Convergent Subsequence of Cauchy Sequence/Metric Space

Theorem
Let $\struct {A, d}$ be a metric space.

Let $\sequence{x_n}_{n \in \N}$ be a Cauchy sequence in $A$.

Let $x \in A$.

Then $\sequence{x_n}$ converges to $x$ it has a subsequence that converges to $x$.

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

Sufficient Condition
Suppose that $\sequence {x_{n_k}}$ is a subsequence of $\sequence {x_n}$ 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:
 * $\displaystyle \forall i, j \in \N: i, j > M \implies \map d {x_i, x_j} < \frac \epsilon 2$

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

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

Therefore, by the triangle inequality:
 * $\displaystyle \forall m \in \N: m > K \implies \map d {x_m, x} \le \map d {x_m, x_K} + \map d {x_K, x} < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon$

That is, $\sequence {x_n}$ converges to $x$.