As everybody knows, the ZFC axioms may serve as a foundation for (almost) all of contemporary mathematics, and it is also well-known that several results are "indecidable" in ZFC, which means that they cannot be proved or disproved within ZFC.

It is therefore natural to look for "new axioms" to add to ZFC and make it a stronger system. But by Godel's second incompleteness theorem, the consistency of ZFC cannot be deduced from ZFC itself.

Therefore, we may add the axiom "ZFC is consistent" and obtain a new system $ ZFC_1 $ consisting of "ZFC+(ZFC is consistent)". We may iterate this, and define $ ZFC_2 $ as "$ZFC_1$+($ZFC_1$ is consistent)", etc, and we may even define $ZFC_{\omega}$, or $ZFC_{\alpha}$ for any ordinal $\alpha$.

This seems a little too easy, so my question to logicians is : is this construction completely irrelevant to logic ans set theory questions ? If so why? Is it true that the results which are classically independent of ZFC are also independent of $ZFC_1$, $ZFC_2$, $ZFC_{\omega}$ etc ?


3 Answers 3


Such constructions are interesting! However, they are often done with PA instead of ZFC (see note). For an interesting discussion, I recommend Torkel Franzén's book Inexhaustibility: a non-exhaustive treatment (Lecture Notes in Logic 16, ASL, 2004). You can also read this excellent blog article by Mike O'Connor.

Note: The following is explained in Mike O'Connor's article, but I think I need to clarify why ZFC is not the ideal base theory to do this and why PA is a better candidate.

The idea is that Con(T) is usually understood as an arithmetical statement. More precisely, given a recursive presentation of the theory T the statement Con(T) is arithmetical formalization of "there is no proof of a contradiction from T" which is encoded using Gödel numbers for proofs and formulas. (This is the messy part of Gödel's Theorem.) This is why PA, or more generally any recursively axiomatizable extension of PA, provides a more natural environment for the analysis of such statements. For example, instead of ZFC, you may as well use the purely arithmetical part of ZFC.

There is also an even more fundamental problem with transfinite iterates. Given a recursive presentation of a theory T, the iterates T0 = T, T1 = T0 + Con(T0), T2 = T1 + Con(T1), etc. Can be continued into the transfinite, but only to a limited extent. It is easy to give a recursive presentation of Tω or Tω+ω+3 but there are only countably many ordinals for which this works. Indeed, these iterates are better defined in terms of ordinal notations than in terms of proper ordinals. Ordinal notations can go pretty far up, but there are clear limitations.

These difficulties and their implications are discussed in great detail in Franzén's book. As Mike O'Connor explains, it is natural to go further and extend these to subsystems of second-order arithmetic, but there and in set theory the appropriate principles to study are reflection principles and large cardinal axioms which have better semantic interpretations and take advantage of their richer surroundings.

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    $\begingroup$ But isn't it true that if the process makes sense with alpha using PA or even second-order arithmetic, then it will also make sense with ZFC for this same ordinal? $\endgroup$ Jan 24, 2010 at 22:52
  • $\begingroup$ Yes, but you're still analyzing arithmetical statements. A more appropriate choice for T is PA + (all arithmetical consequences of ZFC), for example. $\endgroup$ Jan 24, 2010 at 22:58
  • $\begingroup$ But so what? Any large cardianl consistency strength statement is also arithmetic (since it is a consistency statement), but we still analyze them in ZFC rather than PA. My sense of the question is that since we believe ZFC, we naturally also want to include Con(ZFC) etc., as among the strongest theories that we really believe in. Perhaps the situations is that it is the negative results (e.g. on which ordinals are not representable) are stronger/better for PA. Does that sound right? $\endgroup$ Jan 24, 2010 at 23:32
  • $\begingroup$ PA is definitely easier to work with, but which ordinals are representable doesn't really change. (Albeit, some theories do much better than others at understanding certain types of notations, which does have consequences.) The point I was trying to make is to study notions in their appropriate settings. Con(ZFC) has no good semantics, so there is little point in studying it in the context of set theory. On the other hand, ZFC + "there is a wellfounded model of ZFC" is interesting to look at in set theory, but difficult to understand arithmetically. $\endgroup$ Jan 25, 2010 at 0:25
  • $\begingroup$ Are the "representable ordinals of a theory the same thing as the proof-theoretic ordinal? Is this what the proof-theorists mean? $\endgroup$ Jan 25, 2010 at 0:39

What you propose is very reasonable, of course, since when we believe in a theory T, then it is natural for us also to believe that T is consistent. And the axioms that you propose to add to ZFC formalize this process. The (philosophical) question here is, does this process somehow find a completion?

