# Redefinition ofτ-Distance in Metric Spaces

Authors: Tomonari Suzuki
Journal: Journal of Function Spaces
 (2017)

---

## Abstract

The concept ofτ-distance was introduced in 2001; on the other hand, that of τ-function was introduced by Lin and Du. Strongly inspired by τ-function, we introduce a new concept, which is a very slight generalization of τ-distance and is more natural than τ-distance. So we could say that we redefine τ-distance in some sense.


---

## Body

## 1. Introduction

Throughout this paper, we denote byN, Q, and R the sets of all positive integers, all rational numbers, and all real numbers, respectively.In 2001, the concept ofτ-distance was introduced in order to generalize results in [1–8] and others.Definition 1 (see [9]).
Let(X,d) be a metric space. Then a function p from X×X into [0,∞) is called a τ-distance on X if there exists a function η from X×[0,∞) into [0,∞) and the following is satisfied:(τd1)
p(x,z)≤p(x,y)+p(y,z) for any x,y,z∈X.(τd2)
η(x,0)=0 and η(x,t)≥t for any x∈X and t∈[0,∞), and η is concave and continuous in its second variable.(τd3)
limnxn=x and limnsup{η(zn,p(zn,xm)):m≥n}=0 imply p(w,x)≤liminfnp(w,xn) for any w∈X.(τd4)
limnsup{p(xn,ym):m≥n}=0 and limnη(xn,tn)=0 imply limnη(yn,tn)=0.(τd5)
limnη(zn,p(zn,xn))=0 and limnη(zn,p(zn,yn))=0 imply limnd(xn,yn)=0.We note that the metricd is one of τ-distances on X. See [9–15] and references therein for many examples and theorems concerning τ-distance. For instance, using τ-distance, Suzuki [14] gave a simple proof of Zhong’s theorem [7].In 2006, Lin and Du [16] introduced the following very interesting concept of τ-function, which is similar to that of τ-distance. However, both are independent.Definition 2 (Lin and Du [16]).
Let(X,d) be a metric space. Then a function p from X×X into [0,∞) is called a τ-function if the following conditions hold: (τf1)
For anyx,y,z∈X, p(x,z)≤p(x,y)+p(y,z).(τf2)
Ifx∈X and {yn} in X with limnyn=y and p(x,yn)≤M for some M=M(x)>0 then p(x,y)≤M.(τf3)
For any sequence{xn} in X with limnsup{p(xn,xm):m>n}=0, and if there exists a sequence {yn} in X such that limnp(xn,yn)=0, then limnd(xn,yn)=0.(τf4)
Forx,y,z∈X, p(x,y)=0 and p(x,z)=0 imply y=z.In this paper, strongly inspired byτ-function, we introduce a new concept, which is a slight generalization of τ-distance and which is more natural than τ-distance.Before the author wrote the paper [9], he had considered more than twenty definitions. Definition 1 was the most natural definition between them. Therefore, he defined τ-distance as in Definition 1, though Definition 1 was not the weakest. However, strongly inspired by Definition 2, he finds a more natural definition of “τ-distance”; see Definition 7. Now he thinks that we have to change the definition of “τ-distance.”


