ZFC vs ZF

[JOSEPH SHIPMAN]

Good catch. That shows that there is a nonmeasurable set. If you could prove that in ZF you could combine that with Solovay’s result that an inaccessible allows you construct a model where all sets are Lebesgue measurable to conclude that an inaccessible is inconsistent.

So it’s extremely unlikely already that this is provable in ZF but I don’t know whether it’s actually open. The Banach Tarski decomposition seems much stronger than there being a nonmeasurable set, I suspect you might be able to use it to construct a tiny amount of choice (for example choosing from countably infinitely many 2-element sets).

Perhaps this paper of Dougherty and Foreman is relevant:

https://www.ams.org/journals/jams/1994-07-01/S0894-0347-1994-1227475-8/S0894-0347-1994-1227475-8.pdf

Without choice there is a model of ZF in which all sets have the property of Baire, so if Banach-Tarski implied (in ZF) not only nonmeasurable sets but sets without the property of Baire, then we would know that it wasn’t a theorem of ZF, but Dougherty and Foreman seem to block this path because the Banach-Tarski pieces, while not being nice enough to be measurable, can be nice enough to have the property of Baire!

Can anyone provide an example of a theorem of ZFC for which it is not only open whether it is a theorem of ZF, but plausible that it might be? (Inconsistency of an inaccessible rules out “plausible” in my book....)

— JS

Sent from my iPhone

> On Sep 28, 2021, at 8:36 PM, Frode Bjørdal <bjordal.frode at gmail.com> wrote:
> 
> The Banach-Tarski decomposition of a ball into two balls the same size. 
> 
> On Mon, Sep 27, 2021 at 5:10 PM JOSEPH SHIPMAN <joeshipman at aol.com> wrote:
>> 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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