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Unformatted text preview: MATH 202A HOMEWORK 6 Problem 1 . (a) Proof. G is a subspace of X . Suppose G is not connected, there is a separtaion. Hence, ∃ U 1 ,U 2 ⊂ X , s.t. (1) G ⊂ U 1 S U 2 (2) U 1 T G 6 = ∅ (3) U 2 T G 6 = ∅ (4) U 1 T U 2 T G = ∅ By (2), ∃ α 1 s.t. U 1 T C α 1 6 = ∅ . Hence, in C α 1 , C α 1 ⊂ G ⊂ U 1 S U 2 ; U 1 T C α 1 6 = ∅ ; U 1 T U 2 T C α 1 ⊂ U 1 T U 2 T G = ∅ . By the connectedness of C α 1 , U 2 T C α 1 = ∅ and thus C α 1 ⊂ U 1 . Similarly, C α 2 ⊂ U 2 . Since C α 1 and C α 2 has a point in common, ∅ 6 = C α 1 T C α 2 ⊂ U 1 T U 2 , which implies U 1 T U 2 T G 6 = ∅ and thus contradicts to (4). Therefore, G is connected. (b) Proof. Suppose ¯ Y is not connected and thus has a separtaion. Considering the sub space ¯ Y , ¯ Y = A S B , where A ⊂ ¯ Y , B ⊂ ¯ Y , A,B 6 = ∅ . Now that Y is connected in X , it is clear that Y is also connected in ¯ Y . Hence, since A and B are both open in ¯ Y , the sets A T Y and B T Y are open in Y , which forms a separtaion of Y unless one of them is empty. WLOG, Y T B = ∅ and since Y ⊂ ¯ Y = A S B , Y ⊂ A . Hence, ¯ Y ⊂ ¯ A . Since ¯ Y = A S B = ¯ A S ¯ B , we have ¯ A S ¯ B ⊂ ¯ A and thus ¯ B = ∅ , which contradicts to B 6 = ∅ . Therefore, ¯ Y is connected. (c) Proof. Suppose X is locally connected and thus has a base B , consisting of all con nected open sets. Hence, for any component C ⊂ X , ∀ x ∈ C , ∃ B x ∈ B , s.t., x ∈ B x . Since C is a component consisting of x and B x is a connected open set including x , we have B x ⊂ C . Therefore, C = S x ∈ C B x is a union of open sets and thus is also open. (d) 1 2 MATH 202A HOMEWORK 6 Proof. Let S denote the topological sine curve. We claim that for any path σ : I → S started with O (0 , 0), we have σ ( I ) = { O } and thus S is not path connected. Since { O } is a closed set of S and the path is continuous, σ 1 ( O ) is closed in I . WTS σ 1 ( O ) is also an open set in I. ∀ x ∈ σ 1 ( O ), by the continuity of σ and local connectedness of I , ∃ U , U is an connected open set containing x , s.t. σ ( U ) ⊂ S T B ( O, 1 2 ) , O ∈ σ ( U ) and the component of S T B ( O, 1 2 ) containing O is { O } . By the connectedness of σ ( U ), we have σ ( U ) ⊂ { O } and thus U ⊂ σ 1 ( O )....
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This note was uploaded on 04/06/2010 for the course MATH various taught by Professor Tao/analysis during the Spring '10 term at UCLA.
 Spring '10
 tao/analysis
 Math

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