U$ and $(\mu _{n})$ be a sequence in U. Then $\mu _{n}\to \mu$ if and only if $\lim_{n\to \infty }F_{m}(\mu _{n},\mu,t)=1$ for $t\gg \theta$.

• (2) An open ball $B (\mu _{0}, r, t)$ with center $\mu _{0}$ and radius $0< r <1$ can be defined as follows for $t\gg \theta$:

  $$\begin{aligned} B (\mu _{0}, r, t) = \bigl\{ \mu \in U: F_{m} (\mu _{0}, \mu, t) > 1 -r\bigr\} . \end{aligned}$$

  Let

  $$\begin{aligned} T_{fc} = \bigl\{ A \subset U: \mu _{0} \in A \textit{ iff } \exists 0< r< 1 \textit{ and } t \gg \theta \textit{ such that } B (\mu _{0}, r, t) \subset A\bigr\} . \end{aligned}$$

  Then $T_{fc}$ is a topology on U.

We recall the following definitions given in [27].

Definition 2.6

Let $(U,F_{m},\ast )$ be an FCM-space;

• (i) A function $g:U\to \mathbb{R}$ is said to be lower semi-continuous if, for any $(\mu _{i})\subset U$ and $\mu \in U$, $\mu _{i}\to \mu$ implies $g(\mu )\leq \limsup_{i\to \infty }g(\mu _{i})$.

• (ii) A function $g:U\to \mathbb{R}$ is said to be upper semi-continuous if, for any $(\mu _{i})\subset U$ and $\mu \in U$, $\mu _{i}\to \mu$ implies $g(\mu )\geq \limsup_{i\to \infty }g(\mu _{i})$.

• (iii) A multi-valued mapping $G:U\to 2^{U}$ ($2^{U}$ is the collection of all nonempty subsets of a set U) is called upper semi-continuous if, for any $\mu \in U$ and a neighborhood B of $G(\mu )$, there is a neighborhood A of μ such that, for any $\nu \in A$, we have $G(\nu )\subset B$.

• (iv) A multi-valued mapping $G:U\to 2^{U}$ is said to be lower semi-continuous if, for any $\mu \in U$ and a neighborhood B, $G(\mu )\cap B\neq \emptyset$, there is a neighborhood A of μ such that, for any $\nu \in A$, we have $G(\nu )\cap B\neq \emptyset$.

Definition 2.7

Assume that $(U,F_{m},\ast )$ is an FCM-space, $\mu \in U$, and $(\mu _{i})_{i\in \mathbb{N}}$ is a sequence in U. Then:

• (i) a subset $A\subseteq U$ is closed if, for every convergent sequence $(\mu _{i})$ in A such that $\mu _{i}\to \mu$, we have $\mu \in A$.

• (ii) a subset $A\subseteq U$ is compact if every sequence in A has a convergent subsequence in A.

Throughout this paper, $\mathbb{K}(U)$ represents the set of all compact subsets of a set U and $\mathbb{P}(U)$ represents the set of all nonempty subsets of a set U.

3 Some properties and a Hausdorff fuzzy metric in FCM-spaces

Proposition 3.1

Let $(U,F_{m},\ast )$ be an FCM-space. Then $F_{m}$ is continuous on $U^{2}\times \operatorname{int}(P)$ for every $t\gg \theta$ (i.e., $t\in \operatorname{int}(P)$).

Proof

Let $\mu,\nu \in U$, $t\gg \theta$, and $(\mu _{i},\nu _{i},t_{i})_{i}$ be a sequence in $X^{2}\times \operatorname{int}(P)$ converging to $(\mu,\nu,t)$. Since $(F_{m}(\mu _{i},\nu _{i},t_{i}))_{i}$ is a sequence in $(0,1]$, there is a sub-sequence $(\mu _{i_{n}},\nu _{i_{n}},t_{i_{n}})_{n}$ of the sequence $(\mu _{i},\nu _{i},t_{i})_{i}$ such that $(F_{m}(\mu _{i_{n}},\nu _{i_{n}},t_{i_{n}}))_{n}$ converges to a point in $[0,1]$. Fix any $\varepsilon >0$ such that $\varepsilon <\frac{t}{2}$, so there is $i_{0}\in \mathbb{N}$ such that $|t-t_{i}|<\varepsilon$ for all $i\ge i_{0}$. Then we have

$$\begin{aligned} F_{m}(\mu _{i_{n}},\nu _{i_{n}},t_{i_{n}})&\geq F_{m} (\mu _{i_{n}}, \mu,\varepsilon )\ast F_{m} ( \mu,\nu,t-2\varepsilon )\ast F_{m} (\nu,\nu _{i_{n}},\varepsilon ) \\ &\to 1\ast F_{m} (\mu,\nu,t-2\varepsilon )\ast 1=F_{m} ( \mu,\nu,t-2\varepsilon ), \quad\text{as } i\to \infty, t\gg \theta, \end{aligned}$$

and

$$\begin{aligned} F_{m}(\mu,\nu,t+2\varepsilon )&\geq F_{m} (\mu,\mu _{i_{n}}, \varepsilon )\ast F_{m} (\mu _{i_{n}},\nu _{i_{n}},t_{i_{n}} )\ast F_{m} (\nu _{i_{n}},\nu, \varepsilon ) \\ &\to 1\ast F_{m} (\mu _{i_{n}},\nu _{i_{n}},t_{i_{n}} ) \ast 1=F_{m} (\mu _{i_{n}},\nu _{i_{n}},t_{i_{n}} ), \quad\text{as } i\to \infty, t\gg \theta. \end{aligned}$$

Therefore, by the continuity of the function $t\longmapsto F_{m}(\mu,\nu,t)$, we can deduce that

$$\begin{aligned} F_{m}(\mu,\nu,t)=\lim_{i\to \infty }F_{m} (\mu _{i_{n}},\nu _{i_{n}},t_{i_{n}} ) \quad\text{for } t\gg \theta. \end{aligned}$$

Thus, $F_{m}$ is continuous on $U^{2}\times \operatorname{int}(P)$. □

Lemma 3.2

Let $(U,F_{m},\ast )$ be an FCM-space such that

$$\begin{aligned} \ast _{j=i}^{\infty }F_{m}\bigl(\mu, \nu,tb^{j}\bigr)\to 1, \quad\textit{as } i \to \infty, \end{aligned}$$

(3.1)

for all $\mu,\nu \in U$, $t\gg \theta$, and $b>1$. Let $(\mu _{i})$ be a sequence in U such that

$$\begin{aligned} F_{m}(\mu _{i},\mu _{i+1},at)\geq M(\mu _{i-1},\mu _{i},t) \end{aligned}$$

for all $i\in \mathbb{N}$ and $a\in (0,1)$. Then $(\mu _{i})$ is a Cauchy sequence in U.

