We provide a construction for the completion of a dislocated metric space (abbreviated d-metric space); we also prove that the completion of the metric associated with a d-metric coincides with the metric associated with the completion of the d-metric.
1. Introduction
Completion of a metric space via Cauchy sequences can be achieved because of certain convergence properties enjoyed by the metric and the property that convergent sequences are Cauchy sequences. Lack of some of these properties in weaker forms of metric spaces comes in the way of completion process in the above lines. In semimetric spaces several new ways of completeness were invented, for example, Cauchy completeness, McAuley notions of strong and weak completeness [1], Moore completeness [2], and so on. Moshokoa [3] introduced the notion of convergence completeness for semimetric spaces and discussed completion on these lines.
For d-metric spaces adoption of Van der-Waerdens completion process through Cauchy sequences is possible but is not routine, the difficulty being the mischief created by the isolated points. Here we show how to overcome this problem.
In his study of programming languages, Hitzler [4] associated a metric d′ with every d-metric by defining (1)d′a,b=da,b,ifa≠b0,ifa=b.
We establish that the metric associated with the completion of a d-metric is the completion of the metric associated with d.
We recall [4] where a distance function on a set X is said to be a d-metric on X if
d(x,y)=d(y,x);
d(x,y)=0⇒x=y;
d(x,z)≤d(x,y)+d(y,z) for all x,y,z in X.
If d is a d-metric on X then (X,d) is called a d-metric space. Many authors (see, e.g. [5–9]) have studied fixed point theorems in d-metric spaces but topology and topological aspects on this space are discussed by Sarma and Kumari [10].
The class B={Vϵ(x)/x∈X and ϵ>0} is an open base for the topology Jd induced by d, where Vϵ(x)={y/d(x,y)<ϵ}∪{x}. In what follows whenever we talk about topological properties of a d-metric space, we refer to the topology Jd.
In [11], the authors highlighted some convergence properties and covers a huge range of implications and nonimplications among them. By using these convergence axioms many authors (see, e.g. [12–15]) have proved fixed point theorems in certain spaces.
The presence of the triangle inequality lends the Haussdorff property for d and some nice properties to (X,d). In particular (X,d) satisfies properties C1 through C5:
limd(xn,yn)=0=limd(xn,x)⇒limd(yn,x)=0;
limd(xn,x)=0=limd(yn,x)⇒limd(xn,yn)=0;
limd(xn,yn)=0=limd(yn,zn)⇒limd(xn,zn)=0;
limd(xn,x)=0⇒limd(xn,y)=d(x,y);
limd(xn,x)=limd(xn,y)=0⇒x=y; for all x,y∈X.
Above mentioned convergence axioms can be found in [11]. If the triangular inequality is deleted from the axioms on d then it is difficult to define the concept of completion of the resulting distance space. In such an amorphous space, even constant sequences may fail to converge. This and related difficulties compel us to retain the triangle inequality in the discussion of completeness.
Definition 1.
Let (X,d) and (Y,d′) be distance spaces. A map f:X→Y is called an isodistance if for any x,y∈X one has dfx,fy=d′x,y.
2. Completion
In what follows, d is a d-metric on a nonempty set X. A complete d-metric space is a d-metric space in which every Cauchy sequence converges. “Cauchy sequences” in d-metric spaces are defined exactly as in metric spaces.
Lemma 2.
x is an isolated point of X if and only if X=x or infy≠xd(x,y)>0.
Proof.
Suppose x is an isolated point of X. Then there exists r>0 such that y≠x⇒d(x,y)>r⇒X={x} or infy≠xd(x,y)≥r>0. Conversely suppose X={x} or infy≠xd(x,y)>0. If X={x}, then clearly x is an isolated point of X. If X≠{x}, then infy≠xd(x,y)=r>0 which implies that d(x,y)≥r>r/2 for all y≠x. Hence x is an isolated point of X.
Corollary 3.
If d(x,x)>0, then x is an isolated point of X.
Proof.
If y≠x, then d(x,x)≤d(x,y)+d(y,x)=2d(x,y) and so (1/2)d(x,x)≤d(x,y) for all y in X. So X={x} or infy≠xd(x,y)≥(1/2)d(x,x)>0.
Theorem 4.
