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Let $M$ be a countable set. I can take a finite set $F$ of n-ary operations and construct a minimal set $M'$ for which:

  • $M \subset M'$
  • For each $f \in F$ and $m_1,m_2…m_n \in M$ where $n$ is the arity of $f$: $f(m_1,m_2…m_n) \in M$

$M'$ is a set which is closed against all operations in $F$. It can be shown that $M'$ is also countable. Each element of $M'$ can be described by a finite string using elements from $M$ and operations from $F$.

Now let $M$ be $\Bbb{Q}^3$ and $F$ be a set of all rotations needed in Banach-Tarski paradox (i.e. four rotations) and translation. Then ${\Bbb{Q}^3}'$ is a countable set. Now I can perform the Banach-Tarski paradox in this space the same way as in $\Bbb{R}^3$ with one exception: I don't need the axiom of choice because I use only countable sets. ${\Bbb{Q}^3}'$ is countable so it can be well ordered. For each subset of ${\Bbb{Q}^3}'$, I can pick the least element of this ordering.

Is there anything I missed?

Hume2
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  • How do you define volume in your example? – Fabio Somenzi May 19 '19 at 15:45
  • I don't need to define volume here. I take one sphere and get two spheres, regardless of their volumes. – Hume2 May 19 '19 at 15:49
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    Then, I'm not sure most people would consider the result a paradox. See, for instance, page 1 of this expository paper for an example where you get a proper superset of a set in a way reminiscent of Banach-Tarski, but without reaching a paradox. – Fabio Somenzi May 19 '19 at 15:56
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    The whole point of the Banach-Tarski "paradox" is that it teaches us that volume cannot be defined for all subsets of $\Bbb R^3$ without violating either invariance under isometry or summability. So what does your countable version teach us? – Paul Sinclair May 19 '19 at 19:32
  • One sphere is a subset of two spheres. So it tells us that one sphere can be divided into finite number of parts and composed into two copies. And it has a geometric representations. It doesn't tell anything about the volume, only that the two spheres are union of two original spheres. – Hume2 May 20 '19 at 09:31

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