Lecture Notes Topology (2301631) Phichet Chaoha Department of ...

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34 Topology (2301631) let {X n } n∈N be a countable collection of second countable spaces and B n a countable basis of X n . Then B = ∪ ( ∏ n B i × ∏ ) {X i } n∈N i=1 i>n can be shown to be a countable basis for the product space ∏ n∈N X n. Notice the use of the product topology here! □ Exercise 2.8. Prove that a countable product of a separable space is also separable. Example 2.9. R ω satisfies all countability axioms by the previous theorem and exercise. The next example shows that a subspace of a separable space may not be separable. Example 2.10. Since R l is separable (Q is also dense in R l ), then from the previous exercise, so is R 2 l . However, the subspace L = {(x, y) | x + y = 1} of R 2 l is not separable because the topology on L is discrete (i.e., for each (x, y) ∈ L, we have ([x, x + 1) × [y, y + 1)) ∩ L = {(x, y)} which is open in L). Therefore, the uncountable space L cannot have a countable dense subset. As we have seen in Theorem 2.3, the separability is generally weaker than second-countability. However, these two concepts are equivalent when the space is metrizable. Theorem 2.11. A metric space is first-countable, and it is separable iff it is second-countable. Proof. Let X be a metric space. Then, for each x ∈ X, the collection {B(x; 1 n ) | n ∈ N} forms a countable basis at x. Now, if X is separable, we let D be a countable dense subset of X. To see that the collection {B(p; 1 n ) | p ∈ D and n ∈ N} forms a countable basis for X, let U be an open subset of X and x ∈ U. Then there is n ∈ N such that B(x; 1 1 n ) ⊆ U. Since D is dense in X, there exists p ∈ D∩B(x; 2n ). Now it is easy to verify that x ∈ B(p; 1 2n ) ⊆ B(x; 1 n ) ⊆ U. The converse follows directly from Theorem 2.3. □ Corollary 2.12. R l is not metrizable. Theorem 2.13. The converse of the sequence lemma holds for a first-countable space (e.g. a metric space). Proof. Let X be a first-countable space, A ⊆ X and a ∈ A. Let B a = {B 1 , B 2 , . . . } be a countable basis at a. For each n ∈ N, let B n ′ = B 1 ∩ · · · ∩ B n . Clearly, each B n ′ is open and B n+1 ′ ⊆ B n ′ ⊆ B n . Since a ∈ A, we can form a sequence (a n ) in A by choosing a n ∈ B n ′ ∩ A for each n. To see that the sequence (a n ) converges to a, we let U be a neighborhood of a. Since B a is a basis at a, there is N ∈ N such that a ∈ B N ⊆ U. Then for n ≥ N, we have a n ∈ B n ′ ⊆ B N ′ ⊆ B N ⊆ U as desired. □ Corollary 2.14. The space S Ω is not first-countable. Exercise 2.15. Is S Ω separable? Justify your answer. Exercise 2.16. Which one of our three countability axioms does S Ω satisfy? Justify your answer.
Phichet Chaoha 35 Definition 2.17. Let X be a space. 2. The Separation Axioms (1) X is a T 0 −space if for any two distinct points in X, we can find a neighborhood of one of the point not containing the other. (2) X is a T 1 −space or a quasi-separated space if for any two distinct points in X, we can find a neighborhood of each point not containing the other. (3) X is a T 2 −space or a separated space or a Hausdorff space if any two distinct points of X have disjoint neighborhoods. (4) X is a T 3 −space or a regular space if it is T 1 and any point and any closed subset not containing that point have disjoint neighborhoods. (5) X is a T 3 1 −space or a completely regular space if it is T 1 and any closed 2 subset A of X and any point x /∈ A, there exists a continuous map f : X → [0, 1] such that f(x) = 0 and A ⊆ f −1 ({1}). (Note that if A = ∅, then f is the zero map) (6) X is a T 4 −space or a normal space if it is T 1 and any two disjoint closed subsets have disjoint neighborhoods. (7) X is a T 5 −space or a completely normal space if it is T 1 and any two separated subsets A, B of X (i.e. A ∩ B = A ∩ B = ∅) have disjoint neighborhoods. Remark 2.18. We need X to be T 1 in the definition of T i for i ≥ 3 because the space {a, b} with indiscrete topology satisfies the other part of the definition, but it is not even Hausdorff! Example 2.19. X = {a, b} with the topology {∅, {a}, {a, b}} is clearly T 0 , but not T 1 . Exercise 2.20. Prove that X is T 0 iff {x} ̸= {y}, for any distinct points x, y in X. Theorem 2.21. A space X is T 1 iff every singleton is closed. Proof. (⇒) Suppose X is T 1 and let x ∈ X. For each y ≠ x, we can find a neighborhood U y not containing x. Then X − {x} = ∪ y≠x U y is open, and hence {x} is closed. (⇐) Suppose that every singleton is closed. Then, for distinct points x, y ∈ X, the open sets X − {y} and X − {x} are the desired neighborhoods of x and y, respectively. □ Corollary 2.22. A space X is T 1 iff every finite subset of X is closed. Remark 2.23. Using the previous theorem, it is not difficult to see that a T i -space is also a T i−1 -space for i = 1, 2, 3, 4, 5. Moreover, it is straightforward to verify that a T 3 1 -space is a T 3-space. Leter on, we will also see that a T 2 4 -space is a T 3 1 -space. 2 Theorem 2.24. A space X is Hausdorff iff the diagonal is closed in X × X. ∆ = {(x, x) | x ∈ X}
Page 1: Lecture Notes Topology (2301631) Ph
Page 5 and 6: Preliminaries Sets and Maps We assu
Page 7 and 8: CHAPTER 1 Basic Concepts in Topolog
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Page 11 and 12: Phichet Chaoha 11 2. Basis for a To
Page 13 and 14: Phichet Chaoha 13 Exercise 1.34. Fo
Page 15 and 16: Phichet Chaoha 15 { 1 n Exercise 1.
Page 17 and 18: Phichet Chaoha 17 Let X and Y be (n
Page 19 and 20: Phichet Chaoha 19 Exercise 1.70. Le
Page 21 and 22: Phichet Chaoha 21 5. Sequences and
Page 23 and 24: Phichet Chaoha 23 6. Product Topolo
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Page 27 and 28: Phichet Chaoha 27 (1) The 1-sphere
Page 29 and 30: Phichet Chaoha 29 8. Metric Topolog
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Page 41 and 42: CHAPTER 3 Connectedness Definition
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