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A metric space $(X,d)$ is called ultrametric if strict triangle inequality holds: $$d(x,y)\le \max\{ d(x,z), d(z,y)\}.$$

In the book $p$-adic Lie groups by Schneider, the author states following theorem (p.6):

A sequence $(x_n)$ in $X$ is a Cauchy sequence if and only if $\lim_{n\rightarrow \infty} d(x_n,x_{n+1})=0$.

I confused here whether this is true only for ultrametric space $(X,d)$? With usual metric on $\mathbb{R}$, we can take $x_n=1+\frac{1}{2}+\cdots + \frac{1}{n}$, which is not a Cauchy. Am I missing something?

There are questions on mathstack where the condition in above statement is replaced by $d(x_n,x_{n+1})<Cr^n$ for some fixed $C\in\mathbb{R}$ and $0<r<1$ for all $n\ge 1$. Then $(x_n)$ is Cauchy.

It seems that this condition in terms of $C$ and $r$ is weaker than that in above quoted statement. Am I right? In other words, we have only one way implication below but not conversely: $$ \Big{(}d(x_{n+1},x_n)<Cr^n \mbox{ for some $r$, $0<r<1$ and fixed $C\in\mathbb{R}$}\Big{)}\Rightarrow \lim_{n\rightarrow\infty}d(x_{n+1},x_n)=0. $$

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  • $\begingroup$ In an ultrametric you will have $d(x_{n+p},x_n)\leq\max(d(x_{n+p},x_{n+p-1}),...,d(x_{n+1},x_n))$. If $d(x_{n+1},x_n)\to0$, then for $n>N$, $d(x_{n+1},x_n)<\epsilon$. Therefore, $d(x_{n+p},x_n)<\epsilon$. $\endgroup$ – logarithm May 6 '19 at 14:18
  • $\begingroup$ It is true that the hypothesis of $d$ being an ultrametric is important because in general the condition needs not hold, as you've pointed out. The discussion in the last paragraph is not too easy to follow: in my opinion, the phrasing is a bit imprecise. You may want to share the the original. $\endgroup$ – Saucy O'Path May 6 '19 at 14:19
  • $\begingroup$ Ah, I see. So it was $0<r<1$, not $0<r<n$. $\endgroup$ – Saucy O'Path May 6 '19 at 14:25
  • $\begingroup$ Oh, sorry! It was my typing mistake. Thanks @Saucy $\endgroup$ – Beginner May 6 '19 at 14:26

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