Exercise 3.14

Proof. $\text{(⇒)}$ Suppose that $C$ is symmetric. Then we simply have

which shows that $C=D$.

$\text{(⇐)}$ Now suppose that $C=D$ and consider any $(a,b)\in C$. Then $(b,a)\in D$ by definition. Hence also $(b,a)\in C$ since $C=D$, which shows that $C$ is symmetric. □

Proof. Suppose that $C$ is an order relation. Since clearly, $D$ is a relation on $A$, we need only show that it has the three required properties:

• (Comparability) Consider any distinct $a,b\in A$ so that $\mathit{aCb}$ or $\mathit{bCa}$ since $C$ has comparability. Hence either $\mathit{bDa}$ or $\mathit{aDb}$, respectively, by definition so that $a$ and $b$ are comparable in $D$ as well.
• (Nonreflexivity) Consider any $a\in A$. Then $(a,a)\notin C$ since it is nonreflexive, thus also $(a,a)\notin D$ since, if it were, it would also be that $(a,a)\in C$ by definition. Hence $D$ is also nonreflexive since $a$ was arbitrary.
• (Transitivity) Suppose that $\mathit{aDb}$ and $\mathit{bDc}$. Then by definition, we have $\mathit{bCa}$ and $\mathit{cCb}$. That is, $\mathit{cCb}$ and $\mathit{bCa}$ so that $\mathit{cCa}$ since $C$ is transitive. Therefore $\mathit{aDc}$ by definition, which shows that $D$ is also transitive.

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

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