Exercise 11.1

Proof. Since $A$ is nonempty, there is an $a\in A$. Clearly, $\text{∅}$ is vacuously simply ordered. However, it cannot be maximal since clearly the set $\left\{a\right\}$ properly contains $\text{∅}$ as a subset but is also clearly vacuously simply ordered by $\prec$. Hence, since $B$ is maximal it must be that $B\ne \varnothing$ as desired. □

Proof. First consider any $a\in ℝ$ so that $a-a=0$, which is not positive and hence it is not true that $a\prec a$. Therefore $\prec$ is nonreflexive. Now consider $a,b,c\in ℝ$ where $a\prec b$ and $b\prec c$. Then we have that $x=b-a$ and $y=c-b$ are positive and rational. It then clearly follows that

is also rational and positive since both $x$ and $y$ are. Thus $a\prec c$, which shows that $\prec$ is transitive. Since $\prec$ was shown to be nonreflexive and transitive, this shows that it is a strict partial order as desired. □

Proof. Suppose that $B$ is the set of maximally simply ordered subsets of $ℝ$. Then we show that $A=B$.

To show that $A\subset B$ consider any $X\in A$ so that $X=A_x$ for some $x\in ℝ$. Now consider any distinct $y$ and $z$ in $X=A_x$ so that by definition $x-y$ and $x-z$ are both rational so that $z-x=-(x-z)$ is also rational. Then clearly $z-y=(z-x)+(x-y)$ is rational as is $y-z=-(z-y)$. Since $y$ and $z$ are distinct, we have that $z-y$ and $y-z$ are nonzero and that either $y<z$ or $z<y$. In the former case we have that $z-y$ is a positive rational number and in the latter $y-z$ is. Thus either $y\prec z$ or $z\prec y$, which shows that $X=A_x$ is simply ordered since $y$ and $z$ were arbitrary. Now consider any $y\in A_x$ and $z\notin A_x$ so that $x-y$ is rational but $x-z$ is irrational so that $z-x=-(x-z)$ is also irrational. Since a rational added to an irrational is also irrational (which is trivially easy to prove), it follows that $z-y=(z-x)+(x-y)$ is irrational as is $y-z=-(z-y)$. Hence it cannot be that either $y\prec z$ or $z\prec y$. Since $y\in X$ and $z\notin X$ were arbitrary, this show that $X$ is a maximal simply ordered set so that $X\in B$. This shows that $A\subset B$ since $X$ was arbitrary.

Now suppose that $X\in B$ so that $X$ is a maximal simply ordered set. It follows from Lemma 1 that $X$ is nonempty so that there is an $x\in X$, and we claim that in fact $X=A_x$. So consider any $y\in X$. Clearly if $y=x$ then $x-y=x-x=0\in \mathbb{Q}$ so that $y\in A_x$. If $y\ne x$ then either $x\prec y$ or $y\prec x$ since $X$ is simply ordered by $\prec$. In the former case, we have that $y-x$ is positive and rational so that $x-y=-(y-x)$ is negative and rational, and hence $y\in A_x$. In the latter case, we have that $x-y$ is positive and rational so that clearly again $y\in A_x$. Since $y$ was arbitrary this shows that $X\subset A_x$. Now consider any $y\in A_x$ so that $x-y\in \mathbb{Q}$. If $y=x$ then clearly $y\in X$. If $y\ne x$ then either $y-x$ or $x-y$ is positive, and also clearly rational since $x-y$ is rational. Hence either $x\prec y$ or $y\prec x$. It then follows from the fact that $X$ is maximally simply ordered that $y$ must be in $X$ since otherwise, $y$ would not be comparable with $x$. Since again, $y$ was arbitrary this shows that $A_x\subset X$. Hence $X=A_x$ so that clearly $X\in {\left\{A_x\right\}}_{x\in ℝ}=A$. Since $X$ was arbitrary this shows that $B\subset A$.

Therefore we have shown that $A=B$, which shows that $A$ is exactly the complete set of maximally simply ordered subsets. □