
> 
> **Question:** *Does* **Con**($ZF$) *imply* **Con**($ZF$ + $DC$ + "*there is no paradoxical Banach-Tarski decomposition of the unit ball*")?
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Here **Con**($X$) is the consistency of $X$; $DC$ is dependent choice.


**Motivation for the Question**: Since the "paradoxical" sets in the Banach-Tarski theorem have to be non-measurable, Solovay's model of "every subset of reals is measurable" shows: 


**Theorem**. **Con**($ZF$+ **Inacc**) *implies* **Con**($ZF$ + $DC$ + "*there is no paradoxical Banach-Tarski decomposition of the unit ball*"). 


In the above **Inacc** is the statement "there is an inaccessible cardinal". 


Note that Shelah's model [Israel Journal of Math, 1984], constructed only
from **Con**($ZF$), in which $DC$ holds and all sets of reals have the Baire
property, is of no help in answering the above question since Dougherty and
Foreman [ J. Amer. Math. Soc., 1994] have shown that there are paradoxical
decompositions of the unit ball using pieces which *have the property of
Baire*. I do not know how much choice is needed in their construction.


This question was posed a while ago on an [FOM-posting of mine](http://cs.nyu.edu/pipermail/fom/2004-September/008500.html), but remained unanswered.



Paradoxical Decompositions 
A positive answer is proved in S. Wagon's book *"The Banach-Tarski Paradox",* Theorem 13.2. Specifically, the statement proved there is:


Con(ZF) $\leftrightarrow$ Con(ZF + DC + GM),


where GM is the existence of an isometry-invariant measure on all subsets of $\mathbb R^n$ taking the value $1$ on the unit cube.


