Exercise 3.11

Proof. $\text{(⇒)}$ Suppose that $a<b$. If it were the case that $a=b$ then we would have $a<b=a$, which would violate the nonreflexivity property of the order. If it were the case that $b<a$ then we would have $a<b$ and $b<a$ so that $a<a$ by the transitive property of the order. This again violates nonreflexivity. Hence neither $a=b$ nor $b<a$ so that it is not true that $b\le a$.

($\Leftarrow$) Now suppose that it is not true that $b\le a$. Then neither $b=a$ nor $b<a$. Since $b\ne a$, it must be that either $a<b$ or $b<a$ by the comparability property. However, we know that cannot be that $b<a$, so it must be that $a<b$. □

Proof. In what follows Suppose that $A$ is a set with order $<$.

First let $a$ be an element of $A$ and suppose that $b_1$ and $b_2$ are both immediate successors of $a$, which of course means that $a<b_1,b_2$. Then by definition the open intervals $(a,b_1)$ and $(a,b_2)$ are both empty. Now, suppose that $b_1$ and $b_2$ are distinct so that they must be comparable since $<$ is an order. Without loss of generality we can assume that $b_1<b_2$, but then we have $a<b_1<b_2$ so that $b_1\in (a,b_2)$. This is a contradiction since we know that $(a,b_2)$ is empty, so it has to be that $b_1=b_2$. This of course shows that the immediate successor is unique. An analogous argument shows that the immediate predecessor, if it exists, is also unique.

Now suppose that $A_0$ is a subset of $A$ with smallest elements $a_1$ and $a_2$. If $a_1$ and $a_2$ were to be distinct then they must be comparable so that we can assume $a_1<a_2$. However, then it is not true that $a_2\le a_1$ by Lemma, but this means that $a_2$ cannot be a smallest element of $A_0$ since $a_1\in A_0$. As this is a contradiction, it must be that $a_1=a_2$, which shows that the smallest element is unique if there is one. An analogous argument shows any largest element is also unique. □