Functions that preserve totally bounded sets vis-á-vis stronger notions of continuity
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Functions that preserve totally bounded setsvis-´a-vis stronger notions of continuity
Lipsy Gupta · S. Kundu
Received: date / Accepted: date
Abstract
A function between two metric spaces is said to be totally boundedregular if it preserves totally bounded sets. These functions need not be con-tinuous in general. Hence the purpose of this article is to study such functionsvis-´a-vis continuous functions and functions that are stronger than the con-tinuous functions such as Cauchy continuous functions, some Lipschitz-typefunctions etc. We also present some analysis on strongly uniformly continuousfunctions which were first introduced in [8] and study when these functionsare stable under reciprocation.
Keywords
Totally Bounded regular (TB-regular) function · Cauchycontinuous function · strongly uniformly continuous function · Lipschitz-typefunctions · continuous function · complete metric space Mathematics Subject Classification (2010) · · · · A subset A of a metric space ( X, d ) is called totally bounded if for every ǫ >
0, there exists { x , x , ...x n } ⊆ X such that A ⊆ n [ i =1 B ( x i , ǫ ). We call afunction f : ( X, d ) → ( Y, ρ ) between two metric spaces to be totally bounded
Lipsy GuptaDepartment of Mathematics, Indian Institute of Technology Delhi,New Delhi, 110016, IndiaE-mail: [email protected]. KunduDepartment of Mathematics, Indian Institute of Technology Delhi,New Delhi, 110016, IndiaE-mail: [email protected] Lipsy Gupta, S. Kundu regular or TB-regular for short, if it preserves totally bounded sets, that is,for every totally bounded subset A of X , f ( A ) is also totally bounded. It iswell-known that a metric space is totally bounded if and only if each sequencein it has a Cauchy subsequence. Intuitively, TB-regular functions can have aclose relation with the functions that preserve Cauchy sequences called Cauchycontinuous functions. The class of Cauchy continuous functions is well-studied[6,15,20,21] and lies between the class of uniformly continuous functions andthat of continuous functions. TB-regular functions were first considered in [9],where the authors proved that a function is TB-regular if and only if it takesevery Cauchy sequence to a sequence having a Cauchy subsequence. Thus it isimmediate that every Cauchy continuous function (and hence every uniformlycontinuous) is TB-regular. TB-regular functions were then considered in [8],where the authors introduced an interesting class of functions that are strongerthan the uniformly continuous functions, called the strongly uniformly contin-uous functions. TB-regular functions played a key role in the analysis of stronguniform continuity, particularly to determine the structure of the bornology ofthe subsets on which a (continuous) function is strongly uniformly continuous.The main aim of this article is to view TB-regular functions in terms ofcontinuous functions and some well-known stronger notions of continuity. Wediscuss some subclasses of the class of continuous functions from a metricspace ( X, d ) to another metric space (
Y, ρ ) such that f : ( X, d ) → ( Y, ρ ) isTB-regular if and only if whenever f is followed in a composition by anyfunction from those subclasses, it is again a TB-regular function (see Theo-rem 4). We note that these subclasses include the class of Cauchy continuousfunctions, some Lipschitz-type functions, but not the class of uniformly con-tinuous functions. We have already seen that every Cauchy continuous (anduniformly continuous) function is TB-regular. Here we study the conditionsunder which the reverse implications are also true (see Theorem 6). In theprocess of studying relations between TB-regular functions and continuousfunctions, we establish many new useful characterizations of complete metricspaces (e.g., see Theorem 2 and Theorem 5). Finally, we give some applica-tions of strong uniform continuity to some metric spaces possessing strongerproperties than the complete metric spaces (Theorem 10 and Theorem 11) andcharacterize the condition under which every non-vanishing strongly uniformlycontinuous function is stable under reciprocation (Theorem 13). Let us first record the precise definitions of TB-regular and Cauchy continuousfunctions.
