A remark on hereditarily nonparadoxical sets

Call a set A ⊆ R paradoxical if there are disjoint A0, A1 ⊆ A such that both A0 and A1 are equidecomposable with A via countabbly many translations. X ⊆ R is hereditarily nonparadoxical if no uncountable subset of X is paradoxical. Penconek raised the question if every hereditarily nonparadoxical set X ⊆ R is the union of countably many sets, each omitting nontrivial solutions of x − y = z − t. Nowik showed that the answer is ‘yes’, as long as |X| ≤ אω. Here we show that consistently there exists a counterexample of cardinality אω+1 and it is also consistent that the continuum is arbitrarily large and Penconek’s statement holds for any X. Since the celebrated Banach-Tarski decomposition theorem, there have been several results on paradoxical decompositions. One common theme of these results is that for certain variants, if no paradoxical decompositon exists, then there can be only one possible cause for the nonexistence. In the case investigated by Penconek in [5], A,B ⊆ R are equidecomposable, in sign: A ∼ B, if there exist partitions A = ⋃ {Ai : i < ω} and B = ⋃ {Bi : i < ω} such that Bi is a translated copy of Ai. A frequently discovered fact is that I ∼ I ′ if I, I ′ ⊆ R are intervals (cf. e. g. [3]). A set A is paradoxical, if there exist disjoint A0, A1 ⊆ A with A ∼ A0, A ∼ A1. Finally, X is hereditarily nonparadoxical, if no uncountable A ⊆ X is paradoxical. A simple example of hereditarily nonparadoxical set is any set with no repeated differences, i.e., one in which if x − y = z − t, then x = z and y = t. The question whether every hereditarily nonparadoxical set has a countable partition into sets with ∗Research supported by the Hungarian National Research Grant OTKA K 81121.