Abstract
Recently, Khamsi and Hussain (Nonlinear Anal. 73:3123-3129, 2010) discussed a natural topology defined on any metric type space and noted that this topology enjoys most of the metric topology like properties. In this paper, we define topologically complete type metrizable space and prove that being of metrizability type is preserved under a countable Cartesian product and establish the fact that any set in a complete metric type space is a topologically metrizable type space. Next, we introduce the concept of wt-distance on a metric type space and prove some fixed point theorems in a partially ordered metric type space with some weak contractions induced by the wt-distance.
MSC:47H09, 47H10.
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1 Preliminaries
The concept of metric type or b-metric space was introduced and studied by Bakhtin [1] and Czerwik [2]. Since then several papers have dealt with fixed point theory for single-valued and multivalued operators in b-metric and cone b-metric spaces (see [3–12] and references therein). Khmasi and Hussain [13] and Hussain and Shah [14] discussed KKM mappings and related results in metric and cone metric type spaces.
Definition 1.1 Let X be a set. Let be a function which satisfies
-
(1)
if and only if ;
-
(2)
, for any ;
-
(3)
, for any points , for some constant .
The pair is called a metric type space.
Definition 1.2 Let be a metric type space.
-
(1)
The sequence converges to if and only if .
-
(2)
The sequence is Cauchy if and only if .
is complete if and only if any Cauchy sequence in X is convergent.
Example 1.3 Let X be the set of Lebesgue measurable functions on such that
Define by
Then D satisfies the following properties:
-
(1)
if and only if ;
-
(2)
, for any ;
-
(3)
, for any functions .
Example 1.4 Let be metric space. Define
-
(1)
for any ;
-
(2)
, for any .
Then , are metric type spaces with .
Definition 1.5 Let be a metric type space. A subset is said to be open if and only if for any , there exists such that the open ball . The family of all open subsets of X will be denoted by τ.
Theorem 1.6 ([13])
τ defines a topology on .
Theorem 1.7 ([13])
Let be a metric type space and τ be the topology defined above. Then for any nonempty subset we have
-
(1)
A is closed if and only if for any sequence in A which converges to x, we have ;
-
(2)
if we define to be the intersection of all closed subsets of X which contains A, then for any and for any , we have
Theorem 1.8 ([13])
Let be a metric type space and τ be the topology defined above. Let . The following properties are equivalent:
-
(1)
A is compact;
-
(2)
For any sequence in A, there exists a subsequence of which converges, and .
Definition 1.9 The subset A is called sequentially compact if and only if for any sequence in A, there exists a subsequence of which converges, and . Also A is called totally bounded if for any there exits such that
Theorem 1.10 ([13])
Let be a metric type space and τ be the topology defined above. Let . The following properties are equivalent:
-
(1)
A is compact if and only if A is sequentially compact.
-
(2)
If A is compact, then A is totally bounded.
Corollary 1.11 Every closed subset of a complete metric type space is complete.
Theorem 1.12 ([15])
Let be a metric type space and suppose that and converge to , respectively. Then we have
-
(1)
In particular, if , then .
-
(2)
Moreover, for each , we have
2 Topologically complete metrizable type spaces
Lemma 2.1 Let be a metric type space and let then there exists a metric type E on X such that , for each , and E and D induce the same topology on X.
Proof We define . We claim that E is metric type on X. The properties (1) and (2) are immediate from the definition. For the triangle inequality, suppose that . Then and so when either or . The only remaining case is when and . But and and so . Thus E is a metric type on X. It only remains to show that the topology induced by E is the same as that induced by D. But we have if and only if if and only if , and we are done. □
The metric type E in the above lemma is said to be bounded by λ.
Definition 2.2 Let be a metric type space, and . We define
Definition 2.3 A topological space is called a (topologically complete) metrizable type space if there exists a (topologically complete) metric type D inducing the given topology on it.
Example 2.4 Let . The metric type space is not complete because the Cauchy sequence in this space is not convergent. Now, if we consider . It is straightforward to show that is complete. Since tend to x with respect to the metric type if and only if if and only if tend to x with respect to the metric type , then and are equivalent. Hence the metric type space is topologically complete metrizable type.
Lemma 2.5 Metrizability type is preserved under countable Cartesian product.
