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On the topology and wtdistance on metric type spaces
Fixed Point Theory and Applications volume 2014, Article number: 88 (2014)
Abstract
Recently, Khamsi and Hussain (Nonlinear Anal. 73:31233129, 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 ${G}_{\delta}$ set in a complete metric type space is a topologically metrizable type space. Next, we introduce the concept of wtdistance 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 wtdistance.
MSC:47H09, 47H10.
1 Preliminaries
The concept of metric type or bmetric space was introduced and studied by Bakhtin [1] and Czerwik [2]. Since then several papers have dealt with fixed point theory for singlevalued and multivalued operators in bmetric and cone bmetric 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 $D:X\times X\to [0,\mathrm{\infty})$ be a function which satisfies

(1)
$D(x,y)=0$ if and only if $x=y$;

(2)
$D(x,y)=D(y,x)$, for any $x,y\in X$;

(3)
$D(x,y)\le K(D(x,z)+D(z,y))$, for any points $x,y,z\in X$, for some constant $K\ge 1$.
The pair $(X,D)$ is called a metric type space.
Definition 1.2 Let $(X,D)$ be a metric type space.

(1)
The sequence $\{{x}_{n}\}$ converges to $x\in X$ if and only if ${lim}_{n\to \mathrm{\infty}}D({x}_{n},x)=0$.

(2)
The sequence $\{{x}_{n}\}$ is Cauchy if and only if ${lim}_{n,m\to \mathrm{\infty}}D({x}_{n},{x}_{m})=0$.
$(X,D)$ 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 $[0,1]$ such that
Define $D:X\times X\to [0,\mathrm{\infty})$ by
Then D satisfies the following properties:

(1)
$D(f,g)=0$ if and only if $f=g$;

(2)
$D(f,g)=D(g,f)$, for any $f,g\in X$;

(3)
$D(f,g)\le 2(D(f,h)+D(h,g))$, for any functions $f,g,h\in X$.
Example 1.4 Let $(\mathbb{R},\cdot )$ be metric space. Define

(1)
${D}_{1}(x,y)={xy}^{2}$ for any $x,y\in X$;

(2)
${D}_{2}(x,y)={xy}^{2}+{\frac{1}{x}\frac{1}{y}}^{2}$, for any $x,y\in X$.
Then $(\mathbb{R},{D}_{i})$, $i=1,2$ are metric type spaces with $K=2$.
Definition 1.5 Let $(X,D)$ be a metric type space. A subset $A\subset X$ is said to be open if and only if for any $a\in A$, there exists $\epsilon >0$ such that the open ball ${B}_{o}(a,\epsilon )\subset A$. The family of all open subsets of X will be denoted by τ.
Theorem 1.6 ([13])
τ defines a topology on $(X,D)$.
Theorem 1.7 ([13])
Let $(X,D)$ be a metric type space and τ be the topology defined above. Then for any nonempty subset $A\subset X$ we have

(1)
A is closed if and only if for any sequence $\{{x}_{n}\}$ in A which converges to x, we have $x\in A$;

(2)
if we define $\overline{A}$ to be the intersection of all closed subsets of X which contains A, then for any $x\in \overline{A}$ and for any $\epsilon >0$, we have
$${B}_{o}(a,\epsilon )\cap A\ne \mathrm{\varnothing}.$$
Theorem 1.8 ([13])
Let $(X,D)$ be a metric type space and τ be the topology defined above. Let $\mathrm{\varnothing}\ne A\subset X$. The following properties are equivalent:

(1)
A is compact;

(2)
For any sequence $\{{x}_{n}\}$ in A, there exists a subsequence $\{{x}_{{n}_{k}}\}$ of $\{{x}_{n}\}$ which converges, and ${lim}_{{n}_{k}\to \mathrm{\infty}}{x}_{{n}_{k}}\in A$.
Definition 1.9 The subset A is called sequentially compact if and only if for any sequence $\{{x}_{n}\}$ in A, there exists a subsequence $\{{x}_{{n}_{k}}\}$ of $\{{x}_{n}\}$ which converges, and ${lim}_{{n}_{k}\to \mathrm{\infty}}{x}_{{n}_{k}}\in A$. Also A is called totally bounded if for any $\epsilon >0$ there exits ${x}_{1},{x}_{2},\dots ,{x}_{n}\in A$ such that
Theorem 1.10 ([13])
Let $(X,D)$ be a metric type space and τ be the topology defined above. Let $\mathrm{\varnothing}\ne A\subset X$. 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 $(X,D)$ be a metric type space and suppose that $\{{x}_{n}\}$ and $\{{y}_{n}\}$ converge to $x,y\in X$, respectively. Then we have

