# The Definition of Ordinals and the Axiom of Regularity

In the first edition of his set theory textbook, Pinter defines an ordinal number to be a set which is transitive and (strictly) well-ordered by the membership relation $\in$. I believe this is the most common way of defining ordinals. Then he proves many wonderful things without invoking the axiom of regularity (foundation).

However, in the revised edition from Dover, his definition changes! He drops the well-orderedness and says

A set $A$ is called an ordinal if $A$ is transitive and ordered by $\in$.

where, by 'ordered' he means 'strictly ordered'. Moreover, his first theorem on ordinals goes like

9.27 Theorem Let $\alpha$ and $\beta$ be ordinals. If $\alpha\ne\beta$, then either $\alpha\in\beta$ or $\beta\in\alpha$.

and his proof makes use of the axiom of regularity on the following subclass $A$ of $\text{OR}$, where $\text{OR}$ denotes the class of all ordinals. $$A=\{x\in\text{OR}:\exists y\in\text{OR},\text{ such that }x\text{ and }y\text{ are incomparable}\}.$$

The problem is, we don't know whether $A$ is a set or a proper class in this situation, so that the axiom of regularity might not be applicable. The book just says "Let $A$ be the following subSET of $\text{OR}$..." and I'm very confused here.

So my questions are

1. Is the class $A$ above necessarily a set?
2. Is there a short proof that dropping the well-orderedness on the definition of ordinals does not affect the theory? Perhaps assuming the axiom of regularity if necessary?

• If he treats A only as a formula, that is, in the same way OR is a formula ($x\in OR$ is just an abbrevation) then in what he does, it won't matter to his conclusion whether A is a set or not. Still, it's bad form. – DanielWainfleet Aug 7 '17 at 16:09
• @DanielWainfleet He uses the axiom of regularity on A to find an element of A disjoint from A. (assuming A is nonempty, of course) – Dilemian Aug 8 '17 at 0:26

Eric Wofsey's answer shows that regularity for sets implies regularity for classes, provided you have enough other axioms (like replacement) available. In the particular case you asked about, where the class $A$ consists of transitive sets, you can avoid part of Eric's argument, namely the part where he produces a transitive set $T$.
In detail: Given a nonempty class $A$ of transitive sets, begin by considering some $y\in A$. If you're very lucky, $y\cap A=\varnothing$ and you're done. So suppose from now on that $y\cap A$ is a nonempty set. (It's a set, not a proper class, because $y$ is a set.) By regularity for sets, $y\cap A$ has an element $z$ with $z\cap y\cap A=\varnothing$. So we have $z\in A$, and if we can show $z\cap A=\varnothing$ then we'll be done.
Now we use that all elements of $A$ are transitive. In particular $y$ is transitive, so, from $z\in y$, we get $z\subseteq y$ and thus $z\cap y=z$. But then what we already knew, that $z\cap y\cap A=\varnothing$, simplifies to $z\cap A=\varnothing$, and the proof is complete.
Well, given that the result is valid, $A$ turns out to be empty, so it is always a set. But you are right that there is something wrong with this argument.
Here's the trick which makes it valid to apply Regularity to classes, not just sets. Let $C$ be any nonempty class. We will prove there exists $x\in C$ such that no element of $x$ is in $C$.
Since $C$ is nonempty, choose some $y\in C$. Recursively define a sequence of sets $y_0=\{y\}$, $y_{n+1}=\bigcup y_n$ for $n\in\mathbb{N}$. Let $T=\bigcup_{n\in\mathbb{N}}y_n$ (this uses Replacement to say the set $\{y_n:n\in\mathbb{N}\}$ exists and then Union). The set $T$ is transitive: if $a\in y_n$ for some $n$ and $b\in a$ then $b\in y_{n+1}$. (In fact, $T$ is the smallest transitive set containing $\{y\}$, called the transitive closure of $\{y\}$.) Now consider $S=T\cap C$, which is a set by Separation. Since $y\in S$, $S$ is nonempty. By Regularity, there exists $x\in S$ such that no element of $x$ is in $S$. Then $x\in C$, and no element of $x$ is in $C$, since every element of $x$ is in $T$ by transitivity.