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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 $L_2$?

Itay4
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  • Yes, I know. The question is if there exist more than one – Itay4 Feb 16 '17 at 14:59
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    Not if $L_1\not= L_2$, as if $X_n$ is convergent, there is only one possible limit. – M. Winter Feb 16 '17 at 15:08
  • @M.Winter But $X_n$ is not convergent – Itay4 Feb 16 '17 at 15:09
  • 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. – M. Winter Feb 16 '17 at 15:11

7 Answers7

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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 $x$.

M. Winter
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    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. – Michael Anderson Feb 17 '17 at 07:08
  • By $U_{\epsilon}(x)$ you mean some "neighborhood" around $x$? – Anthony Mar 29 '17 at 17:14
  • @AnthonyHernandez I mean the $\epsilon$-neighborhood around $x$, i.e. $U_\epsilon(x)={y\mid d(x,y)<\epsilon }$. – M. Winter Mar 30 '17 at 09:04
  • @MichaelAnderson Right, an this is all you can show, e.g. $(-1)^n$ as only two different sublimits. – M. Winter Mar 30 '17 at 09:05
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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.

Ben Grossmann
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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).

moonknight
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  • Yes, but given any bounded sequence, can I say that from BW theorem we have two convergent subsequences? – Itay4 Feb 16 '17 at 15:11
  • See here: http://math.stackexchange.com/questions/1457080/prove-that-a-bounded-sequence-has-two-convergent-subsequences – moonknight Feb 16 '17 at 15:13
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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.

user247327
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  • I think you made a small mistake - the second sequence converges to 1. – Dason Feb 16 '17 at 15:35
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    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. – 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.

Emilio Novati
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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).

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

Ben Grossmann
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M. Winter
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