1873_solutions

In the industry the diamonds are hot b given that sup

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Unformatted text preview: ed above we know that that A is bounded above. Finally, since sup B, being an upper bound of B, must be an upper bound of A, and since sup A is the least upper bound of A we have sup A sup B. 3. Given that A is a nonempty bounded set of numbers, explain why inf A sup A. Using the fact that A is nonempty we choose a member a of A. We know that inf A a sup A. 4. It is given that A and B are nonempty bounded sets of real numbers, that for every x A there exists y such that x  y and for every y B there exists x A such that y  x. Prove that sup A  sup B. B Solution: We need to show that the two sets A and B have exactly the same upper bounds and for this purpose we shall show that a number fails to be an upper bound of A if and only if it fails to be an upper bound of B. Suppose that u fails to be an upper bound of A. Choose a member x of A such that u  x. Using the given property of A and B we now choose a member y of the set B such that x  y and, since u  x  y, we conclude that u can’t be an upper bound of B. We can show similarly that a number that fails to be an upper bound of B must also fail to be an upper bound of A. 5. Suppose that A and B are nonempty sets of real numbers and that for every x A and every y B we have x  y. Prove that sup A inf B. Give an example of sets A and B satisfying these conditions for which sup A  inf B. Given any member y of the set B it follows from the fact that x  y for every x A that y must be an upper bound of A. In other words, every member of B is an upper bound of A and the fact that B is nonempty shows that A is bounded above. A similar argument shows that every member of A is a lower bound of B and the fact that A is nonempty guarantees that B is bounded below. Thus sup A and inf B exist. Given any member y of B, the fact that y is an upper bound of A and sup A is the least upper bound of A tells us that sup A y. In other words, sup A is a lower bound of B. Since inf B is the greatest lower bound of B we deduce that sup A inf B. 6. Suppose that A and B are nonempty sets of real numbers and that sup A  inf B. Prove that for every number   0 it is possible to find a member x of A and a member y of B such that x    y. Solution: Suppose that   0. Using the fact that sup A    inf B    inf B and that sup A   is therefore not a lower bound of B, choose a member y of the set B such that sup A    y. From the fact that y   sup A we deduce that y  is not an upper bound of A and, using this fact, we choose a member x of A such that y   x. In this way we have found x A and Y B such that x    y. 80 inf B y−δ x sup A sup A + δ y 7. Suppose that A and B are nonempty sets of real numbers, that sup A inf B and that for every number   0 it is possible to find a member x of A and a member y of B such that x    y. Prove that sup A  inf B. Solution: To obtain a contradiction, assume that sup A  inf B and define We observe that   0. Now for all x that   inf B sup A. A and y B,...
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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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