It is well known that generalized metric spaces in the sense of Branciari might not be Hausdorff and, hence, there may exist sequences in them having more than one limit. Thus, in most of the fixed point results obtained recently in such spaces, Hausdorffness was additionally assumed. We show in this article that, nevertheless, most of these results remain valid without this additional assumption.
A lot of generalizations of metric spaces exist. Most of them were introduced in an attempt to extend some fixed point theorems known from the metric case. One of the fruitful generalizations of this kind was given by Branciari in 2000  who replaced the triangular inequality by a more general one, which was later usually called a rectangular or a quadrilateral inequality. These new spaces became known as generalized metric spaces (g.m.s., for short) or rectangular spaces. Several authors (e.g., [2–29]) proved various (common) fixed point results is such spaces.
In some of the first papers which dealt with fixed point theorems in g.m.s., it was sometimes implicitly assumed that the respective topology is Hausdorff and/or that the generalized metric is continuous (see, e.g., [1, 5, 11, 19]). However, as shown by examples in [26, 27], a generalized metric need not be continuous, neither the respective topology need to be Hausdorff. Hence, in further articles, usually one or both of these conditions were additionally assumed (see, e.g., [4, 7–10, 12, 14, 19, 20, 25, 27]).
In this paper, we show that a number of these results are nevertheless valid without assumptions of generalized metric being continuous or the respective topology being Hausdorff. In particular, we prove (common) fixed point results that are extensions of Geraghty-type results, results using altering distance or admissible functions and results in generalized cone metric spaces. References to some further results can be found in .
Preliminaries and auxiliary results
The following definition was given by Branciari in .
Let be a nonempty set, and let be a mapping such that for all and all distinct points , each distinct from and :
Then is called a generalized metric space (g.m.s.).
Convergent and Cauchy sequences in g.m.s., completeness, as well as open balls , can be introduced in a standard way. However, the following example, presented by Sarma et al. in [27, Example 1.1] (see also ), shows several possible properties of generalized metrics, different than in the standard metric case.
 Let , , . Define as follows:
Then is a complete g.m.s. However, it was shown in [27, Example 1.1] that:
there exists a convergent sequence in which is not a Cauchy sequence;
there exists a sequence in converging to two distinct points;
there is no such that ;
but ; hence is not a continuous function.
As shown in the previous example, a sequence in a g.m.s. may have two limits. However, there is a special situation where this is not possible, and this will be useful in some proofs. The following lemma is a variant of [31, Lemma 1.10].
Letbe a g.m.s. and letbe a Cauchy sequence insuch thatwhenever. Thencan converge to at most one point.
Suppose, to the contrary, that , and . Since and are distinct elements, as well as and , it is clear that there exists such that and are different from for all . For , the rectangular inequality implies that
Taking the limit as , it follows that , i.e., . Contradiction.
The following lemma is a g.m.s. modification of a result which is well-known in metric spaces (see, e,g,. [32, Lemma 2.1]). Using it, many known proofs of fixed point results in g.m. spaces become much shorter.
Letbe a g.m.s. and letbe a sequence inwith distinct elements (for).Suppose that and tend toasand thatis not a Cauchy sequence. Then there existand two sequencesandof positive integers such thatand the following four sequences tend toas:
Since is not a Cauchy sequence, there exist and two sequences and of positive integers such that , and is the smallest integer satisfying this inequality, i.e., for .
Let us prove that the first of the sequences in (2.1) tends to as . Note that, by the assumption, and as . Hence, it is impossible that or (because in either of these cases it would be impossible to have ). Thus, we can apply the rectangular inequality to obtain
as , implying that as .
In order to prove that the second sequence in (2.1) tends to as , consider the following two rectangular inequalities:
which, together with imply that as .
The proof for the other two sequences can be done in a similar way, using the following rectangles:
In the following Geraghty-type result , we will use the class of real functions satisfying the condition
Note that we neither assume that the space is Hausdorff, nor that the mapping is continuous.
Letbe a g.m.s. and letbe two self maps such that,one of these two subsets ofbeing complete. If, for some function,
holds for all,thenandhave a unique point of coincidence . Moreover, for each, a corresponding Jungck sequencecan be chosen such that.
If, moreover, andare weakly compatible, then they have a unique common fixed point.
We will prove first that and cannot have more than one point of coincidence. Suppose to the contrary that there exist such that , and for some . Then (3.1) would imply that
which is impossible.
