Topological Preliminaries

2 Topological Preliminaries This chapter is motivated by the prominent role the topology plays in measure theory, and so in probability theory which may be viewed as a branch with specific problems and applications of measure theory. At the same time, comparing Chapter 2 with Chapter 3 shows the existence of certain similarities, that cannot be ignored, between some topological notions and some measure-theoretic notions, such as between topological space and measurable space, and between continuous function and measurable function. In this chapter we introduce and discuss most topological concepts we use in this book. Section 4 deals with methods of constructing topological spaces. In Section 5 and Section 6 we examine important properties of topological spaces that will be used throughout the following chapters. Section 5 is concerned with such properties that do not involve the notion of a metric, while Section 6 is devoted to metric spaces. The content of Section 6 includes some deep results such as Urysohn’s embedding theorem, the Stone-Weierstrass theorem and the Arzelà-Ascoli theorem.

4

Construction of Some Topological Spaces

In this section we introduce the notion of a topological space, and we present three important methods of constructing topological spaces. A usual method of constructing a topological space X involves assigning to each x ∈ X a family of subsets, satisfying certain properties, that will be the system of overneighborhoods of X . Another method consists in describing the topology generated by an arbitrary family of subsets. A third method we present here shows how one can build new topological spaces from others originally given. This method enables us to define the concepts of a relative topology and of a product topology. Definitions 4.1. A topology for X is any family T ⊂ P(X ) satisfying: (i) ∅ ∈ T and X ∈ T ; (ii) U, V ∈ T implies U ∩ V ∈ T ; (iii) U ⊂ T implies ∪U ∈ T . A pair (X, T ), where X is a set and T is a topology for X , is called a topological space. When no confusion seems possible, we will call X itself a topological space. The elements of T are called open sets (relative to T ), and the complements of the open sets are called closed sets (relative to T ). Examples 4.2. (a) Let X be a set. The family T = {∅, X } is a topology for X called the trivial topology for X . Analysis and Probability. http://dx.doi.org/10.1016/B978-0-12-401665-1.00002-3 © 2013 Elsevier Inc. All rights reserved.

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(b) The family P(X ) is a topology for X called the discrete topology for X . (c) Let T be the family of all sets U ⊂ R with the property that for each x ∈ U there is ε > 0 such that ]x − ε, x + ε [ ⊂ U . Then T is a topology for R called the usual topology for R. For a, b ∈ R such that a < b, the interval ]a, b[ is an open set, and for a, b ∈ R such that a < b, the interval [a, b] is a closed set. (d) Let T be the family of all sets U ⊂ C with the property that for each x ∈ U there is ε > 0 such that {y ∈ C : |x − y| < ε} ⊂ U . Then T is a topology for C called the usual topology for C. Obviously, the sets {y ∈ C : |x − y| < ε}, x ∈ C, ε > 0, are open relative to this topology. We will always assume that C is endowed with its usual topology unless the contrary is specifically stated. (e) The family U = {∅, ] − ∞, ∞]} ∪ {]a, ∞] : a ∈ R} is a topology for ] − ∞, ∞]. Remarks 4.3. (a) Let X be a topological space. According to De Morgan’s laws (2.17), it follows immediately that any union of a finite number of closed subsets of X is a closed set, and any intersection of closed subsets of X is a closed set; also, X and ∅ are closed sets. (b) Any intersection of topologies for X is also a topology for X . Before introducing and examining other concepts relative to the basic notion of an open set, we will present several methods of constructing a topological space. To do this, we need the next definitions. Definitions 4.4. Let X be a topological space and x ∈ X . A neighborhood of x is any open set U such that x ∈ U . An overneighborhood of x is any subset of X that contains a neighborhood of x. The family of all overneighborhoods of x is called the system of overneighborhoods of x. Theorem 4.5. Let X be a topological space, and let V(x) be the system of overneighborhoods of x ∈ X . Then: (i) x ∈ V for each V ∈ V(x); (ii) V1 , V2 ∈ V(x) implies V1 ∩ V2 ∈ V(x); (iii) V ∈ V(x) and V ⊂ W imply W ∈ V(x); (iv) if V ∈ V(x), then there is U ∈ V(x) such that U ⊂ V and U ∈ V(y) for each y ∈ U. Proof. Properties (i) and (iii) are obvious. If V1 , V2 ∈ V(x), then there are two open sets U1 and U2 such that x ∈ U1 ⊂ V1 and x ∈ U2 ⊂ V2 . Since U1 ∩ U2 is open and x ∈ U1 ∩ U2 ⊂ V1 ∩ V2 , it follows that V1 ∩ V2 ∈ V(x). This proves (ii). Since any open set is an overneighborhood of each of its points, assertion (iv) follows at once from the definition of an overneighborhood of x. Theorem (4.5) indicates what properties must satisfy a family of subsets associated to each point x of an arbitrary set X in order to form the system of overneighborhoods of x relative to some topology for X . As (4.7) will show, these properties are sufficient to construct a topology for X . First we will prove the next technical result. Theorem 4.6. Let (X, T1 ) and (X, T2 ) be topological spaces, and let Vi (x) be the system of overneighborhoods of x ∈ X relative to Ti , i = 1, 2. If V1 (x) = V2 (x) for any x ∈ X , then T1 = T2 .

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23

Proof. It suffices to prove that V1 (x) ⊂ V2 (x), x ∈ X , implies T1 ⊂ T2 . Let U ∈ T1 and x ∈ U . Then U ∈ V1 (x) and so U ∈ V2 (x). Therefore, there is Vx ∈ T2 such that x ∈ Vx ⊂ U . Since U = ∪x∈U Vx , it follows that U ∈ T2 . Theorem 4.7. Let X be a set, and let V be a mapping that assigns to each element x ∈ X a nonempty family V(x) ⊂ P(X ) satisfying properties (i)–(iv) of (4.5). Then there exists a unique topology for X such that the system of overneighborhoods of x coincides with V(x) for any x ∈ X . Proof. Let T = {U ∈ P(X ) : U ∈ V(x) for each x ∈ U }. Let us show that T is a topology for X . It is trivial that ∅ ∈ T . For any x ∈ X , (iii) implies that X ∈ V(x), and so X ∈ T . If U, V ∈ T and x ∈ U ∩ V , then (ii) implies that U ∩ V ∈ V(x), and so U ∩ V ∈ T . If U ⊂ T and x ∈ ∪ U, then there is U ∈ U such that U ∈ V(x). Hence (iii) implies that ∪ U ∈ V(x), and so ∪ U ∈ T . Consequently, T is a topology for X . Let O(x) be the system of overneighborhoods of x ∈ X relative to T . We show now that O(x) = V(x) for any x ∈ X . If V ∈ O(x), then there is U ∈ T such that x ∈ U ⊂ V , and so (iii) implies V ∈ V(x). Conversely, if V ∈ V(x), then, by (iv), there is U ∈ T such that x ∈ U ⊂ V , and so V ∈ O(x). By virtue of Theorem (4.6), T is the unique topology for X such that O(x) = V(x), x ∈ X . Example 4.8. For x = (x1 , . . . , xn ) ∈ R n and ε > 0, define Uε (x) = {(y1 , . . . , yn ) ∈ R n : |xi − yi | < ε for i = 1, . . . , n}. Let V(x) = {V ∈ P(R n ) : there is ε > 0 such that Uε (x) ⊂ V }, x ∈ R n . It is easily verified that V satisfies properties (4.5.i)– (4.5.iv), and so V defines a topology for R n called the usual topology for R n . From the proof of Theorem (4.7) it follows that a set U ⊂ R n is open relative to this topology if and only if for each x ∈ U there is ε > 0 such that Uε (x) ⊂ U . Particularly, the sets Uε (x), x ∈ R n , ε > 0, are open. We will always suppose that R n is equipped with its usual topology unless the contrary is specifically stated. Another class of topological spaces is the one in which the topologies are generated by families of subsets of given sets. The exact definition follows. Definition 4.9. For an arbitrary set X and a family S ⊂ P(X ), let τ (S) enote the intersection of all topologies for X that contain S. Obviously, τ (S) is a topology for X called the topology generated by S. τ (S) is the smallest topology for X containing S, that is τ (S) is included in each topology for X that contains S. The next theorem describes the topology generated by a family of subsets. Theorem 4.10. For S ⊂ P(X ) define B = {∩A : A ⊂ S and A is f inite}. Then τ (S) = {∪C : C ⊂ B}. Proof. Denote T = {∪C : C ⊂ B}. Since any topology for X containing S contains the family T , we have τ (S) ⊃ T . Therefore, to complete the proof, it will suffice to show that T is a topology for X such that S ⊂ T . Since ∅ ⊂ B, we get ∅ = ∪∅ ∈ T . Since ∅ ⊂ S and ∅ is finite, by the convention of (2.18), we have X = ∩∅ ∈ B, and so X ∈ T . If Ci ⊂ B, i = 1, 2, then (∪C1 ) ∩ (∪C2 ) = ∪C, where C = {C1 ∩ C2 : C1 ∈ C1 and C2 ∈ C2 } ⊂ B. Hence T satisfies (4.1.ii). Since the union of the sets

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of any subfamily of T is a union of sets belonging to B, property (4.1.iii) is verified. Consequently, T is a topology for X . Finally, it is clear that S ⊂ B ⊂ T . Example 4.11. Let S be the family of all intervals of the form [−∞, a[ or ]b, ∞], where a, b ∈ R. Then τ (S) is a topology for R called the usual topology for R. By virtue of Theorem (4.10), a subset of R is open relative to this topology if and only if it is a union of intervals of the form [−∞, a[, ]a, b[, or ]b, ∞], where a, b ∈ R. If not otherwise stated, we will assume throughout that R is endowed with its usual topology. Definitions 4.12. Let (X, T ) be a topological space. A family B ⊂ T is called a base for T if for any U ∈ T there is C ⊂ B such that U = ∪ C. We say that X has a countable base if there is a base for T which is a countable family. A family S ⊂ T is called a subbase for T if the family of all finite intersections of sets in S is a base for T . Examples 4.13. (a) Notation is as in (4.10). Then B is a base for τ (S), and S is a subbase for τ (S). (b) The family of all open intervals ]a, b[, where a, b ∈ R, is a base for the usual topology for R, and the family of all intervals of the form ] − ∞, a[ or ]b, ∞[, where a, b ∈ R, is a subbase for the usual topology for R. Remarks 4.14. (a) Let (X, T ) be a topological space, and let B be a base for T . Every family A such that B ⊂ A ⊂ T is a base for T too. Hence it follows at once that each base for T is a subbase for T . Therefore, there are several different bases and subbases for T . (Notice that B − {∅} and B ∪ {∅} are distinct bases for T .) (b) As Example (4.13.b) shows, not every subbase for a topology is a base for that topology. Also, it is easily seen that not every subfamily of a topology is a subbase for that topology, and so not every subfamily of a topology is a base for that topology. Furthermore, not every family of subsets of X can be a base for a topology for X . In contrast with Theorems (4.7) and (4.10), in what follows we construct new topological spaces from old ones. Theorem 4.15. Let (Y, T ) be a topological space, and let f : X → Y be a function. Then the family f −1 (T ) is a topology for X . Proof.

Applying (2.19.i) and (2.19.ii), this statement is immediate.

Example 4.16. Let (X, T ) be a topological space and A ⊂ X . Then A∩T = i −1 A (T ) is a topology for A called the relative topology on A induced by T . Thus a set U ⊂ A is open relative to this topology if and only if U = A ∩ V for some V ∈ T . We will always assume that a subset of a topological space is equipped with the relative topology on it. Theorem 4.17. τ ( f −1 (S)).

Let f : X → Y be a function and S ⊂ P(Y ). Then f −1 (τ (S)) =

Proof. By virtue of Theorem (4.15), f −1 (τ (S)) is a topology for X that contains f −1 (S), and so f −1 (τ (S)) ⊃ τ ( f −1 (S)). Now let U = {U ∈ τ (S) : f −1 (U ) ∈ τ ( f −1 (S))}. Clearly,

Topological Preliminaries

25

S ⊂ U ⊂ τ (S).

(1)

Since τ (S) and τ ( f −1 (S)) are topologies, by making use of (2.19.i) and (2.19.ii), it is easily verified that U is a topology for Y . From (1) we get τ (S) = U, and so f −1 (τ (S)) ⊂ τ ( f −1 (S)). Remark 4.18. It is visible that another proof of Theorem (4.17) may be given on account of Theorems (4.10) and (2.19). We have preferred the above proof, since it emphasizes a type of reasoning which is applicable in other contexts too (see (7.13)). Definition 4.19. Let {(Yi , Ti ) : i ∈ I } be an indexed family of topological spaces, and let f i : X → Yi , i ∈ I , be functions. The topology τ (∪i∈I f i−1 (Ti )) is a topology for X called the topology generated by { f i : i ∈ I }. Example 4.20. Let I be a nonempty set, and let {(X i , Ti ) : i ∈ I } be an indexed family of topological spaces. The topology generated  by {πi : i ∈ I } (see (2.23)) is a  X i called the product topology on i∈I X i . By virtue of Theorem topology for i∈I  (4.10), a subset of i∈I X i is open relative to this topology if and only if it is a union where Ui ∈ Ti , i ∈ I , and {i ∈ I : Ui = X i } is finite. of sets of the form i∈I Ui ,  We will always suppose that i∈I X i is endowed with the product topology unless the contrary is specifically stated. The following theorem has important applications. Theorem 4.21. Let I be a  nonempty set, let {(Yi , Ti ) : i ∈ I } be a family of topological spaces, and let f : X → i∈I Yi be a function. Let T denote the product topology  on i∈I Yi . Then f −1 (T ) coincides with the topology generated by {πi ◦ f : i ∈ I }, where πi stands for the projection from i∈I Yi onto Yi . Proof.

On account of (4.17), (2.38), and (2.6), we have        −1 −1 −1 −1 −1 f (T ) = τ f πi (Ti ) =τ f (πi (Ti )) i∈I

i∈I

   =τ (πi ◦ f )−1 (Ti ) . i∈I

Corollary 4.22. Let {(Yi , Ti ) : i ∈ I } be a nonempty family of topologicalspaces, and let f i : X → Yi , i ∈ I , be functions. Let T denote the product topology on i∈I Yi . −1 Then ( f i )i∈I (T ) coincides with the topology generated by { f i : i ∈ I }. Proof. As noted in (2.25), we have π j ◦ ( f i )i∈I = f j for each j ∈ I , and so the corollary follows from (4.21). Lemma 4.23. τ (∪a∈A Sa ).

Let X be a set, and let Sa ⊂ P(X ), a ∈ A. Then τ (∪a∈A τ (Sa )) =

Proof. ∪a∈A τ (Sa ) ⊃ ∪a∈A Sa , and so τ (∪a∈A τ (Sa )) ⊃ τ (∪a∈A Sa ). Conversely, ∪a∈A τ (Sa ) ⊂ τ (∪a∈A Sa ), and so τ (∪a∈A τ (Sa )) ⊂ τ (∪a∈A Sa ).

