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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 and $\lnot$B)

Let {$x_n$} be bounded, and every convergent subsequence converges to L.

Assume that $$lim_{x\to\infty}(x_n)\ne L$$

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

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

By Bolzano Weiertrass 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 subsequence of the original sequence $x_n$, then this is a contradiction since every convergent subsequence of xn converges to L.

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

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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 7 '12 at 4:08
    
Thank you very much, Dr. Scott! –  Akaichan Nov 7 '12 at 4:11
3  
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 7 '12 at 4:12
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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 7 '12 at 4:41
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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 7 '12 at 5:29

1 Answer 1

I revise your proof.

Let {$x_n$} be bounded, and every subsequence converges 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 $\{ x_{\Large{n_k}} \}$ such that $|x_{\Large{n_k}} - L|\ge \epsilon \quad \color{red}{(♫)}$

1. How to presage proof by contradiction? Why not a direct proof?

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

By Bolzano Weiertrass Theorem $\{ x_{\Large{n_k}} \}$ has a convergent subsequence $\{ x_{\Large{n_{k_l}}} \}$ that doesn't converge to L. This is a contradiction.

Why? $\{ x_{\Large{n_{k_l}}} \}$ is a sub sequence of the sub sequence $x_{\Large{n_k}} $, which was posited to converge to L.
By the agency of p 57 q2.5.1, every convergent sub sequence of $x_n$ converges to the same limit as the original sequence. So $\{ x_{\Large{n_{k_l}}} \} \to L$.

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