1873_solutions

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

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.

Ask a homework question - tutors are online