## 2. New Definition

In this section, we introduce a new definition of the notion ofτ-distance. Before introducing it, we state something on τ-distance, which are strongly connected with the new definition.Definition 3 (see [9]).
Letp be a τ-distance on a metric space (X,d). Then a sequence {xn} in X is called p-Cauchy if there exist a function η from X×[0,∞) into [0,∞) satisfying (τd2)–(τd5) and a sequence {zn} in X such that limnsup{η(zn,p(zn,xm)):m≥n}=0.Lemma 4 (see [9]).
Letp be a τ-distance on a metric space (X,d). If {xn} is a p-Cauchy sequence, then {xn} is a Cauchy sequence in the usual sense.Lemma 5 (see [9]).
Letp be a τ-distance on a metric space (X,d). If a sequence {xn} in X satisfies limnp(z,xn)=0 for some z∈X, then {xn} is a p-Cauchy sequence. Moreover, if a sequence {yn} in X also satisfies limnp(z,yn)=0, then limnd(xn,yn)=0. In particular for x,y,z∈X, p(z,x)=0 and p(z,y)=0 imply x=y.Lemma 6 (see [9]).
Letp be a τ-distance on a metric space (X,d). If a sequence {xn} in X satisfies limnsup{p(xn,xm):m>n}=0, then {xn} is a p-Cauchy sequence. Moreover, if a sequence {yn} in X satisfies limnp(xn,yn)=0, then {yn} is also a p-Cauchy sequence and limnd(xn,yn)=0.Now we give the new definition.Definition 7.
Let(X,d) be a metric space and let p be a function from X×X into [0,∞). Then p is called a τ′-distance on X if the following holds: (τ′1)
p(x,z)≤p(x,y)+p(y,z) for any x,y,z∈X.(τ′2)
Iflimnsup{p(zn,zm):m>n}=0 and limnp(zn,xn)=0, then limnd(zn,xn)=0. Moreover, if {xn} converges to some x∈X, then p(w,x)≤liminfnp(w,xn) for any w∈X.(τ′3)
Iflimnp(z,xn)=0, then limnd(xn,xn+1)=0 holds. Moreover, if {xn} converges to some x∈X, then p(w,x)≤liminfnp(w,xn) for any w∈X.The concept ofτ′-distance is slightly weaker than that of τ-distance. However, the author considers that the notions of both are the same. Indeed, we can prove τ′-distance versions of all the existence theorems in [9, 11–14] using the same proof.Proposition 8.
Letp be a τ-distance on a metric space (X,d). Then p is a τ′-distance.Proof.
(τd1) and (τ′1) are the same. (τ′2) follows from Lemma 6 and (τd3). (τ′3) follows from Lemmas 4 and 5 and (τd3).Remark 9.
Aτ′-distance need not be a τ-distance. However, the only example we have is quite complicated and we have chosen not to include it.


