ON THE TOPOLOGY OF D-METRIC SPACES AND GENERATION OF D-METRIC SPACES FROM METRIC SPACES

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PII. S0161171204311257 http://ijmms.hindawi.com

ON THE TOPOLOGY OF D-METRIC SPACES AND GENERATION OF D-METRIC SPACES FROM METRIC SPACES

S. V. R. NAIDU, K. P. R. RAO, and N. SRINIVASA RAO Received 20 November 2003

An example of aD-metric space is given, in whichD-metric convergence does not define a topology and in which a convergent sequence can have infinitely many limits. Certain methods for constructingD-metric spaces from a given metric space are developed and are used in constructing (1) an example of aD-metric space in whichD-metric convergence defines a topology which isT1but not Hausdorff, and (2) an example of aD-metric space in whichD- metric convergence defines a metrizable topology but theD-metric is not continuous even in a single variable.

2000 Mathematics Subject Classiﬁcation: 54E99.

1. Introduction. Dhage [2] introduced the notion ofD-metric spaces and claimed thatD-metric convergence defines a Hausdorff topology and that theD-metric is (sequentially) continuous in all the three variables. Many authors (see [1,2,3,4,5,6,7, 8,9,10,11,12,13,14]) have taken these claims for granted and used them in proving fixed point theorems inD-metric spaces.

In this paper, we give examples to show that in aD-metric space (1) D-metric convergence does not always deﬁne a topology,

(2) even whenD-metric convergence defines a topology, it need not be Hausdorff, (3) even when D-metric convergence defines a metrizable topology, the D-metric

need not be continuous even in a single variable.

In fact, we develop certain methods for constructing D-metric spaces from a given metric space and obtain from them, as by-products, examples illustrating the last two assertions. We also introduce the notions of strong convergence, and very strong convergence in aD-metric space and study in a decisive way the mutual implications among the three notions of convergence, strong convergence, and very strong convergence.

Throughout this paper,Rdenotes the set of all real numbers,R⁺the set of all non- negative real numbers,Nthe set of all positive integers, and(R⁺)³^∗ = {(t1,t2,t3)∈ (R⁺)³:t1≤t2+t3, t2≤t3+t1,t3≤t1+t2}.

Note1.1. If(X,d)is a metric space, then(d(x,y),d(y,z),d(z,x))∈(R⁺)³^∗ for all x,y,z∈X.

Definition1.2[2]. LetX be a nonempty set. A functionρ:X×X×X→[0,∞)is called aD-metric onXif

(i) ρ(x,y,z)=0 if and only ifx=y=z(coincidence),

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(ii) ρ(x,y,z)=ρ(p(x,y,z))for allx,y,z∈Xand for any permutationp(x,y,z) ofx,y,z(symmetry),

(iii) ρ(x,y,z)≤ρ(x,y,a)+ρ(x,a,z)+ρ(a,y,z)for allx,y,z,a∈X(tetrahedral inequality).

IfXis a nonempty set andρisD-metric onX, then the ordered pair(X,ρ)is called a D-metric space. When theD-metricρis understood, we say thatXis aD-metric space.

Definition1.3[2,8]. A sequence{xn}in aD-metric space(X,ρ)is said to be convergent (orρ-convergent) if there exists an elementxofXwith the following property:

givenε >0, there exists anN∈Nsuch thatρ(xm,xn,x) < εfor allm,n≥N. In such a case,{xn}is said to converge toxandxis called a limit of{xn}.

Definition 1.4[2, 8]. A sequence {xn}in a D-metric space (X,ρ)is said to be Cauchy (orρ-Cauchy) if, givenε >0, there exists anN∈Nsuch thatρ(xm,xn,xp) < ε for allm,n,p≥N.

Remark 1.5. The definition ofρ-Cauchy sequence as given by Dhage [2] appears to be slightly different fromDefinition 1.4, but it is actually equivalent to it. It can be shown that in aD-metric space every convergent sequence is Cauchy.

Definition1.6[2,8]. AD-metric space(X,ρ)is said to be complete (orρ-complete) if everyρ-Cauchy sequence inXisρ-convergent inX.

Notation1.7. For a subsetEof aD-metric space(X,ρ),E^cdenotes{x∈X: there is a sequence inEwhich converges to xunder theD-metricρ}. For any setX,P (X) denotes the power set ofX, that is, the collection of all subsets ofX.

We now give an example of a completeD-metric space in whichD-metric convergence does not deﬁne a topology and in which there are convergent sequences with inﬁnitely many limits.

Example1.8. LetX=A∪B∪{0}, whereA= {1/2ⁿ:n∈N}andB= {2ⁿ:n∈N}.

Deﬁneρ:X×X×X→R⁺as follows:

(i) ρ(x,y,z)=0 ifx=y=z,

(ii) ρ(x,y,z)=min{max{x,y},max{y,z},max{z,x}}ifx,y,z∈A∪{0},0 does not occur more than once amongx,y,z, and at least two amongx,y,zare distinct, (iii) ρ(x,y,z)=1 if 0 and at least one element ofBoccur amongx,y,z, or 0 occurs

exactly twice amongx,y,z,

(iv) ρ(x,y,z)=min{x,y,z}if x,y,z∈A∪B and exactly one element ofB occurs exactly once amongx,y,z,

(v) ρ(x,y,z)=min{max{1/x,1/y},max{1/y,1/z},max{1/z,1/x}}, ifx,y,z∈A∪B and exactly one element ofAoccurs exactly once amongx,y,z,

(vi) ρ(x,y,z)= |1/x−1/y|+|1/y−1/z|+|1/z−1/x|ifx,y,z∈B.