(1)
$$\frac{1}{{K}^{2}}D(x,y)\le \underset{n}{lim\hspace{0.17em}inf}D({x}_{n},{y}_{n})\le \underset{n}{lim\hspace{0.17em}sup}D({x}_{n},{y}_{n})\le {K}^{2}D(x,y).$$
In particular, if $x=y$, then ${lim}_{n}D({x}_{n},{y}_{n})=0$.

(2)
Moreover, for each $z\in X$, we have
$$\frac{1}{K}D(x,z)\le \underset{n}{lim\hspace{0.17em}inf}D({x}_{n},z)\le \underset{n}{lim\hspace{0.17em}sup}D({x}_{n},z)\le KD(x,y).$$
2 Topologically complete metrizable type spaces
Lemma 2.1 Let $(X,D)$ be a metric type space and let $\lambda \in (0,1)$ then there exists a metric type E on X such that $E(x,y)\le \lambda $, for each $x,y\in X$, and E and D induce the same topology on X.
Proof We define $E(x,y)=min\{\lambda ,D(x,y)\}$. 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 $x,y,z\in X$. Then $E(x,z)\le \lambda $ and so $E(x,z)\le E(x,y)+E(y,z)$ when either $E(x,y)=\lambda $ or $E(y,z)=\lambda $. The only remaining case is when $E(x,y)=D(x,y)<\lambda $ and $E(y,z)=D(y,z)<\lambda $. But $D(x,z)\le K(D(x,y)+D(y,z))$ and $E(x,z)\le D(x,z)$ and so $E(x,z)\le K(E(x,y)+E(y,z))$. 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 $E({x}_{n},x)\u27f60$ if and only if $min\{\lambda ,D({x}_{n},x)\}\u27f60$ if and only if $D({x}_{n},x)\u27f60$, and we are done. □
The metric type E in the above lemma is said to be bounded by λ.
Definition 2.2 Let $(X,D)$ be a metric type space, $x\in X$ and $\mathrm{\varnothing}\ne A\subseteq X$. 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 $X=(0,1]$. The metric type space $(X,{D}_{1})$ is not complete because the Cauchy sequence $\{1/n\}$ in this space is not convergent. Now, if we consider $(X,{D}_{2})$. It is straightforward to show that $(X,{D}_{2})$ is complete. Since ${x}_{n}$ tend to x with respect to the metric type ${D}_{1}$ if and only if ${{x}_{n}x}^{2}\u27f60$ if and only if ${x}_{n}$ tend to x with respect to the metric type ${D}_{2}$, then ${D}_{1}$ and ${D}_{2}$ are equivalent. Hence the metric type space $(X,{D}_{1})$ 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 $\{({X}_{n},{D}_{n}):n\in \mathbb{N}\}$ be a collection of metrizable type spaces. Let ${\tau}_{n}$ be the topology induced by ${D}_{n}$ on ${X}_{n}$ for $n\in \mathbb{N}$ and let $(X,\tau )$ be the Cartesian product of $\{({X}_{n},{\tau}_{n}):n\in \mathbb{N}\}$ 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 ${D}_{n}$ is bounded by ${2}^{n}$ for all $n\in \mathbb{N}$, otherwise we replace ${D}_{n}$ by another metric type which induces the same topology and which is bounded by ${2}^{n}$. Points of $X={\prod}_{n\in N}{X}_{n}$ are denoted as sequences $x=\{{x}_{n}\}$ with ${x}_{n}\in {X}_{n}$ for $n\in \mathbb{N}$. Define $D(x,y)={\sum}_{n=1}^{\mathrm{\infty}}{D}_{n}({x}_{n},{y}_{n})$, for each $x,y\in X$. First note that D is well defined since ${\sum}_{n=1}^{i}{2}^{n}$ is convergent. Also D is a metric type on X because each ${D}_{n}$ is of a metric type. Let $\mathcal{U}$ be the topology induced by the metric type D. We claim that $\mathcal{U}$ coincides with τ. If $G\in \mathcal{U}$ and $x=\{{x}_{n}\}\in G$, then there exists $r>0$ such that $B(x,r)\subset G$. Now choose $N\in \mathbb{N}$ such that ${\sum}_{n=1}^{i}{2}^{n}<\frac{r}{2}$. For each $n=1,2,\dots ,{N}_{0}$, let ${V}_{n}=B({x}_{n},\frac{r}{2N})$, where the ball is with respect to the metric type ${D}_{n}$. Let ${V}_{n}={X}_{n}$ for $n>{N}_{0}$. Put $V={\prod}_{n\in N}{V}_{n}$, then $x\in V$ and V is an open set in the product topology τ on X. Furthermore $V\subset B(x,r)$, since for each $y\in V$
Hence $V\subset B(x,r)\subset G$. Therefore G is open in the product topology. Conversely suppose G is open in the product topology and let $x=\{{x}_{n}\}\in G$. Choose a standard basic open set V such that $x\in V$ and $V\subset G$. Let $V={\prod}_{n\in N}{V}_{n}$, where each ${V}_{n}$ is open in ${X}_{n}$ and ${V}_{n}={X}_{n}$ for all $n>{N}_{0}$. For $n=1,2,\dots ,{N}_{0}$, let ${r}_{n}={\mathrm{\Delta}}_{n}({x}_{n},{X}_{n}{V}_{n})$, if ${X}_{n}\ne {V}_{n}$, and ${r}_{n}={2}^{n}$, otherwise, let $r=min\{{r}_{1},{r}_{2},\dots ,{r}_{{N}_{0}}\}$. We claim that $B(x,r)\subset V$. If $y=\{{y}_{n}\}\in B(x,r)$, then $D(x,y)={\sum}_{n=1}^{\mathrm{\infty}}{D}_{n}({x}_{n},{y}_{n})<r$ and so ${D}_{n}({x}_{n},{y}_{n})<r\le {r}_{n}$ for each $n=1,2,\dots ,{N}_{0}$. Then ${y}_{n}\in {V}_{n}$, for $n=1,2,\dots ,{N}_{0}$. Also for $n>{N}_{0}$, ${y}_{n}\in {V}_{n}={X}_{n}$. Hence $y\in V$ and so $B(x,r,t)\subset V\subset G$. Therefore G is open with respect to the metric type topology and $\tau \subset \mathcal{U}$. Hence τ and $\mathcal{U}$ coincide. □
Theorem 2.6 An open subspace of a complete metrizable type space is a complete metrizable type space.
Proof Let $(X,D)$ 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 $f:G\u27f6{R}^{+}$ by $f(x)=\frac{1}{\mathrm{\Delta}(x,XG)}$ (f is undefined if $XG$ is empty, but then there is nothing to prove). For $x,y\in G$ 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 $\{{x}_{n}\}$ converges to $x\in X$, $D({x}_{n},{x}_{m})\u27f60$ if and only if $E({x}_{n},{x}_{m})\u27f60$. Since $E(x,y)\ge D(x,y)$ for all $x,y\in G$, $D({x}_{n},{x}_{m})\u27f60$ whenever $E({x}_{n},{x}_{m})\u27f60$. To prove the converse, let $D({x}_{n},{x}_{m})\u27f60$, and using Theorem 1.12, we have
