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Assuming the Bolzano-Weierstrass Theorem that every sequence $x_n$ has either a convergent subsequence or a divergent subsequence. My question is can both occur?

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  • $\begingroup$ that is a misuse of notation. no sequence is converging to infinity. $\endgroup$ – Max Mar 4 '14 at 19:58
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    $\begingroup$ Every bounded divergent sequence has both. $\endgroup$ – Joshua Pepper Mar 4 '14 at 20:00
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    $\begingroup$ Every sequence in any topological space has a convergent or a divergent subsequence. No Bolzano theorem needed for that. $\endgroup$ – Christian Blatter Mar 4 '14 at 20:23
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Sure. Take $x_n=(-1)^n$. Then $\{x_n\}$ is divergent but has $\{x_{2n}\}$ as a convergent sub-sequence.

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Let the sequence $(a_n)$ defined by $$a_{n}=1\;\text{if $n$ is odd}\quad;\quad a_{n}=n\;\text{if $n$ is even}$$ so $(a_{2n+1})$ is a convergent subsequence and $(a_{2n})$ is a divergent subsequence.

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