#### Abstract

The purpose of this paper is to prove the existence of fixed points of contractive mapping defined on -metric space where the completeness is replaced with weaker conditions. Moreover, we showed that these conditions do not guarantee the completeness of -metric spaces.

#### 1. Introduction

The study of metric fixed point theory has been researched extensively in the past decades, since fixed point theory plays a major role in mathematics and applied sciences, such as optimization, mathematical models, and economic theories.

Different mathematicians tried to generalize the usual notion of metric space such as Gähler [1, 2] and Dhage  to extend known metric space theorems in more general setting, but different authors proved that these attempts are unvalid (for detail see ).

In 2005, Mustafa and Sims introduced a new structure of generalized metric spaces (see ), which are called -metric spaces as generalization of metric space , to develop and introduce a new fixed point theory for various mappings in this new structure. The -metric space is as follows.

Definition 1.1 (see ). Let be a nonempty set, and let , be a function satisfying the following: (G1) if (G2) for all with (G3) for all with (G4), (symmetry in all three variables), (G5), for all , (rectangle inequality).
Then the function is called a generalized metric , or, more specifically a - metric on , and the pair is a -metric space.
Clearly these properties are satisfied when is the perimeter of the triangle with vertices at , and in ; moreover taking in the interior of the triangle shows that (G5) is the best possible.
If is an ordinary metric space, then can define -metrics on by
(), ().

Proposition 1.2 (see ). Let be a -metric space. Then for any and it follows that(1)if , then ,(2),(3),(4),(5),(6).

Proposition 1.3 (see ). Every -metric space will define a metric space by

Definition 1.4 (see ). Let be a -metric space. Then for , , the -ball with center and radius is

Proposition 1.5 (see ). Let be a -metric space. Then for any and one has(1)if then ,(2)if then there exists a such that .

Proof. follows directly from (G3), while follows from (G5) with .

It follows from of the above proposition that the family of all -balls, , is the base of a topology on , the -metric topology.

Definition 1.6 (see ). Let be a -metric space, let be sequence of points of , a point is said to be the limit of the sequence if , and we say that the sequence is -convergent to .
Thus, if in a -metric space , then for any there exists such that , for all , (through this paper we mean by the set of all natural numbers).

Proposition 1.7 (see ). Let be a -metric space. The sequence is -convergent to if it converges to in the -metric topology, .

Proposition 1.8 (see ). Let be a -metric space. Then for a sequence and a point the following are equivalent(1) is -convergent to (2), as (3), as (4), as

Definition 1.9 (see ). Let and be -metric spaces and let be a function, then is said to be -continuous at a point if and only if, given , there exists such that ; and implies . A function is -continuous at if and only if it is -continuous at all .

Proposition 1.10 (see ). Let , be -metric spaces. Then a function is -continuous at a point if and only if it is -sequentially continuous at ; that is, whenever is -convergent to one has is -convergent to .

Proposition 1.11 (see ). Let be a -metric space. Then the function is jointly continuous in all three of its variables.

Definition 1.12 (see ). Let be a -metric space. Then the sequence is said to be -Cauchy if for every there exists such that for all .

Definition 1.13 (see ). A -metric space is said to be -complete (or complete -metric space) if every -Cauchy sequence in is -convergent in .

#### 2. The Main Results

In this section we will prove several theorems in each of which we have omitted the completeness property of -metric space and we have obtained the same conclusion as in complete -metric space, but with assumed sufficient conditions.

Theorem 2.1. Let be a -metric space and let be a mapping such that satisfies that(A1) for all where (A2) is -continuous at a point , (A3) there is has a subsequence -converges to . Then is a unique fixed point (i.e., ).

Proof. -continuity of at implies that -convergent to . Suppose , consider the two -open balls and where .
Since and , then there exist such that if implies and . Hence our assumption implies that we must have
On the other hand we have from (A1), but, by axioms of -metric (G3), we have So, from (2.3) and (2.4), we see (2.2) becomes where and , since .
Hence (2.3) and (2.5) implies that
For and by repeated application of (2.6) we have So, as we have which contradict (2.1), hence .
Suppose there is , then from (A1), we have This prove the uniqueness of .

In  we have proved the following theorem.

Theorem 2.2 (see ). Let be a complete -metric space and let be a mapping satisfies the following condition for all : where , then has a unique fixed point, say , and is -continuous at .

We see that if we take , the following theorem becomes a direct result.

Theorem 2.3. Let be complete -metric space and let be a mapping satisfies for all where , then has a unique fixed point, say u, and is -continuous at .

If we compare Theorem 2.3 with Theorem 2.1, we see that in Theorem 2.1 we have omitted the completeness property of the -metric space and instead we have assumed conditions and . However, the following examples support that conditions and in Theorem 2.1 do not guarantee the completeness of the -metric space.

Example 2.4. Let and . Then is -metric space but not complete, since the sequence is -cauchy which is not -convergent in . However, conditions and in Theorem 2.1 are satisfied.

Theorem 2.5. Let be a -metric space and let be a -continuous mapping satisfies the following conditions:(B1) for all where is an every where dense subset of (with respect the topology of -metric convergence) and ,(B2) there is such that . Then is unique fixed point.

Proof. The proof will follow from Theorem 2.1, if we can show that condition (A1) in Theorem 2.1 holds for any .
Let be any elements of .
Case 1. If , let , , and be a sequences in such that , and . From (G5) we have also and by (B1), we have again by (G5) we have So, from (2.13) and (2.14) we see that (2.12) becomes by the same argument we deduce that
Hence, by (2.15) and (2.16), we have
Now letting in the above inequality and using the fact that is -continuous we get

Case 2. If and , let be a sequence in such that , then by (G5), we have but by (B1), we have and by (G5), we have Again by (B1), we have Also, by (G5), we have So, from (2.21), (2.22), and (2.23), we see that (2.19) becomes
Now letting in the above inequality, we get
Case 3. If and , let and be a sequences in such that and , but by (G5), we have also, from (B1), we have and from (G5), we have So, by (2.28) and (2.29), we have then from (2.27) and (2.30) we have By the same argument we deduce that Then, from (2.31) and (2.32), we see (2.26) becomes
Now letting in the above inequality and using the fact that is -continuous we get
So, in all cases we have for all where and since , then by Theorem 2.1, has a unique fixed point.

Corollary 2.6. Let be -metric space and let be a mapping such that satisfies that(C1) for all where (C2) is -continuous at a point , (C3) there is has a subsequence -converges to . Then is a unique fixed point.

Proof. Let in condition (A1), then we see that every mapping satisfies condition (C1) will satisfy condition (A1), so the proof follows from Theorem 2.1.

Corollary 2.7. Let be -metric space and let be a -continuous mapping satisfies that(D1) for all where is an every where dense subset of (with respect the topology of -metric convergence) and ,(D2) there is such that . Then is unique fixed point.

Proof. Let in condition (B1), then we see that every mapping satisfies condition (D1) will satisfy condition (B1), so the proof follows from Theorem 2.5.

Corollary 2.8. Let be -metric space and let be a mapping such that satisfies that(E1) for all where (E2) is -continuous at a point , (E3) there is has a subsequence -converges to . Then is a unique fixed point.

Proof. By axioms of -metric (G5), we have so, from (2.36), we see that (E1) becomes then will satisfy the following condition for all , where , and , since .
So, condition (C1) is satisfied and the proof follows from Corollary 2.6.

#### Acknowledgment

The authors are very grateful to the referees for their helpful suggestions.