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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? ${}{}$

apnorton
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Idonknow
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  • No, since all the terms of the sequence might be $\leqslant x-\epsilon_0$. – Did Feb 09 '13 at 16:30
  • No, this is not valid. What tells you that there are infinitely many terms in each of the two intervals of the union? – Julien Feb 09 '13 at 16:30
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    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... – David Mitra Feb 09 '13 at 16:36
  • I haven learn subsequential limit point of a sequence, so I don know what are you talking about.@DavidMitra – Idonknow Feb 10 '13 at 04:55
  • @julien: so what guarantee that I can find infinitely many terms in the two intervals ? – Idonknow Feb 10 '13 at 13:29
  • You don't need that to prove the result. See Herng Yi's answer. – Julien Feb 10 '13 at 14:33
  • From what I understand, suppose the subsequence $(x_{n_i})$converge to $x$. Then by definition of convergence, we have $\epsilon>0$, there exists $N \in \mathbb{N}$ such that for all $n_i > N , |x_{n_i}-x|<\epsilon$. Then I get blur – Idonknow Feb 10 '13 at 14:58

3 Answers3

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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.

Riemann
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Herng Yi
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  • ahh yes, let me insure that now, thanks! :) – Herng Yi Feb 09 '13 at 17:02
  • $j^2(1) = j(j(1))$, which I guess is an example of what you meant by a "$j_m$". – Herng Yi Feb 09 '13 at 17:19
  • Ah, sorry. I thought it was a power. – David Mitra Feb 09 '13 at 17:20
  • Actually I not quite understand the answer – Idonknow Feb 10 '13 at 14:35
  • can @HerngYi elaborate ? – Idonknow Feb 10 '13 at 15:35
  • @Idonknow, I'll infer the object of your confusion from your comments to your question. As you said, my sequence $x_1, x_2, \dotsc$ converges to $a$ if for every $\epsilon > 0$ there exists some $N \in \mathbb{N}$ such that for all $i > N$, $|x_i - a| < \epsilon$. Intuitively, within any tiny neighborhood (radius $\epsilon$) of $a$, if I go far enough into the sequence (past $x_N$), the subsequent terms will all lie within that tiny neighborhood. – Herng Yi Feb 10 '13 at 16:22
  • However, $x_1, x_2, \dotsc$ does not converge to $a$ so there exists some neighborhood of $a$ (radius $\varepsilon$, from my answer) such that no matter how far up the sequence I run, there will always be "erratic" terms that "jump" outside the neighborhood. My sequence $x_{j(1)}, x_{j^2(1)}, x_{j^3(1)}, \dotsc$ consists of some of these erratic terms. – Herng Yi Feb 10 '13 at 16:23
  • but how can we make sure the two intervals $[-M,a-\epsilon]$ and $[a+\epsilon,M]$, one of them is not empty ? – Idonknow Feb 10 '13 at 16:33
  • Sorry if I myself got confused (lol), but my elaboration was for my answer. From what I understand of your approach (which is in the context of $\mathbb{R}$), your "$x_n$" are the "erratic terms" I mentioned that jump outside $(a - \epsilon, a + \epsilon)$. It is possible that the erratic terms all lie in one of your intervals $[-M, a - \epsilon]$ and $[a + \epsilon, M]$ while the other is empty. But it's ok because we only need one sequence from one of those intervals that converges to, say, $b$, because we already have a sequence that converges to $a \neq b$! – Herng Yi Feb 10 '13 at 16:41
  • i thought if we want to use Bolzano-Weierstrass theorem, we can only use it on sequence, but not subsequence? From what I know, sequence is defined as a function $f$ with domain $\mathbb{N}$ – Idonknow Feb 10 '13 at 16:45
  • The subsequence $x_{j(1)}, x_{j^2(1)}, \dotsc$ is a sequence because it is a function $f$ with domain $\mathbb{N}$, where $f(i) = x_{j^i(1)}$. In essence, any subsequence can be turned into your "$f$" by "re-indexing" the terms back to $1, 2, 3, \dotsc$. – Herng Yi Feb 11 '13 at 01:34
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  • 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 $x'\ne x$.

Hagen von Eitzen
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  • I think the OP wants to prove the result without assuming your second bullet point, sort of a "from first principles" thing. – Herng Yi Feb 09 '13 at 17:08
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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 $x$.

Berci
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    Sorry but who is $x$? – Did Feb 09 '13 at 16:36
  • 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. – Julien Feb 09 '13 at 16:36