How to measure the strength of Zermelo over bounded Zermelo? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T08:31:01Z http://mathoverflow.net/feeds/question/117371 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/117371/how-to-measure-the-strength-of-zermelo-over-bounded-zermelo How to measure the strength of Zermelo over bounded Zermelo? Colin McLarty 2012-12-27T23:48:58Z 2012-12-28T18:50:57Z <p>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 $\aleph_n$." </p> <p>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.</p>