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The following theorem may be viewed as an associativity property of the product topology. Theorem 4.24. Let {(X i , Ti ) : i ∈ I } be a family of topological spaces, and let {Ja : a ∈ A} be a partition of I such that Ja = ∅, a ∈ A. For each a ∈ A, let Uadenote the product topology on j∈Ja X j . Let W be the product topology on a∈A ( j∈Ja X j ),  −1 and let T be the product topology on i∈I X i . Then (π Ja )a∈A (W) = T . −1 Proof. Corollary (4.22) shows that (π Ja )a∈A (W) = τ (∪a∈A π J−1 (Ua )). For any a ∈ a −1 −1 A, by virtue of (2.26) and (4.22), we get π Ja (Ua ) = τ (∪ j∈Ja π j (T j )). Therefore, applying (4.23), we have

⎛ −1 (W) = τ ⎝ (π Ja )a∈A



a∈A

⎛ τ⎝



⎞⎞

⎛

⎠⎠ = τ ⎝ π −1 j (T j )

j∈Ja



a∈A

=τ

 

⎛ ⎝



⎞⎞ ⎠⎠ π −1 j (T j )

j∈Ja

 πi−1 (Ti )

=T.

i∈I

Exercise 4.25. Let X be a set. Show that the family T = {∅} ∪ {A ∈ P(X ) : Ac is finite} is a topology for X . Exercise 4.26. Let S be the family of all closed intervals [a, b], where a, b ∈ R. Show that the topology for R generated by S is the discrete topology for R. Exercise 4.27. Prove that there is a countable base for a topology if and only if there is a countable subbase for that topology. Exercise 4.28. Let (X, T ) be a topological space and S ⊂ P(X ). Show that S is a subbase for T if and only if T = τ (S). Exercise 4.29. Let X = {0, 1, 2}, and let A = {{0, 1}, {1, 2}}. Show that A cannot be a base for a topology for X . Exercise 4.30. Notation is as in (4.16). A set B ⊂ A is closed with respect to the relative topology on A if and only if B = A ∩ F, where F is closed with respect to T . Exercise 4.31. Let (X, T ) be a topological space and B ⊂ A ⊂ X . Prove that the relative topology on B induced by T coincides with the relative topology on B induced by A ∩ T . Exercise 4.32. Prove that the relative topology on R induced by the usual topology for C is the usual topology for R. Exercise 4.33. Show that the relative topology on R induced by the usual topology for R is the usual topology for R. Exercise 4.34. Let α, β ∈ R be such that α < β. Use (4.28) and (4.17) to prove that the family of all intervals of the form [α, a[ or ]b, β], where α < a  β and α  b < β, is a subbase for the relative topology on [α, β] induced by the usual topology for R.

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Exercise 4.35. Notation is as in (4.20). Let I = {1, . . . , n} and let (X i , Ti ) = (R, T ), 1  i  n, where T is the usual topology for R. Show that the product topology on R n coincides with the usual topology for R n . Exercise 4.36. Notation is as in (4.20).  (a) Use (2.45.a) and (2.30.b) to prove that every section of an open subset of i∈I X i is open.  (b) Find a set A ⊂ i∈I X i such that every section of A is open, but A is not open. [Hint. Let (X, T ) be the topological space in (4.25), where X is infinite, and let (X i , Ti ) = (X, T ), i = 1, 2. Then every section of D is closed, but the diagonal D is not closed.] Exercise 4.37. Let {(X i , Ti ) : i ∈ I } be a nonempty family of topological spaces, relative and let Ai ⊂ X i , i ∈ I . For each i ∈ I , assume that Ai is equipped with the to prove that the relative topology on i∈I Ai topology on Ai induced by Ti . Use (4.21) coincides with the product topology on i∈I Ai . Exercise 4.38. Notation is as in (4.19). Assume that I is countable, and there is a countable base for Ti , i ∈ I . (a) Use (4.27), (4.28), (4.17) and (4.23) to show that there is a countable base for : i ∈ I }. the topology generated by { f i  (b) Apply (a) to prove that i∈I Yi has a countable base. Exercise 4.39. Let X be a set and x ∈ X . Show that the family T = {A ∈ P(X ) : Ac is finite or x ∈ Ac } is a topology for X .

5

General Properties of Topological Spaces

This section contains properties of topological spaces that will be needed in the sequel. Only those useful properties which do not involve the notion of a metrizable topological space are included here. The special class of topological spaces in which the topology is derived from a metric will be discussed in the next section. We begin with the notion of a limit of a sequence in a topological space. Definitions 5.1. Let X be a topological space, let {xn : n ∈ N } ⊂ X and x ∈ X . We say that {xn : n ∈ N } converges to x [has limit x] if for each neighborhood U of x there is n(U ) ∈ N such that n  n(U ) implies xn ∈ U . We will write limn xn = x [xn → x] if {xn : n ∈ N } converges to x; also, we will write xn  x if {xn : n ∈ N } does not converge to x. Let A be a set, and let f n : A → X, n ∈ N , f : A → X be functions. We say that the sequence { f n : n ∈ N } converges to f [has limit f ] on B ⊂ A if f n (b) → f (b) for any b ∈ B. We will write limn f n = f [ f n → f ] if { f n : n ∈ N } converges to f on A. Let ]a, b[ ⊂ R, t0 ∈ [a, b], and let f t : A → X, t ∈]a, b[, f : A → X be functions. We say that { f t : t ∈ ]a, b[} converges to f as t → t0 , and we write limt→t0 f t = f [ f t → f as t → t0 ], if f tn → f whenever {tn : n ∈ N } ⊂ ]a, b[ − {t0 } is such that tn → t0 .

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Remarks 5.2. (a) A sequence {xn : n ∈ N } converges to x if and only if any subsequence of {xn : n ∈ N } converges to x. (b) Let {x(n) : n ∈ N } = {(x1 (n), . . . , xm (n)) : n ∈ N } ⊂ R m and x = (x1 , . . . , xm ) ∈ R m . From (4.8) it follows that x(n) → x if and only if for each ε > 0 there is n(ε) ∈ N such that n  n(ε) implies |xi (n) − xi | < ε, i = 1, . . . , m. (c) Let (X, T ) be a topological space, where T is the trivial topology for X , and let {xn : n ∈ N } ⊂ X . Then xn → x for any x ∈ X . This shows that a sequence in a topological space may converge to several different points. Now we introduce a class of topological spaces in which the limit of a sequence is unique whenever it exists. Definition 5.3. A topological space is called a Hausdorff space if whenever x and y are distinct points of X there exist a neighborhood U of x and a neighborhood V of y such that U ∩ V = ∅. Example 5.4.

The space R n with its usual topology is a Hausdorff space.

Remark 5.5.

Each sequence in a Hausdorff space converges to at most one point.

Theorem 5.6. Let (X, T ) be a topological space, and let S be a subbase for T . For {xn : n ∈ N } ⊂ X and x ∈ X , the following assertions are equivalent: (i) xn → x; (ii) whatever V ∈ S with x ∈ V there is n(V ) ∈ N such that n  n(V ) implies xn ∈ V . Proof. It is obvious that (i) implies (ii). Suppose that (ii) holds, and let U be a neighborhood of x. Then there is a finite family A ⊂ S such that x ∈ ∩A ⊂ U . For any V ∈ A, let n(V ) ∈ N be such that n  n(V ) implies xn ∈ V . Since A is finite, there is n(U ) ∈ N (n(U )  n(V ), V ∈ A) such that n  n(U ) implies xn ∈ ∩A ⊂ U . Thus (ii) implies (i). Corollary 5.7.  Let {(X i , Ti ) : i ∈ I } be a nonempty family of topological spaces, with the product topology. For {xi (n)i∈I : n ∈ and assume that i∈I X i is equipped   N } ⊂ i∈I X i and (xi )i∈I ∈ i∈I X i , the following assertions are equivalent: (i) (xi (n))i∈I → (xi )i∈I ; (ii) xi (n) → xi for each i ∈ I . Proof. Assume that (i) holds, and consider U j ∈ T j such that x j ∈ U j . Since π −1 j (U j ) is a neighborhood of (x i )i∈I , there is n(U j ) ∈ N such that n  n(U j )

implies (xi (n))i∈I ∈ π −1 j (U j ). Therefore, n  n(U j ) implies x j (n) ∈ U j . Thus (ii) follows from (i). Suppose now that (ii) holds. For any j ∈ I , and any U j ∈ T j such that (xi )i∈I ∈ π −1 j (U j ), we have x j ∈ U j . Consequently, there exists

n(U j ) ∈ N such that n  n(U j ) implies x j (n) ∈ U j , and so (xi (n))i∈I ∈ π −1 j (U j ).  −1 Since ∪i∈I πi (T ) is a subbase for the product topology on i∈I X i , applying (5.6), we see that (ii) implies (i).

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In what follows we describe various important results which are related to the notion of a limit of a sequence in R. The reader is assumed to know that every bounded1 monotone sequence in R has a limit in R. From this it follows at once that every monotone sequence in R is convergent. Theorem 5.8. any A ⊂ R.

If  is the usual ordering on R, then there exist sup A and inf A for

Proof. If sup A exists for any A ⊂ R, then it is easily verified that − sup(−A) = inf A, where −A = {−x : x ∈ A}. Therefore, it suffices to prove that sup A exists for any A ⊂ R. If A = ∅ or A = {−∞}, then each element of R is an upper bound for A, and so −∞ = sup A. If ∅ = A ⊂ ] − ∞, ∞], then ∞ is an upper bound for A. In case ∞ is the only upper bound for A, then clearly ∞ = sup A. Thus the case when ∅ = A ⊂ R and there is an upper bound b ∈ R for A remains to be analyzed. In this case choose a ∈ R such that a < b and [a, b] ∩ A = ∅. Denote l = b − a. We define inductively a sequence of intervals {[an , bn ] : n  1} such that, whatever n  1, an  an+1 < bn+1  bn , bn − an = l/2n−1 ,

(1)

where bn is an upper bound for A, and [an , bn ] ∩ A = ∅. Put [a1 , b1 ] = [a, b]. Let [a2 , b2 ] be one of the intervals [a, (a + b)/2] and [(a + b)/2, b] such that b2 is an upper bound for A and [a2 , b2 ] ∩ A = ∅. Evidently, a1  a2 < b2  b1 and b2 − a2 = l/2. If the intervals [a1 , b1 ], . . . , [an , bn ] have been specified, then let [an+1 , bn+1 ] be one of the intervals [an , (an + bn )/2] and [(an + bn )/2, bn ] such that bn+1 is an upper bound for A and [an+1 , bn+1 ] ∩ A = ∅. It is clear that an  an+1 < bn+1  bn and bn+1 − an+1 = l/2n . By virtue of (1), there is α ∈ R such that an → α and bn → α. For each x ∈ A, we have x  bn , n  1, whence x  limn bn = α. Hence α is an upper bound for A. Now let β be an arbitrary upper bound for A. Since [an , bn ] ∩ A = ∅, n  1, it follows an  β, n  1, and so α = limn an  β. Consequently, α = sup A. Corollary 5.9.

X

Let X be a set, and let  be the partial ordering on R introduced X

in (3.19.b). Then there exist sup F and inf F for any F ⊂ R . Proof. It is easily verified that the function sup F : X → R defined by (sup F)(x) = sup{ f (x) : f ∈ F} for any x ∈ X is the supremum of F. Similarly, the function inf F : X → R defined by (inf F)(x) = inf{ f (x) : f ∈ F} for any x ∈ X is the infimum of F. Definitions 5.10. Let {xn : n ∈ N } ⊂ R. The number lim supn xn = inf n1 (supkn xk ) is called the limit superior of {xn : n ∈ N }, and the number lim inf n xn = supn1 (inf kn xk ) is called the limit inferior of {xn : n ∈ N }. X Let X be a set, and let { f n : n ∈ N } ⊂ R . The function lim supn f n = inf n1 (supkn f k ) is called the limit superior of { f n : n ∈ N }, and the function 1 A set A ⊂ R n is said to be bounded if there is k ∈ N such that A ⊂ U (0), where U (0) was defined in k k

(4.8).

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lim inf n f n = supn1 (inf kn f k ) is called the limit inferior of { f n : n ∈ N }. In accordance with (5.9), we have (lim supn f n )(x) = lim supn f n (x) and (lim inf n f n )(x) = lim inf n f n (x) for any x ∈ X . Theorem 5.11. Let {xn : n ∈ N } ⊂ R, and let L = {x ∈ R : limk xn k = x for some subsequence {xn k : k ∈ N } of {xn : n ∈ N }}. Then we have lim supn xn ∈ L , lim supn xn = sup L, and lim inf n xn ∈ L , lim inf n xn = inf L. Proof. We prove only the former assertion, the proof of the latter one being similar. Denote x = lim supn xn , and set yn = supkn xk , n  1, so that x = inf n1 yn . Therefore, for a < x, the set {n ∈ N : a < xn } is infinite, and for x < b, the set {n ∈ N : b  xn } is finite. If x = ∞, then there exists a subsequence {xn k : k ∈ N } of {xn : n ∈ N } such that xn k > k, k ∈ N , and so xn k → ∞. Actually, {n ∈ N : 1 < xn } is infinite, and so there is n 1 ∈ N such that xn 1 > 1. If n 1 , . . . , n k have been chosen, then, since {n ∈ N : k + 1 < xn } is infinite, there is n k+1 ∈ N ; such that n k+1 > n k and xn k+1 > k + 1. Consequently, x = ∞ ∈ L, and so x = sup L. If x ∈ R, then {n ∈ N : x − 1/k < xn < x + 1/k} is infinite for any k ∈ N . As above, we may choose

inductively a subsequence {xn k : k ∈ N } of {xn : n ∈ N } such that

xn − x < 1/k, k ∈ N , and so xn → x. Thus x ∈ L. Since {n ∈ N : b  xn } is k k finite for any b > x, it follows immediately that x = sup L. If x = −∞, then, whatever b ∈ R, there is n(b) ∈ N such that n  n(b) implies xn < b. Therefore, xn → −∞, and so {−∞} = L (see (5.2.a)). The set L in (5.11) is called the set of limit points of {xn : n ∈ N }. Corollary 5.12. A sequence {xn : n ∈ N } ⊂ R has limit x if and only if x = lim supn xn = lim inf n xn . Proof.

Exercise.

A useful property of topological spaces that have countable bases is presented in what follows. We begin with some definitions. Definitions 5.13. Let A be a set. A family of sets A is called a cover of A if A ⊂ ∪A. If A and B are covers of A such that B ⊂ A, then B is called a subcover of A. Let (X, T ) be a topological space and A ⊂ X . A cover A of A is said to be open if A⊂T. Theorem 5.14 (Lindelöf). Let (X, T ) be a topological space that has a countable base, and let A ⊂ X . Then any open cover of A contains a countable subcover of A. Proof. Let A be a open cover of A, and let B be a countable base for T . Since any set in A is a union of sets in B, there is a cover C ⊂ B of A such that each set in C is included in some set in A. For every C ∈ C, we choose DC ∈ A such that C ⊂ DC . Denote D = {DC : C ∈ C}. By virtue of (3.14) and (3.38), the family D is countable. Since C ⊂ D ⊂ A, D is a subcover of A. Another interesting property of topological spaces that have countable bases will be proved in (5.20). For this we proceed to new definitions. Definitions 5.15. Let X be a topological space and A ⊂ X . Let A− denote the intersection of all closed subsets of X containing A. In accordance with (4.3), A− is a

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closed set containing A which is called the adherence [closure] of A. Obviously, A− is the smallest closed set containing A, and A is closed if and only if A = A− . The set ∂(A) = A− ∩ (Ac )− is called the boundary of A. A point x ∈ X is said to be adherent to A if A ∩ U = ∅ for any neighborhood U of x. Examples 5.16. (a) Let (X, T ) be a topological space, where T is the discrete topology for X . Then A− = A, ∂(A) = ∅, and {x : x is adherent to A} = A for any A ⊂ X. (b) Let I ⊂ R be an interval of the form ]a, b[, [a, b[, or ]a, b]. Then I − = [a, b]. Particularly, R − = R. (c) Relative to the usual topology for R, ∂(] − ∞, a[) = ∂(] − ∞, a]) = {a} for any a ∈ R. (d) We assume the reader knows that for each x ∈ R there exists {xn : n ∈ N } ⊂ Q such that xn → x. From this it follows at once that {x ∈ R : x is adherent to Q} = R. Theorem 5.17. Let X be a topological space and A ⊂ X . Then A− = {x : x is adherent to A} Proof. Denote B = {x : x is adherent to A}. If x ∈ (A− )c , then, since (A− )c is a neighborhood of x and (A− )c ∩ A = ∅, we have x ∈ B c . Thus A− ⊃ B. Conversely, for each x ∈ B c we choose a neighborhood Ux of x such that Ux ⊂ Ac . Since B c = ∪x∈B c Ux ⊂ Ac , B is a closed set containing A, and so B ⊃ A− . Definitions 5.18. Let X be a topological space. A set A ⊂ X is said to be dense in X if A− = X . We say that X is separable if there is a countable subset dense in X . Examples 5.19. (a) Let (X, T ) be a topological space, where X is an uncountable set and T is the discrete topology for X . In accordance with (5.16.a), for each countable set A ⊂ X , we have A− = A = X , and so X is not separable. (b) The space R with its usual topology is separable. Actually, according to (3.41), Q is a denumerable set, and from (5.17) and (5.16.d) it follows that Q − = R. Analogously, using (5.17) and (5.7), it follows that (Q n )− = R n . By virtue of (3.42), Q n is denumerable, and so R n is separable. (c) By making use of (3.11), it is easily seen that the topological space (X, T ) in (4.25) is separable. Theorem 5.20. Any topological space that has a countable base is separable. Proof. Let B be a countable base of the topological space X . Then, in accordance with (4.14.a), the family B = B − {∅} is a countable base of X formed from nonempty sets. For each B ∈ B select x B ∈ B. By virtue of (3.38), the set A = {x B : B ∈ B } is countable. Moreover, A ∩ U = ∅ for any open nonempty set U , and so (5.17) shows that A is dense in X . Remark 5.21. There exist separable spaces that have no countable bases. For example, let (X, T ) be the topological space in (4.25), where X is an uncountable set. As stated in (5.19.c), X is separable. Assume that there is a countable base B for T . For a fixed point x ∈ X , let Ux denote the family of all neighborhoods of x, and put Bx = {B ∈ B : x ∈ B}. Then it is clear that ∩Bx = ∩ Ux = {x}. From this it follows

