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?


2 Answers 2


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.

  • $\begingroup$ This is awesome! $\endgroup$ May 13, 2019 at 17:59
  • $\begingroup$ Is my proof okay, though? $\endgroup$ May 13, 2019 at 17:59
  • $\begingroup$ Although it is not generalisable to arbitrary metric spaces. $\endgroup$ May 13, 2019 at 18:02
  • $\begingroup$ Ignore previous comment, I think your proof is fine. $\endgroup$ May 13, 2019 at 18:03
  • $\begingroup$ Yes. I should have mentioned that more specifically. $\endgroup$ May 13, 2019 at 18:04

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.

  • $\begingroup$ That is why I mentioned, "ANY MONOTONICALLY INCREASING SUBSEQUENCE". not just a particular one. $\endgroup$ May 13, 2019 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. $\endgroup$ May 13, 2019 at 16:24
  • 1
    $\begingroup$ 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, 2019 at 16:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.