[FOM] Eliminating AC
Craig Smorynski
smorynski at sbcglobal.net
Mon Mar 25 17:10:21 EDT 2013
I think you are asking for the impossible. The only reasonable proofs that choice is not needed involve inner models containing the natural numbers in which choice is true. If he already knows predicate calculus and the axioms of ZFC, the construction of L is not that complex as constructions in mathematics go.
On Mar 24, 2013, at 8:47 PM, Joe Shipman wrote:
> This still doesn't help me. Please assume that my mathematical interlocutor understands the axiomatic method, predicate calculus, and what the ZFC axioms actually are and how to derive "ordinary mathematics" from them, but he does not know any technical set-theoretical results about constructibility, absoluteness, etc.
>
> He knows what a proof from ZFC is, he knows what a proof from ZF is, he prefers the latter, and he wants insight into how, if he has a proof of an arithmetical statement from ZFC, he can find a proof of that statement from ZF that he could then present to his colleagues instead of the proof from ZFC that he currently has.
>
> -- JS
>
> Sent from my iPhone
>
> On Mar 23, 2013, at 4:34 PM, Colin McLarty <colin.mclarty at case.edu> wrote:
>
> Well this gives a global well ordering "definable" if you take all ordinals as parameters, so in general I think this is not what someone would mean by eliminating choice from your reasoning. You replace choice by appeal to a moderately arcane idea, not closer to ordinary arithmetic than choice is.
>
> I look at it the other way, as proving that choice is unobjectionable here since it is not a genuine assumption in proofs of arithmetic statements -- its availability is a theorem of ZF (without choice).
>
> To use Nik's terms, whatever extravagance is involved in using ZF, say to handle sheaves in a cohomological argument, it is not more extravagant to use ZFC.
>
> Colin
>
>
>
> On Fri, Mar 22, 2013 at 11:06 PM, Joe Shipman <JoeShipman at aol.com> wrote:
> How would one go about transforming that explanation into an explicit procedure for eliminating AC from a proof of an arithmetical statement? By replacing every use of an axiom of ZFC with the relativization of that axiom to L?
>
> I think this way of explaining it would work if the relativization of an axiom was easy to state. Instead of L, either HOD or M (the strongly constructible sets) could play the same role in the proof, while possibly having an easier and more intuitive construction than L.
>
> -- JS
>
> Sent from my iPhone
>
> On Mar 21, 2013, at 6:30 PM, Colin McLarty <colin.mclarty at case.edu> wrote:
>
>
> Tell the mathematician that Gödel showed we can restrict set theory to just the sets that are definably constructed in a certain way, and even if we do not assume choice in general it holds for these definable sets, since we can actually define a choice function for any definably constructed problem. But finite sets are all definable in this way (indeed they are all definable in the most naive way). So all of arithmetic can be done in this restricted set theory where we can get choice without making it an assumption.
>
> Colin
>
>
> On Thu, Mar 21, 2013 at 2:47 PM, Joe Shipman <JoeShipman at aol.com> wrote:
> I am looking for something that I can explain in a few minutes to a mathematician who is unfamiliar with Godel constructibility, so that he will feel he understands why the result is true.
>
> -- JS
>
> Sent from my iPhone
>
> On Mar 21, 2013, at 12:17 AM, Craig Smorynski <smorynski at sbcglobal.net> wrote:
>
> The original proof by noticing that the natural numbers in V and L are the same must surely be as simple as possible, which of course is not to say that it is simple.
>
> On Mar 20, 2013, at 7:52 PM, Joe Shipman wrote:
>
>> What is the simplest way to see that any arithmetical consequence of ZFC is a consequence of ZF?
>>
>> -- JS
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> Craig
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Craig
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