1873_solutions

# V that is open in the metric space x d such that u v

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Unformatted text preview: c. Give an example of a set S of real numbers such that if U is the set of interior points of S then U S. We could take S to be a singleton like 3 or it could be the set of all integers. It could also be the set of all rational numbers between 0 and 1. d. Give an example of a subset S of the interval 0, 1 such that S  0, 1 but if U is the set of interior points of S then U 0, 1 . Take S to be the set of rational numbers that lie in the interval 0, 1 . The set of interior points of S is empty. 2. Given that 1 n S n Z , evaluate S. Hint: Show that 1 n Z . n First show that 0 S. Then observe that every negative number belongs to the set any positive number then x belongs to the interval 1 ,1 n1 n for some positive integer n. S 0 Þ Ý, 0 and that if x is 3. Given that S is a subset of a metric space, that H is a closed set and that S H, prove that S H. From the elementary properties of closure we know that S H and we know from the relationship between closure and the property of being closed that H  H. 4. Given two subsets A and B of a metric space, prove that A Þ B  A Þ B. Solution: Since A A and B B we have A Þ B  A Þ B. and therefore, since the union of the two closed sets A and B is closed we have A Þ B A Þ B. On the other hand, since A is included in the closed set A Þ B we have A AÞB and, similarly we can see that B AÞB and so A Þ B A Þ B. Therefore A Þ B  A Þ B. 5. Given two subsets A and B of a metric space, prove that A B A B. Do the two sides of this inclusion have to be equal? What if A and B are open? What if they are closed? From the fact that A B A we deduce that A B A and we see similarly that A B B. Therefore A B A B. In the metric space R, if A  0, 1 and B  1, 2 then A and B are open and A B A B. On the other hand, if A and B are closed subsets of a metric space then 122 A BA BA B. 6. Prove that if S is any subset of a metric space X then the set X S is the set of interior points of the set X S. Given any point x X, the condition x X S means that there exists a number   0 such that B x,  S  and this condition says that there exists a number   0 such that B x,  X S. The latter assertion says that x is an interior point of the set X S. 7. Given that  is an upper bound of a given set S of real numbers, prove that the following two conditions are equivalent: a. We have   sup S. b. We have  S. To prove that condition a implies condition b we assume that   sup S. We need to show that  S. Suppose that   0. Using the fact that  is the least upper bound of S and that     we choose a member x of S such that    x. Since x  ,    S we have  ,    S . To prove that condition b implies condition a we assume that  S. We need to show that  is the least upper bound of S. Suppose that p  . Since the set p, Ý is a neighborhood of  we have p, Ý S . Thus, since  is an upper bound of S and since no number p   can be an upper bound of S we conclude that  is the least upper bound of S. 8. Is it tru...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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