Some Fixed-Point Theorems in Proximity Spaces with Applications

: Considering the ω -distance function deﬁned by Kosti´c in proximity space, we prove the Matkowski and Boyd–Wong ﬁxed-point theorems in proximity space using ω -distance, and provide some examples to explain the novelty of our work. Moreover, we characterize Edelstein-type ﬁxed-point theorem in compact proximity space. Finally, we investigate an existence and uniqueness result for solution of a kind of second-order boundary value problem via obtained Matkowski-type ﬁxed-point results under some suitable conditions.


Introduction and Preliminaries
The basic concepts of proximity spaces were initially developed by Frigyes Riesz [1] in 1908, and later on this theory was revived and axiomatized by Efremovich [2] in 1934, which was published in 1951.Over the years, a lot of studies on proximity spaces have been carried out [3][4][5][6].Smirnov [5] established the link between the proximity relation and topological spaces.In addition, he was the first to introduce the relationship between proximities and uniformities.
Let δ be a relation on a set 2 Ω .Then, the pair (Ω, δ) is said to a proximity space if the following hold: for all U, V, W ∈ 2 Ω , where 2 Ω the power set of Ω For all ξ ∈ Ω and U ⊆ Ω, we will use the notation ξδU and Uδξ instead of {ξ}δU and Uδ{ξ}, respectively.A proximity space (Ω, δ) is called separated if ξδη implies ξ = η for all ξ, η ∈ Ω.The properties of proximity spaces are generalizations of uniform properties and continuity properties of metric and topology, respectively.Every proximity relation on a nonempty set Ω induces a topology τ δ via the Kuratowski closure operator.The Kuratowski closure operator using proximity relation can be defined by cl(U) = {ξ ∈ Ω:ξδU} for all U ⊆ Ω.In this case, the topology τ δ is always completely regular, further, it is Tychonoff if (Ω, δ) is separated.If (Ω, τ) is a topological space and δ is a proximity on Ω such that τ δ = τ, then τ and δ are said to be compatible.Every completely regular topology on a nonempty set Ω has a compatible proximity.Additionally, if a sequence {ξ n } converges to a point ξ ∈ Ω with respect to induced topology τ δ , then we obtain ξδ{ξ n }.Furthermore, every uniform space (Ω, U ) has an associated proximity structure defined by for all U, V ⊆ Ω, UδV if (U × V) ∩ W = ∅ for all W ∈ U .For further information, see [7,8].Now we state some examples of proximity spaces.
Example 1.Let (Ω, ρ) be a metric space.Consider the relation δ on Then δ is a proximity on Ω.In addition, the metric topology τ ρ and δ are compatible.
Example 2. Let (Ω, τ) be a T 4 (both normal and T 1 ) topological space.Consider the relation δ on Then δ is a proximity on Ω.In addition, τ and δ are compatible.
Example 3. Let (Ω, τ) be a completely regular topological space.Consider the relation δ on 2 Ω as UδV if and only if there does not exist a continuous function f : Then δ is a proximity on Ω.In addition, τ and δ are compatible.
Considering that fixed-point theory on proximity space will bring a new direction and interesting results, Kostić [9] defined the concept of ω-distance and ω 0 -distance inspired by [10] (see [11] for more information about ω-distance) in proximity space as follows and obtained the proximity space version of Banach fixed-point theorem.In addition, a ω-distance on a proximity space (Ω, δ) is called ω 0 -distance if the following axioms hold: w3) ω is lower semicontinuous in both variables with respect to τ δ , i.e., for all ξ, η ∈ Ω, we have where U ξ is a base of neighborhoods of the point ξ ∈ Ω.

Example 4.
Let Ω = R be endowed with the usual metric and the proximity δ defined in Example 1. Define then both ω 1 and ω 2 are ω 0 -distance on Ω.

Example 5.
Let Ω = [0, ∞) be endowed with the lower limit topology τ l and the proximity δ defined in Example 2. It is well-known that (R, τ l ) is not metrizable but normal topological space.

Example 6.
Let Ω = C[0, 2] be endowed with the metric ρ defined by and the proximity δ defined in Example 1. Define ω : Let ε > 0. Then by (1) there exist f ∈ U and g ∈ V such that Hence we have and hence UδV.Therefore, (w1) holds, and so ω is a ω-distance on Ω.It is clear that (w2) holds.Now, let f , g ∈ Ω, then we have that is, ω is lower semicontinuous in the first variable.On the other hand, let V ∈ U f and f ∈ V, then there exists and so we obtain that is, ω is lower semicontinuous in second variable.
In this study, we will use some auxiliary functions in the contraction inequality to generalize Kostić's theorem.Thus, we will obtain equivalents in proximity space of the famous Boyd-Wong [12] and Matkowski [13] fixed-point theorems known in metric fixedpoint theory.In addition, for the equivalent of Edelstein's theorem [14], in this space, we will consider that the space is compact according to the topology induced by proximity.