Then(X,ρ)is a completeD-metric space. Butρ-convergence does not deﬁne a topology onX.

Proof. Clearlyρ is symmetric in all the three variables andρ(x,y,z)=0 if and only ifx=y=z. We note thatρ(x,y,z)≤1 for allx,y,z∈X. Letx,y,z,u∈X.

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D-METRIC SPACES...

Case(i). x=y=z.

Thenρ(x,y,z)=0≤ρ(u,y,z)+ρ(x,u,z)+ρ(x,y,u).

Case(ii). x,y,z∈A∪{0}, 0 does not occur more than once amongx,y,z, and at least two amongx,y,zare distinct.

We may assume thatx≥y≥z. Ifu∈A∪ {0}, thenρ(x,y,z)=y≤ρ(u,y,z)+ ρ(x,u,z)+ρ(x,y,u), since when u > y, ρ(u,y,z) =y; whenu=y, ρ(x,y,z)= ρ(x,u,z); and whenu < y,ρ(x,y,u)=y. Ifu∈B, then

ρ(x,y,z)=y=min{x,y,u} =ρ(x,y,u)

≤ρ(u,y,z)+ρ(x,u,z)+ρ(x,y,u). (1.1)

Case(iii). 0 occurs exactly twice amongx,y,z.

We may assume thatx=y=0. Thenz≠0. Ifu∈X\{0}, then ρ(x,y,z)=ρ(0,0,z)=1=ρ(0,0,u)=ρ(x,y,u)

≤ρ(u,y,z)+ρ(x,u,z)+ρ(x,y,u). (1.2)

Ifu=0, then

ρ(x,y,z)=ρ(0,0,z)=1=ρ(u,y,z)

≤ρ(u,y,z)+ρ(x,u,z)+ρ(x,y,u). (1.3)

Case(iv). 0 and at least one element ofBoccur amongx,y,z. We may assume thatx=0 andy∈B. Then

ρ(x,y,z)=ρ(0,y,z)=1=ρ(0,y,u)=ρ(x,y,u)

≤ρ(u,y,z)+ρ(x,u,z)+ρ(x,y,u). (1.4)

Case(v). x,y,z∈A∪Band exactly one element ofBoccurs exactly once amongx, y,z.

We may assume thatx∈B. Theny,z∈A. We may also assume thaty≥z. Ifu∈B, then

ρ(x,y,z)=min{x,y,z} =z=ρ(u,y,z)

≤ρ(u,y,z)+ρ(x,u,z)+ρ(x,y,u). (1.5) Ifu∈A∪{0}, then

ρ(x,y,z)=min{x,y,z} =z≤ρ(u,y,z)+ρ(x,u,z)+ρ(x,y,u), (1.6) since whenu < z,ρ(u,y,z)=z; whenu=z,ρ(x,y,z)=ρ(x,y,u); and whenu > z, ρ(u,y,z)=min{u,y} ≥z.

Case(vi). x,y,z∈A∪Band exactly one element ofAoccurs exactly once amongx, y,z.

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We may assume thatx∈A. Theny,z∈B. We may also assume thaty≥z. Ifu∈A, then

ρ(x,y,z)=max 1

y,1 z

= 1

z=ρ(u,y,z)

≤ρ(u,y,z)+ρ(x,u,z)+ρ(x,y,u).

(1.7)

Ifu=0, then

ρ(x,y,z)=max 1

y,1 z

=1 z≤1

2<1=ρ(u,y,z)

(1.8)

Ifu∈B, then

ρ(x,y,z)=max 1

y,1 z

= 1 z

≤max 1

u,1 z

=ρ(x,u,z)

(1.9)

Case(vii). x,y,z∈B.

We may assume that 1/x≥1/y≥1/z. Ifu=0, then

ρ(x,y,z)=2 1

x−1 z

<1=ρ(u,y,z)

(1.10)

Ifu∈A, then

ρ(x,y,z)=2 1

x−1 z

≤ 1 x+1

x

=ρ(x,u,z)+ρ(x,y,u)

(1.11)

Ifu∈B, then

ρ(x,y,z)= 1

x−1 y

+ 1

y−1 z

+ 1

z−1 x

≤ρ(u,y,z)+ρ(x,u,z)+ρ(x,y,u),

(1.12)

since|1/x−1/y| ≤ρ(x,y,u), |1/y−1/z| ≤ρ(u,y,z), and|1/z−1/x| ≤ρ(x,u,z). Thus, for all x,y,z,u∈X, we have ρ(x,y,z) ≤ρ(u,y,z)+ρ(x,u,z)+ρ(x,y,u). Henceρis aD-metric onX.

To show that(X,ρ)isD-complete.

Let{xn}be a Cauchy sequence inX.

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Case1. There existsN∈Nsuch thatxn=xN for alln≥N. In this case, evidently{xn}converges toxN.

Case2. GivenN∈N, there existi,j∈Nsuch thati > N,j > N, andxi≠xj. Then there existsN0∈Nsuch thatxi≠0 for eachi≥N0, sinceρ(0,0,x)=1, for all x∈X\{0}, and{xn}is Cauchy.

Subcase(i). There existsN1∈Nsuch thatN1≥N0andxi∈Afor alli≥N1. Claim1.9. xn→0asn→ ∞in the usual sense.

