Claim: Every bounded and divergent sequence $(a_n)$ contains only convergent subsequences.

A bounded, divergent sequence $(a_n)$ implies that there is at least one convergent subsequence (by Bolzano Weierstrass) $(a_{nk}) \rightarrow l$. Now, because $(a_n)$ diverges, there exists some subsequence $(a_{jk})$ which isn't in the $\epsilon -$ neighborhood of $l$. This means that either $(a_{jk})$ is divergent or converges to some other point (as it cannot be unbounded). If it converges to some other point $(a_{jk}) \rightarrow k$ we have shown that there exist two limit points and we are done. If, however, $(a_{jk})$ is divergent itself, I can apply Bolzano Weierstrass again and basically have a recursive loop until I am only left with only convergent subsequences that all converge to a different limit.

Is this valid? I'm not really sure if this is sufficient or even correct...

  • 4
    $\begingroup$ Not sure I understand the claim. The original sequence is, of course, a subsequence of itself. $\endgroup$
    – lulu
    Mar 23 at 14:53
  • $\begingroup$ Or just remove finitely many elements from the sequence; it still diverges. $\endgroup$
    – saulspatz
    Mar 23 at 14:55
  • $\begingroup$ If the question is whether a bounded divergent sequence contains two subsequences that converge to different numbers, the answer is yes. However, since any sequence is a sub-sequence of itself, not every subsequence of a divergent sequence converges. $\endgroup$ Mar 23 at 14:59
  • $\begingroup$ Right, my bad! What I am asking is if I have a divergent, bounded sequence, are then all proper subsequences convergent (so all subsequences excluding the sequence itself)? $\endgroup$ Mar 23 at 15:03
  • $\begingroup$ As the comments have indicated already, the answer is "no": removing any finite number of elements still leaves you with a divergent subsequence. $\endgroup$
    – NickD
    Mar 23 at 15:05

As the comments already mentioned, the claim is incorrect (The sequence itself is a subsequence for example). The flaw in your reasoning is in your recursive loop. You implicitly assume this loop will end in finite steps. This is by no means clear, since we can have infinitely many different subsequences

  • $\begingroup$ Thanks a lot. That was what I worried about. However, I googled around and found iuuk.mff.cuni.cz/~andrew/MAex5s.pdf where they claim that "...a subsequence can fail in two ways to converge to a limit $l$. Either it is unbounded or it converges to another limit." But what about the case in which the subsequence itself is divergent? I thought that my claim is how they came to that reasoning... $\endgroup$ Mar 23 at 15:11
  • $\begingroup$ @focusonimprove333. ?? "...a subsequence can fail in two ways to converge to a limit $l.$ Either it is unbounded or it converges to another limit." NO. A subsequence fails to converge to $l$ either because (i) it is an unbounded subsequence or (ii) it has a sub-subsequence converging to some $m\ne l$ or (iii) both (i) and (ii). $\endgroup$ Mar 23 at 22:10
  • $\begingroup$ Yeah, this is the case that I was missing as well. That's why I tried it with recursion. But I guess I found a way that is much easier per contradiction. But thanks! This thread really helped me a lot! $\endgroup$ Mar 24 at 0:18

No. Consider the bounded sequence $1,2,3,1,2,3,1,2,3,\cdots $. One of its proper subsequences is $1,2,1,2,1,2,\cdots$ and it does not converge.

  • $\begingroup$ Okay, I see. I totally forgot that case. But since 1,2,1,2 is again not convergent, I could use Bolzano Weierstrass to split them again into two subsequences that are convergent, right? The problem I wanted to solve was the following: Assume that $(a_n)$ is a bounded sequence s.t. every convergent subsequence converges to the same limit $a$. Show that $(a_n)$ must converge to a. $\endgroup$ Mar 23 at 15:21
  • $\begingroup$ @focusonimprove333: You can prove that by contradiction also. The point is that in $\mathbb R$, if you have a bounded sequence, it will always have convergent subsequence. $\endgroup$
    – Koro
    Mar 23 at 15:23
  • $\begingroup$ Okay cool. Thanks for the tip! $\endgroup$ Mar 23 at 15:27

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