1873_solutions

# A subset of a metric space that l s and that 0 prove

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Unformatted text preview: ball B y,  contains more than one point of the set E. Therefore y is a limit point of E. 132 7 Limits of Sequences Some Exercises on Subsequences Decide whether each of the following assertions is true or false. If it is true, prove it. If it is false, illustrate this fact by giving an example. 1. If a sequence x n is eventually in a given set S then every subsequence of x n is eventually in S. The assertion is true. Suppose that x n is a sequence that is eventually in a given set S and that x n i is a subsequence of x n . Choose an integer N such that the condition x n S holds whenever n N. Now choose an integer j such that the condition n i N holds whenever i j. We see that x n i S whenever i j. 2. If a sequence x n is frequently in a given set S then every subsequence of x n is frequently in S. The assertion is false. If we define x n  1 n for every positive integer n then, although x n is frequently in the set 1 , the subsequence x 2n fails to be frequently in the set 1 . 3. If every subsequence of a sequence is frequently in a given set S then x n is frequently in S. The statement is obviously true because every sequence is a subsequence of itself. 4. If every subsequence of a sequence x n is frequently in a given set S then x n is eventually in S. The assertion is true. We shall show that if x n fails to be eventually in S then x n must have a subsequence that is not frequently in S. Suppose that x n is a sequence that fails to be eventually in a given set S. We know that x n is frequently in the set R S and from the preceding theorem we deduce that x n has a subsequence in the set R S. Such a subsequence cannot be frequently in S. 5. If x n is a sequence of real numbers and S R, and if x n is not eventually in S then x n has a subsequence that is eventually in R S. Since a sequence that is not eventually in S must be frequently in R S, the desired result follows at once from a theorem on subsequences. Some Exercises on Limits and Partial Limits 1. Given that xn  3  1 n for each positive integer n, prove that 3 is a limit of x n . We need to prove that for every number  0 the sequence x n will eventually be in the interval 3 , 3  . So we start: Suppose that  0. Before we go any further we need to ask ourselves what it means to say that x n 3 , 3  . We observe that the inequality 3  x n  3  is equivalent to the assertion that 3  3 1  3 n which holds when 1/n  . This tells how how to continue: Using the fact that the number 1/ is not an upper bound of the set Z of integers we choose an integer N such that N  1/ , in other words, 1. N Then, whenever n N we have 3 1  3 3  3  3 1 n N 133 and so we have shown that x n is eventually in 3 ,3  . 2. Given that xn  3  2 n for each positive integer n, prove that 3 is a limit of x n . We need to prove that for every number  0 the sequence x n will eventually be in the interval 3 , 3  . So we start: Suppose that  0. This time we make the observation that the inequality 3  3...
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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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