If a subsequence of a Cauchy sequence converges, then the whole sequence converges.

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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