(Let me quibble with your remark that we can formalize ZFCα for any ordinal α. We need that the ordinal α is somehow representable in the theory in order for the assertion Con(ZFCα) to be expressible. Of course, in a countable language, we have only countably many statements, and so we must eventually run out of representable ordinals.)

The answer is that your axioms are the pre-beginnings of the large cardinal hierarchy, as hinted at by Kristal Cantwell and Dorais. If there is a (strongly) inaccessible cardinal κ, then Vκ is a model of ZFC, and so your theory ZFC1 holds.

But I claim much more, and from a weaker hypothesis. One doesn't need an inaccessible cardinal even to know that all the expressible ZFCα are consistent.

I claim that if there is an ω model of ZFC, then all the expressible ZFCα are true and consistent.

To see this, suppose that M is an ω model of ZFC. This means that M has the standard natural numbers. From this, it follows that the ordinals of M are well-founded for some distance above ω, but may become ill-founded much higher up. Since M has the same natural numbers as we do in the meta-theory, it follows that M has exactly the same formulas in the language of set theory and, more importantly, exactly the same proofs. Thus, for any theory T that exists in M, it will be consistent in M if and only if it is consistent.

This is enough to perform an interesting ramping-up argument. Namely, since M is a model of ZFC, it follows that ZFC is consistent for us, and so M agrees, and so M is a model of ZFC+Con(ZFC), which is to say, of ZFC1. Thus, ZFC1 is consistent, and so M agrees that ZFC1 is consistent, and so M is a model of ZFC2. Thus, ZFC2 is consistent, and so M agrees, and so ZFC3 is consistent, and so on. Do you see how it works? If ZFCα is consistent, then M will agree (if α is in M), and so ZFCα+1 is also consistent. (And limit stages are basically free, since proofs are finite.)

So the scheme of theories ZFCα forms a hierarchy of consistency strength that sits very low below the beginning of the large cardinal hierarchy. I think much of the sense of your question is this:

  • We know by the Incompleteness theorem that no theory can prove its own consistency, and so we want to consider theories that transcend this consistency in the way you describe.

And this is exactly what the large cardinal hierarchy provides. Each level of the large cardinal hierarchy implies the consistency of the lower levels, and the consistency of the consistency and so on, iterating in the style of your questions. But the large cardinals are able to jump higher than these small steps of consistency, by finding natural axioms that imply the consistency of all iterations of the consistency process that you describe for the lower levels.

I just noticed the bit at the end of your question, about whether independence results also hold for ZFCα. This is a very interesting question, and the answer is Yes, they all work just the same. The reason is that all the independence results, proved either by forcing or by the method of inner models, have the property that the resulting models have the same arithmetic truths as the original model. Since the consistency statements you are considering are arithemtic statements, they are not affected by forcing or inner models. In particular, Cohen's proof that Con(ZFC) implies Con(ZFC+¬CH) turns directly into a proof that Con(ZFCα) implies Con(ZFCα+¬CH). If one formalizes a version of (ZFC+¬CH)α, it follows that it will be equivalent to ZFCα+¬CH. And the same holds for all the other indpendence results of which I am aware.


If ZFC is consistent then it can be cannot be axiomatized with a finite number of axioms. On the other hand the consistency of ZFC follows from the existence of a weakly inaccessible cardinal. These results are from the Wikipedia article on Zermelo–Fraenkel set theory.

From the wikipedia article on the continuum hypothesis it says that so far it is independent of all large cardinal axioms in the context of ZFC.

  • $\begingroup$ I don't see how this actually answers the question, which is not about arbitrary new axioms but rather specifically iterated consistency principles. $\endgroup$ Sep 1, 2018 at 19:43

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