# Is the sequence $A_1 \leq A_2 \leq \ldots \leq A_n \leq \ldots \leq B_n \leq \ldots \leq B_1$ Cauchy?

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?

• Yes. Bounded above increasing sequences and bounded below decreasing sequences of real numbers are Cauchy. – Pedro Tamaroff Feb 26 '18 at 23:10

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.