# Alternate definition of ordinals ?

Hello, recently I came upon some personal notes I'd made several years ago while reviewing some basic set theory (ordinals, transfinite recursion, inaccessible cardinals etc.), and I stumbled upon a loose thread which I obviously had not resolved at the time, and which I would like to lay to rest: Assuming some standard set theory (say ZF, even though I prefer NBG), without the Axiom of Foundation (preferably), one may define an ordinal $\alpha$ (von Neumann's definition) as a transitive set whose elements are well-ordered with respect to the membership relation $\in$. This is seen to be equivalent to the statement that $\alpha$ is transitive, all its $\beta\in\alpha$ are transitive too, and (as we cannot rely on foundation) for each non-empty $x\subseteq\alpha$ there exists some $\beta\in x$ such that $x\cap\beta=\emptyset$ (except for the last condition, this is as in Schofield's book on Mathematical Logic). One then goes on to prove that the class of all ordinals is well-ordered with respect to membership etc.; along the way a useful intermediate step is to prove that any ordinal $\alpha$ is (ad hoc definition) $\textbf{strange}$ in the sense that one has $x\in\alpha$ for any transitive $x\subsetneq\alpha$. My question finally (as this would provide an alternate definition of ordinal sets): are elements of strange sets themselves strange, or at least transitive ? Thanks in advance for any useful comments ! Kind regards, Stephan F. Kroneck.

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## 3 Answers

Strange sets are the same thing as ordinals. Given a strange set $\alpha$, let $\beta$ be the smallest ordinal such that $\beta\notin\alpha$. Then $\beta\subseteq\alpha$. If $\beta\subsetneq\alpha$, then $\beta\in\alpha$ as $\alpha$ is strange, which contradicts the definition of $\beta$. Thus $\beta=\alpha$.

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Emil, it looks like we hit on the same observation. – Joel David Hamkins Aug 16 '11 at 14:53
Indeed. And I find the observation somewhat interesting in that it provides a definition of ordinals in ZF which works in the absense of the foundation axiom, despite that it does not refer to well-foundedness (or well-order). (It does, however, quantify over subsets of $\alpha$, which is to be expected, as ordinals are not definable by a bounded formula in ZF minus foundation.) – Emil Jeřábek Aug 16 '11 at 15:04
Well, the way that well-foundedness works its way in is in the definition of ordinal. I have seen alternative definitions of ordinal in the anti-foundation context simply as "hereditarily transitive" sets, dropping the well-founded part of the definition. I'm not sure, but I think that some intuitionists use something like this also, and perhaps they (or you?) can tell us. – Joel David Hamkins Aug 16 '11 at 15:12
@Joel: I think we are talking past each other. I’m not interested in a definition of alternative ordinals (such as hereditarily transitive sets, which do not have any useful properties without the foundation axiom), but alternative definitions of the standard class of ordinals. Its usual definitions explicitly involve the condition of well-foundedness in one way or another (such as by demanding that the set be well-ordered by ∈), which is rather inelegant. In contrast the definition of a strange set does not invoke well-foundedness in any way, it only refers to its transitive subsets. ... – Emil Jeřábek Aug 16 '11 at 15:37
... As for intuitionistic set theories: the usual ones (IZF, CZF) include the foundation axiom (in the form of $\in$-induction). – Emil Jeřábek Aug 16 '11 at 15:38

Theorem. Every strange set is an ordinal.

Proof. Suppose that $\alpha$ is strange. Let $\beta$ be the smallest ordinal such that $\beta\notin\alpha$. Such a $\beta$ exists, because no set can contain all the ordinals, and this does not require the foundation axiom to prove. It follows that $\beta\subset\alpha$ and $\beta$ is transitive. Thus, if $\beta\neq\alpha$, we would have $\beta\in\alpha$, contradicting the choice of $\beta$. Hence $\beta=\alpha$ and $\alpha$ is an ordinal. QED

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Update: I haven't given the whole matter terribly much thought recently, apart from the fact that I'm able to prove that strange sets are necessearily hereditarily transitive (what a mouthful !) and that no element of a strange set has itself as a member (directly using the definition, avoiding the whole ordinal machinery), and (what is then quite easy) that strange sets are ordered linearly w.r.t. subset inclusion. What remains is the original loose end: avoiding ordinals, just using the definition, why are elements of strnage sets again strange ? (ctd. below) – Stephan F. Kroneck Sep 8 '11 at 13:21
(ctd.) Thus I wish to leave the question open (and prefer not to hand out an "answered" just yet). Kind regards to all who have contributed their time and insights so far ! S.F. Kroneck. – Stephan F. Kroneck Sep 8 '11 at 13:22

Would like to close this question, as I've answered its parallel on Stackexchange: "Alternate definition of ordinal sets". Kind regards - Stephan F. Kroneck.

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Why would you close it? Other people may benefit from it or have something to say later, even if you are satisfied with the answers obtained so far. I would suggest that instead you click the check mark next to one of the answers, so the questions does not appear as "unsolved" any longer. – Andrés E. Caicedo Sep 11 '11 at 21:09
@ Andreas Caicedo: Thank you for your comment ! It just seems a bit weird ticking my own answer ... however, it is the only one up till now that satisfies my admittedly "strange" standards. Kind regards - Stephan F. Kroneck. – Stephan F. Kroneck Sep 12 '11 at 7:56