Main Result
Here, we present our main theorems.
Proof.Let ξ 0 ∈ Ω be arbitrary.Consider the corresponding Picard sequence {ξ n } constructed by Since φ is non-decreasing, and by (i), we have Similarly, we can obtain By (φ 2 ), we have lim and lim On the other hand, since φ ∈ Φ, we have φ( ) < for all > 0. Hence by (2), there exists a natural number N satisfying Since ω is a ω 0 -distance and φ is non-decreasing, then we have Similarly, Continuing this process, we obtain ω(ξ N , ξ N+k ) < for all k = 1, 2, • • • and similarly by (3) we obtain ω(ξ N+k , ξ N ) < for all k = 1, 2, • • • .Thus, for all m, n > N, we have By (ii), there exists a subsequence {ξ n k } of the sequence {ξ n } which is convergent with respect to τ δ to some ζ ∈ Ω, and by (w3), Remark 1 and Inequality (4), we have and symmetrically, we obtain for all n k > N. By Inequality (6), we have for all n k > N. Finally, by Inequalities ( 5) and ( 7), we have for all n k > N. Since, by (φ 2 ), φ n k (ω(ξ 0 , ξ 1 )) → 0 as k → ∞, it follows that ω(ζ, Tζ) = 0. On the other hand, by (w3) and Remark 1, we have a contradiction.Hence, the fixed point of T is unique.
Example 8. Let (Ω, δ) be the proximity space and ω be the ω 0 -distance given in Example 5.
Hence, Theorem 1 cannot be applied.
Proof.Let ξ 0 ∈ Ω be arbitrary.Consider the corresponding Picard sequence {ξ n } constructed by for all n ∈ N.For simplicity, call ω n = ω(ξ n , ξ n+1 ).If there exists n 0 ∈ N such that ω n 0 = ω n 0 +1 = 0, then by triangular inequality we have and so by Lemma 1 (i) we have ξ n 0 +1 = ξ n 0 +2 .This shows that ξ n 0 +1 is a fixed point of T. Now assume that any of consecutive terms of the {ω n } are not zero.In this case, there exists a subsequence {ω n k } of {ω n } such that ω n k > 0 for all k ∈ N. By (i), we have Since the index sequence n k is strictly increasing, we have n k+1 > n k and so n k+1 ≥ n k + 1.Therefore, from (8), we have It follows that {ω n k } is a decreasing sequence which is also bounded below.Thus, there exists γ ≥ 0 such that lim k→∞ ω n k = γ.If γ > 0, then by ( 9), (φ 3 ) and (φ 4 ), we have a contradiction.Therefore, γ = 0, and consequently, lim Similarly, it can be obtained that the limit of all subsequences of {ω n } whose all terms are greater than zero, is zero.Hence we have Now, we claim that for every > 0, there exists a natural number N satisfying Here, we can choose m k as the smallest positive integer that satisfies ω(ξ m k , ξ n k ) ≥ .Therefore, we obtain ω(ξ m k −1 , ξ n k ) < , and since ω is a ω 0 -distance, we obtain By Equation (11), it follows that lim Taking limit k → ∞ and by (φ 4 ), we have ≤ φ( ), a contradiction.Thus, for every > 0, there exists N ∈ N, such that for all m > n > N. Similarly, we obtain for all m > n > N. By (ii), there exists a subsequence {ξ n k } of sequence {ξ n } which is convergent with respect to τ δ to some ζ ∈ Ω, and by (ω 3 ) and Inequality (12), we have for all n k ∈ N and symmetrically by (13), we obtain for all n k ∈ N.Then, we have and consequently, ω(ζ, ζ) = 0. On the other hand, we have Since φ(0) = 0 and φ is right continuous at 0, we have ω(ζ, Tζ) = 0. Thus, by Lemma 1 (i), we have ζ = Tζ.
To check the uniqueness, let v ∈ Ω with v = ζ be another fixed point of mapping T.
a contradiction.Thus, the fixed point of T is unique.
Finally, in the theoretical part, we present a fixed-point result, taking into account the compactness of the space and a strict contraction inequality.Here, we will use the following well-known lemma.