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Define a point $ x $ in a metric space $ X $ to be a limit point of a sequence $ (x_{n})_{n \in \mathbb{N}} $ if there exists some subsequence $ (x_{n_{k}})_{k \in \mathbb{N}} $ of $ (x_{n})_{n \in \mathbb{N}} $ that converges to $ x $.

I need to show that if $ x $ is a limit point of a Cauchy sequence $ (x_{n})_{n \in \mathbb{N}} $, then $ (x_{n})_{n \in \mathbb{N}} $ converges to $ x $.

Here is my start: Let $ (x_{n})_{n \in \mathbb{N}} $ be a Cauchy sequence with $ x $ a limit point. Thus there exists some subsequence $ (x_{n_{k}})_{k \in \mathbb{N}} $ of $ (x_{n})_{n \in \mathbb{N}} $ that converges to $ x $. I need to show that $ (x_{n})_{n \in \mathbb{N}} $ converges to $ x $... Thinking now.

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HINT: Let $\langle x_{n_k}:k\in\Bbb N\rangle$ be your subsequence converging to $x$. For any $\epsilon>0$ there is an $m_\epsilon$ such that $d(x,x_{n_k})<\frac{\epsilon}2$ whenever $k\ge m_\epsilon$. There is also an $m_\epsilon'\in\Bbb N$ such that $d(x_k,x_\ell)<\frac{\epsilon}2$ whenever $k,\ell\ge m_\epsilon'$. Put this information together properly with the triangle inequality, and you’ll find of tail of the original sequence within $\epsilon$ of $x$.

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@ Brian Thank you,I just arrived at the same conclusion in my paper. –  Klara Feb 28 '13 at 1:49
    
@Klara: Good! And you’re welcome. –  Brian M. Scott Feb 28 '13 at 1:52

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