Therefore
On the other hand, there exists a ${y}_{0}\in XG$, the positive sequence $\{{a}_{n}\}$ converges to zero, and ${n}_{0}\in \mathbb{N}$ such that for every $n\ge {n}_{0}$ we have
Then
and
By (1) and (2), we have
This implies $f({x}_{n})f({x}_{m})\u27f60$. Hence $E({x}_{n},{x}_{m})\u27f60$. Therefore E and D are equivalent. Next we show that E is a complete metric type. Suppose that $\{{x}_{n}\}$ is a Cauchy sequence in G with respect to E. Since for each $m,n\in \mathbb{N}$, $E({x}_{m},{x}_{n})\ge D({x}_{m},{x}_{n})$, therefore $\{{x}_{n}\}$ is also a Cauchy sequence with respect to D. By completeness of $(X,D)$, $\{{x}_{n}\}$ converges to point p in X. We claim that $p\in G$. Assume otherwise, then for each $n\in \mathbb{N}$, if $p\in XG$ and $D({x}_{n},p)\ge \mathrm{\Delta}({x}_{n},XG)$. Therefore
That is
Therefore as $n\u27f6\mathrm{\infty}$, we get $f({x}_{n})\u27f6\mathrm{\infty}$. On the other hand, $f({x}_{n})f({x}_{m})\le E({x}_{m},{x}_{n})$, for every $m,n\in \mathbb{N}$, that is, $\{f({x}_{n})\}$ is a bounded sequence. This contradiction shows that $p\in G$. Hence $\{{x}_{n}\}$ converges to p with respect to E and $(G,E)$ is a complete metrizable type space. □
Theorem 2.7 (Alexandroff)
A ${G}_{\delta}$ set in a complete metric type space is a topologically complete metrizable type space.
Proof Let $(X,D)$ be a complete metric type space and G be a ${G}_{\delta}$ set in X, that is, $G={\bigcap}_{n=1}^{\mathrm{\infty}}{G}_{n}$, where each ${G}_{n}$ is open in X. By the above theorem, there exists a complete metric type ${D}_{n}$ on ${G}_{n}$ and we may assume that ${D}_{n}$ is bounded by ${2}^{n}$. Let ℋ be the Cartesian product ${\prod}_{n=1}^{\mathrm{\infty}}{G}_{n}$ with the product topology. Then ℋ is a complete metrizable type space. Now, for each $n\in \mathbb{N}$ let ${f}_{n}:G\u27f6{G}_{n}$ be the inclusion map. So the evaluation map $e:G\u27f6\mathcal{H}$ is an embedding. Image of e is the diagonal $\mathfrak{D}G$ which is a closed subset of ℋ and by Corollary 1.11, $\mathfrak{D}G$ is complete. Thus $\mathfrak{D}G$ is a complete metrizable type space and so is G which is homeomorphic to it. □
3 wtDistance
Kada et al. [16] introduced in 1996, the concept of wdistance on a metric space and proved some fixed point theorems. In this section, we introduce the definition of a wtdistance and we state a lemma which we will use in the main sections of this work.
Definition 3.1 Let $(X,D)$ be a metric type space with constant $K\ge 1$. Then a function $P:X\times X\u27f6[0,\mathrm{\infty})$ is called a wtdistance on X if the following are satisfied:

