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Proyecto Fin de Máster en Investigación Matemática Facultad de Ciencias Matemáticas Universidad Complutense de Madrid Bornologies in Metric Spaces Ana Soledad Meroño Moreno Directora Prof. Mª Isabel Garrido Carballo 2009-2010 Abstract. In the same way that the Samuel Compactication, it can be constructed the, also called, Samuel Realcompactication, but taking all the real-valued uniformly continuous functions, not necessarily bounded. To construct the Samuel Realcompactication of a metric space (X, d) we study the case in which all the real-valued uniformly continuous functions f ∈ U C(X) are bounded in the closed metric balls of (X, d). In particular, this functions are bounded in all the bounded subsets of (X, d). Taking these metric spaces, we have that the Samuel Realcompactication is the quotient of a subspace of the Stone-ech compactication of X , which can be written as the (countable) union of the closure in βX of the closed metric balls in (X, d) of radius n ∈ N and center in a xed point x0 ∈ X . This problem of characterization of the Samuel Realcompactication, for these type of metric spaces, is equivalent to solve the following problem. Let B a subset of (X, d), then it is said bounded in the sense of Bourbaki if and only if every real-valued uniformly continuous function f ∈ U C(X) is bounded for it. These types of subsets form what is called a bornology for the metric space (X, d), that is, they form a cover of X stable by inclusions and nite unions. We have that given two uniformly continuous metrics, d and d0 , the bounded sets in sense of Bourbaki for (X, d) and (X, d0 ) are the same. If we solve the problem of nding a uniformly equivalent metric d0 for (X, d) such that all the bounded sets in the sense on Bourbaki of (X, d) concide with the bounded metric sets of (X, d0 ), then we have that all the real-valued uniformly continuous functions f ∈ U C(X) are bounded for the bounded metric sets of (X, d0 ). In this way (X, d0 ) has the Samuel Realcompactication developed before and the Samuel Realcompactication of (X, d) and (X, d0 ) is the same because both metric spaces have the same uniformity. 1 Resumen. Similarmente a la Compacticación de Samuel puede construirse la Realcompacticación, también llamada de Samuel, pero tomando todas las funciones reales uniformemente continuas no necesariamente acotadas. Para construir la Realcompacticación de Samuel de un espacio métrico (X, d) tomamos el caso particular en el que todas las funciones reales uniformemente continuas f ∈ U C(X) son acotadas para las bolas métricas cerradas en (X, d). En particular estas funciones son acotadas para todos los subconjuntos acotados de (X, d). Tomando estos espacios métricos tenemos que la Realcompacticación de Samuel es el cociente de un subespacio de βX , la compacticación de Stone-ech de X , que puede escribirse como la unión (numerable) de las clausuras en βX de las bolas cerradas de radio n ∈ N y centro un punto jo x0 ∈ X . Este problema de caracterización de la Realcompacticación de Samuel para éste tipo de espacios métricos, es equivalente a resolver el siguiente problema. Sea B un subconjunto de (X, d), entonces éste se dice que es acotado en el sentido de Bourbaki si y sólo si toda función real uniformemente continua f ∈ U C(X) es acotada para B . Éste tipo de subconjuntos forman lo que se llama una bornología para el espacio (X, d) i. e. forman un recubrimiento de X estable para las inclusiones y las uniones nitas. Además, dadas dos métricas uniformemente equivalentes d y d0 los acotados en el sentido de Bourbaki para (X, d) y para (X, d0 ) coinciden. Si resolvemos el problema de encontrar una métrica uniformemente equivalente d0 para (X, d) tal que los subconjuntos acotados en el sentido de Bourbaki de (X, d) coinciden con los acotados métricos de (X, d0 ), entonces tenemos que todas las las funciones reales uniformemente continuas f ∈ U C(X) son acotadas para los acotados métricos de (X, d0 ). De esta manera (X, d0 ) tiene la Realcompacticación de Samuel descrita anteriormente y la Realcompacticación de Samuel de (X, d0 ) y (X, d) es la misma ya que los dos espacios métricos tienen la misma uniformidad. 2 Key words: Metric spaces, Bornologies, Bounded sets, Uniformly continuous, Realcompactication, Bourbaki bounded, Totally bounded, Samuel compactication, Heine-Borel property. Palabras clave: Espacios métricos, Bornologías, Conjuntos acotados, Uniformemente continuo, Realcompacticación, Bourbaki acotado, Compacticación de Samuel, Propiedad de Heine-Borel. MSC2000: - 54C35 - Metric spaces - Metrizability; - 54C30 - Real-valued functions; - 54D60 - Realcompactness and realcompactication. 3 La abajo rmante, Ana Soledad Meroño Moreno, matriculada en el Máster en Investigación Matemática de la Facultad de Ciencias Matemáticas, autoriza a la Universidad Complutense de Madrid (UCM) a difundir y utilizar con nes académicos no comerciales y mencionando expresamente a su autor el presente Trabajo Fin de Master:"Bornologies in Metric Spaces", realizado durante el curso académico 2009-2010 bajo la dirección de M Isabel Garrido Carballo en el Departamento de Geometría y Topología, y a la Biblioteca de la UCM a depositarlo en el Archivo Institucional E-Prints Complutense con el objeto de incrementar la difusión, uso e impacto del trabajo en Internet y garantizar su preservación y acceso a largo plazo. a 4 Contents 1 Introduction. 6 2 Some basic notions about bornologies. 8 2.1 Some examples of bornologies. . . . . . . . . . . . . . . 8 2.2 Metrization of bornological universes. . . . . . . . . . . 10 2.3 The Heine-Borel property. . . . . . . . . . . . . . . . . 15 3 Bornologies in uniform spaces. 3.1 Totally bounded sets and bounded sets in the sense of Bourbaki. . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Characterization of the bounded sets in the sense of Bourbaki. . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Uniform metrization of bornologies. . . . . . . . . . . . 3.4 The Heine-Borel property for uniformly equivalent metrics. 18 18 21 24 27 4 Realcompactication v.s compactication. 30 5 Samuel Realcompactication of a metric space. 45 4.1 Compactication of a topological space. . . . . . . . . . 32 4.2 Realcompactication of a topological space. . . . . . . . 35 4.3 Realcompactications as sets of lattice homomorphisms. 39 5.1 Samuel Realcompactication and Compactication. . . . 45 5.2 Conclusion. . . . . . . . . . . . . . . . . . . . . . . . . 48 5 Chapter 1 Introduction. The notion of bounded set has been very important in the theory of metric spaces. So that, new notions of bounded sets have been dened in other types of spaces as vector topological spaces, uniform spaces and more generally, topological spaces. In this way we can see that there exists new notions of bounded sets, very dierent between them which can be viewed as an extension of the classic notion of bounded set. All of them have the common property of forming a bornology on the considered space X . We say that a family of subsets of a set X is a bornology of X if it is a cover of X and if it is stable under inclusions and nite unions. If we take a topological space X and a bornology B on X , then the pair (X, B) is called a bornological universe. A rst interesting question raised in this report, is when a bornological universe (X, d) is metrizable or uniformly metrizable, that is , when there exists an admissible metric d such that d denes the same topology or uniformity of X and the metric bounded sets for (X, d) coincide with the bounded sets of the bornology B. Second, as an application of these results, the theorem have been applied to the compact bornology for a T2 topological space X , that is, the bornology of all the subsets in X such that their closure is compact. As an answer of this application we have a way to nd admissible proper metrics for the topology or the uniformity of X , that is, metrics with the Heine-Borel property. 6 Finally we can applied the theorem of uniform metrization to a metric space (X, d) with the bornology of the bounded sets in the sense of Bourbaki, that is, the subsets of X which are bounded for every real-valued uniformly continuous function f ∈ U C(X). Thanks to this result we have a useful characterization of the Samuel Realcompactication of the metric space (X, d). Recall that the Samuel Realcompactication of a metric space is dened as the smallest realcompactication of (X, d) such that every real-valued uniformly continuous function f ∈ U C(X) can be extended to a real-valued continuous function on the realcompactication. After all these months of work, I want to thank Prof. M Isabel Garrido Carballo who has helped me to the understanding of this interesting theme and to the hard redaction of this report. a 7 Chapter 2 Some basic notions about bornologies. 