Cauchy sequence with a convergent subsequence converges

Proposition. Let $(X,d)$ be a metric space. If $(x_n)$ is a Cauchy sequence and has a subsequence $(x_{n_k})$ that converges to some $a\in X$ , then the entire sequence converges to $a$ .

Remarks

Context. This result is often used to prove convergence once one can identify a candidate limit via a subsequence. It is also a standard step in showing completeness-type statements.

Proof sketch. Fix $\varepsilon>0$ . Since $(x_n)$ is Cauchy, choose $N$ such that $d(x_n,x_m)<\varepsilon/2$ for all $m,n\ge N$ . Since $x_{n_k}\to a$ , choose $K$ such that $d(x_{n_k},a)<\varepsilon/2$ for all $k\ge K$ . Pick $k\ge K$ with $n_k\ge N$ . Then for all $n\ge N$ ,

d(x_n,a)\le d(x_n,x_{n_k})+d(x_{n_k},a)<\varepsilon/2+\varepsilon/2=\varepsilon,

so $x_n\to a$ .