MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In comments on Quora (see, for example, here, here, here), Ron Maimon has repeatedly expressed the strong opinion that Hilbert's program was not killed by Gödel's results in the way typically claimed. More precisely, he argues that the consistency of axioms sufficient for all meaningful mathematics should be provable from "intuitively self-evident" statements about various computable ordinals being well-defined. Here, of course, he points to Gentzen's 1936 consistency proof, which proves Con(PA) from primitive recursive arithmetic plus induction up to the ordinal ε0; as well as more recent results from the field of ordinal analysis, which show that the consistency of various weaker set theories than ZF (for example, Aczel's constructive ZF and Kripke-Platek set theory) can also be reduced to the well-definedness of various computable ordinals. Maimon then goes on to say that Con(ZF) should similarly be reducible to the well-definedness of some "combinatorially-specified," computable ordinal, although the details haven't been worked out yet. (And indeed, the Wikipedia page on ordinal analysis says that it hasn't been done "as of 2008.") This sounds amazing, especially since I'd never heard anything about this problem before! So, here are my questions:

  • Is there a general theorem showing that Con(ZF) must be reducible to the well-definedness of some computable ordinal, i.e. something below the Church-Kleene ordinal (even if we don't know how to specify such an ordinal "explicitly")?

  • Is finding an "explicit description" of a computable ordinal whose well-definedness implies Con(ZF) a recognized open problem in set theory? Do people work on this problem? Or is there some reason why it's generally believed to be impossible, or possible but uninteresting? Or does it just come down to vagueness in what would count as an "explicit description"?

Addendum: There's a connection here to a previous MO question of mine, about the existence of Π1-statements independent of ZF with lots of iterated consistency axioms. In particular, using an observation from Turing's 1938 PhD thesis, I now see that it's indeed possible to define a "computable ordinal" whose well-definedness is equivalent to Con(ZF), but only because of a "cheap encoding trick." Namely, one can define the ordinal ω via a Turing machine which lists the positive integers one by one, but which simultaneously searches for a proof of 0=1 in ZF, and which halts and outputs nonsense if it ever finds one. What I suppose I'm asking for, then, is a computable ordinal α such that Con(ZF) can be reduced to the statement that there's some Turing machine that correctly defines α.

share|cite|improve this question
Scott, I think what you're asking is for the relationship between the proof-theoretic ordinal of ZF and the consistency strength of ZF. While there is often a close relationship between the proof-theoretic ordinal and the consistency strength, there don't seem to be general meta-theorems of the type you're asking for. See the answer to this related MO question: – Timothy Chow Apr 23 '14 at 20:48
To answer the question of why the least ordinal not provably well-founded must be recursive: one way to ask this is, Is there a recursive ordinal $\alpha$ which has a notation $n\in\mathcal{O}$ which ZFC cannot prove corresponds to a well-ordering? Under the assumption that ZFC "proves only true things," the answer must be yes: the set of $n$ such that ZFC proves that $n$ corresponds to a well-ordering is c.e., and so a bounded subset of $\mathcal{O}$. I'm pretty sure any reasonable version of the question will allow an argument like this. – Noah Schweber Apr 23 '14 at 21:13
Noah S's comments at… helped a lot, by confirming for me that there is indeed an unsolved technical problem here, but that it's one that Noah says he doesn't expect to see solved in his lifetime. What's astonishing to me is that I'd never heard of this problem before---despite its seeming like possibly today's best candidate for the P vs. NP or Riemann Hypothesis of set theory and foundations! Big, conceptually-interesting, and not already solved in the 60s -- what's not to like? – Scott Aaronson Apr 23 '14 at 21:24
One difference from P vs NP or the Riemann hypothesis is that there's no crisp, mathematical precise conjecture that one can state here. A satisfactory ordinal analysis of ZF is something that we would probably recognize as such if we saw it, but we don't have a precise way of saying exactly what is desired. – Timothy Chow Apr 23 '14 at 21:52
Scott: Ideally, the ordinal analysis would yield a finitary description of a computable total ordering of the integers. You would have to convince yourself that the ordering was well-founded (i.e., was an ordinal), and that induction holds for such an ordinal. For a set-theory skeptic, the advantage is that each of these statements is "concrete" and involves only countable infinities. But the sticking point is likely to be the well-foundedness. Why should you believe that the given ordering is a well-ordering when you can't prove it, even when you allow yourself the full power of ZF? – Timothy Chow Apr 24 '14 at 15:50
up vote 23 down vote accepted

Rom Maimon is describing the program of proof-theoretic ordinal analysis.

