Proof by contradiction and ZF set theory

Question

The following is quoted from Sir Michael Atiyah's draft proof of Riemann Hypothesis, Section 5, stating that 'To be explicit, the proof of RH in this paper is by contradiction and this is not accepted as valid in ZF, it does require choice.' Does this means that a proof by contradiction is only valid in ZFC, not in ZF, provided the assumption that Axiom of Choice can not inferred by the axioms of ZF?

Answer 1

Does this means that a proof by contradiction is only valid in ZFC, not in ZF, provided the assumption that Axiom of Choice can not inferred by the axioms of ZF?

No. It's not entirely clear to me what Atiyah is claiming, although I think that's the most direct reading, but that statement above is false. Moreover, we can prove in a precise sense that choice plays no essential role in the Riemann hypothesis, and that the "provided ..." bit of your question is unnecessary.

Specifically, the key points are:

• Proof by contradiction is perfectly valid in ZF. The validity or invalidity of proof by contradiction is an issue about the underlying logic, not the theory, and ZF and ZFC use the same underlying logic (namely, classical first-order logic).

• It is known that ZF cannot prove the axiom of choice (at least, as long as ZF is consistent in the first place); this was proved by Cohen (following a much simpler proof, due to Godel, that ZF cannot disprove AC unless ZF is inconsistent.)

• Most interestingly, any proof of RH from ZFC can be turned into a proof of RH from ZF, even if the original proof seems to fundamentally use the axiom of choice. This is a consequence of an absoluteness theorem: absoluteness theorems state that certain axioms (e.g. choice) can be removed from proofs of sentences which are particularly "simple" in form, and it turns out that RH can be expressed in a sufficiently simple manner for an appropriate absoluteness result to apply.

Note that my second and third bulletpoints definitely require proof. They're absolutely (hehe) not obvious, and it's perfectly reasonable for mathematicians outside of logic to not be familiar with them (especially the third). That said, they do demonstrate that the set-theoretic concerns raised are void.