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Find and example of a sequence $\{x_n\}$ which does not contain any subsequences having a finite limit.

I've been thinking of the following seqeunce: $$ x_n = \sin(n)\cdot\sin(\sqrt{3}n) $$

But is that true? Also what are other examples of such sequences?

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    $\begingroup$ Natural numbers $1,2,3,...$? No subsequence is Cauchy, hence convergent? $\endgroup$ – астон вілла олоф мэллбэрг Dec 10 '18 at 15:46
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    $\begingroup$ @астонвіллаолофмэллбэрг oh my, shame on me $\endgroup$ – roman Dec 10 '18 at 15:48
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    $\begingroup$ Remember, every bounded sequence of real numbers has a convergent subsequence. Therefore, your sequence being bounded does not have a convergent subsequence. The best bet is to take an unbounded sequence where the distance between any two terms is greater than some fixed number. $\endgroup$ – астон вілла олоф мэллбэрг Dec 10 '18 at 15:50
  • $\begingroup$ @астонвіллаолофмэллбэрг thank you! it's crystal clear now $\endgroup$ – roman Dec 10 '18 at 15:52
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    $\begingroup$ Great. You can now answer this question yourself and close it. $\endgroup$ – астон вілла олоф мэллбэрг Dec 10 '18 at 15:52
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Any sequence $\{x_n\}$ with $|x_n| \to \infty$ will do.

By Bolzano-Weierstrass Theorem others wont: if $|x_n| \not \to \infty$, there is $M$, such that you can find a subsequence of $\{x_n\}$ that is bounded by $M$. Then you can find a converging subsequence of that subsequence (and hence of the original sequence).


EDIT: By the way, if we think of the one point compactification of $\mathbb{R}$ where we add $\infty$ in, then every sequence has a converging subsequence in this space (which is homeomorphic to circle). So this also shows that the only way to have no subsequence with finite limit is to make that converging subsequence that is quaranteed to exists in the compactified space to have limit $\infty$. Well I guess you still have to make the above argument to show that we must have $|x_n|\to \infty$ (i.e $x \to \infty$ in the compactified space).

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The example in the OP is wrong. This can be shown by Bolzano-Weierstrass theorem which states that every bounded sequence contains a convergent subsequence.

Now if we consider: $$ x_n = \sin(n)\sin(\sqrt{3}n) $$ its clear that the sequence is bounded by $[-1, 1]$. Hence it contains a convergent subsequence.

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  • $\begingroup$ Upvoted. Good job. $\endgroup$ – астон вілла олоф мэллбэрг Dec 10 '18 at 16:00
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    $\begingroup$ Take for example $n=1, 8,77, 146, 721, 5270, 19492 , 33714 ,148357, 263000, 674357$ and you get $ 0.8305543, 0.9506199, 0.9883301, 0.9962933, 0.9996635, 0.9997120, 0.9998930, 0.9999708, 0.9999954, 0.9999987, 0.9999999$ $\endgroup$ – Henry Dec 10 '18 at 23:00

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