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Let $x=(x_n)$ be a Cauchy sequence in a metric space $X$ having a convergent subsequene $x\sigma=(x_{\sigma_n})$ where $\sigma:\mathbb N\to\mathbb N$ is strictly increasing. Then $(x_n)$ is convergent.

  • My attempt: Let $(x\sigma)(n)\to l$ in $X.$ For $\epsilon>0~\exists~k\in\mathbb N$ such that $d(x(\sigma(n)),l)<\epsilon/2~\forall~n\ge k$ and $d(x(m),x(n))<\epsilon/2~\forall~m,n\ge k.$ Since $\sigma(n)\ge n~\forall~n$ it follows that $$d(x(\sigma(n)),x(n))<\epsilon/2~~\forall~n\ge k\\d(x(\sigma(n)),l)<\epsilon/2~\forall~n\ge k$$

By triangular inequality, $d(x(n),l)<\epsilon~\forall~n\ge k.$ Hence Proved.

Am I correct?

share|cite|improve this question
It seems fine to me ! – jibounet Aug 5 '13 at 14:23
In the second line you want $d(x(\sigma(n)),l)$, not $d(x(\sigma(n))-l)$, but otherwise it’s correct. – Brian M. Scott Aug 5 '13 at 14:24
Thanks so much. – Sriti Mallick Aug 5 '13 at 14:31

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