Does every element in a well-ordered set have a successor element?

Question

I am currently reading Faticoni's The Mathematics of Infinity (2nd edition), and I think there is an error in his writing. He defines a well-ordered set by the following:

Definition 6.1.1 Let $A$ be a nonempty set. We say that $A$ is a well-ordered set if it satisfies the following two properties:

1. $A$ satisfies the Trichotomy Property. That is, given $x, y \in A$ then $x$ and $y$ satisfy exactly one of the following options, $x < y$, $y < x$ or $x = y$.
2. $A$ satisfies the Minimum Property. That is, each nonempty subset of $A$ contains a unique least element. Equivalently, to each element $x \in A$, there is a unique element $x^+ \in A$ such that (given $y \in A$ such that $x < y$, then $x^+ \leq y$.) We call $x^+$ the successor of $x$.

The problem is that in condition 2, I don't think the two properties are equivalent; there can be a greatest element in the set which has no successor. Later, he argues that the set $A = \{\emptyset, \{a\}, \{a, b\}, \{a, b, c\}\}$ is not a well-ordered set because $\{a, b, c\}$ does not have a successor element. But I think this is wrong since $A$ satisfies both the Trichotomy Property and the Minimum Property so $A$ should be well-ordered? I feel like this is such a big error that reading this book further won't be possible without resolving it. Am I missing something obvious?

Answer 1

Assuming $\lt$ is meant to represent a strict order (irreflexive and transitive), then the definition is only correct if you delete everything from "Equivalently" onwards in clause 2. That part is not just wrong, it is really wrong...

First: finite totally ordered sets are well-ordered under the usual definition, but the author apparently does not want to consider finite totally ordered sets as well-ordered. That's not necessarily a deal breaker, but it is definitely idiosyncratic, at best.

Second: It is clear that the desired "equivalence" is supposed to come from defining $x^+$ to be the least element of $\{y\in A\mid x\lt y\}$. However, this set can be empty even in infinite well-ordered sets: consider a well ordered set of the form $\mathbb{N}\cup\{*\}$, where we order the natural numbers as usual, and let $n\lt *$ for every $n\in\mathbb{N}$. This is well-ordered, and the set $\{y\in A\mid *\lt y\}$ is empty.

Well, that need not be an obstacle: we can just define $*^+=*$, since the condition "for all $y$, if $*\lt y$ then $*\leq y$" is satisfied.

Third: But this shows a third problem: if we define $x^+=x$ for any $x$, then this satisfies the given condition; but so will the least element of $\{y\in A\mid x\lt y\}$ when the set is not empty, so the element $x^+$ will not be unique as required. For example, in $\mathbb{N}$, both $0^+=0$ and $0^+=1$ have the property that if $0\lt y$, then $0^+\leq y$.

Fourth. We could try to fix the "equivalently" clause by adding that $x^+$ would be strictly larger than $x$ when $x$ is not the maximum, and equal to $x$ otherwise; that would fix the problems outlined above. It could also be fixed if the defining property of $x^+$ were a biconditional: $x\lt y$ if and only if $x^+\leq y$ whenever $x$ is not the maximum element of $A$, and $x^+=x$ if $x=\max A$. But that is still incorrect: consider $\mathbb{Z}$ with its usual order, and define $n^+=n+1$ for every integer. This set is trichotomic, and every element has an immediate successor, but is not well-ordered.

So this (incorrect) definition is a mess, even if we restrict it to infinite sets.

You need to delete "Equivalently, to each element $x\in A$, there is a unique element $x^+\in A$ such that (given $y\in A$such that $x\lt y$, then $x^+\leq y$). We call $x^+$ the successor of $x$."

Instead, add the following additional nomenclature:

If $A$ is well ordered and $x\in A$, then if the set $\{y\in A\mid x\lt y\}$ is not empty, we define $x^+$ to be its least element, and call $x^+$ "the successor of $x$".