Convergent Subsequences of Cauchy Sequences

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

Sufficient Condition

If $(a_k)$ converges, it trivially follows that it has a convergent subsequence.

Necessary Condition

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 $j \ge j_0$, and

$d(a_k, a_{k'}) < \varepsilon / 2$

whenever $k, k' \ge K_0$.

Let $j_1 \ge j_0$ be minimal with $k_{j_1} \ge K_0$. $k \ge K_0$,

$d (a_k,a) \le d(a_k,a_{k_{j_1}}) + d(a_{k_{j_1}},a) < \varepsilon / 2 + \varepsilon / 2 = \varepsilon$.

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

$\blacksquare$