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that (∩Bx )c is an uncountable set. On the other hand, in view of (2.17) and (3.17), the set (∩Bx )c is countable. This contradiction shows that X has no countable base. Now we introduce a very important class of topological spaces. Definitions 5.22. A topological space (X, T ) is said to be compact if each open cover of X contains a finite subcover of X . A set A ⊂ X is said to be compact if (A, A ∩ T ) is compact. By virtue of (4.16), A is compact if and only if every cover of A by sets in T has a finite subcover of A. Before examining some useful properties of topological spaces, in (5.24) and (5.25) we present two characterizations of compactness. Definition 5.23. A family of sets A has the finite intersection property if ∩B = ∅ for each finite family B ⊂ A. Theorem 5.24. A topological space X is compact if and only if ∩F = ∅ for any family F of closed subsets of X which has the finite intersection property. Proof. According to De Morgan’s laws (2.17), A is an open cover of X if and only if B = {Ac : A ∈ A} is a family of closed sets such that ∩B = ∅. Therefore, X is compact if and only if each family B of closed sets having the property ∩B = ∅ contains a finite subfamily C such that ∩C = ∅, and so if and only if each family F of closed sets having the finite intersection property is such that ∩F = ∅. Alexander’s theorem 5.25. Let (X, T ) be a topological space, and let S be a subbase for T . Then the following assertions are equivalent: (i) X is compact; (ii) each cover of X by sets in S contains a finite subcover of X . Proof. It is clear that (i) implies (ii). Assume now that (ii) holds and X is not compact. Then the collection A of all open covers of X that do not admit finite subcovers is a nonempty collection which is partially ordered by inclusion. The union of the elements of any chain in A is obviously an element of A. Applying (3.49), Zorn’s lemma (3.31) shows that A has a maximal element A. Set B = A ∩ S. Since A does not contain finite subcovers of X , it follows that B does not admit finite subcovers of X . Thus, in view of (ii), ∪B = X . For a point x ∈ X − (∪B), choose a set A in the cover A such that x ∈ A. Since S is a subbase, there are n ∈ N and S1 , . . . , Sn ∈ S such that n S ⊂ A. Since x ∈ / ∪B, we have Si ∈ / A, i = 1, . . . , n. Hence, A being a x ∈ ∩i=1 i maximal cover, the cover A ∪ {Si } admits a finite subcover of X for each i. Therefore, for any i = 1, . . . , n, there exists a set Ai , which is a union of a finite number of sets n A ) ∪ A ⊃ (∪n A ) ∪ (∩n S ) = X , it in A, such that Ai ∪ Si = X . Since (∪i=1 i i=1 i i=1 i follows that X is covered by a finite number of sets in A. This contradicts the assertion that A ∈A. We now examine several useful properties of compact spaces. Theorem 5.26.

Any closed subset of a compact space is compact.

Proof. Let X be a compact space, let A ⊂ X be a closed set, and let A be an open cover of A. Then A∪{Ac } is an open cover of X . Since X is compact, A∪{Ac } contains

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33

a finite subfamily B which covers X . Then B − {Ac } is a finite subfamily of A that covers A, and so A is compact. Theorem 5.27.

Any compact subset of a Hausdorff space is closed.

Proof. Let X be a Hausdorff space, let A ⊂ X be a compact set, and let y ∈ Ac . Since X is Hausdorff, for each x ∈ A, there exist a neighborhood Ux of x and a neighborhood Vx of y such that Ux ∩ Vx = ∅.

(1)

The family {Ux : x ∈ A} is plainly an open cover of A. Since A is compact, there are n U . Denote V (y) = ∩n V . Using n ∈ N and x1 , . . . , xn ∈ A such that A ⊂ ∪i=1 xi i=1 xi (1), we get A ∩ V (y) = ∅, and so V (y) is an open set such that V (y) ⊂ Ac . Hence Ac = ∪ y∈Ac V (y) , and so A is closed. Tihonov’s theorem 5.28.  If {(X i , Ti ) : i ∈ I } is a nonempty family of compact topological spaces, then i∈I X i is a compact topological space.  −1 −1 Proof. Let A be a cover of i∈I X i such  that A ⊂ ∪i∈I πi (Ti ). Since ∪i∈I πi (Ti ) is a subbase for the product topology on i∈I X i , applying Alexander’s theorem (5.25), it suffices to show that A contains a finite subcover of i∈I X i . Let Ai = {A ∈ Ti : πi−1 (A) ∈ A}, i ∈ I . Suppose that ∪Ai = X i for any  i ∈ I and select xi ∈ / ∪A, that is A is not a cover of i∈I X i . This contradiction X i −(∪Ai ). Then (xi )i∈I ∈ shows that there is j ∈ I such that ∪A j = X j . Since X j is compact, there are −1 A1 , . . . , An ∈ A j such that ∪nk=1 Ak = X j . Consequently, {π −1 j (A1 ), . . . , π j (An )}  is a finite subfamily of A which covers i∈I X i . We will use Tihonov’s theorem to characterize the compact subsets of R n . We first prove the next lemma. Lemma 5.29.

Any closed interval [a, b] in R is compact.

Proof. Let A be an open cover of [a, b], and let c = sup{x ∈ [a, b] : [a, x] is covered by a finite subfamily of A}. Since any open set containing a contains an interval [a, a + ε] too, it follows that a < c  b. Choose U ∈ A such that c ∈ U , and choose d ∈]a, c[ such that [d, c] ⊂ U . By the definition of supremum, there is a finite family B ⊂ A such that [a, d] ⊂ ∪B. Therefore, B ∪ {U } is a finite subfamily of A which covers [a, c]. Since there is an interval [c, c + ε] ⊂ U , it follows that c = b. Thus B ∪ {U } covers [a, b]. Theorem 5.30 (Heine-Borel-Lebesgue). Let A ⊂ R n . Then A is compact if and only if A is closed and bounded. Proof. Assume that A is compact. Since R n is a Hausdorff space, applying (5.27), it follows that A is closed. Obviously, A ⊂ ∪ j∈N U j (0) = R n . Since A is compact, there is k ∈ N such that A ⊂ Uk (0), that is A is bounded. Conversely, assume that A is closed and bounded. Since A is bounded, there is k ∈ N such that A ⊂ Uk (0) ⊂ [−k, k]n . Applying (5.29), using Tihonov’s theorem (5.28), and taking into account (4.37), it follows that [−k, k]n is a compact subset of R n .

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In accordance with (4.30), A is closed with respect to the relative topology on [−k, k]n . Thus, by virtue of (5.26), A is a compact subset of [−k, k]n . In view of (4.31), A is a compact subset of R n . A useful version of compactness will be introduced in what follows. We need the next definitions. Definitions 5.31. Let X be a topological space and A ⊂ X . A point x ∈ X is a limit point of A if A ∩ (U − {x}) = ∅ for each neighborhood U of x. Let Ad denote the set of all limit points of A. Ad is sometimes called the derived set of A. Obviously, each limit point of A is adherent to A, but a point adherent to a set may fail to be a limit point of that set. Examples 5.32. (a) Let (X, T ) be a topological space, where T is the discrete topology for X . Then Ad = ∅ for any A ⊂ X . (b) Let X be a Hausdorff space. Then Ad = ∅ for any finite set A ⊂ X . (c) Relative to the usual topology for R, we have Z d = ∅, {1/n : n ∈ N }d = {0}, and I d = I − for any interval I ⊂ R. Theorem 5.33. only if Ad ⊂ A.

Let X be a topological space and A ⊂ X . Then A is closed if and

Proof. If A is closed, then, using (5.17), we have Ad ⊂ A− = A. Conversely, assume that Ad ⊂ A. Since A− − A ⊂ Ad ⊂ A, it follows that − A − A = ∅. Therefore, A = A− , and so A is closed. Definitions 5.34. A topological space (X, T ) is said to be Fréchet compact if Ad = ∅ for any infinite set A ⊂ X . A set B ⊂ X is said to be Fréchet compact if (B, B ∩ T ) is Fréchet compact. Obviously, B is Fréchet compact if and only if every infinite subset of B has a limit point in B relative to T . Theorem 5.35.

Every compact topological space is Fréchet compact.

Proof. Let X be a compact space. Assume that there is an infinite set A ⊂ X such that Ad = ∅. Then, according to (5.33), A is closed. Moreover, for each a ∈ A, there exists a neighborhood Ua of a such that A ∩ Ua = {a}. Therefore, {Ua : a ∈ A} ∪ {Ac } is an open cover of X which contains no finite subcover of X . The contradiction we get shows that X is Fréchet compact. The concept of continuity we will now discuss is extremely important. Definitions 5.36. Let X and Y be topological spaces, and let f : X → Y be a function. We say that f is continuous at x ∈ X if for each neighborhood V of f (x) there is a neighborhood U of x such that f (U ) ⊂ V . We say that f is continuous on X if f is continuous at every x ∈ X . Theorem 5.37. Let X, Y and f be as in (5.36), let x ∈ X , and let {xn : n ∈ N } ⊂ X be such that xn → x. If f is continuous at x, then f (xn ) → f (x).

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Proof. Consider a neighborhood V of f (x), and choose a neighborhood U of x such that f (U ) ⊂ V . Since xn → x, there is n(U ) ∈ N such that n  n(U ) implies xn ∈ U . Therefore, n  n(U ) implies f (xn ) ∈ f (U ) ⊂ V . Thus f (xn ) → f (x). Theorem 5.38. Let X, Y and Z be topological spaces, let f : X → Y and g : Y → Z be functions, and let x ∈ X . If f is continuous at x, and g is continuous at f (x), then g ◦ f is continuous at x. Proof. Let W be a neighborhood of (g ◦ f )(x) = g( f (x)). Since g is continuous at f (x), there is a neighborhood V of f (x) such that g(V ) ⊂ W . Since f is continuous at x, there is a neighborhood U of x such that f (U ) ⊂ V . Hence (g ◦ f )(U ) = g( f (U )) ⊂ g(V ) ⊂ W , and so g ◦ f is continuous at x. Corollary 5.39. Let X, Y, Z , f and g be as in (5.38). If f is continuous on X , and g is continuous on Y , then g ◦ f is continuous on X . Theorem 5.40. Let (X, T ) and (Y, U) be topological spaces, and let f : X → Y be a function. Then f is continuous on X if and only if f −1 (U) ⊂ T . Proof. Assume that f is continuous on X . Let V ∈ U and x ∈ f −1 (V ). Since f is continuous at x, there is a neighborhood Ux of x such that f (Ux ) ⊂ V . Hence, by (2.33.c), Ux ⊂ f −1 (V ). Therefore, f −1 (V ) = ∪x∈ f −1 (V ) Ux ∈ T . Thus f −1 (U) ⊂ T . Conversely, suppose that f −1 (U) ⊂ T . Then, for any x ∈ X and any neighborhood V of f (x), f −1 (V ) is a neighborhood of x. According to (2.33.b), f ( f −1 (V )) ⊂ V , and so f is continuous at x. Corollary 5.41. Notation is as in (4.19). Then the topology generated by { f i : i ∈ I } is the smallest topology for X relative to which f i , i ∈ I , are continuous on X . Corollary 5.42. Let (X, T ), (Y, U) and f be as in (5.40), and let S be a subbase for U. Then f is continuous on X if and only if f −1 (S) ⊂ T . Proof. If f is continuous on X , then f −1 (S) ⊂ f −1 (U) ⊂ T . Conversely, if f −1 (S) ⊂ T , then, using (4.28) and (4.17), we have f −1 (U) = f −1 (τ (S)) = τ ( f −1 (S)) ⊂ T , and so f is continuous on X . Remarks 5.43. (a) Let (X, T ), (Y, U) and f be as in (5.40), and let A ⊂ X . Assume that f is continuous on X . Then the restriction of f to A is continuous on A, where A is endowed with the relative topology A ∩ T . Actually, using (2.31), (2.6) and (5.40), −1 (U)) ⊂ i −1 (T ). we have ( f ◦ i A )−1 (U) = i −1 A (f A (b) Let (X, T ), (Y, U) and f be as in (5.40), and let f (X ) ⊂ B ⊂ Y . Suppose that f is continuous on X . Then the function g : X → B defined by g(x) = f (x) for any x ∈ X is continuous on X , where B is equipped with the relative topology B ∩ U. Actually, i B ◦ g = f , and so g −1 (i B−1 (U)) = f −1 (U) ⊂ T . ∈ I } be a nonempty family of topological spaces, and Theorem 5.44. Let {(Yi , Ui ) : i  let U be the product topology on i∈I Yi . Let (X, T ) be a topological space, and let  f : X → i∈I Yi be a function. Then f is continuous on X if and only if πi ◦ f, i ∈ I , are continuous on X .

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 Proof. According to (5.41), U is the smallest  topology for i∈I Yi relative to which the projections πi , i ∈ I , are continuous on i∈I Yi . Consequently, if f is continuous on X , then, applying (5.39), the functions πi ◦ f, i ∈ I are continuous on X . Conversely, assume that T is a topology for X relative to which πi ◦ f, i ∈ I , are continuous on X . Then, using (4.21) and (5.41), we have f −1 (U) ⊂ T , and so f is continuous on X . Theorem 5.45. Let X, Y and f be as in (5.36). If X is compact, and f is continuous on X , then f (X ) is compact. Proof. Let B be an open cover of f (X ). Then f −1 (B) is an open cover of X . Since X is compact, there is a finite family C ⊂ B such that X = ∪ f −1 (C) = f −1 (∪C). Therefore, f (X ) ⊂ ∪C, and so f (X ) is compact. Definitions 5.46. A set A ⊂ K is said to be bounded if there exists M > 0 such that |a|  M for any a ∈ A. A function f : X → K is said to be bounded if f (X ) is bounded. Theorem 5.47. Let X be a compact space, and let f : X → R be a function which is continuous on X . Then f is bounded, and there exist a, b ∈ X such that f (a) = supx∈X f (x) and f (b) = inf x∈X f (x). Proof. By virtue of (5.45), f (X ) is a compact subset of R. Thus, according to (5.30), f (X ) is bounded and closed. Therefore, f is bounded. Denote α = sup f (X ) and β = inf f (X ). Since f (X ) is closed, we have α, β ∈ f (X ). Consequently, there are a, b ∈ R such that f (a) = α and f (b) = β. The following notions play an important role in studying topological spaces. Definitions 5.48. Let X and Y be topological spaces. A function f : X → Y is called a homeomorphism from X onto Y if f is bijective, f is continuous on X , and the inverse function f −1 is continuous on Y . We say that X and Y are homeomorphic if there exists a homeomorphism from X onto Y . Example 5.49. Let a, b ∈ R be such that a < b. Then ]a, b[ and R are homeomorphic. Actually, it is easily verified that the function f : ]a, b[→ R defined by f (x) = 1 1 x−a + x−b for any x ∈ ]a, b[ is a homeomorphism from ]a, b[ onto R. Remarks 5.50. (a) Let X, Y and Z be topological spaces. Then the identity function i X is a homeomorphism from X onto X . If f is a homeomorphism from X onto Y , then f −1 is a homeomorphism from Y onto X . If f is a homeomorphism from X onto Y , and g is a homeomorphism from Y onto Z , then g ◦ f is a homeomorphism from X onto Z . (b) If X and Y are homeomorphic topological spaces, then to each statement concerning X corresponds a statement concerning Y . For example, X has a countable base if and only if Y has a countable base; also, X is separable if and only if Y is separable, and X is compact if and only if Y is compact. Definition 5.51. Let X and Y be topological spaces. A function f : X → Y is called a homeomorphism from X into Y if the function g : X → f (X ) defined by g(x) = f (x) for any x ∈ X is a homeomorphism from X onto f (X ), where f (X ) is endowed with

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the relative topology. According to (5.48) and (5.43.b), f is a homeomorphism from X ˙ into Y if f is injective, f is continuous on X , and the inverse of g is continuous on f (X ). Exercise 5.52. Let (X, T ) be a topological space, let {xn : n ∈ N } ⊂ A and x ∈ A. Prove that xn → x relative to T if and only if xn → x relative to A ∩ T . Exercise 5.53. (a) Let {xn : n ∈ N } be a sequence, and let {N1 , . . . , Nm } be a partition of N such that Ni , 1  i  m, are denumerable. Show that {xn : n ∈ N } converges to x if and only if each subsequence {xn : n ∈ Ni }, 1  i  m, has limit x. (b) Construct a sequence {xn : n ∈ N } and a denumerable partition {Ni : i  1} of N such that Ni , i  1, are denumerable, each subsequence {xn : n ∈ Ni } has limit x, but {xn : n ∈ N } fails to converge to x. Exercise 5.54. A sequence {xn : n ∈ N } converges to x if and only if every subsequence {xn k : k ∈ N } contains a further subsequence which converges to x. Exercise 5.55.

