# Bounded sequence and every convergent subsequence converges to L

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

#### Solutions Collecting From Web of "Bounded sequence and every convergent subsequence converges to L"

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