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
AB 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...
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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.
 Fall '08
 STAFF
 Math, Calculus

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