Question:
($a_n$) and ($b_n$) are non-decreasing function with the property that for each postiive integer $n$, there are integers $q$ and $p$ such that ($a_n$) $\leq$ ($b_p$) and ($b_n$) $\leq$ ($a_q$).
Show that both diverge to infinity.
IDEA: I have some ideas but I don't know how to do the following steps or I don't know how to expand my ideas into a proof.
Since ($a_n$) is a non-decreasing sequence, then it is bounded below. Suppose ($a_n$) is also bounded above, then ($a_n$) is bounded.
If ($a_n$) is bounded, then there exist a number $M$ such that $|a_n| \le M$, then I am thinking that maybe I can use proof by contradiction to show that $|a_n| \ge M$ and prove that ($a_n$) diverges to infinity.
Idea 2:
Proof ($a_n$) is not Cauchy, I am thinking applying the definition of Cauchy sequence and do the following steps. But, I don't where to start or what relationship I can use to start my proof.
Thanks a lot! Any help will be very appreciated.
-JY
The statement is false even if we change $\le$ into $<$ in the given condition.
Consider $$ a_n=1-\frac{1}{2n},\qquad b_n=1-\frac{1}{2n+1} $$ Both sequences are increasing and, for each $n$, $$ a_n<b_n<a_{n+1} $$ so we can take $p=n$ and $q=n+1$. However, both sequences converge to $1$.
What is true is that two sequences with that property are either both divergent or both convergent (to the same limit).
Let's see why. Suppose $(a_n)$ is convergent to $l$; then $$ l=\sup\{a_n:n\ge1\} $$ Let $n$ be any positive integer; then $b_n\le a_q$, for some $q$; therefore $b_n\le l$ and so the sequence $(b_n)$ is bounded by $l$. As a consequence it converges to $m=\sup\{b_n:n\ge1\}$ and $m\le l$.
Now, for every $n$, $a_n\le b_p$, for some $p$; therefore $a_n\le m$. As before, $l\le m$.
Hence $l=m$.
By symmetry, if one of the two sequences diverges, also the other one does.