Proof Without loss of generality we may assume that the index set is ℕ. Let be a collection of metrizable type spaces. Let be the topology induced by on for and let be the Cartesian product of with product topology. We have to prove that there is a metric type D on X which induces the topology τ. By the above lemma, we may suppose that is bounded by for all , otherwise we replace by another metric type which induces the same topology and which is bounded by . Points of are denoted as sequences with for . Define , for each . First note that D is well defined since is convergent. Also D is a metric type on X because each is of a metric type. Let be the topology induced by the metric type D. We claim that coincides with τ. If and , then there exists such that . Now choose such that . For each , let , where the ball is with respect to the metric type . Let for . Put , then and V is an open set in the product topology τ on X. Furthermore , since for each
Hence . Therefore G is open in the product topology. Conversely suppose G is open in the product topology and let . Choose a standard basic open set V such that and . Let , where each is open in and for all . For , let , if , and , otherwise, let . We claim that . If , then and so for each . Then , for . Also for , . Hence and so . Therefore G is open with respect to the metric type topology and . Hence τ and coincide. □
Theorem 2.6 An open subspace of a complete metrizable type space is a complete metrizable type space.
Proof Let be a complete metric type space and G an open subspace of X. If the restriction of D to G is not complete we can replace D on G by another metric type as follows. Define by (f is undefined if is empty, but then there is nothing to prove). For define
It is clear that D is of metric type on G.
We show that E and D are of the type of equivalent metrics on G. We do this by showing that for arbitrary sequence converges to , if and only if . Since for all , whenever . To prove the converse, let , and using Theorem 1.12, we have
Therefore
On the other hand, there exists a , the positive sequence converges to zero, and such that for every we have
Then
and
By (1) and (2), we have
This implies . Hence . Therefore E and D are equivalent. Next we show that E is a complete metric type. Suppose that is a Cauchy sequence in G with respect to E. Since for each , , therefore is also a Cauchy sequence with respect to D. By completeness of , converges to point p in X. We claim that . Assume otherwise, then for each , if and . Therefore
That is
Therefore as , we get . On the other hand, , for every , that is, is a bounded sequence. This contradiction shows that . Hence converges to p with respect to E and is a complete metrizable type space. □
Theorem 2.7 (Alexandroff)
A set in a complete metric type space is a topologically complete metrizable type space.
Proof Let be a complete metric type space and G be a set in X, that is, , where each is open in X. By the above theorem, there exists a complete metric type on and we may assume that is bounded by . Let ℋ be the Cartesian product with the product topology. Then ℋ is a complete metrizable type space. Now, for each let be the inclusion map. So the evaluation map is an embedding. Image of e is the diagonal which is a closed subset of ℋ and by Corollary 1.11, is complete. Thus is a complete metrizable type space and so is G which is homeomorphic to it. □
3 wt-Distance
Kada et al. [16] introduced in 1996, the concept of w-distance on a metric space and proved some fixed point theorems. In this section, we introduce the definition of a wt-distance and we state a lemma which we will use in the main sections of this work.
Definition 3.1 Let be a metric type space with constant . Then a function is called a wt-distance on X if the following are satisfied:
-
(a)
for any ;
-
(b)
for any , is K-lower semi-continuous;
-
(c)
for any , there exists such that and imply .
Let us recall that a real-valued function f defined on a metric type space X is said to be lower K-semi-continuous at a point in X if either or , whenever for each and [17].
Let us give some examples of wt-distance.
Example 3.2 Let be a metric type space. Then the metric D is a wt-distance on X.
Proof (a) and (b) are obvious. To show (c), for any , put . Then we see that and imply . □
Example 3.3 Let and . Then the function defined by for every is a wt-distance on X.
Proof (a) and (b) are obvious. To show (c), for any , put . Then we have
□
Example 3.4 Let and . Then the function defined by for every is a wt-distance on X.
Proof (a) and (b) are obvious. To show (c), for any , put . Then we have
□
Lemma 3.5 Let be a metric type space with constant and P be a wt-distance on X. Let and be sequences in X, let and be sequences in converging to zero, and let . Then the following hold:
-
(1)
If and for any , then . In particular, if and , then ;
-
(2)
if and for any , then ;
-
(3)
if for any with , then is a Cauchy sequence;
-
(4)
if for any , then is a Cauchy sequence.
Proof The proof is similar to [16]. □
4 Fixed point theorems
We introduce first the following concept.
Definition 4.1 Suppose is a partially ordered set and be a self mapping on X. We say f is inverse increasing if for ,
Our first main result is a fixed point theorem for graphic contractions on a partially ordered metric space endowed with a wt-distance.
Theorem 4.2 Let be a partially ordered set and let be a metric type on X such that is a complete metric type space with constant . Suppose that P is a wt-distance in . Let be a non-decreasing mapping and there exists such that
and . Suppose also that:
-
(i)
for every with
(5) -
(ii)
there exists such that .
Then A has a fixed point in X.