Let (X,d) be a d-metric space. Then there exists a complete d-metric space (X∗,d∗) and an isodistance T:(X,d)→(X∗,d∗) such that T(X) is dense in X∗.
Proof.
Let I be the collection of isolated points of X and let J=X-I. Let I¯ be the collection of sequences in X which are ultimately a constant element lying in I and J¯ denote the class of Cauchy sequences in J. We define relations RI and RJ, respectively, on I¯ and J¯ as follows.
If (xn)(yn) are sequences in I¯ then (xn)RI(yn) iff the ultimately constant value of (xn) coincides with that of (yn).
If (xn)(yn) are sequences in J¯ then (xn)RJ(yn) iff limn→∞d(xn,yn)=0. Clearly RI is an equivalence relation. We verify that RJ is an equivalence relation. Suppose (xn)∈J¯ and ϵ>0. Since (xn) is a Cauchy sequence in J, d(xn,xn)=0 and hence RJ is reflexive.
Suppose (xn)RJ(yn) for (xn),(yn)∈J¯. Then limn→∞d(xn,yn)=limn→∞d(yn,xn)=0. Hence RJ is symmetric.
If (xn),(yn),(zn)∈J¯,(xn)RJ(yn) and (yn)RJ(zn). If ϵ>0, then there exists an integer n1 such that d(xn,yn)<ϵ/2 and d(yn,zn)<ϵ/2, if n>n1. Consider(2)dxn,zn≤dxn,yn+dyn,zn<ϵ2+ϵ2=ϵifn>n1.This proves that RJ is transitive and hence an equivalence relation. Let X¯=I¯∪J¯. Then ~=RI∪RJ is an equivalence relation on X¯.
Let X∗ denote X¯/~. If xn∈X¯,[(xn)] denotes the equivalence class in X∗ containing the sequence (xn).
If x∈X let (x) be the constant sequence (xn) where xn=x, ∀n and x^=[(x)], the equivalence class containing (x).
If (xn)∈J¯, (yn)∈J¯, it follows from the triangle inequality that |d(xn,yn)-d(xm,ym)|≤d(xn,xm)+d(yn,ym). Since (xn),(yn) are Cauchy sequences, given that ϵ>0, there exists a positive integer n0 such that d(xn,xm)<ϵ/2 and d(yn,ym)<ϵ/2 for all n,m≥n0.
This implies that |d(xn,yn)-d(xm,ym)|<ϵ proving that (d(xn,yn)) is a Cauchy sequence of real numbers. By the completeness of R this sequence converges. The definition of RJ makes it obvious that limn→∞d(xn,yn) is independent of the choice of the representative sequences (xn),(yn), respectively, from the classes xn,[(yn)].
We can prove similarly if x∈X and (yn)∈J¯, (zn)∈J¯, limd(x,yn), limd(x,zn) exists and or equal. Provided (yn) and (zn) belong to the same equivalence class.
We define d∗:X∗×X∗→[0,∞) as follows:
d∗(xn,[(yn)])=d(x,y) if (xn),(yn)∈I¯ and x,y are, respectively, the ultimately constant terms of (xn),(yn).
d∗(xn,[(yn)])=limn→∞d(x,yn) if (xn)∈I¯, (yn)∈J¯ and xn=x eventually.
If (xn)∈J¯, (yn)∈I¯, then define d∗(xn,[(yn)])=d∗(Yn,[(Xn)]).
If (xn)∈J¯, (yn)∈J¯, then define d∗(xn,[(yn)])=limn→∞d(xn,yn).
Verification That d∗ Is a d-Metric on X∗. Clearly d∗(x∗,y∗)≥0 and d∗(x∗,y∗)=d∗(y∗,x∗) for x∗,y∗∈X∗.
Suppose d∗(x∗,y∗)=0. Let (xn)∈x∗ and (yn)∈y∗. We first see that (xn),(yn) either are both in I¯ or are both in J¯.
Suppose, on the contrary, (xn)∈I¯ and (yn)∈J¯. Let x be the ultimately constant value of (xn). Consider(3)0≤dx,x≤2dx,yn∀n,⟹0=d∗x∗,y∗=limn→∞dx,yn.Hence 0≤d(x,x)≤limn→∞2d(x,yn)=0, contrary to the fact that x∈I.
