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Let $(X,d)$ be a metric space, and say $(x_n)$ is a Cauchy sequence such that it has a convergent subsequence $(x_{n_k})$ that converges to $x$. We show $x_n \to x$. Let $\epsilon > 0$. Take $N >0$ such that for all $n,m > N$, we have

$$d(x_n,x_m) < \frac{\epsilon}{2}.$$

By hypothesis, we can take also $K >0$ such that for all $n_k > K$, we have

$$ d(x_{n_k},x) < \frac{\epsilon}{2}.$$

Put $M = \max \{N,K\}$. Therefore, for all $n,m,n_k > M$, we have

$$ d(x_n,x) \leq d(x_n, x_{n_k}) + d( x_{n_k},x) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$

Hence, $x_n \to x$ as desired.

Is this a correct approach? Thank you very much in advance.

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    $\begingroup$ Excellent approach. $\endgroup$ – drhab Feb 3 '14 at 18:05
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Very nice.

As an alternative, you could say that $d(x_{n_k},x)<\epsilon/2$ whenever $k>K.$ Then, you want to put $M=\max\{N,n_K\}.$ You know that $k>K$ if and only if $n_k>n_K,$ which allows you to draw the desired conclusion.

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