1873_solutions

# 9 x 1 becomes which says that x when x 23 and x 1

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Unformatted text preview: must be an upper bound of A, suppose that x is an upper bound of A a . Using the fact that a is not the largest member of A, choose a member b of A such that a  b. Since x b we see at once that x  a and so x is an upper bound of the set A. Note that the assumption that A has no largest member isn’t really needed here. All we need to know is that the specific number a is not the largest member of A. 2. a have exactly the b. Give an example of a set A that has a largest member a such that the sets A and A exactly the same upper bounds. We can look at the set 0, 1 Þ 3 . 3. a. Give an example of a set A that has a largest member a such that the sets A and A same upper bounds. We can look at the interval 0, 1 . a do not have a. Given that S is a subset of a given interval a, b explain why, for every member x of the set S we have |x | |a |  |b a |. Solution: Suppose that x S. We see that |x |  |x Now since a x a  a| a |  |a |. |x b we see that |x a|  x a b a and so |x | |a |  |b a |. b. Given that a set S of numbers is bounded and that T  |x | x S , prove that the set T must also be bounded. Choose a lower bound a of S and an upper bound b of S and then observe from part a that the number |a |  |b a | is an upper bound of T. 4. Given that A is a set of real numbers and that sup A A, explain why sup A must be the largest member of A. Solution: This assertion is obvious. We are given that inf A is a member of A and we know that no member of A can be less than inf A, and so inf A is the least member of A. 5. Given that A is a set of real numbers and that inf A A, explain why inf A must be the smallest member of A. This assertion is obvious. We are given that sup A is a member of A and we know that no member of A can be larger than sup A, and so sup A is the largest member of A. 6. Is it possible for a set of numbers to have a supremum even though it has no largest member? Solution: You bet it’s possible! That’s the whole point of this chapter! 7. Given that  is an upper bound of a set A and that  A, explain why   sup A. We are given that  is an upper bound of A. Now if w is any number less than  then, because  A, the number w can’t be an upper bound of A. Therefore  must be the least upper bound of A. 8. Explain why the empty set does not have a supremum. Since every number is an upper bound of the empty set, there can’t be a least upper bound of the empty set. 9. Explain why the set 1, Ý does not have a supremum. The interval 1, Ý is unbounded above. This set doesn’t have any upper bounds and so it can’t 78 have a least one. 10. Given that two sets A and B are bounded above, explain why their union A Þ B is bounded above. Using the fact that A is bounded above, choose an upper bound u of A. Now, using the fact that B is bounded above, choose an upper bound v of B. We now define w to be the larger of the two numbers u and v. Given any member x of the set A Þ B, there are two possibilities: Either x A, in which case x u...
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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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