Let $(X,d)$ be a metric space and $(x_n)$ is a sequence in $X$. Then
$\sup_{p\ge 1} d(x_n,x_{n+p}) \rightarrow 0$ as $n\rightarrow \infty$ implies $(x_n)$ is a Cauchy sequence.
Any hints or counterexample to prove or disprove this implication? Thanks in advance.
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Neon
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3I would just go straight to the definition. Fix $\epsilon>0$ and try to use the hypothesis to show there is some $N$ such that $|x_m-x_n|<\epsilon$ for all $m,n\ge N$. Notice what you are given shows there exists some $M$ such that $|x_n-x_{n+p}|<\epsilon$ for all $n\ge M$, $p\ge 1$. Is it now sufficient to take $N=M$? – Jared May 21 '13 at 18:31
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Given $\epsilon>0$ there is $N$ such that $\sup_{p\ge 1}d(x_nx_{n+p})<\epsilon$ for all $n>N$. Then for all $m>n>N$, we have $d(x_n,x_m)\le \sup_{p\ge 1}d(x_n,x_{n+p})<\epsilon$.
Hagen von Eitzen
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