1873_solutions

# inf x g x 0 prove that if a and b are any real

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Unformatted text preview: um values, respectively, of the function f. 84 4. Prove that if f and g are bounded above on a nonempty set S then sup f  g sup f  sup g. Solution: Given any member x of the set S, since f x sup f and g x sup g, we have fg x  f x g x sup f  sup g. Thus the number sup f  sup g is an upper bound of the function f  g and so sup f  g sup f  sup g. 5. Give an example of two bounded functions f and g on the interval 0, 1 such that sup f  g  sup f  sup g. 6. Given that f is a bounded function on a nonempty set S and that c is a real number prove that c sup f if c  0 sup cf  c inf f if c  0 Solution: We give the solution here for the case c  0 and leave it to you to handle the case c  0. Suppose that c  0. Given any member x of the set S, it follows from the fact that inf f f x that cf x c inf f and we conclude that c inf f is an upper bound of the function cf. To show that c inf f is the least upper bound of cf we need to show that no number less than c inf f can be an upper bound of cf. Suppose that p  c inf f. Since p/c  inf f we know that p/c fails to be a lower bound of f. Using this fact, we choose a member x of S such that p c fx. Since p  cf x we have shown, as promised, that p fails to be an upper bound of the function cf. 7. Prove that if f is a bounded function on a nonempty set S then |sup f | sup|f |. Solution: For every x S we have fx |f x | sup|f | and so sup f Futhermore, for every x sup|f |. S we have fx sup|f |. |f x | Using the fact that the set S is nonempty, choose a member x of the set S. We see that sup|f | f x sup f and so sup|f | sup f sup|f | and we have shown that |sup f | sup|f |. Exercises on Sequences of Sets 1. Evaluate 85 Ý  1 ,1 . n n1 0, 1 . Now given any number x We observe first that if n is a positive integer then 1 , 1 n 1 1 there exist positive integers n for which n  x and for such n we have x n , 1 . Therefore Ý  n1 1 ,1 n 0, 1  0, 1 . 2. Evaluate Ý  n1 1  1 ,5 n 2. n Using an argument similar to the one used in Exercise 1 we can see that Ý  n1 1  1 ,5 n 2 n  1, 5 . 3. Evaluate Ý  1 ,2  1 . n n 1 n1 Using an argument similar to the one used in Exercise 1 we can see that Ý  n1 1 1 ,2  1 n n  1, 2 . 4. Explain why if A n is an expanding sequence of subsets of R then the sequence R A n is a contracting sequence. Suppose that A n is an expanding sequence of subsets of R. For each n it follows at once from the fact that A n A n1 that R A n 1 R An. 5. Suppose that A n is a sequence of sets. a. Prove that if we define Bn  Ý  Aj jn for each n then the sequence B n is a contracting sequence of sets. Given any positive integer n, the inequality Ý Ý  Aj jn1  Aj jn Ý A jn1 j follows at once from the observation that if x  then there is an integer j which x A j and so there must certainly exist an integer j n for which x A j . b. Prove that if we define Bn  Ý  Aj jn for each n then the sequence B n is an expanding sequence of sets. Given any positive integer...
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