Hello, 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) $\textbf{strange}$ 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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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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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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Would like to close this question, as I've answered its parallel on Stackexchange: "Alternate definition of ordinal sets". Kind regards - Stephan F. Kroneck. |
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