# Alternate definition of ordinal sets

Going through some personal notes from several years ago 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: are elements of strange sets themselves strange ? Now one can prove that strange sets are precisely the ordinals, whence the answer is clear; what I'm looking for is a direct proof from the definition, $\textbf{without}$ using traditionally defined ordinals, regularity, transfinite induction/recursion etc. (e.g. I can prove them to be hereditarily transitive, and ordered linearly with respect to inclusion). Thanks in advance for any useful comments ! Kind regards, Stephan F. Kroneck.

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For additional details: mathoverflow.net/questions/72986/… – Andrés E. Caicedo Sep 8 '11 at 5:04
Whoops, I accidentally casted a vote to close. Please ignore this vote, if anyone thinks this should be closed please cancel out my vote by leaving a comment. – Asaf Karagila Sep 8 '11 at 5:40
I'm not sure about one thing. This is not an elementary issue, and there are far less set theorists here than MathOverflow. What is the point posting this (and the other question too) on MSE? – Asaf Karagila Sep 8 '11 at 5:41
@Asaf: I don’t see any legitimate reason to close it, but if I were the OP, I think that I’d withdraw it here and concentrate on the discussion in MO. It’s unlikely to get any more useful answers here and may well end up stuck on the Unanswered list. – Brian M. Scott Sep 8 '11 at 5:58
@Brian: Yes, it happened by mistake, which is why I posted as a comment, that if anyone else would think that it needs to be closed he should first cancel out my accidental vote instead of casting an additional. I also very much agree with your comment (which is essentially what I wrote in my second comment). – Asaf Karagila Sep 8 '11 at 6:05

@ Dear all, without spelling out all the details, the following works to my satisfaction:

Using the definition of "strange" sets, it seems best first to show that strange classes are hereditarily transitive, and contain no elements which are members of themselves (indirectly, in one go, by considering the union of all such subsets in a given strange class), quickly giving $\bigcap X\in X$ and $\bigcap X\cap X=\emptyset$ for any nonempty class $X$ of strange sets.

Next one has either $\xi\in\eta$, or $\xi=\eta$, or $\xi\ni\eta$ for any strange sets $\xi,\eta$ (after observing $\xi\cap\eta\in\{\xi,\eta\}$ and going from there), whence an obvious argument shows that transitive classes of strange sets are strange (note that no use of power sets or replacement/separation is made up to this point, just Gödel's axiom groups A and B; without separation, though, the class of strange sets cannot be formed).

All this was clear beforehand, so concerning my actual question: Assuming some form of separation, or just the existence of the class of all strange sets, one easily proves that elements of strange sets are strange:

Given a strange set $\xi$, let $\eta$ be the subset of hereditarily (!) strange elements of $\xi$; clearly $\eta$ is transitive and hereditarily strange by the above, so assume $\eta\subsetneq\xi$ (as one is done otherwise) and thus $\eta\in\xi$, therefore $\eta\in\eta$ by definition of $\eta$, a contradiction (by the above).

Thanks to all contributors, kind regards - Stephan F. Kroneck.

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I tried to format your answer to improve readability. I'm not so sure about the use of "whence" there, but since I am not a native English speaker I can't really judge for myself. Regarding the last edited part, forcing is somewhat of a "reserved word" in set theory, and it has a particular meaning which is unfit here altogether. While semantically it was an okay use (and perhaps in algebra or analysis no one would have complained) - this is triumphed by the fact this is a set theoretic proof, so unless fitting (in the set theoretic sense) it is best to avoid the use of "forcing". – Asaf Karagila Sep 11 '11 at 21:41
@ Asaf Karagila: Thank you for formatting and editing my reply - I didn't manage to enter new paragraphs. Also, I see that you've already removed the offensive (low-level use of the word) "forcing" :-) There remains the matter of closing the question (so that everybody can get back to real work !), as it really was just a mote in my (sleepless) eye, hardly justifying the attention and bother it seems to have generated. Kind regards - Stephan F. Kroneck. – bonnbaki Sep 12 '11 at 8:20
You can accept your own answer, as it is a better solution than closing. – Asaf Karagila Sep 12 '11 at 8:36