## 3. Condition (CL)

We have introduced the concept ofp-Cauchy for τ-distance; see Definition 3. Instead of this, for τ′-distance, we introduce Condition (CL), which is named after Cauchy and lower semicontinuity.Definition 10.
Letp be a τ′-distance on a metric space (X,d). Let {xα:α∈D} be a net in X. Then {xα} is said to satisfyCondition (CL) if the following holds:(CL1)
{xα} is a Cauchy net in the usual sense.(CL2)
Either of the following holds:
(i)
{xα} does not converge.(ii)
If{xα} converges to x, then p(w,x)≤liminfαp(w,xα) holds for any w∈X.The following lemma is important in this section.Lemma 11.
Let(X,d) be a metric space and let {xα:α∈D} be a net in X. Assume that {xα} is not Cauchy. Let {βn} be a sequence in D and let h be a mapping on D. Then there exist a positive number ε and a sequence {f(n)} in D such that (1)βn≤fn,hfn≤fn+1,dxfn,xfn+1≥εfor any n∈N.Proof.
Since{xα} is not Cauchy, there exist a positive number ε and mappings g1 and g2 on D such that α≤g1(α), α≤g2(α), and d(xg1(α),xg2(α))≥2ε for any α∈D. Put f(1)=β1 and choose γ2∈D with γ2≥β2 and γ2≥h(f(1)). Then either (2)dxf1,xg1γ2≥εordxf1,xg2γ2≥εholds. So we can put f(2)=g1(γ2) or f(2)=g2(γ2) such that d(xf(1),xf(2))≥ε. Continuing this argument, we can obtain the desired result.Remark 12.
{
x
f
(
n
)
} is not necessarily a subnet of {xα}.We do not use the concept of net in Definition7. However, we can prove something on net as follows.Lemma 13.
Letp be a τ′-distance on a metric space (X,d). Let {xα:α∈D} be a net in X satisfying limαp(z,xα)=0 for some z∈X. Then the following holds: (i)
{xα} satisfies Condition (CL).(ii)
If a net{yα:α∈D} in X also satisfies limαp(z,yα)=0, then limαd(xα,yα)=0 holds.Proof.
We choose a sequence{βn} in D such that p(z,xα)≤1/n for α≥βn. Arguing by contradiction, we assume {xα} is not Cauchy. Let h be an arbitrary mapping from D into itself. Then by Lemma 11, there exist a positive number ε and a sequence {f(n)} in D satisfying (3)βn≤fn,dxfn,xfn+1≥εfor any n∈N. Since (4)limn→∞pz,xfn≤limn→∞1n=0,from (τ′3), we have (5)0<ε≤limn→∞dxfn,xfn+1=0,which is a contradiction. Hence, {xα} is Cauchy. Next, we assume {xα} converges to x. Fix w∈X and let η∈R with η>liminfαp(w,xα). Then we can choose a sequence {γn} in D such that (6)γn≥βn,dxγn,x<1n,pw,xγn<ηfor any n∈N. Since {p(z,xγn)} converges to 0, we have from (τ′3) (7)pw,x≤liminfn→∞pw,xγn≤η.Since η is arbitrary, we obtain p(w,x)≤liminfαp(w,xα). Therefore, we have shown (i). Let us prove (ii). Define a directed set E and an order ≤ in E(8)E=D×0,1,α,i≤β,j⟺α≤β.Define a net {zα,j:(α,j)∈E} by (9)zα,j=xαifj=0yαifj=1.Then since lim(α,j)p(z,z(α,j))=0, {z(α,j)} is Cauchy by (i). This implies limαd(xα,yα)=0.As a corollary of Lemma13, we obtain the following sequential version. Compare Lemma 14 with Lemma 5.Lemma 14.
Letp be a τ′-distance on a metric space (X,d). If a sequence {xn} in X satisfies limnp(z,xn)=0 for some z∈X, then {xn} satisfies Condition (CL). Moreover, if a sequence {yn} in X also satisfies limnp(z,yn)=0, then limnd(xn,yn)=0 holds. In particular, for x,y,z∈X, p(z,x)=0 and p(z,y)=0 imply x=y.Lemma 15.
Let(D,≤) be a directed set and let h be a mapping from D into itself such that α≤h(α) for any α∈D. Let {xα:α∈D} be a net in a set X. Then {xh(α):α∈D} is a subnet of {xα}.Proof.
Obvious.Lemma 16.
Letp be a τ′-distance on a metric space (X,d). Let D be a directed set such that for any α∈D there exists β∈D with β>α. Let {zα:α∈D} be a net in X satisfying limαsup{p(zα,zβ):β>α}=0. Then the following holds: (i)
{zα} satisfies Condition (CL).(ii)
If a net{xα:α∈D} in X satisfies limαp(zα,xα)=0, then {xα} satisfies Condition (CL) and limαd(zα,xα)=0 holds.Proof.
We choose a sequence{βn} in D such that (10)supβ>αpzα,zβ≤1n,pzα,xα≤1nfor any α≥βn. We consider the following two cases: (a)
There existsζ∈D such that ζ≥α for any α∈D.(b)
For anyα∈D, there exists β∈D with α≱β.