2.1 Some examples of bornologies. In a metric space (X, d) we have the classical notion of metric bounded set, that is, a subset B is bounded if there is a positive number K > 0 such that for all x, y ∈ B , d(x, y) ≤ K . The collection of all the metric bounded sets in (X, d) has the following properties: they form a cover of X , every subset of a bounded set is bounded, and the nite union of bounded sets is also bounded. In general, for a non empty set X , if we have a collection of subsets of X with the same properties than the metric bounded sets of a metric space, we have what is called a bornology on X . Denition 1. Let X be a non empty set, a bornology on it is a non empty collection of subsets {Bi} of X such that: 1. S Bi = X ; 2. every subset of a bounded set is bounded; 3. nite unions of bounded sets are bounded. Remark 1. It follows immediately that: 8 • ∅ is a bounded set; • every x ∈ X is a bounded set; • nite or not intersections of bounded sets are bounded sets. With this denition we have an axiomatic extension of the concept of bounded subset on a generic set X . Obviously the metric bounded sets of a metric space (X, d) form a bornology on X called trivially the metric bornology for d. We are going to denote it by Bd(X). Remark 2. Let B and B0 be two bornologies on a set X . Then, B and B0 coincide, that is, B = B0 i for every B ∈ B there exists B 0 ∈ B0 such that B ⊆ B 0 and for every B 0 ∈ B0 there exists B ∈ B such that B0 ⊆ B. Example 1. For a non empty set X the collection of all the subsets in X , P(X), is a bornology on X called the trivial bornology. Obviously if we have a bornology B on a set X and X ∈ B, that is, X is bounded, then B is the trivial bornology. Example 2. For a metric space (X, d) the metric bornology for d coincides with the trivial bornology i d is a bounded metric. If d is not a bounded metric then the trivial bornology conicides with the metric bornology for d∗, where d∗ = min{1, d} is a bounded metric uniform equivalent to d. Example 3. Let F be the collection of all the nite subsets on X . Then it is a bornology on X called the nite bornology. Example 4. We can have dierent bornologies generated in the next way. Let A be a family of subsets in a set X such that it forms a cover of X and let B be the family of subsets of X which consists of the totality of the subsets of the nite unions of the family A. Then B is the weakest bornology generated by the cover A. Example 5. As in metric spaces it should be natural in topological spaces that the compact sets be bounded. Let A be the collection of all 9 the compact sets of a topological space X , then it generates a bornology as in the previous example. This is the compact bornology denoted by CB. Recall that B ∈ CB i B is compact. 2.2 Metrization of bornological universes. Example 6. Let CBd(X) denote the compact bornology on a metric space (X, d). It coincides with the metric bornology of d i (X, d) has the Heine-Borel Property that is every closed bounded set in (X, d) is compact. In this case d is said to be a proper metric. If X is a compact space then trivially the compact bornology coincides with the trivial bornology. As in the last example it is an interesting problem to know when, in a metric space (X, d) with a given bornology B, the bornology B coincide with the metric bornology for d, Bd (X). A more general problem has been solved by Hu in [7] and [8], for a metrizable topological space. So we are going to present this result in the same way as Hu did it. Consider the next denition: Denition 2. A bornological universe is a pair (X, B) where X is a topological space and B is a given bornology on X . We want to know when there exists a metric d on (X, B) such that X is metrizable by d and the bornology B concide with the metric bornology Bd (X) on (X, d). In this case we say that the bornological universe is metrizable. Denition 3. A base for a bornology B is a subfamily A of B such that every element in B is contained in some element of A. Denition 4. A characteristic function of a bornological universe (X, B) is a real-valued non negative continuous function 10 χ : X → [0, ∞) such that a subset E ⊂ X is bounded i χ(E) is bounded in [0, ∞), that is, B = {E ⊂ X : ∃K > 0, χ(E) ≤ K}. Denition 5. A bornological universe (X, B) is normal i its topological space is normal, that is, every pair of disjoint closed sets in X has disjoint neighborhoods. Lemma 1. A normal bornological universe (X, B) admits a characteristic function χ i B has a countable base and for each B1 ∈ B there ◦ is B2 ∈ B such that B1 ⊆ B 2. Proof. ⇒) Assume that (X, B) is a bornological universe which admits a characteristic function χ. Now, for each positive integer n, let Bn denote the subset of X which consists of all points x ∈ X satisfying χ(x) ≤ n. Then A = {Bn : n ∈ N} is a countable sequence of bounded sets. Let B denote an arbitrary bounded set of X and choose an integer n satisfying n ≥ χ(B), then we have B ⊂ Bn . Hence A is a countable base of the bornology B. Next let B denote an arbitrary bounded set. Then there exists a real number K > 0 with K ≥ χ(B). By continuity of χ, we have that χ(B) ⊂ χ(B) ≤ K < K + 1. Then again by continuity, BK+1 = {x ∈ X : χ(x) < K + 1} is a bounded open set and B ⊂ BK+1 . ⇐) Conversely, assume that X is a normal universe satisfying the two conditions of the theorem. If X is bounded, then the constant zero function on X is a characteristic function of X . So, suppose that X is not bounded. So this implies that X cannot has a maximal bounded set, that is, it doesn't exist a bounded set B ∗ ∈ B such that for every B ∈ B, B ⊆ B ∗ because every point in X is bounded and X is not bounded. First we are going to nd a base G for the bornology which consists of a strictly increasing sequence of open bounded sets G1 , G2 , ..., Gn , ... satisfying Gn ⊂ Gn+1 for every n = 1, 2, ... . Indeed let D = {D1 , D2 , ..., Dn , ...} be a countable base for B. By hypothesis there 11 ◦ exists an open bounded set G1 = B, B ∈ B such that D1 ⊆ D1 ⊆ G1 . Assume that bounded open sets G1 , G2 , ..., Gn have been choosen in such a way that Gi ⊃ Gi−1 ∪ Di and Gi 6= Gi−1 for each i = 2, ..., n. By hypothesis Gn is a bounded set because there exists a B ∈ B such ◦ that Gn ⊆ B . As B has no maximal bounded set then there exists a set B0 ∈ B which is not a subset of Gn . Again by hypothesis there exists an open bounded set Gn+1 ∈ B such that Gn ∪ Dn+1 ∪ B0 ⊂ Gn+1 . Hence Gn+1 ⊃ Gn ∪ Dn+1 , and Gn+1 6= Gn . This completes the inductive construction of a strictly increasing sequence G = {G1 , G2 , ..., Gn , ...} of bounded open sets, satisfying Gn ⊂ Gn+1 for every n = 1, 2, .... Finally, let B a bounded set, since D is a base of B, there exists a positive integer n satisfying B ⊂ Dn ⊂ Gn . Hence G is a base of B. Since Gn and X −Gn+1 are two disjoint closed sets of a normal space X , it follows from Urysohn's lemma that there exists a continuous function φn : X → I satisfying φn (x) = 0 for every x ∈ Gn and φn (x) = 1 for every x ∈ X − Gn+1 . Now dene a function χ : X → R by taking : 1 χ(x) = φ1 (x), ∀x ∈ G1 , χ(x) = (n − 1) + φn (x), ∀x ∈ Gn+1 − Gn , ∀n > 1. The function χ is obviously real-valued and nonnegative. The continuity of χ follows from the fact that φn−1 (x) = 1 and φn (x) = 0 hold for every point x ∈ Gn − Gn and every n > 1. Now let E denote an arbritary subset of X . If E is bounded, then there exists a positive integer n satisfying E ⊂ Gn+1 . This implies χ(x) ≤ n for every x ∈ E . Conversely, assume that χ(E) < K . Choose a positive integer n so large that χ(x) < n holds for every x ∈ E . Then we have E ⊂ Gn+1 and hence E is bounded. Therefore χ is a characteristic function of X. 1 Urysohn's Lemma If A and B are two disjoint closed sets in a normal space a continuous function φ:X→I such that φ(A) = 0 and φ(B) = 1. 