Definition 1
A function f : ( X, d ) → ( Y, ρ ) between two metric spaces issaid to be
TB-regular if it preserves totally bounded sets, that is, for everytotally bounded subset A of X , f ( A ) is also totally bounded. itle Suppressed Due to Excessive Length 3 Definition 2
A function f : ( X, d ) → ( Y, ρ ) between two metric spaces issaid to be Cauchy-continuous if ( f ( x n )) is Cauchy in ( Y, ρ ) for every Cauchysequence ( x n ) in ( X, d ).Throughout our analysis on TB-regular functions we use the following fact:a function is TB-regular if and only if it takes every Cauchy sequence to asequence having a Cauchy subsequence [9, Proposition 5.7 (1)]. The followingexample shows that a TB-regular function need not be continuous.
Example 1
Let X = { n , n ∈ N } and consider the function f : X → R with f (0) = 0 , f ( n ) = 1 for odd n and f ( n ) = 2 for even n . Clearly, f isTB-regular but not even continuous. On the other hand, if we define a function g on Y = { n : n ∈ N } such that g (cid:16) n (cid:17) = n ∀ n ∈ N , then g is continuous butnot TB-regular.Recall that a family B of non-empty subsets of a metric space ( X, d ) issaid to be bornology if ( i ) B forms a cover of X ( ii ) B is hereditary, that is,if A ∈ B and B ⊆ A , then B ∈ B ( iii ) B is closed under finite unions (see [9]and the references therein). A family of subsets B o is said to be a base for B if ∀ B ∈ B , ∃ B o ∈ B o such that B ⊆ B o . If each member of a base is a closedsubset of X , then the base is called a closed base . As an example, note that thesmallest bornology on X is the family of finite subsets of X and the largest oneis the power set P ( X ) of X . Note that the family of totally bounded subsets of X also forms a bornology. So in other words, a function f : ( X, d ) → ( Y, ρ ) isTB-regular if the function image of the bornology of totally bounded subsetsof (
X, d ) is contained in the bornology of totally bounded subsets of (
Y, ρ ). Proposition 1
Let f : ( X, d ) → ( Y, ρ ) be a function between two metricspaces. Then the set of subsets on which f is TB-regular forms a bornology B T . Moreover, if f is continuous, then B T has a closed base.Proof We will prove that if f is TB-regular on A and B , where A and B aresubsets of X , then f is TB-regular on A ∪ B . Let ( x n ) be a Cauchy sequencein A ∪ B . Thus there exists a subsequence ( x n k ) of ( x n ) which lies either in A or in B . Since f is TB-regular on both A and B , there exists a Cauchysubsequence of ( f ( x n )) in ( Y, ρ ). Thus f is TB-regular on A ∪ B .Next, let f be continuous. Let f be TB-regular on a subset A of X . Toprove that f is TB-regular on the closure of A , say A , let ( x n ) be a Cauchysequence in A . Since f is continuous, for each n ∈ N , ∃ y n ∈ A such that d ( x n , y n ) < n and ρ ( f ( x n ) , f ( y n )) < n . So ( y n ) ⊆ A is a Cauchy sequence.Thus ( f ( y n )) has a Cauchy subsequence, which further implies that ( f ( x n ))has a Cauchy subsequence too. Hence we are done. ⊓⊔ In this article, we cast light on the function between metric space that areTB-regular, that is, the case when P ( X ) = B T . In [8], the authors introducedan interesting class of functions that are stronger than the uniform continuousfunctions, called the strongly uniformly continuous functions defined as follows. Lipsy Gupta, S. Kundu
Definition 3
Let (
X, d ) and (
Y, ρ ) be metric spaces and let B be a subset of X . A function f : X → Y is said to be strongly uniformly continuous on B if ∀ ǫ >
