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Let $\{x_n\}$ be a bounded sequence such that every convergent subsequence converges to $L$. Prove that $$\lim_{n\to\infty}x_n = L.$$

The following is my proof. Please let me know what you think.

Prove by contradiction: ($A \wedge \lnot B$)

Let {$x_n$} be bounded, and every convergent sub-sequence converges to $L$.

Assume that $$\lim_{n\to\infty}x_n\ne L$$

Then there exists an $\epsilon>0$ such that $|x_n - L|\ge \epsilon$ for infinitely many n.

Now, there exists a sub-sequence $\{x_{n_{k}}\}$ such that $|x_{n_{k}} - L| \ge\epsilon$.

By Bolzano-Weierstrass Theorem $x{_{n{_k}}}$ has a convergent subsequence $x_{n_{k{_{l}}}}$ that does not converge to $L$.

$x_{n_{k_{l}}}$ is also a sub-sequence of the original sequence $x_n$, then this is a contradiction since every convergent sub-sequence of $x_n$ converges to $L$.

Hence the assumption is wrong. So $\lim_{n\to\infty}x_n = L.$

Math1000
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Akaichan
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    It could be expressed a bit better, but it’s basically fine. Note, though, that it’s not correct to say that $\lim_{n\to\infty}x_n$ ‘does not go to’ $L$: the limit is a number, and it’s not going anywhere. You simply want to assume that $\lim_{n\to\infty}x_n\ne L$. – Brian M. Scott Nov 07 '12 at 04:08
  • Thank you very much, Dr. Scott! – Akaichan Nov 07 '12 at 04:11
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    You’re welcome. While I’m here, you might find this MathJax tutorial helpful; you posts will be a lot easier to read if you can manage at least basic formatting. – Brian M. Scott Nov 07 '12 at 04:12
  • I fixed it a little bit. Thank you for the website. It's a big help. I haven't figure out how to write xnk or xnkl yet though. – Akaichan Nov 07 '12 at 04:39
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    Subscripts nested three deep are a pain, and hard to read, but you can do them: x_{n_{k_\ell}}, for $x_{n_{k_\ell}}$. – Brian M. Scott Nov 07 '12 at 04:41
  • I got it. Thank you very much for your time. – Akaichan Nov 07 '12 at 04:48
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    @wj32, it's a bit different from that earlier question, which talks of metric spaces and compact sets --- topics which someone interested in the current question may not know about. – Gerry Myerson Nov 07 '12 at 05:29
  • Just a wording suggestion: writing $\lim_{n\to\infty}x_n\neq L$ implies that a limit esists. A more rigorous wording could be "$\lim_{n\to\infty}x_n$ does not exist or it is different from $L$". In either case, your proof works. – bartgol Jul 23 '15 at 17:25
  • The Q as stated is trivial because a sequence is a subsequence of itself, and for any $m$ the sequence $(x_n){n>m}$ is a subsequence of $(x_n){n\in N}.$ It would be non-trivial to ask whether ($x_n){n\in N}$ converges to $L$ if $(x{f(n)})_{n\in N}$ converges to $L$ whenever $f:N\to N$ is strictly increasing and $N$ \ ${f(n):n\in N}$ is infinite. – DanielWainfleet Sep 03 '16 at 02:31
  • Since you need Bolzano-Weierstrass, you should make the ambient space more explicit than what the notation $|x_{n_{k}} - L| \ge\epsilon$ suggests: it is restricted to $\Bbb R,$ or at least $\Bbb R^n.$ – Anne Bauval Dec 24 '22 at 10:34

1 Answers1

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I revised your proof.

Let {$x_n$} be bounded, and let every subsequence of {$x_n$} converge to L. Assume that $\lim_{n\to\infty}(x_n)\ne L$. Then there exists an $\epsilon$ such that infinitely many $n \in N \implies |x_n - L|\ge \epsilon $ Now, there exists a subsequence $\{ x_{{n_k}} \}$ such that $|x_{{n_k}} - L|\ge \epsilon \quad \color{red}{(♫)}$. Two questions follow:

1. How can we tell that we should do a proof by contradiction? Why not prove this directly?

2. Where does $\color{red}{(♫)}$ come from?

By Bolzano-Weierstrass, $\{ x_{{n_k}} \}$ has a convergent subsequence $\{ x_{n_{k_l}} \}$ that doesn't converge to L. This is a contradiction, as $\{ x_{{n_{k_l}}} \}$ is a subsequence of the subsequence {$x_{{n_k}}$}, which we assumed to converge to L. By p 57 q2.5.1, every convergent subsequence of $x_n$ converges to the same limit as the original sequence, so it must also be the case that $\{ x_{{n_{k_l}}} \} \to L$.

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