[FOM] Eliminating AC
Colin McLarty
colin.mclarty at case.edu
Sat Mar 23 16:34:05 EDT 2013
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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