Uniform Convergence and Uniform Continuity in Generalized Metric Spaces


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1 Int. Journal of Math. Analysis, Vol. 5, 2011, no. 6, Uniform Convergence and Uniform Continuity in Generalized Metric Spaces Abdul Mohamad Department of Mathematics and Statistics Sultan Qaboos University, Oman Abstract The class of generalized metric spaces is introduced and studied in [9]. In this paper, the concepts of uniform convergence, uniform continuity and BolzanoWeierstrass property in this class are investigated. We show that many theorems in classic metric spaces about these concepts are still valid for this class of spaces. Mathematics Subject Classification: 54A05, 54A20, 54E70, 54H25 Keywords: Generalized metric space; Probabilistic metric space; Uniform Convergence; Uniform Continuity 1 Introduction In the paper [9] we introduced the class of generalized metric spaces. These spaces simultaneously generalize standard metric spaces, probabilistic metric spaces and fuzzy metric spaces. We show that every generalized metric space is, naturally, a uniform space. Thus we can use standard topological techniques to study, for example, probabilistic metric spaces. We illustrate this by proving a Fixed Point Theorem for uniform spaces, and interpret this result in the context of probabilistic metric spaces. A uniform structure (or uniformity) on a set X is a non empty set U of subsets of X X which satisfies the following axioms: 1. Every subset of X X which contains a set of U belongs to U. 2. Every finite intersection of sets of U belongs to U.
2 286 A. Mohamad 3. Every set of U is a reflexive relation on X (i.e. contains the diagonal). 4. If V belongs to U, then V = {(y, x) :(x, y) V } belongs to U. 5. If V belongs to U, then exists V in U such that, whenever (x, y), (y, z) V, then (x, z) V (i.e. V V V ). The sets of U are called entourages or vicinities. The set X together with the uniform structure U is called a uniform space. Recall that a uniformity U on a set X has the Lebesgue property provided that for each open cover G of X there is U Usuch that {U(x) :x X} refines G, and U is called equinormal if for each pair of disjoint nonempty closed subsets A and B of X there is U Usuch that U(A) B =. A metric d on X has the Lebesgue property provided that the uniformity U d, induced by d, has the Lebesgue property and d is equinormal provided that U d have it. 2 Generalized Metric Spaces A range set (G, ) is a partially ordered set with minimum element 0 so that G \{0} is downwards directed. A triangle function is a binary operation τ : G G G which is commutative, associative, has 0 as it s identity, is increasing in each component, and is continuous at 0 for all a>0 there is a b>0 such that τ(p, q) <afor all p, q < b. (Note that if we topologise G so that 0 has neighbourhoods of the form N a = {b : b<a}, where a>0, then the condition τ continuous at 0 is indeed equivalent to τ being continuous at 0.) Definition 1 Let S be a nonempty let (G, ) be a range set, and let τ be a triangle function over G. A function F : S S G is a generalized quasimetric [7] if the following conditions are satisfied for all p, q, r S. GQM1: F(p, p) =0; GQM2: if F(p, q) =0and F(q, p) =0, then p = q; GQM3: F(p, q) τ(f(p, r), F(r, q)) If F is a generalized quasimetric then the triple (S, F,τ) is called a generalized quasimetric space (abbreviated GQM space). A generalized quasimetric F is called a generalized metric if it satisfies the symmetrycondition. GQM4: F(p, q) =F(q, p) for all p, q in S. In the latter case, (S, F,τ) is a generalized metric spaces (abbreviated GM space).