(a)
$P(x,z)\le K(P(x,y)+P(y,z))$ for any $x,y,z\in X$;

(b)
for any $x\in X$, $P(x,\cdot ):X\u27f6[0,\mathrm{\infty})$ is Klower semicontinuous;

(c)
for any $\epsilon >0$, there exists $\delta >0$ such that $P(z,x)\le \delta $ and $P(z,y)\le \delta $ imply $D(x,y)\le \epsilon $.
Let us recall that a realvalued function f defined on a metric type space X is said to be lower Ksemicontinuous at a point ${x}_{0}$ in X if either ${lim\hspace{0.17em}inf}_{{x}_{n}\to {x}_{0}}f({x}_{n})=\mathrm{\infty}$ or $f({x}_{0})\le {lim\hspace{0.17em}inf}_{{x}_{n}\to {x}_{0}}Kf({x}_{n})$, whenever ${x}_{n}\in X$ for each $n\in \mathbb{N}$ and ${x}_{n}\to {x}_{0}$ [17].
Let us give some examples of wtdistance.
Example 3.2 Let $(X,D)$ be a metric type space. Then the metric D is a wtdistance on X.
Proof (a) and (b) are obvious. To show (c), for any $\epsilon >0$, put $\delta =\frac{\epsilon}{2K}$. Then we see that $P(x,z)\le \delta $ and $P(z,y)\le \delta $ imply $D(x,y)\le \epsilon $. □
Example 3.3 Let $X=\mathbb{R}$ and ${D}_{1}(x,y)={(xy)}^{2}$. Then the function $P:X\times X\u27f6[0,\mathrm{\infty})$ defined by $P(x,y)={x}^{2}+{y}^{2}$ for every $x,y\in X$ is a wtdistance on X.
Proof (a) and (b) are obvious. To show (c), for any $\epsilon >0$, put $\delta =\frac{\epsilon}{2}$. Then we have
□
Example 3.4 Let $X=\mathbb{R}$ and ${D}_{1}(x,y)={(xy)}^{2}$. Then the function $P:X\times X\u27f6[0,\mathrm{\infty})$ defined by $P(x,y)={y}^{2}$ for every $x,y\in X$ is a wtdistance on X.
Proof (a) and (b) are obvious. To show (c), for any $\epsilon >0$, put $\delta =\frac{\epsilon}{2}$. Then we have
□
Lemma 3.5 Let $(X,D)$ be a metric type space with constant $K\ge 1$ and P be a wtdistance on X. Let $\{{x}_{n}\}$ and $\{{y}_{n}\}$ be sequences in X, let $\{{\alpha}_{n}\}$ and $\{{\beta}_{n}\}$ be sequences in $[0,\mathrm{\infty})$ converging to zero, and let $x,y,z\in X$. Then the following hold:

(1)
If $P({x}_{n},y)\le {\alpha}_{n}$ and $P({x}_{n},z)\le {\beta}_{n}$ for any $n\in \mathbb{N}$, then $y=z$. In particular, if $P(x,y)=0$ and $P(x,z)=0$, then $y=z$;

(2)
if $P({x}_{n},{y}_{n})\le {\alpha}_{n}$ and $P({x}_{n},z)\le {\beta}_{n}$ for any $n\in \mathbb{N}$, then $D({y}_{n},z)\to 0$;

(3)
if $P({x}_{n},{x}_{m})\le {\alpha}_{n}$ for any $n,m\in \mathbb{N}$ with $m>n$, then $\{{x}_{n}\}$ is a Cauchy sequence;

(4)
if $P(y,{x}_{n})\le {\alpha}_{n}$ for any $n\in \mathbb{N}$, then $\{{x}_{n}\}$ 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 $(X,\le )$ is a partially ordered set and $f:X\to X$ be a self mapping on X. We say f is inverse increasing if for $x,y\in X$,
Our first main result is a fixed point theorem for graphic contractions on a partially ordered metric space endowed with a wtdistance.
Theorem 4.2 Let $(X,\le )$ be a partially ordered set and let $D:X\times X\to [0,\mathrm{\infty})$ be a metric type on X such that $(X,D)$ is a complete metric type space with constant $K\ge 1$. Suppose that P is a wtdistance in $(X,D)$. Let $A:X\to X$ be a nondecreasing mapping and there exists $r\in [0,1)$ such that
and $Kr<1$. Suppose also that:

(i)
for every $x\in X$ with $x\le Ax$
$$inf\{P(x,y)+P(x,Ax)\}>0,\phantom{\rule{1em}{0ex}}\mathit{\text{for every}}y\in X\mathit{\text{with}}y\ne Ay;$$(5) 
(ii)
there exists ${x}_{0}\in X$ such that ${x}_{0}\le A{x}_{0}$.
Then A has a fixed point in X.
Proof If $A{x}_{0}={x}_{0}$, then the proof is finished. Suppose that
Since ${x}_{0}\le A{x}_{0}$ and A is nondecreasing, we obtain
Hence, for each $n\in \mathbb{N}$ we have
Then, for $n\in \mathbb{N}$ with $m>n$, we successively have
By Lemma 3.5(3), we conclude that $\{{A}^{n}{x}_{0}\}$ is Cauchy sequence in $(X,D)$. Since $(X,D)$ is a complete metric type space, there exists $z\in X$ such that
Let $n\in \mathbb{N}$ be an arbitrary but fixed. Then since $\{{A}^{n}{x}_{0}\}$ converges to z in $(X,D)$ and $P({A}^{n}{x}_{0},\cdot )$ is Klower semicontinuous, we have
Assume that $z\ne Az$. Since ${A}^{n}{x}_{0}\le {A}^{n+1}{x}_{0}$, by (5), we have
This is a contradiction. Therefore, we have $z=Az$. □
Another result of this type is the following.
Theorem 4.3 Let $(X,\le )$ be a partially ordered set, let $D:X\times X\to [0,\mathrm{\infty})$ be of a metric type on X such that $(X,D)$ is a complete metric type space with constant $K\ge 1$. Suppose that P is a wtdistance in $(X,D)$. Let $A:X\to X$ be a nondecreasing mapping and there exists $r\in [0,1)$ such that
and $Kr<1$. Assume that one of the following assertions holds:

