What is the error in the following reasoning regarding the Riemann Hypothesis?

Question

I encountered the following earlier today and was wondering where, or if there are, errors in the reasoning.

Assume the Riemann Hypothesis is independent of ZFC.

Assert that the RH is false, though due to its assumed independence, it is unprovably false.

If the RH is false, a non-trivial zero of $\zeta(s)$ exists that is not located on the critical line.

However, the existence of such a point could be used as proof of the falsehood, which would contradict the assumption that RH is independent and false. Therefore, the Riemann Hypothesis is either independent and true, or decidable.

Correct me if I'm wrong, but I believe that the non-trivial zero not located on the critical line would have to be verifiably so in order for the reasoning to hold. In other words, assuming RH is independent and false, let $p$ be a zero of $\zeta(s)$ not located on the critical line. If the proposition $\zeta(p)$ is decidable, then the above reasoning should hold.

The reason I speculate on this condition is because similar reasoning could be applied to well-known undecidable propositions such as the Continuum Hypothesis:

Assume the Continuum Hypothesis is independent of ZFC.

Assert that the CH is false.

If the CH is false, there exists a set $S$ such that $\aleph_0 < |S| < c$.

The existence of such a set can be used as proof of the falsehood of the CH. Therefore, the CH is either independent and true or decidable.

I'm guessing I am wrong on several points here, and I would greatly appreciate any clarification/explanation/thoughts.

Answer 1

You raise two different questions, with rather different associated issues.

With regard to the Riemann Hypothesis, the decidability question is essentially the same as with many other currently undecided statements, e.g., "there are no odd perfect numbers". If there is such a number, then one can verify it with a computation, and the computation can be carried out "inside" ZF--indeed, inside Peano Arithmetic (PA). As pointed out in the comments, and as you correctly guessed, RH has the same character: if it's false, it's "computably false".

This feature of PA is known as $\Sigma_1$ completeness. Any statement that can be expressed in the form $\exists x\;\psi(x)$, where $\psi$ is a so-called recursive (or computable) predicate, is, if true for the "standard natural numbers", provable from PA.

Not all number-theory assertions take this form. For example, consider "there are infinitely many prime pairs". If true, then there is an $N$ with no prime pairs larger than $N$. But how would you computably know that that was true about $N$? The statement takes the form $$\exists N\;\forall p\!\!>\!\!N\; \psi(p)$$ where $\psi(p)$ is computable. The $\exists\forall$ prefix makes this not $\Sigma_1$.

CH is like this, only worse. Its negation can be expressed in the form $$\exists S\subseteq\mathbb{R}\;\forall f\; \psi(f)$$ where $\psi(f)$ says, "$f$ is neither a bijection between $S$ and $\mathbb{R}$, nor between $S$ and $\mathbb{N}$".

So again we have the $\exists\forall$ prefix. But this time, $\psi(f)$ doesn't look particularly "computable".

Answer 2

Provably false means that a statement is false in all models of $\mathsf {ZFC}$.

Let's use a simpler example. The statement $\exists x~(x^2+1=0)$ is undecidable in the theory of fields with characteristic $0$. That just means there are some models of the theory (such as $\Bbb C$) that contain a solution (typically known as $i$) and other models of the theory (such as $\Bbb R$) that contain no solution.

That's what's going on here. For the continuum hypothesis, there are some models that contain a set with cardinality strictly between $\aleph_0$ and $2^{\aleph_0}$, but that set doesn't exist in all models of $\mathsf{ZFC}$. Similarly, if the Riemann Hypothesis is undecidable, it means that there's an off-critical-line solution of $\zeta(s)=0$ in some models but that particular $s$ isn't present in all models.