In order to prove that and have a coincidence point, take an arbitrary and, using that , choose sequences and in such that
Moreover, if for some , then we choose (and hence also ).
If for some , then is a coincidence point of and , and is their (unique) point of coincidence.
Suppose now that for each . Then, using (3.1), we get that
Hence, is a strictly decreasing sequence of positive real numbers, tending to some . Suppose that . Then, since
taking the limit as , we get that . But this implies that , a contradiction. Hence,
In a similar way, one can prove that
Suppose now that for some (and hence, by the way ’s are chosen, for ). Then, (3.1) implies that
a contradiction. Thus, in what follows, we can assume that for .
In order to prove that is a Cauchy sequence, suppose that it is not. Then, by Lemma 2, using (3.2) and (3.3), we conclude that there exist and two sequences and of positive integers such that and the sequences (2.1) tend to as . Using (3.1) with and , one obtains
Letting , it follows that , implying that , a contradiction.
Suppose, e.g., that the subspace is complete (the proof when is complete is similar). Then is a Cauchy sequence, tending to some , i.e., for some . In order to prove that , suppose that . Then, by Lemma 1, it follows that differs from both and for sufficiently large. Hence, we can apply the rectangular inequality to obtain
as . It follows that is a point of coincidence of and .
In the case when and are weakly compatible, a well-known result implies that and have a unique common fixed point.
The following example is inspired by [9, Example 2.4].
Let and be defined by:
where , i.e., . Then it is easy to check that is a g.m.s. which is not a metric space since
Consider the following mappings .
Then . Take the function defined by for and . Let us check that satisfy contractive condition (3.1) of Theorem 1. Let with and consider the following possible cases:
. Then and . Hence, (3.1) trivially holds.
, . Then , and ; , and . Hence,
, . Then , and ; , and . Hence, the inequality (3.1) is again satisfied.
All the conditions of Theorem 1 are satisfied and and have a unique point of coincidence (which is ). is also their unique common fixed point.
Taking , we get the following variant of Geraghty-theorem in generalized metric spaces.
Letbe a complete g.m.s. and letbe a self map. If, for some function,
holds for all,thenhas a unique fixed point .Moreover, for each,the corresponding Picard sequenceconverges to .
Taking in the previous corollary, we get the Banach contraction principle in generalized metric spaces, proved without the assumption that the space is Hausdorff and/or that the function is continuous. In a similar way, most of the results from the papers [1, 4, 7–10, 12, 14, 19, 20, 25, 27] can be proved without the assumption of Hausdorffness.
It is easy to see that the result of Theorem 1 remains valid if the inequality (3.1) is replaced by the following one
Altering distance functions
Recall (see ) that a mapping is called an altering distance function if:
is increasing and continuous,
The following theorem is a g.m.s. version of the main result from . Its proof completely follows the lines of proof of Theorem 1 and hence it is omitted.
Letbe a g.m.s. and letbe two self maps such that,one of these two subsets ofbeing complete. If, for some altering distance functionand some,
holds for all,thenandhave a unique point of coincidence. If, moreover, andare weakly compatible, then they have a unique common fixed point.
Letbe a complete g.m.s. and letbe a self map. If, for some altering distance functionand some,
holds for all,thenhas a unique fixed point.
It is easy to see that the result of Theorem 2 remains valid if the inequality (3.4) is replaced by the following one
In what follows, we will denote by the family of nondecreasing functions such that for each , where is the -th iterate of . Note that for and . Following , we adopt the following terminology:
Let be a nonempty set, , and .
is said to be -admissible if
If is a metric space, then is called --contractive if(3.5)
for all .
Samet et al. in , as well as several other authors proved various fixed point theorems for -admissible mappings. We will prove one such result in the context of generalized metric spaces (as a modification of [35, Theorem 2.1, Theorem 2.2]).
Letbe a complete g.m.s. andbe an--contractive mapping (for someand)satisfying the following conditions:
there existssuch thatand;
is continuous, or
if is a sequence insuch thatfor allandas,thenfor all .
Thenhas a fixed point.
Starting from given in (2), construct the sequence as , . If for some , then is a fixed point of . Assume further that for each .
Since is -admissible, it follows from (2) that
By induction, we get
Applying the inequality (3.5) and using the previous relations, we get
and by induction,
In a similar way, we obtain
Using the properties of function , we get that both sequences and are nonincreasing and tend to as . Suppose that for some , . Then
a contradiction. Hence, all elements of the Picard sequence are distinct.