Suppose x∗,y∗∈I¯, (xn)∈x∗, and (yn)∈y∗ with a,b the ultimately constant values of (xn) and (yn), respectively.
Then d∗(x∗,y∗)=0⇒d(a,b)=0⇒a=b⇒(xn)~(yn)⇒x∗=y∗.
Suppose x∗,y∗∈J¯, (xn)∈x∗ and (yn)∈y∗. Consider (4)d∗x∗,y∗=0⟹limn→∞dxn,yn=0⟹xn~yn⟹x∗=y∗.
Verification of the triangular inequality is routine.
Embedding of X in X∗. Define T:X→X∗ by T(x)=x^. It is clear that T is an isodistance. We now verify that T(x) is dense in X∗. Let [(xn)]∈X∗ and ϵ>0.
Case (i) ((xn)∈I¯). In this case let “a” be the ultimately constant value of (xn).
Then by the definition of T,a^=[(xn)]∈T(X).
Then a^=[(xn)]. Thus [(xn)]∈T(X) in this case.
Case (ii) ((xn)∈J¯). There exists a positive integer n0 such that d(xn,xm)<ϵ if n,m≥n0. Let xn0=a. Then since a∈J, d(a,a)=0, (5)d∗x,a^=limn→∞dxn,a≤ϵ.Hence T(X) is dense in X∗.
(X∗,d∗) Is Complete. Let (xn∗) be a Cauchy sequence in X∗, and ϵ>0. There exists n0 such that n≥m≥n0 implies d∗(xn∗,xm∗)<ϵ/3.
There is no harm in assuming that n0>ϵ/3. Since T(X) is dense in X∗, for each positive integer n, there exists zn in X such that d(xn∗,z^n)<1/n.
Hence(6)d∗z^n,z^m≤d∗z^n,xn∗+d∗xn∗,xm∗+d∗xm∗,z^m<1n+1m+ϵ3<ϵ3+ϵ3+ϵ3=ϵifn,m≥n0.Hence (z^n) is a Cauchy sequence in T(X). Since T is an isodistance, (zn) is a Cauchy sequence in X.
Moreover, d(zn,zm)=d∗(z^n,z^m)<ϵ, if n≥m≥n0.
Let z∗ denote [(zn)], by the triangle inequality:
<(1/n)+limmd(zn,zm)<(2ϵ/3)<ϵ for n≥n0;
d∗(xn∗,z∗)≤d∗(xn∗,z^n)+d∗(z^n,z∗);
⇒limmd∗(xn∗,z∗)=0 proving that (X∗,d∗) is complete.
Definition 5.
Let (X,d) and (X1,d1) be d-metric spaces. (X1,d1) is said to be a completion of (X,d) if
(X1,d1) is complete;
there is an isodistance T:(X,d)→(X1,d1) such that T(X) is dense in X1.
Note. If (X,d) is a complete metric space then its completion is (X,d) itself.
Lemma 6.
Let (X,d) be a d-metric space and let (X1,d1) be a completion of (X,d). Let T:X→X1 be isodistance embedding X in X1 with T(X) dense in X1. Then a point y of X1 is an isolated point if and only if y=T(X) for some isolated point x of X1.
Proof.
Suppose y is an isolated point of X1. If y is not in T(X), then since T(X) is dense in X1, there exists a sequence T(xn) in T(X) such that limn→∞d(T(xn),y)=0.
By Lemma 2, it follows that y is not an isolated point of X1, a contradiction so that y=T(X) for some x∈X. Hence Tx is an isolated point of X1 and hence that of T(X). Since X and T(X) are isometric, x is an isolated point of X.
Conversely, suppose x is an isolated point of X. If T(X) is not an isolated point of X1, then for each positive integer k, there exists xk in X1 such that 0<d1(xk,T(x))≤1/2k. Since xk∈X1, either xk∈T(X) or there exists yn in T(X) such that 0<d1((yk),xk)<d1(xk,T(x)).
Now(7)0<d1yk,Tx≤d1yk,xk+d1xk,Tx≤12k+12k=1k.
Also yk≠x since d1((yk),xk)<d1(xk,T(x)).