First, as you observed in your addendum, it isn't interesting to find some encoding of an ordinal whose well-foundedness implies Con(ZFC), but rather an ordinal such that the well-foundedness of any representation implies Con(ZFC). One hopes that it suffices to consider natural representations of the ordinal, which has been true in practice, but is unproven (and probably unprovable, given the difficulty of making precise what counts as a natural representation).

It's possible to prove that the smallest ordinal which ZFC fails to prove well-founded is computable by noticing that the computable notations for ordinals provably well-founded in ZFC are a $\Sigma_1$ subset of the computable notations for ordinals, so certainly $\Sigma^1_1$, and by a result of Spector, any $\Sigma_1^1$ subset of the computable notations for ordinals is bounded.

As pointed out in the answer Timothy Chow links to above, it's typically true that this notion of proof-theoretic ordinal ends up being the same as ordinals with other nice properties (like implying Con(ZFC)), but there's no proof that that will always happen (and can't be, since there are defective examples that show it's not always true), nor a proof that covers ZFC.

However it's generally believed that for "natural" theories, including ZFC, the different notions of proof-theoretic ordinal will line up.

Finding an explicit description of the ordinal for ZFC is an active problem in proof theory, but progress has been very slow. The best known results are by Rathjen and Arai (separately) at the level of $\Pi^1_2$-comprehension (a subtheory of second order arithmetic, so much, much weaker than ZFC), and after nearly 20 years, those remain unpublished. The results in the area got extremely difficult and technical, and didn't seem to provide insight into the theories the way the smaller ordinals had, so it's not nearly as active an area as it once was. Wolfram Pohlers and his students still seem to working in the area, and some other people seem to be thinking about other approaches rather than directly attacking it (Tim Carlson and Andreas Weiermann and their students come to mind).

share|cite|improve this answer
Thanks so much; that's exactly the context I was looking for! – Scott Aaronson Apr 23 '14 at 21:27
In particular, the claim that (for example) all arithmetical (or even $\Pi_1$) consequences of ZF can be deduced from some axiom in the first-order language of arithmetic with the same general form as Gentzen's axiom, just with $\epsilon_0$ replaced by "the proof-theoretic ordinal of ZF," is at present still, as Ron Maimon admits, an "article of faith." – Timothy Chow Apr 23 '14 at 21:46
@HenryTowsner The claim that the ZFC provably well-founded computable notations for ordinals form a $\Sigma^1_1$ set seems to assume that ZFC proves only true instances of well-foundedness, doesn't it? This claim implies Con(ZFC), but it seems to be strictly stronger. – Joel David Hamkins Apr 23 '14 at 21:48
(For example, in a model of ZFC+Con(ZFC)+ notCon(Con(ZFC)), there are programs that ZFC proves to halt, that don't really halt in that model, and one can use them to build denotations of ordinals in that model that ZFC proves are well-ordered, even though they aren't well-ordered in that model.) So what is the right background theory for the $\Sigma^1_1$-boundedness argument? – Joel David Hamkins Apr 23 '14 at 21:48
Henry: Just to check my understanding, the proof that there exists a computable ordinal $\alpha$ that ZF doesn't prove to be well-founded is completely nonconstructive? I.e., starting from the ZF axioms, we can't write down a Turing machine (even a weird, incomprehensible one) that computes the order relation of that $\alpha$? Also, is it correct that we currently have no proof, not even a nonconstructive one, that there exists a computable ordinal $\alpha$ such that the well-foundedness of $\alpha$ (under any encoding scheme, not just a contrived one) implies Con(ZF)? – Scott Aaronson Apr 23 '14 at 22:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.