1873_solutions

# Be a lower bound of h using this fact we choose a

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Unformatted text preview: e interval 0, 1 such that S  0, 1 but if U is the set of interior points of S then U 0, 1 . Once again, take the set of all rational numbers between 0 and 1. 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 set of real numbers, that H is a closed set and that S H, prove that S We could argue that S H and that, because H is closed, H  H. H. 4. Given two sets A and B of real numbers, 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 sets A and B of real numbers, 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? Since A B A we have A B A and similarly that A B B. Thus A B A B. Now observe that if A  0, 1 and B  1, 2 then A B  0, 1 1, 2  1 and A B  . Of course, we could give more spectacular examples like A  Q and B  R Q. 6. Prove that if S is any set of real numbers then the set R S is the set of interior points of the set R S. Most students should be encouraged to write two separate arguments here. The first task is to show that every member of the set R S must be an interior point of R S. Then one should show that every interior point of R S must belong to R S. 100 On the other hand, a strong student could be permitted to observe that if x is any given number then the statement that x does not belong to S is the statement that there exists a number   0 such that x , x   S  , and that the latter equation is just the condition that x , x   R 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 true that if A and B are sets of real numbers and ABR then A B  R? The answer is no. Look at A  Q and B  R Q. 9. Prove that if A and B are open sets and then A ABR B  R. What if only one of the sets A and...
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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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