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Let $(x_n)$ and $(y_n)$ be Cauchy sequences in a metric space $(X,d)$ defined in $\mathbb{R}$. Show that the sequence $(d(x_n,y_n))_n$ is convergent.

Since the set of $\mathbb{R}$ is complete then $(x_n)$ and $(y_n)$ are convergent. What I need to do (I think) then is show that the sequence of $(d(x_n,y_n))_n$ is a Cauchy sequence too.

So I need to show that for every $\epsilon>0$, there exists an $N\in\mathbb{N}$ such that whenever $m,n\geq N$ follows that $$|(d(x_n,y_n))_n-(d(x_m,y_m))_m|<\epsilon$$

How I can proof it?

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1 Answer 1

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Hint: $d(x_n,y_n) \leq d(x_n,x_m) + d(x_m,y_m) + d(y_m,y_n) $, so

$d(x_n, y_n) - d(x_m, y_m) \le d(x_n, x_m) + d(y_m,y_n)$ now reverse the rôles of $n$ and $m$. We get

$|d(x_n, y_n) - d(x_m, y_m)| \le d(x_n, x_m) + d(y_m,y_n)$ You can take it from here?

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  • $\begingroup$ From where it come from? $\endgroup$
    – Roland
    Jun 9, 2017 at 22:29
  • $\begingroup$ @Roland Simply apply the triangle inequality twice. $\endgroup$
    – Demophilus
    Jun 9, 2017 at 22:29

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