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Let $x_{n}$ be a bounded but not convergent sequence. Prove that $x_{n}$ has two subsequences converging to different limits.

Proving by contradiction. Just one question. It doesent necessarily have to be two subsequence that converge to different limits? There can be in fact infinitely many of them? They can be found by applying the Bolozano Weierstrass theorem infinetly many times to get subsequence of the subsequence of the subsequence etc infinetly many times that will converge to different limits? Could anyone explain. Thanks

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    $\begingroup$ Possible duplicate of Prove that a bounded sequence has two convergent subsequences. $\endgroup$ – Kal S. Feb 10 '17 at 8:49
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    $\begingroup$ The assumption in proof by contradiction is not the negative of the original statement. For proof by contradiction you need to assume that all convergent subsequences, converge to the same limit. Check also this: math.stackexchange.com/questions/231888/… $\endgroup$ – Kal S. Feb 10 '17 at 8:52
  • $\begingroup$ Thanks i got it. Just one question. It doesent necessarily have to be two subsequence that converge to different limits. ? There can be in fact infinitely many of them? They can be found by applying the Bolozano Weierstrass theorem infinetly many times to get subsequence of the subsequence of the subsequence etc infinetly many times that will converge to different limits? $\endgroup$ – ys wong Feb 10 '17 at 11:01
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Hint: Bolzano-Weierstrass ensures that there there is a convergent subsequence $a_{n_k} \to r$. But $a_n$ does not converge, so every subsequence cannot converge to $r$.

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  • $\begingroup$ Thanks i got it. Just one question. It doesent necessarily have to be two subsequence that converge to different limits. ? There can be in fact infinitely many of them? They can be found by applying the Bolozano Weierstrass theorem infinetly many times to get subsequence of the subsequence of the subsequence etc infinetly many times that will converge to different limits? $\endgroup$ – ys wong Feb 10 '17 at 10:59
  • $\begingroup$ No, I don't think so. Consider (-1)^n and pick your firstsubsequence as the even numbers $\endgroup$ – Andres Mejia Feb 11 '17 at 7:36
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A different way to proof this would be the following:

Since $\lim_n x_n$ does not exist we have $$\alpha:= \liminf_n x_n \neq \limsup_n x_n=:\beta$$ and $\alpha$ and $\beta$ exist since $(x_n)_n$ is bounded. Therefore we have by definition of $\liminf $ and $\limsup$ two subsequence $(x_j)_j$ and $(x_k)_k$ of $(x_n)_n$ with $\lim_j x_j = \alpha$ and $\lim_k x_k= \beta$.

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