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Let ($x_{n})$ be an unbounded sequence.

Prove that for all j $\in N$, the subsequence $x_{nj}$ is also unbounded.

Prove by contradiction.

Suppose that there exists a subsequence $x_{nj}$ that is bounded then by the Bolozano weierstrass theorem there exists another subsequence $x_{ni}$ that is convergent.(In fact there are two convergent subsequences) Let it be convergent to a limit $a$ and since this $x_{ni}$ convergers to a limit $a$ then the original sequence also converges to a limit of $a$ which implies that $x_{n}$ is a bounded sequence thus a contradiction. Is my proof correct. Could anyone explain . Thanks.

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    This is incorrect. Consider the sequence $1,1,1,2,1,3,1,4,1,5,1,6,1,7,1,8,\dots$. – 2017-02-12
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    a subsequence converging to $a$ only implies the whole sequence converges to $a $ when you know the whole sequence does indeed converge to something (which must be $a $). If the sequence *doesn't* converge nothing can be concluded by the subsequences. We can obviously have the even terms converge and the odd terms bounce all over the place. I suspect what you are being asked to prove is that some, but not all, of those subsequences are unbounded. – 2017-02-12

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