[FOM] Eliminating AC

[Joe Shipman]

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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