3 Uniform convergence and uniform continuity 287 We often write F pq in place of F(p, q). Standard Metric Spaces If we take G to be the nonnegative real numbers, R +, with the standard order, and τ to be addition, then the generalized (quasi )metric spaces over G and τ are precisely the standard (quasi)metric spaces. Probabilistic Metric Spaces In [6], Menger introduced and studied the class of probabilistic metric spaces. A probabilistic metric space is a generalized metric space appropriate to the study of situations in which distances are measured in terms of distribution functions rather than nonnegative real numbers. Let + be the class of all distance distribution functions. A function F :[0, ) [0, 1] is called a distance distribution function if F satisfies the following conditions: 1. F (0)=0 and lim F (x) =1 x 2. F is nondecreasing 3. F is left continuous on [0, ). The class + is equipped with the modified Levy distance. The modified Levy distance [11] is the function d L defined on + + by d L (F, G) = inf{h >0:G(x) F (x + h) +h and F (x) G(x + h) +h, x (0, 1 )}. h The distance d L is a metric on +. The class + is ordered by the relation F G F (x) G(x), x 0. Let ɛ 0 be the element of + defined by { 0, if x =0 ɛ 0 = 1, if x>0 A probabilistic (quasi)metric space is precisely a generalized (quasi)metric space where the range set (G, )is( +, ) with zero element ɛ 0, and a triangle function. We abbreviate probabilistic (quasi)metric space to P(Q)M. In [5], P.S. Marcus gives an example of PQM space based on stationary Markov chains which is not a PM space. An excellent treatment of PM spaces is given in [11]. Fuzzy Metric Spaces Let S be a nonempty set. Then fuzzy metric spaces (see [2]) are precisely those generalized metric spaces where the range set G is the set of all fuzzy sets on S, and the triangle function comes from a continuous tnorm. A binary operation :[0, 1] [0, 1] [0, 1] is a continuous tnorm if satisfies the conditions:
4 288 A. Mohamad 1. is associative and commutative. 2. a 1=a for every a [0, 1] 3. a b c d whenever a c and b d, a, b, c, d [0, 1]. A tnorm gives rise to a triangle function as follows: F (F, G)(t) =F (t) G(t). Example 2 The four basic examples of tnorms are: 1. The Lukasierviez tnorm: L: I I I,a L b = max{a + b 1, 0}. 2. The product tnorm: a P b = ab. 3. The minimum tnorm: a M b = min{a, b}. 4. The weakest tnorm, the drastic product: { min{a, b}, if max(a, b) =1 D (a, b) = 0, otherwise Using pointwise ordering, we have the inequalities D < L < P < M. 3 Uniform Topologies of GM Spaces Theorem 3 Let (S, F,τ) be a GM space. The filter U = {U S S : U U(a) for some 0 <a G}, where U(a) ={(p, q) S S : F pq <a}, is a uniformity. Proof. We check off the defining properties of a uniformity: 1. Each U in U contains the diagonal because F pp = 0 for all p. 2. The family U is upwards closed by definition (the U(a) s form a basis). 3. It is clear that if U belongs to U, then U 1 = {(q, p) :(p, q) U} belongs to U because F(p, q) =F(q, p) for all p, q in S. 4. The family U is closed under finite intersections because G\{0} is downwards directed.
5 Uniform convergence and uniform continuity It remains to verify that if U is in U, then there is a V in U such that V V U. We may assume U = U(a) for some a>0. By continuity of τ at 0, there is a b>0such that if p,q <bthen τ(p,q ) <a. Let V = U(b). Take any (p, q) inv V. So there is an r such that (p, r), (r, q) U(b). Which means F pr <band F rq <b. Now we see that F pq τ(f pr, F rq ) <a in other words, (p, q) U(a), as required. Fix a GM space (S, F,τ). From the uniformity, we can obtain the topology σ associated with F which has for its neighborhood base at p S the collection {U p (a) :U U,t>0} where U p (a) ={q S :(p, q) U(a)}. Lemma 4 Let (S, F,τ) be a GM, with associated uniformity U and topology σ. 1. (S, σ) is T 0 2. (S, σ) is T 1 iff F pq =0 p = q (GQM2 ) 3. (S, σ) is metrizable iff GQM2 and (G, ) has countable cofinality (there is a countable C G \{0} such that for all a>0 there is a c C such that c<a). Proof. For 1 just note that (S, σ) being T 0 is exactly the content of axiom GQM2. Part 2 is similar. For 3 we recall that a space is metrizable iff it is T 1 and has a compatible uniformity with a countable base. A countable base for the uniformity U associated with (S, F,τ) corresponds directly, by the definition of the associated uniformity, to a countable cofinal set for (G, ). We now establish a dictionary connecting properties of generalized metric spaces and properties of the associated uniformities. Definition 5 Let (X, U) be a uniform space. 1. A sequence (x n ) n N in X is called UCauchy sequence if for each U U there exists k N such that (x r,x s ) U for all s r k. 2. A sequence (x n ) n N in X is called convergent to x X if for all U U there exists k N such that (x, x n ) U for all n k. 3. (X, U) is called Ucomplete if every UCauchy sequence is Uconvergent.