12 X , there exists Theorem 2. Let (X, B) a bornological universe. Then B is metrizable i the following three conditions are satised: 1. the topological space X is metrizable; 2. the bornology B has a countable base; ◦ 3. for each B1 ∈ B there is B2 ∈ B such that B1 ⊂ B 2. Proof. ⇒) Condition 1. is clear. For 2., choose a point z ∈ X . For each positive integer n the open balls Bn (z) = {x ∈ X : d(z, x) < n} form a countable base for B. As for every B ∈ B there exists a Bn (z) such that B ⊂ Bn (z) so B ⊂ Bn (z) ⊂ Bn+1 (z) and we have 3.. ⇐) Assume that (X, B) satises the conditions 1., 2., 3.. Since X is metrizable then there exists a bounded metric e : X 2 → R which denes the topology of X . Again by 1. X is normal and by 2. and 3. and the last theroem we have a characteristic function χ : X → R on the universe (X, B). Now dene the function d : X 2 → R by taking d(x, y) = e(x, y) + |χ(x) − χ(y)| for every x, y ∈ X . In fact d is a metric on X . To prove that d denes the same topology of X , it is enough to show that d and e are equivalent. For this purpose, consider any point p ∈ X and an arbitrary positive real number r. Since χ is continuous and since e denes the same topology of X , there exists a positive real number s < r/2 such that |χ(p) − χ(x)| < r/2 holds for every point x ∈ X satisfying e(p, x) < s. It follows that, for every x ∈ X , e(p, x) < s implies d(p, x) = e(p, x) + |χ(p) − χ(x)| < s + r/2 < r/2 + r/2 = r. On the other hand, d(p, x) < s obviusly implies e(p, x) < s. So we have the topological equivalence between the metrics e and d. 13 It remains to verify that the metric d denes the bornology B. For this purpose, choose a positive real number k such that e(x, y) < k holds for every point (x, y) of X 2 . Let E denote any subset of X . If E is a member of B, then it follows from the denition of a characteristic function that χ(E) < K . Hence we obtain d(x, y) = e(x, y) + |χ(x) − χ(y)| < k + K for all x, y ∈ E . This proves that E is bounded for the metric d. Conversely, assume that E is a member of the metric bornology generated by d. Then there exists a positive real number r such that d(x, y) < r holds for all points x, y ∈ E . Without loss of generality, assume E is not empty. Choose a point p ∈ E . Then we have χ(x) ≤ χ(p) + |χ(p) − χ(x)| ≤ χ(p) + d(p, x) < χ(p) + r for all x ∈ E . This implies E ∈ B. Hence the bornologies are equivalent, that is, d denes the given bornology B of X . Now we give an example of application of this theorem. Corollary 3. Let (X, F) a bornological universe where X is a T2 and F is the nite bornology, then we have the following equivalence: 1. (X, F) is metrizable; 2. X is countable and discrete. Proof. 1. ⇒ 2.) If (X, F) is metrizableSthen by the previous theorem F has a coumerable base such that X = n∈N Bn where Bn are nite sets. So that X is countable. Also let x ∈ X then, as (X, F) is metrizable ◦ ◦ then there exists F ∈ F such that {x} = {x} ⊆ F and obviously F is a nite set. As X is T2 then there exists an open neigborhood G of {x} ◦ such that G ∩ F = {x}. Then we have that {x} is open and closed, because X is T2 , ∀x ∈ X . ◦ 1. ⇐ 2.) If X is discrete then for every F ∈ F we have F = F = F , so we have that for every F1 ∈ F there exists a F2 ∈ F such that 14 ◦ F1 ⊆ B . Now if X is countable then let x0 ∈ X be xed, and dene B0 = {x0 }, B1 = B0 ∪ {x1 } with x1 ∈ / B0 and Bn = Bn−1 ∪ {xn }, xn ∈ / Bn−1 . This is possible because X is numerable. In this way we have a numerable base of F because every F is nite. 2.3 The Heine-Borel property. Let (X, d) a metric space. If C is a compact subset of it then it is closed and bounded. Now we ask for the converse, that is, when (X, d) has the Heine-Borel property: every closed and bounded subsets is compact. This problem can be solved more generally for a Hausdor space X , that is, we can nd, under some hypothesis, and admissible metric d for X such that (X, d) has the Heine-Borel property. In this case we say that d is a proper metric. Theorem 4. Let X be a T2 topological space. The next cases are equivalent: 1. X is metrizable by a proper metric d; 2. (X, CB) is metrizable; 3. the topology of X is metrizable, CB has a countable base and for ◦ every B1 ∈ CB there exists a B2 ∈ CB such that B1 ⊆ B 2; 4. X is locally compact and second countable. Proof. 1. ⇔ 2. If X is metrizable by a proper metric d, that is, (X, d) has the Heine-Borel property, then we have that by Example 6. CB = Bd(X). Conversely if CB = Bd(X) by an admisible metric d, then obviously X is metrizable, and for every closed B ∈ Bd (X) there exists a C ∈ CB such that B ⊂ C so B is compact. 2. ⇔ 3. This equivalence follows from the last Theorem 2.. 3. ⇒ 4. As every point in (X, d) is a bounded set of the compact bornology CB, then there exists a bounded set B ∈ CB such that x ∈ 15 ◦ ◦ B and as B is compact we can concluded that (X, d) is a locally compact space. Now, we take a countable base of CB, {B1 , B2 , ..., Bn , ...}. Obviously, by denition of CB, for every j ∈ N, Bj is a compact subset of X . Given an open cover {Gi }i∈I of X , every Bj is covered by a nite subcover of {Gi }i∈I by compactness. For every j ∈ N take its respective nite cover {Gik j }k=1,...,nk , Then {{Gik j }k=1,...,nk }j∈N is an open countable subcover of X . So (X, d) is a Lindelöf space, and because it is metrizable then it is second countable. 4. ⇒ 3. Conversely suppose that (X, d) is locally compact and second countable. Then X is metrizable because it is regular and second countable . As X is locally compact, for every x ∈ X there exists an open neighborhood V x such that V x is compact. Let B1 ∈ CB then its closure is compact so B1 is covered by a nite subcover V xk . If we ◦ S put B2 = k V xk then it is an open set in CB, such that B1 ⊂ B 2 . Now suppose that X is bounded, then obviously CB has a countable base given by X itself. So suppose that X is not bounded. Then recall that CB doesn't have a maximal bounded set. Remember that X is metrizable and second countable, so it is Lindelöf. Take the open cover of X given by V x , then there exists a countable subcover of X , {V xj }. Let B1 ∈ CB. As B1 is compact, then exists a nite subcover Sn1 there xjk xjk {V }k=1,...,n1 of B1 . Maybe B1 = k=1 V but as {V x } is a cover of X and X is not bounded then there S exists a V x1∗ such that it is 1 not contained in B1 . So if we pick B2 = nk=1 V xjk ∪ V x1∗ ∪ V x1 then B1 is strictly contained in B2 and B2 ∈ CB. Repeating this process we have a strictly increasing sequence of sets {B1 , ..., Bn , ...} such that V xj−1 ⊂ Bj , j ≥ 2. Then let B ∈ CB then its closure is compact so B is covered by a nite subcover {V xjk }k=1,...n . By construction of the Bj0 s, we can take the biggest BjK , K ∈ {1, ..., n} such that V xjk ⊆ BjK , ∀k = 1, ..., n. So {B1 , ..., Bn , ...} is countable base of (X, CB). 2 3 2X is a regular space if every point x∈X 3 Urysohn's metrization theorem. a) X is regular and second countable; b) X is separable and metrizable; c) X can be embedded as subspace of has a nhood base consisting of closed sets. The following are equivalent for a the Hilbert cube I 16 N . T1 space: Remark 3. Regarding in the proof of Theorem 2, and applying it to a unbounded metric space (X, d) and to its compact bornology CBd(X) we have a way to construct a proper metric from d. As in the proof it is given by d0 (x, y) = e(x, y) + |χ(x) − χ(y)| where e = min{d, 1} and χ : X → [0, ∞) is the characteristic function associated to a countable base of CBd(X). To have more information about this theme please consult the articles [13] and [10]. 17 Chapter 3 Bornologies in uniform spaces. 3.1 Totally bounded sets and bounded sets in the sense of Bourbaki. Now, we take a uniform space (X, U) and we look for some natural denition of bounded sets on it. To make easier the work we are going to suppose that U generates a T2 topology on X . Denition 6. A subset B of a uniform space (X, U) is called bounded in the sense of Bourbaki if for each U ∈ U, we can nd an n ∈ N and a nite set F ⊆ X such that B ⊆ U n (K) where U n = n z }| { U ◦ U ◦ ... ◦ U . If n = 1 then we say that B is precompact) . totally bounded (or Remark 4. In these denitions we can suppose that our entourages are from a base of U and that they are symmetric, that is, they satisfy U = U −1 . Obviously the bounded sets in the sense of Bourbaki and the totally bounded sets form a bornology which we are going to denote by BB (X) and TB (X), respectively. U U Remark 5. If a space X has two equivalent uniformities then the Bourbaki bounded sets and the totally bounded sets of both uniform 18 spaces coincide respectively and they dene the same bornology because the entourages, which are needed to be bounded in the sense of Bourbaki or totally bounded, are the same in both uniformities. Remark 6. To be totally bounded implies to be bounded in the sense of Bourbaki, but the converse is not true in general, as we will see in the next example. Example 7. Let (X, k.k) an innite dimensional Banach space. The closed ball of radius 1, B1(x), is bounded in the sense of Bourbaki because, as we will see later, its image under every real-valued uniformly continuous function f ∈ U C(X), f (B1(x)) is bounded in R. If BBk.k(X) = TBk.k(X), that is, BBk.k(X) ⊂ TBk.k(X), this implies that B1(x) is totally bounded and then compact because it was already complete. But this is impossible because (X, k.k) is innite dimensional. For a metric space (X, d) ina easy way we can say that a set B is totally bounded if ∀ > 0, can be covered by a nite union of disks of , that is, ∀ > 0 there exist x1 , x2 , ..., xn such that B ⊆ Snradius i=1 B (xi ), where the B (xi ) are the open metric balls of radius and center xi of (X, d). If B is bounded in the sense of Bourbaki in (X, d) then its visualization it's a little more dicult. Let > 0 and let U ∈ U such that U [x] = {y ∈ X : d(x, y) < }. Then U 2 [x] = {y ∈ X : ∃z ∈ X, d(x, z) < , d(z, y) < } and U 3 [x] = {y ∈ X : ∃z ∈ X, d(x, z) < , z ∈ U 2 [y]} = = {y ∈ X : ∃z, w ∈ X, d(x, z) < , d(y, w) < , d(w, z) < }. More generally U n [x] = {y ∈ X : ∃z ∈ Xd(x, z) < , z ∈ U n−1 [y]}. So, if B bounded in the sense of Bourbaki then ∀ > 0 ∃n ∈ N and there exist x1 , x2 , ..., xm ∈ X such that B⊆ m [ U n [xi ], i=1 19 where we can imagine U n [xi ] as a sequence of at most n balls in X of radius at most each one, needed to join xi and another point y ∈ X in a path of length at most n. Also for a metric space (X, d) we have that each bounded set in the sense of Bourbaki for the uniformity generated by the metric d is bounded for the metric, and so that each totally bounded set for d is also bounded for d. The converses are not true in general. Example 8. Let R with the Euclidean metric d2. We know that it is uniformly equivalent to the bounded metric d∗ = min{1, d2}, so they dene the same uniformity and then BBd (X) = BBd (X) TBd (X) = TBd (X). ∗ 2 ∗ 2 Also we know that to be totally bounded in (R, d2) is equivalent to be bounded for the metric by completeness of it , so we have that to be bounded in the sense of Bourbaki for the uniformity generated by d2 is equivalent to be bounded and totally bounded for d2. That is, Bd (X) = TBd (X) = BBd (X). 2 2 2 Now, as d∗ is a bounded metric we have that every set is bounded by d∗ in R, but this doesn't happen to d2 , so the bornology generated by d∗ is bigger but not equal to the bornology generated by d2 although if they generate the same uniformity. So Bd (X) 6= Bd (X) ∗ 2 and BBd (X) = TBd (X) = Bd (X) ⊂ Bd (X) ∗ ∗ 2 20 ∗ 3.2 Characterization of the bounded sets in the sense of Bourbaki. We have the following results for a uniform space (X, U). Lemma 5. [9] Metrization Lemma. Let {Un}n∈N be a sequence of subsets of X × X such that U0 = X × X , each Un contains the diagonal, and Un+1 ◦ Un+1 ◦ Un+1 ⊂ Un for each n. Then there is a non-negative real-valued function d on X × X such that 1. d(x, y) + d(y, z) ≥ d(x, z) for all x, y and z ; and 2. Un ⊂ {(x, y) : d(x, y) < 1/2n} ⊂ Un−1 for each positive integer n. If each Un is symmetric, then there is a pseudometric d satisfying condition 2.. Denition 7. We say that a pseudometric d is uniformly continuous in a uniform space (X, U) if it denes a uniformity contained in U. Theorem 6. [6] Let (X, U) be a uniform space. The following statements are equivalent: 1. B is bounded in the sense of Bourbaki; 2. each real valued uniformly continuous map is bounded on B ; 3. B is bounded for each pseudometric d uniformly continuous on (X, U). Proof. 1. ⇒ 2.) Let f : X → R be an uniformly continuous function, then let > 0 there exists a U ∈ U such that for every x, y ∈ X if (x, y) ∈ U then |f (x) − f (y)| < . If B ⊂ X is bounded in the sense of Bourbaki, there exists a nite set K and a natural number n such that U n (K) ⊃ B . If z ∈ f (B), there exists y ∈ B such that f (y) = z . As y ∈ U n (K), there exist y = y1 , ..., yn ∈ X , yn+1 ∈ K 21 so that (yi , yi+1 ) ∈ U for i = 1, ..., n. Then |f (yi ) − f (yi+1 )| < and z ∈ Bn (f (K)). 2. ⇒ 3.) If d is a uniformly continuous pseudometric on X and b ∈ B , dene, for all x ∈ X f (x) = d(x, b). Then f : X → R is uniformly continuous and B is bounded for d. 3. ⇒ 1.)Let B ⊂ X be a set given. Let U ∈ U be a symmetric entourage. We dene a relation on X as x ≡ y i there exists a n ∈ N such that x ∈ U n (y). The relation ≡ is an equivalent relation S on X . Therefore it denes a decomposition X = α∈A Xα . By the Metrization Lemma there exists a uniformly continuous pseudometric d on X such that 1 {(x, y) : d(x, y) < 1} ⊂ U. Now we want to construct another pseudometric, dening it rst on each set Xα . Recall that for every x, y ∈ Xα there exists a nite sequence or a chain of points xi ∈ Xα , i = 1, ..., n + 1 such that Pn x1 = x, xn+1 = y, (xi , xi+1 ) ∈ U for i = 1, ..., n. The number i , di+1 ) will be called the length of this chain. Put φ(x, y) = i=1 Pd(x n inf i=1 d(xi , di+1 ) where the inmun is taken over all the chains on U connectingSx and y. It is easy to see that φ is a nite non negative function on α∈A (Xα × Xα )and that in fact φ is a pseudometric on each Xα . It is uniformly continuous because it is equal to d on U . Now we shall extend the pseudometric onto the hole space X . First we dene the function µ on the set M . Put AB = {α ∈ A : Xα ∩ B 6= Φ}. If AB is nite then we put µ(α) = 1 for each α ∈ AB . If AB is innite, there exists an innite sequence {αn }∞ n=1 where αn ∈ AB , αm 6= αn if m 6= n and we put µ(αn ) = n, µ(α) = 1 for α ∈ AB , α 6= αn . Choose wα ∈ Xα for each α ∈ A. If x ∈ Xα , y ∈ Xβ we put ρ(x, y) = φ(x, y) if α = β , 1 The same demostration can be done for an uniformly continuous function any uniform space (Y, V). If n=1 f from (X, U) then we have the same result for a totally bounded set. 22 to ρ(x, y) = φ(x, wα ) + φ(y, wβ ) + µ(α) + µ(β) if α 6= β. It easy to prove that ρ is a pseudo-metric on X . We are going to prove the triangle inequality. Suppose that x ∈ Xα , y ∈ Xβ and x ∈ Xγ . There are only three possible cases: 1. if α = β = γ then ρ = φ and this is clear; 2. if α 6= β 6= γ then ρ(x, y) ≤ φ(x, wα ) + φ(z, wγ ) + µ(α) + µ(γ) < φ(x, wα ) + φ(y, wβ ) + µ(α) + µ(β) + φ(y, wβ ) + φ(z, wγ ) + µ(β) + µ(γ) = ρ(x, y) + ρ(y, z); 3. if α 6= β = γ then wβ = wγ and we have φ(wβ , y) + φ(y, z) ≥ φ(wβ , z), φ(x, wα ) + φ(wβ,y ) + µ(α) + µ(β) + φ(y, z) ≥ φ(x, wα ) + φ(wβ , z) + µ(α) + µ(β), ρ(x, y) + ρ(y, z) ≥ ρ(y, z). If ρ(x, y) < 1, then x and y belong to the same class Xα and ρ(x, y) = φ(x, y), therefore ρ is also uniformly continuous. Now let B be bounded for the pseudometric ρ. Then AB must be nite and φ(x, wα ) is a bounded function for x ∈ B ∩ Xα . Write b = maxα∈A (supx∈B∩Xα φ(x, wα )) < ∞. Let m > b, m ∈ N. Consider an element α ∈ AB . For each point x ∈ B ∩ Xα there exists an irreducible chain on U conecting x with wα . Therefore, therePexists a sequence x = x1 , x2 , ..., xn+1 = wα such that (xi , xi+1 ) ∈ U , ni=1 d(xi , xi+1 ) < m. As this chain is irreducible, the inequality d(xi , xi+1 ) < 21 cannot hold for two consecutive indices i. It means that d(xi , xi+1 ) < 12 holds at most for 21 (n + 1) indices i and therefore d(xi , xi+1 ) ≥ 21 holds at least for 12 (n − 1) indices. Then the length of the chain is at least 41 (n − 1), which implies n < 4m + 1. 23 therefore B ⊂ U 4m+1 (wα , α ∈ AB ) and the boundedness in sense of Bourbaki of B is proved. Corollary 7. [6] Let (X, U) be a uniform metrizable space. Then B ⊆ X is bounded in the sense of Bourbaki i B is bounded for each metric on X dening the same uniformity as U (uniformly equivalent). 3.3 Uniform metrization of bornologies. As in the rst chapter, now we are interested in the study of the problem of when a uniform space (X, U) with a bornology B is metrizable, that is, there exists an admissible metric d such that it denes the same uniformity than U and the bornology B coincide with the metric bornology Bd (X). If we suppose that the uniform space is metrizable, that is, it has a numerable base for the uniformity and the topology generated by the uniformity is T2 , then we can solve the problem directly for a metric space (X, d): for a metric space (X, d) with a bornology B we want to nd a uniformly equivalent metric d0 such that the bornology B coincides with the metric bornology for d0 , that is, B = Bd0 (X). We have two cases: 1. suppose that (X, d) is a metric space such that the bornology B contains X . Then every subset of X is bounded. If we put d0 = min{1, d}, d0 is a bounded uniformly equivalent metric, so B = Bd0 (X); 2. now suppose that X ∈/ B. Denition 8. Let A ⊂ X and x ∈ X , then d(x, A) is the distance from x to A and if A = ∅ then we agree that d(x, A) = ∞. We denote the -enlargement of A by A where A = {x ∈ X : d(x, A) < } = [ x∈A 24 B (x) and B(x)is the open ball of radius around x. Lemma 8. Let (X, d) be a metric space and B a bornology on it such that X ∈/ B. If B has a countable base {Bn : n ∈ N} such that ∃δ > 0, with Bn δ ⊆ Bn+1 , ∀n ∈ N, then there exists a uniformly continuous characteristic func- tion χ : X → [0, ∞) such that B = {E ⊂ X : ∃K > 0, χ(E) < K}. Proof. For n ∈ N dene φn : X → [0, 1] by φn(x) = min{1, 1δ d(x, Bn)}. Then φn (Bn ) = 0, φn (X − Bn+1 ) = 1, 0 ≤ φn ≤ 1 andPφn is uniformly continuous. Put χ = φ1 + φ2 + φ3 + .... Then χ | Bn = n−1 i=2 φi , so that: 1. χ | Bn is uniformly continuous ∀n ∈ N; 2. χ(Bn ) ⊆ [0, n − 1] ∀n ∈ N. Moreover χ is uniformly continuous: ∀ > 0 ∃δ 0 > 0 such that δ 0 < δ and δ 0 < δ , and ∀x, y ∈ X with d(x, y) < δ 0 , then ∃n ∈ N such that x ∈ Bn − Bn−1 and y ∈ Bnδ , then 2 |χ(x) − χ(y)| = |φn (x) − φn (y) + φn+1 (x) − φn+1 (y)| < d(x, y) < 2. δ Now let E ∈ B then there exists an n ∈ N such that E ⊂ Bn . So Pn−1 χ(E) = i=1 φi (E) ⊆ [0, n − 1]. Conversely if χ(E) is bounded then for some n ∈ N, E ⊆ Bn because otherwise for every n ∈ N there exists an x ∈ E such that χ(x) ≥ n. Theorem 9. Let B a bornology on (X, d). The following conditions are equivalent: 1. B = Bd (X) for some uniformly equivalent metric d0 on (X, d) 0 2. B has a countable base {Bn : n ∈ N} such that ∃δ > 0, with Bn δ ⊆ Bn+1 , ∀n ∈ N. 25 Proof. 