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?

  • $\begingroup$ Your set $A$ is well-preserved under inclusion. $\endgroup$
    – Jay
    May 9, 2022 at 14:54
  • $\begingroup$ @SassatelliGiulio Yeah that's also something concerning. The usual definition of a well-ordered set that I know is that it is a set with a total order whose non-empty subsets have least elements. But I am having difficulty showing that the definition in the book and the usual one are equivalent. $\endgroup$ May 9, 2022 at 14:59
  • $\begingroup$ Condition 2 is not well written. You can take $x^+=x$, which satisfies the given condition, and in tbe case where $x$ has strict upper bounds, the minimum of the strict upper bounds. The problem is not necessarily with a maximum, as you can let $x^+=x$. Or in fact, anything, since the implu ation would be satisfied by vacuity. $\endgroup$ May 9, 2022 at 15:01
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    $\begingroup$ @ArturoMagidin Yes, you are right: I did say it was a poor definition. It would need to add $x\lt x^{+}$ included to give uniqueness in the non-empty case. And if it ends up suggesting that finite sets cannot be well-ordered, that would be unhelpful - even if (given the title) only infinite sets were in contemplation. It is easy to forget how important good definitions are until a case like this comes along. $\endgroup$ May 9, 2022 at 16:08
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    $\begingroup$ @MarkBennet: Everything you said was right. I am emphatically agreeing with you, and pointing even more ways this is a mess. ;-) $\endgroup$ May 9, 2022 at 16:55

1 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 $x$".

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    $\begingroup$ "the author apparently does not want to consider finite totally ordered sets as well-ordered" I think the author not merely doesn't want to consider finite sets, but also bounded ones. $\endgroup$ May 10, 2022 at 2:03
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    $\begingroup$ @Acccumulation but even so restricted, his claimed equivalence is no equivalence. $\endgroup$ May 10, 2022 at 2:57

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