= Solution
For a nonempty <set> $S\subseteq\mathbb R$, a <real number> $s$ is its <supremum> if every $x\in S$ satisfies $x\leq s$ and every <upper bound> $u$ of $S$ satisfies $s\leq u$. Equivalently, $s$ is an <upper bound> and, for every $\varepsilon>0$, some $x\in S$ has $s-\varepsilon<x\leq s$. The <least upper bound axiom> states that every nonempty set of <real numbers> bounded above has a real <supremum>.
Suppose $(a_n)$ is nondecreasing and bounded above. Apply the <least upper bound axiom> to $S=\{a_n:n\geq1\}$, obtaining $s=\sup S$. For any $\varepsilon>0$, $s-\varepsilon$ cannot be an <upper bound>, so there is $N$ with $a_N>s-\varepsilon$. For every $n\geq N$, monotonicity gives
$$
s-\varepsilon<a_N\leq a_n\leq s,
$$
so $|a_n-s|<\varepsilon$. This is precisely the definition of a <convergent sequence>. If $(a_n)$ is nonincreasing and bounded below, apply the proved case to $(-a_n)$; its limit is the negative of the <infimum> of the original terms. Thus the <bounded monotone sequence theorem> follows directly from the axiom:
$$
\boxed{a_n\to\sup_n a_n\ \text{in the nondecreasing case},\qquad
a_n\to\inf_n a_n\ \text{in the nonincreasing case}.}
$$
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