Suppose the claim does not hold. Then there exists a positive real numberεsuch thatxn≥εfor inﬁnitely manyn∈N. GivenN∈N, we can choosei,j,k∈Nsuch that k > j > i >max{N,N1},xi≥ε,xj≥ε, andxk≠xj. Then

ρ

xi,xj,xk

=min max

xi,xj ,max

xj,xk ,max xk,xi

≥ε. (1.13)

This is a contradiction since{xn}is Cauchy. Hence the claim.

Form,n≥N1anda∈A∪{0}, we have ρ

a,xm,xn

=min max

a,xm ,max

xm,xn ,max xn,a

≤max

xm,xn →0 asm,n → ∞. (1.14)

Hence{xn}converges toafor anya∈A∪{0}. It can also be shown that{xn}converges tobfor anyb∈B.

Subcase(ii). There existsN2∈Nsuch thatN2≥N0andxi∈Bfor alli≥N2. Claim1.10. xn→ +∞asn→ +∞.

Suppose the claim does not hold. Then there exists a positive real numberM such thatxn≤Mfor inﬁnitely manyn∈N. GivenN∈N, we can ﬁndi,j,k∈Nsuch that k > j > i >max{N,N2}, xi≤M, xj≤M, andxj≠xk. Then ρ(xi,xj,xk)≥ |1/xj− 1/xk| ≥1/2xj≥1/2M. This is a contradiction since{xn}is Cauchy. Hence the claim.

Form,n≥N2anda∈A, we have

ρ

a,xm,xn

=max 1

xm, 1 xn

→0 asm,n → ∞. (1.15)

Hence{xn}converges toafor anya∈A.

Subcase(iii). GivenN∈N, there existi,j∈Nsuch thati > N,j > N,xi∈A, and xj∈B.

Claim1.11. Any element ofAoccurs only ﬁnitely many times in the sequence{xn}. Suppose the claim does not hold. Then there existsa0∈Asuch thata0=xn for inﬁnitely manyn∈N. LetN∈N. Then there existi,j,k∈Nsuch thatk > j > i > N,xi= xj=a0, andxk∈B. Thenρ(xi,xj,xk)=min{xi,xj,xk} =a0. This is a contradiction since{xn}is Cauchy. Hence the claim.

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Claim1.12. Any element ofBoccurs only ﬁnitely many times in the sequence{xn}. Suppose the claim does not hold. Then there existsb∈B such thatb=xnfor in- ﬁnitely manyn∈N. LetN∈N. Then there existi,j,k∈Nsuch thatk > j > i > N, xi=xj=b, andxk∈A. Thenρ(xi,xj,xk)=1/b. This is a contradiction since{xn}is Cauchy. Hence the claim.

Letc∈A. Letε >0. FromClaim 1.11, it follows that there existsN3∈Nsuch that xn<min{ε,c}whenevern≥N3andxn∈A. FromClaim 1.12, it follows that there ex- istsN⁴∈Nsuch thatxn>1/εwhenevern≥N⁴andxn∈B. LetN⁵=max{N0,N³,N⁴}.

Letm,n∈Nbe such thatm≥N5andn≥N5. Thenxn,xm∈A∪B. If bothxn,xm∈A, thenρ(c,xn,xm)=max{xn,xm}< ε. If bothxn,xm∈B, thenρ(c,xn,xm)=max{1/xn, 1/xm}< ε. Suppose that one ofxn,xm belongs toAand the other belongs toB. We may assume that xn∈A and xm∈B. Then ρ(c,xn,xm)=min{c,xn,xm} =xn< ε. Thusρ(c,xn,xm) < εfor alln,m≥N5. Hence{xn}converges toc.

Thus, in any case, {xn}is convergent inX with respect to the D-metricρ. Hence (X,ρ)is a completeD-metric space.

To show that(B^c)^c≠B^c.

Letp∈B^c. Then there exists a sequence{xn}inB such that{xn}converges top. Hence {xn}is Cauchy. If there existsN∈Nsuch that xk=xN for all k≥N, then ρ(p,xN,xN)=0 and hencep=xN∈B.

Suppose that such anNdoes not exist. Then givenN∈N, there existi,j∈Nsuch that i > N,j > N, andxi≠xj. As in Subcase (ii) ofCase 2in the proof of the completeness ofρ, it can be shown thatxn→ +∞asn→ ∞and that{xn}converges to xfor any x∈A.

For anyx∈B, ρ

x,xn,xm

= 1

x− 1 xn

+ 1

xn− 1 xm

+ 1

xm−1 x

→ 2

x asm,n → ∞. (1.16) Hence {xn}does not converge tox for anyx ∈B. We have ρ(0,xn,xm)=1 for all m,n∈N. Therefore{xn}does not converge to 0. Hencep∈A. ThusB^c⊆B∪A. Clearly B⊆B^c.{2ⁿ}converges toxfor anyxinA. HenceA⊆B^c. ThereforeA∪B=B^c. Since {1/2ⁿ}is a sequence inAand it converges toxfor anyx∈X,(B^c)^c=X. Since 0∉A∪B, (B^c)^c≠B^c. Therefore the functionf:P (X)→P (X)deﬁned asf (E)=E^cfor allE∈P (X) is not a closure operator. Henceρ-convergence does not deﬁne a topology onX.

Definition1.13. Let(X,ρ)be aD-metric space and{xn}a sequence inX.{xn}is said to converge strongly to an elementxofXif

(i) ρ(x,xm,xn)→0 asm,n→ ∞,

(ii) {ρ(y,y,xn)}converges toρ(y,y,x)for allyinX. In such a case,xis said to be a strong limit of{xn}.