Proof

For every $i\in \mathbb{N}$ and $t\gg \theta$, we have that

$$\begin{aligned} F_{m}(\mu _{i},\mu _{i+1},t)&\geq F_{m} \biggl(\mu _{i-1},\mu _{i}, \frac{1}{a}t \biggr)\geq F_{m} \biggl(\mu _{i-2},\mu _{i-1}, \frac{1}{a^{2}}t \biggr) \geq \cdots \geq F_{m} \biggl(\mu _{0},\mu _{1}, \frac{1}{a^{i}}t \biggr). \end{aligned}$$

Thus, for all $i\in \mathbb{N}$ and $t\gg \theta$, we have

$$\begin{aligned} F_{m}(\mu _{i},\mu _{i+1},t)\geq F_{m} \biggl(\mu _{0},\mu _{1}, \frac{1}{a^{i}}t \biggr). \end{aligned}$$

Now, we choose a constant $b>1$ and $l\in \mathbb{N}$ such that $ab<1$ and $\sum_{j=l}^{\infty }\frac{1}{b^{j}}=\frac{1/b^{l}}{1-(1/b)}<1$. Hence, for $k\geq i$ and $t\gg \theta$, we have that

$$\begin{aligned} &F_{m}(\mu _{i},\mu _{k},t)\\ &\quad\geq F_{m} \biggl(\mu _{i},\mu _{k}, \biggl( \frac{1}{b^{l}}+ \frac{1}{b^{l+1}}+\cdots +\frac{1}{b^{l+k}} \biggr)t \biggr) \\ &\quad\geq F_{m} \biggl(\mu _{i},\mu _{i+1}, \frac{1}{b^{l}}t \biggr)\ast F_{m} \biggl(\mu _{i+1},\mu _{i+2},\frac{1}{b^{l+1}}t \biggr)\ast \cdots \ast F_{m} \biggl(\mu _{k-1},\mu _{k},\frac{1}{b^{l+k}}t \biggr) \\ &\quad\geq F_{m} \biggl(\mu _{0},\mu _{1}, \frac{1}{a^{i-1}b^{l}}t \biggr) \ast F_{m} \biggl(\mu _{0},\mu _{1},\frac{1}{a^{i} b^{l+1}}t \biggr) \ast \cdots \ast F_{m} \biggl(\mu _{0},\mu _{1}, \frac{1}{a^{k-2}b^{l+k-i-2}}t \biggr) \\ &\quad\geq F_{m} \biggl(\mu _{0},\mu _{1}, \frac{1}{(ab)^{i-1}}t \biggr)\ast F_{m} \biggl(\mu _{0},\mu _{1},\frac{1}{(ab)^{i}}t \biggr)\ast \cdots \ast F_{m} \biggl(\mu _{0},\mu _{1},\frac{1}{(ab)^{k-2}}t \biggr) \\ &\quad\geq \ast _{j=i}^{\infty }F_{m} \biggl(\mu _{0},\mu _{1}, \frac{1}{(ab)^{j-1}}t \biggr)\to 1, \quad\text{as } i\to \infty. \end{aligned}$$

This proves that $(\mu _{i})$ is a Cauchy sequence in U. □

Lemma 3.3

Let $(U,F_{m},\ast )$ be an FCM-space. Then, for every $\mu \in U$, $A\in \mathbb{K}(U)$ and $t\gg \theta$, there exists $a_{0}\in A$ such that

$$\begin{aligned} F_{m}(\mu,A,t)=F_{m}(\mu,a_{0},t). \end{aligned}$$

Proof

Let $\mu \in U$, $A\in \mathbb{K}(U)$, and $t\gg \theta$. Then, by Proposition 3.1, the function $\nu \longmapsto F_{m}(\mu,\nu,t)$ is continuous. Thus, by the compactness of A, there exists $a_{0}\in A$ such that

$$\begin{aligned} \sup_{a\in A}F_{m}(\mu,a,t)=F_{m}( \mu,a_{0},t), \end{aligned}$$

that is,

$$\begin{aligned} F_{m}(\mu,A,t)=F_{m}(\mu,a_{0},t). \end{aligned}$$

 □

Lemma 3.4

Let $(U,F_{m},\ast )$ be an FCM-space. Then, for all $\mu \in U$ and $A\in \mathbb{K}(U)$, the function $t\longmapsto F_{m}(\mu,A,t)$ is continuous on $\operatorname{int}(P)$, where $t\gg \theta$.

Proof

Since $F_{m}(\mu,A,t)=\sup_{a_{0}\in A}F_{m}(\mu,a_{0},t)$ and for every $a_{0}\in A$, the function $t\longmapsto F_{m}(\mu,a_{0},t)$ is continuous on $\operatorname{int}(P)$, it follows that $t\longmapsto F_{m}(\mu,A,t)$ is lower semi-continuous on $\operatorname{int}(P)$. Now, we prove that $t\longmapsto F_{m}(\mu,A,t)$ is upper semi-continuous on $\operatorname{int}(P)$.

Let $t\gg \theta$ and $(t_{j})_{j}$ be a sequence in $\operatorname{int}(P)$ which converges to t. By Lemma 3.3, there exists $a_{j}\in A$ such that, for all $j\in \mathbb{N}$,

$$\begin{aligned} F_{m}(\mu,A,t_{j})=F_{m}(\mu,a_{j},t_{j}). \end{aligned}$$

Since $A\in \mathbb{K}(U)$, there are a subsequence $(a_{j_{n}})_{n}$ of the sequence $(a_{j})_{j}$ and a point $a^{*}\in A$ such that $a_{j_{n}}\to a^{*}$ in $(U,F_{m},\ast )$. Hence,

$$\begin{aligned} F_{m}(\mu,a_{j_{n}},t_{j_{n}})\to F_{m} \bigl(\mu,a^{*},t\bigr), \quad\text{as } n\to \infty, \end{aligned}$$

for $t\gg \theta$. Now, by Proposition 3.1, we have that

$$\begin{aligned} F_{m}(\mu,A,t_{j_{n}})\to F_{m}\bigl( \mu,a^{*},t\bigr)\leq F_{m}(\mu,A,t), \quad\text{as } n\to \infty, \end{aligned}$$

for $t\gg \theta$. Consequently, the function $t\longmapsto F_{m}(\mu,A,t)$ is upper semi-continuous on $\operatorname{int}(P)$, which concludes the required proof. □

Lemma 3.5

Let $(U,F_{m},\ast )$ be an FCM-space. Then, for every $A\in \mathbb{K}(U)$ and $B\in \mathbb{P}(U)$, there exists $a^{*}\in A$ such that

$$\begin{aligned} \inf_{a_{0}\in A}F_{m}(a_{0},B,t)=F_{m} \bigl(a^{*},B,t\bigr) \end{aligned}$$

for $t\gg \theta$.

Proof

By putting $\beta =\inf_{a_{0}\in A}F_{m}(a_{0},B,t)$, there is a sequence $(a_{j})_{j}$ in A such that $\beta +\frac{1}{j}>F_{m}(a_{j},B,t)$ for all $j\in \mathbb{N}$. Since $A\in \mathbb{K}(U)$, there are a subsequence $(a_{j_{n}})_{n}$ of $(a_{j})_{j}$ and a point $a^{*}\in A$ such that $a_{j_{n}}\to a^{*}$ in $(U,F_{m},\ast )$. Here, we choose an arbitrary point $b_{0}\in B$. Now, by Proposition 3.1, we have

$$\begin{aligned} F_{m}(a_{j_{n}},b_{0},t)\to F_{m} \bigl(a^{*},b_{0},t\bigr), \quad\text{as } n \to \infty, \end{aligned}$$

for $t\gg \theta$. Since for all $n\in \mathbb{N}$ and $\beta +\frac{1}{j_{n}}>F_{m}(a_{j_{n}},b_{0},t)$. Then, by taking the limit $n\to \infty$, we get

$$\begin{aligned} \beta \geq F_{m}\bigl(a^{*},b_{0},t\bigr) \quad\Rightarrow\quad \beta =F_{m}\bigl(a^{*},b_{0},t\bigr) \quad\text{for } t\gg \theta. \end{aligned}$$

 □

Proposition 3.6

Let $(U,F_{m},\ast )$ be an FCM-space. Then, for every $A,B\in \mathbb{K}(U)$, $t\longmapsto \inf_{a^{*}\in A}F_{m}(a^{*},B,t)$ is a continuous function in $\operatorname{int}(P)$, where $t\gg \theta$.

