Basic Analysis – Gently Done Topological Vector Spaces

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4: Separation 33 We claim that U and V are disjoint. Indeed, if x ∈ U ∩V, then x ∈ U a ∩V b for some a ∈ A and b ∈ B. This means that d(x,a) < 1 2 ε a d(a,b) ≤ d(x,a)+d(x,b) < 1 2 (ε a +ε b ) ≤ max{ε a ,ε b } and d(x,b) < 1 2 ε b which is a contradiction. We conclude that U ∩V = ∅, as required. . Hence We can give an alternative proof using the continuity of the distance between a point and a set in a metric space. We recall that if x is any point in a metric space (X,d) and A is any non-empty subset of X, the distance between x and A is given by dist(x,A) = inf{d(x,a) : a ∈ A}. One shows that x ↦→ dist(x,A) is a continuous mapping from X into R and that x ∈ A if and only if dist(x,A) = 0. Now suppose that A and B are disjoint closed non-empty sets in X. Put U = {x ∈ X : dist(x,B)−dist(x,A) > 0} and V = {x ∈ X : dist(x,B)−dist(x,A) < 0}. Then U is the inverse image of the open set {t ∈ R : t > 0} under the continuous map x ↦→ dist(x,B) − dist(x,A) and so is open in X. Similarly, V is open, being the inverse image of the open set {t ∈ R : t < 0} under the same continuous map. It is clear that U ∩V = ∅. Now, if a ∈ A, then dist(a,A) = 0. If dist(a,B) were 0, we would conclude that a ∈ B. But B is closed so that B = B and we know that a /∈ B. Hence dist(a,B) > 0 and so a ∈ U. Thus A ⊆ U. Similarly, B ⊆ V and the result follows. Theorem 4.4 Every compact Hausdorff space is normal. Proof Suppose that (X,T) is a compact Hausdorff topological space. We shall first show that if z ∈ X and F is a closed set not containing z, then z and F have disjoint neighbourhoods. For any x ∈ F, there are disjoint open sets U x and V x such that z ∈ U x and x ∈ V x (by the Hausdorff property). The sets {V x : x ∈ F} form an open cover of F. Now, F is a closed set in a compact space and so is compact. Hence there is a finite set J in F such that F ⊆ � x∈J V x . Put U = � x∈J U x and V = � x∈J V x . Then U and V are both open, z ∈ U, F ⊆ V and for any x ∈ J, U ⊆ U x so that U ∩V x = ∅. Hence U ∩V = ∅. Now let A and B be any pair of disjoint non-empty closed sets in X. For each a ∈ A there are disjoint open sets U a and V a such that a ∈ U a and B ⊆ V a , by the above argument. The sets {U a : a ∈ A} form an open cover of the compact set A and so there is a finite set S in A such that A ⊆ � a∈S U a V = � a∈S V a . Put U = � a∈S U a and . Then U and V are both open and A ⊆ U and B ⊆ V. Furthermore, U a ∩V ⊆ U a ∩V a = ∅ and so U and V are disjoint and the proof is complete.
34 <strong>Basic</strong> <strong>Analysis</strong> Next we show that normal spaces are characterized by the property that any open neighbourhood of a closed set contains the closure of an open neighbourhood of the closed set. Theorem 4.5 The topological space (X,T) is normal if and only if for any closed set F and open set V with F ⊆ V, there is an open set U such that F ⊆ U ⊆ U ⊆ V. Proof For notational convenience, let us denote the complement X\A of any set A by A c . Suppose that (X,T) is normal and let F be any closed set and V any open set with F ⊆ V. Then F ∩ V c = ∅. Now, V c is closed and so there are disjoint open sets U and W such that F ⊆ U and V c ⊆ W. Thus W c ⊆ V. But to say that U and W are disjoint is to say that U ⊆ W c . Since W is open, W c is closed and therefore U ⊆ W c . Piecing all this together, we have F ⊆ U ⊆ U ⊆ W c ⊆ V. In particular, F ⊆ U ⊆ U ⊆ V. For the converse, let A and B be any pair of disjoint closed sets. Then B c is open and A ⊆ B c . By hypothesis, there is an open set U such that A ⊆ U ⊆ U ⊆ B c . Put V = U c . Then V is open and U ⊆ B c implies that B ⊆ U c = V. It is clear that U ∩V = ∅ and we conclude that (X,T) is normal. The next result, Urysohn’s lemma, ensures a plentiful supply of continuous functions on a normal space. Theorem 4.6 (Urysohn’s lemma) For any pair of non-empty disjoint closed sets A and B in a normal space (X,T) there is a continuous map f : X → R with values in [0,1] such that f ↾ A = 0 and f ↾ B = 1. Proof Suppose that A and B are non-empty disjoint closed sets in the normal space (X,T). Let U 1 = X \B. Then U 1 is open and A ⊆ U 1 . Hence there is an open set U 0 , say, such that A ⊆ U 0 ⊆ U 0 ⊆ U 1 . We shall construct a family {U q : q ∈ Q} of open sets, labelled by the rationals, Q, such that U p ⊆ U q whenever p < q. To do this, we first consider the rational numbers in the interval [0,1]. This is a countable set, so we may list its members in a sequence, and we may begin with the rationals 0,1 as the first two terms: Q∩[0,1] = {p 1 = 0,p 2 = 1,p 3 ,p 4 ,...}. Department of Mathematics King’s College, London
Page 1 and 2: Basic Analysis - Gently Done Topolo
Page 3 and 4: Contents 1 Topological Spaces . . .