Definition1.14. Let(X,ρ)be aD-metric space and{xn}a sequence inX.{xn}is said to converge very strongly to an elementxofXif

(i) ρ(x,xm,xn)→0 asm,n→ ∞,

(ii) {ρ(y,z,xn)}converges toρ(y,z,x)for any elementsy,zofX. In such a case,xis said to be a very strong limit of{xn}.

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Remark 1.15. Let {xn}be a sequence in a D-metric space X and x∈X. If {xn} converges very strongly to x, then{xn}converges strongly to x. If{xn}converges strongly tox, then it converges tox with respect toρ. Examples1.21,1.39, and1.40 show that the converse statements are false.

Proposition1.16. Let(X,ρ)be aD-metric space. Let{xn}be a sequence inXcon- verging to an elementxofX. Then{ρ(x,x,xn)}is convergent.

Proof. Since{xn}is convergent, it isD-Cauchy. We haveρ(x,x,xn)≤ρ(xm,x,xn)+

ρ(x,xm,xn)+ρ(x,x,xm). Henceρ(x,x,xn)−ρ(x,x,xm)≤2ρ(x,xm,xn). Similarly, we haveρ(x,x,xm)−ρ(x,x,xn)≤2(x,xn,xm). Hence |ρ(x,x,xn)−ρ(x,x,xm)| ≤ 2ρ(x,xn,xm). Since this inequality is true for allm,n∈Nand{xn}converges toxun- der theD-metricρ, it follows that{ρ(x,x,xn)}is a Cauchy sequence of real numbers and hence convergent.

Remark1.17. Example 1.21shows that the hypothesis ofProposition 1.16does not ensure that the limit of{ρ(x,x,xn)}isρ(x,x,x).

Proposition1.18. In aD-metric space, every strongly convergent sequence has a unique strong limit.

Proof. Let(X,ρ)be aD-metric space and{xn}a strongly convergent sequence in X. Lety,zbe strong limits of{xn}. Then{ρ(y,y,xn)}converges to bothρ(y,y,y) andρ(y,y,z). Henceρ(y,y,z)=ρ(y,y,y)=0. Hencey=z.

Theorem1.19. Let(X,d)be a metric space,x0∈X, and letAbe a nonempty subset ofX\{x0}. Deﬁneρ:A×A×A→R⁺as

ρ(x,y,z)=











0 ifx=y=z,

min max

d x0,x

,d x0,y , max

d x0,y

,d x0,z , max

d x⁰,z

,d x⁰,x

otherwise.

(1.17)

Then(A,ρ)is a completeD-metric space andρ-convergence deﬁnes a topologyτonA. IfA∩{x∈X:d(x0,x) < r0} =φfor somer0∈(0,∞), thenτis the discrete topology on X; otherwiseτ= {φ}∪{E⊆A:{x∈A:d(x0,x) < r} ⊆Efor somer∈(0,∞)}and, in particular,τisT¹but not Hausdorﬀ.

Let{xn} ⊆A. Then{xn}converges toxwith respect toρfor somex∈Aandxn≠x for all suﬃciently largen⇒d(xn,x0)→0asn→ ∞ ⇒ {xn}converges toxwith respect toρfor eachxinA. IfAhas at least two elements, there does not exist a sequence inA which is strongly convergent with respect toρ.

Proof. Clearlyρ is symmetric in all the three variables andρ(x,y,z)=0 if and only ifx=y=z. Letx,y,z,u∈A. We may assume thatd(x0,x)≥d(x0,y)≥d(x0,z). Irrespective of whetherd(x⁰,u) < d(x⁰,y)ord(x⁰,u)≥d(x⁰,y), we have

ρ(x,y,z)=d x0,y

≤ρ(x,y,u). (1.18)

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Henceρ(x,y,z)≤ρ(u,y,z)+ρ(x,u,z)+ρ(x,y,u)for allx,y,z,u∈A. Henceρis a D-metric onA. Let{xn}be a sequence inAandx∈A. Ifxn≠x, we have

ρ

x,xn,xn

=min max

d x0,x

,d

x0,xn , max

d x0,xn

,d

x0,xn , max

d x0,xn

,d

x0,x

≥d x0,xn

.

(1.19)

That is, ρ(x,xn,xn)≥d(x0,xn) if xn ≠x. Henced(x0,xn)→0 as n→ ∞ if {xn} converges toxwith respect toρfor somex∈Aandxn≠xfor all suﬃciently largen. We have

ρ

xn,xm,x

≤min max

d x⁰,xn

,d

x⁰,xm , max

d x0,xm

,d x0,x , max

d x0,x

,d x0,xn

≤max d

x0,xn

,d

x0,xm ∀n,m∈N.

(1.20)

Thus{xn}converges toxwith respect toρfor eachxinAifd(x0,xn)→0 asn→ ∞. Ifxm≠xn, then we have

ρ

xn,xm,xn

=min max

d x0,xn

,d

x0,xm , max

d x0,xm

,d

x0,xn , max

d x0,xn

,d x0,xn

≥d x0,xn

.

(1.21)

Henced(x0,xn)→0 asn→ ∞if{xn}isρ-Cauchy and there does not exist anN∈N such thatxn=xN for alln > N. If there existsN∈Nsuch thatxn=xN for alln > N, then, evidently,{xn}converges toxN with respect toρ.

If such anN does not exist and {xn} isρ-Cauchy, thend(x0,xn)→0 asn→ ∞, and hence{xn}converges toxwith respect toρfor anyxinA. Hence everyρ-Cauchy sequence inAis convergent with respect toρ. Therefore(A,ρ)isD-complete. Ifxn≠x, we have

ρ x,x,xn

=min max

d x0,x

,d x0,x , max

d x0,x

,d

x0,xn , max

d x0,xn

,d x0,x

≥d x0,x

.

