A New Concept of Fixed Point in Metric and Normed Interval Spaces

The main aim of this paper is to propose the concept of so-called near fixed point and establish many types of near fixed point theorems in the set of all bounded and closed intervals in R . The concept of null set will be proposed in order to interpret the additive inverse element in the set of all bounded closed intervals. Based on the null set, the concepts of metric interval space and normed interval space are proposed, which are not the conventional metric and normed spaces. The concept of near fixed point is also defined based on the null set. In this case, we shall establish many types of near fixed point theorems in the metric and normed interval spaces.


Introduction
We denote by I the set of all bounded and closed intervals in R. The element in I is denoted by [a, b] where a, b ∈ R with a ≤ b. It is clear to see that the real number a ∈ R can also be regarded as an element [a, a] ∈ I. The topic of interval analysis has been studied for a long time. The detailed discussion can refer to the monographs [1][2][3][4][5].
Let T : I → I be a function from I into itself. We say that [a, b] ∈ I is a fixed point if and only if T([a, b]) = [a, b]. The well-known Banach contraction principle presents the fixed point of function T when (I, d) is taken to be a metric space. However (I, d) may not be a metric space for some special kinds of metric d. Then (I, d) is not a metric space. In this paper, we shall propose the so-called metric interval space (I, d) as shown above. The main purpose is to study the so-called near fixed point theorem in metric interval space (I, d). The concept of near fixed point proposed in this paper is the first attempt for studying the fixed point theory in interval space. Therefore, based on the limited knowledge of author, it seems that there are no appropriate articles that can be cited.
On the other hand, we shall also propose the concept of normed interval space. The Hahn-Banach theorem in normed interval spaces has been studied in Wu [6]. Recall that the (conventional) normed space is based on the vector space by referring to the monographs [7][8][9]. However, I cannot form a real vector space. The main reason is that there is no additive inverse element for each element in I. Therefore we cannot consider the (conventional) noremd space (I, · ). In this paper, we shall propose the normed interval space (I, · ) based on the concept of null set, although I is not a real vector space. We shall also study the near fixed point theorem in the normed interval space (I, · ).
In Section 2, since any element in I cannot have the additive inverse element, we propose the concept of null set to play the role for interpreting the additive inverse element. In Sections 3 and 4, we introduce the concepts of metric interval space (I, d) and normed interval space (I, · ). We also present many useful properties. In Section 5, we propose the concepts of limit and Cauchy sequences in (I, d) and (I, · ). In Sections 6 and 7, based on the concept of null set defined in Section 2, we can define the so-called near fixed point. We also present many kinds of near fixed point theorems in the metric interval space (I, d) and normed interval space (I, · ).

Interval Space
Let I be the set of all closed intervals in R. The addition is given by and the scalar multiplication is given by It is easy to see that I is not a (conventional) vector space under the above addition and scalar multiplication. The main reason is that the inverse element does not exist for any non-degenerated closed interval, which will be more clear from the following discussion.
It is clear to see that [0, 0] ∈ I is a zero element. However, for any [a, b] ∈ I, the substraction is not a zero element. In other words, the inverse element of [a, b] does not exist. In this paper, instead of considering the zero element, we define the null set as follows It is clear to see that We also see that Ω is generated by [−1, 1] based on the nonnegative scalar multiplication as shown below Ω = {k[−1, 1] : k ≥ 0}.
In this case, we say that [−1, 1] is a generator of the null set Ω. Now we have the following observations.

•
The distributive law for scalar addition doe not holds true in general; that is, for any [a, b] ∈ I and α, β ∈ R. • The distributive law for positive scalar addition holds true; that is, for any [a, b] ∈ I and α, β > 0.

