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
-
1For additional details: http://mathoverflow.net/questions/72986/alternate-definition-of-ordinals – 2011-09-08
-
0Whoops, 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. – 2011-09-08
-
0I'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? – 2011-09-08
-
0@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. – 2011-09-08
-
0@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). – 2011-09-08
-
0@Asaf: Yes, I was basically just agreeing with you on both counts. – 2011-09-08
-
0@Asaf Karagila & Brian Scott: Thank you both for your comments ! Just quickly my reasons for adding my question here to StackExchange: 1.) The question is certainly not of a research nature or value, it really is just a loose end (making it more a question for here than MO), so that my conscience plagued me whether I should leave it on Overflow at all; 2.) I would consider it to be an elementary issue; I've certainly seen far more advanced questions than this small one on StackExchange ! Kind regards - Stephan F. Kroneck. – 2011-09-08
-
0It may be a question of crowd. Most, if not all, set theorists on this site hang out on MO. The converse is not true. Asking this question on MO is more than reasonable. I also cannot see what in the MO answers is lacking. – 2011-09-08
-
0@Asaf Karagila: just quickly - what's lacking in my eyes is that the comments offered up till now all require the use of ordinals; ideally I would like to replace the traditional definition with the "strange" one given there (as it seems to require less "overhead" to formulate, besides being quite natural, when looked at closely). The point where I get stuck is proving that the property is hereditary ... Kind regards ! Stephan F. Kroneck. – 2011-09-08
-
0Just discovered another, similar posting here on StackExchange: – 2011-09-08
-
0(don't know how to add a link, I'm afraid) "A transitive set of ordinals is an ordinal", which deals with almost exactly the same problem (for instance I wasn't aware of the fact that my "strange" sets are Bourbaki's definition of ordinals, shame on me), though the question goes the other way around (i.e. "up" - that transitive sets of Bourbaki ordinals are again such) - have to sort out whether my question is answered there in effect or not. – 2011-09-08
-
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.
-
0I 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". – 2011-09-11
-
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