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

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20 Topology (2301631) Example 1.78. Let S 1 = {(x, y) | x 2 +y 2 = 1} be a subspace of R 2 . The continuous bijection f : [0, 1) → R 2 given by f(t) = (cos 2πt, sin 2πt) is not an imbedding because f −1 : S 1 → [0, 1) is not continuous. Notice that (f −1 ) −1 ([0, 1 2 )) = f([0, 1 2 )) is not open in S 1 . However, f| (0,1) is an imbedding. Exercise 1.79. Prove that f : N ∪ {0} → R given by f(n) = (−1) n n 2 is an imbedding. What about g : N ∪ {0} → R given by g(0) = 0 and g(n) = 1 n for n > 0?
Phichet Chaoha 21 5. Sequences and Nets Definition 1.80. A sequence in a space X is a map (x n ) : N → X. We usually denote (x n )(i) by x i . Definition 1.81. The sequence (x n ) in X is said to converge to x ∈ X, abbreviated by (x n ) → x, if for each neighborhood U of x, there exists N ∈ N such that x n ∈ U for all n ≥ N. When (x n ) → x, we will call x a limit of (x n ). Example 1.82. In the space {a, b}, where a ≠ b, with the indiscrete topology, the constant sequence (a) converges to both a and b. Hence, a limit of a convergent sequence may not be unique. Theorem 1.83. A convergent sequence in a Hausdorff space has a unique limit. Proof. Let (x n ) be a convergent sequence in X. Suppose (x n ) → x and (x n ) → x ′ . If x ≠ x ′ , there exist disjoint neighborhoods U and V of x and x ′ , respectively. Hence, there is a large enough N ∈ N such that x n ∈ U ∩ V for all n ≥ N (can you see why?). This is impossible because U ∩ V = ∅. Hence, we must have x = x ′ . □ Definition 1.84. Let (x n ) and (y n ) be sequences in a space X. We say that (y n ) is a subsequence of (x n ) if there is a strictly increasing map f : N → N such that y i = x f(i) for all i ∈ N. Exercise 1.85. Let (y n ) be a subsequence of a convergent sequence (x n ) in a Hausdorff space. Prove that if (y n ) converges to y ∞ ∈ X, then so is (x n ). What happens if (x n ) is not assumed to be convergent? Theorem 1.86 (Sequence Lemma). Let X be a space, x ∈ X and A ⊆ X. If there is a sequence in A converging to x, then x ∈ A. Proof. Let (a n ) be a sequence in A converging to x, and U a neighborhood of x. Then, by the definition of convergence, there exists N ∈ N such that a n ∈ U for all n ≥ N. In particular, a N ∈ U ∩ A. Hence, U ∩ A ≠ ∅; i.e, x ∈ A. □ The converse of the above theorem is not true in general as we will see in the next example. Example 1.87. The space S Ω does not satisfy the converse of the sequence lemma because there is no sequence in S Ω converging to Ω. To see this, let (x n ) be a sequence in S Ω . Since C = {x n | n ∈ N} is a countable subset of S Ω , it has an upper bound in S Ω , says α. Now it is easy to see that C ∩ (α, Ω] = ∅ and hence (x n ) does not converge to Ω.
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Page 5 and 6: Preliminaries Sets and Maps We assu
Page 7 and 8: CHAPTER 1 Basic Concepts in Topolog
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Phichet Chaoha 71 subset Y of R n b
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