Exercise 10.2

Proof. Suppose that $A$ is well-ordered by $<$ and consider any $a\in A$ where $a$ is not the largest element. It then follows that there is some $x\in A$ where $a<x$ since otherwise, $a$ would be the largest element of $A$. Let $X=\left\{y\in A\mid a<y\right\}$ so that clearly $X\subset A$ and $x\in X$. Thus $X$ is a nonempty subset of $A$ and so has a smallest element $b$ since $<$ well-orders $A$. We claim that $b$ is the immediate successor of $a$. To see this suppose that there is a $z\in A$ such that $a<z<b$, noting that clearly $a<b$ since $b\in X$. Then we would have that $z\in X$ but $z<b$ so that it is not true that $b\le z$, which contradicts the definition of $b$ as the smallest element of $X$. So it must be that no such $z$ exists, which shows that $b$ is indeed the immediate successor of $a$. □

Proof. First, clearly $\mathbb{Z}$ is not well-ordered since, for example, the set of negative integers is a nonempty subset of $\mathbb{Z}$ but has no smallest element. Also, for any $n\in \mathbb{Z}$, clearly $n+1$ is the immediate successor of $n$, which was shown back in Corollary 1 of Exercise 4.9. □