Convergent Subsequence of Cauchy Sequence

Lemma
Let $$\left({X,d}\right)$$ be a metric space, and let $$(a_k)$$ be a Cauchy sequence in $$X$$.

Then $$(a_k)$$ converges iff $$(a_k)$$ has a convergent subsequence.

Proof
The "only if" direction is trivial.

So suppose that $$(a_{k_j})$$ is a subsequence that has some limit $$a \in X$$.

Let $$\epsilon > 0$$.

Then, by definition of convergence and of Cauchy sequences, there exist $$K_0$$ and $$j_0$$ such that
 * $$d(a_{k_j},a)< \varepsilon/2$$

whenever $$j\geq j_0$$, and
 * $$d(a_k,a_{k'})< \varepsilon/2$$

whenever $$k,k'\geq K_0$$.

Let $$j_1\geq j_0$$ be minimal with $$k_{j_1}\geq K_0$$. $$k\geq K_0$$,
 * $$ d(a_k,a) \leq d(a_k,a_{j_1}) + d(a_{j_1},a) < \varepsilon/2 + \varepsilon/2 = \varepsilon$$.

So $$(a_k)$$ converges to $$a$$, as required.