6 290 A. Mohamad Definition 6 Let (S, F,τ) be GM space. A sequence {x n } n N is called a F Cauchy sequence if for each a>0 there exists k N such that F xrxs <afor all s r k. {x n } n N is called σconvergent to x S if for every 0 <a G there exists k N such that F xxn <afor all n k. Definition 7 We say that a GM space (S, F,τ) is F complete if every F Cauchy sequence is σconvergent. It is easy to prove the following lemma. Lemma 8 Let U F be the uniformity determined by (S, F,τ). Then (S, F,τ) is Fcomplete iff (X, U) is Ucomplete. 4 Uniform Continuity Definition 9 Let (S 1, F 1,τ 1 ) and (S 2, F 2,τ 2 ) be generalized metric spaces with range sets (G 1,< 1 ) and (G 2,< 2 ) respectively. A function f : S 1 S 2 is called uniformly continuous if for every g 2 G 2, there exists g 1 G 1 such that if F 1 (p, q) <g 1 then F 2 (f(p),f(q)) <g 2. It easy to show that every uniformly continuous function is continuous. However, the converse may not be true. We shall see next that these two concepts, as in the standard metric, agree on certain kind of generalized metric spaces called compact. Theorem 10 Let (S, F,τ) and (S, F,τ ) be generalized metric spaces with range sets (G, <) and (G,< ) respectively. If the function f : S S is continuous and S is compact then f is uniform continuous. Proof. Let g G, be given, then there exists h G such that τ (h,h ) < g. Since f is continuous, for each p S, there extis h G, such that if F(p, q) <hthen F (f(p),f(q)) <h q S. But h G and therefore there extis g G such that τ(g, g) <h. From compactness of S there exist s 1,s 2,..., s n S such that S = n i=1 B(s i,g si ). Let g 0 = min{g si : i = 1, 2,..., n}. Now for any p, q S, iff(p, q) <g 0 then F(p, q) <g si. Since p S, there exists an s i such that p B(s i,g 0 ). This implies that F(p, s i ) < g si. Therefore F (f(p),f(s i )) <h. Now F(p, s i ) τ(f(p, q), F(q, s i )) τ(g si,g si ) <h si. Therefore F (f(p),f(s i )) <h. But
7 Uniform convergence and uniform continuity 291 F (f(p),f(q)) τ (F (f(p),f(s i )), F (f(s i ),f(q))) τ (h,h ) < g. completes the proof, i.e. f is uniformly continuous. This The Proof of the following theorem is similar as in classic (standard) metric spaces. Theorem 11 Let (S, F,τ) and (S, F,τ ) be generalized metric spaces. If f : S S is uniformly continuous and {x n } is Cauchy sequence in S, then {f(x n )} is also a Cauchy sequence in S. Definition 12 Suppose that X is a nonempty set and (Y,F,τ ) is a generalized metric space. A sequence {f n } of functions from X to Y is converge uniformly to a function f : X Y if for each r G, there exists n 0 N such that F (f n (x),f(x)) <rfor all n n 0 and for all x X. Definition 13 Let (X, F,τ) and (Y,F,τ ) be generalized metric spaces. A family of functions F = {f} from X to Y is called equicontinuous if for each r G, there exists r G such that if F(p, q) <rthen F (f(p),f(q)) <r, for all f F and for all p, q X. Definition 14 A real valued function f on the generalized metric space (X, F,τ) is Runiformly continuous provided that for each ɛ>0 there exists r G such that if F(x, y) <rthen f(x) f(y) <ɛ. Definition 15 A generalized metric space (X, F,τ) is called equinormal if for each pair of disjoint nonempty closed subsets A and B of (X, T F ) such that sup{f(a, b) :a A, b B} > 0. Definition 16 A generalized metric space (X, F,τ) has the Lebesgue property if for each open cover G of (X, T F ) there exist r G such that {B F (x, r) :x X} refines G. The proofs of the following lemmas and theorems are similar to those lemmas and theorems in [1] and [3]. Lemma 17 Let (X, F,τ) and (Y,F,τ ) be generalized metric spaces and let F = {f n } be an equicontinuous sequence of functions from X to the complete space Y.IfF = {f n } converges for each point of a dense subset D of X, then F = {f n } converges at each point of X and the limit function is continuous.