(1.22)

Sincex ∈A⊆ X\{x0}, d(x0,x) >0. Hence{ρ(x,x,xn)}does not converge to 0 if xn≠xfor inﬁnitely manyn. Consequently,{xn}is not stronglyρ-convergent ifAhas at least two elements. LetEbe a subset ofA. Clearly,E⊆E^c.

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Case1. E∩{x∈X:d(x,x0) < r} =φfor somer∈(0,∞).

Letz∈E^c. Then there exists a sequence{xn}inE such that{xn}converges toz with respect toρ. Suppose thatxn≠zfor eachn. Thend(x0,xn)→0 asn→ ∞. Since E∩{x∈X:d(x,x0) < r} =φ, we haved(x,x0)≥r for allx∈E. Henced(xn,x0)≥ r for alln∈N. Therefore {d(xn,x0)}does not converge to 0. Thus we arrived at a contradiction. Consequently,xn=zfor somen∈N. Hencez∈E. ThereforeE^c⊆E. ThusE=E^c.

Case2. Case 1is false.

ThenE∩{x∈X:d(x,x0) <1/n}≠φfor eachn∈N. Hence there exists a sequence {un}inEsuch thatd(un,x⁰) <1/nfor alln∈N. Therefored(un,x⁰)→0 asn→ ∞.

Hence{un}converges toxwith respect toρfor eachxinA. HenceE^c=A. Case(I). A∩{x∈X:d(x,x0) < r0} =φfor somer0∈(0,∞).

In this case, for any subsetEofA, we haveE^c=E, and hence(E^c)^c=E^c=E. Thus the functionf deﬁned on the power setP (A)ofAasf (E)=E^c for allE∈P (A)is a closure operator. Thereforeρ-convergence deﬁnes a topologyτ onAin which every subset ofAis closed. Henceτ= {E:E⊆A}. Consequently,τis the discrete topology onX.

Case(II). A∩{x∈X:d(x,x0) < r}≠φfor anyr∈(0,∞).

In this case, for a subsetEofX, we haveE^c=EorAaccording to whetherCase 1or Case 2holds. Hence(E^c)^c=E^cfor allE∈P (A). Therefore the functionf deﬁned on P (A)asf (E)=E^cfor allE∈P (A)is a closure operator. Thusρ-convergence deﬁnes a topologyτonAwith respect to which a subsetBofAis closed if and only ifB=E^c for someE∈P (A). Hence

τ=

A\E^c:E∈P (A)

= {φ}∪

E∈P (A):

x∈A:d x0,x

< r ⊆Efor somer∈(0,∞) . (1.23) IfU1,U2are nonempty open sets inτ, thenU1∩U2≠φsince there existr1,r2∈(0,∞) such that

x∈A:d x0,x

< ri ⊆Ui, i=1,2. (1.24) Henceτis not Hausdorﬀ. Letp,qbe distinct elements ofA. Sincex0∉A,d(p,x0)and d(q,x0)are positive real numbers. Let 0< r <min{d(p,x0),d(q,x0)}. LetV0= {x∈ A:d(x,x0) < r}. ThenV0∪ {p}is aτ-open subset ofAcontainingpbut notq, and V⁰∪ {q}is aτ-open subset ofAcontainingqbut notp. Hence the topologyτ isT¹.

Example1.20. LetX=R with the usual metric,x0=0, andA=[1,2]. Then the functionρdeﬁned inTheorem 1.19onA×A×Areduces to the following:

ρ(x,y,z)=





0 ifx=y=z,

min

max{x,y},max{y,z},max{z,x} otherwise. (1.25) From Theorem 1.19it follows that (A,ρ) is a complete D-metric space and that ρ- convergence deﬁnes a topologyτonA, which is the discrete topology onA.

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Example1.21. LetX=Rwith the usual metric,x0=0, andA= {1/n:n∈N}. Then the functionρdefined inTheorem 1.19onA×A×Ahas the same form as that given in Example 1.20. FromTheorem 1.19it follows that(A,ρ)is a completeD-metric space, any sequence in A which converges to zero in the usual sense converges tox with respect toρfor eachx inA, and thatρ-convergence defines a topologyτ onAwith respect to which nonempty subsetEofAis open if and only ifEcontains{1/n:n∈N and n≥N} for someN∈N. Further, τ is T¹ but not Hausdorff. Letxn=1/nfor alln∈Nandx0=1/2. Then{xn}converges to 1/2 under theD-metricρ. We have ρ(x0,x0,xn)=ρ(1/2,1/2,1/n)=1/2 for all n∈N\{2}. Hence {ρ(x0,x0,xn)}does not converge to 0=ρ(x⁰,x⁰,x⁰). We note that{1/n}does not converge strongly even though it converges to every element ofX.

Theorem1.22. Let(X,d)be a metric space,x0∈X, and let{xn}be a convergent sequence inX\{x0}with limitx0,Aa proper subset ofX\{x0}containing{xn}, andB a subset ofX\{x0}which containsAproperly. Deﬁneρ:B×B×B→R⁺as

ρ(x,y,z)=











0 ifx=y=z,

min max

d x0,x

,d x0,y , max

d x0,y

,d x0,z , max

d x0,z

,d x0,x

otherwise.

