# Prove that if $(x_n)$ has a Cauchy subsequence, then for any decreasing sequence

Let $$(X,d)$$ be a metric space and let $$(x_n)$$ be a sequence in $$X$$ . Prove that if $$(x_n)$$ has a Cauchy subsequence, then for any decreasing sequence of $$\epsilon_k$$ -> $$0$$, there is a subsequence $$(x_{n_k})$$ of $$(x_n)$$ such that

$$d(x_{n_k}, x_{n_l}) \leqslant \epsilon_k$$ for all $$k \leqslant l.$$

I have proved that $$(x_n)$$ is convergent and Cauchy, and from that proved in a separate case that its subsequences are also convergent Cauchy, $$(x_{n_l})$$, yet I can't finish the proof for the any decreasing sequence $$\epsilon_k$$.

• How can you have proved that $(x_n)_{n\in\mathbb N}$ converges only from the fact that it has a Cauchy subsequence? – José Carlos Santos Sep 25 '18 at 10:18

For any $$k$$ we can choose $$n_k$$ such that$$d(x_j,x_l) \leq \epsilon_k$$ for $$j,l \geq n_k$$. Since $$n_k$$'s can be replaced by any larger number we can inductively choose these integers such that $$n_k$$ is increasing in $$k$$. We then have $$d(x_{n_k}, x_l) \leq \epsilon_k$$ for all $$l \geq n_k$$ and we can take $$l =n_j$$ as long as $$j \geq k$$.
The proof for metric spaces carries over with virtually no change from the result in $$\mathbb R$$, so lets use $$(\mathbb R,|\cdot|)$$. Assume without loss that $$x_n$$ is Cauchy. Recall that for each $$\epsilon>0$$, we can find $$N(\epsilon)>0$$ such that $$n,m\ge N(\epsilon ) \implies |x_n - x_m| < \epsilon$$
let me tell you something that does not work: define the sequence $$n_k$$ via
$$n_k := N(\epsilon_k)$$ Then for any $$l,k>0$$ with $$k,
$$|x_{n_k} - x_{n_l}| \le \epsilon_k$$ since $$n_k = N(\epsilon_k) \le N(\epsilon_l) = n_l$$.