Tarski’s circle squaring problem has a constructive solution

You need to know: Euclidean plane {\mathbb R}^2, notation A+v for the translation of set A\subset {\mathbb R}^2 by a vector v\in {\mathbb R}^2, open subset of {\mathbb R}^2, countable union, countable intersection, and set difference (B \setminus A = \{x\in B, \, x\not\in A\}) of sets. You also need to know what is the Axiom of choice to fully understand the context.

Background: A subset A\subset {\mathbb R}^2 is a Borel set if it can be formed from open sets through the operations of countable union, countable intersection, and set difference. We call two sets A,B \subset {\mathbb R}^2 equidecomposable by translations if there are partitions A = A_1 \cup \dots \cup A_m and B = B_1 \cup \dots \cup B_m, such that B_i = A_i + v_i, i=1,\dots,m, for some vectors v_1, \dots, v_m \in {\mathbb R}^2. If, moreover, all A_i (and thus B_i) are Borel sets, we say that A and B are equidecomposable by translations with Borel parts.

The Theorem: On 17th December 2016, Andrew Marks and Spencer Unger submitted to arxiv a paper in which they proved that a circle and a square of the same area on the plane are equidecomposable by translations with Borel parts.

Short context: In 1990, Laczkovich, answering a 1925 question of Tarski, proved that circle and a square of the same area are equidecomposable by translations. In a paper submitted in 2015, Grabowski, Máthé, and Pikhurko proved that this is possible even using only Lebesgue measurable pieces (that is, those having a well-define area). However, both results use axiom of choice and the resulting pieces A_i are impossible to construct explicitly. The Theorem states that the circle can be squared with only Borel pieces. The proof does not use the axiom of choice. If such a proof is possible, we say that a problem has a constructive solution.

Links: Free arxiv version of the original paper is here, journal version is here.

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