hmwk6sol_a

hmwk6sol_a - MATH 202A HOMEWORK 6 Problem 1 . (a) Proof. G...

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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), 1 s.t. U 1 T C 1 6 = . Hence, in C 1 , C 1 G U 1 S U 2 ; U 1 T C 1 6 = ; U 1 T U 2 T C 1 U 1 T U 2 T G = . By the connectedness of C 1 , U 2 T C 1 = and thus C 1 U 1 . Similarly, C 2 U 2 . Since C 1 and C 2 has a point in common, 6 = C 1 T C 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 : I S started with O (0 , 0), we have ( I ) = { O } and thus S is not path connected. Since { O } is a closed set of S and the path is continuous, - 1 ( O ) is closed in I . WTS - 1 ( O ) is also an open set in I. x - 1 ( O ), by the continuity of and local connectedness of I , U , U is an connected open set containing x , s.t. ( U ) S T B ( O, 1 2 ) , O ( U ) and the component of S T B ( O, 1 2 ) containing O is { O } . By the connectedness of ( U ), we have ( U ) { O } and thus U - 1 ( O )....
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hmwk6sol_a - MATH 202A HOMEWORK 6 Problem 1 . (a) Proof. G...

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