Convergent Subsequence of Cauchy Sequence

Theorem
Let $\left({A, d}\right)$ be a metric space.

Let $\left\langle{x_n}\right\rangle_{n \in \N}$ be a Cauchy sequence in $A$.

Let $x \in A$.

Then $\left\langle{x_n}\right\rangle$ converges to $x$ it has a subsequence that converges to $x$.

Necessary Condition
If $\left\langle{x_n}\right\rangle$ converges to $x$, it trivially follows that $\left\langle{x_n}\right\rangle$ is a subsequence of itself that converges to $x$.

Sufficient Condition
Suppose that $\left\langle{x_{n_k}}\right\rangle$ is a subsequence of $\left\langle{x_n}\right\rangle$ that converges to $x$.

Let $\epsilon$ be a strictly positive real number.

By the definition of a Cauchy sequence, there exists a real number $M$ such that:
 * $\displaystyle \forall i, j \in \N: i, j > M \implies d \left({x_i, x_j}\right) < \frac \epsilon 2$

By the definition of convergence, there exists a real number $N$ such that:
 * $\displaystyle \forall k \in \N: k > N \implies d \left({x_{n_k}, x}\right) < \frac \epsilon 2$

By the Archimedean principle, there exists a natural number $K > \max \left\{{M, N}\right\}$.

By Strictly Increasing Sequence of Natural Numbers, we have $n_K \ge K > N$.

Therefore, by the triangle inequality:
 * $\displaystyle \forall m \in \N: m > K \implies d \left({x_m, x}\right) \le d \left({x_m, x_{n_K}}\right) + d \left({x_{n_K}, x}\right) < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon$

That is, $\left\langle{x_n}\right\rangle$ converges to $x$.