A question on Cauchy sub-sequences in a metric space $(X,d)$

Question

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 positive $\epsilon_k \rightarrow 0$, there is a subsequence $(x_{n_k})$ of $(x_n)$ such that $$ d(x_{n_k},x_{n_l})\leq \epsilon_k \hspace{5mm} \text{for all } k\leq l.$$

Answer 1

What Brian is saying is that you are given that a sequence $(x_n)_{n=1}^\infty$ has a Cauchy subsequence, which means that there are $n_1, n_2, ...$ such that $(x_{n_k})_{k=1}^\infty$ is a Cauchy sequence, so by relabeling, we can just assume that $(x_n)_{n=1}^\infty$ is your Cauchy sequence.

So given the sequence of $(\varepsilon_k)_{k=1}^\infty$ we know there is an $N_1$ for $\varepsilon_1$ such that if $m>n \geq N_1$ then $d(x_m, x_n) < \varepsilon_1$. Similarly, there is an $N_2$ for $\varepsilon_2$ such that if $m>n \geq N_2$ then $d(x_m, x_n) < \varepsilon_2$. Inductively, we have an $N_k$ for $\varepsilon_k$ such that if $m>n \geq N_k$ then $d(x_m, x_n) < \varepsilon_k$.

So then our subsequence is $(x_{N_k})_{k=1}^\infty$ and then we have that $d(x_{N_k}, x_{N_l}) < \varepsilon_k$ for $l \geq k$. (Here if $l > k$ then $l$ corresponds to the $m$ above and $k$ corresponds to $n$, if $l = k$ then $d(x_{N_k}, x_{N_l}) = 0$.)