PII. S0161171204311257 http://ijmms.hindawi.com
© Hindawi Publishing Corp.
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 meth- ods 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 (se- quentially) 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 con- vergence 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),
(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 con- vergent (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),Ecdenotes{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.
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.
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)
≤ρ(u,y,z)+ρ(x,u,z)+ρ(x,y,u).
(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)
≤ρ(u,y,z)+ρ(x,u,z)+ρ(x,y,u).
(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)
≤ρ(u,y,z)+ρ(x,u,z)+ρ(x,y,u).
(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)
≤ρ(u,y,z)+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.
D-METRIC SPACES...
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.
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(Bc)c≠Bc.
Letp∈Bc. 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. ThusBc⊆B∪A. Clearly B⊆Bc.{2n}converges toxfor anyxinA. HenceA⊆Bc. ThereforeA∪B=Bc. Since {1/2n}is a sequence inAand it converges toxfor anyx∈X,(Bc)c=X. Since 0∉A∪B, (Bc)c≠Bc. Therefore the functionf:P (X)→P (X)defined asf (E)=Ecfor 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}.
D-METRIC SPACES...
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)
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⊆Ec.
D-METRIC SPACES...
Case1. E∩{x∈X:d(x,x0) < r} =φfor somer∈(0,∞).
Letz∈Ec. 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. ThereforeEc⊆E. ThusE=Ec.
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. HenceEc=A. Case(I). A∩{x∈X:d(x,x0) < r0} =φfor somer0∈(0,∞).
In this case, for any subsetEofA, we haveEc=E, and hence(Ec)c=Ec=E. Thus the functionf defined on the power setP (A)ofAasf (E)=Ec 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 haveEc=EorAaccording to whetherCase 1or Case 2holds. Hence(Ec)c=Ecfor allE∈P (A). Therefore the functionf defined on P (A)asf (E)=Ecfor allE∈P (A)is a closure operator. Thusρ-convergence defines a topologyτonAwith respect to which a subsetBofAis closed if and only ifB=Ec for someE∈P (A). Hence
τ=
A\Ec: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.
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≤ti+ti 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.
D-METRIC SPACES...
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.
(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 continu- ous 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)}con- verge 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.
D-METRIC SPACES...
Lemma1.27. Letp∈[1,∞). Then(a+b)p≥ap+bpfor alla,b∈R+.
Proof. Definef :R+ →Rasf (t)=(1+t)p−1−tp 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+tp ∀t∈R+. (1.31)
Leta,b∈R+. We may assume thata≥b. Ifa=0, thenb=0 and(a+b)p=0=ap+bp. 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+bp
ap. (1.33)
Hence(a+b)p≥ap+bp.
Corollary1.28. Letp∈[1,∞). Then(a+b+c)p≥ap+bp+cpfor 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+cp≥ap+bp+cp. (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
p1/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)
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≥ap+bp+cp
= 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
=21/pΨ(t).
(1.39)
Letεbe a positive real number. Chooseδ=21/pΨ(ε). Thenδ >0 sinceΨ(t)=0 implies t=0.
Φ(0,t,t) < δ ⇒21/pΨ(t) <21/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)=[tp1+t2p+t3p]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)
D-METRIC SPACES...
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)