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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 $\epsilon_k$.

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

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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 $k<l$,

$$ |x_{n_k} - x_{n_l}| \le \epsilon_k$$ since $n_k = N(\epsilon_k) \le N(\epsilon_l) = n_l$.

Why isn't this a proof, and what can be done about it?

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