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Math Help - Convergent subsequence

  1. #1
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    Convergent subsequence

    Prove or disprove:

    Let x_n \in R and suppose the sequence \{\sin(x_n)\} converges. The sequence \{{x_n}\} must contain a convergent subsequence.

    I am not being able to figure out on how to prove or disprove this. Any help will be appreciated.
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  2. #2
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    Quote Originally Posted by serious331 View Post
    Prove or disprove:

    Let x_n \in R and suppose the sequence \{\sin(x_n)\} converges. The sequence \{{x_n}\} must contain a convergent subsequence.

    I am not being able to figure out on how to prove or disprove this. Any help will be appreciated.
    False:

    Let  x_k = 2\pi k
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  3. #3
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    Quote Originally Posted by southprkfan1 View Post
    False:

    Let  x_k = 2\pi k
    Thanks southparkfan.

    does the sequence \{sin(2\pi k)\} converge to 0? because it is 0 for all values of k?

    and how can we show that the subsequence of \{(2\pi k)\} doesn't converge?
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  4. #4
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    Quote Originally Posted by serious331 View Post
    how can we show that the subsequence of \{(2\pi k)\} doesn't converge?
    You are joking arenít you? You cannot be serious.
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  5. #5
    MHF Contributor Bruno J.'s Avatar
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    A convergent sequence is always bounded!
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  6. #6
    Senior Member Pinkk's Avatar
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    Given any M>0, let N=\frac{M}{2\pi}. Then k>N implies 2\pi k > M. And so \lim_{k\to \infty} 2\pi k = +\infty. It's pretty clear why any subsequence cannot converge since the tail of any subsequence will still tend to +\infty.
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  7. #7
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    Quote Originally Posted by Plato View Post
    You are joking arenít you? You cannot be serious.
    Okay, my question was retarded. Make as much fun as you can!
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  8. #8
    Senior Member Pinkk's Avatar
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    Quote Originally Posted by serious331 View Post
    Okay, my question was retarded. Make as much fun as you can!
    I have to say Plato, I did think that was a tiny bit arrogant to ask. It should be obvious, but better to ask than to just assume!

    If you wanna formalize why there can be no convergent subsequence, we can assume to the contrary. Let [tex]s_{n} = 2\pi n[/math[ and assume that there is a subsequence (s_{n_{k}}) such that given \epsilon > 0, there exists N such that n_{k} > N implies |s_{n_{k}} - a| < \epsilon, where a is the finite number that we suppose the subsequence will converge to. And so this means for all n_{k} > N, s_{n_{k}} <  \epsilon + a (assume that a >  0 because if it wasn't, then clearly after some point the sequence would have to be negative). But since \epsilon + a > 0, we know for all n >  \frac{\epsilon + a}{2\pi}, 2\pi n > \epsilon + a, and so rises the contradiction.
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  9. #9
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    Quote Originally Posted by Pinkk View Post
    I have to say Plato, I did think that was a tiny bit arrogant to ask. It should be obvious, but better to ask than to just assume!

    If you wanna formalize why there can be no convergent subsequence, we can assume to the contrary. Let [tex]s_{n} = 2\pi n[/math[ and assume that there is a subsequence (s_{n_{k}}) such that given \epsilon > 0, there exists N such that n_{k} > N implies |s_{n_{k}} - a| < \epsilon, where a is the finite number that we suppose the subsequence will converge to. And so this means for all n_{k} > N, s_{n_{k}} <  \epsilon + a (assume that a >  0 because if it wasn't, then clearly after some point the sequence would have to be negative). But since \epsilon + a > 0, we know for all n >  \frac{\epsilon + a}{2\pi}, 2\pi n > \epsilon + a, and so rises the contradiction.
    Thank you for the detailed explanation Pinkk. Not everyone is as smart as Mr. Plato.

    I have another question. how can we prove that if the sequence \{|x_n|\} converges, then the sequence \{x_n\} has a convergent subsequence.
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  10. #10
    Senior Member Pinkk's Avatar
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    Since |x_{n}| converges to let's say a \ge 0 (obviously the limit must be nonnegative since |x_{n}| \ge 0 by virtue of the absolute value), then we have some N such that n > N implies ||x_{n}| - a| <  1, so |x_{n}| is bounded by 1+a for all n > N. So let M = \max\{|x_{1}|, |x_{2}|, ... , |x_{N}|, 1+a\}. Then we have |x_{n}|\le M for all n, so by definition, (x_{n}) is a bounded sequence. Thus, by the Bolzano-Weierstrass Theorem, (x_{n}) has a convergent subsequence.
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