8 292 A. Mohamad Theorem 18 AscoliArzela Theorem Let (X, F,τ) and (Y,F,τ ) be generalized metric spaces and suppose that X is compact and Y is complete. Let A = {f} be an equicontinuous family of functions from X to Y and let F = {f n } be a sequence in A such that {f n (x)} is a compact subset of Y for each x X. Then there exists a continuous function f : X Y and a subsequence {g n } of F such that g n converges uniformly to f on X. Theorem 19 Let X be any nonempty set and (Y,d) be a metric space and let (Y,F d,τ ) be the induced generalized metric space. Then a sequence of functions {f n } from X to Y converges uniformly to a function f from X to Y with respect to the metric d iff {f n } converges uniformly to f with respect to the generalized metric F d. Theorem 20 Suppose that (X, d) and (Y,d ) are metric spaces. Let (X, F d,τ) and (Y,F d,τ ) be the corresponding induced generalized metric spaces. Then a family F of functions from X to Y is equicontinuous with respect to the metric iff F is equicontinuous with respect to the generalized metric. F d. Corollary 21 Let (X, d) and (Y,d ) be metric spaces and suppose that X is compact and Y is complete. Let A = {f} be an equicontinuous family of functions from X to Y and let F = {f n } be a sequence in A such that {f n (x)} is a compact subset of Y for each x X. Then there exists a continuous function f : X Y and a subsequence {g n } of F such that g n converges uniformly to f on X. Theorem 22 Suppose that (X, d) is a metric space and let (X, F d,τ) induced generalized metric spaces. Then X is equinormal (has Lebesgue property) with respect to the metric iff X is equinormal (has Lebesgue property) with respect to the generalized metric. Theorem 23 Let (X, F,τ) be a generalized metric space. Then following are equivalent. 1. For each generalized metric space (Y,F,τ ), any continuous mapping from (X, T F ) to (Y,T F ) is uniformly continuous as a mapping from (X, F,τ) to (Y,F,τ ). 2. Every real valued continuous function on (X, T F ) is Runiformly continuous on (X, F,τ).
9 Uniform convergence and uniform continuity Every real valued continuous function on (X, T F ) is uniformly continuous on (X, U F ). 4. (X, F,τ) is an equinormal. 5. U F is an equinormal uniformity on X. 6. The uniformity U F has the Lebesgue property. 7. (X, F,τ) has the Lebesgue property. Consider a generalized metric space (X, F,τ) with the range set (G, <). For any x X and any subset D of X we denote by F(x, D) the distance from x to D, that is, d(x, D) = inf{f(x, d) :d D}. ByF(x) we denote the distance from x to X x. Recall that a point x X is called an accumulation point of a subset S of X if F(x, S x) = 0. The accumulation points of X is denoted by A, is the set {x X : F(x) =0}. The set of isolated points is the set {x X : F(x) > 0}. Theorem 24 Let (X, F,τ) be a generalized metric spaces. statement are equivalent: The following 1. Every continuous function f : X R is uniformly continuous. 2. Every sequence {x n } in X with lim n F(x n ) = 0 has a convergent subsequence. 3. The set A is compact, and for every r 1 G there exists r 2 G such that for all x X with F(x, A) >r 1 we have F(x) >r 2. Proof. The proof is similar to that one of Theorem 1 [4]. Theorem 25 Let (X, F,τ) be a generalized metric spaces with (G, <) as a range set. Then X is compact if and only if every continuous function f : X R is uniformly continuous, and for every 0 <r G the set {x X : F(x) >r} is finite. Proof. Suppose that X is a compact space. Since on a compact space every continuous function is uniformly continuous by Theorem 10, the first condition is necessary to provide compactness of X. If for some 0 <r G the set {x X : F(x) >r} were infinite, then the family {B(x, r) ={y X : F(x, y) <r}} x X would be an open cover of X that has no finite subcover. This shows that also the second condition is necessary for compactness of X.
