Completeness via Cauchy sequences

Statement

Let $(X, d)$ be a metric space. Then $X$ is complete if and only if each Cauchy sequence in $X$ has a convergent subsequence.

Remarks

The key fact: if a Cauchy sequence $(x_n)$ has a subsequence that converges to $x$, then $x_n \to x$. For $\varepsilon > 0$, choose $N$ for $\varepsilon / 2$ in the Cauchy condition and a term $x_{n_k}$ with $n_k \ge N$ and $d(x_{n_k}, x) < \varepsilon / 2$. A consequence: a metric space in which each sequence has a convergent subsequence is complete. Thus each sequentially compact metric space is complete.

Explanations

L1 human

To show that a space is complete, you need not find the limit of a whole Cauchy sequence. It is enough to find a limit of some subsequence, and then the whole sequence follows it.