1. ⇒ 2.) If B = Bd0 (X), xed a certain z ∈ X the basis {Bn = {x ∈ X : d(x, z) < n}, n ∈ N} and δ < 12 satises the thesis. 2. ⇒ 1.)The case X ∈ B is of course trivial because B = P(X) and we have d0 = min{1, d}. So that, suppose X ∈/ B and take the bounded metric d∗ = min{1, d}, which is uniformly equivalent to d. Applying the previous lemma we dene a new metric d0 : X 2 → R, d0 (x, y) = d∗ (x, y) + |χ(x) − χ(y)| where χ is the characteristic function associated to the base of the bornology. To show that d0 and d are uniformly equivalent it is enough to proof that d0 and d∗ = min{1, d} are uniformly equivalent. On one hand d∗ ≤ d0 , so if d0 (x, y) < then d∗ (x, y) < and then Ud is coarser than Ud∗ . On the other hand, using the uniform continuity of χ, ∀ > 0, ∃δ > 0 such that ∀x, y ∈ X with d∗ (x, y) < δ ⇒ d0 (x, y) = d∗ (x, y) + |χ(x) − χ(y)| < + δ, and we have that the equivalence of the uniformities follows and also the uniform equivalence of the metrics. Now we want to show that B ∈ B i B ∈ Bd0 (X). If B ∈ B then it is clearly bounded for the metric d∗ and χ(B) is bounded in R by the lemma, so obviously B ∈ Bd0 (X). If B ∈ Bd0 (X) then there exists a k ∈ R, k > 0 such that ∀x, y ∈ B then d0 (x, y) < k . As d∗ is bounded metric then χ(B) is bounded in R so then B ∈ B by the lemma. 26 3.4 The Heine-Borel property for uniformly equivalent metrics. As an application of the last theorem now we come back to the study of the compact bornology CB and to the study of proper metrics as in the second chapter. We have (X, U) a uniform such that its uniformity generates a T2 topology. Denition 9. [13] [1] The uniform space (X, U) is uniformly locally compact if there is an entourage U ∈ U such that U [x] is a compact subspace of X for every x ∈ X . This denition implies that X with the topology generated by U is locally compact because U generates a T2 topology. Now we ask when for a metric space (X, d) there exists an uniform equivalent proper metric d0 , that is, (X, d0 ) has the Heine-Borel property. As in the second chapter we give a similar answer: Theorem 10. Let (X, d) a metric space. The following cases are equivalent: 1. there exists a uniform equivalent proper metric d0; 2. there exists a uniform equivalent metric such that CBd(X) = Bd0 (X); 3. CBd(X) has a countable base {Bn : n ∈ N} such that ∃δ > 0, ∀n ∈ N with Bnδ ⊆ Bn+1 ; 4. (X, d) is uniformly locally compact and second countable. Proof. Everything is similar to the proof of Theorem 4. in the second chapter. We have to control only the fact of the uniformly locally compactness. S 3. ⇒ 4.) Recall that B δ = x∈B Bδ (x). As for every x ∈ X there exists nx ∈ N such that x ∈ Bnδ x then Bδ (x) ⊂ Bnδ x so Bδ (x) ∈ CBd (X) 27 so that Bd (x) is compact. Then as (X, d) is a T2 space then it is uniformly locally compact. 4. ⇒ 3.) If (X, d) is uniformly locally compact then there exists a entourage V ∈ U, such that V [x] is compact for every x ∈ X . By denition of base of a uniformity there exists a δ > 0 such that U = {(x, y) ∈ X 2 : d(x, y) < δ} ⊂ V , so that U [x] are compact neighborhoods of every x. Let B1 ∈ CB then its closure S is compact so B1 is covered by a nite subcover U [xk ]. If we put B2 = k U [xk ] then ◦ δ it is an open set in CB, such that B1 ⊂ B 2 . The rest of the proof follows the homologous proof in the second chapter. It is a known result that C is a compact subset of (X, U) i it is totally bounded and complete. Now I want to give a similar result but for the more general notion of bounded set in the sense of Bourbaki. This is going to be done comparing the compact bornology with bornology of the bounded sets in the sense of Bourbaki. Lemma 11. [1] Let (X, U ) be a uniform space not necessarily T2 then if (X, U) is uniformly locally compact, then A is bounded in the sense of Bourbaki X i A is a compact subset of X . Proof. We know that there exists an entourage V ∈ U such that V [x] is compact for every x ∈ X and suppose that B ⊂ X is compact. Take a symmetric entourage U ∈ U such that U 2 ⊆ V then there exists a nite subset F ⊂ X with B ⊆ U [F ]. So U [B] ⊆ U 2 [F ] ⊆ V [F ]. As V [F ] is compact then it is also U 2 [F ]. By induction we can conclude that U n [F ] is compact in X for every n ∈ N and every nite subset F of X . So we can conclude with the next theorem Theorem 12. Let (X, U) a uniformly locally compact space then the compact bornology CBU(X) coincides with BBU(X), the Bourbaki bornology. 28 Soon, in the last chapter, we are going to study the problem of metrization of the bornology of bounded sets in the sense of Bourbaki, that is, when, for a metric space (X, d), we have a uniformly equivalent metric d0 such that BBd (X) = Bd0 (X). Obviously by the theorem of uniform metrization of bornologies, we already have the answer but as the Bourbaki bounded sets aren't easily visualized, then to work with a base of this sets is not very practic. So we want to see other equivalents solutions that maybe are easier to work with. In this last part of this chapter we have a nice answer: Theorem 13. Let (X, d) be a uniformly locally compact metric space. Then there exists a uniformly equivalent metric d0, such that, (X, d0) has the Heine-Borel property i BBd(X) = Bd (X). 0 Remark 7. If (X, d) is a uniformly locally compact then BBd(X) = Bd (X) i d is a proper metric. 29 Chapter 4 Realcompactication v.s compactication. Now we are going to introduce the notions of compactication and realcompactication of a topological space to compare them. But rst we are going to give some basic notions about a Tychono space because as we will see later we can speak only about compactication or realcompactication of a Tychono space. All this notions can be studied in more detail in citar [12] and [3]. Denition 10. A Tychono space is topological space which is both T1 and completely regular, where a topological space X is completely regular i whenever F is a closed set in X and x ∈/ F there is a realvalued continuous function f : X → [0, 1] such that f (x) = 0 and f (F ) = 1. Remark 8. Subspaces of a Tychono space are Tychono and non empty product of Tychono spaces is Tychono. Remark 9. All the T4 spaces that is normal spaces that are also T1 are Tychono spaces. Denition 11. Let {fα : α ∈ A} be a collection of functions on X to spaces Xα , then {fα : α ∈ A} separates points and closed sets in X if given a closed subset F of X and x ∈/ F there exists some f ∈ {fα : α ∈ A} such that f (x) ∈ / f (F ). 30 Theorem 14. Immersion's lemma. If X is a T1 space and {fα : α ∈ A} is a collection of continuous functions on X to spaces Xα , which separates points from closed sets, then the evaluation map e:X→ Y Xα α∈A dened as e(x) = (fα (x))α∈A is a homeomorphic embedding. Corollary 15. A topological space X is a Tychono space i it is homeomorphic to some subspace of some cube, that is, of some product of closed real intervals. Proof. In fact as X is a Tychono space then the family of real-valued bounded continuous functions on X , C ∗ (X), separates points from closed sets. As f (X) is a bounded set in R for every f ∈ C ∗ (X), take If a closed interval in R such that f (X) ⊆ If . Obviously C ∗ (X) = {f ∈ C ∗ (X) : f : X → If }. So by the Q Inmersion's lemma we can conclude that the evaluation map e : X → f ∈C ∗ (X) If dened as e(x) = (f (x))f ∈C ∗ (X) is an homeomorphic embedding of X in the cube f ∈C ∗ (X) If . The converse is obvious because the interval [0, 1] is a Tychono space. Q The way in which has been proved the previous corollary, that is, using the Inmersion's lemma, is going to be repeated to construct compactications and realcompactications, as we will see in the following sections. 