0, there exists δ > d ( x, y ) < δ and { x, y } ∩ B = ∅ , then ρ ( f ( x ) , f ( y )) < ǫ .It is interesting to note that a function f on a metric space ( X, d ) is continu-ous if and only if it is strongly uniformly continuous on { x } for each x ∈ X .Also, if f is a continuous function between any two metric spaces and K is acompact subset of the domain, then f is strongly uniformly continuous on theset K . In [8], TB-regular functions played a key role in the analysis of stronglyuniformly continuous functions.Other than the above mentioned classes of continuous functions, we alsostudy some relations of TB-regular functions with some Lipschitz-type func-tions. Much analysis on these functions and their applications can be foundin [4,5,6,7,12,13]. Let us note their precise definitions in order of increasingsize. Definition 4
For metric spaces (
X, d ) and (
Y, ρ ), any function f : ( X, d ) → ( Y, ρ ) is called:(a)
Lipschitz if there exists k > ρ ( f ( x ) , f ( x ′ )) ≤ kd ( x, x ′ ) ∀ x, x ′ ∈ X .(b) uniformly locally Lipschitz if there exists δ > x ∈ X ,there exists k x > ρ ( f ( u ) , f ( w )) ≤ k x d ( u, w ), whenever u, w ∈ B ( x, δ ).(c) Cauchy-Lipschitz if f is Lipschitz when restricted to the range of eachCauchy sequence ( x n ) in X .( e ) locally Lipschitz if for each x ∈ X , there exists δ x > f restrictedto B ( x, δ x ) is Lipschitz.Note that the function g of Example 1 shows that a locally Lipschitz func-tion need not be TB-regular, while, since every Cauchy-Lipschitz function isCauchy continuous, every Cauchy-Lipschitz function is TB-regular. We would like to begin with the necessary and sufficient conditions under whichevery TB-regular function is continuous, Cauchy-continuous and uniformlycontinuous. The routine proof of the following result has been omitted.
Theorem 1
Let ( X, d ) be a metric space.(a) Every TB-regular function from ( X, d ) to any other metric space ( Y, ρ ) iscontinuous if and only if ( X, d ) is discrete.(b) Every TB-regular function from ( X, d ) to any other metric space ( Y, ρ ) isCauchy-continuous if and only if ( X, d ) is complete and discrete.(c) Every TB-regular function from ( X, d ) to any other metric space ( Y, ρ ) isuniformly continuous if and only if ( X, d ) is uniformly discrete. itle Suppressed Due to Excessive Length 5 It is known that a metric space (
X, d ) is complete if and only if eachreal-valued continuous function defined on it is Cauchy continuous. The nextresult shows that the TB-regularity of continuous functions also characterizescomplete metric spaces.
Theorem 2
Let ( X, d ) be a metric space. The following are equivalent.(a) ( X, d ) is complete.(b) Each continuous function from ( X, d ) to any other metric space ( Y, ρ ) isTB-regular.(c) Each locally Lipschitz function from ( X, d ) to any other metric space ( Y, ρ ) is TB-regular.(d) Each real-valued locally Lipschitz function on ( X, d ) is TB-regular.Proof The implications ( a ) ⇒ ( b ) ⇒ ( c ) ⇒ ( d ) are all immediate.( d ) ⇒ ( a ): Suppose ( X, d ) is not complete. Thus there exists a Cauchysequence ( x n ) of distinct points in ( X, d ) such that it does not converge. Let( b X, ˆ d ) be the completion of ( X, d ) and let ( x n ) converges to ˆ x ∈ b X . Define thefollowing function f on X . f ( x ) = 1ˆ d ( x, ˆ x ) ∀ x ∈ X The function f is not TB-regular as { x n : n ∈ N } is totally bounded but theits image is not even bounded. To see that f is locally Lipschitz, note thatthe function x ˆ d ( x, ˆ x ) is Lipschitz and thus locally Lipschitz. It is easy tosee that if a real-valued function (which is never zero) on a metric space islocally Lipschitz, then so is its reciprocal. Hence f is locally Lipschitz but notTB-regular, a contradiction. ⊓⊔ The following result characterizes the class of complete metric spaces interms of the Cauchy-continuity of a subclass of continuous functions.
