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Unformatted text preview: M 453 S A 2 S 20, 2004 Exercise 1 from Sections 12 & 13, page 83 We will show that A is open by exhibiting it as a union of open sets. For each x A , let U x be the open set containing x such that U x A . It is easy to see that A = S x A U x , so A is open. Exercise 4 from Sections 12 & 13, page 83 (a) Let T = T T . To show that T is a topology, we have to verify that T satisfies the three properties in the definition of a topology: (1) Are and X in T ? Yes, because and X are in T for each . (2) Let { U } be a collection of open sets in T . Since T is the intersection of the topologies T , { U } is a collection of open sets in T for each . Hence their union S U is in T for each , and so S U T . (3) Starting with a finite collection of open sets in T , the argument is as in (2) above....
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This note was uploaded on 02/10/2010 for the course MATH 205 taught by Professor Smith during the Spring '05 term at Adrian College.
 Spring '05
 SMITH
 Sets

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