31 4.1 Compactication of a topological space. Denition 12. Let X be a topological space. Then a compactication of it is a compact space cX and a map c : X → cX which is an homeomorphic embedding of X in cX in such a way that c(X) = cX . Theorem 16. A topological space X has a compactication i X is a Tychono space. Proof. ⇒) If X has a compactication then X is homeomorphic to a subspace of a compact space cX . As cX is compact then it is a Tychono space so X is a Tychono space. ⇐) Let X be a Tychono space then it is homeomorphic to a subspace of a cube. Every cube is compact so if we take the closure of X in the cube then it is a compactication of X . Denition 13. Given two compactications of X we say that c1X ≤ c2 X i there exists a continuous mapping h : c2 X → c1 X such that h ◦ c2 X = c1 X . Thus the inequality c1 X ≤ c2 X means that c2 X can be mapped onto c1X in such a way that every point of the space X , considered as a subspace of both c2 and c1, is mapped onto himself. Two compactications c1X and c2X are equivalent compactications i h is a homeomorphism. This is equivalent to say that c1X ≤ c2X and c2X ≤ c1X . The previous denitions gives us an order ≤ is the familly of all the compactications of a Tychono space. In particular we have a largest element in this familly. Denition 14. Let X be a Tychono. Then its biggest compactication is the Stone-ech Compactication, denoted by βX . Theorem 17. Every continuous function f : X → Z of a Tychono space X to a compact space Z is extendable to a mapping F : βX → Z . 32 Proof. By the Immersion's lemma c = β × f : X → βX × Z is a homeomorphic embedding, so that cX = c(X) ⊂ βX × Z is a compactication of X . By the maximality of βX there exists a mapping h : βX → cX such that h ◦ β = c. Let p : cX → Z be the restriction of the projection of βX × Z onto Z to cX and let F = p ◦ h : βX → Z . Since F β = phβ = pc = f , the mapping F is an extension of f . Theorem 18. If every continuous function of a Tychono space X to a compact space is continuously extendable over a compactication cX of X , then cX is equivalent to the Stone-ech Compactication of X . Proof. If a compactication of X has the property of the hypothesis then there exists an extension B : cX → βX of the embedding β : X → βX . We have then that B ◦ c = β , that is, βX ≤ cX . As βX is the biggest compactication of X then cX = βX . Recall that when we have proved that every Tychono space is homeomorphic to some subspace of some cube we have construct indeed the Stone-ech Compactication of X . We know that e(X) is a compactication of X as it has been set when it have proved that a Tychono space has a compactication. There the evaluation map ∗ e : X → RC (X) was given by e(x) = (f (x))f ∈C ∗ (X) . Now we are going to prove that every bounded continuous function f : X → R can be extended to a continuous function f˜ : e(X) → R. For every f ∈ C ∗ (X) let f˜ : e(X) → R be the restriction of the continuous projection pf : R C ∗ (X) →R such that pf ((y)f ∈C ∗ (X) ) = y ∈ R, pf (e(x)) = f (x). Then f˜ is also continuous and then f˜ ◦ e = pf ◦ e = f . By the previous theorem, as every bounded continuous function f ∈ C ∗ (X) is 33 a continuous function from X to the compact interval in R, If ⊃ f (X), we have that βX = e(X). In particular the continuous extension f˜ : βX → R is also a bounded function. Remark 10. Obviously X is a compact i βX = X . Repeating this argument for any subfamily Γ ⊆ C ∗ (X) of continuous bounded real-valued functions separating points from closed sets we have always a compactication of the Tychono space X . Theorem 19. Let X be a Tychono space and let Γ ⊆ C ∗(X) be a family of continuous bounded real-valued functions separating points from closed sets. Then e(X) ⊂ RΓ, where e : X → RΓ is the evaluation map given by e(x) = (f (x))f ∈Γ , is the smallest compactication such that every f ∈ Γ can be extended to continuous bounded function f˜ : e(X) → R. Proof. We have to prove only that in fact it is the smallest. Suppose that there exists another compactication cX with the same property of extendability in the hypothesis, then we have to prove that cX ≥ e(X). In fact there exists a continuous mapping h : cX → e(X) such that h ◦ c = e, dened in the next way: c h(y) = (f (y))f ∈Γ ∈ Y Rf = RΓ f ∈Γ where f c : cX → R is the continuous extension of f ∈ Γ. Then for every f ∈ Γ let pf : RΓ → R be the continuous projections dened by pf ((y)f ∈Γ ) = y ∈ R, pf (e(x)) = f (x). In particular pf ◦ h = f c so h(y) ∈ h(c(X)) ⊂ h ◦ c(X) = e(X). And we have it. 34 Denition 15. Let Γ = U C ∗(X) the family of all the uniformly continuous bounded real-valued functions from a metric space (X, d). Then as in the previous example sXe(X) ⊂ RU C (X) is a compactication of (X, d) called Samuel Compactication. Then sX can be characterized as the smallest compactication of (X, d) such that every uniformly continuous bounded real-valued function from f : X → R is extended to a continuous (bounded) function f˜ : sX → R. ∗ 4.2 Realcompactication of a topological space. Denition 16. A topological space X is called a realcompact space if X is a Tychono space and there is no Tychono space X̃ which satises the following two conditions: 1. there exists a homeomorphic embedding α : X → X̃ such that α(X) 6= α(X) = X̃ ; 2. for every continuous real-valued function f : X → R there is a continuous function f˜ : X̃ → R such that f˜ ◦ α = f . Denition 17. A topological space X is called pseudocompact if X is a Tychono space and every continuous real-valued function dened on X is bounded. Theorem 20. A topological space is compact i it is a pseudocompact realcompact space. Proof. ⇒) This is not dicult to prove from the denitions. ⇐) Suppose that X is not compact but realcompact. Then, since X 6= βX , there exists a function f : X → R which cannot be continuously extended over βX by condition 2. of the denition of realcompact space, as βX is Tychono. Clearly the function is not bounded, because if it were bounded then f : X → f (X) ⊂ If ⊂ R where If is a closed and bounded interval in R. So If is compact and then f : X → If ⊂ R 35 should be extendable to a continuous function f˜ : βX → If ⊂ R. We can concluded then that X is not pseudocompact. Theorem 21. A topological space is realcompact i it is homeomorphic to a closed subspace of a power RΛ of the real line. Proof. Let X be a realcompact space. As X is a Tychono space then we can apply the last Inmersion's lemma to the set of all the continuous real-valued functions on X , C(X). So the evaluation map Y e:X→ Rf f ∈C(X) is a homeomorphic embedding of X in Q f ∈C(X) Rf := RC(X) . Let X̃ = e(X) ⊂ RC(X) . For every continuous real-valued function f : X → R there exists a function f˜ : X̃ → R such that f˜Q e = f . Infact, it is the restriction f˜ = pf | X̃ of the projection pf : f ∈C(X) Rf → Rf . Obviously f˜ is continuous functions by the continuity of the projections maps pf Now, by denition of realcompact space, we have that X̃ = e(X) and then, the space X is homeomorphic to the closed subspace e(X) of RC(X) . Q Conversely, let X be a closed subspace of RΛ := λ∈Λ Rλ . Obviously it is a Tychono space and suppose that it is no realcompact. Then it has an extension X̃ and α a homeomorphic embedding α : X → X̃ with α satisfying the condition 2. of the denition of realcompact space, that is, α(X) 6= α(X) = X̃ . For every λ ∈ Λ there exists a function p̃λ : X̃ → Rλ = R such that p̃λ α = pλ , where pλ (x) = pλ ((xλ )λ∈Λ ) = xλ ∈ Rλ = R. Let F : X̃ → Y Rλ λ∈Λ dened as F (y) = (p̃λ (y))λ∈Λ . 36 Since F α(x) = x for every x ∈ X , we have that F (X̃) = F (α(X)) ⊂ F α(X) = X = X, then F maps X̃ onto X . For every x ∈ X we have that αF (α(x)) = α(x), so αF : X̃ → X̃ when restricted to α(X) coincides with idα(X) . Since α(X) is dense in X̃ then αF = idX̃ . As αF (X̃) ⊂ α(X), it follows that α(X) = X̃ . Hence there is no Tychono space X̃ satisfying 1. and 2., i.e., X is a realcompact space. Remark 11. As an immediate corollary R with the usual topology is realcompact and so any power of it, RΛ. Corollary 22. Every closed subspace of a realcompact space is realcompact Q Corollary 23. The non empty product s∈S Xs is realcompact i Xs is realcompact for every s ∈ S . Denition 18. In the same way that for compactication, by realcompactication of a topological space X we intend a realcompact space αX and a map α : X → αX which is an homeomorphic embedding of X in αX in such a a way that α(X) = αX . Remark 12. As every compact space is realcompact then every compactication of a space is a realcompactication. But there exist also realcompactications which are not compactications. Theorem 24. A topological space has a realcompactication i it is a Tychono space. Proof. Suppose that X has a realcompactication X̃ . Then it is home- omorphic to a subspace of X̃ which is a Tychono space. Then X is also a Tychono space. Conversely suppose that X is a Tychono space. Then we can apply the Immersion's lemma to C(X) and then X ∼ = e(X) ⊂ e(X) ⊂ RC(X) , where the evaluation map e : X → RC(X) , given by e(x) = (f (x))f ∈C(X) 37 is an homeomorphic embedding of X in RC(X) . Obviously e(X) is a realcompactication of X because is a closed subspace of RC(X) . Denition 19. The realcompactication constructed in the previous theorem is the Hewitt Realcompactication