Theorem 3
Let ( X, d ) be a metric space. The following are equivalent.(a) ( X, d ) is complete.(b) Each continuous TB-regular function from ( X, d ) to any other metric space ( Y, ρ ) is Cauchy continuous.(c) Each real-valued continuous TB-regular function on ( X, d ) is Cauchy con-tinuous.Proof Only the implication ( c ) ⇒ ( a ) requires some justification. Supposethere exists a Cauchy sequence ( x n ) of distinct points in ( X, d ) such that itdoes not cluster. Define the following function on the set A = { x n : n ∈ N } . f ( x n ) = (cid:26) n is odd2 : n is evenBy Theorem 5.1 in [11, p. 149], f can be extended to a function F : X → (0 , F is continuous. Thus F is continuous and TB-regular, but notCauchy-continuous. ⊓⊔ Lipsy Gupta, S. Kundu
Theorem 4
Let f : ( X, d ) → ( Y, ρ ) be a function between two metric spaces.The following are equivalent.(a) f is TB-regular.(b) If ( Z, µ ) is a metric space and g : ( Y, ρ ) → ( Z, µ ) is a TB-regular function,then g ◦ f is TB-regular.(c) If ( Z, µ ) is a metric space and g : ( Y, ρ ) → ( Z, µ ) is a Cauchy-continuousfunction, then g ◦ f is TB-regular.(d) If ( Z, µ ) is a metric space and g : ( Y, ρ ) → ( Z, µ ) is a Cauchy-Lipschitzfunction, then g ◦ f is TB-regular.(e) If ( Z, µ ) is a metric space and g : ( Y, ρ ) → ( Z, µ ) is a uniformly locallyLipschitz function, then g ◦ f is TB-regular.(f ) Whenever g : ( Y, ρ ) → R is a uniformly locally Lipschitz function, g ◦ f isTB-regular.Proof The implications ( a ) ⇒ ( b ) ⇒ ( c ) ⇒ ( d ) ⇒ ( e ) ⇒ ( f ) are all immediate.( f ) ⇒ ( a ): Suppose f is not TB-regular. Thus there exists a Cauchy se-quence ( x n ) in ( X, d ) such that ( f ( x n )) has no Cauchy subsequence. Thus thereexists ǫ > ρ ( f ( x n ) , f ( x m )) >ǫ ∀ n, m ∈ N . Define a function g : ( Y, ρ ) → R as follows: g ( y ) = ( n − nǫ d ( y, f ( x n )) : y ∈ B (cid:16) f ( x n ) , ǫ (cid:17) for some n ∈ N otherwise The function g is uniformly locally Lipschitz because ∀ y ∈ Y , B ( y, ǫ ) in-tersects at most one of the balls B ( f ( x m ) , ǫ ) and g restricted to each ball B ( f ( x m ) , ǫ ) is Lipschitz. Now for each n ∈ N , ( g ◦ f )( x n ) = n . Thus ( x n ) isa Cauchy sequence such that ( g ◦ f )( x n ) has no Cauchy subsequence, whichimplies that g ◦ f is not TB-regular, a contradiction. ⊓⊔ Remark 1
Note that even if every uniformly continuous function is TB-regular,the above result need not be true if we replace Cauchy continuous functions byuniformly continuous functions. As an example, let X = { n : n ∈ N } . Considerthe real Hilbert space l . Let Y ⊆ l be the closed ball centered at 0 of radius2. If we denote the induced metric on Y by ρ , then ( Y, ρ ) is a finitely chainablemetric space. It is well-known that every real-valued uniformly continuousfunction on a metric space (
Y, ρ ) is bounded if and only if (
Y, ρ ) is finitelychainable [1]. Define the following function f on X . f (cid:16) n (cid:17) = e n ∀ n ∈ N where { e n : n ∈ N } denotes the standard orthonormal basis of l . Clearly, f is not TB-regular. Now, if we take any uniformly continuous function g :( Y, ρ ) → R , then g would be bounded, which further implies that for everysuch g , g ◦ f is TB-regular. Theorem 5