In order to prove that is a Cauchy sequence in , we consider the distance in the cases and . In the first case, using the rectangular inequality and (3.6), we get that
The last expression tends to as , since the series converges, by the properties of function .
In the case , using (3.7) we obtain in a similar way that
Again, by the properties of function , it follows that as . Hence, is a Cauchy sequence that converges to some in the complete g.m.s. .
In the case (3) when the function is continuous, it immediately follows that , since a Cauchy sequence with distinct elements in cannot have two limits, by Lemma 1.
Assume now that condition holds. Using that differs from and for sufficiently large, we get that for such ,
The first two terms on the right-hand side tend to as , and for the third one we have that
as . Hence, , i.e., is a fixed point of .
Let and define by
Then it is easy to see that is a complete g.m.s. which is not a metric space.
Consider now given as
and given as
Finally, take defined by , . We will prove that:
is an --contractive mapping;
there exists such that and ;
if is a sequence in such that for all and as , then for all .
The only nontrivial case to check is when and (or vice versa). Then
It was proved in [36, Example 2.7].
For , we have .
Let be a sequence in such that for all and as . From definition of , it follows that
Then, the only possibility is that . Thus we have for all . Hence, all the conditions of Theorem 3 are satisfied, and has a fixed point (which is ).
Cone rectangular metric spaces
Cone metric spaces were defined by L. G. Huang and X. Zhang in . Following this paper, a huge number of articles appeared where various fixed point results in such spaces were proved. However, later it became clear that a lot of these results can be reduced to their standard metric counterparts using various methods. These include, among others, the so-called scalarization method  and the method of Minkowski functional .
Generalized cone metric spaces and fixed point results in them were treated in [6, 21, 22, 31]. We will show that some of these results can be deduced from the respective g.m.s. results using Minkowski functionals, similarly as in .
Let be a real Banach space with as the zero element and a solid cone with the respective order . Let be a nonempty set and satisfy the following
for all and if and only if ;
for all ;
for all and for all distinct points , both distinct from and .
Then is called a cone rectangular metric on and is called a cone rectangular metric space (or a cone g.m.s.).
Recall also the following (see, e.g., ).
If is an absolutely convex and absorbing subset of , its Minkowski functional is defined by
It is a semi-norm on . If is an absolutely convex neighborhood of in , then is continuous and
Let . Then is an absolutely convex neighborhood of ; its Minkowski functional will be denoted by .
The following theorem can be proved in a very similar way as [39, Theorem 3.2].
Letbe a cone g.m.s. over a solid coneand let.Letbe the corresponding Minkowski functional of.Thenis a (real-valued) rectangular metric on .Moreover,
For a sequencein, inif and only ifin.
is a-Cauchy sequence if and only if it is a-Cauchy sequence.
is complete if and only ifis complete.
As a consequence, most of the results from the papers [6, 21, 22, 31] can be proved by reducing them to respective known results in (standard) g.m.s. As a sample, we will state the result for the quasicontraction (in the sense of Ćirić ). It is clear that the result will remain valid for several other contractive conditions listed in the well-known Rhoades’s paper .
Recall the following recent result proved in [28, Theorem 6]:
Letbe a complete partial rectangular metric space andbe a quasicontraction, i.e., there existssuch that
for all.Thenhas a unique fixed point (and).
Since each g.m.s. is also a partial rectangular metric space, the previous theorem remains valid in generalized metric spaces. Now, applying Theorem 4, we get the following.
Letbe a complete cone g.m.s. and lethas the property that for someand for allthere exists
such that.Thenhas a unique fixed point in .
As a consequence, we get that several other fixed point results (as, e.g., Kannan’s, Chatterjea’s, Zamfirrescu’s and others, listed in ) can be proved in g.m.s. without using the assumption of Hausdorffness.
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The authors are indebted to the referees of this paper who helped us to improve its presentation. The authors are thankful to the Ministry of Education, Science and Technological Development of Serbia.
Conflict of interest
The authors declare that they have no competing interests.
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Cite this article
Kadelburg, Z., Radenović, S. Fixed point results in generalized metric spaces without Hausdorff property. Math Sci 8, 125 (2014). https://doi.org/10.1007/s40096-014-0125-6
- Generalized metric space
- Rectangular space
- Geraghty condition
- Altering distance
- Cone metric space
Mathematics Subject Classification (2010)