Real Analysis, 2nd Edition, G.B.Folland Chapter 4 Point Set Topology Yung-Hsiang Huang 4.1 Topological Spaces 1. If card(x) 2, there is a topology on X that is T 0 but not T 1. 2. If X is an infinite set, the cofinite topology on X is T 1 but not T 2, and is first countable iff X is countable. 3. Every metric space is normal. 4. Let X = R, and let T be the family of all subsets of R of the form U (V Q) where U, V are open in the usual sense. Then T is a topology that is Hausdroff but not regular. (In view of Exercise 3, this shows that a topology stronger than a normal topology need not be normal or even regular.) 5. Every separable metric space is second countable. Last Modified: 2017/10/13. Department of Math., National Taiwan University. Email: d04221001@ntu.edu.tw 1
6. Let E = {(a, b] : < a < b < }. (a) E is a base for a topology T on R in which the members of E are both open and closed. (b) T is first countable but not second countable. (If x R, every neighborhood base at x contains a set whose supremum is x.) (c) Q is dense in R with respect to T. (Thus the converse of Proposition 4.5 is false.) 7. If X is a topological space, a point x X is called a cluster point of the sequence {x j } if for every neighborhood U of x, x j U for infinitely many j. If X is first countable,, x is a cluster point of {x j } iff some subsequence of {x j } converges to x. 8. If X is an infinite set with the cofinite topology and {x j } is a sequence of distinct points in X, then x j x for every x X. 9. If X is a linearly order set, the topology T generated by the sets {x : x < a} and {x : x > a}(a X) is called the order topology. (a) If a, b X and a < b, there exist U, V T with a U, b V, and x < y for all x U and y V. the order topology is the weakest topology with this property. (b) If Y X, the order topology on Y is never stronger than, but may be weaker than, the relative topology on Y induced by the order topology on X. (c) The order topology on R is the usual topology. 10. A topological space X is called disconnected if there exist nonempty open sets U, V such that U V = and U V = X; otherwise, X is connected/ When we speak of connected or disconnected subsets of X, we refer to the relative topology on them. 2
(a) X is connected iff and X are the only subsets of X that are both open and closed. (b) If {E α } α A is a collection of connected subsets of X such that α A E α, then α A E α is connected. (c) If A X is connected, then A is connected. (d) Every point x X is contained in a unique maximal connected subset of X, and this subset is closed. (It is called the connected component of x) (a) is trivial. (b) Let E := α A E α. Given O 1, O 2 open in X such that O 1 E, O 2 E are non-empty, O 1 O 2 E. Choose x α A E α, and WLOG assume x O 1. Choose y O 2 E, then y E β for some β A. Note that (O 1 E β ) (O 2 E β ) = (O 1 O 2 ) E β = E β and x O 1 E β, y O 2 E β. By the connectedness of E β, O 1 O 2. Since O 1, O 2 are arbitrary chosen, E is connected. (c) Given O 1, O 2 open in X such that O 1 A, O 2 A are non-empty, O 1 O 2 A A. Given x O 1 A, since O 1 is an open neighborhood of x, O 1 A. Similar for O 2 A. By the connectedness of A, O 1 O 2. Since O 1, O 2 are arbitrary chosen, A is connected. (d) Let C := {A X : A is connected and contains x}. Let C = A C A. By (b), C is connected. Then it s obvious maximal, unique. The closedness follows from (c). 11. If E 1, E n are subsets of a topological space, then n 1E j = n 1E j. (It s not true for intersection, e.g. E 1 = Q and E 2 = Q c ; it s also not true for infinitely many E j, e.g. E j = { 1 j }.) This is easy, we omit it. 12. Let X be a set. A Kuratowski closure operator on X is a map A A from P(X) to itself satisfying (i) =, (ii) A A for all A, (iii) (A ) = A for all A, and (iv) (A B) = A B for all A, B. (a) If X is a topological space, the map A A is a Kuratowski closure operator. (b) Conversely, given a Kuratowski closure operator, let F = {A X : A = A } and T = {U X : U c F }. Then T is a topology, and for any set A X, A is its closure with respect to T. 3
13. If X is a topological space, U is open in X and A is dense in X, then U = U A. (It s not true if U is not open, e.g. X = R, U = R \ Q and A = Q.) Of course, U U A. Given x U, if x A, then x U A. If x A, then x acc(a). Given V open in X, since U is open, V U is open in X and hence there is a point y lying in A (V U) = V (A U) and hence y x. Since V is given, x acc(a U). Consequently, x U A. Since x is given, U U A. Since U A is closed, U U A. 4.2 Continuous Maps 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 4
28. Let X be a topological space equipped with an equivalence relation, X the set of equivalence classes, π : X X the map taking each x X to its equivalence class, and T = {U X : π 1 (U) is open in X}. (a) T is a topology on X. (It is called the quotient topology.) (b) If Y is a topological space, f : X Y is continuous iff f π is continuous (c) X is T 1 iff every equivalence class is closed in X. (a) Since π 1 ( X) = X and π 1 ( X) = X, X and X are in T. Given {U α } T, then π 1 ( α U α ) = α π 1 U α hence α U α T. Similarly n 1U i T for every finite set {U 1, U n } T. (b) By definition, π is continuous. So f π is continuous provided f is. Conversely, if f π is continuous, then for every open set V Y, f 1 (V ) is open in X since π 1 f 1 (V ) = (f π) 1 (V ) is open in X. Therefore, f is continuous. (c) Suppose X is T 1. Given any equivalence class π 1 ({[x]}) for some [x] X, since the singleton {[x]} is closed, by continuity, π 1 ({[x]}) is closed. Conversely, suppose every equivalence class is closed. Given [x] X, then π 1 ({[x]}) is closed. ( c Since π 1 ({[x]} c ) = π ({[x]})) 1 is open, {[x]} c T. So {[x]} is closed. Hence X is T 1. 29. 4.3 Nets 30. 31. 32. 33. 34. 35. 36. 5
4.4 Compact Spaces 37. 38. 39. 40. 41. 42. 43. 44. 45. 4.5 Locally Compact Hausdorff Spaces 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 6
4.6 Two Compactness Theorems 58. 59. 60. 61. 62. 63. 64. By Arezla-Ascoli Theorem. 65. 4.7 The Stone-Weierstrass Theorem 66. 67. 68. 69. 70. 71. References 7