[FOM] Eliminating AC

[Colin McLarty]

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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