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On the topology and wt-distance on metric type spaces
- Nawab Hussain^{1},
- Reza Saadati^{2}Email author and
- Ravi P Agrawal^{1, 3}
https://doi.org/10.1186/1687-1812-2014-88
© Hussain et al.; licensee Springer. 2014
- Received: 17 January 2014
- Accepted: 21 March 2014
- Published: 3 April 2014
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 ${G}_{\delta}$ 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.
Keywords
- metric type space
- topologically complet
- Alexandroff theorem
- wt-distance
- fixed point theorem
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.
- (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.
- (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.
- (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 $f,g,h\in X$.
- (1)
${D}_{1}(x,y)={|x-y|}^{2}$ for any $x,y\in X$;
- (2)
${D}_{2}(x,y)={|x-y|}^{2}+{|\frac{1}{x}-\frac{1}{y}|}^{2}$, 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])
- (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])
- (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 ${lim}_{{n}_{k}\to \mathrm{\infty}}{x}_{{n}_{k}}\in A$.
Theorem 1.10 ([13])
- (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])
- (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 2.3 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.
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.
It is clear that D is of metric type on G.
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 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.
- (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 K-lower semi-continuous;
- (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 real-valued function f defined on a metric type space X is said to be lower K-semi-continuous 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 wt-distance.
Example 3.2 Let $(X,D)$ 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 $\epsilon >0$, 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 ${D}_{1}(x,y)={(x-y)}^{2}$. 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 wt-distance on X.
□
Example 3.4 Let $X=\mathbb{R}$ and ${D}_{1}(x,y)={(x-y)}^{2}$. 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 wt-distance on X.
□
- (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.
Our first main result is a fixed point theorem for graphic contractions on a partially ordered metric space endowed with a wt-distance.
- (i)for every $x\in X$ with $x\le Ax$$inf\{P(x,y)+P(x,Ax)\}>0,\phantom{\rule{1em}{0ex}}\mathit{\text{for every}}y\in X\mathit{\text{with}}y\ne Ay;$(5)
- (ii)
there exists ${x}_{0}\in X$ such that ${x}_{0}\le A{x}_{0}$.
Then A has a fixed point in X.
This is a contradiction. Therefore, we have $z=Az$. □
Another result of this type is the following.
- (i)for every $x\in X$ with $x\le Ax$$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.
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.
□
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.
- (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.
then, in general, g may be not continuous in $(X,D)$.
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.
- (a)
$g(X)\subseteq f(X)$;
- (b)
g is f-non-decreasing 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 $P(u,u)=0$.
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.
which implies that $P(g(y),g(y))=0$.
□
Put $K=2$. Then all conditions of Theorem 5.4 hold and $u=0$ is the common fixed point of f and g and $P(0,0)={|0|}^{2}=0$.
Declarations
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.
Authors’ Affiliations
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