Exercise 10.1

Proof. Suppose that $A$ is a set with well-ordering $<$, and that $B$ is some nonempty subset of $A$ with upper bound $a\in A$. Let $C$ then be the set of upper bounds of $B$, which is not empty since clearly $a\in C$. Then $C$ is a nonempty subset of $A$ and so has a smallest element $c$ since $A$ is well-ordered. Clearly then $c$ is the least upper bound of $B$ by definition. This shows that $A$ has the least upper bound property since $B$ was arbitrary. □