Let ( X, d ) be a metric space. The following are equivalent. itle Suppressed Due to Excessive Length 7 (a) ( X, d ) is complete.(b) Each continuous TB-regular function from ( X, d ) to any other Banachspace ( Y, k . k ) can be uniformly approximated by Cauchy-Lipschitz func-tions.(c) Each real-valued continuous TB-regular function can be uniformly approx-imated by Cauchy-Lipschitz functions.Proof ( a ) ⇒ ( b ): This follows from Theorem 4.5 in [6] which says that everyCauchy continuous function from ( X, d ) to any other Banach space ( Y, k . k )can be uniformly approximated by Cauchy-Lipschitz functions.( b ) ⇒ ( c ): This is immediate.( c ) ⇒ ( a ): Suppose ( X, d ) is not complete. By Theorem 3, there existsa real-valued continuous TB-regular function f on X which is not Cauchy-continuous. Thus there exists a Cauchy sequence ( x n ) in ( X, d ) such that( f ( x n )) is not Cauchy. Hence for some ǫ >
0, by passing to a subsequence, weget | f ( x k ) − f ( x k +1 ) | > ǫ ∀ k ∈ N . By hypothesis, there exists a Cauchy-Lipschitz function g : ( X, d ) → R such that sup x ∈ X | f ( x ) − g ( x ) | < ǫ . Since( x n ) is Cauchy, there exists M > | g ( x n ) − g ( x m ) | ≤ M d ( x n , x m ) ∀ n, m ∈ N . Choose n o ∈ N such that n o < ǫ M . Also, there exists k o ∈ N such that ∀ k > k o , d ( x k , x k +1 ) < n o . Now ∀ k > k o we have | f ( x k ) − f ( x k +1 ) | ≤ | f ( x k ) − g ( x k ) | + | g ( x k ) − g ( x k +1 ) | + | g ( x k +1 ) − f ( x k +1 ) | < ǫ We get a contradiction, thus (
X, d ) is complete. ⊓⊔ The following example shows that on a complete metric space, the set ofCauchy-Lipschitz functions need not exhaust continuous TB-regular functionson the space.
Example 2
Let X = { n , n ∈ N } . Define the following real-valued functionon X . f ( x ) = (cid:26) n : x = n for some n ∈ N x = 0Clearly, f is continuous and TB-regular, but since n − n +112 n − n +1 = n +1 n + n , f is notCauchy-Lipschitz.We have noticed that every Cauchy-continuous function is TB-regular butthe converse is not true. The next result adds an extra condition on TB-regular functions to make it Cauchy-continuous. For proving the result, weneed Efremovic Lemma [2, p. 92] which says that if ( x n ) and ( y n ) are twosequences in a metric space ( X, d ) such that d ( x n , y n ) > ǫ ∀ n ∈ N , then thereexists an infinite subset N of N such that d ( x k , y l ) > ǫ ∀ k, l ∈ N . Theorem 6
Let f : ( X, d ) → ( Y, ρ ) be a function between two metric spaces.Then the following are equivalent.(a) f is Cauchy-continuous. Lipsy Gupta, S. Kundu (b) f is TB-regular and the set of subsets on which f is Cauchy-continuous isclosed under finite unions.(c) f is TB-regular and the set of subsets on which f is Cauchy-continuousforms a bornology.Proof The implications ( a ) ⇒ ( b ) ⇒ ( c ) are immediate.( c ) ⇒ ( a ): Suppose f is not Cauchy-continuous. Thus there exists a Cauchysequence ( x n ) in ( X, d ) such that ( f ( x n )) is not Cauchy. It follows that f is not uniformly continuous on the set A = { x n : n ∈ N } . So for some ǫ >
0, there exists sequences ( y n ) and ( z n ) in A such that d ( y n , z n ) < n but ρ ( f ( y n ) , f ( z n )) > ǫ ∀ n ∈ N . By the Efremovic Lemma, we can assumethat { y n : n ∈ N }∩{ z n : n ∈ N } = ∅ . Since ( y n ) ⊆ A , which is totally bounded,by passing to a subsequence we can assume that ( y n ) is Cauchy. Hence thecorresponding sequence ( z n ) is also Cauchy. Since f is TB-regular, there existsan infinite subset N of N such that ( f ( y n )) n ∈ N is Cauchy in Y and similarly,there exists an infinite subset N of N such that ( f ( z n )) n ∈ N is Cauchy in Y . Thus f is Cauchy-continuous on { y n : n ∈ N } and { z n : n ∈ N } butnot on their union because if we enumerate the elements of N in the increas-ing order (by the well-ordering principle) say n , n , n , ... , then the sequence y n , z n , y n , z n , ... is Cauchy but its image is not. We get a contradiction.Hence f is Cauchy-continuous. ⊓⊔ We have the following similar characterization for the class of uniformlycontinuous functions.
