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Can I conclude from Bolzano–Weierstrass theorem that there is more than one convergent subsequence, or the theorem tells me that there's only one ?

To be more clear, given a bounded sequence $X_n$, not ecessarily converges, can I conclude there are two different subsequences $X_{n_k}$ that converges to $L_1$ and $X_{n_l}$ that converges to $L_2$?

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  • $\begingroup$ Yes, I know. The question is if there exist more than one $\endgroup$ – Itay4 Feb 16 '17 at 14:59
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    $\begingroup$ Not if $L_1\not= L_2$, as if $X_n$ is convergent, there is only one possible limit. $\endgroup$ – M. Winter Feb 16 '17 at 15:08
  • $\begingroup$ @M.Winter But $X_n$ is not convergent $\endgroup$ – Itay4 Feb 16 '17 at 15:09
  • $\begingroup$ But it can! And this is why Bolzano-Weierstrass cannot (in general) let you conclude two such sequencies. But you can add the assumption that $X_n$ should not converge and I am already looking for an answer. $\endgroup$ – M. Winter Feb 16 '17 at 15:11
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If you sequence $X_n$ is not convergent then you will indeed find at least two limits. Bolzano-Weierstrass ensures one, say $x$. As the sequence does not itself converge to $x$, there is an $\epsilon$, so that infinitely many elemets of the sequence are outside of $U_\epsilon(x)$. These elements make up new subsequence of $X_n$ which itself is bounded and by Bolzano-Weierstrass has a (sub-)limit which, now, cannot be $x$.

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    $\begingroup$ Might be obvious, but I feel it is worth noting that even though $X_n$ does not converge, the elements outside $U_\epsilon(x)$ may converge, so we only get the guarantee of 2, not infinitely many convergent subsequences. $\endgroup$ – Michael Anderson Feb 17 '17 at 7:08
  • $\begingroup$ By $U_{\epsilon}(x)$ you mean some "neighborhood" around $x$? $\endgroup$ – Antonio Hernandez Maquivar Mar 29 '17 at 17:14
  • $\begingroup$ @AnthonyHernandez I mean the $\epsilon$-neighborhood around $x$, i.e. $U_\epsilon(x)=\{y\mid d(x,y)<\epsilon \}$. $\endgroup$ – M. Winter Mar 30 '17 at 9:04
  • $\begingroup$ @MichaelAnderson Right, an this is all you can show, e.g. $(-1)^n$ as only two different sublimits. $\endgroup$ – M. Winter Mar 30 '17 at 9:05
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This does not quite follow from BW, but we have the following:

Proposition: A sequence in $\Bbb R$ will converge if and only if all subsequences converge to the same limit.

So: for any non-convergent bounded sequence, you will be able to find subsequences with distinct limits. For any convergent sequence, every subsequence will have the same limit as the original sequence.

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The theorem states that, given a bounded sequence, one (or more) convergent subsequence/s exist/s.

Given a sequence that converges to $L \in \mathbb{R}$, all its subsequences converge to $L$.

Example: $(-1)^n$ is our bounded sequence. We can observe two convergent subsequences: $1^n$ and $-(1^n)$. The first one converges to $1$, whereas the second one converges to $-1$. Indeed, the bounded sequence is irregular (it doesn't converge nor diverge).

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The Bolzano-Weierstrass the theorem says that every infinite, bounded sequence has a convergent sub-sequence. I think the question here is whether such a sequence can have more than one such convergent subsequence, converging to a different limit. The answer to that is clearly "yes". Look at the sequence formed by interweaving two sequences converging to two different limits. For example, the sequence 1, 1/2, 1/3, ..., 1/n converges to 0 while the sequence 2, 3/2, 4/3, ..., (n+1)/n converges to 0. The sequence 1, 2, 1/2, 3/2, 1/3, 3/4, ..., alternating terms from the two sequences is a bounded sequence that has two convergent subsequence, one converging to 0, the other converging to 1.

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  • $\begingroup$ I think you made a small mistake - the second sequence converges to 1. $\endgroup$ – Dason Feb 16 '17 at 15:35
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    $\begingroup$ We could make it even simpler - just have a sequence alternating between 0 and 1. It's a bounded infinite sequence and we can choose subsequences that consist of either only 0 or only 1. $\endgroup$ – Dason Feb 16 '17 at 15:44
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An equivalent formulation of the BW theorem is that any bounded sequence has at least one accumulation point. As noted in the answer of Omnomnomnm we also know that for a convergent sequence, all subsequences converge to the same limit.

So, given two convergent sequences $\{a_n\} \to a$ and $\{b_n\} \to b$ with $a \ne b$ , the sequece $\{c_n\}$ with $c_{2k}=a_k$ and $c_{2k+1}=b_k$ is bounded ( because the two starting sequeces are bounded) and has two different accumulation points.

Genarizing this we can construct a bounded sequence with many accumulation points.

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Just to add a counter-example:

We know that we can enumerate all rationals between 0 and 1 in a sequence $x_1, \ldots, x_n, \ldots$

So $(x_n)$ is real and bounded and therefore BW.

But for every real $a \in (0,1)$ (rational or not), we can find a subsequence of $(x_n)$ that converges to $a$ (by a simple density argument).

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All subsequences of a convergent sequence are also convergent. So if you found one, you found infinitely many.

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  • $\begingroup$ I want to look at two subsequences that each converge to different limit. $\endgroup$ – Itay4 Feb 16 '17 at 14:58
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    $\begingroup$ @Itay4 say so in your question, then $\endgroup$ – Omnomnomnom Feb 16 '17 at 14:59
  • $\begingroup$ Ok, will edit. Thanks $\endgroup$ – Itay4 Feb 16 '17 at 14:59
  • $\begingroup$ @Itay4 For a convergent sequence there is only one possible limit. So you cannot guarantee more than one. $\endgroup$ – M. Winter Feb 16 '17 at 15:00
  • $\begingroup$ @M.Winter The BW theorem does not require a convergent sequence, it just requires a bounded sequence. So as I understand it the OP wants to know if you have a bounded sequence, can you find 2 subsequences converging to different limits? $\endgroup$ – Ovi Feb 16 '17 at 15:07

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