Proof

By Lemma 3.4, $t\longmapsto F_{m}(a^{*},B,t)$ is a continuous function in $\operatorname{int}(P)$. Therefore, $t\longmapsto \inf_{a^{*}\in A}F_{m}(a^{*},B,t)$ is an upper semi-continuous function in $\operatorname{int}(P)$.

Now, we prove that $t\longmapsto \inf_{a^{*}\in A}F_{m}(a^{*},B,t)$ is lower semi-continuous in $\operatorname{int}(P)$. Let $(t_{j})_{j}$ be any sequence in $\operatorname{int}(P)$ such that $(t_{j})_{j}\to t$ in $\operatorname{int}(P)$, where $t\gg \theta$. By Lemma 3.5, there exists $a_{j}\in A$ such that, for all $j\in \mathbb{N}$,

$$\begin{aligned} F_{m}(a_{j},B,t_{j})=\inf_{a^{*}\in A}F_{m} \bigl(a^{*},B,t_{j}\bigr). \end{aligned}$$

Since $A\in \mathbb{K}(U)$, there are a subsequence $(a_{j_{n}})_{n}$ of $(a_{j})_{j}$ and a point $a_{1}\in A$ such that $a_{j_{n}}\to a_{1}$ in $(U,F_{m},\ast )$. Then, by Lemma 3.3, there exists $b_{1}\in B$ such that

$$\begin{aligned} F_{m}(a_{1},b_{1},t)=F_{m}(a_{1},B,t) \quad\text{for } t\gg \theta. \end{aligned}$$

Now, by Proposition 3.1,

$$\begin{aligned} F_{m}(a_{j_{n}},b_{1},t_{j_{n}})\to F_{m}(a_{1},b_{1},t), \quad\text{as } n\to \infty. \end{aligned}$$

Therefore, for given $\delta >0$, there exists $n_{0}\in \mathbb{N}$ such that, for all $n\geq n_{0}$,

$$\begin{aligned} F_{m}(a_{1},b_{1},t)< \delta +F_{m}(a_{j_{n}},b_{1},t_{j_{n}}). \end{aligned}$$

Hence,

$$\begin{aligned} \inf_{a^{*}\in A}F_{m}\bigl(a^{*},B,t\bigr)\leq F_{m}(a_{1},b_{1},t)< \delta +F_{m}(a_{j_{n}},B,t_{j_{n}})= \delta +\inf_{a^{*}\in A}F_{m}\bigl(a^{*},B,t_{j_{n}} \bigr) \end{aligned}$$

for all $n\geq n_{0}$. Consequently, $t\longmapsto \inf_{a^{*}\in A}F_{m}(a^{*},B,t)$ is a lower semi-continuous function in $\operatorname{int}(P)$. It completes the proof. □

Remark 3.7

Note that Proposition 3.6 showed that, for any $A,B\in \mathbb{K}(U)$, $t\longmapsto \inf_{b^{*}\in B}F_{m}(A,b^{*},t)$ is a continuous function in $\operatorname{int}(P)$.

Hausdorff fuzzy cone metric on $\mathbb{K}(U)$: Let $(U,F_{m},\ast )$ be an FCM-space. Then we define a function $F_{H}$ on $\mathbb{K}(U)\times \mathbb{K}(U)\times \operatorname{int}(P)$ by

$$\begin{aligned} F_{H}(A,B,t)=\min \Bigl\{ \inf_{b\in B}F_{m}(A,b,t), \inf_{a\in A}F_{m}(a,B,t) \Bigr\} \end{aligned}$$

(3.2)

for all $A,B\in \mathbb{K}(U)$ and $t\gg \theta$.

Lemma 3.8

Let $(U,F_{m},\ast )$ be an FCM-space, $\mu \in U$, $A\in \mathbb{K}(U)$, $B\in \mathbb{P}(U)$, and $s,t\gg \theta$. Then

$$\begin{aligned} F_{m}(\mu,B,t+s)\geq F_{m}(\mu,a_{\mu },t)\ast F_{m}(a_{\mu },b,s), \end{aligned}$$

where $a_{\mu }\in A$ satisfies $F_{m}(\mu,A,t)=F_{m}(\mu,a_{\mu },t)$.

Proof

First, we note that an element $a_{\mu }\in A$ satisfying $F_{m}(\mu,A,t)=F_{m}(\mu,a_{\mu },t)$ exists by Lemma 3.3. Now, for every $b\in B$, we have that

$$\begin{aligned} F_{m}(\mu,B,t+s)\geq F_{m}(\mu,b,t+s)\geq F_{m}(\mu,a_{\mu },t)\ast F_{m}(a_{\mu },b,s). \end{aligned}$$

Thus, by the continuity of ∗,

$$\begin{aligned} F_{m}(\mu,B,t+s)\geq F_{m}(\mu,a_{\mu },t)\ast F_{m}(a_{\mu },b,s) \quad\text{for } s,t\gg \theta. \end{aligned}$$

 □

Theorem 3.9

Assume that $(U,F_{m},\ast )$ is an FCM-space. Then $(\mathbb{K}(U),F_{H},\ast )$ is an FCM-space.

Proof

Suppose that $A,B,C\in \mathbb{K}(U)$ and $s,t\gg \theta$. Then, by Lemma 3.5, there exist $a^{*}\in A$ and $b^{*}\in B$ such that

$$\begin{aligned} \inf_{a_{0}\in A}F_{m}(a_{0},B,t)=M \bigl(a^{*},B,t\bigr) \end{aligned}$$

and

$$\begin{aligned} \inf_{b_{0}\in B}F_{m}(A,b_{0},t)=F_{m} \bigl(A,b^{*},t\bigr) \end{aligned}$$

for $t\gg \theta$. Thus, $F_{H}(A,B,t)>0$.

In addition, we know that $F_{H}(A,B,t)=1$ if and only if $A=B$, and hence $F_{H}$ is symmetric, that is,

$$\begin{aligned} F_{H}(A,B,t)=F_{H}(B,A,t) \quad\text{for } t\gg \theta. \end{aligned}$$

Moreover, we note that, by Lemma 3.8 and by the continuity of ∗, we have that

$$\begin{aligned} \inf_{a_{0}\in A}F_{m}(a_{0},C,t+s)\geq \inf _{a_{0}\in A}F_{m}(a_{0},B,t) \ast \inf _{a_{0}\in A}F_{m}(b_{a_{0}},C,s) \end{aligned}$$

for $s,t\gg \theta$. Since $\{b_{a_{0}}: a_{0}\in A\}\subseteq B$ such that

$$\begin{aligned} \inf_{a_{0}\in A}F_{m}(b_{a_{0}},C,s)\geq \inf _{b_{0}\in B}F_{m}(b_{0},C,s) \end{aligned}$$

for $s\gg \theta$, we have

$$\begin{aligned} \inf_{a_{0}\in A}F_{m}(a_{0},C,t+s)\geq \inf _{a_{0}\in A}F_{m}(a_{0},B,t) \ast \inf _{b_{0}\in B}F_{m}(b_{0},C,s) \end{aligned}$$

for $s,t\gg \theta$. Similarly, we get that

$$\begin{aligned} \inf_{c_{0}\in C}F_{m}(A,c_{0},t+s)\geq \inf _{b_{0}\in B}F_{m}(A,b_{0},,t) \ast \inf _{c_{0}\in C}F_{m}(B,c_{0},s). \end{aligned}$$

It follows that

$$\begin{aligned} F_{H}(A,C,t+s)\geq F_{H}(A,B,t)\ast F_{H}(B,C,s). \end{aligned}$$

Finally, the continuity of the function $t\longmapsto F_{H}(A,B,t)$ on the cone is a direct consequence of Proposition 3.6 and Remark 3.7. We conclude that $(\mathbb{K}(U),F_{H},\ast )$ is an FCM-space. □

4 Set-valued mapping results in FCM-spaces

In this section, we prove a fixed point theorem for set-valued mappings in FCM-spaces.

