ZFC vs ZF

[JOSEPH SHIPMAN]

But we have metatheorems that Choice can be eliminated from most such sentences (Shoenfield Absoluteness etc.).

I want an example which has been proven with AC but not in ZF, and doesn’t satisfy any of the known conditions for eliminating AC, and also hasn’t been shown to require some use of Choice.

— JS

Sent from my iPhone

> On Sep 27, 2021, at 5:55 PM, Buzzard, Kevin M <k.buzzard at imperial.ac.uk> wrote:
> 
> 
> What about an arbitrary statement in modern mathematics like the Poincare conjecture or the local Langlands conjectures? Anything which uses analysis (which these statements do) will almost certainly use at least countable dependent choice (at least in the presentation of the results in the literature, because the mathematicians working in these areas will not in general be interested in whether their uses of AC are essential). 
> 
> Kevin
> From: FOM <fom-bounces at cs.nyu.edu> on behalf of JOSEPH SHIPMAN <joeshipman at aol.com>
> Sent: 27 September 2021 05:16
> To: Foundations of Mathematics <fom at cs.nyu.edu>
> Subject: ZFC vs ZF
>  
> 
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> Is there any well-known theorem of ZFC for which it is an open question whether it is a theorem of ZF?
> 
> Nothing involving large cardinals, please, I already know it is open whether you need choice to refute certain statements about embedding ranks into themselves.
> 
> — JS
> 
> Sent from my iPhone
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