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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) 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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  • $\begingroup$ @Soham Unlike other Stackexchange sites, salutations and signatures are fine here. $\endgroup$
    – Andrés E. Caicedo
    Commented Jul 24, 2017 at 4:35
  • $\begingroup$ @AndrésE.Caicedo ah, my apologies. Thank you for reverting the edit. $\endgroup$
    – Soham Chowdhury
    Commented Jul 24, 2017 at 4:45
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    $\begingroup$ Also on Math.SE: math.stackexchange.com/questions/62716/… $\endgroup$
    – Nate Eldredge
    Commented Jul 25, 2017 at 17:57

2 Answers 2

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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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  • $\begingroup$ 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) $\endgroup$
    – Stephan F. Kroneck
    Commented Sep 8, 2011 at 13:21
  • $\begingroup$ (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. $\endgroup$
    – Stephan F. Kroneck
    Commented Sep 8, 2011 at 13:22
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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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  • $\begingroup$ Emil, it looks like we hit on the same observation. $\endgroup$
    – Joel David Hamkins
    Commented Aug 16, 2011 at 14:53
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    $\begingroup$ 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.) $\endgroup$
    – Emil Jeřábek
    Commented Aug 16, 2011 at 15:04
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    $\begingroup$ @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. ... $\endgroup$
    – Emil Jeřábek
    Commented Aug 16, 2011 at 15:37
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    $\begingroup$ ... As for intuitionistic set theories: the usual ones (IZF, CZF) include the foundation axiom (in the form of $\in$-induction). $\endgroup$
    – Emil Jeřábek
    Commented Aug 16, 2011 at 15:38
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    $\begingroup$ A contrary point is that saying "$\alpha$ is well-ordered by $\in$'' is about the same complexity as "$\alpha$ is strange", and quantifies over the same collection of subsets of $\alpha$, so it isn't clear if one is really saving any set-theoretical baggage this way. $\endgroup$
    – Joel David Hamkins
    Commented Aug 16, 2011 at 18:12

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