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.
Alternate definition of ordinal sets
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0@Asaf Karagila: Assuming your comment to my other question actually belonged here - one reason might simply be sheer obtuseness :-), or rather the fact that I can nearly (!) prove everything desired in a few lines (i.e. hereditary transitivity, linear ordering, least element etc.), without bringing in well-orderings ... of course my "salvation" doesn't depend on finding an answer - it's just a loose end ! – 2011-09-09
1 Answers
@ 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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0@ 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. – 2011-09-12