Theorem 7
Let f : ( X, d ) → ( Y, ρ ) be a function between two metric spaces.Then the following are equivalent.(a) f is uniformly continuous.(b) f is TB-regular and the set of subsets on which f is uniformly continuousis closed under finite unions.(c) f is TB-regular and the set of subsets on which f is uniformly continuousforms a bornology.Proof Using Theorem 4.9 in [8], this can be proved in a manner similar to theproof of Theorem 6. ⊓⊔ In view of Theorem 6, let us note a modified characterization of completemetric spaces.
Theorem 8
Let ( X, d ) be a metric space. Then the following are equivalent.(a) ( X, d ) is complete.(b) If ( Y, ρ ) is a metric space and f : ( X, d ) → ( Y, ρ ) is continuous, then theset of subsets on which f is Cauchy-continuous forms a bornology.(c) If f : ( X, d ) → R is continuous, then the set of subsets on which f isCauchy-continuous forms a bornology. itle Suppressed Due to Excessive Length 9 Proof
The implication ( a ) ⇒ ( b ) follows from the fact that every continuousfunction on a complete metric space is Cauchy-continuous. The implication( b ) ⇒ ( c ) is immediate.( c ) ⇒ ( a ): Suppose ( X, d ) is not complete. Thus, in (
X, d ), there exists aCauchy sequence ( x n ) of distinct points with no cluster point. Let A = { x n : n is odd } and B = { x n : n is even } . Since A and B are closed and so is A ∪ B ,by Tietze’s extension theorem, let f : ( X, d ) → R be a continuous functionsuch that f ( A ) = 0 and f ( B ) = 1. Now, f is Cauchy-continuous on A and B ,but not on A ∪ B . We get a contradiction. ⊓⊔ Let us note some more characterizations of complete metric spaces in termsof strong uniform continuity. Note that this characterization has originatedfrom Proposition 4.11 in [8].
Theorem 9
Let ( X, d ) be a metric space. Then the following are equivalent.(a) ( X, d ) is complete.(b) If ( Y, ρ ) is a metric space and f : ( X, d ) → ( Y, ρ ) is continuous, then f isstrongly uniformly continuous on totally bounded subsets of ( X, d ) .(c) If ( Y, ρ ) is a metric space and f : ( X, d ) → ( Y, ρ ) is continuous, then f isstrongly uniformly continuous on each Cauchy sequence of ( X, d ) .(d) If f : ( X, d ) → R is continuous, then f is strongly uniformly continuouson each Cauchy sequence of ( X, d ) .Proof The implication ( a ) ⇒ ( b ) follows from Proposition 4.11 in [8] whichsays that every Cauchy-continuous function on ( X, d ) is strongly uniformlycontinuous on totally bounded subsets of (
X, d ). The implications ( b ) ⇒ ( c ) ⇒ ( d ) are immediate.( d ) ⇒ ( a ): Suppose ( X, d ) contains a Cauchy sequence ( x n ) of distinctpoints without any cluster point. By Tietze’s extension theorem, let f : ( X, d ) → R be a continuous function such that f ( x n ) = n ∀ n ∈ N . Now, ( x n ) n ∈ N is aCauchy sequence on which f is not strongly uniformly continuous. Because oth-erwise, for ǫ = , there would exist δ > n , if d ( x n , y ) < δ ,then ρ ( f ( x n ) , y ) < ǫ , which is not possible. ⊓⊔ Now we turn to metric spaces which possess stronger properties than thecomplete metric spaces. Cofinally complete metric spaces and UC spaces aresome popular examples of such spaces.
