Exercise 11.2

Proof. We show that $\preccurlyeq$ satisfies the three partial order axioms:

(i) Consider any $a\in A$. Since obviously $a=a$ we have by definition that $a\preccurlyeq a$.

(ii) Suppose that $a\preccurlyeq b$ and $b\preccurlyeq a$. Then either $a\prec b$ or $a=b$, and either $b\prec a$ or $b=a$. So suppose that $a\ne b$ so that it must be that $a\prec b$ and $b\prec a$. Since $\prec$ is a strict partial order, it is transitive so that $a\prec a$ since $a\prec b$ and $b\prec a$. But this contradicts the non-reflexivity of $\prec$. Hence it must be that $a=b$ as desired.

(iii) Suppose that $a\preccurlyeq b$ and $b\preccurlyeq c$. Hence either $a\prec b$ or $a=b$, and either $b\prec c$ or $b=c$.

Case: $a\prec b$. If $b\prec c$ then clearly $a\prec c$ since $\prec$ is transitive (since it is a strict partial order). If $b=c$ then we have that $a\prec b=c$.

Case: $a=b$. If $b\prec c$ then we have that $a=b\prec c$. If $b=c$ then we have that $a=b=c$.

Hence in all cases and sub-cases we have that $a\prec c$ or $a=c$, and thus $a\preccurlyeq c$ by definition. □

Proof. We show that $S$ satisfies the two strict partial order axioms:

Nonreflexivity. Consider any $a\in A$. Since $a=a$ it follows that it is not true that $a\ne a$ and hence not true that $\mathit{aSa}$. Thus $S$ is nonreflexive since $a$ was arbitrary.

Transitivity. Suppose that $\mathit{aSb}$ and $\mathit{bSc}$. Hence by definition $\mathit{aPb}$ and $a\ne b$, and $\mathit{bPc}$ and $b\ne c$. Then, by the transitivity property of the partial order axioms, which is property (iii), we have that $\mathit{aPc}$. Suppose for a moment that $a=c$. Then we would have $\mathit{aPb}$ and $\mathit{bPa}$ (since $\mathit{bPc}$ and $c=a$). Then by partial order axiom (ii) we have that $a=b$, which contradicts the fact that $a\ne b$. So it must be that $a\ne c$. Thus $\mathit{aPc}$ and $a\ne c$ so that $\mathit{aSc}$, which shows that $S$ is transitive. □