•
The distributive law for negative scalar addition holds true; that is, for any [a, b] ∈ I and α, β < 0.
This shows [a, b] Ω = [e, f ], since Ω is closed under the addition. This completes the proof.
Proposition 1 says that the classes defined in (2) form the equivalence classes. In this case, the family I is called the quotient set of I. We also have that [c, d] In other words, the family of all equivalence classes form a partition of the whole set I. We also remark that the quotient set I is still not a (conventional) vector space. The reason is

Metric Interval Space
Let I be the set of all bounded closed intervals with the null set Ω. To study the near fixed point, we are going to consider the metric d defined on I × I.
Example 1. Let us define a nonnegative real-valued function d : Then (I, d) is not a (conventional) metric space. However, we are going to claim that (I, d) is a metric interval space such that d satisfies the null equality.
We are going to claim [a, b] and Let Therefore, from (5) and (6), we obtain ∈ Ω for some positive k 1 and k 2 .
(iv) For any [a, b], [c, d] ∈ I and k 1 , The verification is complete.

Normed Interval Space
Different kinds of normed interval spaces are proposed below. We shall also study the near fixed point theorems in the normed interval spaces.

Definition 2.
Given a nonnegative real-valued function · : I → R + , we consider the following conditions: We say that · satisfies the null condition when condition (iii) is replaced by [a, b] = 0 if and only if [a, b] ∈ Ω.

•
We say that (I, · ) is a pseudo-seminormed interval space if and only if conditions (i • ) and (ii) are satisfied. We also consider the following definitions: (using the null super-inequality for m times) [e, f ] (using the triangle inequality).
This completes the proof.
Proposition 3. The following statements hold true.
(i) Let (I, · ) be a pseudo-seminormed interval space such that · satisfies the null equality. For any Therefore, using the null equality, we have for some ω 3 ∈ Ω. Using the null super-inequality, null condition and (7), we have This completes the proof.

Cauchy Sequences
In this section, we are going to introduce the concepts of Cauchy sequences and completeness in the metric interval space and normed interval space.

Cauchy Sequences in Metric Interval Space
We first introduce the concept of limit in the metric interval space. The element [a, b] is called the limit of the sequence {[a n , b n ]} ∞ n=1 .
Then, by the triangle inequality (iii) in Definition 1, we have

Proposition 4. Suppose that d satisfies the null equality
Using the null equality, we obtain This completes the proof.
Inspired by the above result, we propose the following definition.
This completes the proof. The following result is not hard to prove.

Proposition 6.
Every convergent sequence in a metric interval space is a Cauchy sequence.

Example 3.
Continued from Example 1, we see that d satisfies the null equality, where the metric d is defined in (3). Now we want to show that this space is also complete. Suppose that {[a n , b n ]} ∞ n=1 is a Cauchy sequence in the metric interval space (I, d). Then we have for sufficiently large n and m. Let c n = a n + b n . Then (9) shows that {c n } ∞ n=1 is a Cauchy sequence in R. Since R is complete, there exists c ∈ R such that |c n − c| < for sufficiently large n. Now we can define a closed interval [a, b] such that a + b = c. Therefore we have for sufficiently large n. This shows that the sequence {[a n , b n ]} ∞ n=1 is convergent, i.e., the space (I, d) is complete.

