Talk:Axiom of projective determinacy

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Independence[edit]

Article states that PD is undecidable in ZFC, but as far as I know (and consulting the references given) it is merely unknown whether it is consistent with ZFC (but most mathematicians think it is). Don't have a reference to hand on this, though. Fish-Face (talk) 21:36, 3 May 2011 (UTC)

Depends on what you mean by "known". ZFC+PD has strictly higher consistency strength than ZFC, and therefore it is impossible to prove in ZFC that PD is consistent. (Unless, of course, ZFC is inconsistent, in which case ZFC does prove that ZFC+PD is consistent, and ZFC proves everything else as well).
But it's misleading to say that it's "unknown". It's "known" in the same sense that, say, ZFC itself is "known" to be consistent, or even something like Peano arithmetic is "known" to be consistent. It's not a matter of proof, because that gets you into an infinite regress where you have to justify the reliability of the proof system first. It's more of an empirical fact. --Trovatore (talk) 21:59, 3 May 2011 (UTC)
And still, the phrase "The axiom is independent of ZFC (assuming that it is consistent with ZFC), unlike the full axiom of determinacy (AD), which contradicts the Axiom of Choice" looks strange for me. PD does not follow from ZFC (and hopefully does not contradict it); AD does contradict (and hopefully does not follow...). The "unlike" is a bit misleading, and "independent (assuming consistency)" is hardly more clear than just "does not follow from" (or "is not a theorem of"). Boris Tsirelson (talk) 12:15, 17 December 2012 (UTC)

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