Page 5 and 6: 2 Basic Analysis Definition 1.3 A t
Page 7 and 8: 4 Basic Analysis Proof The statemen
Page 9 and 10: 6 Basic Analysis Proposition 1.20 L
Page 11 and 12: 8 Basic Analysis Theorem 1.27 Suppo
Page 13 and 14: 10 Basic Analysis Proposition 1.36
Page 15 and 16: 12 Basic Analysis Thepreviouspropos
Page 17 and 18: 14 Basic Analysis ample, the proble
Page 19 and 20: 16 Basic Analysis Proposition 2.8 L
Page 21 and 22: 18 Basic Analysis Theorem 2.13 Let
Page 23 and 24: 20 Basic Analysis A point z is a cl
Page 25 and 26: 22 Basic Analysis family {U x : x
Page 27 and 28: 24 Basic Analysis and so � A∩B
Page 29 and 30: 26 Basic Analysis Remark 3.2 Let G
Page 31 and 32: 28 Basic Analysis Proposition 3.7 S
Page 33 and 34: 30 Basic Analysis Proof (version 2)
Page 35: 4. Separation The Hausdorff propert
Page 39 and 40: 36 Basic Analysis Q x contains no r
Page 41 and 42: 38 Basic Analysis that g 0 (x) =
Page 43 and 44: 5. Vector Spaces We shall collect t
Page 45 and 46: 42 Basic Analysis Zorn’s lemma ca
Page 47 and 48: 44 Basic Analysis Finally, suppose
Page 51 and 52: 48 Basic Analysis Nextweconsiderthe
Page 53 and 54: 50 Basic Analysis and so, multiplyi
Page 55 and 56: 52 Basic Analysis so that Λ ↾ M
Page 57 and 58: 54 Basic Analysis subsets of X. We
Page 59 and 60: 56 Basic Analysis Remark 6.7 The co
Page 63 and 64: 60 Basic Analysis Corollary 6.20 Su
Page 65 and 66: 62 Basic Analysis For each 1 ≤ i
Page 67 and 68: 64 Basic Analysis Corollary 6.29 Le
Page 69 and 70: 7. Locally Convex Topological Vecto
Page 71 and 72: 68 Basic Analysis To show that each
Page 73 and 74: 70 Basic Analysis vector topology o
Page 75 and 76: 72 Basic Analysis at 0 is to say th
Page 77 and 78: 74 Basic Analysis For the last part
Page 79 and 80: 76 Basic Analysis are equivalent; s
Page 81 and 82: 78 Basic Analysis C = {f ∈ X : p
Page 83 and 84: 80 Basic Analysis Hence, for x ∈
Page 85 and 86: 82 Basic Analysis Thus, for t ∈ K
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8. Banach Spaces In this chapter, w
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86 Basic Analysis 5. The Banach spa
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88 Basic Analysis We shall apply th
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90 Basic Analysis Proposition 8.7 F
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92 Basic Analysis Proposition 8.10
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94 Basic Analysis and we deduce tha
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96 Basic Analysis Theorem 8.16 Supp
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98 Basic Analysis for x ∈ X = ℓ
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100 Basic Analysis and so �ϕ x
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102 Basic Analysis The next result
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104 Basic Analysis Now we consider
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106 Basic Analysis Theorem 9.13 For
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108 Basic Analysis Definition 9.19
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110 Basic Analysis According to the
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10. Fréchet Spaces Inthischapter w
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114 Basic Analysis and this is sepa
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116 Basic Analysis For any m,n > N,
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118 Basic Analysis Theorem 10.18 (B
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120 Basic Analysis because X = �
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122 Basic Analysis Corollary 10.23
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124 Basic Analysis Remark 10.27 It
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126 Basic Analysis Define P : X →
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Basic Analysis – Gently Done Topological Vector Spaces

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