Theorem 4.1

Let $(U,F_{m},\ast )$ be a complete FCM-space and $G: U\to U$ be a set-valued mapping with nonempty compact values such that, for all $\mu,\nu \in U$ and $t\gg \theta$,

$$\begin{aligned} F_{H}\bigl(G\mu,G\nu,\delta \bigl(d(\mu,\nu,t)\bigr)t \bigr)\geq F_{m}(\mu,\nu,t) \ast F_{m}(\nu,G\mu,t), \end{aligned}$$

(4.1)

where $\delta:\operatorname{int}(P)\to [0,1)$ satisfies

$$\begin{aligned} \limsup_{r\to t^{+}}\delta (r)< 1 \quad\textit{for all } t\in [0, \infty ] \end{aligned}$$

and $d(\mu,\nu,t)=\frac{t}{F_{m}(\mu,\nu,t)}-t$. Moreover, we suppose that $(U,F_{m},\ast )$ satisfies (3.1) for some $\mu _{0}\in U$ and $\mu _{1}\in G\mu _{0}$. Then G has a fixed point in U.

Proof

First, we notice that, if A and B are nonempty compact subsets of a set U and $a\in A$, then by Lemma 3.3, there exists $b\in B$ such that

$$\begin{aligned} F_{H}(A,B,t)\leq \sup_{b\in B}F_{m}(a,b,t)=F_{m}(a,b,t) \end{aligned}$$

for $t\gg \theta$. Thus, given $\delta \leq F_{H}(A,B,t)$, there exists a point $b\in B$ such that $\delta \leq F_{m}(a,b,t)$.

Now, let us fix $\mu _{0}$ in U and $\mu _{1}\in G\mu _{0}$. If $G\mu _{0}=G\mu _{1}$, then $\mu _{1}\in G\mu _{1}$ and $\mu _{1}$ is a fixed point of G. The proof is completed. Otherwise, we may assume that $G\mu _{0}\neq G\mu _{1}$. Then, from (4.1), we have

$$\begin{aligned} F_{H}\bigl(G\mu _{0},G\mu _{1},\delta \bigl(d( \mu _{0},\mu _{1},t)\bigr)t\bigr)\geq F_{m}( \mu _{0},\mu _{1},t)\ast F_{m}(\mu _{1},G \mu _{0},t)\geq F_{m}(\mu _{0}, \mu _{1},t) \end{aligned}$$

for $t\gg \theta$. Since $\mu _{1}\in G\mu _{0}$ and G is a compact-valued mapping, then again by Lemma 3.3, there exists $\mu _{2}\in G\mu _{1}$ such that

$$\begin{aligned} F_{m}(\mu _{1},\mu _{2},t)&\geq F_{m} \bigl(\mu _{1},\mu _{2},\delta \bigl(d( \mu _{0}, \mu _{1},t)\bigr)t\bigr) \\ &=\sup_{r\in G\mu _{1}}F_{m}\bigl(\mu _{1},r,\delta \bigl(d(\mu _{0},\mu _{1},t)\bigr)t\bigr) \\ &\geq F_{H}\bigl(G\mu _{0},G\mu _{1},\delta \bigl(d(\mu _{0},\mu _{1},t)\bigr)t\bigr) \geq F_{m}(\mu _{0},\mu _{1},t) \end{aligned}$$

for $t\gg \theta$. Similarly,

$$\begin{aligned} F_{m}(\mu _{2},\mu _{3},t)\geq F_{m}( \mu _{1},\mu _{2},t) \quad\text{for } t\gg \theta. \end{aligned}$$

By induction, we choose a sequence $(\mu _{n})_{n\geq 0}$ in U such that $\mu _{n}\in G\mu _{n-1}$. If $G\mu _{n-1}=G\mu _{n}$ for some n, then $\mu _{n}\in G\mu _{n}$, and so $\mu _{n}$ is a fixed point of G. The proof is completed. Otherwise, we may assume that $G\mu _{n-1}\neq G\mu _{n}$. Then from (4.1) we have

$$\begin{aligned} F_{m}(\mu _{n},\mu _{n+1},t)&\geq F_{m}\bigl(\mu _{n},\mu _{n+1},\delta \bigl(d( \mu _{n-1},\mu _{n},t)\bigr)t\bigr) \\ &=\sup_{r\in G\mu _{n}}F_{m}\bigl(\mu _{n},r,\delta \bigl(d(\mu _{n-1},\mu _{n},t)\bigr)t\bigr) \\ &\geq F_{H}\bigl(G\mu _{n-1},G\mu _{n},\delta \bigl(d(\mu _{n-1},\mu _{n},t)\bigr)t\bigr) \\ &\geq F_{m}(\mu _{n-1},\mu _{n},t)\ast F_{m}(\mu _{n},G\mu _{n-1},t) \\ &\geq F_{m}(\mu _{n-1},\mu _{n},t) \quad\text{for } t \gg \theta. \end{aligned}$$

Hence, $(F_{m}(\mu _{n},\mu _{n+1},t))_{n}$ is a nondecreasing sequence. Thus, $(d(\mu _{n},\mu _{n+1},t))_{n}$ is a positive nonincreasing sequence, and so it is convergent to some constant, say $\xi \geq 0$. Recall that

$$\begin{aligned} \limsup_{n\to \infty }\delta \bigl(d(\mu _{n},\mu _{n+1},t)\bigr)\leq \limsup_{ \varepsilon \to t^{+}} \delta (\varepsilon )< 1. \end{aligned}$$

(4.2)

Then there are $\beta <1$ and $n_{0}\in \mathbb{N}$ such that

$$\begin{aligned} \delta \bigl(d(\mu _{n},\mu _{n+1},t)\bigr)< \beta, \quad \forall n>n_{0}, t\gg \theta. \end{aligned}$$

(4.3)

Since $F_{m}(\mu,\nu,.)$ is nondecreasing, we have from (4.1) and (4.3) that, for $t\gg \theta$,

$$\begin{aligned} F_{m}(\mu _{n},\mu _{n+1},\beta t)\geq F_{m}\bigl(\mu _{n},\mu _{n+1}, \delta \bigl(d(\mu _{n-1},\mu _{n},t)\bigr)t\bigr)\geq F_{m}(\mu _{n-1},\mu _{n},t). \end{aligned}$$

Thus, we get that

$$\begin{aligned} F_{m}(\mu _{n},\mu _{n+1},\beta t)\geq F_{m}(\mu _{n-1},\mu _{n},t) \quad\text{for } t\gg \theta. \end{aligned}$$

Hence, by Lemma 3.2, we conclude that $(\mu _{n})$ is a Cauchy sequence in U. Since $(U,F_{m},\ast )$ is complete, there exists $u\in U$ such that

$$\begin{aligned} \lim_{n\to \infty }F_{m}(\mu _{n},u,t)=1 \quad\text{for } t\gg \theta. \end{aligned}$$

(4.4)

This implies that

$$\begin{aligned} \lim_{n\to \infty }d(\mu _{n},u,t)=0 \quad\text{for } t\gg \theta. \end{aligned}$$

Therefore,

$$\begin{aligned} \limsup_{n\to \infty }\delta \bigl(d(\mu _{n},u,t)\bigr)\leq \limsup_{ \varepsilon \to 0^{+}}\delta (\varepsilon )< 1. \end{aligned}$$