(i)
for every $x\in X$ with $x\le Ax$
$$inf\{P(x,y)+P(x,Ax)\}>0,\phantom{\rule{1em}{0ex}}\mathit{\text{for every}}y\in X\mathit{\text{with}}y\ne Ay;$$(8) 
(ii)
if both $\{{x}_{n}\}$ and $\{A{x}_{n}\}$ converge to y, then $y=Ay$;

(iii)
A is continuous.
If there exists ${x}_{0}\in X$ with ${x}_{0}\le A{x}_{0}$, 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 $y\in X$ with $y\ne Ay$ such that
Then there exists $\{{z}_{n}\}\in X$ such that ${z}_{n}\le A{z}_{n}$ and
Then $P({z}_{n},y)\u27f60$ and $P({z}_{n},A{z}_{n})\u27f60$. By Lemma 3.5, we have that $A{z}_{n}\u27f6y$. We also have
Again by Lemma 3.5, we get ${A}^{2}{z}_{n}\u27f6y$. Put ${x}_{n}=A{z}_{n}$. Then both $\{{x}_{n}\}$ and $\{A{x}_{n}\}$ converges to y. Thus, by (ii) we have $y=Ay$. Thus (ii) ⟹ (i) holds.
Now, we show that (iii) ⟹ (ii). Let A be continuous. Further assume that $\{{x}_{n}\}$ and $\{A{x}_{n}\}$ 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 $(X,d)$ and let $g:X\u27f6X$ be another mapping, such that the following conditions are satisfied:

(a)
$g(X)\subseteq f(X)$;

(b)
g commutes with f;

(c)
$d(g(x),g(y))\le kd(f(x),f(y))$, for all $x,y\in X$ and for some $0\le k<1$.
Then f and g have a unique common fixed point.
The next example shows that if the mapping $f:X\to X$ is continuous with respect to a metric type D on X and $g:X\to X$ satisfies the condition
then, in general, g may be not continuous in $(X,D)$.
Example 5.2 Let $X:=(\mathbb{R},\cdot )$ be a normed linear space. Consider Example 3.4 with wtdistance defined by
Consider the functions f and g defined by $f(x)=4$ and
Then
Definition 5.3 Let $(X,\le )$ be a partially ordered set and $g,h:X\to X$. By definition, we say that g is hnondecreasing if for $x,y\in X$,
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 wtdistance.
Theorem 5.4 Let $(X,\le )$ be a partially ordered set, let $D:X\times X\to [0,\mathrm{\infty})$ be a metric type on X such that $(X,D)$ is a complete metric type space with constant $K\ge 1$. Suppose that P is a wtdistance on X. Let $f,g:X\u27f6X$ be mappings that satisfy the following conditions:

(a)
$g(X)\subseteq f(X)$;

(b)
g is fnondecreasing and f is inverse increasing;

(c)
g commutes with f and f, g are continuous in $(X,D)$;

(d)
$P(g(x),g(y))\le rP(f(x),f(y))$ for all $x,y\in X$ with $x\le y$ and some $0<r<1$ such that $rK<1$.

(e)
there exists ${x}_{0}\in X$ such that:

(i)
$f({x}_{0})\le g({x}_{0})$ and

(ii)
$f({x}_{0})\le f(g({x}_{0}))$.