Definition 5
A sequence ( x n ) in a metric space ( X, d ) is called cofinallyCauchy if ∀ ǫ >
0, there exists an infinite subset N ǫ of N such that for each n, j ∈ N ǫ , we have d ( x n , x j ) < ǫ. A metric space (
X, d ) is said to be cofinallycomplete if every cofinally Cauchy sequence in X clusters.It is known that a metric space ( X, d ) is cofinally complete if and only ifit is uniformly paracompact [19], where a metric space is said to be uni-formly paracompact if for each open cover V of X, there exists an open re-finement U and δ > x ∈ X , B ( x, δ ) intersects onlyfinitely many members of U [18]. In [3], Beer has characterized the aforesaid spaces in the following way: A metric space ( X, d ) is cofinally complete if andonly if each sequence ( x n ) in X satisfying lim n →∞ ν ( x n ) = 0 clusters, where ν ( x ) = sup { ǫ > cl ( B d ( x, ǫ )) is compact } if x has a compact neighborhood,and ν ( x ) = 0 otherwise. This geometric functional is called the local compact-ness functional on X .Since cofinal completeness is a stronger form of completeness, let us studya family of subsets of cofinally complete metric spaces on which every continu-ous function must be strongly uniformly continuous. Recall that a non-emptyclosed subset A of X is said to be almost nowhere locally compact if ∀ ǫ > { a ∈ A : ν ( a ) ≥ ǫ } is compact [10,14]. In [10], it has been proved thata metric space ( X, d ) is cofinally complete if and only if whenever A is a closedsubset of X and B is almost nowhere locally compact subset such that A and B are disjoint, then d ( A, B ) >
0. Let us denote the set of almost nowherelocally compact subsets by AC ( X ). Theorem 10
Let ( X, d ) be a metric space. Then the following are equivalent.(a) ( X, d ) is cofinally complete.(b) If ( Y, ρ ) is a metric space and f : ( X, d ) → ( Y, ρ ) is continuous, then f isstrongly uniformly continuous on each member of AC ( X ) .(c) If ( Y, ρ ) is a metric space and f : ( X, d ) → ( Y, ρ ) is continuous, then D d ( A, B ) = 0 implies D ρ ( f ( A ) , f ( B )) = 0 , where A ∈ AC ( X ) and B isany non-empty subset of X .Proof ( a ) ⇒ ( b ): Since ( X, d ) is cofinally complete, A ∈ AC ( X ) implies that A is compact [10, Corollary 3.7] and thus every continuous function on X isstrongly uniformly continuous on every A ∈ AC ( X ).( b ) ⇒ ( c ): Let f : ( X, d ) → ( Y, ρ ) be a continuous function and let A and B be subsets of X such that A ∈ AC ( X ) and D d ( A, B ) = 0. To show that D ρ ( f ( A ) , f ( B )) = 0, let ǫ >
0. Since f is strongly uniformly continuous on A , ∃ δ > d ( x, y ) < δ and { x, y } ∩ A = ∅ , then ρ ( f ( x ) , f ( y )) < ǫ .Choose a ∈ A and b ∈ B such that d ( a, , b ) < δ and thus ρ ( f ( a ) , f ( b )) < ǫ .Consequently, D ρ ( f ( A ) , f ( B )) = 0.( c ) ⇒ ( a ): Suppose ( X, d ) is not cofinally complete. By Proposition 5.5 in[10], there exists A ∈ AC ( X ) and a closed subset B of X such that A ∩ B = ∅ but D d ( A, B ) = 0. Thus for each n ∈ N , there exists a n ∈ A and b n ∈ B such that d ( a n , b n ) < n . Since the sequences ( x n ) and ( y n ) cannot have anycluster point, the sets U = { a n : n ∈ N } and V = { b n : n ∈ N } are closed and U being a subset of A belongs to AC ( X ). By Tietze’s extension theorem, let f : ( X, d ) → R be a continuous function such that f ( U ) = 0 and f ( V ) = 1.Now, f is a continuous function such that U ∈ AC ( X ) and D d ( U, V ) = 0 but D ρ ( f ( U ) , f ( V )) = 0, a contradiction. ⊓⊔ Another well-known class of metric spaces which is stronger than the com-plete metric spaces is of UC spaces (also known as Atsuji spaces) [1].