(1.26)

Letρ⁰denote the restriction ofρtoA×A×A. Then(B,ρ)and(A,ρ⁰)are completeD- metric spaces,A⊆B, but{x∈B: there is a sequence{yn}inAwhich converges tox with respect toρ} =B≠A.

Proof. The proof follows fromTheorem 1.19.

Remark1.23. If(X,d)is a metric space,Y ⊆X,d0is the restriction ofdtoY×Y, and(Y ,d0)is complete, then{x∈X: there is a sequence{yn}inY which converges to x} =Y.Theorem 1.22shows that an analogous result does not hold inD-metric spaces.

Theorem1.24. Suppose thatΦ:(R⁺)³^∗→R⁺is (i) symmetric in all the three variables,

(ii) Φ(t1,t2,t3)=0if and only ift1=t2=t3=0,

(iii) Φ(t1,t2,t3)≤Φ(t1,t2,t3)+Φ(t1,t2,t3)+Φ(t1,t2,t3)whenever(t1,t2,t3),(t1,t2, t3),(t1,t2,t3),(t1,t2,t3)∈(R⁺)³^∗ andti≤t_i+t_i for alli=1,2,3.

Letdbe a metric onXand letρ:X×X×X→R⁺be deﬁned as ρ(x,y,z)=Φ

d(x,y),d(y,z),d(z,x)

. (1.27)

Thenρis aD-metric onX. IfΦis continuous at(0,0,0), then (1) anyd-Cauchy sequence inXisρ-Cauchy,

(2) {xn} ⊆X,x∈X, andd(xn,x)→0asn→ ∞ ⇒ {xn}converges toxwith respect to theD-metricρ.

Suppose thatΦhas the following property:

(iv) givenε >0, there existsδ >0such thatt < εwhenevert∈R⁺andΦ(0,t,t) < δ.

(11)

Then

(1) anyρ-Cauchy sequence isd-Cauchy,

(2) {xn}(⊆X)converges tox∈Xwith respect toρ⇒d(xn,x)→0asn→ ∞. IfΦis continuous at(0,0,0),{Φ(0,tn,tn)}converges toΦ(0,t,t)whenevert∈R⁺, and {tn}is a sequence inR⁺ converging tot, then{xn} ⊆X,x∈X, andd(xn,x)→0as n→ ∞ ⇒ {xn}converges strongly toxwith respect to theD-metricρ.

IfΦis continuous at(0,0,0)and is continuous in any two variables, then{xn} ⊆X, x∈X, andd(xn,x)→0asn→ ∞ ⇒ {xn}converges very strongly toxwith respect to theD-metricρ.

Proof. We prove thatρis aD-metric onX. SinceΦ is symmetric in all the three variables, so isρ. From property (ii) of Φ, it follows thatρ(x,y,z)=0 if and only if x=y=z.

Letx,y,z,u∈X. From property (iii) ofΦ, we have ρ(x,y,z)=Φ

d(x,y),d(y,z),d(z,x)

≤Φ

d(x,y),d(y,u),d(u,x) +Φ

d(u,y),d(y,z),d(z,u) +Φ

d(x,u),d(u,z),d(z,x)

(1.28)

sinced(x,y)≤d(u,y)+d(x,u),d(y,z)≤d(y,u)+d(u,z), andd(z,x)≤d(u,x)+ d(z,u). Henceρ(x,y,z)≤ρ(x,y,u)+ρ(u,y,z)+ρ(x,u,z)for allx,y,z,u∈X. Hence ρis aD-metric onX.

Suppose thatΦis continuous at(0,0,0).

(1) Let{xn}be ad-Cauchy sequence inX. Thend(xn,xm)→0 asn,m→ ∞. We have ρ

xn,xm,xk

=Φ d

xn,xm ,d

xm,xk ,d

xk,xn

→Φ(0,0,0)=0 asn,m,k → ∞ (sinceΦis continuous at(0,0,0)).

(1.29) Hence{xn}isρ-Cauchy inX.

(2) Let{xn} ⊆Xand letx∈Xbe such thatd(xn,x)→0 asn→ ∞. We have ρ

x,xn,xm

=Φ d

x,xn

,d xn,xm

,d xm,x

→Φ(0,0,0)=0 asn,m → ∞ (1.30) (since everyd-convergent sequence isd-Cauchy andΦis continuous at(0,0,0)).

Henceρ(x,xn,xm)→0 asn,m→ ∞. Hence{xn}converges toxwith respect toρ. Suppose thatΦhas property (iv).

(1) Let{xn}be a ρ-Cauchy sequence in X. Let εbe a positive real number. Then there existsδ >0 such thatt < εwhenevert∈R⁺and Φ(0,t,t) < δ. Sinceρ(xn,xm, xn)→0 as n,m→ ∞, there existsN∈Nsuch thatρ(xn,xm,xn) < δfor alln,m≥ N. That is, Φ(d(xn,xm),d(xm,xn),d(xn,xn)) < δ for all n,m≥N. In other words, Φ(0,d(xn,xm),d(xn,xm)) < δfor alln,m≥N(sinceΦis symmetric). Henced(xn,xm)

< εfor alln,m≥N. Therefore{xn}isd-Cauchy.

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(2) Let{xn} ⊆Xconverge tox∈Xwith respect to theD-metricρ. Letε >0. Then there existsδ >0 such thatt < εwhenevert∈R⁺andΦ(0,t,t) < δ. Sinceρ(xn,xn,x)→ 0 as n→ ∞, there exists N ∈N such that ρ(xn,xn,x) < δ for all n≥N. That is, Φ(d(xn,xn),d(xn,x),d(x,xn)) < δfor all n≥N. That isΦ(0,d(xn,x),d(xn,x)) < δ for alln≥N. Henced(xn,x) < εfor alln≥N. Therefore{xn}converges tox with respect to the metricd.

