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Suppose I have two sequences of real numbers $\{A_n\}$ and $\{B_n\}$ such that they satisfy the relation $A_1 \leq A_2 \leq \ldots \leq A_n \leq \ldots \leq B_n \leq \ldots \leq B_1.$ Does this relation automatically make the two sequences Cauchy? Clearly, if one of them is not Cauchy, they do not limit to anything. So WLOG, if $\{A_n\}$ is not Cauchy, $A_n$ limits to $\infty$ and has no bound. But the sequences is clearly bounded by $B_1.$ Is this correct?

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  • $\begingroup$ Yes. Bounded above increasing sequences and bounded below decreasing sequences of real numbers are Cauchy. $\endgroup$ – Pedro Tamaroff Feb 26 '18 at 23:10
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Yes. In this case $A_n$ is increasing and bounded above, hence convergent, hence Cauchy. Likewise, $B_n$ is decreasing and bounded below, hence convergent, hence Cauchy.

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Yes, both of them are Cauchy sequnces. This is so, because they are both monotonic and bounded. Therefore they converge and therefore they are both Cauchy sequences.

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