Show that any finite subset of a Hausdorff space is closed.

Exercise 5.56. Let X be a topological space, and let D be the diagonal of the Cartesian product X × X . Prove that X is a Hausdorff space if and only if D is closed. Exercise  5.57. Let {X i : i ∈ I } be a nonempty family of topological spaces. Show that i∈I X i is a Hausdorff space if and only if each X i , i ∈ I , is a Hausdorff space. Exercise 5.58. Use (3.50) to prove the following assertions. (a) If A = ∪i∈I Ai ⊂ R, then sup A = supi∈I (sup Ai ) and inf A = inf i∈I (inf Ai ). X

(b) If F = ∪i∈I Fi ⊂ R , then sup F = supi∈I (sup Fi ) and inf F = inf i∈I (inf Fi ). Exercise 5.59. Let f i : X → R, i ∈ I , and h :  → X be functions. Show that (supi∈I f i ) ◦ h = supi∈I ( f i ◦ h) and (inf i∈I f i ) ◦ h = inf i∈I ( f i ◦ h). Exercise 5.60. Let {xn : n ∈ N }, {yn : n ∈ N } and {z n : n ∈ N } be sequences in R such that xn  yn , n ∈ N , and limn z n = 1. Prove the following. (a) supn1 xn  supn1 yn , inf n1 xn  inf n1 yn . (b) lim supn xn  lim supn yn , lim inf n xn lim inf n yn . (c) inf n1 xn  lim inf n xn  lim supn xn  supn1 xn . (d) lim supn xn z n = lim supn xn , lim inf n xn z n = lim inf n xn . Exercise 5.61. Let {xn : n ∈ N } ⊂ R. (a) If {xn : n ∈ N } is nondecreasing, then limn xn = supn1 xn . (b) If {xn : n ∈ N } is nonincreasing, then limn xn = inf n1 xn . (c) lim supn (−xn ) = − lim inf n xn , lim inf n (−xn ) = − lim supn xn . Exercise 5.62. Let amn ∈ R, m, n ∈ N , be such that, for each m ∈ N , the sequence {amn : n ∈ N } is nondecreasing, and for each n ∈ N , the sequence {amn : m ∈ N } is nondecreasing. Prove that limm limn amn = limn limm amn . X

Exercise 5.63. Let X be a set, let { f n : n ∈ N } ⊂ R , and let a ∈ R. Prove the following. (a) {x ∈ X : { f n (x) : n ∈ N } has limit} = {x ∈ X : (lim supn f n )(x) = (lim inf n f n )(x)}.

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Analysis and Probability

(b) {lim supn f n > a} ⊂ lim supn { f n > a} ⊂ lim supn { f n  a} ⊂ {lim supn f n  a}. (c) {lim inf n f n < a} ⊂ lim supn { f n < a} ⊂ lim supn { f n  a} ⊂ {lim inf n f n  a}. [Use (b) and (5.61.c).] Exercise 5.64. (a) Let A be a family of sets. Show that 1∪A = sup A∈A 1 A and 1∩A = inf A∈A 1 A . (b) Let {An : n  1} be a sequence of sets. Prove that 1lim inf n An = lim inf n 1 An and 1lim supn An = lim supn 1 An . Exercise 5.65. Let (X, T ) be a topological space and B ⊂ A ⊂ X . Use (4.30) to show that the adherence of B relative to A∩T is the intersection of A and the adherence of B relative to T . Exercise 5.66. For subsets A and B of a topological space, prove: (a) (A− )− = A− ; (b) if A ⊂ B, then A− ⊂ B − ; (c) (A ∪ B)− = A− ∪ B − ; (d) (A ∩ B)− ⊂ A− ∩ B − ; (e) ∂(∂(A)) = ∅; (f) ∂(A ∪ B) ⊂ ∂(A) ∪ ∂(B); (g) ∂(A ∩ B) ⊂ ∂(A) ∪ ∂(B). Exercise: Interior of a set 5.67. Let X be a topological space and A ⊂ X . The union of all open subsets of A, written A◦ , is called the interior of A. For A, B ⊂ X , prove: (a) A◦ is the largest open subset of A, and A is open if and only if A = A◦ ; (b) (A ∪ B)◦ ⊃ A◦ ∪ B ◦ ; (c) (A ∩ B)◦ = A◦ ∩ B ◦ ; (d) (A◦ )c = (Ac )− ; (e) ∂(A) = A− − A◦ . Exercise 5.68. Let X and Y be topological spaces, and let A ⊂ X, B ⊂ Y . Prove the following. (a) (A × B)− = A− × B − . (b) ∂(A × B) = (∂(A) × B − ) ∪ (A− × ∂(B)). (c) (A × B)◦ = A◦ × B ◦ . Exercise 5.69. Use (3.14) and (3.12) to show that a subset A of R is dense in R whenever Ac is countable. Exercise 5.70. Let U = ∅ be an open subset of R. Prove that there exists a unique partition of U formed from open intervals, and show that this partition is countable. The following steps may be helpful. (a) For x ∈ U , put ax = inf{a : ]a, x] ⊂ U } and bx = sup{b : [x, b[ ⊂ U }. Then ]ax , bx [ ⊂ U . / U and bx ∈ / U. (b) For x ∈ U, ax ∈ (c) Let I = {]ax , bx [: x ∈ U }. Then any two intervals in I are either identical or disjoint. [Use (b).] (d) ∪I = U .

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39

(e) If J is a partition of U formed from open intervals, then I = J . (f) I is countable. [Use (5.16.d) to assign to each I ∈ J a point x I ∈ Q ∩ I . Show that the function f : I → Q defined by f (I ) = x I , I ∈ I, is injective.] Exercise 5.71. Let I be a nonempty set, and let {X i : i ∈ I } be a family of topological spaces. Prove  the following statements. (a) If i∈I X i is separable, then each X i , i ∈ I , is separable. (b) If I is countable, and X i is separable for every i ∈ I , then i∈I X i is separable. [Hints. For i ∈ I , let Ai be a countable subset dense in X i . Fix a point (yi )i∈I ∈  X . Then {(xi )i∈I : there is a finite set J ⊂ I such that x j ∈  A j , j ∈ J, and xi = i i∈I / J } is a countable set ((3.42), (3.40) and (3.17)) dense in i∈I X i .] yi , i ∈ Exercise 5.72. compact.

If A and B are compact subsets of a topological space, then A ∪ B is

Exercise 5.73.

Use (5.24) to show that R is not a compact space.

Exercise 5.74. is compact.

Use (4.11) and (5.29) to prove that the space R with its usual topology

Exercise 5.75. Let X be a topological space, and let f : X → R be a continuous function on X . Prove the following. (a) The function α f is continuous on X for any α ∈ R. (b) The function | f | is continuous on X . (c) If f (x) = 0 for each x ∈ X , then the function 1f is continuous on X . (d) For any a ∈ R, the sets {x : f (x) < a} and {x : f (x) > a} are open. Exercise 5.76. Let X be a topological space, and let f : X → R and g : X → R be continuous functions on X . (a) Show that the functions f + g and f g are continuous on X . (b) Prove that f ∨ g = ( f + g + | f − g|)/2 and f ∧ g = ( f + g − | f − g|)/2, and infer that the functions f ∨ g and f ∧ g are continuous on X . (c) Show that the set {x : f (x) < g(x)} is open. Exercise 5.77. Prove the following. (a) The function f : R 2 → R defined by f (x, y) = x + y for any x, y ∈ R is continuous on R 2 . (b) The function g : R 2 → R defined by g(x, y) = x − y for any x, y ∈ R is continuous on R 2 . (c) The function h : R 2 → R defined by h(x, y) = x y for any x, y ∈ R is continuous on R 2 . Exercise 5.78. Notation is as in (5.40). Then f is continuous on X if and only if f −1 (B) is closed whenever B is a closed subset of Y . be a nonempty family of topological spaces, let Y Exercise 5.79. Let {X i : i ∈ I }  be a topological space, and letf : i∈I X i → Y be a function. (a) If f is continuous on i∈I X i , then every section of f is continuous. [Use (5.40), (2.45.f ) and (3.46.a).] (b) Show that the converse of the statement (a) is not in general true.

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Exercise 5.80. Let X be a topological space, let f : X → Y be a function, let {Z i : i ∈ I } be a family of topological spaces, and let gi : Y → Z i , i ∈ I , be functions. Assume Y is endowed with the topology generated by the family {gi : i ∈ I }. Prove that f is continuous on X if and only if gi ◦ f, i ∈ I , are continuous on X . Exercise 5.81. Let X and Y be topological spaces, and let f : X → Y and g : X → Y be continuous functions on X . Use (5.56), (5.80) and (5.78) to show that {x : f (x) = g(x)} is closed whenever Y is a Hausdorff space. Exercise 5.82. Let X and Y be topological spaces, let f : X → Y be a continuous function on X , let A be a subset dense in X , and let B be a closed subset of Y . If f (A) ⊂ B, then f (X ) ⊂ B. Exercise 5.83. For f ∈ K X , the number  f u = supx∈X | f (x)| is called the uniform norm of f . Let α ∈ K and f, g ∈ K X . Prove the following. (a) 0u = 0 and  f u > 0 if f = 0. (b) α f u = |α|  f u . (c)  f + gu   f u + gu . (d)  f gu   f u gu . (e) f is bounded if and only if  f u < ∞. (f) For f ∈ C X , f is bounded if and only if Re f and Im f are bounded. Exercise 5.84. Let x ∈ C and r > 0. Use (5.29) and (5.45) to show that the circle {y ∈ C : |x − y| = r } is a compact subset of C. Exercise 5.85. (a) Show that R 2 and C are homeomorphic topological spaces. (b) Let X be a topological space, and let f : X → C be a function. Prove that f is continuous on X if and only if Re f and Im f are continuous on X . Exercise 5.86. Let (X, T ) and (Y, U) be topological spaces, and let f : X → Y be a homeomorphism from X into Y . Show that f −1 (U) = T . Exercise 5.87. Let Y and Z be topological spaces, let g : Y → Z be a homeomorphism from Y into Z , let X be a set, and let f n : X → Y, n ∈ N , f : X → Y be functions. Prove that f n → f if and only if g ◦ f n → g ◦ f . Exercise 5.88. Let X be a set, let T = {∅} ∪ {A ∈ P(X ) : Ac is countable }, and let {xn : n ∈ N } ⊂ X and x ∈ X . Prove the following. (a) T is a topology for X . (b) xn → x relative to T if and only if xn → x relative to P(X ). Exercise 5.89. Let (X, T ) be a topological space, and let T s denote the family of all subsets U of X with the property that x ∈ U and xn → x relative to T imply that there is n(U ) ∈ N such that xn ∈ U whenever n  n(U ). Show that T s is a topology for X . Exercise 5.90. Let X be a topological space and A ⊂ X . Show that {x : 1 A is not continuous at x} = ∂(A). Exercise 5.91. Let x ∈ ]0, 1[ ∩Q. Find a sequence {xn : n ∈ N } ⊂ ]0, 1[ such that n n limn n1 i=1 xi = limn n1 i=1 xii = x.

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41

Exercise 5.92. Define recursively the sequence of functions {Pn : n  1} ⊂ R R by P1 (x) = x for any x ∈ R, and P2n (x) = (x + 1)P2n−1 (x) + x, P2n+1 (x) = (x + 1)P2n (x), n  1, for any x ∈ R. For n  1, let x2n−1 be the least real root of the equation P2n−1 (x) + 1 = 0, and let x2n be the least real root of the equation P2n (x) = 0. Show that the sequence {xn : n  1} is convergent. [Hint. Prove that {xn : n  1} is bounded and monotone.] Exercise 5.93. Let {xn : n ∈ N } ⊂ [0, ∞] and {yn : n ∈ N } ⊂ [0, ∞]. Prove the following. (a) lim inf n xn +lim inf n yn  lim inf n (xn + yn )  lim supn (xn + yn )  lim supn xn + lim supn yn . (b) limn (xn + yn )  lim inf n xn + lim supn yn whenever {xn + yn : n ∈ N } is convergent. Exercise: One-point compactification 5.94. Let (X, T ) be a topological space, and / X . Define T ∗ = {A ⊂ X ∗ : A ∈ T or X ∗ − A ∈ let X ∗ = X ∪ {δ}, where δ ∈ ∗ T and X − A is compact}. Prove the following. (a) T ∗ is a topology for X ∗ . (b) T is the relative topology on X induced by T ∗ . (c) (X ∗ , T ∗ ) is compact. Exercise: Lower semicontinuous function 5.95. Let (X, T ) be a topological space, and let f : X → ] − ∞, ∞] be a function. We say that f is lower semicontinuous at x ∈ X if for every a < f (x) there is a neighborhood U of x such that f (U ) ⊂ ]a, ∞]. f is called lower semicontinuous on X if it is lower semicontinuous at any x ∈ X . Prove the following. (a) If f is continuous at x, then f is lower semicontinuous at x. (b) If f (X ) ⊂ R, and both f and − f are lower semicontinuous at x, then f is continuous at x. (c) If f is lower semicontinuous at x, then α f is lower semicontinuous at x for any α > 0. (d) For U ⊂ X, 1U is lower semicontinuous on X if and only if U ∈ T . (e) f is lower semicontinuous on X if and only if f −1 (U) ⊂ T , where U is as in (4.2.e). (f) If f : X → ] − ∞, ∞] and g : X → ] − ∞, ∞] are lower semicontinuous on X , then f + g and f ∧ g are lower semicontinuous on X . (g) If { f i : i ∈ I } ⊂ ] − ∞, ∞] X is a family of lower semicontinuous functions on X , then supi∈I f i is lower semicontinuous on X . (h) If f : X → ]a, b[ is lower semicontinuous on X , and h : ]a, b[ → ]c, d[ is strictly increasing and surjective, then h ◦ f is lower semicontinuous on X . and let Exercise 5.96. For i ∈ I = ∅, let X i and Yi be topological  spaces, h i : X i → Yi be a homeomorphism from X i into Yi . Define h : i∈I X i → i∈I Yi by h((x (xi ))i∈I . Use (5.80) and (4.37) to show that h is a homeomorphism i )i∈I ) = (h i from i∈I X i into i∈I Yi . Exercise 5.97. Let {bn : n ∈ N } ⊂]0, ∞[ be such that bn → ∞and bn+1 /bn → 1. Prove the following.