Suppose that Φ is continuous at (0,0,0) and {Φ(0,tn,tn)} converges to Φ(0,t,t) whenever t∈R⁺ and {tn} is a sequence in R⁺ converging tot. Let{xn} ⊆X and letx∈X be such thatd(xn,x)→0 asn→ ∞. Lety∈X. Thend(xn,y)→d(x,y) asn→ ∞. Hence{Φ(0,d(xn,y),d(xn,y))} →Φ(0,d(x,y),d(x,y))asn→ ∞. That is, ρ(y,y,xn)→ρ(y,y,x)asn→ ∞. SinceΦ is continuous at(0,0,0), from what we have already proved, it follows that{xn}converges toxwith respect toρ. Hence{xn} converges strongly tox with respect to the D-metricρ. Suppose that Φ is continuous at(0,0,0)and is continuous in any two variables. Let{xn} ⊆Xand letx∈Xbe such thatd(xn,x)→0 asn→ ∞. Lety,z∈X. Then {d(z,xn)}and {d(xn,y)}converge tod(z,x)andd(x,y), respectively. SinceΦis continuous in any two variables, it follows that{Φ(d(y,z),d(z,xn),d(xn,y))}converges to Φ(d(y,z),d(z,x),d(x,y)), that is,{ρ(y,z,xn)} converges toρ(y,z,x). Since Φ is continuous at(0,0,0), {xn} converges toxwith respect toρ. Hence{xn}converges very strongly toxwith respect toρ.

Corollary1.25. Let(X,d)be a metric space and letΦ:(R⁺)³^∗→R⁺be continuous at(0,0,0), and have properties (i), (ii), (iii), and (iv) specified inTheorem 1.24. Letρbe defined as inTheorem 1.24. Thenρis aD-metric onX, a sequence inX isD-Cauchy if and only if it isρ-Cauchy, and a sequence{xn}inXconverges with respect todto an elementxofX if and only if{xn}converges tox with respect toρ. In particular, ρ-convergence defines a topology onXwhich coincides with the metric topology induced by the metricdonX, andXis complete with respect to the metricdif and only if it is complete with respect to theD-metricρ. Further, the following statements are true.

(1)If{Φ(0,tn,tn)}converges toΦ(0,t,t)whenevert∈R⁺and{tn}is a sequence in R⁺converging tot,{xn} ⊆X, andx∈X, then{xn}converges toxwith respect toρif and only if{xn}converges strongly toxwith respect toρ.

(2)IfΦis continuous in any two variables,{xn} ⊆X, andx∈X, then{xn}converges toxwith respect toρif and only if{xn}converges very strongly toxwith respect toρ. (3) IfΦ is continuous on(R⁺)³^∗, then ρis sequentially continuous in all the three variables, that is,{ρ(un,vn,wn)}converges to ρ(u,v,w)whenever u,v,w∈X and {un},{vn}, and{wn}are sequences inXconverging tou,v, andw, respectively with respect toρ.

Note1.26. Corollary 1.25is useful in generating a number ofD-metrics from a given metric on a set.

We now prove a number of propositions which show that the class of functionsΦ: (R⁺)³^∗→R⁺, which are continuous at(0,t,t)for allt∈R⁺and which satisfy properties (i), (ii), (iii), and (iv) speciﬁed inTheorem 1.24, is very rich.

(13)

Lemma1.27. Letp∈[1,∞). Then(a+b)^p≥a^p+b^pfor alla,b∈R⁺.

Proof. Deﬁnef :R⁺ →Rasf (t)=(1+t)^p−1−t^p for allt∈R⁺. Thenf(t)= p(1+t)^p−¹−pt^(p−¹⁾=p[(1+t)^p−¹−t^(p−¹⁾]. Since 1+t≥tfor allt∈R⁺andp−1≥0, we have(1+t)^p−1≥t^(p−1)for allt∈R⁺.

Hencef(t)≥0 for allt∈R⁺. Thereforef is monotonically increasing onR⁺. We havef (0)=0. Hencef (t)≥f (0)for allt∈R⁺. Therefore

(1+t)^p≥1+t^p ∀t∈R⁺. (1.31)

Leta,b∈R⁺. We may assume thata≥b. Ifa=0, thenb=0 and(a+b)^p=0=a^p+b^p. Suppose thata >0. Then, from what we have already proved above, we have

1+b

a p

≥1+ b

a p

, (1.32)

that is,

a+b a

p

≥1+b^p

a^p. (1.33)

Hence(a+b)^p≥a^p+b^p.

Corollary1.28. Letp∈[1,∞). Then(a+b+c)^p≥a^p+b^p+c^pfor alla,b,c∈R⁺. Proof. Leta,b,c∈R⁺. Then, fromLemma 1.27, we have

(a+b+c)^p=

(a+b)+cp

≥(a+b)^p+c^p≥a^p+b^p+c^p. (1.34)

Proposition1.29. Suppose thatΨ:R⁺→R⁺is monotonically increasing andΨ(t)=0 if and only ift=0. Letp∈[1,∞). DeﬁneΦ:(R⁺)³^∗→R⁺as

Φ t1,t2,t3

= Ψ

t1

_p +

Ψ t2

_p +

Ψ t3

_p_1/p

∀ t1,t2,t3

∈ R⁺3^∗

. (1.35)

ThenΦhas properties (i), (ii), (iii), and (iv) speciﬁed inTheorem 1.24. If Ψis continuous at0, thenΦis continuous at(0,0,0), and ifΨis continuous onR⁺, thenΦis continuous on(R⁺)³^∗.