Hence 0<d1(yk,T(x))<1/k which, by Lemma 2, contradicts the fact that T(x) is an isolated point of T(X).
Theorem 7.
Let (X,d) be a d-metric space, (X1,d1) and (X2,d2) completion of (X,d), and Ti:(X,d)→(Xi,di)(i=1,2) isometrics such that Ti(x) is dense in Xi. Then there exists an isodistance T:(X1,d1)→onto(X2,d2) such that following diagram is commutative.
Proof.
Consider the following:(8)
Definition of T. If x∈X1 and x is an isolated point of X1, then T1-1(x) is an isolated point of X; hence T2(T1-1(x)) is an isolated point of X2.
Define T(x)=T2(T1-1(x)). If x∈X1 and is not an isolated point, there exists a sequence (zn) in X such that {T1zn} converges to x in (X1,d1).
Since T1 is an isodistance and {T1zn} is convergent and hence a Cauchy sequence, it follows that {zn} is a Cauchy sequence in X. Since T2 is an isodistance and {zn} is a Cauchy sequence, it follows that {T2zn} is a Cauchy sequence in (X2,d2). Since (X2,d2) is complete, there exists z∈X2 such that limd2(T2zn,z)=0. Clearly this z is independent of the choice of the sequence {zn} in X.
Define T(x)=z. Clearly TT1=T2 and bijection.
T Is an Isodistance. If x,y∈X, T(T1(x))=T2(x) and T(T1(y))=T2(y).
So d2(T(T1(x)),T(T1(y)))=d2((T2(x)),T2(y))=d2(x,y)=d1((T1(x)),T1(y)).
If x,y∈X1-X and x=limT1xn, y=limT1yn where xn,yn∈X, then(9)d2Tx,Ty=d2limT2xn,limT2yn=limd2limT2xn,limT2yn=limdxn,yn=d1limT1xn,limT1yn=d1x,y.The arguments for the cases when x∈X1-X and y∈X or x∈X and y∈X1-X are similar. Hence T is an isodistance. Interchanging the places of X1 and X2, we get in a similar way an isodistance S:X2→X1 such that ST2=T1.
Since ST2=T1 and TT1=T2, we have TST2=TT1 and STT1=ST2=T1.
Since T(X) is dense in X1 and T2(x) in X2, we get TS = identity on X1 and ST is identity on X2.
Hence S and T are bijections.
3. Completion of the Metric Associated with a d-Metric
If d is a d-metric on X then d′ is a metric on X if d′ is defined by d′(x,y)=d(x,y) when x≠y and d′(x,y)=0 for all x,y in X.
Suppose (X¯,d¯) is the completion of (X,d); then d¯ gives rise to a metric d¯′ defined by d¯′(x,y)=d¯(x,y) for all x,y∈X¯ and d¯(x,x)=0 for all x,y∈X¯.
Also, the metric space (X,d′) has a metric space (X0,d0) as its completion. In this section, we prove that the metric spaces (X¯,d¯′) and (X0,d0) are isometric.
Definition 8.
Let X,d be a d-metric space. Define ρ on X×X by(10)ρx,y=dx,y,ifx≠y0,ifx=y.
ρ is a metric on X and is called the metric associated with d.
Clearly 0≤ρ(x,y)≤d(x,y)∀x,y and d(x,x)≤2ρ(x,y) whenever x≠y. If s∈{ρ,d}, r>0 and x∈X.
Write Brs(x)={y/s(x,y)<r}. Then Brρ(x)=Brd(x)∪{x} and Vrs(x)=Brs(x)∪{x}.
The collection {Vrd(x)/x∈X,r>0} and Vrρ(x)={Vrρ(x)/x∈X,r>0} generate the same topology on X. However, convergent sequences in X are not necessarily the same since constant sequences are convergent sequences with respect to ρ, while this holds with respect to d for x with d(x,x)=0 only.
Existence of points with positive self-distance leads to unpleasantness in the extension of the concept of continuity in metric spaces as well. This is evident from the following.
Example 9.
Let d be a d-metric on a set X which is not a metric. So that the set A={x/d(x,x)≠0} is nonempty. If ρ is a metric associated with d then the identity map i:(X,ρ)→(X,d) is continuous in the usual sense. But if x∈A, the constant sequence (x) converges in (X,ρ) while it does not converge in (X,d).
