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 $\lim_{n\to\infty}x_n = L.$


marked as duplicate by Marcus M, Aqua, Rohan, Claude Leibovici, José Carlos Santos Dec 11 '17 at 15:43

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    $\begingroup$ 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 $\lim_{n\to\infty}x_n\ne L$. $\endgroup$ – Brian M. Scott Nov 7 '12 at 4:08
  • $\begingroup$ Thank you very much, Dr. Scott! $\endgroup$ – Akaichan Nov 7 '12 at 4:11
  • 3
    $\begingroup$ 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. $\endgroup$ – Brian M. Scott Nov 7 '12 at 4:12
  • 1
    $\begingroup$ Subscripts nested three deep are a pain, and hard to read, but you can do them: x_{n_{k_\ell}}, for $x_{n_{k_\ell}}$. $\endgroup$ – Brian M. Scott Nov 7 '12 at 4:41
  • 1
    $\begingroup$ @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. $\endgroup$ – Gerry Myerson Nov 7 '12 at 5:29

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 $\{ x_{n_{k_l}} \}$ 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 $\{ x_{\Large{n_{k_l}}} \} \to L$.


Not the answer you're looking for? Browse other questions tagged or ask your own question.