Proof. ClearlyΦis symmetric in all the three variables:

Φ

t¹,t²,t³

=0⇐⇒

Ψ

t¹p+ Ψ

t²p+ Ψ

t³p1/p=0

⇐⇒

Ψ ti

p

=0 ∀i

⇐⇒Ψ ti

=0 ∀i

⇐⇒ti=0 ∀i.

(1.36)

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Lett1,t1,t1,t2,t2,t2,t3,t3,t3∈R⁺. Let

a= Ψ

t1

p

+ Ψ

t2

p

+ Ψ

t3

p1/p

, b=

Ψ t1

p

+ Ψ

t2

p

+ Ψ

t3

p1/p

, c=

Ψ t1

p

+ Ψ

t2

p

+ Ψ

t3

p1/p

.

(1.37)

We have

(a+b+c)^p≥a^p+b^p+c^p

= Ψ

t1

+ Ψ

t2

+ Ψ

t3

+

Ψ t1

_p +

Ψ t2

_p +

Ψ t3

_p +

Ψ t1

p+ Ψ

t2

p+ Ψ

t3p

≥ Ψ

t1p+ Ψ

t2p+ Ψ

t3p

.

(1.38)

Hence a+b+c≥[[Ψ(t1)]^p+[Ψ(t2)]^p+[Ψ(t3)]^p]^1/p. ThereforeΦ has property (iii).

We have

Φ(0,t,t)= Ψ(0)p

+ Ψ(t)p

+

Ψ(t)p1/p

= 2

Ψ(t)p1/p

=2^1/pΨ(t).

(1.39)

Letεbe a positive real number. Chooseδ=2^1/pΨ(ε). Thenδ >0 sinceΨ(t)=0 implies t=0.

Φ(0,t,t) < δ ⇒2^1/pΨ(t) <2^1/pΨ(ε)

⇒Ψ(t) <Ψ(ε)

⇒t < ε (sinceΨ is monotonically increasing).

(1.40)

HenceΦhas property (iv).

Corollary 1.30. Let p ∈ [1,∞). Then the function Φ :(R⁺)³^∗ →R⁺ deﬁned as Φ(t1,t2,t3)=[t^p1+t2^p+t3^p]^1/p for all(t1,t2,t3)∈(R⁺)³^∗ is continuous on(R⁺)³^∗ and has properties (i), (ii), (iii), and (iv) speciﬁed inTheorem 1.24.

Proof. The proof follows fromProposition 1.29by takingΨ(t)=tfor allt∈R⁺.

Proposition1.31. Suppose thatΨ:R⁺→R⁺is monotonically increasing andΨ(t)=0 if and only ift=0. DeﬁneΦ:(R⁺)³^∗→R⁺as

Φ t1,t2,t3

=max Ψ

t1

,Ψ t2

,Ψ

t3 ∀ t1,t2,t3

∈ R⁺3^∗

. (1.41)

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ThenΦhas properties (i), (ii), (iii), and (iv) speciﬁed inTheorem 1.24. If Ψis continuous at0, thenΦis continuous at(0,0,0), and ifΨis continuous onR⁺, thenΦis continuous on(R⁺)³^∗.

Proof. ClearlyΦis symmetric in all the three variables.

Φ

t1,t2,t3

=0⇐⇒max Ψ

t1 ,Ψ

t2 ,Ψ

t3 =0

⇐⇒Ψ ti

=0 ∀i

⇐⇒ti=0 ∀i.

(1.42)

HenceΦhas property (ii).

Lett1,t1,t1,t2,t2,t2,t3,t3,t3∈R⁺. Then

max Ψ

t1

,Ψ t2

,Ψ

t3 ≤max Ψ

t1

,Ψ t2

,Ψ t3

+max Ψ

t1

,Ψ t2

,Ψ t3

+max Ψ

t1

,Ψ t2

,Ψ t3 .

(1.43)

HenceΦhas property (iii).

Letεbe a positive real number. Chooseδ=Ψ(ε). Thenδ >0 sinceΨ(t)=0 implies t=0.

Φ(0,t,t) < δ ⇒max

Ψ(0),Ψ(t),Ψ(t) <Ψ(ε)

⇒Ψ(t) <Ψ(ε)

⇒t < ε (sinceΨ is monotonically increasing).

(1.44)

HenceΦhas property (iv).

Corollary1.32. The functionΦ:(R⁺)³^∗→R⁺ deﬁned asΦ(t1,t2,t3)=max{t1,t2, t3}for all(t1,t2,t3)∈(R⁺)³^∗is continuous on(R⁺)³^∗and has properties (i), (ii), (iii), and (iv) speciﬁed inTheorem 1.24.

Proof. The proof follows fromProposition 1.31by takingΨ(t)=tfor allt∈R⁺.

Proposition1.33. Suppose thatΨ:R⁺→R⁺is monotonically increasing,Ψ(s+t)≤ Ψ(s)+Ψ(t)for alls,t∈R⁺, andΨ(t)=0if and only ift=0. DeﬁneΦ:(R⁺)³^∗→R⁺as

Φ t1,t2,t3

=min max

Ψ t1

,Ψ

t2 ,max Ψ

t2

,Ψ t3 , max

Ψ t3

,Ψ t1

∀ t1,t2,t3

∈ R⁺3^∗

.

(1.45)