While reading Mills' proof of the prime formula that's named after him, I believe the following statement is used:
Let $a_n$ be a strictly increasing sequence and $b_n$ be a strictly decreasing sequence such that $a_n < b_n$ for all $n \in \mathbb{N}$. Then $a =\lim_{n\to \infty} a_n$ exists and $a_n < a < b_n$ for all $n\in \mathbb{N}$.
I attempted to give a more detailed proof of this step as an exercise.
Proof: Since $ a_n <b_n <b_0$, then $a_n$ is strictly increasing and bounded from above, so it converges. We denote the existing limit by $a$.
BWOC we assume that $a\le a_{m}$ for some $m \in \mathbb{N}$. We define $\varepsilon_a = a_{m+1} -a>0$. Since for all naturals $n>m+1$ we have $a_n >a_{m+1} > a$, we get $$|a_n – a| = a_n -a \mathbin{\color{red}{>}}a_{m+1} -a = \varepsilon_a$$
which contradicts the existence of $a$, so
we conclude that $a > a_n$ for all $n \in \mathbb{N}$.
Similarly, BWOC we assume $a \ge b_m$ for some $m \in \mathbb{N}$. We define $\varepsilon_b = a – b_{m+1}>0$. Since for all naturals $n >m+1$ we have
$a \ge b_m>b_{m+1}>b_n >a_n$, we get
$$
|a_n -a| =a – a_n\mathbin{\color{red}{>}}a-b_{m+1} = \varepsilon_b
$$
which contradicts the existence of $a$, so
we conclude that $a < b_n$ for all $n \in \mathbb{N}$.
$$\tag*{$\blacksquare$}$$
Although I believe the previous proof is valid, I didn't like that I had to separate the problem into several smaller proofs. So my question is
Are there any alternative ways to prove the statement? Preferably without separating the proof into cases.
Any and all ideas are welcome. Thank you!
EDIT: I incorrectly used "monotone" to mean "strict increasing/decreasing". I've changed the writing in my question to correct this mistake.
Best Answer
Assume that $a_n$ is a strictly increasing sequence, $b_n$ a strictly decreasing sequence, and $a_n<b_n$ for all $n$.
Note that $a_n$ is bounded by $b_0$ and $b_n$ by $a_0$.
By the Monotone Convergence Theorem, the limit of $a_n$ equals $\sup(a_n)$. Denote this as $a$. For all $i$, $a_i<a_{i+1}\leq \sup(a_n) =a$.
For any positive integer $i,j$ we have that $a_i<a_{i+j}<b_{i+j}<b_{j}$. Therefore, for all $j$, $b_j$ is an upper bound for the sequence $a_n$, so by the definition of $\sup$, $b_j$ is at least $\sup(a_n)=a$.
Applying this to $b_{n+1}$ gives $a \leq b_{n+1}<b_n$ for all $n$. Putting this together with the above, for all $n$, $a_n<a<b_n$ as desired.