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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.
 Fall '08
 STAFF
 Math, Calculus

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