Cauchy Sequences in Normed Interval Space
Therefore the concept of convergence is defined below.  Proof. To prove part (i), we have This completes the proof.
Inspired by part (ii) of Proposition 7, we propose the following concept of limit. Definition 7. Let (I, · ) be a pseudo-seminormed interval space. If the sequence {[a n , b n ]} ∞ n=1 in I converges to some [a, b] ∈ I, then the equivalence class [a, b] is called the class limit of {[a n , b n ]} ∞ n=1 . We also write lim We need to remark that if [a, b] is a class limit and [c, d] ∈ [a, b] then it is not necessarily that the sequence {[a n , b n ]} ∞ n=1 converges to [c, d] unless · satisfies the null equality. In other words, for the class limit [a, b] , if · satisfies the null equality, then part (ii) of Proposition 7 says that Proposition 8. Let (I, · ) be a pseudo-normed interval space such that · satisfies the null super-inequality. Then the class limit is unique.
This completes the proof.
for m, n > N with m = n. If every Cauchy sequence in I is convergent, then we say that I is complete. Proposition 9. Let (I, · ) be a pseudo-seminormed interval space such that · satisfies the null super-inequality. Every convergent sequence is a Cauchy sequence.
Proof. If {[a n , b n ]} ∞ n=1 is a convergent sequence, then, given any > 0, [a n , b n ] [a, b] = [a, b] [a n , b n ] < /2 for sufficiently large n. Therefore, by Proposition 2, we have for sufficiently large n and m, which says that {[a n , b n ]} ∞ n=1 is a Cauchy sequence. This completes the proof. Definition 9. Different kinds of Banach interval spaces are defined below.
• Let (I, · ) be a pseudo-seminormed interval space. If I is complete, then it is called a pseudo-semi-Banach interval space.
• Let (I, · ) be a seminormed interval space. If I is complete, then it is called a informal semi-Banach interval space. • Let (I, · ) be a pseudo-normed interval space. If I is complete, then it is called a pseudo-Banach interval space. • Let (I, · ) be a normed interval space. If I is complete, then it is called a Banach interval space.

Example 4.
Continued from Example 2, we want to show that (I, · ) is complete. Suppose that {[a n , b n ]} is a Cauchy sequence in (I, · ). Then we have for sufficiently large n and m. Let c n = a n + b n . Then (11) shows that {c n } ∞ n=1 is a Cauchy sequence in R. Since R is complete, there exists c ∈ R such that |c n − c| < for sufficiently large n. Now we can define a closed interval [a, b] such that a + b = c. Therefore we have for sufficiently large n. This shows that the sequence {[a n , b n ]} is convergent, i.e., (I, · ) is a Banach interval space.

Near Fixed Point Theorem in Metric Interval Space
Let T : I → I be a function from I into itself. We say that The well-known Banach contraction principle presents the fixed point of function T when (I, d) is taken to be a metric space. Since (I, d) presented in Example 1 is not a metric space, we cannot study the Banach contraction principle on this space (I, d). In other words, we cannot study the fixed point of contractive mappings defined on (I, d) into itself in the conventional way. However, we shall investigate the so-called near fixed point defined below. if and only if there exist k 1 , k 2 ∈ R + such that one of the following equalities is satisfied: Given any initial element [a 0 , b 0 ] ∈ I, we define the iterative sequence {[a n , b n ]} ∞ n=1 using the function T as follows: Under some suitable conditions, we are going to show that the sequence {[a n , b n ]} ∞ n=1 can converge to a near fixed point. If the metric interval space (I, d) is complete, then it is also called a complete metric interval space. Proof. Given any initial element [a 0 , b 0 ] ∈ I, we have the iterative sequence {[a n , b n ]} ∞ n=1 according to (12). We are going to show that {[a n , b n ]} ∞ n=1 is a Cauchy sequence. Since T is a metric contraction on I, we have Since 0 < α < 1, we have 1 − α m−n < 1 in the numerator, which says that This proves that {[a n , b n ]} ∞ n=1 is a Cauchy sequence. Since the metric interval space I is complete, there exists [a, b] ∈ I such that d([a n , b n ], [a, b]) → 0, i.e., [a n , b n ] → [a, b] according to Definition 4 and Proposition 5.
Assume further that d satisfies the null equality. We are going to show that any point Therefore we obtain for some ω i ∈ Ω, i = 1, · · · , 4. Since T is a metric contraction on I and d satisfies the null equality, we obtain It is clear that if T is a metric contraction on I, then it is also a weakly strict metric contraction on I. Assume further that d satisfies the null equality. Then we also have the following properties.

