#### Abstract

We establish some new fixed point results for order-closed multivalued mappings in complete metric spaces endowed with a partial order.

#### 1. Introduction

In 1976, an order relation was defined in metric spaces [1]. Later, many researchers proved various fixed point results in this setting (see [2–11]), while the authors in [4, 5, 9] considered coupled questions for so-called monotone conditions. Zhang [12] proved some (coupled) fixed point theorems for multivalued mappings with monotone conditions in metric spaces with a partial order. Since then, Agarwal and Khamsi [13] extended Caristi’s fixed point to vector-valued metric spaces. Also, Chung [14] considered nonlinear contraction mappings. Nadler [15] defined multivalued contractions, and Assad and Kirk [16] proved fixed point theorems for set-valued mappings (see also the related works [17–22]).

In this paper, we present some fixed point results in ordered metric spaces for order-closed multivalued operators. First, we need some facts.

Lemma 1. *[23] Let be a metric space and be a functional. Consider a nondecreasing, continuous, and subadditive function so that Take the relation “” on given as
**Then, “” is a partial order relation on . Apparently, then .*

Here, we state some definitions. Let be a topological space. Denote by the family of nonempty subsets of . Let be a partial order on .

*Definition 2. *[24] Given two nonempty subsets of .

Note that and are different relations between and (see Remark 114 of [24]). Also, , , and are not partial orders on (see Remark 2 of [24]).

*Definition 3. *If satisfies or , then is called a monotone sequence.

*Definition 4. *A multivalued operator is said to be order-closed if, for monotone sequences and , we have , , and imply .

*Definition 5. *A function is said to be order upper (lower) semicontinuous, if, for a monotone sequence with , we have

Note that an upper (lower) semicontinuous function is an order upper (lower) semicontinuous. But the converse is not true (see Remark 3 of [24]).

#### 2. Main Results

Let be a metric space. For , define the partial order “” on induced by and in Lemma 1.

Theorem 6. *Let be a complete ordered metric space and be a bounded below function. Suppose that is order-closed with respect to “” so that
**Then, there is a monotone sequence , so that for , and converges to , which is a fixed point of . If in addition, is order lower semicontinuous on , then for each .*

*Proof. *By the condition , take . From , there is so that . Again, from , there is with . Continuing this procedure, we have an increasing sequence so that . Therefore,
That is, the real sequence is decreasing, so it is a convergent sequence because is bounded from below on .

Using Remark 3 of [25] and Remark 2 of [26], we find that
provided that is subadditive. Thus,
By (7), there are and so that
Since is nondecreasing, we have for each .

Let . Then
i.e.,
Therefore, we have
which together with (8) implies that
Note that is convergent; then, there is so that for all ,
Moreover, by (12), we get
The convergence of implies that is a Cauchy sequence. By the completeness of , there is so that as . Since is order-closed, is monotone and . Consequently,
i.e., is a fixed point of .

If is order lower semicontinuous on , by definition of “,” then we have for each It yields that . The proof is completed.

The proof of the following theorem carries over in the same manner as for Theorem 6.

Theorem 7. *Let be a complete ordered metric space and be a bounded below function. Suppose that is order-closed with respect to “.” Assume that
**Then, there is a monotone sequence , , , such that converges to , which is a fixed point of . If, in addition, is order upper semicontinuous on , then for all .*

Theorem 8. *Let be a complete ordered metric space and be bounded below function. Suppose that is order-closed with respect to “.” Assume that
*(i)*for each with *(ii)*there exists such that, .**Then, has a fixed point , and there is a sequence with for , such that . Moreover, if is order lower semicontinuous, then for all .*

*Proof. *Since , by , we can choose so that . This implies that , by definition of , there is so that . Continuing this procedure, we can find an increasing sequence such that . The rest of the proof is similar as in Theorem 6.

The following supports Theorem 8.

*Example 9. *Let and , for . The metric space is complete. Consider the multivalued mapping given as
where is the set of rational numbers. Let and . Note that is bounded from below. Take the order induced by , that is, given as follows:
Mention that verifies the following assertions:
(a)for each with , we have (b)(c) is order-closed on .Hence, has a fixed point on , which is .

Theorem 10. *Let be a complete ordered metric space and be a bounded below function. Suppose is order-closed with respect to “.” If the following conditions hold:
*(i)*for each with *(ii)*there exists such that, **then has a fixed point . Also, there is a sequence with for any such that . Moreover, if is order upper semicontinuous, then for all .**The following example illustrates Theorem 10.*

*Example 11. *Let , , for and for each . Here, is a complete metric space, and is a bounded below function. Consider the order induced by :
Clearly, this partial order is the usual order on . Define by . It is obvious that satisfies the following:
(a)for each with (b)(c) is order-closed on .Hence, has a fixed point on , which is .

Now, a multivalued version of Theorem 2 of [27] may be obtained. Here, the considered multivalued mapping is not necessarily continuous.

Theorem 12. *Let be a complete ordered metric space and be a continuous function bounded below. Let be a multivalued mapping. Suppose that
**Then, has a fixed point.*

*Proof. *Set
We claim that has a maximum element. For a directed set , let be a totally ordered subset in . For with , the fact that yields . Due to the fact that is bounded below, is a convergent set in . Consider
As in proof of Theorem 6, is Cauchy in , which is complete, so converges to in . For ,
Hence, for each . By , for each , there is so that . By the compactness of , there is a convergence subset of . Assume that converges to . Take such that implies . One writes
So for all . Also,
So and . Thus, has an upper bound in .

By Zorn’s Lemma, there is a maximum element . also, there is so that . Using , there is so that . Hence, . The element is maximum in , so and . That is, is a fixed point of .

Theorem 13. *Let be a complete ordered metric space and be a continuous function bounded below. Let be a multivalued mapping. Assume that
**Then, has a fixed point.*

*Remark 14. *If is a continuous single-valued mapping in Theorem 6 (resp., Theorem 7), we can replace the condition by
and we can obtain the same result.

*Remark 15. *If in Theorem 8 (resp., Theorem 10), is assumed to be a continuous single-valued mapping, then we get the same result when replacing the conditions and by
(i) is monotone increasing (resp., decreasing), that is, for , we have (ii)there exists with (resp., ).If in addition is order upper (resp., lower) semicontinuous on , then is the smallest (resp., largest) fixed point of in (resp., ).

*Proof. *Let be a fixed point of in , i.e., . Since , we get . Hence, , i.e., . Suppose , then , i.e., . By a mathematical induction, we have for all , then
That is, .

Note that if we omit the continuity of and add above condition on , the results remain true.

#### Data Availability

The data used to support the findings of this study are available from the corresponding author upon request.

#### Conflicts of Interest

The authors declare that they have no competing interests regarding the publication of this paper.

#### Authors’ Contributions

All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.