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Analysis and Probability

(a) If c ∈ ]0, 1[, then there exists a subsequence {m n : n ∈ N } ⊂ N such that m n → ∞, n − m n → ∞ and bm n /bn → c. [Hint. Let n c ∈ N be such that bn−1 /bn  c, n  n c . For n  n c , define m n = inf{m < n : bm /bn  c}.] (b) If c ∈ ]1, ∞[, then there exists a subsequence {m n : n ∈ N } ⊂ N such that m n − n → ∞ and bm n /bn → c. [Define m n = inf{m < n : bm /bn  c}, n ∈ N .]  Exercise 5.98 (Dini). Let {dn : n  1} ⊂]0, ∞[ be such that n1 dn = ∞, set  Dn = nk=1 dk , n  1, and let δ > 0. Prove the following. (a) n1 Ddnn = ∞.  n n (b) n1 d1+δ < ∞. [Show that d1+δ  1δ ( D δ1 − D1δ ), n > 1.] Dn Dn n n−1  dn (c) If Dn 0 > 1, then < ∞. [Show that D (logdnD )1+δ  1+δ nn 0 D (log D ) 1 1 δ ( (log Dn−1 )δ

6

−

1 ), n (log Dn )δ

> n 0 .]

n

n

n

n

Metric Spaces

In Section 5 we have already exhibited several properties of topological spaces. In the present section we extend the study of topological notions by introducing a special class of topological spaces in which the topology is determined by means of a metric. We examine here basic properties of metric spaces which we will encounter in various guises throughout the rest of the book, and we prove some deep results, such as Urysohn’s embedding theorem, the Stone-Weierstrass theorem and the Arzelà-Ascoli theorem. Definitions 6.1. A metric [distance-function] for X is any function d : X × X → R such that for x, y, z ∈ X : (i) d(x, y)  0; (ii) d(x, y) = 0 if and only if x = y; (iii) d(x, y) = d(y, x); (iv) d(x, z)  d(x, y) + d(y, z) (the triangle inequality). A pair (X, d), where X is a set and d is a metric for X , is called a metric space. When confusion appears impossible, we will call X itself a metric space. Examples 6.2. d(x, y) =



(a) Let X be a set. The function d : X × X → R defined by 0 if x = y 1 if x = y

is a metric for X called the discrete metric for X . (b) For x, y ∈ C, define d(x, y) = |x − y|. Then d is a metric for C called the usual metric for C. For x = (x1 , . . . , xn ) ∈ R n and y = (y1 , . . . , yn ) ∈ R n , define d(x, y) = (c) n (xi − yi )2 )1/2 . Then d is a metric for R n called the Euclidean metric for R n . ( i=1 (The triangle inequality (6.1.iv) is a special case of Minkowski’s inequality (13.7).) (d) For x, y ∈ R n , define d(x, y) = sup1in |xi − yi |. Then (R n , d) is a metric space.

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43

(e) Let X be a set, and let B(X ) denote the set of all bounded functions from X into K . For f, g ∈ B(X ), define d( f, g) =  f − gu . Then d is a metric for B(X ) called the uniform metric for B(X ). Unless otherwise stated, we will always assume that B(X ) is equipped with the uniform metric. Definitions 6.3. Let (X, d) be a metric space. For x ∈ X and r > 0, the set S(x, r ) = {y : d(x, y) < r } is called the open ball [sphere] with center x and radius r, and the set S(x, r ) = {y : d(x, y)  r } is called the closed ball [sphere] with center x and radius r. We will write Sd (x, r ) and S d (x, r ) wherever we will want to emphasize that S(x, r ) and S(x, r ) depend on d. Examples 6.4. (a) Notation is as in (6.2.a). Then S(x, 1) = {x}, and S(x, 1) = X for any x ∈ X . (b) Notation is as in (6.2.c). For n = 1, whatever x ∈ R and r > 0, S(x, r ) = ]x − r, x + r [, and S(x, r ) = [x − r, x + r ]. (c) Notation is as in (6.2.d). For x = (x1 , . . . , xn ) ∈ R n and r > 0, S(x, r ) = Ur (x) and S(x, r ) is the cube [x1 − r, x1 + r ] × · · · × [xn − r, xn + r ]. Definition 6.5. Let (X, d) be a metric space, and let Sd = {S(x, r ) : x ∈ X, r > 0}. The topology Td = τ (Sd ) is called the topology generated by d. We will always suppose that X is endowed with the topology Td , unless we make some explicit statement to the contrary. Remarks 6.6. (a) Each metric space is a Hausdorff space. (b) Notation is as in (6.5). Then the following are equivalent: (i) U ∈ Td ; (ii) For each x ∈ U there is ε > 0 such that S(x, ε) ⊂ U . Actually, it is clear that (ii) implies (i). Conversely, let U ∈ Td and x ∈ U . In view of (4.10), there are x1 , . . . , xn ∈ X and r1 > 0, . . . , rn > 0 such that x ∈ n S(x , r ) ⊂ U . Thus d(x , x) < r , 1  i  n. Choose ε = inf ∩i=1 i i i i 1in (ri − d(xi , x)). For y ∈ S(x, ε) and every i = 1, . . . , n, we have d(xi , y)  d(xi , x) + n d(x, y) < d(xi , x) + ε  ri , so that y ∈ S(xi , ri ). Consequently, S(x, ε) ⊂ ∩i=1 S(xi , ri ) ⊂ U . (c) Let (X, d1 ) and (X, d2 ) be metric spaces. If for any x ∈ X and ε > 0 there is δ > 0 such that Sd1 (x, ε) ⊃ Sd2 (x, δ), then, applying (b), it follows that Td1 ⊂ Td2 . Therefore Td1 = Td2 if and only if for each x ∈ X and ε > 0 there is δ > 0 such that Sd1 (x, ε) ⊃ Sd2 (x, δ) and Sd2 (x, ε) ⊃ Sd1 (x, δ). This happens whenever cd1  d2  Cd1 for some c, C ∈ ]0, ∞[. Examples 6.7. (a) Notation is as in (6.2.a). Then Td is the discrete topology for X . (b) The usual metric for C generates the usual topology for C. (c) Let d be the Euclidean metric for R n . Then, for any x ∈ R n and r > 0, we have Ur/√n (x) ⊂ Sd (x, r ) ⊂ Ur (x). Thus, using (6.6.c), (6.6.b), (6.4.c) and (4.8), it follows that both metrics introduced in (6.2.c) and (6.2.d) generate the usual topology for R n . Definition 6.8. A topological space (X, T ) is said to be metrizable if there exists a metric d for X such that T = Td .

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As (6.6.a) indicates, not every topological space is metrizable. Theorems (6.9) and (6.11) below show two simple examples of metrizable topological spaces. Theorem 6.9. metrizable.

Let (X, d) be a metric space and A ⊂ X . Then (A, A ∩ Td ) is

Proof. Let d A be the restriction of d to A × A. Then d A is a metric for A. Moreover, for any x ∈ A and r > 0, Sd A (x, r ) = A ∩ Sd (x, r ).

(1)

Using (4.16), (6.6.b) and (1), it follows that the relative topology on A induced by Td coincides with the topology generated by d A . Notation being as in (6.9), we will always assume that A is endowed with the metric d A . Lemma 6.10. Let (X, d) be a metric space and let d = 1 ∧ d. Then d is a metric for X such that Td = Td . Proof. Properties (6.1.i)–(6.1.iii) are obvious for d . Further, if a, b, c are nonnegative numbers such that a  b + c, then 1 ∧ a  (1 ∧ b) + (1 ∧ c). From this it follows immediately that d satisfies the triangle inequality. Thus d is a metric for X . Now, for each x ∈ X , we have Sd (x, r ) if r  1 Sd (X, r ) = . X if r > 1 Therefore, by virtue of (6.6.c), Td = Td . Theorem 6.11. Let I be a nonempty countable set, and let {(X i , di ) : i ∈ I } be a  family of metric spaces. Then i∈I X i with the product topology on it is metrizable. Proof. We prove this theorem in case I is denumerable. The case when I is finite is simpler and is left to the reader. Without anyloss of generality, assume that I = N. Set dn = 1 ∧ dn , n ∈ N . For x = (xn )n∈N ∈ n∈N X n and y = (yn )n∈N ∈ n∈N X n , yn )/n. Since dn is a metric for X n , n ∈ N , it is easily define d(x, y) = supn∈N dn (xn , verified that d is a metric for n∈N X n . For any  r > 0, and n > 1/r , we have dn (xn , yn )/n  1/n < r . Thus, for each x ∈ n∈N X n and r > 0, {y : d(x, y) < r } = {y : dn (xn , yn ) < r n for n  1/r }. Therefore

πn−1 (Sdn (xn , r n)). (1) Sd (x, r ) = n1/r

Applying (6.10), from (1) it follows that Sd (x, r ) is openrelative to the product topology  on n∈N X n . Consequently, the product topology on n∈N X n includes the topology generated by d. To prove that the two topologies coincide, it will suffice to show that, for any m ∈ N and V ∈ Tdm , πm −1 (V ) ∈ Td . Let x ∈ πm −1 (V ) and xm = πm (x). Then xm ∈ V . By (6.6.b) and (6.10), there is ε ∈]0, 1] such that Sdm (xm , ε) = Sdm (xm , ε) ⊂ V.

(2)

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45

From (1) and (2) we get

πn−1 (Sdn (xn , εn/m)) ⊂ πm−1 (Sdm (xm , ε)) ⊂ πm−1 (V ). Sd (x, ε/m) = nm/ε

By virtue of (6.6.b), this shows that πm−1 (V ) ∈ Td .

 Notation being as in (6.11), we will always suppose that i∈I X i is equipped with the metric d. In (5.20) we proved that each topological space which has a countable base is separable, and in (5.21) we showed that the converse of this statement is not generally true. However, for metric spaces the following theorem holds. Theorem 6.12.

Every separable metric space has a countable base.

Proof. Let (X, d) be a metric space, and let A be a countable subset dense in X . Then, according to (3.41), (3.42) and (3.38), the family B = {S(x, r ) : x ∈ A, r ∈ Q ∩ ]0, ∞[} is countable. We will show that B is a base for Td . Let U ∈ Td and x ∈ U . By (6.6.b), there is ε > 0 such that S(x, ε) ⊂ U . Since A is dense in X , (5.17) shows that there is y ∈ A ∩ S(x, ε/3). Now select r ∈ Q such that ε/3 < r < 2ε/3. Then, for any z ∈ S(y, r ), we have d(x, z)  d(x, y)+d(y, z) < (ε/3)+r < ε, and so z ∈ S(x, ε). At the same time, d(y, x) < ε/3 < r . Thus x ∈ S(y, r ) ⊂ S(x, ε) ⊂ U . Therefore, U is a union of balls in B. For a metric space (X, d), the convergence of sequences in X can describe completely the topology Td . Corollary (6.15) illustrates this statement. Remark 6.13. Let (X, d) be a metric space, {xn : n ∈ N } ⊂ X and x ∈ X . Then, according to (6.6.b), xn → x if and only if d(x, xn ) → 0. Theorem 6.14. Let X be a metric space and A ⊂ X . Then the following are equivalent: (i) x ∈ A− ; (ii) there is a sequence {xn : n ∈ N } ⊂ A such that xn → x. Proof. If x ∈ A− , then, using (5.17), choose xn ∈ A ∩ S(x, 1/n) for any n ∈ N . Applying (6.13), we have xn → x, and so (i) implies (ii). Plainly, (ii) implies (i). Corollary 6.15. Let X be a metric space and A ⊂ X . Then the following are equivalent: (i) A is closed; (ii) for any sequence {xn : n ∈ N } ⊂ A such that xn → x ∈ X , we have x ∈ A. Proof. Assume that A is closed, and consider a sequence {xn : n ∈ N } ⊂ A such that xn → x ∈ X . Then, by (6.14), x ∈ A− = A. Thus (i) implies (ii). Suppose now that (ii) holds, and let x ∈ A− . According to (6.14), there is a sequence {xn : n ∈ N } ⊂ A such that xn → x, and so x ∈ A. Therefore, A = A− , and so A is closed. The above result does not generalize to arbitrary topological spaces. To provide examples confirming this statement, objects more general than sequences are required. In what follows we discuss the concept of compactness for metric spaces.

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Definitions 6.16. A topological space (X, T ) is said to be sequentially compact if each sequence in X contains a convergent subsequence. A set A ⊂ X is said to be sequentially compact if (A, A ∩ T ) is sequentially compact. By virtue of (5.52), A is sequentially compact if and only if every sequence in A contains a subsequence converging to some point of A. Lemma 6.17. Let X be a metric space and A ⊂ X . Then x is a limit point of A if and only if A ∩ U is infinite for any neighborhood U of x. Proof.

Exercise.

Lemma 6.18.

Any sequentially compact metric space is separable.

Proof. Let (X, d) be a sequentially compact metric space. For n ∈ N , consider An = {A ⊂ X : d(x, y)  1/n for x, y ∈ A, x = y}. Then An is a family of finite character, and so, by Tukey’s lemma (3.32), An has a maximal element An . The set An is finite, since otherwise, applying (3.11), there would exist a sequence {xk : k  1} ⊂ An such that d(xk , xl )  1/n for k = l. Therefore, according to (6.13), {xk : k  1} would contain no convergent subsequence. Hence the set A = ∪n∈N An is countable. We will show that A is dense in X . If x ∈ (A− )c , then, since (A− )c is open, there would exist ε > 0 such that S(x, ε) ⊂ (A− )c . Select n ∈ N such that 1/n < ε. Then d(x, y)  1/n for any y ∈ An , and so An ∪ {x} ∈ An . This would contradict the assertion that An is maximal. Consequently, A− = X . Theorem 6.19. Let X be a metric space. Then the following assertions are equivalent: (i) X is compact; (ii) X is Fréchet compact; (iii) X is sequentially compact. Proof. We showed in (5.35) that (i) implies (ii). Assume that (ii) holds, and consider a sequence {xn : n ∈ N } ⊂ X . If {xn : n ∈ N } has a finite number of distinct points, then obviously there exists a strictly increasing sequence {n k : k ∈ N } such that xn k = xn 1 , k ∈ N . In this case the subsequence {xn k : k ∈ N } converges to xn 1 . If the set {xn : n ∈ N } is infinite, then it has a limit point x. We will construct inductively a subsequence {xn k : k ∈ N } of {xn : n ∈ N } such that limk xn k = x. Put xn 1 = x1 . If xn 1 , . . . , xnl have been defined, then, according to (6.17), we may choose nl+1 > nl such that xnl+1 ∈ S(x, 1/(l + 1)). It is clear that xn k → x. Therefore, (ii) implies (iii). Suppose now that (iii) holds, and let A be an open cover of X . Then, from (6.18) and (6.12), it follows that X has a countable base. Hence, according to Lindelöf’s theorem (5.14), A contains a countable subfamily B that covers X . If B is finite, then the proof is n B , n ∈ N. complete. If B is denumerable, then set B = {Bi : i ∈ N }. Let Cn = ∪i=1 i To get a contradiction, assume that Cnc = ∅, n ∈ N , and choose xn ∈ Cnc for each n ∈ N . Then there are x ∈ X and a subsequence {xn k : k ∈ N } such that xn k → x. Since B is a cover of X , there is m ∈ N such that x ∈ Bm ⊂ Cm . Since {Cnc : n ∈ N } / Cm is a nonincreasing sequence of sets, Cm is a neighborhood of x such that xn ∈ for n  m. This contradicts the assertion that xn k → x. Therefore, there exists n ∈ N such that X = Cn , and so X is covered by a finite number of sets in A. Thus (iii) implies (i).

