1873_solutions

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

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
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