# If a sequence $(u_n)$ be such that its every subsequence has a subsequence that converges to $0$, then $\lim u_n= 0$

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?

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.

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

• That is why I mentioned, "ANY MONOTONICALLY INCREASING SUBSEQUENCE". not just a particular one. – Subhasis Biswas May 13 '19 at 16:23
• that one statement covers your case too. But I agree with your opinion that this proof is lacking some formulation/ proper arrangement. – Subhasis Biswas May 13 '19 at 16:24
• 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. – Thorgott May 13 '19 at 16:31