Proof If , then the proof is finished. Suppose that
Since and A is non-decreasing, we obtain
Hence, for each we have
Then, for with , we successively have
By Lemma 3.5(3), we conclude that is Cauchy sequence in . Since is a complete metric type space, there exists such that
Let be an arbitrary but fixed. Then since converges to z in and is K-lower semi-continuous, we have
Assume that . Since , by (5), we have
This is a contradiction. Therefore, we have . □
Another result of this type is the following.
Theorem 4.3 Let be a partially ordered set, let be of a metric type on X such that is a complete metric type space with constant . Suppose that P is a wt-distance in . Let be a non-decreasing mapping and there exists such that
and . Assume that one of the following assertions holds:
-
(i)
for every with
(8) -
(ii)
if both and converge to y, then ;
-
(iii)
A is continuous.
If there exists with , then A has a fixed point in X.
Proof The case (i), was proved in Theorem 4.2.
Let us prove first that (ii) ⟹ (i). Assume that there exists with such that
Then there exists such that and
Then and . By Lemma 3.5, we have that . We also have
Again by Lemma 3.5, we get . Put . Then both and converges to y. Thus, by (ii) we have . Thus (ii) ⟹ (i) holds.
Now, we show that (iii) ⟹ (ii). Let A be continuous. Further assume that and converges to y. Then we have
□
5 Common fixed point theorem for commuting mappings
The following theorem was given by Jungck [18] and it represents a generalization of the Banach contraction principle in complete metric spaces.
Theorem 5.1 Let f be a continuous self mapping on a complete metric space and let be another mapping, such that the following conditions are satisfied:
-
(a)
;
-
(b)
g commutes with f;
-
(c)
, for all and for some .
Then f and g have a unique common fixed point.
The next example shows that if the mapping is continuous with respect to a metric type D on X and satisfies the condition
then, in general, g may be not continuous in .
Example 5.2 Let be a normed linear space. Consider Example 3.4 with wt-distance defined by
Consider the functions f and g defined by and
Then
Definition 5.3 Let be a partially ordered set and . By definition, we say that g is h-non-decreasing if for ,
Our next result is a generalization of the above mentioned result of Jungck [18], for the case of a weak contraction with respect to a wt-distance.
Theorem 5.4 Let be a partially ordered set, let be a metric type on X such that is a complete metric type space with constant . Suppose that P is a wt-distance on X. Let be mappings that satisfy the following conditions:
-
(a)
;
-
(b)
g is f-non-decreasing and f is inverse increasing;
-
(c)
g commutes with f and f, g are continuous in ;
-
(d)
for all with and some such that .
-
(e)
there exists such that:
-
(i)
and
-
(ii)
.
-
(i)
Then f and g have a common fixed point . Moreover, if for all , then .
Proof We claim that for every
for every with . For the moment suppose the claim is true. Let with . By (a) we can find such that . By induction, we can define a sequence such that
Since and , we have
Then from (b),
that means, by (10), that . Again by (b) we get
that is, . By this procedure, we obtain
Hence from (10) and (12) we have and by (3) we have . By induction we get
for . This implies that, for with ,
Thus, by Lemma 3.5, we find that is a Cauchy sequence in . Since is complete, there exists such that . As a result the sequence tends to y as and hence converges to as . However, , by the commutativity of f and g, implies that converges to as . Because limit is unique, we get and, thus, . On the other hand, by K-lower semi-continuity of we have, for each ,
Notice that, by (11), (10), and (9), we obtain and thus, by (9), we get . Then
By (9) we get and thus . Continuing this process we get
and by (3) we get
Using now the condition (d), we have
We will show that . Suppose, by contradiction, that . Then we have
This is a contradiction. Therefore . Thus, . Hence is a common fixed point of f and g.
Furthermore, since for all , we have
which implies that .
Now it remains to prove the initial claim. Assume that there exists with and
Then there exists such that
Since and , by Lemma 3.5, we have
Also, since and , by Lemma 3.5, we have
By (13), (14), and the continuity of g we have
Therefore, , which is a contradiction. Hence, if , then
□
Example 5.5 Let be a normed linear space. Consider Example 3.4 with wt-distance defined by
Consider the functions f and g defined by and . Then
Put . Then all conditions of Theorem 5.4 hold and is the common fixed point of f and g and .
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Acknowledgements
The authors would like to thank the referees for giving useful suggestions and comments for the improvement of this paper. This article was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah. Therefore, the first author acknowledges with thanks DSR, KAU for financial support.
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All authors carried out the proof. All authors conceived of the study and participated in its design and coordination. All authors read and approved the final manuscript.
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Hussain, N., Saadati, R. & Agrawal, R.P. On the topology and wt-distance on metric type spaces. Fixed Point Theory Appl 2014, 88 (2014). https://doi.org/10.1186/1687-1812-2014-88
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DOI: https://doi.org/10.1186/1687-1812-2014-88