1873_solutions

And therefore bounded we choose a number p such that

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Unformatted text preview: 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 166 |x n y n |  |y n | x n yn 1 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 . 15. Given that x n is a sequence in R k and that for every point x in R k we have x n  x 0 as n Ý, prove that the sequence x n converges to the point O. Taking x  1, 0, , 0 we see that the sequence of first coordinates of the points x n converges to zero. We can repeat this argument for each of the other coordinates. Exercises on Sequences and the Topology 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 way is in the set S and it is clear that xn Ý. 2. Suppose that S is a nonempty set of real numbers and that  is an upper bound of S. Prove that the following conditions are equivalent: a. We have   sup S. b. There exists a sequence x n in S such that x n  as n Ý. In Exercise 7 of the exercises on closure we saw that an upper bound  of set S is close to S if and only if  is equal to sup S. We also saw in a recent theorem that a number  is close to a set S if and only if there exists a sequence in S that converges to . The present exercise follows at once from these two facts. Of course, we can write this exercise out directly if we wish. 4. Given a sequence x n that is frequently in a subset S of a metric space X, and given a partial limit x of the sequence x n , is it necessarily true that x S? No, this statement need not be true. If we define x n ...
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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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