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In what follows we introduce the notion of a complete metric space. Definition 6.20. Let (X, d) be a metric space. A sequence {xn : n ∈ N } ⊂ X is said to be a Cauchy [fundamental] sequence if for any ε > 0 there is n ε ∈ N such that m, n  n ε implies d(xm , xn ) < ε. Examples 6.21. (a) Each convergent sequence in a metric space is a Cauchy sequence. (b) Let (X, d) be a metric space, where d is the discrete metric for X , and let {xn : n ∈ N } ⊂ X . Then {xn : n ∈ N } is a Cauchy sequence if and only if there is n 0 ∈ N ; such that xn = xn 0 for n  n 0 . Remarks 6.22. (a) Let (X, d) be a metric space and {xn : n ∈ N } ⊂ X . Then the following are equivalent: (i) {xn : n ∈ N } is a Cauchy sequence; (ii) for any ε > 0 there is n ε ∈ N such that n  n ε implies d(xn ε , xn ) < ε; (iii) supm>n d(xm , xn ) → 0. (b) If (X, d1 ) and (X, d2 ) are metric spaces such that Td1 = Td2 , then a sequence {xn : n ∈ N } ⊂ X may be a Cauchy sequence relative to d1 , but {xn : n ∈ N } may fail to be a Cauchy sequence relative to d2 . For example, let X = {1/n : n ∈ N }, let d1 be the usual metric for X , and let d2 be the discrete metric for X . It is easily seen that both d1 and d2 generate the discrete topology for X . Let xn = 1/n, n ∈ N . Then {xn : n ∈ N } is a Cauchy sequence relative to d1 . However, in view of (6.21.b), {xn : n ∈ N } is not a Cauchy sequence relative to d2 . Definition 6.23. A metric space X is said to be complete if each Cauchy sequence in X is convergent. Example 6.24. According to (6.21.b), any set endowed with the discrete metric is a complete metric space. Remarks 6.25. (a) If d1 and d2 are metrics for X that generate the same topology for X , then X may be complete relative to d1 , but X may fail to be complete relative to d2 . For example, let X = {1/n : n ∈ N }, let d1 be the discrete metric for X , and let d2 be the usual metric for X . As stated in (6.24), (X, d1 ) is complete. Since / X , it follows that (X, d2 ) {1/n : n ∈ N } is a Cauchy sequence relative to d2 and 0 ∈ is not complete. (b) Let (X, d) be a complete metric space and A ⊂ X . Using (6.15) and (6.21.a), it is easily seen that (A, d A ) is complete if and only if A is closed. (c) A metric space X is complete if each Cauchy sequence in X contains a convergent subsequence. Definition 6.26. Let (X, d) be a metric space, and let A be a nonempty subset of X . The number diam(A) = sup{d(x, y) : x, y ∈ A} is called the diameter of A. We define diam(∅) = 0. Theorem (6.28) below is an important characterization of the notion of completeness. We need the next lemma. Lemma 6.27. Let (X, d) be a metric space, and let A be a nonempty subset of X . Then diam(A) = diam(A− ).

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Proof.

Clearly,

diam(A)  diam(A− ).

(1)

For x, y ∈ A− and each ε > 0, there are x ∈ A ∩ S(x, ε/2) and y ∈ S(y, ε/2). Thus, using the triangle inequality, we have d(x, y)  d(x , y ) + ε  diam(A) + ε, and so diam(A− )  diam(A) + ε. Since ε is arbitrary, it follows that diam(A− )  diam(A).

(2)

From (1) and (2) we get diam(A) = diam(A− ). Theorem 6.28 (Cantor). Let (X, d) be a metric space. Then the following are equivalent: (i) X is complete; (ii) for every nonincreasing sequence {An : n ∈ N } of nonempty closed subsets of X such that diam(An ) → 0, there exists x ∈ X such that ∩n∈N An = {x}. Proof. Assume that X is complete, and consider a nonincreasing sequence {An : n ∈ N } of nonempty closed subsets of X such that diam(An ) → 0. Choose xn ∈ An for each n ∈ N . If ε > 0, then there is n ε ∈ N such that diam(An ε ) < ε. Hence, for m, n  n ε , we have d(xm , xn )  diam(An ε ) < ε. Therefore, {xn : n ∈ N } is a Cauchy sequence, and so there is x ∈ X such that xn → x. Whatever m ∈ N , we have xn ∈ Am for n  m, and so, in view of (6.15), x ∈ Am . Consequently, x ∈ ∩n∈N An . If y ∈ ∩n∈N An , then d(x, y)  diam(An ), n ∈ N . Since diam(An ) → 0, it follows that d(x, y) = 0, and so x = y. Therefore, ∩n∈N An = {x}. Thus (i) implies (ii). Suppose now that (ii) holds, and let {xn : n ∈ N } ⊂ X be a Cauchy sequence. Whatever n ∈ N , denote An = {xm : m  n}− . Obviously, {An : n ∈ N } is a nonincreasing sequence of nonempty closed sets. For ε > 0, there is n ε ∈ N such that m, n  n ε implies d(xm , xn ) < ε. Then, by virtue of (6.27), it follows that diam(An ε )  ε.

(1)

Therefore, {An : n ∈ N } being nonincreasing, it follows that diam(An ) → 0. Let {x} = ∩n∈N An . If ε > 0, then, using (1), we have d(x, xn )  diam(An ε )  ε for any n  n ε . Thus xn → x, and so (ii) implies (i). Corollary 6.29.

Any compact metric space is complete.

Proof. Let X be a compact metric space, and let {An : n ∈ N } be a nonincreasing sequence of nonempty closed subsets of X such that diam(An ) → 0. Since {An : n ∈ N } has the finite intersection property, applying (5.24), it follows that ∩n∈N An = ∅. Since diam(An ) → 0, ∩n∈N An contains a single point. Thus, by (6.28), X is complete. Theorem (6.31) below exhibits the relation between compact metric spaces and complete metric spaces. Definitions 6.30. A metric space (X, d) is said to be totally bounded if, for each r > 0, X is covered by a finite number of balls with radius r . A set A ⊂ X is said

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to be totally bounded if (A, d A ) is totally bounded. Obviously, A is totally bounded if and only if, for each r > 0, A can be covered by a finite number of balls in X with radius r . Theorem 6.31 (Hausdorff). Let X be a metric space. Then the following are equivalent: (i) X is compact; (ii) X is totally bounded and complete. Proof. Suppose that X is compact. Then X is complete (6.29). Further, for each r > 0, the family {S(x, r ) : x ∈ X } is an open cover of X which contains a finite subcover, and so X is totally bounded. Assume now that (ii) holds. In view of (6.19), it will suffice to prove that X is Fréchet compact. Let A be an infinite subset of X . Since X is totally bounded, there is a ball S1 with radius 1 such that A1 = A ∩ S1 is infinite. If the infinite sets A1 , . . . , An have been chosen, then, since X is totally bounded, there is a ball Sn+1 with radius 1/(n + 1) such that An+1 = An ∩ Sn+1 is infinite. The − sequence {A− n : n ∈ N } is nonincreasing, and diam(An ) = diam(An )  2/n → 0. Thus, by virtue of Cantor’s theorem (6.28), there is x ∈ X such that ∩n∈N A− n = {x}. For ε > 0, select n ∈ N such that 2/n < ε. Then, for any y ∈ An , we have d(x, y)  diam(An )  2/n < ε, and so An ⊂ S(x, ε). Since A ∩ S(x, ε) ⊃ An ∩ S(x, ε) = An , and An is infinite, it follows that x ∈ Ad (6.17). Therefore, X is Fréchet compact. Theorem 6.32.

The space R n is complete relative to the Euclidean metric.

Proof. Let {xk : k ∈ N } ⊂ R n be a Cauchy sequence. Choose k0 > 1 such that d(xk0 , xk ) < 1 for any k  k0 . Let ρ = 1 ∨ d(xk0 , x1 ) ∨ · · · ∨ d(xk0 , xk0 −1 ) and r = d(0, xk0 ) + ρ. Then d(xk0 , xk )  ρ, k ∈ N , and so {xk : k ∈ N } ⊂ S(0, r ) ⊂ [−r, r ]n . From (5.30) and (6.9) it follows that [−r, r ]n is a compact metric space, and so [−r, r ]n is complete relative to the Euclidean metric (6.29). Since {xk : k ∈ N } is a Cauchy sequence in [−r, r ]n , there is x ∈ [−r, r ]n such that d(x, xk ) → 0. Consequently, R n is complete. Corollary 6.33. Let X be a set, and let Br (X ) be the set of all bounded functions from X into R. Then Br (X ) is a complete space relative to the uniform metric. Proof. Let { f n : n ∈ N } ⊂ Br (X ) be a Cauchy sequence. Then, whatever x ∈ X, { f n (x) : n ∈ N } is a Cauchy sequence in R, and so { f n (x) : n ∈ N } is convergent (6.32). Let f (x) = limn f n (x), x ∈ X . We will show that f ∈ Br (X ) and that f n → f . For ε > 0, choose n ε ∈ N such that m, n  n ε implies  f m − f n u < ε/2. For fixed x ∈ X , select m 

− f m (x)| < ε/2.Then we have | f (x)|

n ε such that | f (x)   | f (x) − f m (x)| + f m (x) − f n ε (x) + f n ε (x) < (ε/2) +  f m − f n ε u +  f n ε u <     ε+  f n ε u . Thus  f u  ε+  f n ε u < ∞, and so f ∈ Br (X ) (5.83.e). If n  n ε , then we get | f (x) − f n (x)|  | f (x) − f m (x)|+| f m (x) − f n (x)| < (ε/2)+ f m − f n u < ε. Therefore,  f − f n u  ε for any n  n ε . Thus f n → f . The concept of continuity is next discussed within the framework of metric spaces. We begin with some remarks.

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Remarks 6.34. Notation is as in (5.36). (a) Assume that the topology for X is generated by d. Then f is continuous at x if and only if for any neighborhood V of f (x) there is δ > 0 such that d(x, y) < δ implies f (x) ∈ V . (b) Suppose that the topology for Y is generated by ρ. Then f is continuous at x if and only if for each ε > 0 there is a neighborhood U of x such that y ∈ U implies ρ( f (x), f (y)) < ε. (c) Assume that the topology for X is generated by d, and the topology for Y is generated by ρ. Then f is continuous at x if and only if for each ε > 0 there is δ > 0 such that d(x, y) < δ implies ρ( f (x), f (y)) < ε. Theorem 6.35. Let X, Y and f be as in (5.36). Assume that X is a metric space, and let x ∈ X . Then the following are equivalent: (i) f is continuous at x; (ii) for any sequence {xn : n ∈ N } ⊂ X such that xn → x, we have f (xn ) → f (x). Proof. We showed in (5.37) that (i) implies (ii). Suppose now that (i) does not hold. Then, according to (6.34.a), there exists a neighborhood V of f (x) such that, for every n ∈ N , there is xn ∈ X with d(x, xn ) < 1/n and f (xn ) ∈ / V . Thus xn → x, but f (xn )  f (x). Therefore, (ii) does not hold. Consequently, (ii) implies (i). Definition 6.36. Let (X, d) be a metric space, let A be a nonempty subset of X , and let x ∈ X . The number d(x, A) = inf{d(x, y) : y ∈ A} is called the distance from x to A. Theorem 6.37. Let (X, d) be a metric space. (i) If ∅ = A ⊂ X , then A− = {x : d(x, A) = 0}. (ii) If ∅ = A ⊂ X , then the function f (x) = d(x, A), x ∈ X , is continuous on X . (iii) If A and B are disjoint closed subsets of X , then there exists a continuous function g : X → [0, 1] such that g(A) = {0} and g(B) = {1}. Proof. If x ∈ A− , then A ∩ S(x, 1/n) = ∅ for any n ∈ N . Let xn ∈ A ∩ S(x, 1/n), n ∈ N . Then d(x, A)  d(x, xn ) < 1/n, n ∈ N , and so d(x, A) = 0. / A− , then there is ε > 0 such Hence A− ⊂ {x : d(x, A) = 0}. Conversely, if x ∈ that A ∩ S(x, ε) = ∅, and so d(x, A)  ε. Therefore, {x : d(x, A) = 0} ⊂ A− . This proves (i). Let f (x) = d(x, A), x ∈ X . For any x, y ∈ X and z ∈ A, we have f (x)  d(x, z)  d(x, y) + d(y, z), whence f (x) − d(x, y)  f (y).

(1)

Upon interchanging x and y, and using (6.1.iii), (1) shows that | f (x) − f (y)|  d(x, y). Thus, according to (6.34.c), f is continuous on X . This proves (ii). If A and B are disjoint closed subsets of X , then, by (i), d(x, A) + d(x, B) > 0 for any x ∈ X . Hence, using (ii), (5.76.a) and (5.76.c), the function g : X → [0, 1] defined by g(x) =

d(x, A) , x ∈ X, d(x, A) + d(x, B)

is continuous on X . Moreover, g(A) = {0} and g(B) = {1}. This proves (iii).

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The following theorem is a special case of Urysohn’s embedding theorem. Theorem 6.38 (Urysohn). Let (X, d) be a separable metric space. Then there exist a compact metric space Y and a homeomorphism f from X into Y . Proof. Let A be a countable subset dense in X . Then there exist a countable set I and a bijective function from I onto A. For each i ∈ I , let ai denote the image of i under this function. Put Y = [0, 1] I . Since [0, 1] is compact relative to the Euclidean metric (5.29), Tihonov’s theorem (5.28) and (6.11) show that Y is a compact metric space. Let h be a homeomorphism from R onto ]0, 1[. Whatever i ∈ I , consider the function f i : X → [0, 1] defined by fi (x) = h(d(x, ai )) for any x ∈ X . By virtue of (6.37.ii) and (5.39), f i is continuous on X for each i ∈ I . Set f = ( f i )i∈I . Then, according to (5.80), f is continuous on X . Now let x ∈ X , and let {xn : n ∈ N } ⊂ X be a sequence such that xn  x. Then there are ε > 0 and a subsequence {xn k : k ∈ N } such that d(x, xn k )  ε for any k ∈ N . Since A is dense in X , there is a j ∈ A such that d(x, a j ) < ε/3. Therefore, by the triangle inequality, d(xn k , a j ) > 2ε/3 for any k ∈ N . Consequently, d(xn k , a j ) > d(x, a j ) + ε/3, k ∈ N . This means that d(xn , a j )  d(x, a j ). Since h −1 is continuous at f j (x), (5.37) shows that f j (xn )  f j (x). Then, according to (5.7), it follows that f (xn )  f (x). Further, let z ∈ X such that z = x, and put xn = z, n ∈ N . Then, since xn  x, we have f (z) = f (xn )  f (x), and so f (z) = f (x). Thus f is injective. Finally, let y ∈ f (X ), and let {yn : n ∈ N } ⊂ f (X ) be such that yn → y. Let x ∈ X be such that y = f (x). For each n ∈ N , let xn ∈ X be such that yn = f (xn ). Since yn → y implies xn → x, (5.50) and (6.35) show that the proof is complete. We next introduce the notion of uniform convergence. Definition 6.39. Let X be a set, and let f n : X → K , n ∈ N , f : X → K be functions. We say that the sequence { f n : n ∈ N } converges uniformly to f if u  f − f n u → 0. We will write f n → f if { f n : n ∈ N } converges uniformly to f . u

Remarks 6.40. (a) For f n ∈ B(X ), n ∈ N , and f ∈ B(X ), f n → f if and only if f n → f relative to the uniform metric. u (b) Let f n , n ∈ N , and f be as in (6.39). If f n → f , then f n (x) → f (x) for all x ∈ X . The converse of this statement does not generally hold. For example, let X = [0, 1], let f n (x) = x n for any x ∈ X, n ∈ N , and let f = 1{1} . It is clear that f n (x) → f (x) for each x ∈ X . However, it can be directly verified, and it follows from (6.41), that { f n : n ∈ N } does not converge uniformly to f . Theorem 6.41.

Let X be a topological space, and let f n ∈ K X , n ∈ N , f ∈ K X u

be functions such that f n → f . If f n is continuous at x ∈ X for any n ∈ N , then f is continuous at x. Proof. For ε > 0, choose n ∈ N such that  f − f n u < ε/3. Since f n is continuous at x, (6.34.b) shows that there is a neighborhood U of x such that y ∈ U implies | f n (x) − f n (y)| < ε/3. Therefore, for y ∈ U , we have | f (x) − f (y)|  | f (x) − f n (x)|+| f n (x) − f n (y)|+| f n (y) − f (y)|  2  f − f n u +| f n (x) − f n (y)| < ε. Thus f is continuous at x.

