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Suppose $\left \{ a_n \right \}$ is a Cauchy sequence, and $\left \{ x_n \right \}$ is a sequence with a number $k>0$ such that $|x_n - x_m|\leq k|a_n - a_m|$ for all $n,m\in \mathbb{N}$. Is $\left\{ x_n \right\}$ necessarily a Cauchy sequence? Either prove or give a counter-example.

My attempt: I think the question is true. So since $\left \{ a_n \right \}$ is a Cauchy sequence, then for $\forall \epsilon >0$, there is an $N$ so that for all $n,m>N$ $|a_n - a_m| \leq \frac{\epsilon}{k} $.

So for any $n,m$, we get $|x_n - x_m|<\epsilon \Rightarrow |x_n - x_m|\leq k|a_n - a_m|$.

Is that it to the proof? Looks quite simple to me.

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    $\begingroup$ Did you mean to say that there is a number $k>0$ such that $|x_{n}-x_{m}|\leq k|a_{n}-a_{m}|$ for all $n,m\in\mathbb{N}$? If so your proof is almost correct -- you just need to switch the direction of the last implication. That is, for any $n,m\geq N$ we have $|x_{n}-x_{m}|\leq k|a_{n}-a_{m}|\Rightarrow |x_{n}-x_{m}|<\epsilon$. $\endgroup$
    – Eric
    Commented Dec 3, 2013 at 19:35
  • $\begingroup$ Simple facts have simple proofs. $\endgroup$
    – Carsten S
    Commented Dec 3, 2013 at 19:37
  • $\begingroup$ So is {$x_n$} necessarily a Cauchy? $\endgroup$
    – user87274
    Commented Dec 3, 2013 at 20:44

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Yeah, that's right. The fact that $|a_n-a_m| \leq \frac{\epsilon}{k}$ is important: by transitivity, we then have $|x_n - x_m| \leq |a_n - a_m | \leq \frac{\epsilon}{k}$, so $|x_n - x_m| \le \epsilon$.

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