10 294 A. Mohamad Now assume both conditions are fulfilled, and let {x n } be any sequence in X. We have to show that {x n } has a convergent subsequence. This is trivial if some point of X occurs infinitely often in the sequence. So we may assume that no point occurs infinitely often in our sequence. Since for every 0 <r G the set {x X : F(x) >r} is finite, this implies lim n F(x n ) = 0. Thus, by Theorem 24, {x n } has a convergent subsequence. Hence, by Theorem 31, X is a compact space 5 BolzanoWeierstrass Property Definition 26 Let (X, F,τ) be a generalized metric spaces. We say that a set A X has BolzanoWeierstrass property if for every sequence {x n } of elements of A has a subsequence that converges to some x A. Definition 27 Let (X, F,τ) be a generalized metric spaces. Then X is called totally bounded if given any 0 <r G there exists a finite collection of open balls B(x 1,r),B(x 2,r),..., B(x n,r) of radius r which covers X. Lemma 28 Let (X, F,τ) be a generalized metric spaces and let K X has BolzanoWeierstrass property. If G = {G λ } is an open cover of K, then there is an 0 <r G such that for every B(k, r),k K, there exits λ Λ such that B(k, r) G λ. Proof. Suppose for every 0 <r G there is an k r K such that B(k r,r)is not subset of any G λ,λ Λ. Thus, for every n N, there is x n K such that B(x n, 1 ) is not contained in any G n λ. Now, by BolzanoWeierstrass property, there exists a subsequence {x λn } that converges to an x 0 K. Since G = {G λ } is a cover of K, there is λ 0 Λ such that x 0 G λ0. But G λ0 is open, so there exists 0 <s G such that B(x 0,s) G λ0 G λ0. By continuity of τ, there is r N such that if x B(x λr, 1 λ r ) then x B(x 0,s). Hence B(x λr, 1 λ r ) G λ, which is lead to a contradiction. Lemma 29 Let (X, F,τ) be a generalized metric spaces. X is totally bounded iff every sequence has a Cauchy subsequence. Proof. The proof is similar to that one in the classic metric spaces. Lemma 30 Let (X, F,τ) be a generalized metric spaces. Then every Cauchy sequence that clusters to a point is converges to that point.
11 Uniform convergence and uniform continuity 295 Proof. The proof is similar to that one in the classic metric spaces. Theorem 31 Let (X, F,τ) be a generalized metric spaces. statement are equivalent: The following 1. X is compact 2. X is complete and totally bounded. 3. X has the BolzanoWeierstrass property. Proof. (1) (2). Let X be a compact space. Every Cauchy sequence in X has a cluster point. So by Lemma 30, the sequence converges to that point. Hence X is complete. Also, for every r G, the open cover {B(x, r) :x X} of X, has a finite subcover. Therefore X is totally bounded. (2) (3). Since X is totally bounded, by Lemma 29, every sequence has a Cauchy subsequence. This subsequence is convergent as X is complete. Therefore X has the BolzanoWeierstrass property. (3) (1). We claim that if X has the BolzanoWeierstrass property then for every r G there exists a finite subset {x 1,x 2,..., x n } of X such that {B(x 1,r),B(x 2,r),..., B(x n,r)} covers X. Suppose not, let x 1 X, then X is not subset of B(x 1,r). Let x 2 X B(x 1,r), then {B(x 1,r),B(x 2,r)} is not covering of X. This argument leads to constrict a sequence {x n } of X such that x n+1 X n i=1 B(x i,r) for all n N. Therefore {x n } has no convergent subsequence which construct with BolzanoWeierstrass property. Hence X is compact. Corollary 32 BolzanoWeierstrass Theorem Every bounded sequence in R n has a convergent subsequence. References [1] A. George and P. Veeramani, Some Theorems in Fuzzy Metric Spaces, J. Fuzzy Math. 3(1995), [2] V. Gregori and S. Romaguera, fuzzy quasimetric spaces, Applied Gen Top 5(2004),
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