Definition 6
A metric space is called a UC space if each real-valued functiondefined on it is uniformly continuous. itle Suppressed Due to Excessive Length 11
The class of UC spaces is contained in the class of cofinally complete metricspaces and it iself contains the class of compact metric spaces. Many attractivecharacterizations of UC spaces can be found in the survey article [16]. One canobserve that R with the usual metric is cofinally complete but not UC. Theorem 11
Let ( X, d ) be a metric space. Then the following are equivalent.(a) ( X, d ) is UC.(b) If ( Y, ρ ) is a metric space and f : ( X, d ) → ( Y, ρ ) is continuous, then f isstrongly uniformly continuous on each subset of X .(c) f ( Y, ρ ) is a metric space and f : ( X, d ) → ( Y, ρ ) is continuous, then f isstrongly uniformly continuous on each closed subset of X .Proof The implication ( a ) ⇒ ( b ) follows from Theorem 5.2 in [8] and theimplication ( b ) ⇒ ( c ) is immediate.( c ) ⇒ ( a ): It is known that a metric spaces ( X, d ) is UC if and only if forany closed subsets A and B of X such that A ∩ B = ∅ , we have D d ( A, B ) > c ) ⇒ ( a ) in Theorem 10. ⊓⊔ Remark 2
Using the fact that each TB-regular function sends Cauchy se-quences to sequences having a Cauchy subsequence, it can be easily seen thatif f and g are two real-valued TB-regular functions on a metric space ( X, d ),then so are the functions f + g and f.g , where f.g denotes the pointwise mul-tiplication of the functions f and g .Using the above remark, we can prove the following result in a manner similarto the proof of Theorem 3.12 in [13]. Theorem 12
Let ( X, d ) be a metric space. Then the following statements areequivalent:(a) ( X, d ) is complete.(b) Whenever f : ( X, d ) → R is a locally Lipschitz TB-regular function suchthat f is never zero, then f is also locally Lipschitz and TB-regular.(c) Whenever f : ( X, d ) → R is a continuous TB-regular function such that f is never zero, then f is also continuous and TB-regular. Here we would like to mention that the stability of never-zero Lipschitz-type functions under reciprocation has been recently studied in [4]. Now westudy the condition under which a non-vanishing strongly uniformly continu-ous function on a set is stable under reciprocation, that is, if f is a never-zeroreal-valued continuous function on ( X, d ) such that f is strongly uniformly con-tinuous on a subset A of X , then under what conditions f would be stronglyuniformly continuous on A ? Recall that a metric space ( X, d ) is UC if andonly if each sequence ( x n ) in X satisfying lim n →∞ I ( x n ) = 0 clusters, where I ( x ) = d ( x, X \ { x } ) measures the isolation of x in the space. Definition 7
A subset A of a metric space ( X, d ) is called a UC set if everysequence ( a n ) in A with lim n →∞ I ( a n ) = 0 has a cluster point in X . In [8], it has been proved that a subset A of X is UC if and only if each real-valued continuous function defined on X is strongly uniformly continuous on A . Theorem 13
Let ( X, d ) be a metric space and let ∅ 6 = A ⊆ X . Then thefollowing statements are equivalent:(a) A is a UC set.(b) Whenever f : ( X, d ) → R is a continuous and never zero function such that f is strongly uniformly continuous on A , then f is also strongly uniformlycontinuous on A.Proof ( a ) ⇒ ( b ): This follows from the fact that every continuous function ona UC set A is strongly uniformly continuous on A [8, Theorem 5.2].( b ) ⇒ ( a ): Suppose A is not a UC set. Thus there exists a sequence ( a n )of distinct points in A such that lim n →∞ I ( a n ) = 0, but ( a n ) has no cluster pointin X . Without loss of generality, let us assume that I ( a n ) < n ∀ n ∈ N . Case 1:
The sequence ( a n ) is pseudo-Cauchy.Let B = { a n : n ∈ N } . Let f be a function on B such that f ( a n ) = n ∀ n ∈ N .Since f is a bounded uniformly continous function on B , it can be extendedto a uniformly continuous function ˆ f on the whole space X [17]. Define thefollowing function. g : X −→ R x ˆ f ( x ) + d ( x, B )It is clear that g is uniformly continuous and is never zero. Thus g is stronglyuniformly continuous on A , but g is not uniformly continuous on A as ( a n ) ispseudo-Cauchy and g ( a n ) = n ∀ n ∈ N . Case 2: