Timeline for If a set contains all its proper transitive subsets as members, do its members as well?
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11 events
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| Aug 16, 2011 at 18:12 | comment | added | Joel David Hamkins | 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. | |
| Aug 16, 2011 at 15:52 | comment | added | Stephan F. Kroneck | slight (copy + paste) typo: (strong version) of Global Choice, not Limitation, naturally ! | |
| Aug 16, 2011 at 15:50 | comment | added | Stephan F. Kroneck | (ctd.) Interestingly, due to Separation, a large corpus of the elementary theory of ordinals remains provable. In fact, my motivation for the whole business was to give a slick exposition of the equivalence of the (strong version) of the Axiom of Limitation of Size (in NBG, be it in injective or surjective form)) with the combined Axioms of Global Choice and Replacement, $\textbf{without}$ use of the Axiom of Power Sets (because Limitation yields this inthe special instances required), and implying (as observed by Azriel Lévy, the Axiom of Union. | |
| Aug 16, 2011 at 15:42 | comment | added | Stephan F. Kroneck | (ctd.) Another remark: the quantization over subsets does not seem (to me) to be such a drawback (subsets are, of course, simply sets whose elements also lie in the putative superset). In fact, an advantage of the definition is that it can be reformulated for classes (consider all $\textbf{subclasses}$ $x\subsetneq\alpha$), and yields the class of all strange/ordinal sets as the unique proper strange class. Naturally, this class formation is not $\textit{per se}$ predicative (and thus not permissible in NBG, though it is in MK), but can be made so due to the Axiom of Separation. (ctd. below) | |
| Aug 16, 2011 at 15:38 | comment | added | Emil Jeřábek | ... As for intuitionistic set theories: the usual ones (IZF, CZF) include the foundation axiom (in the form of $\in$-induction). | |
| Aug 16, 2011 at 15:37 | comment | added | Emil Jeřábek | @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. ... | |
| Aug 16, 2011 at 15:35 | comment | added | Stephan F. Kroneck | $\textbf{Thank you}$ already for your comments; indeed, the reason for considering strange sets is precisely that one avoids the whole notion of well-order, and needs to verify membership of a set only one level (as say compared to the hereditarily transitive version); still, I was hoping for a more direct proof (from the definition) that elements of strange sets are again strange (and avoiding any already developed theory of ordinals in the usual sense) - I should have phrased my question more precisely. (ctd. below) | |
| Aug 16, 2011 at 15:12 | comment | added | Joel David Hamkins | 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. | |
| Aug 16, 2011 at 15:04 | comment | added | Emil Jeřábek | 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.) | |
| Aug 16, 2011 at 14:53 | comment | added | Joel David Hamkins | Emil, it looks like we hit on the same observation. | |
| Aug 16, 2011 at 14:51 | history | answered | Emil Jeřábek | CC BY-SA 3.0 |