# 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

## 2 Answers

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.

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.