In the first case, from the assumption, we can takeω∈D with ω>ζ. Then since ζ≥βn and ω≥βn for any n∈N, we have (11)pzζ,zω=pzω,zζ=pzζ,xζ=pzω,xω=0.So (12)pzζ,xω≤pzζ,zω+pzω,xω=0.Similarly, p(zω,xζ)=0. By Lemma 14, we obtain zω=xω=zζ for ω≥ζ. Therefore, limαd(zα,xα)=0. Since {xα} converges to zζ, thus, {xα} is Cauchy. Also, p(w,zζ)=limαp(w,xα) holds. Therefore, we have shown that {xα} satisfies Condition (CL) and limαd(zα,xα)=0. As in the above proof, we can prove that {zα} satisfies Condition (CL). In the second case, we can define a mapping h from D into itself satisfying (13)α<hα,α≱γfor any γ≥h(α). In fact, for α∈D, there exists β∈D with α≱β. We can choose h(α)∈D satisfying h(α)≥α and h(α)≥β. If there exists γ∈D such that γ≥h(α) and α≥γ, then α≥h(α) and hence α≥β, which is a contradiction. Similarly h(α)=α implies a contradiction. Therefore, we have defined h. We will show that {xα} satisfies Condition (CL). Arguing by contradiction, we assume {xα} is not Cauchy. Then by Lemma 11, there exist a positive number ε and a sequence {f(n)} in D satisfying (14)βn≤fn<hfn≤fn+1,dxfn,xfn+1≥ε.From the definition of h, we note f(m)>f(n) for m,n∈N with m>n. Since (15)limn→∞pzfn,xfn≤limn→∞1n=0,limn→∞pzfn,zfn+1=limn→∞supm>npzfn,zfm≤limn→∞1n=0,from (τ′2), we have (16)limn→∞dzfn,zfn+1=limn→∞dzfn,xfn=0.So (17)0<ε≤limn→∞dxfn,xfn+1≤limn→∞dxfn,zfn+dzfn,zfn+1+dzfn+1,xfn+1=0,which is a contradiction. Hence, {xα} is Cauchy. Next, we assume {xα} converges to x. Fix w∈X and let η∈R with η>liminfαp(w,xα). Then we can choose a sequence {γn} in D such that (18)βn≤γn<hγn≤γn+1,dxγn,x<1n,pw,xγn<η.Since limnsup{p(zγn,zγm):m>n}=limnp(zγn,xγn)=0, we have from (τ′2) that (19)pw,x≤liminfn→∞pw,xγn≤η.Since η is arbitrary, we obtain p(w,x)≤liminfαp(w,xα). Therefore, we have shown that {xα} satisfies Condition (CL). Let us prove limαd(zα,xα)=0. Arguing by contradiction, we assume limsupαd(zα,xα)>0. Fix ε∈R with limsupαd(zα,xα)>ε>0. Then we can choose a sequence {δn} in D such that (20)βn≤δn<hδn≤δn+1,dzδn,xδn>ε.Since limnsup{p(zδn,zδm):m>n}=limnp(zδn,xδn)=0, we obtain from (τ′2) that(21)0<ε≤limn→∞dzδn,xδn=0,which is a contradiction. Therefore, we have shown limαd(zα,xα)=0. We have proved (ii). We shall show that {zα} satisfies Condition (CL). Fix w∈X and let η∈R with η>liminfαp(w,zα). Then we can define a mapping g from D into itself such that (22)α≤hα≤gα,pw,zgα<η.By Lemma 15, we note that {zg(α)} is a subnet of {zα}. We have (23)limα∈Dpzα,zgα≤limα∈Dsupβ>αpzα,zβ=0.So from (ii), {zg(α)} satisfies Condition (CL) and limαd(zα,zg(α))=0 holds. So, {zα} is Cauchy. We assume that {zα} converges to z. Since {zg(α)} also converges to z, we have (24)pw,z≤liminfα∈Dpw,zgα≤η.Since η is arbitrary, we obtain p(w,z)≤liminfαp(w,zα). Therefore, we have shown that {zα} satisfies Condition (CL). We have proved (i).As a corollary of Lemma16, we obtain the following sequential version. Compare Lemma 17 with Lemma 6.Lemma 17.
Letp be a τ′-distance on a metric space (X,d). If a sequence {zn} in X satisfies limnsup{p(zn,zm):m>n}=0, then {zn} satisfies Condition (CL). Moreover, if a sequence {xn} in X satisfies limnp(zn,xn)=0, then {xn} satisfies Condition (CL) and limnd(zn,xn)=0 holds.Lemma 18.
Letp be a τ′-distance on a metric space (X,d) and let {zn} and {xn} be sequences in X satisfying limnsup{p(zn,zm):m>n}=0. Assume that, for any subsequence {f(n)} of the sequence {n} in N, there exist subsequences {f∘g(n)} of {f(n)} and {h(n)} of {n} satisfying limnp(zh(n),xf∘g(n))=0. Then {xn} satisfies Condition (CL) and limnd(zn,xn)=0 holds.Proof.
We note that{zn} is Cauchy from Lemma 17. Let {f(n)} be an arbitrary subsequence of {n}. Then from the assumption, there exist subsequences {f∘g(n)} of {f(n)} and {h(n)} of {n} satisfying limnp(zh(n),xf∘g(n))=0. We note limnsup{p(zh(n),zh(m)):m>n}=0. From (τ′2), we have limnd(zh(n),xf∘g(n))=0 and hence (25)limn→∞dzf∘gn,xf∘gn≤limn→∞dzf∘gn,zhn+dzhn,xf∘gn=0.Since {f(n)} is arbitrary, we obtain limnd(zn,xn)=0. {zn} is Cauchy and so is {xn}. Next, we assume that {xn} converges to x∈X, fix w∈X, and choose a subsequence {f′(n)} of {n} satisfying limnp(w,xf′(n))=liminfnp(w,xn). Then from the assumption, there exist subsequences {f′∘g′(n)} of {f′(n)} and {h′(n)} of {n} satisfying limnp(zh′(n),xf′∘g′(n))=0. By (τ′2), we obtain (26)pw,x≤limn→∞pw,xf′∘g′n=liminfn→∞pw,xn.Therefore, {xn} satisfies Condition (CL).


