Can There be a 1 dimensional Banach-Tarski paradox in the absence of choice

Question

Let $\mathbb{R}$ act on itself by translation. Then there is no finite decomposition of a unit interval into pieces which, when translated, yields two distinct unit intervals.

More formally does there exist a partition of the unit interval $[0,1]$ into finitely many sets $S_1,...,S_n$ and a collection of finitely many constants ${r_1,...r_n} \in \mathbb{R}$ such that the the union of $S_a+r_a$ is the set $[0,1]\cup [2,3]$. Where $S_a+r_a$ is the set $S_a$ shifted to the right by $r_a$.

The only proofs I know of this crucially use the invariant mean-characterization of amenability for $\mathbb{R}$. The standard proofs of this characterization use the ultrafilter lemma and are therefore not in ZF+DC.

Is it possible in ZF or (more strongly) ZF+DC for there to be finite decomposition of the unit interval into two pieces. Can we put a bound on how many pieces you would need? (i.e. could you do it with 3 or 4 pieces)?

Answer 1

No. Let $A$ be the free abelian group generated by the $r_a$s. We can view this as a lattice in the real vector space $A \otimes \mathbb R$. Let $n$ be the rank of this group / the dimension of this vector space.

There is a natural evaluation map $f: A \otimes \mathbb R \to \mathbb R$ from this vector space to $\mathbb R$. Applying the assumed bijection between $[2,3] \cup [0,1] $ and $[0,1]$, which is obtained by at each point by subtracting one of the $r_a$s and hence preserved the property of lying in $A$, we obtain a bijection between $A \cap f^{-1} ( [2,3] \cup [0,1])$ and $ A \cap f^{-1} ([0,1])$.

In the case $n=1$, this is already impossible as soon as the lattice intersects $[2,3]$, which we can guarantee if necessary by adding additional $r_a$s.

Otherwise, this bijection involves moving in the lattice a distance at most $1$. Hence for $B_R$ a ball of radius $R$, we obtain an injection $B_R \cap A \cap f^{-1} ([2,3] \cup [0,1]) \to B_{R+1} \cap A \cap f^{-1} ([0,1])$.

We now simply apply lattice point counting to check that the number of lattice points in these sets are asymptotic to the volumes of $B_R \cap f^{-1} ([2,3] \cup [0,1]) $ and $ B_{R+1} \cap f^{-1} ([0,1])$ respectively, which are asymptotic to $2 C R^{n-1}$ and $C R^{n-1}$ respectively, contradicting the claimed existence of an injection.

The only subtlety in the lattice point counting is whether the images of the lattice points under $f$ are equidistributed. However, by Weyl equidistribution, it suffices that $A$ is a dense subset of $\mathbb R$, which is automatic as $n>1$.

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Here is a more general version of my and YCor's argument:

Let $\Gamma$ be a group acting by measurable, volume-preserving transformations on a space $X$. Suppose that $\Gamma$ is amenable, in the sense that for each finite set $F$ and each $\epsilon>0$, there is a subset $S$ of $\Gamma$ such that $| FS| < (1+ \epsilon) |S|$ (Følner subsets).

Let $I$ and $J$ be two measurable subsets of $X$ such that can each be decomposed into finitely many pieces such that each piece of $I$ is a translate under an element of $\Gamma$ of a corresponding piece of $J$. Then the measure of $I$ equals the measure of $J$.

Proof: Let $F$ be the set of elements of $\Gamma$ that appear in the bijection between $I$ and $J$.

Let $f_I(x)$ be the number of $\gamma \in F S$ such that $\gamma(f) \in I$, and let $f_J(x)$ be the number of $\gamma \in S$ such that $\gamma(f) \in J$. Then $f_J(x) \leq f_I(x)$ because for each $\gamma$ with $\gamma(f) \in J$, we have chosen a $g\in F$ with $g \gamma(f) \in I$, and for distinct $\gamma(f)$, these give distinct elements of $I$ and hence distinct pairs $g \gamma$. (For distinct $\gamma$ that give the same $\gamma(f)$, we have the same $g$, and so $g \gamma$ remains distinct.)

So $\int f_J \leq \int f_I$. But exchanging the order of summation, $\int f_I = |FS| \mu(I)$ and $\int f_J = |S| \mu(J)$. Because so $(1+\epsilon) \mu(I) \leq \mu(J)$. Taking $\epsilon$ to $0$, $\mu(I) \leq \mu(J)$.

By symmetry, $\mu(I)=\mu(J)$.

Answer 2

No choice is needed.

First, assuming such a paradoxical decomposition, we get another one in a large circle $C=\mathbf{R}/k\mathbf{Z}$. Now we have, on the subgroup $\Gamma$ generated by the $r_i$ (and by $x\mapsto x+1$), an explicit sequence of Følner subsets, and corresponding normalized $\ell^1$-functions $f_m$.

Start from any Dirac measure on $C$, average it by $f_m$ to have a sequence of (finitely supported) measures $\mu_m$. Then $\mu_m$ is approximately invariant, in the sense that for every $g\in\ell^\infty(C)$ and $s\in\Gamma$, we have $\mu_m(g-sg)\to 0$. Indeed, $$\mu_m(g-sg)=(\mu_m-s\mu_m)g,$$ which tends to 0.

Split $C$ into $k$ intervals $C_1,\dots,C_k$ of length 1 (say, half-open). Then $\mu_m(C_i)-\mu_m(C_j)\to 0$. In particular, we do not have $\mu_m(C_1)\to 0$, since this would force $\mu_m(C_i)\to 0$ for each fixed $i$ and would entail the contradiction $1=\mu_m(C)=\sum_i\mu_m(C_i)\to 0$.

The proof is finished, since the paradoxical decomposition yields a decomposition of the form $\sum_{i=1}^nu_i-r_iu_i=1_{[2,3]}$, and applying $\mu_m$, the left-hand term tends to 0 and not the second one.