Then there exists $\beta <\xi <1$ such that

$$\begin{aligned} \limsup_{n\to \infty }\delta \bigl(d(\mu _{n},u,t)\bigr)< \xi\quad \text{for } t \gg \theta. \end{aligned}$$

Now, we have to show that $u\in Gu$. Since $\mu _{n+1}\in G\mu _{n}$, one writes

$$\begin{aligned} F_{m}(\mu _{n+1},Gu, t)&\geq F_{H}(G\mu _{n},Gu,\xi t) \\ &\geq F_{H}(G\mu _{n},Gu,\beta t) \\ &\geq F_{H}\bigl(G\mu _{n},Gu,\delta \bigl(d(\mu _{n},u,t)\bigr)t\bigr) \\ &\geq F_{m}(\mu _{n},u,t)\ast F_{m}(u,G\mu _{n},t) \\ &\geq F_{m}(\mu _{n},u,t)\ast F_{m}(u,\mu _{n+1},t)\to 1\ast 1=1, \quad\text{as } n\to \infty, \end{aligned}$$

for $t\gg \theta$. Hence, we get that

$$\begin{aligned} \lim_{n\to \infty }\sup_{r\in Gu}F_{m}(\mu _{n+1},r,t)=1 \quad\text{for } t\gg \theta. \end{aligned}$$

Thus, there exists a sequence $(r_{n})$ in Gu such that

$$\begin{aligned} \lim_{n\to \infty }F_{m}(\mu _{n},r_{n},t)=1 \quad\text{for } t\gg \theta. \end{aligned}$$

(4.5)

Now, by Definition 2.3(iii), we have that

$$\begin{aligned} F_{m}(r_{n},u,2t)\geq F_{m}(r_{n}, \mu _{n},t)\ast F_{m} (\mu _{n},u,t) \quad\text{for } t \gg \theta, \end{aligned}$$

(4.6)

for each $n\in \mathbb{N}$. By using (4.4), (4.5) together with (4.6), we can get

$$\begin{aligned} \lim_{n\to \infty }F_{m}(r_{n},u,2t)=1 \quad\text{for } t\gg \theta. \end{aligned}$$

This implies that $\lim_{n\to \infty }r_{n}=u$. Since $r_{n}\to u$ and $r_{n}\in Gu$, using the fact that Gu is closed and compact, we get $u\in Gu$. □

Without δ mapping directly, we can get the following two corollaries from Theorem 4.1.

Corollary 4.2

Let $(U,F_{m},\ast )$ be a complete FCM-space and $G: U\to U$ be a set-valued mapping with nonempty compact values such that, for all $\mu,\nu \in U$ and $t\gg \theta$, it satisfies

$$\begin{aligned} F_{H}(G\mu,G\nu,\beta t)\geq F_{m}(\mu, \nu,t)\ast F_{m}(\nu,G \mu,t), \end{aligned}$$

(4.7)

where $\beta \in (0,1)$. Furthermore, we assume that $(U,F_{m},\ast )$ satisfies (3.1) for some $\mu _{0}\in U$ and $\mu _{1}\in G\mu _{0}$. Then G has a fixed point in U.

Corollary 4.3

Let $(U,F_{m},\ast )$ be a complete FCM-space and $G: U\to U$ be a set-valued mapping with nonempty compact values such that, for all $\mu,\nu \in U$ and $t\gg \theta$, it satisfies

$$\begin{aligned} F_{H}(G\mu,G\nu,\beta t)\geq F_{m}(\mu, \nu,t), \end{aligned}$$

(4.8)

where $\beta \in (0,1)$. Furthermore, we assume that $(U,F_{m},\ast )$ satisfies (3.1) for some $\mu _{0}\in U$ and $\mu _{1}\in G\mu _{0}$. Then G has a fixed point in U.

5 Multi-valued contraction results in FCM-spaces

In this section, we present some fixed point results for multi-valued contractions in FCM-spaces. Further, we present a fixed point theorem for rational type multi-valued contractions. We present some illustrative examples.

Let $G: U\to 2^{U}$ be a multi-valued map. Consider $g(\mu )=F_{m}(\mu,G\mu,t)$ for $t\gg \theta$. For $\alpha \in (0,1)$, we take the set

$$\begin{aligned} J_{\alpha }^{\mu }= \bigl\{ \nu \in G \mu;F_{m}(\mu,\nu,t)\geq F_{m}( \mu,G\mu,\alpha t) \bigr\} . \end{aligned}$$

(5.1)

Theorem 5.1

Let $(U,F_{m},\ast )$ be a complete FCM-space and $G: U\to \mathbb{K}(U)$ be a multi-valued map. If there exists a constant $\beta \in (0,1)$ such that, for any $\mu \in U$, there is $\nu \in j_{\alpha }^{\mu }$, so that

$$\begin{aligned} F_{m}(\nu,G\nu,\beta t)\geq F_{m}(\mu, \nu,t)\ast F_{m}(\nu,G\mu,t) \end{aligned}$$

(5.2)

for $t\gg \theta$. Suppose that $(U,M,\ast )$ verifies (3.1) for some $\mu _{0}\in U$. Then G has a fixed point in U, provided $\beta <\alpha$ and g is upper semi-continuous.

Proof

Since $G(\mu )\in \mathbb{K}(U)$, by Lemma 3.3, $J_{\alpha }^{\mu }$ is nonempty for all μ in U and $\alpha \in (0,1)$. Let us fix $\mu _{0}$ in U, so there exists $\mu _{1}\in J_{\alpha }^{\mu _{0}}$, that is, $\mu _{1}\in G\mu _{0}$ such that

$$\begin{aligned} F_{m}(\mu _{1},G\mu _{1},\beta t)&\geq F_{m}(\mu _{0},\mu _{1},t) \ast F_{m}(\mu _{1},G\mu _{0},t)\geq F_{m}(\mu _{0},\mu _{1},t) \end{aligned}$$

for $t\gg \theta$. Similarly, for $\mu _{1}$ in U, there exists $\mu _{2}\in J_{\alpha }^{\mu _{1}}$, that is, $\mu _{2}\in G\mu _{1}$, which satisfies

$$\begin{aligned} F_{m}(\mu _{2},G\mu _{2},\beta t)&\geq F_{m}(\mu _{1},\mu _{2},t) \ast F_{m}( \mu _{2},G\mu _{1},t)\geq F_{m}(\mu _{1},\mu _{2},t) \end{aligned}$$

for $t\gg \theta$. By induction, we obtain a sequence $(\mu _{i})_{i\geq 0}$ in U such that there exists $\mu _{i+1}\in J_{\alpha }^{\mu _{i}}$, that is, $\mu _{n+1}\in G\mu _{i}$, which satisfies

$$\begin{aligned} F_{m}(\mu _{i+1},G\mu _{i+1},\beta t)&\geq F_{m}(\mu _{i},\mu _{i+1},t) \ast F_{m}(\mu _{i+1},G\mu _{i},t)\geq F_{m}( \mu _{i},\mu _{i+1},t) \end{aligned}$$

(5.3)

for $t\gg \theta$. On the other hand, $\mu _{i+1}\in J_{\alpha }^{\mu _{i}}$, which gives that

$$\begin{aligned} F_{m}(\mu _{i},\mu _{i+1},t)\geq F_{m}(\mu _{i},G\mu _{i},\alpha t) \quad\text{for } t \gg \theta. \end{aligned}$$

(5.4)

From (5.3) and (5.4), we get that

$$\begin{aligned} F_{m}(\mu _{i+1},G\mu _{i+1},\beta t)\geq F_{m}(\mu _{i},G\mu _{i}, \alpha t) \quad\text{for } t \gg \theta, \end{aligned}$$

i.e.,

$$\begin{aligned} F_{m}(\mu _{i+1},G\mu _{i+1},t)\geq F_{m} \biggl(\mu _{i},G\mu _{i}, \frac{\alpha }{\beta } t \biggr) \quad\text{for } t\gg \theta. \end{aligned}$$

