Exercise 3.7

Proof. Suppose that $A$ is a set with order relation $<$. Also let $A_0$ be a subset of $A$ so that $\prec =<\cap (A_0\times A_0)$ is the restriction of $<$ to $A_0$. Clearly $\prec \subset A_0\times A_0$ so that it s a relation on $A_0$. So we must show that $\prec$ satisfies the three properties of an order relation:

• (Comparability) Consider any $x,y\in A_0$ so that also $x,y\in A$ since $A_0\subset A$. Then $x$ and $y$ are comparable in $<$ since it is an order. So, without loss of generality, we can assume that $x<y$ and so $(x,y)\in <$. Since clearly also $(x,y)\in A_0\times A_0$, it follows that $(x,y)\in <\cap (A_0\times A_0)=\prec$ and hence $x\prec y$. This shows that $x$ and $y$ are comparable in $\prec$.
• (Nonreflexivity) Suppose that $x\in A_0$ so that also $x\in A$ since $A_0\subset A$. Then it cannot be true that $x<x$ since $A$ is an order and so is nonreflexive. Thus $(x,x)\notin <$ so that also $(x,x)\notin <\cap (A_0\times A_0)=\prec$. Hence it is not true that $x\prec x$ so that $\prec$ is also nonreflexive since $x$ was arbitrary.
• (Transitivity) Suppose that $x\prec y$ and $y\prec z$. Then of course $(x,y)\in A_0\times A_0$ and $(x,y)\in <$, and similarly $(y,z)\in A_0\times A_0$ and $(y,z)\in <$. Hence $x<y$ and $y<z$ so that $x<z$ since $<$ is an order and therefore transitive. Thus $(x,z)\in <$ so that $(x,z)\in <\cap (A_0\times A_0)=\prec$ since clearly $(x,z)\in A_0\times A_0$. So then $x\prec z$, which shows that $\prec$ is transitive.