## 4. Examples

In this section, in order to show that the concept ofτ′-distance is more natural than that of τ-distance, we give some examples. Compare them with Propositions 4 and 5 in [9] and Proposition 4.2 in [11]. We note that we have not yet proved τ-distance versions of Propositions 19–21 below.The following is connected with the result in Zhong [7]. See also Turinici [17].Proposition 19.
Letp be a τ′-distance on a metric space (X,d). Let b be a nonincreasing function from [0,∞) into (0,∞) with ∫0∞‍b(t)dt=∞; and let Γ be a function from X into [0,∞) with Γ(y)≤Γ(x)+p(x,y) for x,y∈X. Then a function q from X×X into [0,∞) defined by (27)qx,y=∫ΓxΓx+px,ybtdtfor x,y∈X is also a τ′-distance on X.Proof.
Using (τ′1), we have (28)qx,z=∫ΓxΓx+px,zbtdt≤∫ΓxΓx+px,y+py,zbtdt=∫ΓxΓx+px,ybtdt+∫Γx+px,yΓx+px,y+py,zbtdt≤∫ΓxΓx+px,ybtdt+∫ΓyΓy+py,zbtdt=qx,y+qy,zfor any x,y,z∈X and hence (τ′1)q holds. Define a function g from X into (0,∞) by (29)∫ΓxΓx+gxbtdt=1for x∈X. From the assumption of ∫0∞‍b(t)dt=∞, g is well defined. If q(x,y)≤1, then we have p(x,y)≤g(x) from the definition of g. So, (30)bΓx+gxpx,y≤bΓx+px,ypx,y=∫ΓxΓx+px,ybΓx+px,ydt≤∫ΓxΓx+px,ybtdt=qx,y.Hence, we note that(31)qx,y≤1impliespx,y≤qx,ybΓx+gx.Using (31), we next show (τ′2)q. We assume limnsup{q(zn,zm):m>n}=0 and limnq(zn,xn)=0. Without loss of generality, we may assume sup{q(zn,zm):m>n}≤1 and q(zn,xn)≤1 for any n∈N. Then for n∈N with n≥2, we have (32)pz1,zn≤qz1,znbΓz1+gz1≤1bΓz1+gz1and hence (33)Γzn≤Γz1+pz1,zn≤Γz1+1bΓz1+gz1.This implies that {Γ(zn)} is bounded. From the definition of g, {Γ(zn)+g(zn)} is also bounded and so is {1/b(Γ(zn)+g(zn))}. We put (34)M≔sup1bΓzn+gzn:n∈N<∞.Then we have by (31) (35)limn→∞supm>npzn,zm≤Mlimn→∞supm>nqzn,zm=0,limn→∞pzn,xn≤Mlimn→∞qzn,xn=0.By (τ′2), we obtain limnd(zn,xn)=0. Moreover, if {xn} converges to some x∈X, then p(w,x)≤liminfnp(w,xn) for any w∈X. This yields q(w,x)≤liminfnq(w,xn) for any w∈X. Therefore, (τ′2)q holds. We can prove (τ′3)q as in the proof of (τ′2)q.We call that a nonempty subsetA of X is called p-bounded if sup{p(a,b):a,b∈A}<∞. The following is connected with the result in Ume [18].Proposition 20.
Letp be a τ′-distance on a metric space (X,d). Let T be a set-valued mapping on X such that Tx is a nonempty p-bounded subset of X for any x∈X. Assume that limnxn=z, yn∈Txn, and limnyn=z imply Tz={z}. Then a function q from X×X into [0,∞) defined by (36)qx,y=max⁡supa∈Txinfb∈Typa,b,supa∈Txpa,yfor any x,y∈X is also a τ′-distance.Proof.
