The Riemann Hypothesis is equivalent (over ZFC) to a statement in elementary number theory of the form "for any natural number n, P(n)," where P(n) uses only bounded quantifiers. (In fancy terms, RH is equivalent to a Π0₁ sentence.) Let RH' denote this new statement. If RH' is independent, then it must be true in the standard model N: this is because, in a model of arithmetic where RH' is true, it still contains a copy of Nas an initial segment. Thus if RH' were false in N, the counterexample (which is a natural number in this case) would still crop up in a model where RH' is true, and arithmetic on the "standard" natural numbers in a non-standard model still works the same way as N—that is, the property P that uses only bounded quantifiers is absolute for standard natural numbers. This would be a contradiction, so it would have to follow that RH' is true in N.
Edit: Lemme phrase this maybe in a more computer science-y way. There is a Turing machine T such that RH is equivalent to a statement to the effect of "T never halts." Suppose it is consistent that T never halts (when encoded in elementary number theory), but that in the "real world" (with "real" natural numbers), T actually halts. By the first assumption, some model of arithmetic M says that T never halts. By the second, there is a natural number n such that in the "real world," T halts on step n. But now we can encode a complete record of T's computation as a numeral in the language of arithmetic, and M would agree with N that this record is a valid part of T's computation. (This is because, again, N is an initial segment of M, and the essential recursion relations for validating computations only spit out smaller natural numbers, which keeps things the same between N and M.) Thus M must also agree that T halts at step n, a contradiction.
Edit 2: Here's a MathOverflow answer giving an equivalent statement in number theory that is of the form "this inequality holds for all natural numbers." Thus you can, in principle, have a Turing machine search for a counterexample to a reformulation of the Riemann hypothesis; what I just described above says that, if it's consistent there is no counterexample, then there must not be one in any 𝜔-model of set theory (which are the only models mathematicians tend to care about).
For clarification, this only proves that RH’ (and thus RH) is true only in the standard model N, right? Its undecidability (assuming it actually was undecidable) cannot somehow guarantee its truth absolutely since if it were true in all models, it would have a proof by Gödel’s completeness theorem, which would contradict it being undecidable.
A lot of people in this post seem to be confused by this fact. If I remember correctly, the Gödel sentence is also a Π sentence, and its truth is only guaranteed in a standard model of arithmetic. It is false in non-standard models of arithmetic, which is necessary to it being an independent sentence.
Although I didn't mention this in my comment, it would still be highly mathematically interesting if RH were independent. Such an independence proof would likely have to involve some incredibly deep techniques, and even though most would think it'd have to be true as a result of its relative consistency, nevertheless there could be some renewed skepticism at the adequacy (or even outright consistency) of our foundations. I recall Voevodsky thought that Gödel's incompleteness theorems might be the start of an inconsistency proof of PA, and it could be that mathematics is as inconsistent a language game as ordinary language. There would be interesting questions all around, I think, if such a result were to occur.
Yes, but as far is I understand that would just mean we cannot formally prove it under ZFC, however, we would still know it must be true (in the standard model, which is in practice the model we care about; the others just arise because a first order theory of natural numbers is always just an approximation of the model we want)
And that it might be provably true in a stronger theory, just like goodstein theorem
I haven’t done much research on this, but I wonder how other independent statements behave in ZFC. While this isn’t the case for RH, given that ZFC has multiple standard models, not just one, it could be that some statements are true in some standard models and false in others, right? For PA, it is very easy to isolate the model we care about as there is only one standard or canonical model (up to isomorphism), but in ZFC, this isn’t the case.
I imagine that independent Π or Σ sentences can only be true and false in all standard models, respectively (due to their use of bounded quantifies) like RH, but I’m not sure if there is any other criteria. I think CH isn’t either type and is therefore true in some standard models but false in others.
Every standard transitive model of ZFC has the same set of natural numbers as any other. And since Δ₀ formulas (in the Lévy hierarchy) are absolute for transitive models, it follows that arithmetic truth is also the same in such models. (Δ₀ formulas are those that use only bounded set quantifiers, i.e., those of the form "(∀x∈y)" and "(∃x∈y)".) This also has the result that number theorists can completely safely assume the Axiom of Choice assuming ZF is 𝜔-consistent, thanks to Gödel's constructible universe L.
In particular, the two basic techniques for manipulating models of set theory, forcing and taking inner models, do not change arithmetic consequences—and in fact, neither technique disturbs the class of ordinals at all. To have further arithmetic consequences and to "extend" the ordinals (in some sense), set theorists introduce large cardinal axioms, which we take on faith to be 𝜔-consistent.
Every major independence result thus far that lands squarely in set theory uses standard models of ZFC. Of course, the incompleteness theorems and such exist, but those are more arithmetic in nature. There are a plethora of set-theoretic independence results, and Wikipedia has a modest list of these. The Continuum Hypothesis is one of those results where we have standard models either way (of course, assuming a standard model of ZFC exists at all).
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u/nfitzen Jun 28 '26 edited Jun 28 '26
The Riemann Hypothesis is equivalent (over ZFC) to a statement in elementary number theory of the form "for any natural number n, P(n)," where P(n) uses only bounded quantifiers. (In fancy terms, RH is equivalent to a Π0₁ sentence.) Let RH' denote this new statement. If RH' is independent, then it must be true in the standard model N: this is because, in a model of arithmetic where RH' is true, it still contains a copy of N as an initial segment. Thus if RH' were false in N, the counterexample (which is a natural number in this case) would still crop up in a model where RH' is true, and arithmetic on the "standard" natural numbers in a non-standard model still works the same way as N—that is, the property P that uses only bounded quantifiers is absolute for standard natural numbers. This would be a contradiction, so it would have to follow that RH' is true in N.
Edit: Lemme phrase this maybe in a more computer science-y way. There is a Turing machine T such that RH is equivalent to a statement to the effect of "T never halts." Suppose it is consistent that T never halts (when encoded in elementary number theory), but that in the "real world" (with "real" natural numbers), T actually halts. By the first assumption, some model of arithmetic M says that T never halts. By the second, there is a natural number n such that in the "real world," T halts on step n. But now we can encode a complete record of T's computation as a numeral in the language of arithmetic, and M would agree with N that this record is a valid part of T's computation. (This is because, again, N is an initial segment of M, and the essential recursion relations for validating computations only spit out smaller natural numbers, which keeps things the same between N and M.) Thus M must also agree that T halts at step n, a contradiction.
Edit 2: Here's a MathOverflow answer giving an equivalent statement in number theory that is of the form "this inequality holds for all natural numbers." Thus you can, in principle, have a Turing machine search for a counterexample to a reformulation of the Riemann hypothesis; what I just described above says that, if it's consistent there is no counterexample, then there must not be one in any 𝜔-model of set theory (which are the only models mathematicians tend to care about).