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Suppose that $(u_n)$ is unbounded above. Then, we pick any arbitrary monotone increasing subsequence $(v_n)$ of $(u_n)$. But by hypothesis, we can find a subsequence of $(v_n)$ that converges to $0$. Hence, $\lim v_n =0$. Therefore, $(u_n)$ must be bounded above.

Again, let $(u_n)$ be unbounded below. Then we pick any monotone decreasing subsequence $(w_n)$. By similar arguments, $(u_n)$ must be bounded below.

Being bounded, it has a finite $\limsup =l$ and $\liminf =m$.

Hence, there must be convergent subsequences $(a_n)$ and $(b_n)$ that converge to $l$ and $m$ respectively. But every subsequences of the above two must converge to $l$ and $m$ respectively (treating the two as convergent sequences). Again, by hypothesis, they both have subsequences that converge to $0$. Therefore, both $(a_n)$ and $(b_n)$ converge to $0$.

In conclusion, $\limsup u_n = \liminf u_n = 0 =\lim u_n$

Is this alright?

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Here's a more direct approach that also generalizes to limits in any metric space (or any sequential topological space).

Suppose to the contrary that $u_n$ does not converge to zero. Then there is a neighborhood $V$ of $0$ such that infinitely many of the $u_n$ are outside $V$. Consider these infinitely many $u_n$ as a subsequence of the original sequence, $u_{n_1}, u_{n_2}, \dots$. But by assumption, the subsequence $u_{n_k}$ has a further subsequence that converges to 0, which means that infinitely many of the $u_{n_k}$ are in $V$. This is a contradiction.

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  • $\begingroup$ This is awesome! $\endgroup$ – Subhasis Biswas May 13 '19 at 17:59
  • $\begingroup$ Is my proof okay, though? $\endgroup$ – Subhasis Biswas May 13 '19 at 17:59
  • $\begingroup$ Although it is not generalisable to arbitrary metric spaces. $\endgroup$ – Subhasis Biswas May 13 '19 at 18:02
  • $\begingroup$ Ignore previous comment, I think your proof is fine. $\endgroup$ – Nate Eldredge May 13 '19 at 18:03
  • $\begingroup$ Yes. I should have mentioned that more specifically. $\endgroup$ – Subhasis Biswas May 13 '19 at 18:04
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I think the general idea behind your proof is fine, but your argumentation is lacking.

As my misunderstanding shows, I don't think it is quite clear what the exact argument in your answer is supposed to be. Here is a clearly formulated rendition (which I think is in roughly the same spirit as what you have written)

Suppose $(u_n)_n$ is unbounded above. Then there is a subsequence $(u_{n_k})_k$ such that $u_{n_k}\rightarrow0$ as $k\rightarrow\infty$. Since $(u_n)_n$ is monotonically increasing, we must have $u_n\le0$ for all $n\in\mathbb{N}$. This contradicts the unboundedness assumption, hence $(u_n)_n$ is bounded above.

Analogously, you show that $(u_n)_n$ is unbounded below and then you can proceed with the $\limsup,\liminf$ argument. I suggest also rewriting that part in a way to make it clear what the argument, e.g. by contradiction, is.

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  • $\begingroup$ That is why I mentioned, "ANY MONOTONICALLY INCREASING SUBSEQUENCE". not just a particular one. $\endgroup$ – Subhasis Biswas May 13 '19 at 16:23
  • $\begingroup$ that one statement covers your case too. But I agree with your opinion that this proof is lacking some formulation/ proper arrangement. $\endgroup$ – Subhasis Biswas May 13 '19 at 16:24
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    $\begingroup$ Your first paragraph says "an" instead of "any". Guess that is an unfortunate typo then. I'll edit my answer to reflect how I think your argument is supposed to go. Let me know whether the new version will be closer to what you intended. $\endgroup$ – Thorgott May 13 '19 at 16:31

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