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Prove statement on a sequence of real numbers

  1. Dec 3, 2013 #1
    The problem statement, all variables and given/known data.

    Prove that ##\{x_n\}_{n \in \mathbb N} \subset \mathbb R## doesn't have any convergent subsequence iff ##lim_{n \to \infty} |x_n|=+\infty##.

    The attempt at a solution.

    I think I could correctly prove the implication ##lim_{n \to \infty} |x_n|=+\infty \implies## it doesn't exist any convergent subsequence:

    Suppose there exists ##\{x_{n_k}\}_{k \in \mathbb N}## convergent and call the limit ##A##. Then, for ##ε=1##, there exists ##n_0: \forall n≥n_0, |x_{n_k}-A|< 1##.

    This means, ##\forall n≥n_0, |x_{n_k}|-|A|\leq |x_{n_k}-A|< 1 \implies |x_{n_k}|<1+|A| ##.

    On the other hand, ##lim_{n \to \infty} |x_n|=+\infty##, so, for ##M=1+|A|##, there exists ##n_1 : \forall n≥n_1, |x_n|>M##.

    Take ##N=max\{n_0,n_1\}##, then for all ##n_k≥N, \space M<|x_{n_k}|<M##, which is clearly absurd. This proves that with the given hypothesis, it can't exist any convergent subsequence of the original sequence.

    I need help to prove the other implication: If there is no convergent subsequence then ##lim_{n \to \infty} |x_n|=+\infty##.
     
  2. jcsd
  3. Dec 3, 2013 #2

    Dick

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    Try and think about proving the contrapositive. I.e. if it's not true that ##lim_{n \to \infty} |x_n|=+\infty## then ##x_n## has a convergent subsequence. You'll have to think carefully about how to negate the definition of the limit.
     
    Last edited: Dec 3, 2013
  4. Dec 3, 2013 #3
    Suppose it's not true that ##lim_{n \to \infty} |x_n|=+\infty##. Then, there exists ##M>0## ,such that for every natural number ##j##, there is ##n_j>j## : ##|a_{n_j}|\leq M##. This means the subsequence ##\{a_{n_j}\}_{j \in \mathbb N}## is bounded. By the Bolzano Weierstrass theorem, there exits ##\{x_{n_{j_k}}\}_{k \in \mathbb N}## convergente subsequence. But ##\{x_{n_{j_k}}\}_{k \in \mathbb N}##is also a subsequence of the sequence ##\{x_n\}_{n \in \mathbb N}##. From here it follows the contraposivite is true, so if ##\{x_n\}_{n \in \mathbb N}## doesn't have any convergent subsequence, then ##lim_{n \to \infty} |x_n|=+\infty##.

    Thanks, Dick!
     
    Last edited: Dec 3, 2013
  5. Dec 3, 2013 #4

    Dick

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    That's roughly the idea, but I'd stay away from writing things like ##lim_{n \to \infty} |x_n|≠+\infty##, it sort of implies that ##|x_n|## has a limit, but it's just not infinity. ##|x_n|## may have no limit. More seriously, you can't pick ##M>0## arbitrary, some M's may not work. Think about the negation of the limit definition again.
     
  6. Dec 3, 2013 #5
    You're right, ##M## is not arbitrary. I've corrected it. However, I don't understand why saying it's not true ##lim_{n \to \infty} |x_n|=+\infty## is not equivalent to say ##lim_{n \to \infty} |x_n|≠+\infty##. If it is not infinity, then, it is different to infinity, I agree that the limit may not exist (the sequence could oscillate) but why saying ##lim_{n \to \infty} |x_n|≠+\infty## would imply ##|x_n|## has a limit?
     
  7. Dec 3, 2013 #6

    Dick

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    Because if the limit doesn't exist, then you are saying that something that doesn't exist is not equal to infinity. Doesn't that sound funny to you? It does to me.
     
  8. Dec 3, 2013 #7
    Yes, now I see it.
     
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