Exercise 3.13

Proof. Suppose that $A_0$ is any nonempty subset of $A$ that is bounded below so that $b$ is a lower bound of $A_0$. Let $B_0$ be the set of lower bounds of $A_0$ so that $B_0$ is nonempty since $b\in B_0$. Since $A_0$ is nonempty there is an $a\in A_0$. Now, for any $x\in B_0$ we have that $x$ is a lower bound of $A_0$ so that $x\le a$, which shows that $a$ is an upper bound of $B_0$. Hence $B_0$ is a nonempty subset of $A$ that is bounded above, and so has a least upper bound $c$ since $A$ has the least upper bound property. We claim that $c$ is also the greatest lower bound of $A_0$.

Consider any $x\in A_0$ and any $y\in B_0$. Then $y$ is a lower bound of $A_0$ so that $y\le x$ since $x\in A_0$. Since $y\in B_0$ was arbitrary, this shows that $x$ is an upper bound of $B_0$. Thus we have $c\le x$ since $c$ is the least upper bound of $B_0$. Since $x\in A_0$ was arbitrary, this shows that $c$ is a lower bound of $A_0$. If $y$ is any other lower bound then $y\in B_0$ so that $y\le c$ since $c$ is an upper bound of $B_0$. Since $y$ was an arbitrary lower bound, this shows that in fact $c$ is the greatest lower bound of $A_0$. Hence $A$ also has the greatest lower bound property since the nonempty subset $A_0$ was arbitrary. □