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Bounded Zermelo is Zermelo set theory with only bounded separation. It has the same strength as simple type theory or MacLane set theory or ETCS. It is a finitely axiomatized fragment of Zermelo, so Zermelo proves it is consistent. And Mathias proved a paradigmatic example of the difference: Even if we add choice, Bounded Zermelo proves $\aleph_0$ exists, and every $\aleph_{\alpha}$ has a successor cardinal $\aleph_{\alpha+1}$, while BZ does not prove the quantified statement "for every $n\in \mathbb{N}$, there exists $\aleph_n$."

But is there some more quantitative measure of its strength? For example, do Zermelo and bounded Zermelo have meaningful proof theoretic ordinals? I have heard that proof theoretic ordinals do not work well for theories strong enough to prove existence of power sets.

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There is something confusing in the statement about cardinals. Is BZ the same as Bounded Zermelo? If so, is this just a failure of induction? Also, when you write $\aleph_\alpha$ do you mean a wellordered cardinal or the actual initial ordinal? –  François G. Dorais Dec 28 '12 at 1:25
    
Ah, in the absence of choice I should not use $\aleph$s so freely. Mathias shows: Bounded Zermelo proves there are transfinite sets, and any finite list of them can be lengthened, but does not prove the quantified statement "for every $n$ there is a set of $n$ transfinite sets each larger than the last." I would not call it a failure of induction. It is just that induction is stated for subsets of $\mathbb{N}$, and the above statement is not bounded and does not define a subset of $\mathbb{N}$. –  Colin McLarty Dec 28 '12 at 2:33
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Toshiyasu Arai has been working on getting proof theoretic analysis higher up toward ZF. I'm not familiar enough with it to say anything. Arai has a lot of relevant papers on the arxiv - arxiv.org/find/math/1/au:+Arai_T/0/1/0/all/0/1 - and some surveys on his home page - kurt.scitec.kobe-u.ac.jp/~arai –  François G. Dorais Dec 28 '12 at 2:45
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