It is obvious that{x} is p-bounded for any x∈X. For nonempty p-bounded subsets A and B of X, we put (37)δA,B=supa∈Ainfb∈Bpa,b.Then δ(A,B)<∞. Indeed, for a0∈A, we have (38)δA,B≤supa∈Ainfb∈Bpa,a0+pa0,b=supa∈Apa,a0+infb∈Bpa0,b≤supa1,a2∈Apa1,a2+infb∈Bpa0,b<∞.Using δ, we can write (39)qx,y=max⁡δTx,Ty,δTx,y.From (τ′1), we have (40)δA,C≤δA,B+δB,Cfor nonempty p-bounded subsets A, B, and C of X. So, for x,y,z∈X, we obtain (41)qx,z=maxδTx,Tz,δTx,z≤maxδTx,Ty+δTy,Tz,δTx,Ty+δTy,z≤qx,y+qy,z.This is (τ′1)q. We next show (τ′2)q. We assume limnsup{q(zn,zm):m>n}=0 and limnq(zn,xn)=0. We choose a subsequence {h(n)} of the sequence {n} such that sup{q(zh(n),zm):m>h(n)}<2-n for any n∈N. Since δ(Tzh(n),Tzh(n+1))≤q(zh(n),zh(n+1))<2-n, we can choose a sequence {an} in X such that (42)an∈Tzhn,pan,an+1<2-nfor n∈N. We note (43)limn→∞supm>npan,am≤limn→∞supm>n∑j=nm-1paj,aj+1≤limn→∞supm>n∑j=nm-12-j≤limn→∞21-n=0.Let {f(n)} be an arbitrary subsequence of {n} and choose a subsequence {f∘g(n)} of {f(n)} such that h(n)<f∘g(n) for any n∈N. Then we have (44)limn→∞pan,zf∘gn≤limn→∞δTzhn,zf∘gn≤limn→∞qzhn,zf∘gn=0and hence limnd(an,zn)=0 from Lemma 18. We also have (45)limn→∞pan,xf∘gn≤limn→∞δTzhn,xf∘gn≤limn→∞qzhn,xf∘gn≤limn→∞qzhn,zf∘gn+qzf∘gn,xf∘gn=0and hence limnd(an,xn)=0 from Lemma 18. Thus, limnd(zn,xn)=0 holds. If {xn} converges to some x∈X, then {zh(n)} and {an} also converge to x. So from the assumption, Tx={x} holds. Also, p(w,x)≤liminfnp(w,xn) holds for any w∈X from Lemma 18. We obtain (46)qw,x=δTw,x=supb∈Twpb,x≤supb∈Twliminfn→∞pb,xn≤liminfn→∞supb∈Twpb,xn=liminfn→∞δTw,xn≤liminfn→∞qw,xn.We have shown (τ′2)q. Let us prove (τ′3)q. We assume limnq(z,xn)=0. Fix a∈Tz. Then since (47)limn→∞pa,xn≤limn→∞δTz,xn≤limn→∞qz,xn=0,{xn} satisfies Condition (CL) by Lemma 14. We assume that {xn} converges to x. We choose a sequence {yn} in X such that (48)yn∈Txn,pa,yn≤δa,Txn+1nfor n∈N. Since (49)limn→∞pa,yn≤limn→∞δa,Txn≤limn→∞δTz,Txn≤limn→∞qz,xn=0,we have limnd(xn,yn)=0 by Lemma 14. Hence, {yn} converges to x. From the assumption, Tx={x} holds. As in the proof of (τ′2), we can prove q(w,x)≤limnq(w,xn) for any w∈X.We finally prove the following. We note that our proof is very easy while the proof of Proposition4.2 in [11] is difficult.Proposition 21.
Let(X,d) be a metric space. Let {pj:j∈J} be a set of τ′-distances on X. If a function q defined by (50)qx,y=supj∈Jpjx,yis a real-valued function, then q is a τ′-distance on X.Proof.
We have(51)qx,z=supj∈Jpjx,z≤supj∈Jpjx,y+pjy,z≤supj∈Jpjx,y+supj∈Jpjy,z=qx,y+qy,zfor any x,y,z∈X. We next show (τ′2)q. We assume limnsup{q(zn,zm):m>n}=0 and limnq(zn,xn)=0. Then since pj≤q for any j∈J, limnsup{pj(zn,zm):m>n}=0 and limnpj(zn,xn)=0 hold. Thus, from (τ′2)pj, limnd(zn,xn)=0 holds. If {xn} converges to some x∈X, then pj(w,x)≤liminfnpj(w,xn) holds for any w∈X and j∈J by using (τ′2)pj again. Using it, we have (52)qw,x=supj∈Jpjw,x≤supj∈Jliminfn→∞pjw,xn≤liminfn→∞supj∈Jpjw,xn=liminfn→∞qw,xnfor any w∈X. We have shown (τ′2)q. We can similarly prove (τ′3)q.


