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Let $(x_n)$ be a sequence such that $(\forall m,n \in \mathbb{N}), |x_m-x_n|\ge\frac{1}{\min(m,n)}$. Is $(x_n)$ a Cauchy sequence? I know that a sequence is Cauchy if for every $\epsilon$ there exists $n_0\in \mathbb{N}$ such that for $n,m\ge n_0$ it follows that $|x_n-x_m|<\epsilon$. But here we have $|x_m-x_n|\ge \frac{1}{\min(m,n)}$. How can I prove whether this sequence is indeed a Cauchy. I either need to find $n_0$ or another way I guess is to try to prove that it converges. Frankly I'm not sure I even understand the statement.

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  • $\begingroup$ Let $x_n = n$ for all $n$. Then $|x_m-x_n| \ge 1 \ge 1/\min(m,n)$. So we can't say that $(x_n)$ is Cauchy. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 1 '18 at 22:23
  • $\begingroup$ Such a sequence cannot exist because for $m = n$ we have $$|x_n - x_n| = 0 \not\geq \frac{1}{n}$$ $\endgroup$ – mechanodroid Mar 1 '18 at 22:24
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    $\begingroup$ If it's an exercise, especially if to prove $(x_n)$ is Cauchy, then I guess this is a typo only, and they intended to write $\le$ in place of $\ge$. $\endgroup$ – Berci Mar 1 '18 at 22:26
  • $\begingroup$ So I guess the point was to find an example. Thanks, makes sense now. $\endgroup$ – DreaDk Mar 1 '18 at 22:29
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I think the likelihood is that you've misread the claim, or that there is a typo in the claim.

Firstly, it does not specify that $n \neq m$. If $n = m$ then $|x_{n}-x_{m}|=0$ so $ 0 \geq \frac{1}{\min{(n,m)}} $ which is a contradiction, so there are no such sequences as written.

If we refine the condition slightly so that $n\neq m$ then note that $ \min {(n,m)} \geq 1 $ so that $1 \geq \frac{1}{\min{(n,m)}} $ so, as Giuseppe mentioned $x_{n} = n$ works (since $ |n-m| \geq 1 \geq \frac{1}{\min{(n,m)}} $).


Now suppose that we take what's probably the correct interpretation of the claim: if $(x_{n})$ is a sequence such that $ (\forall n,m \in \mathbb{N}) |x_{m}-x_{n}| \leq \frac{1}{\min{(n,m)}} $, then $(x_{n})$ is Cauchy.

In this case, the strategy you outlined works perfectly well. We fix some $ \epsilon > 0 $ and look directly for some $ n_{0} $ for which we can guarantee that $ n,m>n_{0} \implies |x_{n}-x_{m}|<\epsilon $. The simple way to do this is to see that $ n,m > n_{0} \implies \frac{1}{\min{(n,m)}} < \frac{1}{n_{0}}$, so all we need is some $n_{0}$ such that $\frac{1}{n_{0}} < \epsilon $; in other words, $n_{0} > \frac{1}{\epsilon}$ we we know must exist by the Archimedian property of $\mathbb{R}$. (For instance, $n_{0} = \lceil \frac{1}{\epsilon} \rceil$ works.) If you're uncomfortable with this last step, it's worth learning more about the Archimedean principle, which is beyond the scope of this question.

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  • $\begingroup$ It' a past exam question that I have in front of me. I think it's supposed to be more of a find an example that contradicts it. $\endgroup$ – DreaDk Mar 1 '18 at 23:08
  • $\begingroup$ @DreaDk if you're looking for a counter-example to the statement as written, do the two I gave you make sense? $\endgroup$ – Oscar Heath Mar 8 '18 at 1:14
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Indeed, the answer is negative. If you consider the sequence $x_n=n+1$, this satisfies the hypotesis clearly but it is not a Cauchy sequence since it diverges.

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