Let $(x_n)$ be a bounded but not convergent sequence. Prove that $(x_n)$ has two subsequences converging to different limits.

My attempt is: Since the sequence is bounded , there exists $M>0$ such that $x_n \in [-M,M]$ for all $n \in \mathbb{N}$. Since the sequence does not converge to $x$, there exists $\epsilon_0>0$ such that $ \forall N \in \mathbb{N}$, there exists $n \geq N$ such that $|x_n-x| \geq \epsilon_0$.

Then we have $x_n \in [-M,x-\epsilon_0] \cup x_n \in [x+\epsilon_0,M]$. By Bolzano-weierstrass theorem, there exists a convergent subsequence in the two intervals.

Is my proof valid? ${}{}$

  • $\begingroup$ No, since all the terms of the sequence might be $\leqslant x-\epsilon_0$. $\endgroup$ – Did Feb 9 '13 at 16:30
  • $\begingroup$ No, this is not valid. What tells you that there are infinitely many terms in each of the two intervals of the union? $\endgroup$ – Julien Feb 9 '13 at 16:30
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    $\begingroup$ First choose $x$ so that it is a subsequential limit point of $(x_n)$. Then choose a subsequence $(x_{n_k})$ of $(x_n)$ that is bounded away from $x$. Finally, choose a subsequence of $(x_{n_k})$ that converges... $\endgroup$ – David Mitra Feb 9 '13 at 16:36
  • $\begingroup$ I haven learn subsequential limit point of a sequence, so I don know what are you talking about.@DavidMitra $\endgroup$ – Idonknow Feb 10 '13 at 4:55
  • $\begingroup$ @julien: so what guarantee that I can find infinitely many terms in the two intervals ? $\endgroup$ – Idonknow Feb 10 '13 at 13:29

Use Bolzano-Weierstrass to extract a subsequence $x_{i_1}, x_{i_2}, \dotsc$ that converges to some $a$. Since $x_1, x_2, \dotsc$ does not converge to $a$, there exists some $\varepsilon > 0$ such that for each positive integer $N$, there exists some $j(N) > N$ such that $|x_{j(N)} - a| \geq \varepsilon$. Use Bolzano-Weierstrass to extract from $x_{j(1)}, x_{j^2(1)}, x_{j^3(1)}, \dotsc$ a subsequence $x_{k_1}, x_{k_2}, \dotsc$ that converges to some $b$. Clearly, $a \neq b$.

By the way, this method works for any $x_1, x_2, \dotsc$ in $\mathbb{R}^n$, which goes beyond what the OP intended.

  • $\begingroup$ ahh yes, let me insure that now, thanks! :) $\endgroup$ – Herng Yi Feb 9 '13 at 17:02
  • $\begingroup$ $j^2(1) = j(j(1))$, which I guess is an example of what you meant by a "$j_m$". $\endgroup$ – Herng Yi Feb 9 '13 at 17:19
  • $\begingroup$ Ah, sorry. I thought it was a power. $\endgroup$ – David Mitra Feb 9 '13 at 17:20
  • $\begingroup$ Actually I not quite understand the answer $\endgroup$ – Idonknow Feb 10 '13 at 14:35
  • $\begingroup$ can @HerngYi elaborate ? $\endgroup$ – Idonknow Feb 10 '13 at 15:35
  • If a sequence $(x_n)$ is bounded, $a\le x_n \le b$ say, then it has at least one limit point $x$ with $a\le x\le b$ (Bolzano-Weierstraß) and
  • a bounded sequence with exactly one limit point $x$ converges towards that limit point.

Therefore there must exist at least two distinct limit points and we can extract a converging sequence for each.

Just in case the second bullit point above is not clear: If $a\le x_n\le b$ for all $n$ and $x$ is not the limit of $x_n$, then there exists $\epsilon>0$ such that infinitely many $x_n$ are outside $(x-\epsilon,x+\epsilon)$, hence there are infinitely many $x_n>x+\epsilon$ or infinitely many $x_n<x-\epsilon$, leading to at least one limit point $x'$ in $[x+\epsilon,b]$ or $[a,x-\epsilon]$. Thus if $x$ is a limit point but not the limit, there is another limit point $x'\ne x$.

  • $\begingroup$ I think the OP wants to prove the result without assuming your second bullet point, sort of a "from first principles" thing. $\endgroup$ – Herng Yi Feb 9 '13 at 17:08

Yes, valid, but could write it even clearer.

For each $N$ there is an $n$ such that $|x_n-x|\ge \epsilon_0$. Let such an $n$ be chosen for all $N$ and denote it by $k(N)$, for example. Thus we get a subsequence $(x_{k(1)},x_{k(2)},\dots)$ of $(x_n)$, which lies in $[-M,x-\epsilon_0]\cup [x+\epsilon_0,M]$, so this subsequence has a convergent subsequence, and the limit of that cannot be $x$.

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    $\begingroup$ Sorry but who is $x$? $\endgroup$ – Did Feb 9 '13 at 16:36
  • $\begingroup$ This proves that if $x_n$ does not converge to $x$, then there is a convergent subsequence which does not converge to $x$. So you need a little more, i.e. to take $x$ to be a limit of some subsequence of $(x_n)$ in the first place. $\endgroup$ – Julien Feb 9 '13 at 16:36

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