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I have to prove this: if every Cauchy sequence is convergent then if ${(I_n)}_{n\ge 1}$ is a sequence of closed intervals whose lengths tend to zero, so exists an unique $x\in \mathop{\cap}_{n=1}^\infty I_n $. Should i prove that ${(I_n)}_n$ is Cauchy and converge to $x$? Thanks in advance.

PD:the intervals are nested $I_1\supseteq I_2\supseteq I_3\supseteq\ldots$

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    $\begingroup$ No. And the intervals have to be nested: $I_1\supseteq I_2\supseteq I_3\supseteq\ldots$ $\endgroup$ – Michael Greinecker Aug 22 '15 at 14:50
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    $\begingroup$ It does not make sense to say a sequence of intervals is Cauchy. A sequence of points can be Cauchy. $\endgroup$ – GEdgar Aug 22 '15 at 14:52
  • $\begingroup$ @Gedgar: It is a Cauchy filter. But I suppose if the O.P. knew that, he wouldn't have to ask this question… $\endgroup$ – Bernard Aug 22 '15 at 14:53
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Hint:

If $I_n=[a_n,b_n]$, consider the sequences $(a_n)_{n\ge 0}$ and $(b_n)_{n\ge 0}$.

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