(5.5)

Let $a=\frac{\beta }{\alpha }$, then (5.5) can be expressed as follows:

$$\begin{aligned} F_{m}(\mu _{i},\mu _{i+1},t)&\geq F_{m} \biggl(\mu _{i-1},\mu _{i}, \frac{1}{a}t \biggr)\geq F_{m} \biggl(\mu _{i-2},\mu _{i-1}, \frac{1}{a^{2}}t \biggr) \geq \cdots \\ &\geq F_{m} \biggl(\mu _{0},\mu _{1}, \frac{1}{a^{i}}t \biggr) \end{aligned}$$

(5.6)

for $t\gg \theta$, $i\in \mathbb{N}$, and $a\in (0,1)$. Choose a constant $b>1$ such that $ab<1$ and $\sum_{n=0}^{\infty }\frac{1}{b^{n}}<1$, i.e., $\sum_{n=i}^{j-1}\frac{1}{b^{n}}<1$. Then, for all $j>i$, we get that

$$\begin{aligned} \biggl(\frac{1}{b^{i}}+\frac{1}{b^{i+1}}+\cdots + \frac{1}{b^{j-2}}+ \frac{1}{b^{j-1}} \biggr)t< t, \end{aligned}$$

(5.7)

where $i,j\in \mathbb{N}$. Then we have

$$\begin{aligned} F_{m}(\mu _{i},\mu _{j},t)&\geq F_{m} \biggl(\mu _{i},\mu _{j},t \biggl( \frac{1}{b^{i}}+\frac{1}{b^{i+1}}+\cdots +\frac{1}{b^{j-2}}+ \frac{1}{b^{j-1}} \biggr) \biggr) \\ &\geq F_{m} \biggl(\mu _{i},\mu _{i+1}, \frac{t}{b^{i}} \biggr)\ast F_{m} \biggl(\mu _{i+1},\mu _{i+2},\frac{t}{b^{i+1}} \biggr)\ast \cdots \ast F_{m} \biggl(\mu _{j-1},\mu _{j},\frac{t}{b^{j-1}} \biggr) \\ &\geq F_{m} \biggl(\mu _{0},\mu _{1}, \frac{t}{(ab)^{i}} \biggr)\ast F_{m} \biggl(\mu _{0},\mu _{1},\frac{t}{(ab)^{i+1}} \biggr)\ast \cdots \ast F_{m} \biggl(\mu _{0},\mu _{1},\frac{t}{(ab)^{j-1}} \biggr) \\ &\geq \ast _{n=i}^{\infty } \biggl( F_{m} \biggl(\mu _{0},\mu _{1}, \frac{t}{(ab)^{n}} \biggr) \biggr)\to 1, \quad\text{as } i\to \infty, \end{aligned}$$

(5.8)

for $t\gg \theta$. By Lemma 3.2, it is proved that $(\mu _{i})$ is a Cauchy sequence in U. Since U is complete, there exists $\mu \in U$ such that $\mu _{i}\to \mu$ as $i\to \infty$. In view of (5.3) and (5.4), it is clear that $g(\mu _{i})=F_{m}(\mu _{i},G\mu _{i},t)$ is an increasing function and converges to 1. Since g is upper semi-continuous, we have

$$\begin{aligned} 1\geq g(\mu )\geq \limsup_{n\to \infty } g(\mu _{n})= 1. \end{aligned}$$

This implies that $g(\mu )=1$, so that $F_{m}(\mu,G\mu,t)=1$. Hence, by using Lemma 3.3, we get that $\mu \in G\mu$. Hence, the proof is completed. □

Corollary 5.2

Let $(U,F_{m},\ast )$ be a complete FCM-space and $G: U\to \mathbb{K}(U)$ be a multi-valued mapping. If there exists a constant $\beta \in (0,1)$ such that, for any $\mu \in U$, we have $\nu \in j_{\alpha }^{\mu }$ with

$$\begin{aligned} F_{m}(\nu,G\nu,\beta t)\geq F_{m}(\mu, \nu,t) \end{aligned}$$

(5.9)

for $t\gg \theta$. Suppose that $(U,F_{m},\ast )$ satisfies (3.1) for some $\mu _{0}\in U$. Then G has a fixed point in U provided that $\beta <\alpha$ and g is upper semi-continuous.

In a special case, we get the following corollary of Kiany et al. [19].

Corollary 5.3

([19])

Let $(U,F_{m},\ast )$ be a complete FCM-space and $G: U\to \mathbb{K}(U)$ be a multi-valued map. Suppose that there exists $\beta \in (0,1)$ such that

$$\begin{aligned} F_{H}(G\mu,G\nu,\beta t)\geq F_{m}(\mu, \nu,t) \end{aligned}$$

(5.10)

for all $\mu,\nu \in U$ and $t\gg \theta$. Moreover, assume that $(U,F_{m},\ast )$ satisfies (3.1) for some $\mu _{0}\in U$ and $\mu _{1}\in G\mu _{0}$. Then G has a fixed point in U.

Remark 5.4

Corollary 5.2 is the generalized form of Corollary 5.3. Suppose that G satisfies the conditions of Corollary 5.3, and if g is upper semi-continuous, then from (5.10) we obtain, for any $\mu \in U$, $\nu \in G\mu$, and $t\gg \theta$,

$$\begin{aligned} F_{m}(\nu,G\nu,\beta t)\geq F_{H}(G\mu,G\nu,\beta t)\geq F_{m}( \mu,\nu,t) \end{aligned}$$

for $t\gg \theta$. Hence, G verifies the conditions of Corollary 5.2 and the existence of a fixed point has been proved.

In the following (Example 5.5), we show that Corollary 5.2 is the generalized form of Corollary 5.3.

Example 5.5

Let $U= \{ \frac{1}{3},\frac{1}{9},\ldots,\frac{1}{3^{i}},\ldots \} \cup \{0,1\}$ and the fuzzy metric $F_{m}: U^{2}\times (0,\infty )\to [0,1]$ be defined as

$$\begin{aligned} F_{m}(\mu,\nu,t)=\frac{t}{t+d(\mu,\nu )}, \quad\text{where } d(\mu, \nu )= \vert \mu -\nu \vert , \forall \mu,\nu \in U, t>0. \end{aligned}$$

Then $(U,F_{m},\ast )$ is a complete FCM-space, where $\ast:[0,1]^{2}\to [0,1]$ is defined as $a\ast b=ab$.