## 5. Kirk and Saliga’s Fixed Point Theorem

Kirk and Saliga generalized Caristi’s fixed point theorem [2, 3]; see Theorem 2.3 in [19]. The author thinks that the proof in [19] is splendid. In this section, we generalize Kirk and Saliga’s fixed point theorem.Letα be an ordinal number. We denote by α+ and α- the successor and the predecessor of α, respectively. We recall that α isisolated if α- exists. α islimit if α≠0 and α- does not exist. For a set A, we denote by #A cardinal number of A.Theorem 22.
Let(X,d) be a complete metric space with a τ′-distance p. Let T be a mapping on X. Let f be a function from X into R bounded from below such that f(x)≤liminfδf(xδ) whenever a net {xδ:δ∈E} satisfies the following: (a)
{xδ} converges to x.(b)
For anyδ∈E, there exists ε∈E satisfying ε>δ.(c)
limδsup{p(xδ,xε):ε>δ}=0.(d)
f(xδ)≥f(xε) for δ,ε∈E with δ≤ε.Assume f(Tx)≤f(x) for any x∈X, and there exists κ∈N such that (53)fTκx+px,Tx≤fxfor any x∈X. Fix u∈X. Let γ be the first uncountable ordinal number and put D={α:α<γ}. Define a net {uα:α∈D} by (54)uα=uifα=0Tuα-ifαisisolatedlim⁡uβ:β<αifαislimit.Then {uα} is well defined and there exists α∈D satisfying Tuα=uα and p(uα,uα)=0.Proof.
Define a functionF form X into R by (55)Fx=∑j=0κ-1fTjx,where T0 is the identity mapping on X. It is obvious that (56)FTx≤FTx+px,Tx≤Fxholds for any x∈X. We shall show by transfinite induction (Pα) that (a)
uα is well defined,(b)
F(uα)≤F(uα)+p(uβ,uα)≤F(uβ) for β<α,for α<γ. It is obvious that P0 is true. Fix α with 0<α<γ, and assume that Pβ is true for β<α. In the case where α is isolated, it is obvious that uα=Tuα- is well defined. We have (57)Fuα≤Fuα+puα-,uα≤Fuα-.For β<α-, we have (58)Fuα≤Fuα+puβ,uα≤Fuα+puβ,uα-+puα-,uα≤Fuα-+puβ,uα-≤Fuβ.Hence, Pα is true. In the other case, where α is limit, since {F(uβ):β<α} is a nonincreasing and F is bounded from below, {F(uβ):β<α} converges and hence it is Cauchy. We have (59)limβ<αsuppuβ,uδ:β<δ<α≤limβ<αsupFuβ-Fuδ:β<δ<α=0.By Lemma 16, {uβ:β<α} satisfies Condition (CL). Since X is complete, {uβ:β<α} converges. Hence, uα is well defined. For β<α, we have (60)Fuα≤Fuα+puβ,uα≤κfuα+puβ,uα≤limδ<αFuδ+liminfδ<αpuβ,uδ=liminfδ<αFuδ+puβ,uδ≤Fuβ.Hence, Pα is true. Therefore, by transfinite induction, Pα is true for any α<γ. Arguing by contradiction, we assume the following: (a)
F(uα++)<F(uα) for any α∈D.Then there exists s,t∈Q satisfying F(uα++)<s<t<F(uα). So we have (61)#Q=#N<#γ≤#Q,which is a contradiction. Therefore, there exists α∈D satisfying F(uα++)=F(uα). We have (62)Fuα=Fuα++≤Fuα+++puα,uα++≤Fuα,Fuα=Fuα++≤Fuα+≤Fuα++puα,uα+≤Fuα.Hence, p(uα,uα++)=0 and p(uα,uα+)=0 hold. By Lemma 14, uα+=uα++ holds. Therefore, Tuα+=uα+ holds.



---

*Source: 4168486-2017-04-20.xml*