(i)
Then f and g have a common fixed point $u\in X$. Moreover, if $g(v)={g}^{2}(v)$ for all $v\in X$, then $P(u,u)=0$.
Proof We claim that for every $f(x)\le g(x)$
for every $z\in X$ with $g(z)\ne g(g(z))$. For the moment suppose the claim is true. Let ${x}_{0}\in X$ with $f({x}_{0})\le g({x}_{0})$. By (a) we can find ${x}_{1}\in X$ such that $f({x}_{1})=g({x}_{0})$. By induction, we can define a sequence ${\{{x}_{n}\}}_{n}\in X$ such that
Since $f({x}_{0})\le g({x}_{0})$ and $f({x}_{1})=g({x}_{0})$, we have
Then from (b),
that means, by (10), that $f({x}_{1})\le f({x}_{2})$. Again by (b) we get
that is, $f({x}_{2})\le f({x}_{3})$. By this procedure, we obtain
Hence from (10) and (12) we have $f({x}_{n1})\le f({x}_{n})$ and by (3) we have ${x}_{n1}\le {x}_{n}$. By induction we get
for $n=1,2,\dots $ . This implies that, for $m,n\in \mathbb{N}$ with $m>n$,
Thus, by Lemma 3.5, we find that $\{f({x}_{n})\}$ is a Cauchy sequence in $(X,D)$. Since $(X,D)$ is complete, there exists $y\in X$ such that ${lim}_{n\to \mathrm{\infty}}f({x}_{n})=y$. As a result the sequence $g({x}_{n1})=f({x}_{n})$ tends to y as $n\to +\mathrm{\infty}$ and hence ${\{g(f({x}_{n}))\}}_{n}$ converges to $g(y)$ as $n\to +\mathrm{\infty}$. However, $g(f({x}_{n}))=f(g({x}_{n}))$, by the commutativity of f and g, implies that $f(g({x}_{n}))$ converges to $f(y)$ as $n\to +\mathrm{\infty}$. Because limit is unique, we get $f(y)=g(y)$ and, thus, $f(f(y))=f(g(y))$. On the other hand, by Klower semicontinuity of $P(x,\cdot )$ we have, for each $n\in \mathbb{N}$,
Notice that, by (11), (10), and (9), we obtain $f({x}_{0})\le f(f({x}_{1}))$ and thus, by (9), we get $g({x}_{0})\le g(f({x}_{1}))$. Then
By (9) we get $g({x}_{1})\le g(f({x}_{2}))$ and thus $f({x}_{2})\le f(g({x}_{2}))$. Continuing this process we get
and by (3) we get
Using now the condition (d), we have
We will show that $g(y)=g(g(y))$. Suppose, by contradiction, that $g(y)\ne g(g(y))$. Then we have
This is a contradiction. Therefore $g(y)=g(g(y))$. Thus, $g(y)=g(g(y))=f(g(y))$. Hence $u:=g(y)$ is a common fixed point of f and g.
Furthermore, since $g(v)=g(g(v))$ for all $v\in X$, we have
which implies that $P(g(y),g(y))=0$.
Now it remains to prove the initial claim. Assume that there exists $y\in X$ with $g(y)\ne g(g(y))$ and
Then there exists $\{{x}_{n}\}$ such that
Since $P(f({x}_{n}),g({x}_{n}))\u27f60$ and $P(f({x}_{n}),y)\u27f60$, by Lemma 3.5, we have
Also, since $P(g({x}_{n}),y)\u27f60$ and $P(g({x}_{n}),g(g({x}_{n})))\u27f60$, by Lemma 3.5, we have
By (13), (14), and the continuity of g we have
Therefore, $g(y)=g(g(y))$, which is a contradiction. Hence, if $g(y)\ne g(g(y))$, then
□
Example 5.5 Let $X:=(\mathbb{R},\cdot )$ be a normed linear space. Consider Example 3.4 with wtdistance defined by
Consider the functions f and g defined by $f(x)=3x$ and $g(x)=\sqrt{2}x$. Then
Put $K=2$. Then all conditions of Theorem 5.4 hold and $u=0$ is the common fixed point of f and g and $P(0,0)={0}^{2}=0$.
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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 wtdistance on metric type spaces. Fixed Point Theory Appl 2014, 88 (2014). https://doi.org/10.1186/16871812201488
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Keywords
 metric type space
 topologically complet
 Alexandroff theorem
 wtdistance
 fixed point theorem