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Unformatted text preview: Modern Analysis, Homework 3 Possible solutions 0. (2 pts) Rudin, Ch 2, problem 12: Let K âŠ‚ R consist of 0 and the numbers 1 /n for n = 1 , 2 , 3 , Â·Â·Â· . Prove that K is compact directly from the definition (without using the HeineBorel theorem). Solution: Let { U Î± } Î± âˆˆ I be an open cover of K . Then 0 âˆˆ U Î± for some Î± âˆˆ I . Since U Î± is open, there exists Îµ > 0 such that N Îµ (0) âŠ‚ U Î± . By the Archimedean property of R , it follows that for all sufficiently large n (say, greater than some n âˆˆ N ), we have 1 /n < Îµ and therefore 1 /n âˆˆ U Î± . Now for each n = 1 , 2 , Â·Â·Â· ,n , there is some open set, U Î± n such that n âˆˆ U Î± n . Therefore U Î± ,U Î± 1 , Â·Â·Â· ,U Î± 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 âˆˆ 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 â—¦ denote the set of all interior points of a set E ( E â—¦ is called the interior of E ). (a) Prove that E â—¦ is open. Solution: Let x âˆˆ E â—¦ . Then x is an interior point of E , so there exists r > 0 such that N r ( x ) âŠ‚ 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 ) âŠ‚ E â—¦ and E â—¦ is open. (b) Prove that E is open if and only if E â—¦ = E . Solution: By definition, E â—¦ âŠ‚ E and if E is open, then every point of E is an interior point, so that E = E â—¦ . On the other hand, if E = E â—¦ , then by (b) E itself is open (since E â—¦ is). (c) If G âŠ‚ E and G is open, prove that G âŠ‚ E â—¦ . Solution: Let x âˆˆ G . Since G is open, there exists r > 0 such that N r ( x ) âŠ‚ G . But G âŠ‚ E so x is therefore an interior point of E , ie G âŠ‚ E â—¦ . (d) Prove that the complement of E â—¦ is the closure of the complement of E . Solution: Let x âˆˆ ( E â—¦ ) 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 âˆˆ ( E c ) and hence ( E â—¦ ) c âŠ‚ ( E c ). Now let x âˆˆ ( E c ). Then either x âˆˆ E c or x âˆˆ ( E c ) . In the former case, we canâ€™t have x âˆˆ E and therefore canâ€™t have x âˆˆ E â—¦ . 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 ) âŠ‚ ( E â—¦ ) c ....
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 Spring '11
 Peters
 Metric space, General topology, limit point, Archimedean

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