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# hw1 - Math 245C Homework 1 Brett Hemenway Folland Chapter 4...

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Math 245C Homework 1 Brett Hemenway April 18, 2007 Folland Chapter 4 10. Let X be a topological space. a. Suppose X is disconnected, then there exist U, V non-empty open subsets of X such that U V = and U V = X . Thus U c = V in X . V is open by assumption, and U c is closed because U is open, so V is both open and closed. If we have that if , X are the only sets in X that are closed and open, then X must be connected. Conversely, suppose there is an open set = U X that is both open and closed in X , then U c is open, U U c = , and U U c = X . So we have that X is connected iff the only sets that are closed and open in X are and X . b. Let { E α } α A be a collection of connected subsets of X with α A E α = . Suppose E = α A E α is disconnected, then we can find open sets U, V in E with U V = , U V = E . With U, V = . Then, since each E α is connected, we have that U E α = and V E α = E α or vice-versa, for each α A . Let x α A E α , then x U V = α A E α . Without loss of generality, we may assume x U . Thus U E α = for all α . This gives us that V E α = for all α A . This gives us that V = , thus E is connected. c. Suppose A is disconnected. This means that we can find open sets U, V in X such that ( U A ) ( V A ) = U V A = , A U V , and U A = = V A . Now U V A U V A = , and A A U V . So to show that A is disconnected, it remains 1

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to show that U A = = V A . Since U A and V A are nonempty, let u U A and v V A . Then since U, V are neighborhoods of u and v , and u and v are in A , we have that U A = and V A = . Thus if A is connected, so is A . d. Let x X . Consider the set A = { E : E X, E connected, x E } Then by part (b), the set C x = E A E is connected. The set C x is clearly maximal, and by part (c), C x is closed. 13. Let X be a topological space, U X open, and A X dense in X . Then since U A U , we have that U A U To show the other inclusion, let x U . Then for any open neigh- borhood V of x , we have that V U = . Since U is open, we have that U V is open. Since A is dense we have that ( A U ) V = A ( U V ) = . If x A U then we are done. If x A U , then since ( A U ) ( V \{ x } ) = ( A U ) V , we have x acc( A U ). Thus x U A . So U U A and hence U = U A 26. a. If f is continuous on X , and f ( X ) = Y is disconnected, then we can find nonempty open sets U, V Y such that U V = , and U V = Y . Since f is continuous, f - 1 ( U ), and f - 1 ( V ) are open in X , they are nonempty since U , and V are nonempty, and X = f - 1 ( Y ) = f - 1 ( U V ) = f - 1 ( U ) f - 1 ( V ) = f - 1 ( ) = f - 1 ( U V ) = f - 1 ( U ) f - 1 ( V ) . Thus X is disconnected. We conclude that if X is connected, f ( X ) must also be connected.
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hw1 - Math 245C Homework 1 Brett Hemenway Folland Chapter 4...

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