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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 $r_a(S_a)$ is the set $[0,1]\cup [2,3]$. Where $r_a(S_a)$ 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)?

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)?

Let $\mathbb{R}$ act on itself by translation. Then there is no finite decomposition of a unit interval into two 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 $r_a(S_a)$ is the set $[0,1]\cup [2,3]$. Where $r_a(S_a)$ 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)?

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 $r_a(S_a)$ is the set $[0,1]\cup [2,3]$. Where $r_a(S_a)$ 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)?