•
The uniqueness is in the sense that there is a unique equivalence class [a, b] such that any [ā,b] ∈ [a, b] cannot be a near fixed point. Proposition 5. Therefore, given any > 0, there exists an integer N such that d(T n ([a 0 , b 0 ]), [a, b]) < for n ≥ N. Since T is a weakly strict metric contraction on I, we consider the following two cases.
The above two cases say that d(T n+1 ([a 0 , b 0 ]), T([a, b])) → 0. Using the triangle inequality, we obtain Then, using the null equality for d, we obtain • Suppose that [a n−1 , b n−1 ] = [a n , b n ] . By Remark 1, since T is also a weakly strict metric contraction on I, we have • Suppose that [a n−1 , b n−1 ] = [a n , b n ] . Then, by the first condition of Definition 13, , T([a n , b n ])) = 0 ≤ η n−1 .
The above two cases say that the sequence {η n } ∞ n=1 is decreasing. We consider the following cases.
Since the sequence {η n } ∞ n=1 is decreasing, we assume that η n ↓ > 0, i.e., η n ≥ > 0 for all n. There exists δ > 0 such that ≤ η m < + δ for some m, i.e., By the second condition of Definition 13, we have This completes the proof. Assume further that d satisfies the null equality. Then we also have the following properties.
• The uniqueness is in the sense that there is a unique equivalence class [a, b] such that any [ā,b] ∈ [a, b] cannot be a near fixed point. Proof. According to Theorem 2 and Remark 1, we just need to claim that if T is a weakly uniformly strict metric contraction, is not a Cauchy sequence. Then there exists 2 > 0 such that, given any N, there exist m, n ≥ N satisfying d([a m , b m ], [a n , b n ]) > 2 . Since T is a weakly uniformly strict metric contraction on I, there exists δ > 0 such that Let δ = min{δ, }. We are going to claim Indeed, if δ = δ, then it is done, and if δ = , i.e., < δ, then + δ = + < + δ. This proves the statement (14).
Let η n = d([a n , b n ], [a n+1 , b n+1 ]). Since the sequence {η n } ∞ n=1 is decreasing to zero by Lemma 1, we can find N such that η N < δ /3. For n > m ≥ N, we have which says that [a m , b m ] Ω = [a n , b n ]. Since the sequence {η n } ∞ n=1 is decreasing by Lemma 1 again, we obtain For j with m < j ≤ n, using the triangle inequality, we also have We want to show that there exists j with Let γ j = d([a m , b m ], [a j , b j ]) for j = m + 1, · · · , n. Then (15) and (16) say that γ m+1 < and γ n > + δ .

Near Fixed Point Theorems in Banach Interval Space
In this section, we shall study the near fixed point in Banach interval space.  [a n , b n ] = [a n , b n ] [a, b] → 0 as n → ∞ in which the sequence {[a n , b n ]} ∞ n=1 is generated according to (12). Assume further that · satisfies the null equality. We also have the following properties.

•
The uniqueness is in the sense that there is a unique equivalence class [a, b] such that any [ā,b] ∈ [a, b] cannot be a near fixed point. Proof. Given any initial element [a 0 , b 0 ] ∈ I, we are going to show that {[a n , b n ]} ∞ n=1 is a Cauchy sequence. Since T is a norm contraction on I, we have For n < m, using Proposition 2, we obtain Since 0 < α < 1, we have 1 − α m−n < 1 in the numerator, which says that This proves that {[a n , b n ]} ∞ n=1 is a Cauchy sequence. Since I is complete, there exists [a, b] ∈ I such that [a, b] [a n , b n ] = [a n , b n ] [a, b] → 0 as n → ∞.
Assume further that · satisfies the null equality. We are going to show that any point for some ω i ∈ Ω, i = 1, · · · , 4. We obtain , which says that the weakly strict norm contraction is well-defined. We further assume that · satisfies the null super-inequality and null condition. Part (iii) of Proposition 3 says that if T is a contraction on I, then it is also a weakly strict contraction on I. Assume further that · satisfies the null equality. Then we also have the following properties.

