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 Dec20 answered Is there a bounded-quantifier formula in ZF without Foundation that defines the naturals? Dec20 comment Is there a bounded-quantifier formula in ZF without Foundation that defines the naturals? @Asaf Karagila. I am a bit confused. Let us take the axiom foundation: $$\forall z [z\neq \emptyset \rightarrow (\exists y\in z) y\cap z=\emptyset]$$ The formula that says "$x$ is well-founded by $\in$" is: $$(\forall z\subseteq x) [z\neq \emptyset \rightarrow (\exists y\in z) y\cap z=\emptyset]$$ How do you eliminate the $(\forall z\subseteq x)$ at the beginning and get a bounded formula? Dec20 comment Is there a bounded-quantifier formula in ZF without Foundation that defines the naturals? @Asaf Karagila. How do you write "$x$ is well-ordered by $\in$" as a bounded formula over ZF-Foundation? Dec13 comment If $\alpha$ is an indecomposable ordinal, why $\Gamma(\alpha\times{\alpha})=\alpha$? @Camilo Arosemena. You're welcome. Can you please post any errors you find. Dec13 comment An exercise from Levy's Basic Set Theory (Exercise 3.17) Thank you for your help. Unfortunately, I fail to see why $\varphi$ is order-preserving. Let $\sigma = (5,5,5,5)$, $\tau = (1,1)$, $\xi = 1$, $\eta = 3$. Now $\xi < \eta$ but $\sigma^\frown \xi <_l \tau^\frown \eta$ fails to hold. Indeed, $$\tau^\frown \eta = (1,1,3) <_l (5,5,5,5,1) = \sigma^\frown \xi.$$ Dec11 revised If $\alpha$ is an indecomposable ordinal, why $\Gamma(\alpha\times{\alpha})=\alpha$? corrected grammar Dec11 awarded Supporter Dec11 revised If $\alpha$ is an indecomposable ordinal, why $\Gamma(\alpha\times{\alpha})=\alpha$? corrected spelling Dec11 revised If $\alpha$ is an indecomposable ordinal, why $\Gamma(\alpha\times{\alpha})=\alpha$? corrected spelling Dec11 answered If $\alpha$ is an indecomposable ordinal, why $\Gamma(\alpha\times{\alpha})=\alpha$? Dec7 awarded Teacher Dec7 answered Textbooks on set theory Dec5 comment An exercise from Levy's Basic Set Theory (Exercise 3.17) I added his statement to my original post. Dec5 comment An exercise from Levy's Basic Set Theory (Exercise 3.17) Mathematicians use paradox for surprising and unexpected results (for example Banachâ€“Tarski paradox, Skolem's paradox). Milner-Rado paradox fails for finite unions. I guess that is why the infinite case is unexpected. Dec5 revised An exercise from Levy's Basic Set Theory (Exercise 3.17) added the exercise and the hint from Levy's book. Dec5 comment An exercise from Levy's Basic Set Theory (Exercise 3.17) Thank you very much for the clarification about the order. Unfortunately, I am still unable to complete the proof. But still working on it. Nov28 awarded Analytical Nov28 awarded Editor Nov28 comment An exercise from Levy's Basic Set Theory (Exercise 3.17) Fixed it. Thanks. Nov28 revised An exercise from Levy's Basic Set Theory (Exercise 3.17) changed 0004 to 00004