of a Tychono space X . It is denoted by υX . Obviously if X is realcompact i X = υX . Applying the Inmersion's Lemma for a subcollection of continuous functions Γ ⊂ C(X) separating points form closed sets in a Tychono space X ,we can construct other realcompactications of X . Theorem 25. If X is a Tychono space and Γ is a collection of realvalued functions from X which separates points from closed sets then X̃ = e(X) ⊂ RΓ is a realcompactication of X where the evaluation map e : X → RΓ dened as e(x) = (f (x))f ∈Γ is the homeomorphic embedding. In particular every function f ∈ Γ can be extended to a continuous function f˜ : X̃ → R such that f˜ ◦ e = f . Proof. As X̃ is a closed subspace of RΓ it is obviously a realcompact- ication of X . The continuousQextension of f ∈ Γ is f˜ = pf | X̃ where pf is the projection pf : RΓ = f ∈Γ Rf → Rf = R Remark 13. The following cases are equivalent: 1. υX = βX ; 2. υX is pseudocompact; 3. X is pseudocompact; 4. C(X) = C ∗(X). As for the compactications, in the set of all the realcompactications of a given Tychono space X there is an order for which we can say, given two compactications, if one of them is smaller than the other or if they are equivalent. Denition 20. Given two realcompactications of X we say that α1 X ≤ α2 X i there exists a continuous mapping h : α2 X → α1 X 38 such that h ◦ α2 = α1, leaving X xed . Moreover α1X and α2X are equivalent realcompactications i h is a homeomorphism. This is equivalent to say that α1X ≤ α2X and α2X ≤ α1X . Theorem 26. Let Γ ⊂ C(X) a collection of continuous real-valued functions separating points from closed sets of X . Then X̃ = e(X) ⊂ RΓ is the smallest realcompactication of X such that every function f ∈ Γ can be extended to a a continuous function f˜ : X̃ → R. Proof. The proof is analogous that for the compactications. 4.3 Realcompactications as sets of lattice homomorphisms. Now, we are going to give an equivalent way to construct realcompactications. All the theory needed can be found in [4]. Let X again a Tychono space and L ⊂ C(X) be a unital vector lattice, that is: Denition 21. A unital vector lattice is a partially ordered unital vector space together with the two operators supremum and inmum, ∨ and ∧. Denition 22. A lattice homomorphism is a map ϕ : L → R which satises: 1. ϕ(λf + µg) = λϕ(f ) + µϕ(g), for all f, g ∈ L and λ, µ ∈ R; 2. ϕ(|f |) = |ϕ(f )| for all f ∈ Γ 3. ϕ(1) = 1 Remark 14. Please note that in this case ϕ is positive, that is, ϕ(f ) ≥ 0 when f ≥ 0. ϕ preserves the supremum and the inmun of functions, that is, ϕ(f ∨ g) = ϕ(f ) ∨ ϕ(g) and ϕ(f ∧ g) = ϕ(f ) ∧ ϕ(g). 39 Denition 23. The structure space H(L) is the set of all the lattice homomorphisms ϕ : L → R, considered as a topological subspace of the product RL. Theorem 27. If L ⊂ C(X) separates points from closed sets in X then H(L) is the smallest realcompactication of X such that for every f ∈ L can be extended to a continuous function fˆ on H(L), given by fˆ(ϕ) = ϕ(f ) for each ϕ ∈ H(L). Proof. In fact H(L) is homeomorphic to X̃ = e(X) where e : X → RL was the evaluation map given by e(x) = (f (x))f ∈Γ . The homomorphism is given by the next commutative diagrams e X → e(X) ,→ X̃ = e(X) δ X → δ(X) ,→ H(Γ) = δ(X) where δ is dened as δ(x) = δx and δx is the point evaluation map δx (f ) = f (x) for every f ∈ L. We shall prove that δ(X) is in fact dense in H(L). Given ϕ ∈ H(L), f1 , ..., fn ∈ L and > 0 there exists an x ∈ X such that Pn |δx (fi ) − ϕ(fi )| < for all i = 1, ..., n. Otherwise the function g = i=1 |fi − ϕ(fi )| ∈ L would satises g ≥ and ϕ(g) = 0. And this is imposible since ϕ is monotone. As L separates points from closed sets in X and δ is continuous than by the Immersion's Lemma δ : X → H(L) is an homeomorphic embedding of X in H(L). In particular if fˆ : H(L) → R is an extension of f ∈ L ⊂ C(X) such that fˆ(ϕ) = ϕ(f ), then if x ∈ X fˆ(x) = fˆ(δx ) = δx (f ) = f (x) = pf (x) whereQpf : X̃ → Rf is the continuous projection over restricted to X̃ ⊂ f ∈L Rf . If L ⊂ C(X) then L∗ ⊂ L is the unital vector lattice of all the bounded functions of L, and so L ⊂ C ∗ (X). Example 9. By the last theorem it is obvious that H(C(X)) = υX and that H(C ∗(X)) = βX . Corollary 28. If L∗ ⊂ C ∗(X) is a unital vector lattice of continuous real-valued bounded functions separating points from closed sets then H(L∗ ) is a compactication of X . 40 Proof. By the last theorem H(L∗) ≈ X̃ = e(X) with the evaluation map e : X → RL . As The f ∈ L∗ are bounded then f (X) Q ⊂ If where If is a closed interval of R. So really we have that e : X → f ∈L∗ If ⊂ Q Q L∗ R = R . As X̃ is closed in ∗ f f ∈L f ∈L∗ If then it is compact. ∗ Theorem 29. H(L) is a topological subspace of H(L∗). Proof. Consider the restriction map r : H(L) → H(L∗) dened by r(ϕ) = ϕ | L∗ . We are going to see that r is a topological embedding, by means of which we shall consider H(L) as topological subspace of H(L∗ ). First note that if ϕ, ψ ∈ H(L) coincide on L∗ , then ϕ = ψ . Indeed, let f ∈ L and α, β ∈ R such that α < ϕ(f ) < β and α < ψ(f ) < β . Then for g = (f ∨ α) ∧ β ∈ L∗ , we have that ϕ(f ) = ϕ(g) = ψ(g) = ψ(f ), as required, This shows that the map r is one-to one. It is clear that r is continuous since it is the restriction of the natural projection ∗ from RL onto RL . In order to check that r : H(L) → H(L∗ ) is an open map, it is enough to observe that given f1 , ..., fn ∈ L and real numbers α1 < β1 , ..., αn < βn , we have {ϕ ∈ H(L) : αi < ϕ(fi ) < βi , i = 1, ..., n} = = {ϕ ∈ H(L) : αi < ϕ(gi ) < βi , i = 1, ..., n} where gI = (fi ∨ αi ) ∧ βi ∈ L∗ for i = 1, ..., n. Theorem 30. Let X be a Tychono space. Then any realcompactication is of the form H(L) with L ⊂ C(X) a unital vector sublattice separating points from closed sets of X . Proof. Let αX a realcompactication of X where α : X → αX is the embedding. Then if we put L = C(αX) | X Then H(L) is homeomorphic ,that is equivalent to αX . In fact L separates points from closed sets in X because αX is a Tychono space. Again the homomorphism is given by the next squares: α X → α(X) ,→ αX = α(X) 41 δ X → δX ,→ H(L) = δ(X) Corollary 31. Let X be a Tychono space. Then any realcompactication is of the form X̃ = e(X) ⊂ RL where L ⊂ C(X) separates points from closed sets in X and e : X → RL is the evaluation map given by e(x) = (f (x))f ∈L. Theorem 32. Taimanov extension theorem. Lat A a dense subspace of a topological space and f a continuous mapping of A to a compact space Y . The mapping f has a continuous extension over X i for every pair F1 , F2 of disjoint closed subsets of Y the inverse images f −1(F1) and f −1(F2) have disjoint closures in the space X . Remark 15. Please recall for the next theorem that if f ∈ L, we can regard it as a continuous function from X into R ∪ {∞} (the one-point compactication of R). Theorem 33. Every f ∈ L can be extended to a (unique) continuous function f ∗ : H(L∗) → R ∪ {∞}. Moreover, H(L) = {ϕ ∈ H(L∗ ) : f ∗ (ϕ) 6= ∞, ∀f ∈ L}. Proof. Let f : X → R ∪ {∞} be a function in L. By the Taimanov extension theorem, f can be extended to a (unique) f ∗ : H(L∗ ) → R ∪ {∞} i for every a < b < c < d ∈ R, the sets f −1 ([b, c]) and f −1 (R − (a, d)) have disjoint closures in H(L∗ ). Let h : R → R be the polygonal function dened by h(t) = 0 when t ∈ R−(a, d), h(t) = 1 for t ∈ [c, d], h(t) = (t − a)/(b − a) for t ∈ (a, b), and h(t) = (t − d)/(c − d) for t ∈ (c, d). Since L is a unital vector lattice , it is not dicult to check that L is closed under composition with a real polygonal functions, so h ◦ f ∈ L. In addition h ◦ f is bounded so by characterization of H(L∗ ) can be extended to a continuous real function on H(L∗ ). Note that this extension takes the value 1 on f −1 ([b, c]) and 0 on f −1 (R − (a, b), and therefore these sets have disjoint closure in H(L∗ ), as we required. Now, let ϕ ∈ H(L∗ ) be such that f ∗ (ϕ) 6= ∞ for all f ∈ L. Then ϕ can be extended to a homomorphism on L by dening f → f ∗ (ϕ). 42 This shows that {ϕ ∈ H(L∗ ) : f ∗ (ϕ) 6= ∞, ∀f ∈ L} ⊂ H(L). The converse inclusion is clear since for every f ∈ L, the functions f ∗ and fˆ are continuous extensions of f to H(L). Remark 16. By the two last theorems we can conclude that for every continuous function f : X → R admits a unique continuous extension f β : βX → R ∪ {∞}. Let L ⊂ C(X) be a unital vector lattice separating points from closed sets . Now, let δ : X → H(L∗ dened as previously. Then it has a continuous extension δ̃ : βX → H(L∗ ) that is a quotient map by compactness. For every f ∈ L, since f ∗ ◦ δ̃ and f β are continuous extensions of f to βX , by uniqueness, it is clear that f ∗ ◦ δ̃ = f β , Hence we can describe H(L∗ ) as the quotient space of βX given by the following equivalent relations: given ξ, η ∈ βX , δ̃(ξ) = δ̃(η) ⇔ f ∗ (δ̃)(ξ) = f ∗ (δ̃(η))∀f ∈ L∗ ⇔ f β (ξ) = f β (η))∀f ∈ L∗ ⇔ f β (ξ) = f β (η))∀f ∈ L where the last equivalence follows from the fact that, for every continuous real function f and every ξ ∈ βX , we have f β (ξ) = limn→∞ [f ∨ (−n) ∧ n]β (ξ). Theorem 34. For each ϕ ∈ H(L) there exists a ξ ∈ βX such that f β (ξ) 6= ∞ and ϕ(f ) = f β (ξ) for every f ∈ L. Proof. Given ϕ ∈ H(L) ⊂ H(L∗). choose ξ ∈ βX with δ̃(ξ) = ϕ. Then, for every f ∈ Γ since the functions f ∗ and fˆ conincide on H(L), it follows that ϕ(f ) = fˆ(ϕ) = f ∗ (ϕ) = f β (ξ). This means that f β (ξ) 6= ∞, and that the homomorphism ϕ is given by evaluation at the point ξ ∈ βX . 