•
The uniqueness is in the sense that there is a unique equivalence class [a, b] Therefore, given any > 0, there exists an integer N such that T n ([a 0 , b 0 ]) [a, b] < for n ≥ N. Since T is a weakly strict norm contraction on I, we consider the following two cases. Assume further that · satisfies the null equality. Now we are going to claim that each point ⊕ ω 2 for some ω 1 , ω 2 ∈ Ω. Then, using the null equality for · , we obtain   = 0, which says that the weakly uniformly strict norm contraction is well-defined.

Remark 2.
We observe that if T is a weakly uniformly strict norm contraction on I, then T is also a weakly strict norm contraction on I. Proof. For convenience, we write T n ([a, b]) = [a n , b n ] for all n. Let η n = [a n , b n ] [a n+1 , b n+1 ] .
The above two cases say that the sequence {η n } ∞ n=1 is decreasing. We consider the following cases. Using the first condition of Definition 16, we also have Using the similar arguments, we can obtain η m+1 = 0 and [a m+1 , b m+1 ] = [a m+2 , b m+2 ] . Therefore the sequence {η n } ∞ n=1 is decreasing to zero.
Since the sequence {η n } ∞ n=1 is decreasing, we assume that η n ↓ > 0, i.e., η n ≥ > 0 for all n. There exists δ > 0 such that ≤ η m < + δ for some m, i.e., By the second condition of Definition 16, we have This completes the proof. Assume further that · satisfies the null equality. Then we also have the following properties.

•
The uniqueness is in the sense that there is a unique equivalence class [a, b] such that any [ā,b] ∈ [a, b] cannot be a near fixed point. Proof. According to Theorem 5 and Remark 2, we just need to claim that if T is a weakly uniformly strict norm contraction, then {T n ([a 0 , b 0 ])} ∞ n=1 = {[a n , b n ]} ∞ n=1 forms a Cauchy sequence. Suppose that {[a n , b n ]} ∞ n=1 is not a Cauchy sequence. Then there exists 2 > 0 such that, given any N, there exist n > m ≥ N satisfying [a m , b m ] [a n , b n ] > 2 . Since T is a weakly uniformly strict norm contraction on I, there exists δ > 0 such that Indeed, if δ = δ then it is done, and if δ = , i.e., < δ, then + δ = + < + δ. Let η n = [a n , b n ] [a n+1 , b n+1 ] . Since the sequence {η n } ∞ n=1 is decreasing to zero by Lemma 2, we can find N such that η N < δ /3. For n > m ≥ N, we have [a m , b m ] [a n , b n ] > 2 ≥ + δ , which says that [a m , b m ] Ω = [a n , b n ]. Since the sequence {η n } ∞ n=1 is decreasing by Lemma 2 again, we obtain For j with m < j ≤ n, using Proposition 2, we have Let γ j = [a m , b m ] [a j , b j ] for j = m + 1, · · · , n. Then (23) and (24) say that γ m+1 < and γ n > + δ . Let j 0 be an index such that j 0 = max j ∈ [m + 1, n] : γ j ≤ + 2δ 3 .

Conclusions
Owing to the set of all bounded and closed intervals, R cannot form a real vector space, and the concept of null set is proposed for the purpose of endowing a norm to the set of all bounded and closed intervals in R. Although the (conventional) metric space is not necessarily a real vector space, we also endow a metric to the set of all bounded and closed intervals in R based on the concept of null set, which is then called a metric interval space and is different from the conventional metric space.
Since we do not consider the conventional normed and metric space in this paper, the concept of so-called near fixed point is proposed based on the concept of null set. The main contribution of this paper is to establish many types of near fixed point theorems in metric interval space and normed interval space. We also remark that the conventional fixed point theorems cannot be established in metric interval space and normed interval space, since they are not the conventional normed spaces.
Funding: This research received no external funding.

Conflicts of Interest:
The author declares no conflict of interest.