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Let $(x_n)$ be a divergent sequence in a compact subset of $\mathbb R^n$. Prove that there are two subsequences of $(x_n)$ that are convergent to different limit points.

Some ideas that might be helpful:

Heine-Borel theorem states that a subset of $\mathbb R^n$ is compact if and only if it is closed and bounded.

Bolzano-Weierstrass Theorem, every bounded sequence contains a convergent subsequence

A number $c$ is a limit point of $(x_n)$ if there exists a subsequence of $(x_n)$ convergening to $c$

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  • $\begingroup$ You can find some inspiration here: math.stackexchange.com/questions/298817/… You just have to replace the absolute value by the norm in $\mathbb{R}^n$. $\endgroup$ – Julien Mar 2 '13 at 22:13
  • $\begingroup$ @Ludolila: I don't understand why you have removed the part about the Bolzano-Weierstrass theorem from the post. $\endgroup$ – Asaf Karagila Mar 2 '13 at 22:18
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    $\begingroup$ @Ludolila: math.stackexchange.com/posts/318970/revisions $\endgroup$ – Asaf Karagila Mar 2 '13 at 22:20
  • $\begingroup$ @AsafKaragila : you're right... Sorry, OP... have no idea how it happened... I'll put it back... =) $\endgroup$ – Ludolila Mar 2 '13 at 22:21
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By Bolzano Weierstrass you can pull out a convergent subsequence whose limit is some point $c$. Since your original sequence cannot converge, there's going to be a subsequence that doesn't converge to $c$. Now carefully pluck this subsequence so that it stays away at some fixed positive distance away from $c$. But this subsequence also satisfies Bolzano Weierstrass...

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By Bolzano-Weierstrass $(x_{n})$ has a subsequence converging to some $x$. Now we know that $(x_n)$ does not converge to $x$ because it is divergent. Hence for some $\varepsilon>0$ we have $x_{n}\notin B(x,\varepsilon)$ for infinitely many $n$. Take these indices and denote the obtained subsequence by $(x_{n_{k}})$. This sequence is bounded and has a limit point (by Bolzano-Weierstrass) and it is different from $x$.

Hence $(x_n)$ has two different limit points.

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There is a subsequence which converges to limsup of the original sequence and another sequence which converge to liminf of original sequence. As the original sequence diverges, its limsup and liminf must be different.

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  • $\begingroup$ Note that we are in $\mathbb{R}^{n}$ and not in $\mathbb{R}$. How do you define limsup and liminf of a vector valued sequence? $\endgroup$ – T. Eskin Mar 3 '13 at 10:46
  • $\begingroup$ Thanks Thomas! I did not look at the problem carefully. Indeed, my arguments work only when n=1. $\endgroup$ – chandu1729 Mar 3 '13 at 20:50

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