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seen Feb 27 '13 at 10:12

Dec
20
answered Is there a bounded-quantifier formula in ZF without Foundation that defines the naturals?
Dec
20
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?
Dec
20
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?
Dec
13
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.
Dec
13
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, $$\tau^\frown \eta = (1,1,3) <_l (5,5,5,5,1) = \sigma^\frown \xi.$$
Dec
11
revised If $\alpha$ is an indecomposable ordinal, why $\Gamma(\alpha\times{\alpha})=\alpha$?
corrected grammar
Dec
11
awarded  Supporter
Dec
11
revised If $\alpha$ is an indecomposable ordinal, why $\Gamma(\alpha\times{\alpha})=\alpha$?
corrected spelling
Dec
11
revised If $\alpha$ is an indecomposable ordinal, why $\Gamma(\alpha\times{\alpha})=\alpha$?
corrected spelling
Dec
11
answered If $\alpha$ is an indecomposable ordinal, why $\Gamma(\alpha\times{\alpha})=\alpha$?
Dec
7
awarded  Teacher
Dec
7
answered Textbooks on set theory
Dec
5
comment An exercise from Levy's Basic Set Theory (Exercise 3.17)
I added his statement to my original post.
Dec
5
comment An exercise from Levy's Basic Set Theory (Exercise 3.17)
Mathematicians use paradox for surprising and unexpected results (for example Banach–Tarski paradox, Skolem's paradox). Milner-Rado paradox fails for finite unions. I guess that is why the infinite case is unexpected.
Dec
5
revised An exercise from Levy's Basic Set Theory (Exercise 3.17)
added the exercise and the hint from Levy's book.
Dec
5
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.
Nov
28
awarded  Analytical
Nov
28
awarded  Editor
Nov
28
comment An exercise from Levy's Basic Set Theory (Exercise 3.17)
Fixed it. Thanks.
Nov
28
revised An exercise from Levy's Basic Set Theory (Exercise 3.17)
changed 0004 to 00004