1873_solutions

# The fact that x n converges to the number x we choose

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Unformatted text preview: and y n  x n x n y n for each n the present result follows at once from Exercise 2. 11. Suppose that x n and y n are sequences of real numbers, that x n y n a partial limit of at least one of the sequences x n and y n . Prove that xn 1 yn as n 0 and that the number 0 fails to be Ý. Solution: From Exercise 10 we know that x n and y n have the same sets of partial limits and we know, therefore, that 0 is not a partial limit of either of these two sequences. Using the fact that 0 is not a partial limit of y n , choose an integer N 1 and a number   0 such that the inequality |y n |  holds whenever n N 1 . For every n N 1 we see that |x n y n | xn 1  xn yn yn yn  and so the fact that xn 1 yn as n Ý follows from the sandwich theorem. 12. Give an example to show that the requirement in this exercise that 0 not be a partial limit of at least one of the two sequences is really needed. We define x n  2/n and y n  1/n for each n and observe that, even though x n y n 0 as n Ý, x n /y n 2 as n Ý. 13. Suppose that x n and y n are sequences of real numbers, that x n /y n 1 and that at least one of the sequences x n and y n is bounded. Prove that x n y n 0. Give an example to show that the conclusion x n y n 0 can fail if both x n and y n are unbounded. Hint: Make the observation that both of the sequences x n and y n must be bounded. Explain carefully how you arrive at this conclusion. Now make use of the fact that x |x n y n |  |y n | y n 1 n for every n. 14. Suppose that x n and y n are sequences of real numbers, that y n 1 and that for each n we have z n  x n y n . Prove that the sequences x n and z n have the same set of partial limits. It follows from the part of Exercise 3 for which a solution is provided above that if x is an partial limit of x n then the number x 1  x is a partial limit of the sequence z n . At the same time, since x n  z n /y n for all sufficiently large n we know that if x is any partial limit of z n then the number x/1  x is a partial limit of the sequence x n . 142 Exercises on Sequences and the Topology of R 1. Prove that a set S of real numbers is unbounded below if and only if there exists a sequence x n in S such that x n Ý. Solution: We want to show that, for a given set S of real numbers, the following conditions are equivalent: a. The set S is unbounded below. b. There exists a sequence x n in the set S such that x n Ý as n Ý. To show that condition b implies condition a, assume that condition b holds and choose a sequence x n in S such that x n Ý as n Ý. If w is any real number then it follows at once from the fact that the sequence x n is eventually in the interval Ý, w that w is not a lower bound of S. Therefore S is not bounded above. Now to show that condition a implies condition b, assume that condition a holds. For each positive integer n we use the fact that the number n is not a lower bound of S to choose a number that we shall call x n such that x n  n. The sequence x n that we have made in this wa...
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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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