# 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 $x$.

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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## 1 Answer

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