Let the multi-valued mapping $G:U\to \mathbb{K}(U)$ be defined as

$$\begin{aligned} G\mu = \textstyle\begin{cases} \{ \frac{1}{3^{i}},1 \} &\text{if } \mu = \frac{1}{3^{i}} \text{ for } i\geq 0, \\ \{ 0,{\frac{1}{3}} \} &\text{if } \mu =0. \end{cases}\displaystyle \end{aligned}$$

Since

$$\begin{aligned} \lim_{i\to \infty }\ast _{j=i}^{\infty }M\bigl(\mu, \nu,tb^{j}\bigr)=M \biggl( \frac{1}{3^{i}},0,tb^{j} \biggr) =\lim_{i\to \infty }\ast _{j=i}^{ \infty } \frac{tb^{j}}{tb^{j}+\frac{1}{3^{i}}}=1, \end{aligned}$$

this shows that G satisfies (3.1). Moreover,

$$\begin{aligned} F_{H} \biggl(G \biggl(\frac{1}{3^{i}} \biggr),G(0),\beta t \biggr)= \frac{\beta t}{\beta t+H (G (\frac{1}{3^{i}} ),G(0) )}= \frac{\beta t}{\beta t+\frac{1}{3}} \end{aligned}$$

and

$$\begin{aligned} F_{m} \biggl(\frac{1}{3^{i}},0,t \biggr)= \frac{t}{t+d (\frac{1}{3^{i}},0 )}= \frac{t}{t+\frac{1}{3^{i}}}. \end{aligned}$$

There does not exist any $\beta \in (0,1)$ such that Corollary 5.3 is satisfied. If it exists, then we get

$$\begin{aligned} \frac{t}{t+1/3^{i}}\leq \frac{\beta t}{\beta t+1/3}. \end{aligned}$$

This implies that $\beta \geq 3^{i-1}$, which is a contradiction. On the other hand,

$$\begin{aligned} g(\mu )=F_{m}\bigl(\mu,G(\mu ),t\bigr)=\frac{t}{t+d(\mu,G(\mu ))}= \textstyle\begin{cases} \frac{1}{3^{i+1}} &\text{if } \mu =\frac{1}{3^{i}}, \\ 0 &\text{if } \mu =0, \end{cases}\displaystyle \end{aligned}$$

is continuous and so there exists $\nu \in J_{\frac{3}{4}}^{\mu }$ for any μ such that

$$\begin{aligned} d\bigl(\nu,G(\nu )\bigr)=\frac{1}{3} d(\mu,\nu )\leq \frac{2}{3} d(\mu,\nu )\quad \Rightarrow\quad \frac{3}{2} d(\nu,G\nu )\leq d(\mu,\nu ). \end{aligned}$$

Hence, there exists $\beta =\frac{2}{3}<\frac{3}{4}$ such that

$$\begin{aligned} F_{m} \biggl(\nu,G(\nu ),\frac{2}{3}t \biggr) = \frac{\frac{2}{3}t}{\frac{2}{3}t+d(\nu,G(\nu ))}= \frac{t}{t+\frac{3}{2}d(\nu,G(\nu ))}\geq \frac{t}{t+d(\mu,\nu )}=F_{m}( \mu,\nu,t). \end{aligned}$$

Then, by Corollary 5.2, we can get the existence of a fixed point of G in U.

Now, we will deal with rational type multi-valued contractions in FCM-spaces. For this, let $G: U\to 2^{U}$ be a multi-valued map. Define $g(\mu )=F_{m}(\mu,G\mu,t)$ for $t\gg \theta$. For $\alpha \in (0,1)$, define the set

$$\begin{aligned} J_{\alpha }^{\mu }= \biggl\{ \nu \in G\mu; \frac{1}{F_{m}(\mu,\nu,t)}-1 \leq \frac{1}{F_{m}(\mu,G\mu,\alpha t)}-1 \biggr\} . \end{aligned}$$

(5.11)

Theorem 5.6

Let $(U,F_{m},\ast )$ be a complete FCM-space and $G: U\to \mathbb{K}(U)$ be a multi-valued map. If there exists a constant $\beta \in (0,1)$ such that, for any $\mu \in U$, there is $\nu \in j_{\alpha }^{\mu }$ so that

$$\begin{aligned} \frac{1}{F_{m}(\nu,G\nu,\beta t)}-1&\leq \frac{1}{F_{H}(G\mu,G\nu,\beta t)}-1 \\ &\leq \frac{F_{m}(\mu,\nu,t)\ast F_{m}(\nu,G\nu,t)}{F_{m}(\mu,G\mu,t)\ast F_{m}(\mu,G\nu,2t)\ast F_{m}(\nu,G\mu,2t)}-1 \end{aligned}$$

(5.12)

for $t\gg \theta$. Suppose that $(U,F_{m},\ast )$ satisfies (3.1) for some $\mu _{0}\in U$. Then G has a fixed point in U provided that $\beta <\alpha$ and g is upper semi-continuous.

Proof

Since $G(\mu )\in \mathbb{K}(U)$, by Lemma 3.3, we have that $J_{\alpha }^{\mu }$ is nonempty for any μ in U and $\alpha \in (0,1)$. Let us fix $\mu _{0}\in U$, so there exists $\mu _{1}\in J_{\alpha }^{\mu _{0}}$. Then, by (5.12), for $t\gg \theta$,

$$\begin{aligned} \frac{1}{F_{m}(\mu _{1},G\mu _{1},\beta t)}-1&\leq \frac{1}{F_{H}(G\mu _{0},G\mu _{1},t)}-1 \\ &\le \frac{F_{m}(\mu _{0},\mu _{1},t)\ast F_{m}(\mu _{1},G\mu _{1},t)}{F_{m}(\mu _{0},G\mu _{0},t)\ast F_{m}(\mu _{0},G\mu _{1},2t)\ast F_{m}(\mu _{1},G\mu _{0},2t)}-1 \\ &\le \frac{ F_{m}(\mu _{1},\mu _{2},t)}{ F_{m}(\mu _{0},\mu _{2},2t)}-1 \end{aligned}$$

by using Definition 2.3(iii) and $F_{m}(\mu _{0},\mu _{2},2t)\geq F_{m}(\mu _{0},\mu _{1},t)\ast F_{m}( \mu _{1},\mu _{2},t)$ for $t\gg \theta$. After simplification, we get that

$$\begin{aligned} \frac{1}{F_{m}(\mu _{1},G\mu _{1},\beta t)}-1\leq \frac{1}{F_{m}(\mu _{0},\mu _{1}, t)}-1 \quad\text{for } t\gg \theta. \end{aligned}$$

Again for $\mu _{1}\in U$, there exists $\mu _{2}\in J_{\alpha }^{\mu _{1}}$. In view of (5.12),

$$\begin{aligned} \frac{1}{F_{m}(\mu _{2},G\mu _{2},\beta t)}-1 &\leq \frac{1}{F_{H}(G\mu _{1},G\mu _{2},t)}-1 \\ &\le \frac{F_{m}(\mu _{1},\mu _{2},t)\ast F_{m}(\mu _{2},G\mu _{2},t)}{F_{m}(\mu _{1},G\mu _{1},t)\ast F_{m}(\mu _{1},G\mu _{2},2t)\ast F_{m}(\mu _{2},G\mu _{1},2t)}-1 \\ &\le \frac{ F_{m}(\mu _{2},\mu _{3},t)}{ F_{m}(\mu _{1},\mu _{3},2t)}-1. \end{aligned}$$

Again by Definition 2.3(iii), $F_{m}(\mu _{1},\mu _{3},2t)\geq F_{m}(\mu _{1},\mu _{2},t)\ast F_{m}( \mu _{2},\mu _{3},t)$ for $t\gg \theta$. After simplification, we get that

$$\begin{aligned} \frac{1}{F_{m}(\mu _{2},G\mu _{2},\beta t)}-1\leq \frac{1}{F_{m}(\mu _{1},\mu _{2}, t)}-1\quad \text{for } t\gg \theta. \end{aligned}$$

Similarly, by induction, we obtain a sequence $(\mu _{i})_{i\geq 0}$ in U such that there exists $\mu _{i+1}\in J_{\alpha }^{\mu _{i}}$, then by (5.12)

$$\begin{aligned} \frac{1}{F_{m}(\mu _{i+1},G\mu _{i+1},\beta t)}-1 \leq \frac{1}{F_{m}(\mu _{i},\mu _{i+1},t)}-1 \quad\text{for } t\gg \theta. \end{aligned}$$

(5.13)