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Corollary 6.42. Let f n , n ∈ N , and f be as in (6.41). If f n is continuous on X for any n ∈ N , then f is continuous on X . Theorem 6.43. Let X be a topological space, and let Cr (X ) be the set of all bounded continuous functions from X into R. Then Cr (X ) is a complete space relative to the uniform metric. Proof. As Cr (X ) ⊂ Br (X ), and Br (X ) is complete relative to the uniform metric (6.33), it suffices to show that Cr (X ) is a closed subset of Br (X ) (6.25.b). Let f n ∈ Cr (X ), n ∈ N , and let f ∈ Br (X ) be such that f n → f relative to the uniform metric. u Then, by (6.40.a), f n → f . Therefore, in view of (6.42), f ∈ Cr (X ). Thus Cr (X ) is closed. Theorem 6.44 (Dini). Let X be a compact space, let { f n : n ∈ N } ⊂ R X be a monotone sequence, and let f ∈ R X be such that f n (x) → f (x) for any x ∈ X . If f n u is continuous on X for any n ∈ N , and f is continuous on X , then f n → f . Proof. Set gn = | f − f n | , n ∈ N . Then the sequence {gn : n ∈ N } is nonincreasing, and gn (x) → 0 for each x ∈ X . Therefore, for ε > 0, we have X = ∪n∈N {x : gn (x) < ε}. Since gn is continuous on X for any n ∈ N , (5.75.d) shows that {{x : gn (x) < ε} : n ∈ N } is an open cover of X . Since X is compact, and {gn : n ∈ N } is nonincreasing, ε {x : gk (x) < ε} = {x : gn ε (x) < ε}. Consequently, there is n ε ∈ N such that X = ∪nk=1 if n  n ε , then, for each x ∈ X , we have | f (x) − f n (x)|  gn ε (x) < ε. Thus u  f − f n u  ε for any n  n ε , and so f n → f . One of the essential results of abstract analysis is the Stone-Weierstrass theorem which we will prove in (6.57). This theorem illustrates the strong interplay between algebra and contemporary analysis. We begin by defining a few algebraic structures. Definition 6.45. Any function  : X × X → X is called a binary operation on X . Following the usual custom, we will write xy instead of (x, y). Definition 6.46. A group is a pair (G, +), where G is a set, and + is a binary operation on G satisfying: (i) x + (y + z) = (x + y) + z for any x, y, z ∈ G; (ii) there is an element 0 ∈ G such that 0 + x = x + 0 = x for any x ∈ G; (iii) for each x ∈ G there is −x ∈ G such that (−x) + x = x + (−x) = 0. If, moreover, (iv) x + y = y + x for any x, y ∈ G, then (G, +) is called a commutative [Abelian] group. For x, y ∈ G, we will write x − y instead of x + (−y). Definitions 6.47. A linear [vector] space over K is a triple (X, +, ∗), where (X, +) is a commutative group, and ∗ is a function from K × X into X , whose value at (α, x) is denoted αx, such that for α, β ∈ K and x, y ∈ X : (i) α(x + y) = αx + αy; (ii) (α + β)x = αx + βx; (iii) α(βx) = (αβ)x; (iv) 1x = x. We note that α0 = 0x = 0, α ∈ K , x ∈ X , and that (−1)x = −x, x ∈ X .

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53

A set Y ⊂ X is called a linear [vector] subspace of X if x, y ∈ Y implies x + y ∈ Y , and (α, x) ∈ K × Y implies αx ∈ Y . It is easily seen that Y is a linear subspace of X if and only if Y is a linear space over K relative to the restriction of + to Y × Y and the restriction of ∗ to K × Y . Examples 6.48. (a) For x = (x1 , . . . , xn ) ∈ K n and y = (y1 , . . . , yn ) ∈ K n , define x + y = (x1 + y1 , . . . , xn + yn ). For α ∈ K and x ∈ K n , define αx = (αx1 , . . . , αxn ). Then K n is a linear space over K . (b) For α ∈ R and x, y ∈ C n , define x + y as in (a), and put αx = (αx1 , . . . , αxn ). Then C n is a linear space over R. Moreover, R n is a linear subspace of C n . (c) Let X be a set. For α ∈ K and f, g ∈ K X , define f + g and α f as in (2.2). Then K X is a linear space over K , and B(X ) is a linear subspace of K X . If X is a topological space, then Cr (X ) is a linear subspace of Br (X ). (d) Notation is as in (3.43). If A = R, then P(I ) is a linear subspace of R X ; also, for every J ∈ J , P(J ) is a linear subspace of P(I ), and Pn (J ) is a linear subspace of P(J ). Definition 6.49. A ring is a triple (X, +, ·), where (X, +) is a commutative group, and · is a binary operation on X such that for x, y, z ∈ X : (i) x · (y · z) = (x · y) · z; (ii) x · (y + z) = (x · y) + (x · z); (iii) (x + y) · z = (x · z) + (y · z). Definitions 6.50. An algebra over K is a quadruple (X, +, ·, ∗), where (X, +, ·) is a ring and (X, +, ∗) is a linear space over K such that for α ∈ K and x, y ∈ X : (i) α(x · y) = (αx) · y = x · (αy). A set Y ⊂ X is called a subalgebra of X , if x, y ∈ Y implies x · y ∈ Y , and Y is a linear subspace of X . Plainly, Y is a subalgebra of X if and only if Y is an algebra over K relative to the restrictions of + and · to Y × Y and the restriction of ∗ to K × Y. Examples 6.51. (a) For α ∈ K and x, y ∈ K n , define x + y and αx as in (6.48.a), and set x · y = (x1 y1 , . . . , xn yn ). Then K n is an algebra over K . (b) Let X be a set. For α ∈ K and f, g ∈ K X , define f + g, α f and f · g as in (2.2). Then K X is an algebra over K , and B(X ) is a subalgebra of K X . If X is a topological space, then Cr (X ) is a subalgebra of Br (X ). (c) Let X be a topological space, and let Cc (X ) be the set of all bounded continuous functions from X into C. Then Cc (X ) is a subalgebra of C X . (d) Notation is as in (3.43). Suppose that A = R. Then P(I ) is a subalgebra of R X . For J ∈ J , P(J ) is a subalgebra of P(I ), but in general Pn (J ), n  1, is not an algebra over R. If X is a topological space, and f i ∈ Cr (X ) for each i ∈ I , then P(I ) is a subalgebra of Cr (X ). (e) Let X ⊂C − {0}, and let A be the set of all functions from X into C of the form f (x) = nk=−n αk x k , x ∈ X , where αk ∈ C for each k = −n, . . . , 0, . . . , n, and n ∈ N . Then A is an algebra over C. If there are a, b > 0 such that a  |x|  b for any x ∈ X , then A is a subalgebra of Cc (X ). To prove the Stone-Weierstrass theorem, we need the next four lemmas.

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Lemma 6.52. For M > 0, consider the function ϕ : [−M, M] → R defined by ϕ(x) = |x| , x ∈ [−M, M]. Then there exists a sequence { pn : n ∈ N } ⊂ R [−M,M] of u polynomials with real coefficients such that pn (0) = 0, n ∈ N , and pn → ϕ. Proof.

Set p0 (x) = 0, x ∈ [−M, M], and define

pn+1 (x) = pn (x) + (x 2 − pn2 (x))/2M, x ∈ [−M, M], n  0.

(1)

Then, by induction, pn is a polynomial with real coefficients such that pn (0) = 0, n ∈ N , and 0  pn (x)  pn+1 (x)  |x| , x ∈ [−M, M], n  0.

(2)

From (2) it follows that the sequence { pn (x) : n ∈ N } is convergent for each x ∈ [−M, M]. Hence, letting n → ∞ in (1), we get pn2 (x) → x 2 , x ∈ [−M, M]. Therefore, pn → ϕ. Since [−M, M] is compact, and { pn : n ∈ N } is nondecreasing, Dini’s u theorem (6.44) shows that pn → ϕ. Throughout (6.53)–(6.58), X is a topological space, and Cr (X ) is the algebra over R of all bounded continuous functions from X into R.Moreover, Cr (X ) is endowed with the uniform metric. Lemma 6.53.

If A is a subalgebra of Cr (X ), then A− is a subalgebra of Cr (X ).

Proof. According to (6.50) and (6.47), we must show that α ∈ R and f, g ∈ A− imply α f, f + g, f g ∈ A− . Applying (6.14), there are { f n : n ∈ N } ⊂ A and {gn : n ∈ N } ⊂ u u u A such that f n → f and gn → g. Then, by (5.83.b), α f n → α f , and so α f ∈ A− . Furu ther, by (5.83.c), f n +gn → f +g, and so f +g ∈ A− . Using (5.83.b)–(5.83.d), we have  f g − f n gn u   f − f n u gu +  f − f n u g − gn u +  f u g − gn u , n ∈ u N , and so f n gn → f g. Since f n gn ∈ A, n ∈ N , (6.14) shows that f g ∈ A− . Thus A− is a subalgebra of Cr (X ). Lemma 6.54. Let A be a subalgebra of Cr (X ) such that A is a closed subset of Cr (X ). Then, for f, g ∈ A, we have: (i) | f | ∈ A; (ii) f ∨ g ∈ A, f ∧ g ∈ A. Proof. Since f is bounded, there is M > 0 such that f (X ) ⊂ [−M, M]. Let pn , n ∈ N , and ϕ be as in (6.52). Then, since pn (0) = 0, n ∈ N , it follows that u u pn ◦ f ∈ A, n ∈ N . Since pn → ϕ, we have pn ◦ f → ϕ ◦ f = | f |. Therefore, according to (6.14), | f | ∈ A. Thus (i) is proved. The assertion (ii) follows at once from (5.76.b) and (i). Lemma 6.55. Let X be a compact space, and let A be a subset of Cr (X ) such that f, g ∈ A implies f ∨ g ∈ A and f ∧ g ∈ A. If f ∈ Cr (X ) is such that for any x, y ∈ X there is f x,y ∈ A satisfying f x,y (x) = f (x) and f x,y (y) = f (y), then f ∈ A− . Proof. For x, y ∈ X and ε > 0, let Ux,y = {z ∈ X : f (z) − ε < f x,y (z)} and Vx,y = {z ∈ X : f x,y (z) < f (z) + ε}. Since x ∈ Ux,y , (5.76.c) shows that the family

Topological Preliminaries

55

{Ux,y : x ∈ X } is an open cover of X for each y ∈ X . Since X is compact, there are n U n x1 , . . . , xn ∈ X such that X = ∪i=1 xi ,y . For y ∈ X , denote Vy = ∩i=1 Vxi ,y and f y = sup1in f xi ,y . If z ∈ X , then z ∈ Uxi ,y for some i, and so f (z) − ε < f xi ,y (z)  f y (z).

(1)

If z ∈ Vy , then f xi ,y (z) < f (z) + ε, 1  i  n, and so f y (z) < f (z) + ε.

(2)

Since y ∈ Vy , (5.76.c) shows that the family {Vy : y ∈ X } is an open cover of X . Since X is compact, there are y1 , . . . , ym ∈ X such that X = ∪mj=1 Vy j . Define f ε = inf 1 jm f y j . If z ∈ X , then, applying (1), f (z) − ε < f y j (z), 1  j  m, and so f (z) − ε < f ε (z);

(3)

also, if z ∈ X , then z ∈ Vy j for some j, and so, by (2), f ε (z)  f y j (z) < f (z) + ε.

(4)

From (3) and (4) we get  f − f ε u  ε.

(5)

Further, using (5.58.b), we have f y ∈ A, y ∈ X , and f ε ∈ A. Since ε is arbitrary, (6.6.b) and (5) show that f ∈ A− . Definition 6.56. Let X be a set and A ⊂ K X . We say that A separates points of X if whenever x and y are distinct points of X there is f ∈ A such that f (x) = f (y). Stone-Weierstrass theorem 6.57. gebra of Cr (X ) such that: (i) 1 ∈ A; (ii) A separates points of X .

Let X be a compact space, and let A be a subal-

Then A− = Cr (X ). Proof. Since A is a subalgebra of Cr (X ), (6.53) shows A− is a subalgebra of Cr (X ). Therefore, f, g ∈ A− implies f ∨ g ∈ A− and f ∧ g ∈ A− (6.54). Let f ∈ Cr (X ) and x, y ∈ X . If x = y, then the constant function f x,y = f (x) satisfies f x,y (x) = f (x) and f x,y (y) = f (y). Moreover, since 1 ∈ A, f x,y ∈ A ⊂ A− . If x = y, then, since A separates points of X , there is g ∈ A such that g(x) = g(y). Consequently, there are a, b ∈ R (depending on x and y) such that ag(x) + b = f (x) and ag(y) + b = f (y). In this case, set f x,y = ag + b. Evidently, f x,y (x) = f (x) and f x,y (y) = f (y). Since 1 ∈ A, we have f x,y ∈ A ⊂ A− . Thus, by virtue of Lemma (6.55), f ∈ (A− )− = A− . Consequently, A− = Cr (X ). Theorem 6.58. metric space.

Let (X, d) be a compact metric space. Then Cr (X ) is a separable

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Proof. By virtue of (6.19) and (6.18), X is separable, and so there exists a countable set A such that A− = X . For z ∈ A, consider the function dz : X → R defined by dz (x) = d(x, z) for any x ∈ X . Then, by (6.37.ii) and (5.47), dz ∈ Cr (X ) for each z ∈ A. Moreover, the set {dz : z ∈ A} separates points of X . Actually, if x = y, then 0 = dx (x) = dx (y) = d(y, x). Let now A be the set of all polynomials in functions from {dz : z ∈ A} with coefficients in R. Then, according to (6.51.d), A is a subalgebra of Cr (X ). Obviously, 1 ∈ A. Moreover, since {dz : z ∈ A} ⊂ A, the set A separates points of X . Hence, by the Stone-Weierstrass theorem (6.57), A− = Cr (X ). Further, let B be the set of all polynomials in functions from {dz : z ∈ A} with coefficients in Q. Since Q − = R, (6.14) and (5.82) show that for any f ∈ A, there is { f n : n ∈ N } ⊂ B u such that f n → f . Consequently, A ⊂ B− . Since B ⊂ A, (5.66.a) and (5.66.b) show that B− = A− = Cr (X ). Since Q is denumerable, from (3.43) it follows that B is denumerable, and so Cr (X ) is separable. We next prove the complex version of the Stone-Weierstrass theorem. Theorem 6.59. Let X be a compact space, and let A be a subalgebra of Cc (X ) such that: (i) 1 ∈ A; (ii) A separates points of X ; (iii) f ∈ A implies f ∈ A. Then A− = Cc (X ), where Cc (X ) is endowed with the uniform metric. Proof. Let B = { f ∈ A : f (X ) ⊂ R}. Then, using (4.32) and (5.43.b), it follows that B is a subalgebra of Cr (X ). Plainly, 1 ∈ B. Moreover, B separates points of X . Actually, if x and y are distinct points of X , then, by (ii), there is f ∈ A such that f (x) = f (y). Consequently, either (Re f )(x) = (Re f )(y) or (Im f )(x) = (Im f )(y). Since Re f = 21 ( f + f ) and Im f = 2i1 ( f − f ), (iii) shows that Re f ∈ B and Im f ∈ B. Thus B separates points of X . Let now g ∈ Cc (X ). Then, by (5.83.f) and (5.85.b), Re g ∈ Cr (X ) and Im g ∈ Cr (X ). For any ε > 0, by virtue of (6.57), there are f 1 , f 2 ∈ B such that Re g − f 1 u < ε/2 and Im g − f 2 u < ε/2. Therefore, we have g − ( f 1 + i f 2 )u  Re g − f 1 u + Im g − f 2 u < ε. Since f 1 + i f 2 ∈ A, it follows that g ∈ A− , and so A− = Cc (X ). The notion of uniform continuity we introduce now is closely related to the concept of continuity. Definition 6.60. Let (X, d) and (Y, ρ) be metric spaces, and let f : X → Y be a function. We say that f is uniformly continuous if for any ε > 0 there is δ > 0 such that d(x, y) < δ implies ρ( f (x), f (y)) < ε. Remark 6.61. From (6.34.c) it follows at once that each uniformly continuous function is continuous. The converse of this assertion fails in general. For example, the function f : R → R defined by f (x) = x 2 , x ∈ R, is continuous on R but not uniformly continuous. The next theorem presents a special case in which the converse of this statement holds. Theorem 6.62. Let (X, d), (Y, ρ) and f be as in (6.60). If X is compact, and f is continuous on X , then f is uniformly continuous.

Topological Preliminaries

Proof. that

57

Let ε > 0. Then, according to (6.34.c), for each x ∈ X there is δx > 0 such

f (Sd (x, δx )) ⊂ Sρ ( f (x), ε/2).

