# Bolzano–Weierstrass theorem conclusion

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 $L_2$?

• Yes, I know. The question is if there exist more than one – Itay4 Feb 16 '17 at 14:59
• 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

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 $x$.

• 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 7:08
• By $U_{\epsilon}(x)$ you mean some "neighborhood" around $x$? – Antonio Hernandez Maquivar 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 9:04
• @MichaelAnderson Right, an this is all you can show, e.g. $(-1)^n$ as only two different sublimits. – M. Winter Mar 30 '17 at 9:05

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.

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).

• 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
• – moonknight Feb 16 '17 at 15:13

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.

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

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.

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).

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

• I want to look at two subsequences that each converge to different limit. – Itay4 Feb 16 '17 at 14:58
• @Itay4 say so in your question, then – Omnomnomnom Feb 16 '17 at 14:59
• Ok, will edit. Thanks – Itay4 Feb 16 '17 at 14:59
• @Itay4 For a convergent sequence there is only one possible limit. So you cannot guarantee more than one. – M. Winter Feb 16 '17 at 15:00
• @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? – Ovi Feb 16 '17 at 15:07