On the other hand, by (5.11) and $\mu _{i+1}\in J_{\alpha }^{\mu _{i}}$,

$$\begin{aligned} \frac{1}{F_{m}(\mu _{i},\mu _{i+1},t)}-1\leq \frac{1}{F_{m}(\mu _{i},G\mu _{i},\alpha t)}-1 \quad\text{for } t\gg \theta. \end{aligned}$$

(5.14)

From (5.13) and (5.14), we can obtain

$$\begin{aligned} \frac{1}{F_{m}(\mu _{i+1},G\mu _{i+1},\beta t)}-1\leq \frac{1}{F_{m}(\mu _{i},G\mu _{i},\alpha t)}-1 \end{aligned}$$

for $t\gg \theta$, that is,

$$\begin{aligned} \frac{1}{F_{m}(\mu _{i+1},G\mu _{i+1},t)}-1\leq \frac{1}{F_{m} (\mu _{i},G\mu _{i},\frac{\alpha }{\beta } t )}-1 \end{aligned}$$

(5.15)

for $t\gg \theta$. Let $a=\frac{\beta }{\alpha }$, then (5.15) can be expressed as follows:

$$\begin{aligned} \frac{1}{F_{m}(\mu _{i},\mu _{i+1},t)}-1&\leq \frac{1}{F_{m} (\mu _{i-1},\mu _{i},\frac{1}{a}t )}-1 \\ &\leq \cdots \leq \frac{1}{F_{m} (\mu _{0},\mu _{1},\frac{1}{a^{i}}t )}-1 \quad\text{for } t\gg \theta, \end{aligned}$$

(5.16)

for all $i\in \mathbb{N}$ and $a\in (0,1)$. Choose a constant $b>1$ such that $ab<1$ and $\sum_{n=0}^{\infty }\frac{1}{b^{n}}<1$, i.e., $\sum_{n=i}^{j-1}\frac{1}{b^{n}}<1$. Then, for all $j>i$, we get

$$\begin{aligned} \biggl(\frac{1}{b^{i}}+\frac{1}{b^{i+1}}+\cdots + \frac{1}{b^{j-2}}+ \frac{1}{b^{j-1}} \biggr)t< t, \end{aligned}$$

(5.17)

where $i,j\in \mathbb{N}$. Then we have

$$\begin{aligned} \frac{1}{F_{m}(\mu _{i},\mu _{j},t)}-1 &\leq \frac{1}{F_{m} (\mu _{i},\mu _{j},t (\frac{1}{b^{i}}+\frac{1}{b^{i+1}}+\cdots +\frac{1}{b^{j-2}} +\frac{1}{b^{j-1}} ) )}-1 \\ &\leq \frac{1}{F_{m} (\mu _{i},\mu _{i+1},\frac{t}{b^{i}} )\ast F_{m} (\mu _{i+1},\mu _{i+2},\frac{t}{b^{i+1}} )\ast \cdots \ast F_{m} (\mu _{j-1},\mu _{j},\frac{t}{b^{j-1}} )}-1 \\ &\leq \frac{1}{F_{m} (\mu _{0},\mu _{1},\frac{t}{(ab)^{i}} )\ast F_{m} (\mu _{0},\mu _{1},\frac{t}{(ab)^{i+1}} )\ast \cdots \ast F_{m} (\mu _{0},\mu _{1},\frac{t}{(ab)^{j-1}} )}-1 \\ &\leq \frac{1}{\ast _{n=i}^{\infty } F_{m} (\mu _{0},\mu _{1},\frac{t}{(ab)^{n}} )}-1 \end{aligned}$$

(5.18)

for $t\gg \theta$. By Lemma 3.2, we have that

$$\begin{aligned} \lim_{i,j\to \infty }F_{m}(\mu _{i},\mu _{j},t)=1 \quad\text{for } t \gg \theta. \end{aligned}$$

It is proved that $(\mu _{i})$ is a Cauchy sequence in U. Since U is complete, there exists $\mu \in U$ such that $\mu _{i}\to \mu$ as $i\to \infty$. In view of (5.13) and (5.14), it is clear that $g(\mu _{i})=F_{m}(\mu _{i},G\mu _{i},t)$ is an increasing function and converges to 1. Since g is upper semi-continuous, we have

$$\begin{aligned} 1=\limsup_{n\to \infty } g(\mu _{n})\leq g(\mu )\leq 1. \end{aligned}$$

This implies that $g(\mu )=1$, so that $F_{m}(\mu,G\mu,t)=1$. Hence, by using Lemma 3.3, we get that $\mu \in G\mu$. □

Directly from Theorem 5.6, we get the following corollary.

Corollary 5.7

Let $(U,F_{m},\ast )$ be a complete FCM-space and $G: U\to \mathbb{K}(U)$ be a multi-valued map. If there exists a constant $\beta \in (0,1)$ such that

$$\begin{aligned} \frac{1}{F_{m}(\nu,G\nu,\beta t)}-1\leq \frac{1}{F_{m}(\mu,\nu,t)}-1 \end{aligned}$$

(5.19)

for all $\mu,\nu \in U$ and $t\gg \theta$. Moreover, assume that $(U,F_{m},\ast )$ satisfies (3.1) for some $\mu _{0}\in U$ and $\mu _{1}\in G\mu _{0}$. Then G has a fixed point in U.

Example 5.8

Let $U=\{0.4, 0.4^{2},\ldots, 0.4^{i}, \ldots \}\cup \{0,1\}$. Let $G:U\to \mathbb{K}(U)$ be defined as

$$\begin{aligned} G\mu = \textstyle\begin{cases} \{ 0.4^{i},1 \}, &\text{if } \mu =0.4^{i}, \text{ for } i\geq 0, \\ \{ 0,0.4 \}, &\text{if } \mu =0. \end{cases}\displaystyle \end{aligned}$$

Since

$$\begin{aligned} \lim_{i\to \infty }\ast _{j=i}^{\infty }M\bigl(\mu, \nu,tb^{j}\bigr)=M \bigl(0.4^{i},0,tb^{j} \bigr) = \lim_{i\to \infty }\ast _{j=i}^{\infty } \frac{tb^{j}}{tb^{j}+0.4^{i}}=1, \end{aligned}$$

which shows that G satisfies (3.1). By a direct calculation as discussed in Example 5.5, we get $\beta \geq 0.4^{i-1}$ which is a contradiction to the fact that $\beta \geq 0.4^{i-1}\to 0$, as $i\to \infty$, where $\beta \in (0,1)$. On the other hand, we define

$$\begin{aligned} g(\mu )=F_{m}\bigl(\mu,G(\mu ),t\bigr)=\frac{t}{t+d(\mu,G(\mu ))}= \textstyle\begin{cases} 0.4^{i+1}, &\text{if } \mu = 0.4^{i}, \\ 0, &\text{if } \mu =0. \end{cases}\displaystyle \end{aligned}$$

It is continuous and so there exists $\nu \in J_{0.3}^{\mu }$ for any μ such that

$$\begin{aligned} d\bigl(\nu,G(\nu )\bigr)=0.3 d(\mu,\nu )\leq 0.5 d(\mu,\nu ). \end{aligned}$$

That is,

$$\begin{aligned} \frac{ d(\nu,G\nu )}{0.5}\leq d(\mu,\nu ). \end{aligned}$$

(5.20)

Hence, there exists $\beta =0.5< 0.8$, and from (5.20) we get, for $t\gg \theta$,

$$\begin{aligned} \frac{1}{F_{m} (\nu,G(\nu ),0.5 t )}-1 = \frac{d(\nu, G\nu )}{0.5 t}\leq \frac{d(\mu,\nu )}{t}= \frac{1}{F_{m}(\mu,\nu,t)}-1. \end{aligned}$$

Then the existence of a fixed point follows from Corollary 5.7.