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

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    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.2016-09-03

1 Answers 1

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