Observe that one direction is
Theorem 2.1.14, so we’ll only prove the interesting direction, namely that bounded monotone sequences converge.
Here’s the proof for a nondecreasing sequence.
Set
\(L=\sup\left\{x_n\middle\vert n\in\mathbb{N}\right\}\text{.}\) That is,
\(L\) is the supremum of the image of
\(X\text{.}\) We’ll show that
\(x_n\to L\text{.}\)
To this end, we let
\(\epsilon\gt 0\) be given. By the
\(\epsilon\)-characterization of supremum, there is
\(x_K\) so that
\(L-x_K\lt \epsilon\text{.}\) Use this
\(K\in\mathbb{N}.\)
To see that this \(K\) works, consider any \(n\geq K\text{.}\) Because \(X\) is nondecreasing, \(x_n\geq x_K\text{.}\) Therefore
\begin{gather*}
\lvert L-x_n\rvert=L-x_n\leq L-x_K\lt \epsilon
\end{gather*}
which is what we need.