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Unformatted text preview: Modern Analysis, Homework 3 Possible solutions 0. (2 pts) Rudin, Ch 2, problem 12: Let K ⊂ R consist of 0 and the numbers 1 /n for n = 1 , 2 , 3 , ··· . Prove that K is compact directly from the definition (without using the HeineBorel theorem). Solution: Let { U α } α ∈ I be an open cover of K . Then 0 ∈ U α for some α ∈ I . Since U α is open, there exists ε > 0 such that N ε (0) ⊂ U α . By the Archimedean property of R , it follows that for all sufficiently large n (say, greater than some n ∈ N ), we have 1 /n < ε and therefore 1 /n ∈ U α . Now for each n = 1 , 2 , ··· ,n , there is some open set, U α n such that n ∈ U α n . Therefore U α ,U α 1 , ··· ,U α n is a finite open cover of K . 1. (2 pts) Rudin, Ch 2, problem 14: Give an example of an open cover of the segment (0 , 1) which has no finite subcover. Solution: Consider the family { (1 /n, 1) } n ∈ N . By the Archimedean property of R , it is clear that this forms an open cover of (0 , 1). It is also clear that it has no finite subcover: if there were, choose the maximal n such that (1 /n, 1) is in the subcover. Then 1 / ( n + 1) is not covered! 2. (5 pts) Rudin, Ch 2, problem 9: Let E ◦ denote the set of all interior points of a set E ( E ◦ is called the interior of E ). (a) Prove that E ◦ is open. Solution: Let x ∈ E ◦ . Then x is an interior point of E , so there exists r > 0 such that N r ( x ) ⊂ E . We saw in class that balls are open, so each point in N r ( x ) is an interior point of E , ie N r ( x ) ⊂ E ◦ and E ◦ is open. (b) Prove that E is open if and only if E ◦ = E . Solution: By definition, E ◦ ⊂ E and if E is open, then every point of E is an interior point, so that E = E ◦ . On the other hand, if E = E ◦ , then by (b) E itself is open (since E ◦ is). (c) If G ⊂ E and G is open, prove that G ⊂ E ◦ . Solution: Let x ∈ G . Since G is open, there exists r > 0 such that N r ( x ) ⊂ G . But G ⊂ E so x is therefore an interior point of E , ie G ⊂ E ◦ . (d) Prove that the complement of E ◦ is the closure of the complement of E . Solution: Let x ∈ ( E ◦ ) c , ie x is not an interior point of E . This means that for every r > 0, N r ( x ) contains a point not in E . This exactly says that x ∈ ( E c ) and hence ( E ◦ ) c ⊂ ( E c ). Now let x ∈ ( E c ). Then either x ∈ E c or x ∈ ( E c ) . In the former case, we can’t have x ∈ E and therefore can’t have x ∈ E ◦ . In the latter case, every neighborhood of x must contain a point of E c , so x cannot be an interior point of E . In either case, we obtain ( E c ) ⊂ ( E ◦ ) c ....
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This note was uploaded on 10/18/2011 for the course MATH S4061X taught by Professor Peters during the Spring '11 term at Columbia.
 Spring '11
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