43 Now we are going to give a nal result for the characterization of realcompactications. If we collect the ponits in βX taht by evaluation, give rise to ahomomorphism on L, we obtain he space υL X = δ̃ −1 (H(L)) ⊂ βX . Then we have that υL X = {ξ ∈ βX : f β (ξ) 6= ∞∀f ∈ L}. Therefore, υL X is arealcompact space and we have X ⊂ υX ⊂ υL X ⊂ βX. Moreover, it is not dicult to check that the restricted map δ̃ | υL X → H(L) is closed by deniion of υL X , and therefore it is again a quotient map. So we teh following conmutative diagram: υL X ,→ βX δ̃ υL (X) → H(L) H(L) ,→ H(L∗ ) δ̃ βX → H(L∗ ). Finally we have our last characterization of the realcompactications of a Tychono space X . Theorem 35. Every realcompactication of a Tychono space is the quotient, for a certain L ⊂ C(X), of υL(X) by the next equivalent realtion: given f ∈ L and f β : βX → R ∪ {∞} its unique continuous extension, then ξ ∼ η i f β (ξ) = f β (η) ∀f ∈ L. 44 Chapter 5 Samuel Realcompactication of a metric space. 5.1 Samuel Realcompactication and Compactication. Until now we have done a study of the realcompactications only taking into the count the topology of X and now we want to study realcompactications for unifrom spaces. In particular for metric spaces. So considere the following results that can be found in [12]: Denition 24. A collection {fα : α ∈ A} of functions on X will be said to separates points in X i whenever x 6= y in X , then for some α ∈ A, fα (x) 6= fα (y). Theorem 36. Uniform Immersion's lemma . Let Xα be uniform spaces and fα : X → Xα , for each α ∈ A, a map dened on the Q uniform space X . Then the evaluation map e : X → α∈A Xα is a uniform embedding i the maps fα separates points and the uniformity on X is the weak uniformity generated by fα Recall that the weak uniformity generated by the maps fα : X → Xα is the weakest uniformity making each fα uniformly continuous. Let (X, d) be a metric space, and consider the uniformity generated by its metric, Ud . 45 Denition 25. Let (X, d) be a metric space and U C(X) the set of the uniformly continuous real-valued functions on (X, Ud). Then the Samuel Realcompactication of (X, d), srX , is the smallest realcompactication of it such that every uniformly continuous function f ∈ U C(X) can be extended to a continuous function f˜ : srX → R. By the last section we know that in fact srX is a realcompactication of (X, d) and srX = e(X) where e, the evaluation map, is a topological embedding of X in RU C(X) where e(x) = (f (x))f ∈U C(X) . We want to know if this topological embedding is really a uniform embedding of (X, Ud ) in RU C(X) with the usual uniformity. By the Uniform immersion's lemma we have to look for the weak uniformity generated by the set of functions U C(X)([5]). It is the uniformity U0 which has the subbase {(x, y) ∈ X × X : |f (x) − f (y)| < } where f ∈ U C(X) and > 0, and we want to know if it is equivalent to Ud . Let U0 = {(x, y) ∈ X × X : |f (x) − f (y)| < } for a certain f ∈ U C(X) and > 0. Then as f ∈ U C(X) then there exists a δ > 0 such that for every x, y ∈ X satisfying d(x, y) < δ then |f (x) − f (y)| < . So U = {(x, y) ∈ X × X : d(x, y) < δ} ⊂ U0 and U0 ∈ Ud . That is U0 is coarser that Ud . The converse is not satised in general. In fact let d be the discrete metric on a not countable space X , that is, d(x, x) = 0 and d(x, y) = 1 ∀x 6= y . Then every function f : (X, Ud ) → R is uniformly continuous. If U0 is ner than Ud then we will have the equivalent and then U0 will be metrizable. If U0 is metrizable then it has a numerable base for it. But this is impossible because it can be proved that it can't be nd a numerable sequence of f ∈ U C(X) that generates U0 . 46 So we conclude that e : (X, U0 ) → RU C(X) is a uniformly continuous embedding of X ∼ = e(X) in RU C(X) but that e : (X, Ud ) → RU C(X) is only a topological embedding with e a uniformly continuous function, because the inverse function e−1 : (e(X), URU C(X) | e(X) ) don't need to be a uniformly continuous function. Equivalently to the Samuel Realcompactication we have already dened in a previous chapter the Samuel Compactication of a metric space in the next way: Denition 26. Let (X, d) be a metric space and U C ∗(X) be the set of all the bounded uniformly continuous real-valued functions on (X, Ud). Then the Samuel Compactication of (X, d), sX , is the smallest compactication of (X, d) such that every f ∈ U C(X)∗ can be extended to a (bounded)continuous function f˜ : sX → R. Recall that the weak uniformity generated dy U C ∗ (X) is uniformly equivalent to the weak uniformity generated by U C(x), U0 . So have ∗ again the uniform embedding e : (X, U0 ) → RU C (X) We can construct the Samuel Compactication in other way. Let (X̃, Ũ0 ) be the uniform completion of (X, U0 ). Then it is uniformly isomorphic to the Samuel Compactication, because as it is compact then by uniqueness of the completion they have to coincide. Theorem 37. The next cases are equivalent: 1. srX = sx; 2. U C ∗(X) = U C(X) 47 5.2 Conclusion. Remember from the last chapter that if (X, d) is a metric space then its Samuel Realcompactication can be expressed as a quotient of a subspace of the Stone-ech compactication of (X, d): given f ∈ U C(X) and let f β : βX → R ∪ {∞} be its unique continuous extension, then the Samuel Realcompactication of the metric space (X, d) is srX = υU C(X) X/ ∼ where υU C(X) X = {ξ ∈ βX : f β (ξ) 6= ∞, ∀f ∈ U C(X)} ⊂ βX and ξ ∼ η i f β (ξ) = f β (η) ∀f ∈ U C(X). Now suppose that our metric space (X, d) has the property that every uniformly continuous function f ∈ U C(X) is bounded for the closed metric balls of the metric space then we have the nest result: Theorem 38. Let (X, d) be a metric space such that for every f ∈ U C(X) the image of the closed metric balls are bounded in R and let x0 ∈ X . Then υU C(X) X = ∞ [ clβX Bn (x0 ). n=1 Proof. ⇒Let ξ ∈ clβX Bn(x0) for some n ∈ N. For every f ∈ U C(X), f is bounded in Bn (x0 ) so f β (ξ) 6= ∞ and by denition of υU C(X) X , ξ ∈ υU C(X) X . ⇐ Let ξ ∈ υU C(X) X and consider f (y) = d(y, x0 ) ∈ U C(X). Since f β (ξ) 6= ∞ we can choose n > f β (ξ). Then, ξ ∈ clβX Bn (x0 ). Otherwise, there exists an open neighborhood V of ξ in βX such that V ∩ X ⊂ {x ∈ X : d(x, x0 ) > n}. Since ξ ∈ clβX V = clβX (V ∩ X), we have that f β (ξ) > n, which is a contradiction. Example 10. Now we give some examples of spaces that have the porperty that for every f ∈ U C(X) the image of the closed balls in (X, d) are bounded: 48 1. Proper metric spaces, that is, metric spaces for which their metric has the Heine-Borel Property. 2. Normed spaces. 3. More generally, length spaces. 4. For a normed space (X, k.k) then we can construct a metric space (X, d) with d(x, y) = kx − yk + |f (x) − f (y)| with f ∈ U C(X) and such that (X, d) is not necessarily a normed or a length space. So we are looking for metric space such that for every f ∈ U C(X) the image of the closed balls in (X, d) are bounded. Remember from the second section that in a metric space (X, d) we can characterize the bornology of the bounded sets in the sense of Bourbaki of (X, d), BBd(X) in the next way: Let (X, U) be a uniform space and B ⊂ X . Then each real valued uniformly continuos function f ∈ U C(X) is bounded on B i B is bounded in the sense of Bourbaki. So if we ask to a metric space (X, d) that the images of its closed metric balls by all the uniformly-continuos real-valued functions f ∈ U C(X) have to be bounded then, as the closed metric balls form a base for the metric bornology of (X, d), Bd X , this is equivalent to ask that Bd(X) = BBd(X). By the theorem of unifrom metrization in the second section we have the next corollary: 49 Corollary 39. Let BBd(X) the bornology of the bounded sets in the sense of Bourbaki of (X, d). The following conditions are equivalent: 1. BBd(X) = Bd0 (X) for some uniformly equivalent metric d0 on (X, d) 2. BBd(X) has a countable base {Bn : n ∈ N} such that ∃δ > 0 ∀n ∈ N, with Bn δ ⊆ Bn+1 where Bnδ = {x ∈ X : d(x, Bn ) < δn } = [ Bδ (x). x∈Bn Obviously BBd (X) = BBd0 (X) for every uniformly equivalent metric d0 in (X, d) so we have our problem result sumarized in the next theorem: Theorem 40. 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