(1)

Since X is compact, and {Sd (x, δx /2) : x ∈ X } is an open cover of X , there are n S (x , δ /2). Put δ = inf x1 , . . . , xn ∈ X such that X = ∪i=1 d i xi 1in δxi /2. Now let x, y ∈ X be such that d(x, y) < δ. Then x ∈ Sd (xi , δxi /2) ⊂ Sd (xi , δxi ) for some i. Consequently, we have d(xi , y)  d(xi , x) + d(x, y) < (δxi /2) + δ  δxi , and so y ∈ Sd (xi , δxi ). Therefore, applying (1), we get ρ( f (x), f (y))  ρ( f (x), f (xi )) + ρ( f (xi ), f (y)) < (ε/2) + (ε/2) = ε. Thus f is uniformly continuous. If X is a compact metric space, then the Arzelà-Ascoli theorem we will prove in (6.67) characterizes the compact subsets of Cr (X ). We begin with some definitions. Definition 6.63. A set F ⊂ R X is said to be equibounded if there is M > 0 such that  f u  M for all f ∈ F. Definition 6.64. Let (X, d) and (Y, ρ) be metric spaces, and let F ⊂ Y X . We say that F is uniformly equicontinuous if for each ε > 0 there is δ > 0 such that d(x, y) < δ implies ρ( f (x), f (y)) < ε for all f ∈ F. Example 6.65. Let X and Y be metric spaces. Then any finite set of uniformly continuous functions from X into Y is uniformly equicontinuous. To prove the Arzelà-Ascoli theorem, we need the next lemma. Lemma 6.66. Let (X, d) be a metric space and let A ⊂ X . Then the following are equivalent: (i) A− is compact; (ii) each sequence in A contains a convergent subsequence. Proof. If A− is compact, then A− is sequentially compact (6.19). Hence (i) implies (ii). Assume now that (ii) holds, and consider a sequence {xn : n ∈ N } ⊂ A− . For every n ∈ N , according to (6.14), choose xn ∈ A− such that d(xn , xn ) < 1/n. Then there are x ∈ A− and a subsequence {xn k : k ∈ N } such that xn k → x. Since d(x, xn k )  d(x, xn k ) + 1/n k , it follows that xn k → x. Therefore, A− is sequentially compact. Thus, by (6.19), A− is compact, and so (ii) implies (i). Arzelà-Ascoli theorem 6.67. Let (X, d) be a compact metric space and let F ⊂ Cr (X ), where Cr (X ) is endowed with the uniform metric. Then F− is compact if and only if F is equibounded and uniformly equicontinuous. Proof. Assume that F− is compact, and let g ∈ F− . Then, by (6.37.ii), the function dg : F− → R defined by dg ( f ) =  f − gu , f ∈ F− , is continuous on F− . By virtue of (5.47), there is h ∈ F− such that  f − gu  h − gu for any f ∈ F− . Therefore, for each f ∈ F, we have  f u   f − gu + gu  h − gu + gu . This shows that F is equibounded. Further, since F− is compact, (6.31) shows that F− is totally bounded. Consequently, for ε > 0, there are f 1 , . . . , f n ∈ Cr (X )

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Analysis and Probability

such that F− ⊂

n 

{ f ∈ Cr (X ) :  f i − f u < ε/3}.

(1)

i=1

Since f 1 , . . . , f n are continuous on X , (6.62) and (6.65) show that { f 1 , . . . , f n } is uniformly equicontinuous. Therefore, there is δ > 0 such that d(x, y) < δ implies | f i (x) − f i (y)| < ε/3, i = 1, . . . , n.

(2)

If f ∈ F , then, by (1),  f i − f u < ε/3 for some i. Hence, if d(x, y) < δ, then, using (2), we have | f (x) − f (y)|  | f (x) − f i (x)| + | f i (x) − f i (y)| + | f i (y) − f (y)|  2  f i − f u + | f i (x) − f i (y)| < ε. This shows that F is uniformly equicontinuous. Conversely, suppose that F is equibounded and uniformly equicontinuous. Let { f n : n ∈ N } ⊂ F . Then, according to (6.66), it will suffice to show that { f n : n ∈ N } contains a convergent subsequence. Since F is equibounded, there is M > 0 such that | f n (x)|  M, x ∈ X, n ∈ N . Since X is compact, (6.19) and (6.18) show that X is separable, and so there exists a countable set A dense in X . The case in which A is finite is simpler and is left to the reader. If A is denumerable, then put A = {xi : i ∈ N }. We choose inductively a sequence ( f n 11 , f n 12 , . . . , f n 1k , f n 1,k+1 , . . .) ( f n 21 , f n 22 , . . . , f n 2k , f n 2,k+1 , . . .) .. .

(3)

( f n k1 , f n k2 , . . . , f n kk , f n k,k+1 , . . .) ( f n k+1,1 , f n k+1,2 , . . . , f n k+1,k , f n k+1,k+1 , . . .) .. . of subsequences of { f n : n ∈ N } as follows. The first row in (3) is a subsequence of { f n : n ∈ N } so chosen that the sequence { f n 1m (x1 ) : m ∈ N } is convergent. Since [−M, M] is compact, and f n (x1 ) ∈ [−M, M], n ∈ N , (6.19) shows that such a subsequence must exist. If kth row in (3) has been defined, then (k + 1)st row in (3) is a subsequence of { f n km : m ∈ N } so chosen that the sequence { f n k+1,m (xk+1 ) : m ∈ N } is convergent. Such a subsequence must exist, since f n km (xk+1 ) ∈ [−M, M], m ∈ N , and [−M, M] is compact. Set n k = n kk , k ∈ N . Then { f n k : k ∈ N } is a subsequence of { f n : n ∈ N }. Moreover, for each i ∈ N , { f n k : k  i} is a subsequence of { f n im : m ∈ N }, and so the sequence { f n k (xi ) : k ∈ N } is convergent. Let now ε > 0. Since F is uniformly equicontinuous, there is δ > 0 such that d(x, y) < δ implies | f (x) − f (y)| < ε/3,

f ∈ F.

(4)

Further, since X is compact, X is totally bounded (6.31), and so there exists a finite family {S1 , . . . , Sl } of open balls with radius δ/2 such that X = ∪lj=1 S j . Since A is

Topological Preliminaries

59

dense in X , we can select a j ∈ A ∩ S j for each j = 1, . . . , l. Since the sequences { f n k (a j ) : k ∈ N }, 1  j  l, are convergent, (6.21.a) shows that there is kε ∈ N such that p, q  kε implies



f n (a j ) − f n (a j ) < ε/3, j = 1, . . . , l. (5) p q If x ∈ X , then x ∈ S j for some j, and so d(x,

a j ) < δ. Therefore,

for p, q  kε , using

f n (x) − f n (x)  f n (x) − f n (a j ) + f n (a j ) − f n (a j ) + (4) and (5), we have p q p p p q



  ε ε ε

f n (a j ) − f n (x) < = ε. Consequently,  f n p − f n q u q q 3 + 3 + 3  ε for p, q  kε . This means that { f n k : k ∈ N } is a Cauchy sequence in Cr (X ). Since Cr (X ) is complete relative to the uniform metric (6.43), the subsequence { f n k : k ∈ N } is convergent. The process of obtaining the subsequence { f n k : k ∈ N } in (6.67) is called the Cantor diagonal method. Exercise 6.68. Notation is as in (6.3). (a) Prove that S(x, r ) is a closed set. (b) Give an example to show that adherence of S(x, r ) may be different from S(x, r ). Exercise 6.69.

Prove that R with its usual topology is metrizable.

Exercise 6.70. Let (X, d) be a separable metric space and let A ⊂ X . Use (6.12) and (5.20) to show that (A, d A ) is a separable metric space. Exercise 6.71. Use (6.12) and (5.14) to prove that each open subset of a separable metric space is a union of countably many closed balls. Exercise 6.72. Let X be a metric space and let {xn : n ∈ N } ⊂ X . If {xn : n ∈ N } contains no convergent subsequence, then any subset of {xn : n ∈ N } is closed. Exercise 6.73. Let X be a metric space and let A ⊂ X . Show that A is totally bounded if and only if A− is totally bounded. Exercise 6.74.

If (X, d) is a metric space, then d is continuous on X × X .

Exercise 6.75.

If X is a compact metric space, then diam(X ) < ∞.

Exercise 6.76. Let (X, d) be a metric space, and let F be a closed subset of X . Construct a nonincreasing sequence { f n : n ∈ N } ⊂ [0, 1] X of uniformly continuous functions such that f n → 1 F . [For F = ∅, set f n (x) = e−nd(x,F) , x ∈ X, n ∈ N .] Exercise 6.77. Let (X, d) be a metric space. Use (5.41) and (5.40) to prove that the topology generated by d coincides with the topology generated by {dx : x ∈ X }, where dx stands for the section of d at x. Exercise 6.78. Notation is as in (3.37). Show that, with the composition operation ◦, P is a group. Exercise 6.79. Let X be a topological space, and let A ⊂ K X be a set of continuous functions that separates points of X . Prove that X is a Hausdorff space. Exercise 6.80. Let (X, d) be a metric space, and define the function ω : B(X ) × ]0, ∞[ → R by ω( f, δ) = sup{| f (x) − f (y)| : x, y ∈ X, d(x, y) < δ} for any f ∈

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Analysis and Probability

B(X ) and δ ∈ ]0, ∞[. For f ∈ B(X ) and δ ∈]0, ∞[, the number ω( f, δ) is called the δ-oscillation of f . Prove the following. (a) For each f ∈ B(X ), the section of ω at f is nondecreasing. (b) Whatever δ ∈ ]0, ∞[, let ωδ be the section of ω at δ. For f, g ∈ B(X ), |ωδ ( f ) − ωδ (g)|  2  f − gu , and so ωδ is uniformly continuous. Exercise 6.81. Let X be a topological space, let Y be a metric space, and let f : X → Y be a function. For x ∈ X , the number ω(x) = inf{diam f (U ) : U is a neighborhood of x} is called the oscillation of f at x. Prove the following. (a) f is continuous at x if and only if ω(x) = 0. (b) For any a > 0, the set {x : ω(x) < a} is open. (c) The set {x : f is continuous at x} is a countable intersection of open sets. Exercise 6.82.

Let X be a set, let { f n : n ∈ N } ⊂ Br (X ), and let f ∈ Br (X ). Show u

that { f n : n ∈ N } is equibounded whenever f n → f . Exercise 6.83. Let (X, d) and (Y, ρ) be metric spaces, and let F ⊂ Y X . We say that F is equicontinuous if for each x ∈ X and ε > 0 there is δ > 0 (depending on x and ε) such that d(x, y) < δ implies ρ( f (x), f (y)) < ε for all f ∈ F. Mimic the proof of (6.62) to show that if X is compact, then any equicontinuous subset of Y X is uniformly equicontinuous. Exercise 6.84. Let [a, b] ⊂ R be a closed interval, let { f n : n ∈ N } ⊂ R [a,b] be a sequence of nondecreasing functions, and let f ∈ R [a,b] be such that f n (x) → f (x) for u any x ∈ [a, b]. Use (5.29) and (6.62) to prove that f n → f whenever f is continuous on [a, b]. Exercise 6.85. Use (6.37) and (5.75.d) to prove that each closed subset of a metric space is an intersection of countably many open sets. Exercise 6.86. Let (X, d) and (Y, ρ) be metric spaces, and let f : X → Y be a function. For ε, δ > 0, put Aε,δ = {x ∈ X : there exist y, z ∈ Sd (x, δ) such that ρ( f (y), f (z))  ε}. Prove the following. (a) The set Aε,δ is open. (b) {x : f is continuous at x} = ∩m1 ∪n1 Ac1/m,1/n . Exercise 6.87. Let [a, b] ⊂ R, and let A be the set of all polynomials of the form  p(x) = nk=0 αk x k , x ∈ [a, b], where αk ∈ R for k = 0, . . . , n, and n ∈ N . Verify that A is dense in Cr ([a, b]). Exercise 6.88. Let (X, d) be a metric space, and set d = d/(1 + d). (a) Show that d is a metric for X . (b) Use (6.6.c) and (6.10) to prove that Td = Td . [Hint. d /2  d  d , where d is as in (6.10).] Exercise 6.89. Let (X ∗ , T ∗ ) be as in (5.94), where X = R. Show that (X ∗ , T ∗ ) is metrizable. [Hint. Verify that R ∪ {δ} and the circle {(x1 , x2 ) ∈ R 2 : x12 + x22 = 1} (equipped with the Euclidean metric) are homeomorphic topological spaces.]

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Exercise 6.90. If A and B are disjoint closed subsets of a metric space, then there are disjoint open sets U and V with A ⊂ U and B ⊂ V . Let (X, d) be a metric space, let A ⊂ X be closed, let n ∈ N , t if 0  t  1 and set Bn = {x ∈ X : d(x, A)  1/n}. Let ϕ(t) = , and define 1 if t > 1 f n (x) = ϕ(nd(x, A)), x ∈ X . Prove the following. (a) Bn is closed and ∪n∈N Bn = Ac . (b) f n (A) = {0} and f n (B) = {1}. (c) f n is uniformly continuous. [Show that | f n (x) − f n (y)|  nd(x, y), x, y ∈ X .] Exercise 6.91.

Exercise 6.92. Let X be a metric space, and let f : X → R be a continuous function. For a > 0, put A = {x : | f (x)|  a}, and define g = f 1 A + a( f / | f |)1 Ac . Show that g is continuous on X . Exercise 6.93. Let (X, d) be a metric space. For ∅ = A ⊂ X and ε > 0, write Aε = ∪x∈A S(x, ε). Let F denote the family of all closed subsets of X . For A, B ∈ F − {∅}, define d H (A, B) = inf{ε > 0 : Aε ⊂ B and Bε ⊂ A}. Prove that d H is a metric for F − {∅}. d H is called the Hausdorff distance. Exercise 6.94. Let X be a metric space, let {xn : n ∈ N } ⊂ X , and let x ∈ X . Use (6.37.iii) to show that xn → x whenever f (xn ) → f (x) for all f ∈ Cr (X ). Exercise 6.95. Let (X, d) be a metric space. For ∅ = A ⊂ X and ε > 0, write Aε = {x ∈ X : d(x, A) < ε}. (a) Prove that ∩ε>0 Aε = A− . (b) For ε, ε > 0, show that (Aε )ε ⊂ Aε+ε , and that this inclusion may be strict. Exercise 6.96. Let (X, d) be a metric space, and let f : X → R be a function. Prove the following. (a) f is lower semicontinuous at x ∈ X if and only if lim inf n f (xn )  f (x) whenever {xn : n ∈ N } ⊂ X and xn → x. (b) If f is lower semicontinuous on X , then there is a nondecreasing sequence { f n : n ∈ N } ⊂ R X of continuous functions such that f n → f and | f n |   f u , n ∈ N . [Hints. If f (X ) ⊂ [0, ∞[, take f n (x) = inf y∈X ( f (y) + nd(x, y)), x ∈ X . If | f |  M < ∞, take f n = gn − M, where {gn : n ∈ N } ⊂ [0, ∞[ X is a nondecreasing sequence of continuous functions such that gn → f + M. In general, apply (5.95.h) with h(z) = arctan z + π/2, z ∈ R.] Exercise 6.97. Let X be a topological space, let A ⊂ X be compact, and let { f n : n ∈ N } ⊂ ] − ∞, 0] X be a nondecreasing sequence of lower semicontinuous functions on u X such that f n → 0. Prove that f n | A → 0. [Mimic the proof of Dini’s theorem (6.44).] Exercise 6.98. Let (X i , di ), 1  i  n, be metric spaces, and define d(x, y) = sup1in di (xi , yi ) for x = (x1 , . . . , xn ) and y = (y1 , . . . , y n ). n Xi . (a) Show that Td coincides with the product topologyon i=1 n (b) If (X i , di ) is complete for1  i  n, prove that ( i=1 X i , d) is complete.  Exercise 6.99. Let {Yi : i ∈ I } be a countable family of sets, put Y = i∈I Yi ,  and let α = (αi )i∈I ∈]0, ∞[ I be such that i∈I αi < ∞. For y = (yi )i∈I ∈ Y and

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Analysis and Probability

y = (yi )i∈I ∈ Y , define dα,H (y, y ) to be the sum of the values αi over those indices for which yi = yi . Prove the following. (a) dα,H is a metric for Y . dα,H is called the α-Hamming distance. (b) For yn = (yni )i∈I ∈ Y, n ∈ N , and y = (yi )i∈I ∈ Y, dα,H (y, yn ) → 0 if and only if, for each i ∈ I , there is n(i) such that yni = yi , n  n(i). (c) For i ∈ I , consider the topological space (Yi , Tdi ), where di is the discrete metric for Yi . Then Tdα,H coincides with the product topology on Y . [Use (b) and (6.77).] (d) (Y, dα,H ) is separable if and only if Yi is finite for any i ∈ I .