1873_solutions

# An example of a set a that has a largest member a

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: w or x B, in which case x v w. Therefore no member of the set A Þ B can be greater than w and we have shown that A Þ B is bounded above. 11. a. If a man says truthfully that he sells more BMWs than anyone in the Southeast, what can you deduce about him? Solution: He is not in the Southeast. Next problem: "I sell diamonds for less than anyone in the industry." No wonder he can sell them for less. He isn’t in the industry. The diamonds are hot. b. Given that   sup A and that x  , what conclusions can you draw about the number x? Solution: The number x cannot be an upper bound of A. In other words, there must exist a member of A that is larger than x. c. Given that   inf A and that x  , what conclusions can you draw about the number x? The number x cannot be a lower bound of A. In other words, there must exist a member of A that is less than x. 12. If A and B are sets of real numbers then the sets A  B and A B are defined by a A and b B such that x  a  b AB  x and A a. Work out A  B and A B x a A and b B such that x  a b B in each of the following cases: i. A  0, 1 and B  1, 0 . We have A  B  1, 1 . Certainly, the most that a  b can be if a A and b B is 1  0  1 and the least that a  b can be if a A and b B is 0  1  1. Now if x 1, 1 then there are two possibilities: Case 1: 1 x 0. In this case, the fact that 0 A and x B and x  0  x shows that x A  B. Case 2: 0  x 1. In this case, the fact that x A and 0 B and x  x  0 shows that x A  B. ii. A  0, 1 and B  1, 2, 3 . One can use an argument similar to that used for part i to show that 0, 1  1  1, 2 or, perhaps, one could declare this fact to be obvious. Similar remarks apply to 0, 1  2 and to 0, 1  3 . Finally, A  B  0, 1  1, 2, 3  1, 2 Þ 2, 3 Þ 3, 4  1, 4 . iii. A  0, 1 and B  1, 2, 3 . We have A  B  1, 2 Þ 2, 3 Þ 3, 4 . b. Prove that if two sets A and B are bounded then so are A  B and A B. Suppose that A and B are bounded. Choose lower bounds u 1 and u 2 of A and B respectively and choose upper bounds v 1 and v 2 of A and B respectively. To show that v 1  v 2 is an upper bound of A  B, suppose that x A  B. Choose a A and b B such that x  a  b. Since a u 1 and b u 2 we have x  a  b u 1  u 2 . Thus u 1  u 2 is an upper bound of A  B and similar argument show that u 1 v 2 is an upper bound of A B and v 1  v 2 is a lower bound of 79 A  B and v 1 u 2 is a lower bound of A B. Exercises on Supremum And Infimum 1. Suppose that A is a nonempty bounded set of real numbers that has no largest member and that a that sup A  sup A a. You saw in an earlier exercise that the sets A and A A. Prove a have exactly the same upper bounds. 2. Given that A and B are sets of numbers, that A is nonempty, that B is bounded above and that A B, explain why sup A and sup B exist and why sup A sup B. Since A is nonempty and A B we know that B is nonempty. Since A B we know that any upper bound of B must be an upper bound of A. Therefore, since B is bound...
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