are they still smaller than omega-1-CK?what are the notations of them?
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The proof-theoretic ordinal of any theory is less than $\omega_1^{CK}$. No notations are known for second-order arithmetic, let alone ZFC.
Henry Towsner
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1Henry, could you elaborate with further explanation of your first sentence? For example, what exactly would it mean to have the proof-theoretic ordinal of ZFC? (And by the way, thanks again for your talk at CUNY last week.) – Joel David Hamkins Mar 24 '14 at 23:30
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1Is the notion of $\omega_1^{CK}$ absolute? I mean, it's a fairly small ordinal (as countable ordinals go), so certainly it is less than the least height of a transitive model of $\sf ZFC$. But given two transitive models of $\sf ZFC$ (regardless to their size), do they have to agree [with $V$] on $\omega_1^{CK}$? (@Joel, you might know the answer to that as well, so I'm pinging you too.) – Asaf Karagila Mar 25 '14 at 00:32
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1For transitive models, yes, it is absolute, since transitive models have all the same TM programs and they agree on whether a given relation on the natural numbers is well-founded or not. Meanwhile, there are models of ZFC with the same $\mathbb{N}$ that disagree on $\omega_1^{CK}$; this is proved in my recent paper http://jdh.hamkins.org/satisfaction-is-not-absolute/. – Joel David Hamkins Mar 25 '14 at 00:35
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1@JoelDavidHamkins: The proof-theoretic ordinal of a theory T is usually defined to be the smallest ordinal $\alpha$ such that there is no computable (equivalently: primitive recursive) representation of $\alpha$ as an ordering of $\mathbb{N}$ such that T proves the well-foundedness of this ordering. Such an ordinal exists and is less than $\omega_1^{CK}$ because $\omega_1^{CK}$ is, by definition, larger than any ordinal with a computable representation. – Henry Towsner Mar 25 '14 at 03:36
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Roughly what I was saying at the start of that talk was that usually when we talk about a proof-theoretic ordinal, we expect it to have various other properties as well, and sometimes people take one of those to be the definition. But for ZFC (as a "reasonable" theory), they're probably equivalent anyway. – Henry Towsner Mar 25 '14 at 04:15