If (X,d),(Y,ρ) we call f:X→Y sequentially d-continuous if limd(xn,x)=0⇒limρ(f(xn),f(x))=0.
If s∈{ρ,d} and {xn} is a sequence in X, we say that X is s-Cauchy sequence or simply s-Cauchy if {xn} is a Cauchy sequence in (X,s).
Proposition 10.
limρ(xn,x)=0⇔ either
xn=x eventually or
(xn) can be split into subsequences (yn) and (zn) where yn=x for every n, zn≠x for any n and limd(zn,x)=0.
Proof.
Routine.
Proposition 11.
If a sequence (xn) in X is d-Cauchy then (xn) is ρ-Cauchy. Conversely if (xn) is ρ-Cauchy and is not eventually constant, then (xn) is d-Cauchy.
Proof.
Since 0≤ρ(xn,xm)≤d(xn,xm), d-Cauchy ⇒ρ-Cauchy.
Conversely suppose that (xn) is ρ-Cauchy, given ϵ>0∃N(ϵ) such that ρ(xn,xm)<ϵ if n≥N(ϵ) and m≥N(ϵ). So if m≥N(ϵ), n≥N(ϵ), and xn≠xm, then d(xm,xn)<ϵ.
Since (xn) is not eventually constant and n≥N(ϵ), there exists m≥N(ϵ) such that xm≠xn. Then(11)dxn,xn≤dxn,xmdxm,xn=2dxm,xn=2dxn,xn<2ϵ.Thus if (xn) is not eventually constant then for all n≥N(ϵ) and m≥N(ϵ), d(xn,xm)<2ϵ. Hence (xn) is d-Cauchy.
Example 12.
Let X=(0,∞) and d(x,y)=x+y; then(12)ρx,y=x+y,ifx≠y0,ifx=y.
If (xn) is any eventually nonconstant sequence in (0,∞), then (xn) is d-Cauchy if and only if ∀ϵ>o there exists N(ϵ) such that xn+xm<ϵ for n≥m≥N(ϵ). This implies that limxn=0.
However, if limxn=0, then ∀ϵ>0∃N(ϵ), such that xn<ϵ/2 for m≥N(ϵ), n≥N(ϵ).
Hence xn+xm<ϵ for m≥N(ϵ), n≥N(ϵ). However, constant sequences are not d-Cauchy but ρ-Cauchy.
Theorem 13.
Let (X,d) be a metric space, ρ the metric associated with d on X, (X∗,d∗) the completion of (X,d), and ρ∗ the metric associated with d∗ on X∗. Then (X∗,ρ∗) is the completion of (X,ρ). In particular if (X,d) is a complete metric space then (X,ρ) is a complete metric space. We prove that
X is dense in (X∗,ρ∗);
every ρ∗-Cauchy sequence in X∗ is ρ∗-convergent.
Proof of (i).
Let x∗∈X∗-X. Then there exists a sequence (xn) in X such that limd∗(xn∗,x∗)=0 since xn∈X, xn≠x∗∀n. So that limρ∗(xn∗,x∗)=0.
This implies that X is dense in (X∗,ρ∗).
Proof of (ii).
Let (xn∗) be ρ∗-Cauchy in X∗. If (xn∗) is eventually constant, then there exist N and x∗∈X∗ such that xn∗=x∗ for n≥N.
In this case limρ∗(xn∗,x∗)=0 for n≥N; hence (xn∗) is ρ∗-convergent.
Suppose (xn∗) is not eventually constant. Then (xn∗) is a d∗-Cauchy sequence. Since (X∗,d∗) is complete, there exists x∗∈X∗ such that limd∗(xn∗,x∗)=0. Since 0≤ρ∗(xn∗,x∗)≤d∗(xn∗,x∗)=0, limρ∗(xn∗,x∗)=0.
Hence (xn∗) is ρ∗-convergent to x∗. This completes the proof of (ii).
Disclosure
I. Ramabhadra Sarma is a retired professor from Acharya Nagarjuna University, Andhra Pradesh, India.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgments
The authors thank the editor and anonymous referees for their constructive comments and ideas which helped us to improve the paper.
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